NBME 25 EDU OBJ

170 Exam Section 4: Item 20 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 20. A 65-year-old man is brought to the emergency department 12 hours after the onset of chest pain. He dies 5 days later of cardiogenic shock. At autopsy, examination of the heart shows blockage of the coronary arteries and massive amounts of ventricular necrosis. A lack of which of the following best explains the inability of this patient's heart muscle to repair itself? O A) Myoglobin B) Sarcolemma O C) Sarcoplasmic reticulum D) Satellite cells E) T tubules Correct Answer: D. Satellite cells are not present in myocardial tissue and are partly responsible for cardiomyocytes' inability to repair damage sustained during hypoxia. Myocardial infarction refers to the irreversible death of cardiomyocytes. In normal skeletal muscle cells, with exclusion of cardiomyocytes, satellite cells play an important role in muscle fiber repair and regeneration after both routine exercise and catastrophic injury. They reside in a quiescent state between the sarcolemma and basal lamina of muscle fibers. Cytokines released during exercise or injury, such as IL-1, insulin-like growth factor-1, myostatin, and IL-6/MGF, directly activate satellite cells and lead to proliferation. Subsequent to this, satellite cells can aggregate and fuse to form new myofibers or they can interface with existing muscle fibers by donating their nucleus. Cardiac myocytes do not possess satellite cells, and because of this the repair of infarcted cardiac tissue occurs via fibrosis, leading to impaired contractility in the area of infarction. The hypoxic environment induced by myocardial infarction leads to acidosis and increased intracellular calcium, membrane damage, necrosis, and activation of pro-apoptotic pathways, ultimately resulting in cardiomyocyte death. Incorrect Answers: A, B, C, and E. Myoglobin (Choice A) is an iron- and oxygen-binding protein similar to hemoglobin that functions to store oxygen for use in skeletal muscles when oxygen delivery is impaired. It is present in cardiomyocytes and does not play a role in tissue regeneration. The sarcolemma (Choice B), sarcoplasmic reticulum (SR) (Choice C), and T tubule (Choice E) are all components of skeletal and cardiac muscle that are required for muscle contraction. The sarcolemma is the cell membrane of a skeletal or cardiac myocyte, and it contains ion channels. When a neuronal action potential causes depolarization of a cardiomyocyte, this causes activation of voltage gated calcium channels within the sarcolemma. T tubules are invaginations of the sarcolemma and extend deep into the myofibrils, thereby increasing the available surface area and allowing simultaneous activation of calcium channels along the sarcolemma. The calcium influx binds to receptors on the SR and leads to a significantly larger efflux of stored calcium from the SR. This substantial increase in intracellular calcium concentration binds to and causes a conformational change of the troponin-tropomyosin complex, which facilitates myosin binding to actin with subsequent muscle contraction. While these structures are critical for muscle contraction, they play no role in repair or regeneration of the cardiomyocyte.

Objective: Satellite cells are found in skeletal muscle and function to repair damaged myofibrils. They are not found in cardiomyocytes, which are unable to undergo regeneration following myocardial infarction or injury. Previous Next Score Report Lab Values Calculator Help Pause

43 Exam Section 1: Item 43 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 43. A 53-year-old woman comes to the physician because of a 6-month history of increasingly severe abdominal discomfort. She is 170 cm (5 ft 7 in) tall and weighs 104 kg (230 lb); BMI is 36 kg/m2. Abdominal examination shows fullness and tenderness of the right lower quadrant that increases with coughing. A CT scan of the abdomen is shown; the arrow indicates an abnormality. Which of the following structures forms the medial border of this abnormality? A) Arcuate line B) Inguinal ligament C) Rectus abdominis D) Round ligament E) Transversus abdominis Correct Answer: C. The rectus abdominis forms the medial border of this patient's abnormality, which is typical of a Spigelian hernia, a type of ventral hernia that peaks between the fourth and seventh decades of life. The Spigelian aponeurosis lies between the rectus abdominis medially and the linea semilunaris laterally. Most Spigelian hernias occur near the level of the arcuate line. The location and character of pain is variable; and treatment is with surgical reduction. Complications of hernias include incarceration, where hernia contents become trapped within the connective tissue defect often leading to bowel obstruction, and strangulation, where blood supply to the herniated contents becomes compromised, leading to rapid necrosis and often perforation of the involved intestine. Incorrect Answers: A, B, D, and E. The arcuate line (Choice A) lies one-third of the distance between the umbilicus and the pubic symphysis. It defines the transition point where the internal oblique and the aponeurosis of the transversus abdominis (Choice E) muscle run anterior to the rectus abdominis muscle, with the transversalis fascia forming the posterior wall of the transversus abdominis muscle. Hernias in this location are rare, and the rectus abdominis muscle would lie anterior to this hernia, not medially. The inguinal ligament (Choice B) forms the inferior border of the Hesselbach triangle, with the medial border formed by the lateral aspect of the rectus sheath, and the inferior epigastric vessels forming the superolateral border. Hernias through this triangle are called direct inguinal hernias. The round ligament (Choice D) is a fibromuscular band of tissue that extends from the superolateral part of the uterus through the deep inguinal ring and inguinal canal to the mons pubis. While indirect inguinal hernias may occur through the inguinal canal, this is uncommon in females as the processus vaginalis less frequently remains patent.

Objective: A Spigelian hernia occurs at the site of the Spigelian aponeurosis which lies between the rectus abdominis medially and the semilunar line laterally. II Previous Next Score Report Lab Values Calculator Help Pause

8 Exam Section 1: Item 8 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 8. In a study designed to test the effectiveness of a new drug in the treatment of endometriosis, 100 women are randomly assigned to one of two groups: 48 of the women receive the new drug and 52 receive standard therapy. Which of the following is the major benefit of assigning patients in this manner? O A) The effect of confounding variables is decreased B) The likelihood of achieving statistical significance is increased C) The power of the study is increased D) The study groups are equalized Correct Answer: A. When conducting a study, there are many potential biases that could impact the results and lead the investigators to erroneously reject the null hypothesis. One such bias is confounding bias, in which an unmeasured characteristic differs between two groups of subjects and causes a difference in outcome that is incorrectly attributed to the study intervention. For example, if a study on a new drug in the treatment of endometriosis assigns subjects to one of two groups and the groups have a difference in the severity of their disease, this may distort the apparent effect of the drug. In this example, the severity of disease represents a confounding variable. Ideally, when subjects are randomly distributed into groups, their characteristics are equally distributed between the two groups to minimize the effect of confounding variables on the study outcome. Randomization is one method of reducing confounding bias in studies. Incorrect Answers: B, C, and D. Assigning subjects to intervention or placebo groups randomly does not affect the statistical significance of the outcome (Choice B). The presence or absences of biases must be evaluated for every study, regardless of the P-value, which is used to interpret the probability of a type I error. Power can be increased by increasing the size of a study, not by randomly assignation of subjects to two groups (Choice C). If there is a true difference between two groups being compared in a study making the alternative hypothesis true, either the investigators will find that difference or they will miss that difference. If the difference is missed, they have made a type il error. A type Il error threshold is set at the beta level. Power, or the likelihood that the investigators will find the difference if it is there, is equal to one minus the beta level (B). Even with random assignation, the study groups are not equalized (Choice D). Random assignation minimizes the differences between two groups but there is still a possibility that they will have demographic differences. This is unavoidable in a study but is minimized by avoiding bias in selection of participants and evaluating characteristics between the groups.

Objective: A confounding variable is one not controlled by the investigators that impacts both the intervention and outcome outside of the causal pathway. Randomization is one way to minimize confounding variable bias. Previous Next Score Report Lab Values Calculator Help Pause

54 Exam Section 2: Item 4 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 4. A 30-year-old woman has polyhydramnios at 25 weeks' gestation. The fetus has an anatomically normal esophagus. Which of the following abnormalities is the most likely cause of the polyhydramnios? A) Anencephaly B) Caudal dysplasia C) Lumbar myelomeningocele D) Renal agenesis E) Urethral obstruction Correct Answer: A. Polyhydramnios describes the condition of excess amniotic fluid within the amniotic sac during fetal development. It is normally detected on ultrasonography as increased fluid depth around the fetus. It can be caused by several known pathologies but is often idiopathic. Anencephaly is a fetal neural tube defect in which the brain and cranial vault do not form correctly. It is a severe condition that leads to fetal loss or neonatal death. As the pregnancy progresses, anencephaly frequently results in polyhydramnios as the malformations of the face and head do not allow the fetus to appropriately swallow amniotic fluid, thereby increasing the amount of fluid surrounding the fetus in the amniotic sac. Additionally, if the fetal central nervous system does not fully form in anencephaly, the stimulus to swallow along with appropriate neuromuscular pathways required to do so may be absent. An atretic esophagus can be a cause of polyhydramnios, also because of the inability to swallow and therefore clear amniotic fluid, however in this case, the esophagus is normal. Incorrect Answers: B, C, D, and E. Caudal dysplasia (caudal regression syndrome) (Choice B) is a condition in which the lower lumbar spine and sacrum and the corresponding spinal cord and nerve roots are malformed leading to paralysis and dysfunction of the lower extremities, bowel, and urinary bladder. Deformity of the lower extremities of the body often results. It does not necessarily cause increased amniotic fluid volume, and if involving urogenital malformation, may result in decreased amniotic fluid volume as one source of amniotic fluid is fetal urine. This condition is associated with early pregnancy diabetes mellitus. Lumbar myelomeningocele (Choice C) is a condition where the neural crest arch fails to fully form posteriorly in the spinal cord. As a result, the dura and spinal cord protrude posteriorly out of the spinal canal. This condition is associated with an abnormality in folate metabolism or a lack of folic acid to support neural tube and neural crest cell differentiation. The condition results in nerve dysfunction and paralysis distal to the lesion but does not usually result in increased amniotic fluid production. Renal agenesis (Choice D) is a failure of the formation of the kidneys, which leads to minimal fetal urine production, and therefore limited production of amniotic fluid (oligo- or anhydramnios) as fetal urine is a major contributor. Without production of amniotic fluid to distribute stress evenly over the surface of the developing fetus, deformities will result. Fetuses with renal agenesis typically exhibit compressed, flattened facial features and pulmonary hypoplasia caused by inadequate amniotic fluid-induced lung expansion. This is known as the Potter sequence. Urethral obstruction (Choice E) leads to decreased fetal urine output and decreased amniotic fluid production by the fetus. This condition results in oligohydramnios and may also lead to the Potter sequence. This is commonly seen in the setting of posterior urethral valves.

Objective: Amniotic fluid is recycled by fetal swallowing; malformations in the neurologic or craniofacial anatomy required to swallow (eg, anencephaly or esophageal atresia) may result in the accumulation of amniotic fluid, a condition known as polyhydramnios. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

29 Exam Section 1: Item 29 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 29. A67-year-old man is brought to the physician because of a 2-day history of double vision and drooping of his left eyelid. He has no history of trauma to the eye. His vital signs are within normal limits. Physical examination shows ptosis of the left eyelid. When the eyelid is raised manually, the eye is fixed in the out position and the pupil is dilated. The visual acuity of the left eye is within normal limits. Which of the following is the most likely cause of these findings? A) Aneurysm of the posterior communicating artery B) Compression of the superior cervical ganglion O C) Damage to the trochlear nerve D) Occlusion of the scleral venous sinus E) Tumor of the optic nerve Correct Answer: A. This patient's presenting findings of unilateral ptosis, exotropia and hypotropia (inferior and lateral pupil position), and mydriasis (pupillary dilation) are consistent with an oculomotor nerve palsy. Two of the most common causes of acquired oculomotor nerve palsy are microvascular ischemia and nerve compression from an adjacent vascular aneurysm, commonly located at the posterior communicating artery (PCA). Upon leaving the brainstem, the oculomotor nerve courses beneath the posterior cerebral and posterior communicating arteries and is vulnerable to aneurysmal compression at this location. The fibers responsible for control of the pupil are found on the superficial aspect of the oculomotor nerve and are thus first affected by external compression. This distinguishes acquired oculomotor nerve palsy caused by aneurysmal external compression from palsy caused by microvascular ischemia, which typically spares the pupil. The detection of a new oculomotor nerve palsy is therefore a medical emergency, as aneurysmal rupture can lead to a life-threatening subarachnoid hemorrhage. A PCA aneurysm is diagnosed via CT angiography, MR angiography, or conventional angiography. Incorrect Answers: B, C, D and E. Compression of the superior cervical ganglion (Choice B) will lead to Horner's syndrome, which presents with unilateral ptosis, miosis (pupillary constriction), and ipsilateral facial anhidrosis. The position of the eye would not be affected; thus, an inferior and lateral eye position would not be observed. Horner syndrome can also be caused by injury to the brainstem or to descending sympathetic fibers in the neck and upper thorax, such as in the setting of apical pulmonary tumors (Pancoast tumor) or dissections of the carotid artery. Damage to the trochlear nerve (Choice C) presents with binocular diplopia that is most pronounced when the eyes attempt to gaze downward and medially, such as while reading or walking down a flight of stairs. The trochlear nerve innervates the superior oblique muscle, which is responsible for intorsion and depression of the eye. Examination may show ipsilateral hypertropia and compensatory head tilt toward the contralateral shoulder to minimize symptoms of diplopia. Occlusion of the scleral venous sinus (Choice D) also termed the canal of Schlemm, would lead to acute angle closure glaucoma (AAG). AAG classically presents with acute, severe, unilateral ocular pain, and loss of vision, with conjunctival erythema, and a fixed, mid-dilated pupil noted on physical examination. Ptosis and disconjugate gaze would not be observed. Tumors of the optic nerve (Choice E) are most commonly gliomas. Optic nerve gliomas typically present in children with painless proptosis and vision loss. An afferent pupillary defect is usually observed on examination. Optic nerve gliomas are associated with neurofibromatosis type I.

Objective: Aneurysms of the posterior communicating artery may compress the ipsilateral oculomotor nerve, resulting in diplopia, mydriasis, exotropia, and hypotropia. They are diagnosed on CT angiography, MR angiography, or conventional angiography. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

114 Exam Section 3: Item 14 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 14. An analysis of patients admitted to a specific hospital finds that the risk for myocardial infarction is higher in men with chronic obstructive pulmonary disease (COPD) than in men without COPD (odds ratio 5.4, 95% confidence interval: 1.4-10.1). These results can most reasonably be generalized to which of the following populations? A) Men with COPD and myocardial infarction B) Men with COPD in the hospital C) Men with COPD living at home D) Men with myocardial infarction in the hospital E) All patients with COPD and myocardial infarction F) All patients with COPD in the hospital G) All patients with COPD living at home H) All patients with myocardial infarction Correct Answer: B. In order to correctly apply the findings of an investigation, the persons to whom the study is applicable must be considered. In this case, the characteristics of the study population include being male, having COPD, and being hospitalized. This study would be readily generalizable to a similar male, hospitalized population with COPD. However, the study does not tell us what the impact of female gender or outpatient setting would be on the outcome. Thus, these results cannot be generalized to women in a clinic setting with COPD. Ideally, to apply the results of this study to a new population or group of people, the group to which the study is applied should meet the same baseline characteristics as those of the persons studied. Incorrect Answers: A, C, D, E, F, G, and H. The subjects included in the study had only COPD as a comorbidity, not myocardial infarction (Choice A). These results were based on a cohort of hospitalized men, not men at home, and thus cannot be generalized to non-hospitalized patients (Choice C). The study looked at men with COPD, not myocardial infarction or both (Choices A and D), who were hospitalized and calculated their risk for myocardial infarction. Thus, the results cannot be generalized to patients with myocardial infarction who were hospitalized because the risk profile for myocardial infarction differs between these two comorbidities. This study included only men and thus is only generalizable to men (Choice E, F, G, and H). If the study had included all patients regardless of gender, then it would be generalizable to all patients regardless of gender.

Objective: A study is generalizable to patients that are similar to the subjects studied. While results may be used to guide treatment in these cases, they are not necessarily generalizable to patients who do not meet the same demographic criteria as the subjects enrolled in the study. Previous Next Score Report Lab Values Calculator Help Pause

112 Exam Section 3: Item 12 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 12. ITTT Time Serum Osmolality Urine Osmolality Urine Flow Rate (mOsmol/kg) 305 410 650 (mOsmol/kg) 290 285 (mL/min) 3.0 2.0 1.5 1.0 (minutes) 30 60 90 280 275 710 The table shows serum osmolality, urine osmolality, and urine flow rate measured at 30-minute intervals in a 30-year-old man. Which of the following substances was most likely administered at time 0? A) ADH (vasopressin) B) Aldosterone C) Atrial natriuretic peptide D) Furosemide E) Mannitol Correct Answer: A. ADH (vasopressin) acts at the level of the distal convoluted tubule and collecting duct to increase water reabsorption. Water reabsorption in the absence of solute reabsorption, specifically sodium, will lead to a net decrease in urine flow but an increase in urine osmolality. Reabsorption of water also dilutes the serum, leading to a decreased serum osmolality. Release of ADH by the posterior pituitary is stimulated by an increase in plasma osmolality but can also occur without stimulus in the setting of syndrome of inappropriate antidiuretic hormone (SIADH). ADH or its analog, desmopressin, can be administered therapeutically in the treatment of sodium and fluid imbalance (eg, diabetes insipidus). Desmopressin also promotes the release of prothrombotic molecules from the endothelium, and is used therapeutically in the treatment of von Willebrand disease and mild cases of hemophilia A. Incorrect Answers: B, C, D, and E. Aldosterone (Choice B) promotes the reabsorption of sodium in the distal convoluted tubule and collecting ducts, causing the indirect reabsorption of water, although to a lesser extent than the water reabsorption that occurs with ADH. This generally leads to sodium reabsorption in excess of water reabsorption. Administration of aldosterone thus would cause a decrease in urine flow rate, a decrease in urine osmolality, and an increase in serum osmolality. Atrial natriuretic peptide (ANP) (Choice C) is released by atrial myocytes in response to wall stress caused by an increased atrial volume. It leads to vasodilation and natriuresis, thus maintaining blood volume and pressure homeostasis. ANP would cause increased urine flow and increased urine osmolality. Furosemide (Choice D) is a potent diuretic that acts at the loop of Henle to prevent the concentration of urine. Administration would lead to a decrease in urine osmolality, and an increase in urine flow rate and serum osmolality, as it causes more water to be excreted than solute. Mannitol (Choice E) is an osmotic diuretic that pulls water into the renal tubules as a result of increased oncotic pressure. This would lead to an increase in urine output, rather than a decrease as seen in the patient.

Objective: ADH (vasopressin) increases water reabsorption without increasing sodium reabsorption. When administered exogenously, this leads to decreased urine output, increased urine osmolality, and decreased serum osmolality. Previous Next Score Report Lab Values Calculator Help Pause

183 Exam Section 4: Item 33 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 33. A 50-year-old man with chronic back pain has taken a variety of analgesics (including aspirin, acetaminophen, and ibuprofen) daily and in large doses for the past 25 years. He is found dead. A photograph of his kidneys at autopsy is shown. Prior to his death, which of the following changes in urine volume and urine osmolality would be expected after administration of ADH (vasopressin)? Urine Osmolality increased Urine Volume O A) Increased B) Increased no change C) Increased decreased D) No change increased E) No change no change F) No change decreased G) Decreased increased H) Decreased no change I) Decreased decreased Correct Answer: E. ADH (vasopressin) binds V2 receptors in the collecting tubule and through cyclic adenosine monophosphate second messenger pathways. This results in the insertion of aquaporin channels into the luminal membrane. Aquaporins promote the uptake of free water out of the urine, thereby concentrating urine and diluting the bloodstream. The resultant urine has a higher concentration of solutes (eg, sodium, urea) and therefore, an increased osmolality and an overall decreased urine volume. ADH acts upon cells in the collecting tubule of the nephron, which spans the renal medulla and terminates in the renal papilla. Renal papillary necrosis (RPN) occurs following ischemic, inflammatory, infectious, or toxin-mediated damage to renal papillae and describes the sloughing and loss of the papillae including substructures such as the distal collecting tubule. RPN can be triggered by infections (eg, acute pyelonephritis), diabetes mellitus, sickle cell disease, or nonsteroidal anti-inflammatory drugs (NSAIDS). Necrosis and sloughing of the papillae are noted on laboratory evaluation as gross hematuria and proteinuria. In the setting of RPN, ADH is unable to act on the necrotic or sloughed collecting ducts of the nephron, therefore, urine osmolality and urine volume are unchanged. Incorrect Answers: A, B, C, D, F, G, H, and I. ADH regulates the V2 receptors in the collecting tubule and promotes the uptake of free water via aquaporins. An increase in urine volume (Choices A, B, and C) would be expected with decreased concentrations of ADH, along with a decrease in the urine osmolality (Choices C, F, and I). However, in the case of renal papillary necrosis, ADH is ineffective because of the sloughing and necrosis of the collecting ducts in the renal papillae. Conversely, the administration of ADH would typically promote the reabsorption of water via aquaporins into the bloodstream and decrease urine volume overall (Choices G, H, and I). The reabsorption of water through aquaporins would work to increase urine osmolality, as the concentration of solutes in the urine increases (Choices A, D, and G). However, in the case of renal papillary necrosis, ADH is ineffective because of the sloughing and necrosis of the collecting ducts in the renal papillae.

Objective: ADH (vasopressin) regulates the V2 receptors in the collecting tubule and promotes the uptake of free water via aquaporins; the collecting tubules are located within the renal medulla and terminate in the renal papillae. Renal papillary necrosis results in the loss of the renal papillae and associated substructures, which can be triggered by infections, diabetes, sickle cell disease, or NSAIDS. In the setting of renal papillary necrosis, ADH (vasopressin) is unable to act on the collecting ducts of the nephron. ID Previous Next Score Report Lab Values Calculator Help Pause

122 Exam Section 3: Item 22 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 22. A 43-year-old man with a 10-year history of alcoholism comes to the physician because of a change in skin color. The patient says that he has been taking two extra-strength acetaminophen tablets every 4 to 6 hours for the past 3 days for a severe headache. Physical examination shows jaundice. Laboratory studies show an increased prothrombin time and a markedly increased serum AST activity. Alteration in which of the following metabolites within hepatocytes is most likely associated with his illness? A) Decreased glucuronide conjugates B) Decreased glutathione C) Decreased NAD+ D) Decreased NADH E) Increased glucuronide conjugates O F) Increased glutathione G) Increased NAD+ H) Increased NADH Correct Answer: B. Acetaminophen poisoning presents with nonspecific symptoms of fatigue, malaise, and abdominal pain followed by signs and symptoms of liver failure, including jaundice, altered mental status and coagulopathy. Serum transaminases are markedly increased. Acetaminophen poisoning is more common in patients with underlying risk factors, including exposure to alcohol and hepatotoxic medications. Chronic alcohol use increases the risk for hepatotoxicity in patients who take high doses of acetaminophen. When taken at therapeutic doses, acetaminophen is safely metabolized through phase Il conjugations, including glucuronidation and sulfation. In the setting of acetaminophen overdose, saturation of phase Il metabolic pathways leads to excess acetaminophen metabolized by CYP-mediated reactions to N-acetyl-p-benzoquinoneimine (NAPQI), which has strong oxidizing properties and is directly hepatotoxic. The antioxidant molecule glutathione conjugates NAPQI, allowing it to be safely excreted. The depletion of glutathione is a hallmark of acetaminophen toxicity. Acetaminophen toxicity is treated by increasing hepatic stores of glutathione through intravenous or oral administration of N-acetylcysteine. Incorrect Answers: A, C, D, E, F, G, and H. Decreased glucuronide conjugates (Choice A) is incorrect in the setting of acetaminophen overdose. Rather, increased glucuronide conjugates (Choice E) would be observed since the primary mechanism of acetaminophen metabolism is through phase II metabolic reactions including glucuronidation. However, glucuronide conjugates are harmlessly excreted. They are not responsible for the liver injury that is associated with acetaminophen overdose. The conversion of NAD+ to NADH plays a role in several aspects of liver metabolism in alcoholics. The metabolism of ethanol to acetaldehyde by alcohol dehydrogenase converts NAD* to NADH, leading to decreased NAD* (Choice C) and increased NADH (Choice H). As a result of Le Chatelier's principle, the excess NADH is responsible for an increased activity of numerous metabolic pathways that utilize NADH as a cofactor, including the production of lactate, malate, and triglycerides. Medications such as fomepizole are useful for treating overdoses of methanol or ethylene glycol by inhibiting alcohol dehydrogenase and preventing the production of toxic products such as oxalate, glycolate, and formic acid. However, decreased NAD* and increased NADH do not play a direct role in acetaminophen-mediated hepatotoxicity. Decreased NADH (Choice D) and increased NAD+ (Choice G) are observed in neither acetaminophen overdose nor poisoning from methanol or ethylene glycol. Increased glutathione (Choice F) is not responsible for hepatotoxicity in the setting of acetaminophen overdose.

Objective: Acetaminophen is normally metabolized by phase II conjugation to sulfates or glucuronides. When this pathway is overwhelmed, acetaminophen is metabolized by CYP-mediated reactions to NAPQI, a strong oxidizer which is directly hepatotoxic. The depletion of glutathione is a hallmark of acetaminophen toxicity. I3D Previous Next Score Report Lab Values Calculator Help Pause

38 Exam Section 1: Item 38 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 38. A 48-year-old woman comes to the physician because of a 1-year history of progressive ringing in her right ear. She also has felt dizzy while exercising. Neurologic examination shows dysmetria of the right upper and lower extremities. Muscle strength and somatosensory function testing of all extremities shows no abnormalities. Audiometry shows moderate hearing loss in the right ear. An MRI of the brain is most likely to show a mass compressing which of the following labeled structures in the photograph of a cross section of the brain stem? A C D E Right Left A) B) C) D) O E) O F) Correct Answer: A. This patient most likely has an acoustic neuroma (vestibular schwannoma) compressing the right inferior cerebellar peduncle, pictured above as a large black-stained area labeled "A" (representing white matter as myelin stains black). Acoustic neuromas are benign tumors consisting of Schwann cells that arise from the vestibulocochlear nerve (cranial nerve VIII). Lesions of the vestibulocochlear nerve typically disrupt ipsilateral hearing and the sensation of head and body position, which could lead to dizziness with exercise as in this patient. Mass effect from acoustic neuromas can compress the neighboring cerebellar peduncles, which can lead to ipsilateral dysmetria. Depending on patient age, tumor size, and symptom burden, acoustic neuromas can be managed with observation, radiation, or surgery. Incorrect Answers: B, C, D, E, and F. The left inferior cerebellar peduncle (Choice B) would not be compressed by a right-sided acoustic neuroma. A left-sided acoustic neuroma may compress the left cerebellar peduncles, causing dysmetria of the left extremities. The medial lemnisci (Choices C and D) would not be compressed by an acoustic neuroma. The medial lemnisci are white matter tracts located medial and anterior in the medulla compared to the posterolateral location of the vestibulocochlear nerves. The medial lemniscus carries fibers from the dorsal column tract, which is responsible for the sensation of pressure, fine touch, and proprioception. The medullary pyramids (Choices E and F) would not be compressed by acoustic neuromas. The medullary pyramids are white matter tracts located at the anterior-most region of the medulla compared to the posterolateral-located vestibulocochlear nerves. The medullary pyramids carry the fibers of the lateral corticospinal tracts.

Objective: Acoustic neuromas, benign tumors of Schwann cells that arise from the vestibulocochlear nerve, can compress the cerebellar peduncles. Acoustic neuromas can cause ipsilateral hearing loss and positional dizziness and, if compressing the cerebellar peduncles, can lead to ipsilateral dysmetria. Previous Next Score Report Lab Values Calculator Help Pause

2 Exam Section 1: Item 2 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 2. New reagents are being designed to treat graft-versus-host disease in HLA-unrelated bone marrow transplant recipients. Reagents should be designed that selectively deplete which of the following donor cells? A) B lymphocytes O B) Dendritic cells C) Macrophages D) Neutrophils E) T lymphocytes Correct Answer: E. T lymphocytes, which are responsible for the immune attack against the recipient, should be the target of agents used to treat graft-versus-host disease (GVHD) in HLA-unrelated bone marrow transplant recipients. Prior to grafting of allogeneic stem cells in a bone marrow transplant, recipient native marrow must be completely ablated. It is then replaced with donor marrow, which recolonizes the marrow cavity. Successful transplantation results in reconstitution of the immune system with the donor's immune cells. GVHD occurs when the donor's immune system recognizes the host's tissues as foreign and mounts an immune response against them. Manifestations of acute GVHD include rash, ranging from maculopapular to blistering, diarrhea, abdominal pain, and hepatitis with hyperbilirubinemia. Diagnosis should be suspected in patients with any of these symptoms and a recent bone marrow transplant. Medications that are used in the treatment of GVHD act either by directly inhibiting the production of T lymphocytes or by suppressing the normal function of these cells. Examples include mycophenolate mofetil, which inhibits the synthesis of nucleotides needed for lymphocyte proliferation; etanercept, a TNF-alpha receptor inhibitor; pentostatin, a purine analog that inhibits Ť lymphocyte proliferation; sirolimus, which inhibits mTOR to suppress T lymphocyte proliferation; and anti-thymocyte globulin, which directly targets T lymphocytes. Incorrect Answers: A, B, C, and D. B lymphocyte depletion (Choice A) is the result of several medications that target CD20, a surface marker found on B lymphocytes. The most used monoclonal antibody targeting this site is rituximab. It is currently used as a primary component of chemotherapy regimens to treat a variety of B-cell lymphomas. It is also used to treat vasculitis such as granulomatosis with polyangiitis, microscopic polyangiitis, neurologic disorders such as myasthenia gravis and autoimmune encephalitis, and refractory immune thrombocytopenia. It is well tolerated, but patients must be tested for hepatitis B prior to treatment, as reactivation of hepatitis B is a known side effect. Dendritic cell depletion (Choice B) occurs indirectly through targeted medications against the inflammatory cytokines produced by dendritic cells. One example of this is anakinra, an IL-1 receptor antagonist. It is used for the treatment of gout, rheumatoid arthritis, and neonatal-onset multisystem inflammatory disease (NOMID). Macrophage cell depletion (Choice C) occurs indirectly through inhibiting the release of inflammatory cytokines such as TNF-a and through the inhibition of antigen presentation and activation of T lymphocytes, thereby dampening the overall immune response. Examples of anti-TNF-a therapies include infliximab, adalimumab, and eculizumab. Medications that target antigen presentation include abatacept. These medications have broad use, often to treat autoimmune diseases such as rheumatoid and psoriatic arthritis. Neutrophil depletion (Choice D) is exceedingly common and is typical of many medications. Neutrophil suppression is typified by neutropenia and can be seen for example with the use of methimazole, clozapine, and dapsone, among others. Some nonsteroidal anti-inflammatory drugs also impair neutrophil function without causing neutropenia.

Objective: Acute GVHD may occur in HLA-unrelated bone marrow transplant recipients as a result of donor T lymphocytes recognizing and targeting recipient cells of the gastrointestinal tract, skin, and liver. Treatment is with medications that directly or indirectly prevent T lymphocyte proliferation, activation, and function. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

94 Exam Section 2: Item 44 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 44. A 37-year-old woman comes to the physician for a follow-up examination 3 months after receiving a kidney transplant from a cadaver donor. She feels well, and her condition has been stable since the transplant. She receives standard immunosuppressive medications. Her temperature is 37°C (98.6°F), pulse is 84/min, respirations are 14/min, and blood pressure is 150/90 mm Hg. Physical examination shows a well-healed surgical scar over the lower abdomen. Serum studies show a creatinine concentration of 2 mg/dL; baseline creatinine concentration was 1.4 mg/dL 1 week post transplant, and 2 weeks ago. Serum concentrations of the immunosuppressive drugs are in the low therapeutic range. Examination of a renal biopsy specimen shows lymphocytes in the tubules and arterial walls. Which of the following best explains these findings? A) Acute antibody-mediated rejection B) Acute T-lymphocyte-mediated rejection C) Adverse effects of immunosuppressant medications D) Chronic allograft rejection E) Hyperacute rejection Correct Answer: B. Acute T-lymphocyte-mediated rejection (TCMR) is characterized by transplant allograft organ dysfunction (eg, increased creatinine), and transplant organ histology (in the case of the kidney) demonstrating lymphocytes in the tubules and arterial walls, often with inadequate immunosuppressive drug concentrations. This most commonly occurs within the first 6 months after transplant. TCMR is caused by host T-lymphocytes reacting to donor major histocompatibility complexes (MHC) present on cells within the glomeruli, tubules, interstitium, and blood vessels of the allograft. Biopsy is required for the evaluation of graft dysfunction because it can distinguish between acute rejection and other causes of kidney injury, including acute tubular necrosis, interstitial nephritis, or viral infections (eg, cytomegalovirus or BK virus). Primary histologic changes include interstitial infiltration with lymphocytic cells, in addition to obliteration of the tubular basement membrane. Treatment is based upon the degree of rejection and allograft dysfunction but includes steroids and lymphocyte-depleting agents. Incorrect Answers: A, C, D, and E. Acute antibody-mediated rejection (Choice A) is caused by the binding of antibodies to donor graft antigens leading to a local inflammatory response and subsequent graft dysfunction. There is often evidence of arteritis, thrombosis, and tubular injury. It may result in graft failure. Adverse effects of immunosuppressant medications (Choice C) include those from the calcineurin inhibitors, cyclosporine, and tacrolimus. Acute tubular necrosis is the classic allograft histologic finding in the setting of immunosuppressant toxicity, as they induce vascular constriction of the afferent arteriole. Treatment consists of reducing the dose of immunosuppressive medications. Prolonged toxicity can lead to interstitial fibrosis of the allograft. Immunosuppressant medications are also associated with side effects such as neurotoxicity, hirsutism, gingival hyperplasia, and hypertension. Chronic allograft rejection (Choice D) of a transplanted organ occurs over months to years. It is secondary to a CD4+ T lymphocyte response against donor peptides such as MHC. T lymphocyte activation leads to cytokine production and humoral and cellular hypersensitivity reactions (type Il and IV, respectively). These reactions result in vascular arteriosclerosis and smooth muscle proliferation with parenchymal fibrosis and atrophy. Hyperacute rejection (Choice E) occurs within minutes following allograft transplantation. The transplanted kidney will appear mottled and cyanotic. It occurs secondary to preformed recipient antibodies reacting to donor antigens (type Il hypersensitivity), which activate complement and cause thrombosis of the allograft vasculature.

Objective: Acute T-lymphocyte-mediated rejection (TCMR) typically occurs within months following transplantation and presents with allograft dysfunction. Kidney biopsy demonstrates lymphocytes in the interstitium, tubules, glomeruli, and blood vessels. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

100 Exam Section 2: Item 50 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 50. A 30-year-old man comes to the office because of a 1-week history of nausea and yellowing of the whites of his eyes. He has no history of serious illness and takes no medications. Vital signs are within normal limits. Abdominal examination shows hepatomegaly; there is tenderness to palpation over the right upper quadrant. Serologic testing is positive for hepatitis A virus IgM antibodies. A high-power photomicrograph of a liver biopsy specimen obtained from a patient with a similar condition is shown. Which of the following is the most likely mechanism of hepatocyte cell death in this patient? A) Apoptosis caused by activation of the death receptor extrinsic pathway B) Apoptosis caused by activation of the mitochondrial intrinsic pathway C) Autophagy caused by inflammation D) Necrosis caused by free radical injury E) Necrosis caused by lack of blood supply Correct Answer: A. The liver photomicrograph of this patient with acute viral hepatitis A infection demonstrates ballooning degeneration of hepatocytes and an intensely eosinophilic apoptotic body (Councilman body). Apoptotic bodies represent hepatocytes in a state of apoptosis or necrosis, and their distinct appearance histologically often reflects contrast from surrounding normal, or ballooning hepatocytes. Apoptotic bodies are most frequently observed in viral hepatitis caused by hepatitis A, although they may also be observed in association with yellow fever. Viral hepatitis A infection leads to hepatocyte death primarily through apoptosis caused by activation of the death receptor extrinsic pathway. While the intrinsic and extrinsic pathways of apoptosis both converge on caspase-mediated cell death, the extrinsic pathway is dependent on external cellular signals as with the binding of Fas and Fas-ligand or, in this case, the release of perforin and granzyme B from cytotoxic CD8+ T Lymphocytes responding to intracellular viral pathogens. Lymphocytic infiltrates are frequently seen on liver histology in the setting of an acute hepatitis infection. Incorrect Answers: B, C, D, and E. Apoptosis caused by activation of the mitochondrial intrinsic pathway (Choice B) is triggered by cellular damage such as radiation, oxidative damage, ischemia, or toxin exposure, and leads to release of cytochrome C from mitochondria. Cytochrome C in turn leads to the activation of caspase enzymes. Viral hepatitis primarily leads to apoptosis through the extrinsic rather than the intrinsic pathway. Autophagy caused by inflammation (Choice C) involves the regulated digestion of intracellular proteins using lysosomes and autophagosomes. It is a part of cell turnover in states of normal physiology and in pathology. Inflammation can lead to increased autophagy during repair of intracellular damage; however, autophagy is not the prevailing mechanism responsible for the histologic findings of ballooning degeneration and apoptotic bodies. Viral hepatitis is classically associated with T-cell mediated injury and apoptosis. Necrosis caused by free radical injury (Choice D) can be caused by medication-induced liver injury. Common offending agents include acetaminophen, isoniazid, HMG-CoA reductase inhibitors, iron, carbon tetrachloride, and hormonal medications. Viral hepatitis does not lead to a significant degree of free radical injury. Necrosis caused by lack of blood supply (Choice E) may occur secondary to any category of shock (eg, cardiogenic, hypovolemic, distributive, obstructive). For example, a patient with persistent, refractory shock following the rupture of an abdominal aortic aneurysm may develop liver failure i to 3 days after the inciting event, as necrosis occurs within the ischemic cells. Viral infection with hepatitis A would be unlikely to produce hepatic ischemia.

Objective: Acute hepatic inflammation in the setting of viral hepatitis A or yellow fever may cause apoptosis of hepatocytes. This occurs via the extrinsic pathway of apoptosis and is seen on histology as ballooning hepatocytes and apoptotic bodies. II Previous Next Score Report Lab Values Calculator Help Pause

136 Exam Section 3: Item 36 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 36. A 23-year-old woman at 32 weeks' gestation comes to the emergency department because of a 1-day history of left flank pain and fever. Her temperature is 39.1°C (102.3°F), pulse is 104/min, respirations are 14/min, and blood pressure is 120/72 mm Hg. Physical examination shows prominent tenderness over the left costovertebral angle. A photomicrograph of a renal biopsy specimen from a similar patient is shown. Which of the following is the most likely diagnosis? A) Acute pyelonephritis B) Acute renal infarction C) Acute tubulointerstitial nephritis D) Crescentic glomerulonephritis O E) Hemolytic uremic syndrome Correct Answer: A. Acute pyelonephritis is a urinary tract infection that has spread to the kidneys, which is characterized by flank pain, fever and chills, nausea, vomiting, and costovertebral tenderness on physical examination. Urinary tract infections are much more common in women because of the shorter urethra and favorable regional environment for bacterial growth. If untreated, bacteria can travel retrograde up the ureter to infect the kidney, resulting in pyelonephritis. The biopsy specimen shows many leukocytes within the tubules, including polymorphonuclear lymphocytes (PMNS also known as neutrophils), which are characteristic of pyelonephritis. Urinalysis typically shows leukocyte esterase and nitrites along with many leukocytes and leukocyte casts on microscopy. Treatment includes antibiotics covering common pathogens (eg, Escherichia coli). Incorrect Answers: B, C, D, and E. Acute renal infarction (Choice B) would cause flank pain but would not be characterized by fever or WBC on biopsy. Lymphocytic infiltrates are usually because of infectious or inflammatory states, whereas signs of coagulation necrosis would be expected with a renal infarction on histology. Acute tubulointerstitial nephritis (Choice C) is usually caused by a hypersensitivity reaction to an offending medication (eg, nonsteroidal anti-inflammatory drugs, diuretics, and sulfonamides) and is characterized by a rash and eosinophilia. On histology, there are typically scattered plasma cells, eosinophils, and macrophages, with occasional non-necrotizing granulomas. Crescentic glomerulonephritis (Choice D) is characterized by crescents of eosinophilic fibrin and plasma proteins on biopsy of the glomerulus, which are adjacent to glomerular parietal cells, monocytes, and macrophages. It is seen histologically in the setting of Goodpasture syndrome, granulomatosis with polyangiitis, and microscopic polyangiitis. Hemolytic uremic syndrome (HUS) (Choice E) is typically seen in children, caused by Shiga toxin-producing E. coli (0157:H7), and is usually accompanied by bloody diarrhea. HUS is characterized by a classic triad of thrombocytopenia, acute renal failure, and microangiopathic hemolytic anemia.

Objective: Acute pyelonephritis typically presents with a fever, chills, myalgias, and systemic symptoms such as nausea and vomiting, most commonly in a female patient following symptoms of a lower urinary tract infection. On histology, lymphocytic and neutrophilic infiltrates are seen adjacent to and within the tubules, and urinalysis may show pyuria and white blood cell casts. Previous Next Score Report Lab Values Calculator Help Pause

21 Exam Section 1: Item 21 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 21. This graph best depicts the natural history of which of the following human disorders? A A) Creutzfeldt-Jakob disease B) Guillain-Barré syndrome C) HIV infection D) Subacute sclerosing panencephalitis E) Yellow fever Time Clinical disease threshold Course of disease Active production of infectious agent Correct Answer: E. The graph of clinical symptoms shows the active production of an infectious agent for a distinct period with a discrete endpoint. The clinical course of disease begins with the production of the infectious agent and is followed by a short asymptomatic period during which the clinical course is below the clinical disease threshold. This is followed by increasing clinical symptoms that peak and sharply decrease, eventually falling below the clinical disease threshold to resolution. Shortly after symptoms resolve, the production of the infectious agent ceases. Such a pattern of disease with an acute symptomatic onset and resolution, as well as initiation and cessation of infectious agent production is classic for many viral infections such as yellow fever. Yellow fever is a positive-sense RNA flavivirus. The disease classically presents with a short duration viral syndrome that includes fever, chills, myalgias, and headache. In some patients, it causes acute hepatitis, which would present with nausea, vomiting, jaundice, and hepatosplenomegaly, along with signs of liver dysfunction (eg, hypoalbuminemia, increased prothrombin time). The disease is also associated with Councilman bodies within the cytoplasm of hepatocytes on histology. Increases in the serum concentrations of bilirubin lead to the jaundice observed. The virus is transmitted by the Aedes aegypti mosquito. A live attenuated vaccine is available for the prevention of yellow fever and is recommended for people traveling to endemic areas. Incorrect Answers: A, B, C, and D. Creutzfeldt-Jakob disease (Choice A) is a prion disease, caused by a misfolded protein. Disease progression occurs through the aggregation and additional misfolding of prion peptides in the central nervous system, leading to progressive neurologic deterioration including myoclonus followed by rapid decline and death usually within a year of symptom onset. In contrast to yellow fever, which often has a self-limited disease course, Creutzfeldt-Jakob disease presents with progressively worsening clinical symptoms. Guillain-Barré syndrome (Choice B) is an inflammatory, demyelinating polyneuropathy that occurs idiopathically or following infection with a virus or bacteria, classically a gastrointestinal illness from a pathogen such as Campylobacter jejuni. It typically presents as ascending paralytic weakness starting in the lower extremities and may eventually involve respiratory muscles. It generally self-resolves but can persist for months. Symptoms commonly occur after the active production of the infectious agent, not during the same time as indicated in the graph. HIV infection (Choice C) can manifest as an asymptomatic initial infection or it can present as an acute retroviral syndrome in the weeks following inoculation. The production of infectious agent does not stop; it is a chronic viral infection, and in spite of symptom resolution, production of infectious agent will continue. Late manifestations of HIV are those of acquired immune deficiency syndrome which is characterized by opportunistic infections such as Pneumocystis pneumonia, Cryptosporidia diarrhea, or Candida esophagitis. Subacute sclerosing panencephalitis (Choice D) is a delayed manifestation of infection from measles. It occurs typically 5 to 15 years after an acute measles infection and presents with intellectual regression and personality changes, and eventually motor deterioration and death.

Objective: Acute, self-resolving viral syndromes (eg, yellow fever) follow a course that includes an asymptomatic period while the infectious agent is reproduced, followed by a symptomatic period that peaks, then improves with resolution of the infection. II Previous Next Score Report Lab Values Calculator Help Pause Clinical symptoms

26 Exam Section 1: Item 26 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 26. A 38-year-old woman who is undergoing intensive chemotherapy for acute myelogenous leukemia develops pancytopenia. She also has pain, tingling, and itching of the skin of the right lower lip. One day later, a group of vesicles containing clear fluid appears on the lower lip. The most appropriate pharmacotherapy for this patient is a drug that inhibits which of the following? O A) DNA polymerase B) Neuraminidase O C) Protease D) Pyrophosphatase E) Reverse transcriptase Correct Answer: A. This patient's immunocompromised state plus prodrome of pain, tingling, and itching of the skin on the lip, followed by a vesicular eruption is consistent with the reactivation of herpes simplex virus in the form of herpes labialis. Treatment for herpetic infections involves drugs that inhibit viral DNA polymerase, classically by guanosine analogs such as acyclovir, valacyclovir, and famciclovir. Prior to exerting their antiviral effects, guanosine analogs must be phosphorylated by the viral enzyme thymidine kinase. They are then able to inhibit the viral DNA polymerase by terminating the nascent DNA chain during replication. These drugs are effective against herpes simplex virus and varicella zoster virus, weakly effective against Epstein-Barr virus, and not effective against cytomegalovirus. Development of a mutation in the viral thymidine kinase enzyme preventing drug phosphorylation confers resistance to guanosine analogs. Incorrect Answers: B, C, D, and E. Neuraminidase (Choice B) is involved in the release of newly developed virus from an infected cell. This enzyme is targeted by oseltamivir and zanamivir, both agents used to treat influenza. Protease (Choice C) cleaves the initial polypeptides produced by the translation of viral RNA into smaller, functional parts. Protease inhibitors are used in the treatment of HIV and include darunavir, indinavir, ritonavir, and saquinavir. Pyrophosphatase (Choice D) cleaves pyrophosphate into two phosphate ions in numerous biochemical pathways. It is used to render pathways generally irreversible as a great amount of energy is required to reverse this cleavage. It is not the target of guanosine analogs. Reverse transcriptase (Choice E) transcribes viral DNA from viral MRNA for incorporation into the host cell's DNA. This enzyme is inhibited by both nucleoside reverse transcriptase inhibitors, such as abacavir and didanosine, and non-nucleoside reverse transcriptase inhibitors, such as efavirenz and nevirapine. These medications are used to treat HIV.

Objective: Acyclovir, famciclovir, and valacyclovir are guanosine analogs that inhibit viral DNA polymerase used to treat herpes simplex virus infections. Previous Next Score Report Lab Values Calculator Help Pause

37 Exam Section 1: Item 37 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 37. A 49-year-old woman comes to the emergency department because of a 3-day history of fever, shortness of breath, and confusion. She is a postal worker. Her temperature is 38.4°C (101.2°F), respirations are 28/min, and blood pressure is 100/60 mm Hg. Physical examination shows nuchal rigidity. Breath sounds are decreased on the right side of the chest. A lumbar puncture is done. Analysis of cerebrospinal fluid (CSF) shows: Glucose Total protein Leukocyte count Segmented neutrophils Monocytes RBC 18 mg/dL 138 mg/dL 638/mm3 87% 13% 2300/mm3 A Gram stain of CSF shows large, gram-positive, spore-forming bacilli. A chest x-ray shows marked widening of the mediastinum. Which of the following extracellular virulence factors most likely enables the causal organism to evade phagocytosis? O A) Alginate B) Glucuronoxylomannan C) Hyaluronic acid D) Polyglutamic acid E) Polyribitol phosphate Correct Answer: D. Anthrax is the disease caused by the bacterium Bacillus anthracis, which manifests with pulmonary, cutaneous, or gastrointestinal syndromes. The hallmarks of pulmonary anthrax include cough, shortness of breath, and chest pain within weeks to months after inhalation. Pulmonary anthrax is often fatal, which may occur via hemorrhagic mediastinitis. Anthrax also causes cutaneous lesions, often painless, necrotic eschars that are generally easily treated with antibiotics. Ingestion of anthrax-infected meat can cause diarrhea and abdominal pain. The pathogen is a gram-positive rod and facultative anaerobe, which arranges in chains on microscopy. The bacterium has two principal virulence factors, which are its polyglutamic acid capsule and toxin production. The polyglutamic acid capsule serves to prevent phagocytosis. Specific toxins produced by the bacterium include lethal factor (LF) and edema factor (EF). LF leads to apoptosis of macrophages, while EF alters cyclic adenosine monophosphate signaling pathways within the immune cells. Incorrect Answers: A, B, C, and E. Alginate (Choice A) is the ionized or salt form of allodynic acid. It is the monomer of a polysaccharide which produces a viscous gel when hydrated, and is a key component of the biofilm produced by Pseudomonas aeruginosa, which facilitates antibiotic resistance for this organism. Glucuronoxylomannan (Choice B) is a polysaccharide constituent of the capsule of Cryptococcus species, interfering with leukocyte motility and preventing the phagocytosis of this pathogen. In immunocompetent hosts, Cryptococcus may colonize the respiratory tract without causing symptoms, however, in patients with an immunocompromised state (eg, AIDS, chemotherapy), it can lead to life-threatening infections such as meningitis or disseminated cryptococcosis. Hyaluronic acid (Choice C) is a glycosaminoglycan that is a key component of human extracellular matrix. It is a major component of synovial fluid as the long polymers serve to increase the viscosity of synovial fluid. It is not a known virulence factor of anthrax. It is, however, a polysaccharide that makes up the capsule of Group A Streptococcus. Polyribitol phosphate (Choice E) is found in high concentration in the cell walls of gram-positive bacteria. It is also a known virulence factor for Haemophilus influenzae type B. It mainly functions to provide rigidity to the cell wall of these bacteria and prevent phagocytosis.

Objective: Anthrax is a pathogenic bacterium, which uses a polyglutamic acid capsule to evade phagocytosis. Many other bacteria produce polysaccharide or glycosaminoglycan capsules that act in a similar fashion to protect the bacteria and increase virulence. I3D Previous Next Score Report Lab Values Calculator Help Pause

67 Exam Section 2: Item 17 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 17. A 45-year-old man who works in a government building is brought to the emergency department because of a 2-hour history of severe shortness of breath with obvious bronchoconstriction and cough productive of copious secretions. He also has had nausea, vomiting, and diarrhea. At least 20 other people in his office building were affected by similar symptoms at approximately the same time. Two of these coworkers died on the way to the hospital. Only a few of the affected employees ate together in the government cafeteria that day. His respirations are 36/min. Physical examination shows excessive lacrimation. Which of the following is the most likely cause of the sudden disease outbreak? A) Anthrax infection B) Anticholinesterase poisoning C) Botulinum toxin poisoning D) Staphylococcus endotoxin poisoning E) Viral pneumonitis and gastroenteritis Correct Answer: B. Anticholinesterase poisoning is caused by the inhibition of acetylcholinesterase (ACHE), which is an enzyme that acts to degrade acetylcholine (ACh) in the neuromuscular junction. ACh acts on both nicotinic and muscarinic ACh receptors. Nicotinic ACh receptors are found in autonomic ganglia, adrenal medulla, and skeletal muscle, whereas muscarinic ACh receptors are found in heart, brain, smooth muscle, exocrine glands, and sweat glands. Inhibitors of acetylcholinesterase inhibit ACh degradation and leads to an excess concentration of ACh in the neuromuscular junction. This can result in vomiting, diarrhea, urination, diaphoresis, salivation, lacrimation, bronchorrhea, bronchospasm, bradycardia, seizures, and muscle weakness. Common causes of anticholinesterase poisoning include organophosphates, which are commonly found in insecticides and exposure can occur transcutaneously, or via inhalation or ingestion. Treatment involves decontamination along with atropine to antagonize the muscarinic receptors, and 2- pralidoxime to regenerate functional acetylcholinesterase. Incorrect Answers: A, C, D, and E. Anthrax (Choice A) or Bacillus anthracis is a gram-positive rod that forms spores and produces a toxin. There are several manifestations of anthrax. Cutaneous anthrax is characterized by a painless necrotic ulcer and is generally treatable and nonfatal. Pulmonary anthrax causes a flu-like illness, fever, mediastinitis, pulmonary hemorrhage, and if it leads to septic or hemorrhagic shock is often fatal. Intestinal anthrax presents with nausea, vomiting, and diarrhea. It does not cause bradycardia, bronchoconstriction, or bronchorrhea, and lacrimation would also be uncommon. Botulinum toxin (Choice C) is synthesized by Clostridium botulinum and prevents the release of acetylcholine into neuromuscular junctions, causing flaccid paralysis. In contrast with anticholinesterase poisoning, there would be a decreased concentration of ACh in the neuromuscular junction. Staphylococcus endotoxin poisoning (Choice D) is not commonly recognized. Endotoxin is generally used to describe a toxin that is an innate part of a bacterium, not something that is secreted. An example includes lipopolysaccharide, which is found in gram- negative bacteria. Staphylococcus species, especially S. aureus, is associated with the secretion of several exotoxins, including enterotoxin, exfoliatin, and toxic shock syndrome toxin. Enterotoxins commonly cause nausea, vomiting, diarrhea, and intestinal cramping after ingestion of food contaminated by S. aureus. It would not cause salivation, lacrimation, bronchorrhea, or bronchoconstriction. Viral pneumonitis and gastroenteritis (Choice E) would cause a presentation of cough, sputum production, nausea, vomiting, and diarrhea. It would not present with lacrimation and is seldom fatal.

Objective: Anticholinesterase poisoning is caused by the inhibition of acetylcholinesterase, which is an enzyme that acts to degrade acetylcholine (ACh) in the neuromuscular junction. Excess ACh can lead to vomiting, diarrhea, urination, diaphoresis, salivation, lacrimation, bronchorrhea, bronchospasm, bradycardia, seizures, and muscle weakness. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

142 Exam Section 3: Item 42 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 42. A 22-year-old woman comes to the physician for a follow-up examination. One year ago, she was diagnosed with a pulmonary embolism. Two years ago, she delivered a female stillborn at 23 weeks' gestation. Physical examination today shows no abnormalities. Laboratory studies show a platelet count of 250,000/mm3, a prothrombin time within the reference range, and an increased partial thromboplastin time. The findings in this patient are most consistent with which of the following conditions? O A) Antiphospholipid antibody syndrome B) Factor V Leiden mutation C) Increased factor VIII (antihemophilic factor) concentration D) Protein C deficiency E) Prothrombin G20210A mutation Correct Answer: A. The combination of unprovoked venous thromboembolic disease, fetal demise, and prolonged partial thromboplastin time in a young woman raises suspicion for antiphospholipid antibody syndrome (APS). APS is an autoimmune disease characterized by the presence of one of more circulating antiphospholipid antibodies (eg, Lupus anticoagulant, anticardiolipin antibody, and anti-beta-2-glycoprotein-1 antibody). This may occur as a primary condition, or secondary to an additional autoimmune disease, most commonly systemic lupus erythematosus. It typically presents with a history of unprovoked arterial or venous thromboembolic disease and/or spontaneous abortions. The prolonged aPTT does not indicate an increased risk for bleeding but is falsely prolonged in the presence of antiphospholipid antibodies, which interfere with the diagnostic laboratory study. Treatment typically requires therapeutic anticoagulation. Incorrect Answers: B, C, D, and E. Factor V Leiden deficiency (Choice B) is an inherited condition that predisposes to venous thromboembolism and is caused by a point mutation in the F5 gene of factor V. Factor V augments the action of thrombin and is normally inactivated by protein C. This mutation renders factor V resistant to inactivation from protein C, thereby promoting thrombosis. Pregnancy loss is not a typical feature of this disease. Increased factor VIII (antihemophilic factor) concentration (Choice C) is found in conditions including pregnancy, malignancy, and hyperthyroid states. Increased concentrations may predispose to coronary artery thrombosis or stroke, but the association is weak. Protein C deficiency (Choice D) results in an increased risk for venous thromboembolism as a result of the loss of its normal inhibitory action on factors Va, VIlla, and X. Thrombosis can occur at unusual sites such as the mesenteric or portal veins, and this disease also predisposes to warfarin-induced skin necrosis. Pregnancy loss and prolonged PTT are not characteristic. Prothrombin G20210A mutation (Choice E) is a gain-of-function mutation that occurs as a result of substitution of an adenine for guanine at position 20210 of the prothrombin gene. This results in increased circulating concentrations of prothrombin, which predisposes to venous thromboembolism.

Objective: Antiphospholipid antibody syndrome is an autoimmune condition in which the presence of circulating antiphospholipid antibodies predisposes to venous and arterial thrombosis. Recurrent pregnancy loss is also a common presenting feature. Previous Next Score Report Lab Values Calculator Help Pause

89 Exam Section 2: Item 39 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 39. A 28-year-old woman, gravida 4, para 0, aborta 3, with systemic lupus erythematosus comes to the physician at 12 weeks' gestation for an initial prenatal examination. She is concerned that she will not be able to carry her current pregnancy to term. The presence of which of the following is most likely increasing the risk for complications during pregnancy in this patient? O A) Antimitochondrial antibodies B) Antineutrophil cytoplasmic antibodies O C) Antiphospholipid antibodies D) Antistreptolysin O antibodies E) Rheumatoid factor Correct Answer: C. Antiphospholipid syndrome (APLS), which is an autoimmune disorder that can be primary or secondary to an additional autoimmune disorder (eg, systemic lupus erythematosus), often presents with recurrent miscarriages, along with arterial or venous thromboembolic disease. Diagnosis of APLS requires the presence of a clinical event such as thrombosis or stillbirth, along with blood tests separated by at least three months that detect antiphospholipid antibodies including anticardiolipin antibodies, anti- apolipoprotein antibodies, or lupus anticoagulant factor. Pregnancy-related complications involving the fetus include stillbirth, intrauterine growth restriction, and preterm labor. Maternal complications are frequent and include deep venous thrombosis, or arterial or venous blood clots in nearly any organ. Strokes and pulmonary emboli are common in pregnant women with APLS. Clots involving the placenta may lead to infarction, which is one hypothesized mechanism for the pathogenicity related to the fetus. The antibodies that are produced in APLS often bind to or interfere with the action of proteins C and S, shifting the balance of hemostasis in favor of hypercoagulability. Incorrect Answers: A, B, D, and E. Antimitochondrial antibodies (Choice A) are associated with primary biliary cholangitis, which affects the liver and biliary tree. Symptoms include pruritus, weight loss, early satiety, and nausea, with eventual progression to jaundice, hepatic dysfunction, and end- stage liver disease. Antineutrophil cytoplasmic antibodies (Choice B) are associated with vasculitides such as granulomatosis with polyangiitis, which classically presents with sinopulmonary inflammation and nephritic syndrome. Antistreptolysin O antibodies (Choice D) are classically associated with rheumatic fever and poststreptococcal glomerulonephritis. Rheumatoid factor (Choice E) is classically associated with rheumatoid arthritis but may be nonspecifically increased in multiple rheumatologic disease states (eg, mixed connective tissue disease).

Objective: Antiphospholipid syndrome, an autoimmune disorder that can be primary or secondary to an additional autoimmune disorder (eg, systemic lupus erythematosus), often presents with recurrent miscarriages along with arterial or venous thromboembolic disease. The presence of antiphospholipid antibodies confirms the diagnosis when paired with a known, associated clinical event. Previous Next Score Report Lab Values Calculator Help Pause

1 Exam Section 1: Item 1 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 1. A75-year-old man has decreased exercise tolerance and occasional syncopal episodes related to exertion. Left-heart catheterization discloses this pressure tracing (Ao -aorta; LV -left ventricle; LA - left atrium). The most frequently encountered mechanism underlying this disorder is which of the following? ЕCG A) Amyloid deposition 150- B) Calcification C) Cystic medial necrosis D) Pericardial effusion 100- O E) Septal hypertrophy Ao 50- LV LA Time Time- Correct Answer: B. The pressure tracing demonstrates a large pressure gradient between the left ventricle and aorta during systole, which is consistent with the diagnosis of aortic stenosis. In the setting of aortic stenosis, the left ventricle must generate higher pressures in order to overcome the opening resistance of the stenotic valve. This increased pressure is not fully transmitted across the stenotic valve to the aorta, which causes a delay in the aortic peak pressure. In a normal aortic valve, the pressure tracing in the left ventricle and aorta follow a similar curve during systole while the valve is open. Calcification of the valve is a pathologic consequence of mechanical stresses on the heart valves, and results from repetitive microtrauma from the opening and closing of valve leaflets with associated chronic inflammation. Many people will develop some degree of valve stenosis because of chronic inflammation with resultant calcification and fibrosis over time, with the aortic valve most commonly affected. Early-onset aortic stenosis can occur in the setting of bicuspid aortic valve or chronic rheumatic heart disease. Patients may report fatigue, shortness of breath, cough, diminished exercise tolerance, or syncope with exertion. Physical examination findings include a crescendo-decrescendo systolic murmur best heard at the upper right sternal border, and pulsus parvus et tardus (weak and delayed) may be noted on examination of peripheral pulses. As a result of the chronic increased afterload from a fixed obstruction by the valve, left ventricular hypertrophy, and resultant diastolic dysfunction can occur. Incorrect Answers: A, C, D, and E. Amyloid deposition (Choice A) occurs in cardiac amyloidosis, which is an infiltrative cardiomyopathy caused by the deposition of amyloid protein throughout the extracellular space of the heart. It does not directly cause aortic stenosis. Cystic medial necrosis (Choice C) is a disorder of the large arteries (particularly the aorta) which is associated with connective tissue diseases such as Marfan syndrome, Ehlers-Danlos syndrome, and annuloaortic ectasia. It is characterized by basophilic ground substance deposition in the tunica media with cyst-like lesions. It increases the risk for dissection, aneurysm, or rupture of the aorta, but it is not associated with aortic stenosis. Pericardial effusion (Choice D) refers to the presence of an abnormal amount of fluid in the pericardium. If severe enough, it can potentially progress to cardiac tamponade with compression of the cardiac chambers and obstructive shock. It typically causes the equalization of pressures amongst the four heart chambers. Septal hypertrophy (Choice E) is present in hypertrophic cardiomyopathy. If there is severe hypertrophy, the cardiac pressure tracing is like that seen in aortic stenosis with a large pressure gradient between the left ventricle and the aorta. This is because of left ventricular outflow tract obstruction from the septal hypertrophy. However, aortic stenosis is a far more common disorder, especially in older adults.

Objective: Aortic stenosis can be identified on a cardiac pressure tracing by the presence of a large pressure gradient between the left ventricle and the aorta. It is most commonly caused by age-related fibrosis and calcification. II Next Score Report Lab Values Calculator Help Pause Pressure (mm Hg)

73 Exam Section 2: Item 23 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 23. A 25-year-old woman comes to the physician because of a 9-month history of fatigue, excessive sweating, heat intolerance, palpitations, and nervousness; she has had a 9-kg (20-lb) weight loss during this period. Her pulse is 105/min. Physical examination shows moist skin and a goiter. Which of the following is most likely decreased in this patient? O A) lodide uptake by the thyroid gland B) Number of thyroid-stimulating hormone receptors C) Peripheral conversion of thyroxine (T) to triiodothyronine (T3) D) Recycling of intrathyroid iodide E) Thyrotropin-releasing hormone concentration Correct Answer: E. Thyrotropin-releasing hormone (TRH) concentration is most likely decreased in this patient with symptoms of hyperthyroidism. Symptoms of hyperthyroidism include heat intolerance, sweating, hair loss, warm, flushed skin, amenorrhea, diarrhea, palpitations, and weight loss. Vital signs may disclose hyperthermia, tachycardia, tachypnea, and hypertension. Examination may show an enlarged thyroid, or a thyroid nodule. There are many etiologies of hyperthyroidism, the most common being Graves disease, in which a thyroid-stimulating antibody is produced that increases the production of thyroid hormone by stimulating the thyroid-stimulating hormone (TSH) receptor in place of TSH itself. Patients are often younger, female, and present with a diffusely enlarged thyroid gland. As the thyroid produces thyroxine and triiodothyronine, these hormones exert negative feedback on the hypothalamus and pituitary, resulting in downregulation of TRH and TSH in primary hyperthyroid states. Incorrect Answers: A, B, C, and D. lodide uptake by the thyroid gland (Choice A) is incorrect, as the production of thyroid hormone requires iodide. Similarly, recycling of intrathyroid iodide (Choice D) would be increased as well because of the increased synthesis of thyroid hormone, an iodide- dependent process. A radioactive iodide uptake scan would likely reflect diffusely increased uptake in this case of a patient with a large, non-nodular goiter. Number of thyroid-stimulating hormone receptors (Choice B) may or may not be definitively affected depending on the cause of hyperthyroidism. Peripheral conversion of thyroxine (T4) to triiodothyronine (T3) (Choice C) would be increased in states of hyperthyroidism, as the excess production of T4 by the thyroid provides peripheral deiodinases with excess substrate.

Objective: As the thyroid produces thyroxine and triiodothyronine, these hormones exert negative feedback on the hypothalamus and pituitary, resulting in the decreased production of TRH and TSH in primary hyperthyroid states. Previous Next Score Report Lab Values Calculator Help Pause

120 Exam Section 3: Item 20 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 20. In a clinical study of a molecular variant hypothesized to be related to schizophrenia, 200 subjects are examined for a particular single nucleotide polymorphism. The results are shown in the table. Genotype AA AG GG Number of Controls 50 80 70 If A is one allele and G is the other, which of the following is the frequency of allele A? O A) 130/200 B) 130/400 O C) 140/200 D) 140/400 E) 180/200 F) 180/400 Correct Answer: F. In this study, there are 200 individuals who each have two alleles at the single nucleotide site of interest, bringing the total number of alleles in the population to 400. Each individual with homozygosity for allele A (AA) has two A alleles and each individual with heterozygosity at the site (AG) has one A allele, bringing the total number of A alleles in the population to 180. The frequency of allele A is then 180/400. Assuming Hardy-Weinberg equilibrium, the sum of the frequency of two alleles in a population is equal to one. The sum of the total number of homozygous dominant subjects, heterozygous subjects, and homozygous recessive subjects also is equal to one in Hardy-Weinberg equilibrium. This is expressed by the equations p + q = 1 and p2 + 2pg + q? = 1, where p and q are the allele frequencies of the two alleles in question. Incorrect Answers: A, B, C, D, and E. 130/200 (Choice A), 140/200 (Choice C), and 180/200 (Choice E) utilize the total number of subjects in the denominator rather than the total number of alleles in the population, which is 400. 130/400 (Choice B) uses the appropriate denominator regarding the total number of alleles in the population, but for the numerator sums the total number of individuals with an A allele, rather than the total number of A alleles within the study population. 140/400 (Choice D) uses the appropriate denominator with regard to the total number of alleles in the population, but the numerator only accounts for the number of G alleles in the homozygous GG subjects.

Objective: Assuming Hardy-Weinberg equilibrium, the frequency of a dominant and recessive allele within a population can be determined using the equations p + q = 1 and p² + 2pq + q? = 1, where p and q are the allele frequencies of the two alleles in question. The total number of alleles at a single given nucleotide in a population is equal to two times the number of individuals in the population. Previous Next Score Report Lab Values Calculator Help Pause

59 Exam Section 2: Item 9 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 9. An 18-year-old man has a 1-hour history of wheezing, chest tightness, and shortness of breath. He weighs 60 kg (132 Ib). Minute ventilation is 14 L/min (N=5-10 L/min). Which of the following sets of arterial blood gas findings is most likely present in this patient? 111 Pco 2 (mm Hg) Po 2 (mm Hg) pH A) 7.30 70 50 B) 7.30 39 103 C) 7.40 30 75 D) 7.40 39 103 E) 7.48 30 75 F) 7.48 70 50 Correct Answer: E. This patient's presenting symptoms of wheezing, chest tightness and dyspnea as well as hyperventilation are suggestive of an asthma exacerbation. Asthma is characterized by the reversible obstruction to airflow as a result of spasm of smooth muscle in the bronchi and small airways. It is often triggered by an allergen (eg, smoke, pollen), exercise, or temperature change. Dyspnea and hypoxia stimulate an increase in respiratory drive, which in turn leads to tachypnea or increased tidal volume; if the airway resistance remains low enough such that ventilation is not entirely compromised, hyperventilation occurs leading to respiratory alkalosis as acidic carbon dioxide is eliminated. Characteristic arterial blood gas findings in early asthma include hypoxia and respiratory alkalosis (increased pH, decreased PCO,). If pH begins to fall or CO, begins to rise during an asthma exacerbation with ongoing respiratory distress, or if respiratory acidosis is identified, decompensation and a severe, life-threatening exacerbation may be occurring. Mild acute exacerbations of asthma are treated with inhaled B-adrenergic agonists, antimuscarinic agents, and glucocorticoids. Incorrect Answers: A, B, C, D, and F. The arterial blood gas findings in Choice A include acidosis, hypercapnia, and hypoxia, and are suggestive of respiratory acidosis. This condition commonly occurs because of decreased central respiratory drive secondary to opioid use, acute respiratory distress syndrome, neuromuscular diseases (eg, myasthenia gravis or myasthenic [Lambert-Eaton] syndrome), or because of chronic conditions such as chronic obstructive pulmonary disease or obesity hypoventilation syndrome. If a respiratory acidosis of this degree with hypoxia is seen in asthma, a severe, life-threatening exacerbation may be occurring. The arterial blood gas findings in Choice B include acidosis and normal arterial partial pressures of carbon dioxide and oxygen and may suggest metabolic acidosis or a mixed derangement (eg, concomitant respiratory derangement). The differential diagnosis of metabolic acidosis is broad; examples include ketoacidosis, lactic acidosis, ingestion of toxins such as ethylene glycol or methanol, and diarrhea or renal tubular acidosis. Evaluation of the anion gap can help to narrow the differential of metabolic acidosis. An asthma exacerbation does not generally lead to metabolic acidosis unless a superimposed lactic acidosis is also present because of severe hypoxia. The arterial blood gas findings in Choice C include a normal pH, hypocapnia, and hypoxia, suggestive of an early stage of respiratory alkalosis or a mixed derangement. This patient presents with an increased minute ventilation and is likely to have more significant alkalosis rather than a normal pH. These arterial blood gas findings in Choice D are within normal limits, which is unlikely given the patient's presentation with an acute asthma exacerbation and abnormally increased minute ventilation. The arterial blood gas findings in Choice F include alkalosis, hypercapnia, and hypoxia, and are suggestive of metabolic alkalosis. Metabolic alkalosis commonly results from the loss of hydrogen ions, such as in the setting of diarrhea or vomiting, or from volume contraction. Respiratory compensation for metabolic alkalosis leads to retention of carbon dioxide. Asthma exacerbation is not a common cause of metabolic alkalosis.

Objective: Asthma exacerbations, secondary to a reversible obstruction to airflow, typically present with respiratory alkalosis because of hyperventilation when mild. If severe, symptoms of respiratory distress will remain present (eg, shortness of breath, tachycardia, tachypnea, hypoxia, poor air movement, biphasic wheezing, accessory muscle use) and the arterial blood gas analysis may appear within reference ranges or demonstrate developing or worsening respiratory acidosis. II Previous Next Score Report Lab Values Calculator Help Pause

79 Exam Section 2: Item 29 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment Vmer [X] max v = K +[X] m 29. The formula shown describes the rate of metabolism of Drug X where v is the rate of metabolism, Vmax is the maximal rate of metabolism, and Michaelis constant [K] is the plasma concentration at one half the maximum rate of metabolism. If the concentration of Drug X is 100 times greater than the Km of the metabolic enzyme for Drug X, which of the following conclusions is most appropriate? A) First-order kinetics governs the rate of the reaction B) Further metabolism of Drug X is inhibited C) Metabolism of Drug X involves covalent bonding of the enzyme D) The velocity of the reaction is proportional to the concentration of Drug X E) A zero-order rate of metabolism governs the reaction Correct Answer: E. The Michaelis-Menten equation is a non-linear equation used to model the pharmacokinetics of enzymatic elimination (metabolism) of drugs in which the rate of the elimination reaction is dependent on the concentration of available drug. This model assumes a finite number of enzymes; at low concentrations of the drug, the reaction rate increases rapidly as more drug is added. However, as the enzymes become saturated, the relative increase in the rate of metabolism slows, with minimal differential change in the metabolic rate with increased drug concentration. The Michaelis constant Km is the concentration of the drug at which metabolism proceeds at half the maximal rate. At 100 times the Km, the elimination enzymes will likely be completely saturated. In this scenario, the equation can be reduced to v = Vmax, and a zero-order rate of metabolism governs the reaction. Zero-order kinetics refers to a constant rate of elimination, regardless of differential changes in drug concentration, as the rate of elimination is capacity-limited and dependent on the amount of enzyme present. The most common examples of medications with zero-order elimination kinetics are alcohol, aspirin, and phenytoin. Incorrect Answers: A, B, C, and D. First-order kinetics governs the rate of the reaction (Choice A) is not correct as it applies when the rate of drug elimination is directly proportional to the concentration of the drug being metabolized. A constant percentage of drug will be eliminated per unit time. Further metabolism of Drug X is inhibited (Choice B) would not occur, as metabolism will continue at a rate which will approach Vmax given the enzymes are completely saturated with Drug X. Metabolism of Drug X involves covalent bonding of the enzyme (Choice C) is also incorrect, as this is usually seen in irreversible inhibitor binding, which inactivates the enzyme by changing its structure. Aspirin is an example of an irreversible inhibitor which becomes covalently bound to the cyclooxygenase enzyme. The velocity of the reaction is proportional to the concentration of Drug X (Choice D) will occur at low concentrations, when the velocity of reactions will typically follow Michaelis-Menten kinetics and be proportional to the concentration of Drug X. In the setting of enzymatic saturation from excess Drug X, the system will behave with zero-order kinetics.

Objective: At high concentrations of Drug X, reactions which follow Michaelis-Menten pharmacokinetics will demonstrate zero-order kinetics because of enzymatic saturation. The maximal rate of reaction in a completely saturated environment will remain constant, independent of further changes in concentration. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

117 Exam Section 3: Item 17 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 17. A 42-year-old man comes to the physician because of progressive weakness of his hands during the past 10 months. Physical examination shows atrophy of the hand muscles with fasciculations. There are also fasciculations of the tongue and trapezius muscles. Electromyography confirms the diagnosis of amyotrophic lateral sclerosis. Which of the following initial statements by the physician is most appropriate? A) "You have amyotrophic lateral sclerosis, also called Lou Gehrig disease, which is a terminal illness. You will eventually need to go on a respirator. Have you thought about advance directives?" B) "You have an incurable illness. I'm sorry to say there is nothing I can do for you. Are you religious?" C) "You have Lou Gehrig disease. Have you ever heard anything about this disease before?" D) "You have a mild case of amyotrophic lateral sclerosis, also called Lou Gehrig disease. We'll check it every few months just to see if it is progressing." E) "You may have amyotrophic lateral sclerosis. We'll check every few months and if it progresses, then we'll know for sure." Correct Answer: C. Bad news should be delivered honestly, directly, and using small pieces of information at a time. This patient now has a confirmed diagnosis of amyotrophic lateral sclerosis, so the physician should inform him of the diagnosis. To deliver information that is relevant to the patient and to address any misconceptions the patient may have, the physician should open the conversation by asking the patient what he already knows about the diagnosis. Since Lou Gehrig disease is the more well-known term for the disease among people outside of the medical community, the physician should use this term when asking the patient about his preconceptions. Incorrect Answers: A, B, D, and E. Immediately informing the patient he has a "terminal illness" (Choices A and B) may overwhelm the patient with too much information. Given the gravity of this information, the physician should deliver the news in small pieces to allow the patient to process the information. Choices A and B also eliminate the opportunity to inquire about the patient's preconceptions about the illness so that further discussion can be tailored to the patient's knowledge and fears. Even though there is not a cure for amyotrophic lateral sclerosis, physicians can provide palliative treatment and emotional support and should not discharge the patient from their care (Choice B). Implying that this confirmed terminal diagnosis is mild or unclear (Choices D and E) is not recommended. According to the ethical principle of autonomy, patients who have a confirmed terminal diagnosis should be informed of the diagnosis. Having this information will empower the patient to make medical decisions related to the diagnosis and begin thinking about end-of-life options.

Objective: Bad news should be delivered honestly, directly, and using small pieces of information at a time. To tailor the discussion to the patient's preconceptions and fears, the physician should open the conversation by asking the patient what they already know about the diagnosis. Previous Next Score Report Lab Values Calculator Help Pause

177 Exam Section 4: Item 27 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 27. A 1-year-old girl is brought to the physician because of multiple bacterial infections since birth. She is at the 40th percentile for length and 40th percentile for weight. She has white hair, pale skin, blue irises, and prominent red pupils. A complete blood count shows neutropenia. A peripheral smear shows giant granules in the neutrophils. This patient's disorder is most likely due to which of the following pathogenetic mechanisms? A) Bone marrow suppression B) Defect in leukocyte adhesion C) Defect in phagolysosome function D) Myeloperoxidase deficiency E) NADPH oxidase deficiency Correct Answer: C. A defect in phagolysosome function is the pathogenetic mechanism underlying Chediak-Higashi syndrome. Chediak-Higashi syndrome is a rare, autosomal recessive disorder of the immune system caused by mutations in the lysosomal trafficking regulator gene (LYST). The protein encoded by the LYST gene is essential for normal formation and transportation of lysosomes within the cell. Disruption of normal phagolysosome function leads to immunodeficiency as giant, abnormal granules accumulate within neutrophils and other granulocytes. Symptoms are often apparent in infancy or early childhood. The clinical manifestations include frequent bacterial infections, oculocutaneous albinism, peripheral neuropathy, and progressive neurologic dysfunction. Some patients may present with hemophagocytic lymphohistiocytosis (HLH), with fever, pancytopenia, and coagulopathy. Incorrect Answers: A, B, D, and E. Bone marrow suppression (Choice A) presents with pancytopenia and increased risk for infection. There are multiple acquired and inherited causes. Acquired causes include adverse effects of medication, toxins, myelodysplastic syndrome, and viral infections. A common inherited cause is Fanconi anemia, characterized by short stature, microcephaly, developmental delay, and hyperpigmented maculae skin lesions. A defect in leukocyte adhesion (Choice B) caused by abnormal LFA-1 integrin (CD18) protein is the defining feature of leukocyte adhesion deficiency type I. It is usually characterized by recurrent bacterial infections and delayed detachment of the umbilical cord. In contrast to Chediak-Higashi syndrome, patients will have increased concentrations of neutrophils in the blood. Myeloperoxidase deficiency (Choice D) is an autosomal recessive immune disorder caused by mutations in the MPO gene on chromosome 17. It is associated with fungal infections. NADPH oxidase deficiency (Choice E) is the mechanism for chronic granulomatous disease. Impairment of the neutrophil respiratory burst leads to an increased susceptibility to catalase-positive organisms.

Objective: Chediak-Higashi syndrome is a rare immune disorder caused by mutations in the LYST gene leading to abnormal phagolysosome function. Giant granules in neutrophils and platelets are characteristic of the disease. The syndrome can be recognized early in life with recurrent bacterial infections, oculocutaneous albinism, and neurologic dysfunction. Previous Next Score Report Lab Values Calculator Help Pause

78 Exam Section 2: Item 28 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 28. A 65-year-old man comes to the physician 2 weeks after noticing a lesion on his neck. He is an avid golfer, and he always wears a hat but does not always use sunscreen while playing. Examination of the neck shows a 10-mm, pearly pink, raised nodule. This patient's condition most likely originated in which of the following? O A) Epidermis B) Papillary dermis C) Reticular dermis D) Subcutaneous tissue E) Deep fascia Correct Answer: A. Basal cell carcinoma is the most common form of skin cancer and is derived from the basal cells of the epidermis as a result of mutations in the sonic hedgehog pathway or p53 tumor suppressor. The primary risk factor for developing basal cell carcinoma is ultraviolet light exposure, though age, fair skin, and family history also contribute. Basal cell carcinomas typically present as pink, pearly papules or nodules with rolled borders and central ulceration most commonly on the sun exposed areas of the head and neck. Subtypes of basal cell carcinoma include those that are less invasive, such as superficial and nodular, and those that are more invasive, such as infiltrative and morpheaform. In all subtypes, the risk for metastasis is extremely low, though the tumors can invade locally and cause significant tissue damage. Treatment of basal cell carcinoma requires surgical removal, either via excision or Mohs micrographic surgery. Incorrect Answers: B, C, D, and E. The dermis, located directly inferior to the epidermis, is made up of the superficial papillary dermis (Choice B) and deep reticular dermis (Choice C). Both the papillary and reticular dermis will contain the basaloid islands found in basal cell carcinoma but are not the origin of the tumor. The dermis contains fibroblasts, capillaries, post-capillary venules, and small nerves. Pathologic proliferations of these cell types include dermatofibromas, hemangiomas, and neurofibromas. The subcutaneous tissue (Choice D) is located inferior to the dermis and made up of lobules of adipocytes separated by fibrous septae. Tumors of the adipocytes, such as lipomas, originate in the subcutaneous tissue. The deep fascia (Choice E) is a thick and dense fibrous tissue layer, which serves to protect and support muscles and other adjacent soft tissue structures. It also serves to provide a protective barrier against the spread of infection into the muscle compartments.

Objective: Basal cell carcinoma is the most common form of skin cancer and is derived from the basal cells of the epidermis. Previous Next Score Report Lab Values Calculator Help Pause

118 Exam Section 3: Item 18 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 18. A 50-year-old woman comes to the physician because of progressive shortness of breath during the past 2 years. Her respirations are 20/min. Physical examination shows cyanosis and ankle edema. Her pulmonary artery pressure, pulmonary vascular resistance, and right atrial pressure are increased; her pulmonary capillary wedge pressure is 9.3 mm Hg (N=8-16). Her ventilation improves when inhaled nitric oxide is administered. Treatment with oral bosentan is begun for long-term therapy. Which of the following mediators of pulmonary resistance in this patient will most likely be antagonized by this drug? A) Adenosine B) Calcitonin gene-related peptide C) C-reactive protein D) Endothelin E) Prostacyclin (PGI,) Correct Answer: D. Bosentan is a competitive endothelin-1 antagonist that is used for the treatment of primary pulmonary arterial hypertension. Two clinically relevant subtypes of endothelin receptors are the endothelin A (ETA) and the endothelin B (ETB) receptors. Binding of endothelin-1 to ETÁ results in vasoconstriction, whereas binding of endothelin-1 to ETB results in production of nitric oxide and vasodilation. In the context of pulmonary arterial hypertension, the aggregate effect of endothelin-1 results in vasoconstriction and an increase in pulmonary vascular resistance. Bosentan competitively antagonizes the binding of endothelin-1 to both ETA and ETB receptors. However, its greater affinity for ETA results in a vasodilatory effect and decreased pulmonary vascular resistance, limiting pulmonary hypertension, and relieving strain on the right side of the heart. Bosentan is hepatotoxic; patients treated with bosentan should be monitored for elevations of serum transaminases. Incorrect Answers: A, B, C, and E. Adenosine (Choice A) is responsible for a variety of physiologic effects, including vasodilation mediated through the effects of the adenosine receptor on adenylate cyclase. Adenosine is commonly used in the treatment of paroxysmal supraventricular tachycardia because of its antagonistic effect on the atrioventricular node. Inhibitors of the adenosine receptor result in vasoconstriction and include caffeine and theophylline. Bosentan is not known to interact with adenosine. Calcitonin gene-related peptide (CGRP) (Choice B) is a peptide molecule that plays a role in vasodilation and nociception. CGRP may play a role in the pathogenesis of migraines as a result of these properties. Bosentan does not target CGRP or its receptor. C-reactive protein (CRP) (Choice C) is an acute phase reactant that is produced by hepatocytes in response to interleukin-6. It is diagnostically useful in the context of inflammatory and infectious conditions and is often used to monitor and guide therapy for inflammatory bowel disease (eg, Crohn disease), septic arthritis, and osteomyelitis. Bosentan has no known direct effect on CRP. Prostacyclin (PG|2) (Choice E) is a prostaglandin molecule that acts to decrease both platelet aggregation and vascular tone. Synthetic prostacyclin, such as epoprostenol, are useful for the treatment of pulmonary arterial hypertension through their vasodilatory effects. However, bosentan is not a member of this drug class.

Objective: Bosentan is a competitive antagonist of endothelin-1 and causes reduced pulmonary vascular resistance. It is effective in the treatment of pulmonary arterial hypertension. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

159 Exam Section 4: Item 9 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 9. A 63-year-old woman comes to the physician because of fatigue and weakness for the past 3 months. She has not had abdominal pain or a change in bowel habits. Findings on physical examination are unremarkable. Laboratory studies show: 9.5 g/dL 29% 70 μm3 20 pg/cell Hemoglobin Hematocrit Mean corpuscular volume Mean corpuscular hemoglobin Adenocarcinoma of the colon is suspected. Which of the following best describes the most likely location of the primary tumor in the large intestine? A) Ascending O B) Descending C) Rectum D) Splenic flexure E) Synchronous multiple sites F) Transverse Correct Answer: A. The ascending colon is the second most common location of colorectal carcinoma (CRC) after the sigmoid colon. Right-sided (ascending) colon carcinomas tend to be more insidious in presentation, with minimal to no abdominal pain or stool changes present. As well, ascending colon masses are more likely to be exophytic, rather than infiltrative as in the descending colon. Because of this, diagnosis typically occurs later, increasing the risk that the CRC will be advanced and poorly differentiated. Over time, the carcinoma typically exhibits slow bleeding, resulting in a progressive, gradual-onset microcytic anemia caused by iron deficiency from the chronic blood loss. If a male or postmenopausal female presents with unexplained iron deficiency anemia, the patient should undergo colonoscopic evaluation for a potential ČRC. Incorrect Answers: B, C, D, E, and F. Descending-colon CRC (Choice B) is more likely to produce symptoms of abdominal cramping and change of stool caliber as the mass is more likely to be infiltrative in nature, which increases the risk for obstruction. As a result of this, left-sided CRC is occasionally detected earlier than right-sided CRC, although the implementation of routine colonoscopy offers the potential for early detection of both varieties. Malignancy of the rectum (Choice C) typically presents with tenesmus, hematochezia, and a potential palpable mass on digital rectal examination. It would less likely present with insidious onset microcytic anemia. The splenic flexure (Choice D) is an uncommon location of CRC and represents less than 10% of all cases. Synchronous multiple sites (Choice E) of CRC is uncommon in patients without a predisposing risk factor such as familial adenomatous polyposis or Lynch syndrome. This patient has no features to suggest such a family history, and typically the presentation of CRC in these cases is usually at a much younger age. Transverse colon (Choice F) malignancy is less common compared to the incidence of left or right-sided CRC.

Objective: CRC of the ascending colon is typically exophytic in nature, making it less likely to present with symptoms of obstruction, such as abdominal pain or stool caliber change. During its insidious growth, it can result in chronic, slow blood loss, which can manifest as progressive, gradual-onset microcytic anemia. Previous Next Score Report Lab Values Calculator Help Pause

167 Exam Section 4: Item 17 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 17. The initial event in muscle contraction is the release of calcium ion from the sarcoplasmic reticulum. Which of the following is most likely responsible for buffering the concentration of calcium ion in the sarcoplasmic reticulum? O A) Calcium-dependent protein kinase O B) Calmodulin C) Calsequestrin D) Myosin light chain O E) Troponin Correct Answer: C. Calsequestrin is a calcium-binding protein found in the sarcoplasmic reticulum and anchored to the ryanodine receptor. Each calsequestrin molecule can hold up to 50 calcium ions and thus increases the calcium storing potential of the sarcoplasmic reticulum. When calcium ions increase in the sarcoplasmic reticulum, such as when an action potential triggers calcium ions to enter through L-type calcium channels, calsequestrin binds these calcium ions and loosens its hold on the ryanodine receptor and allows the receptor to open and release calcium into the cytoplasm of the cell. This initiates the excitation-contraction coupling process. In contrast, when the concentration of calcium ions in the sarcoplasmic reticulum decreases, calsequestrin binds the ryanodine receptor more tightly and prevents it from opening and releasing further calcium ions. In this way, it buffers the concentration of calcium ions in the sarcoplasmic reticulum. Incorrect Answers: A, B, D, and E. Calcium-dependent protein kinase (Choice A) is a protein kinase regulated by calcium and calmodulin. It phosphorylates several of the regulatory proteins involved in excitation-contraction coupling, the process by which an action potential is converted into muscle contraction. It does not play a role in calcium ion buffering in the sarcoplasmic reticulum. Calmodulin (Choice B) is a calcium-binding messenger protein. It activates myosin light chain kinase, which phosphorylates the myosin head, allowing myosin to bind to actin and cause muscle contraction. When bound to calcium, calmodulin can also inhibit the calcium channels on the sarcoplasmic reticulum and regulate excitation-contraction coupling. Myosin light chain (Choice D) binds to actin filaments at the myosin-binding groove and has a second binding site for adenosine triphosphate (ATP). When ATP is bound, myosin detaches from actin. When ATP is hydrolyzed, the energy generated moves the myosin head into a cocked position. If the actin binding sites are uncovered, myosin will form a cross-bridge with actin. Phosphate is released, and the myosin head moves through a power stroke, pulling the actin fiber to shorten the sarcomere. Troponin (Choice E) is a calcium-binding protein which, along with tropomyosin, blocks the myosin-binding groove on actin. When the sarcoplasmic reticulum releases calcium, troponin binds calcium ions and causes a conformational change in tropomyosin. This change moves tropomyosin from the myosin-binding groove, allowing myosin to form a cross-bridge with actin.

Objective: Calcium is the central element in excitation-contraction coupling. It is stored in the sarcoplasmic reticulum, where it is buffered by calsequestrin. When calcium is released, it results in displacement of tropomyosin and triggers sarcomere contraction through myosin and actin cross-bridging. Previous Next Score Report Lab Values Calculator Help Pause

109 Exam Section 3: Item 9 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 9. A 68-year-old man with Parkinson disease is being treated with levodopa and carbidopa. He is doing well and has few adverse effects. Concurrent therapy with levodopa and carbidopa is useful because of which of the following characteristics of carbidopa? A) Blocks muscarinic receptors B) Does not cross the blood-brain barrier C) Enhances activity of the transporter for levodopa D) Increases intestinal metabolism of levodopa E) Inhibits monoamine oxidase Correct Answer: B. Carbidopa does not cross the blood-brain barrier (BBB), which indirectly increases dopamine availability in the basal ganglia, reversing the symptoms of Parkinson disease. Levodopa is converted to dopamine by the enzyme dopa decarboxylase in the peripheral circulation and the brain. Carbidopa inhibits dopa decarboxylase peripherally. Consequently, more levodopa is available to cross into the brain, which is necessary because dopamine itself cannot cross the BBB. In the brain, levodopa can be converted to dopamine without inhibition by carbidopa. Additionally, by preventing peripheral dopamine formation, carbidopa prevents peripheral dopamine side effects such as nausea and vomiting. Incorrect Answers: A, C, D, and E. Muscarinic receptors (Choice A), in the context of Parkinson disease therapy, are antagonized by benztropine. Acetylcholine and dopamine have opposing effects in the basal ganglia related to movement such that decreasing acetylcholine has the same effect as increasing dopamine. Thus, benztropine can improve some Parkinsonian symptoms such as tremor and rigidity. Enhances activity of the transporter for levodopa (Choice C) is incorrect, as an active levodopa transporter made of amino acids is responsible for levodopa absorption in the gut and transport across the BBB, but its activity is not altered by levodopa or carbidopa. Diets rich in amino acids that optimize intestinal absorption and BBB crossing have been proposed; however, there are no current Parkinson disease medications that utilize this mechanism. Increasing intestinal metabolism of levodopa (Choice D) would decrease luminal levodopa concentrations, decreasing the amount of levodopa absorbed from the intestine and subsequently transported into the brain to be converted to dopamine. Increasing intestinal metabolism of levodopa would increase availability of dopamine and 3-0-methyldopa, levodopa's main metabolites. Neither metabolite can cross the BBB to increase dopamine signaling in the basal ganglia. A substance with this mechanism would therefore not improve Parkinson symptoms. Medications that inhibit monoamine oxidase (Choice E) decrease the conversion of dopamine to 3,4-dihydroxyphenylacetic acid in the brain, thus increasing dopamine availability. Selegiline and rasagiline utilize this mechanism of action and are therefore used as adjunctive agents to levodopa.

Objective: Carbidopa cannot cross the blood-brain barrier and therefore increases the bioavailability of levodopa without inhibiting the conversion of levodopa to dopamine in the brain. Carbidopa also decreases peripheral dopamine concentrations, reducing the likelihood of dopamine-related side effects such as nausea. Previous Next Score Report Lab Values Calculator Help Pause

186 Exam Section 4: Item 36 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 36. A 58-year-old woman comes to the physician because of a 3-week history of shortness of breath after walking 2 blocks. She has been receiving doxorubicin as part of chemotherapy for breast cancer for 6 months. Her pulse is 110/min, and respirations are 32/min. Inspiratory crackles are heard over both lung fields. Physical examination shows 2+ pitting edema of the ankles. A chest x-ray shows cardiomegaly and interstitial edema. Cardiac studies show a decreased ejection fraction of the left ventricle. Which of the following is the most likely cause of the cardiac findings in this patient? A) Decreased oxidative phosphorylation B) Impairment of B-oxidation of fatty acids C) Inhibition of acetylcholinesterase activity D) Inhibition of metal-dependent enzyme synthesis E) Lipid peroxidation of myocyte membranes Correct Answer: E. The patient's presentation is consistent with decompensated heart failure, characterized by dyspnea on exertion, pulmonary crackles, tachycardia, tachypnea, and peripheral edema on physical examination. The diagnostic evaluation further supports this diagnosis with cardiomegaly, pulmonary interstitial edema, and a decreased ejection fraction indicating abnormal systolic function. Heart failure is a clinical syndrome with a broad range of causes. The most likely cause in this case is anthracycline-induced dilated cardiomyopathy secondary to doxorubicin exposure. Doxorubicin (and other anthracyclines) have a well-established cardiotoxicity adverse effect. Myocyte damage occurs because of the increased production of toxic reactive oxygen species that increases oxidative stress leading to lipid peroxidation of myocyte membranes and myocyte apoptosis. Anthracycline-induced dilated cardiomyopathy confers a poor prognosis. Patients managed with doxorubicin should be monitored for cardiotoxicity with routine echocardiography. Incorrect Answers: A, B, C, and D. Decreased oxidative phosphorylation (Choice A) leads to cell injury and death by impairing mitochondrial ATP synthesis. This may be seen in the setting of myocardial infarction, and with substances such as carbon monoxide, cyanide, and azide, which are potent inhibitors of the electron transport chain. Impairment of B-oxidation of fatty acids (Choice B) leads to the decreased catabolism of fatty acids into acetyl-CoA. It is the pathologic mechanism of several congenital disorders such as medium-chain acyl-CoA dehydrogenase deficiency. Doxorubicin is not associated with this pathway. Inhibition of acetylcholinesterase activity (Choice C) leads to increased concentrations of acetylcholine in synaptic clefts. Organophosphates are poisons that can cause cholinergic toxicity through this mechanism. Carbamates such as physostigmine and neostigmine are used therapeutically to reverse these effects. Inhibition of metal-dependent enzyme synthesis (Choice D) does not occur with doxorubicin. Approximately one-third of known enzymes in the human body are metalloenzymes dependent on metals for proper function.

Objective: Cardiotoxicity leading to dilated cardiomyopathy is a well-known adverse effect of doxorubicin chemotherapy. This occurs secondary to an increased concentration of reactive oxygen species and subsequent lipid peroxidation of myocyte membranes. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

168 Exam Section 4: Item 18 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 18. A study is conducted to assess the relationship between a history of cigarette smoking and the development of transitional cell carcinoma. In 10,000 patients with transitional cell carcinoma of the bladder, 6750 had a history of cigarette smoking; in 10,000 patients without transitional cell carcinoma, 1250 had a history of cigarette smoking. Which of the following best describes this study design? A) Case-control study B) Case series study O C) Cohort study D) Cross-sectional study E) Randomized control study Correct Answer: A. This study compares one group of patients with an outcome under study (transitional cell carcinoma) (cases) against a second matched group without that outcome (controls) and identifies the associated exposure within each group. This study design is known as a case-control study. Case-control studies can be conducted in a prospective or retrospective manner but are always observational studies. By grouping patients by outcome and comparing differences in the odds of exposure, case-control studies can detect associations between exposure and outcome, such as exposure to cigarette smoking and outcome of developing transitional cell carcinoma in this study. This is described statistically as an odds ratio (OR). Two unrelated variables will have an OR of 1.0, whereas positive association between an exposure and an outcome will have an OR greater than 1.0 and negatively associated variables will have an OR less than 1.0. Case-control studies are therefore capable of establishing association between exposure and outcome, but do not establish causality. Incorrect Answers: B, C, D and E. A case series study (Choice B) is a descriptive study that describes the history, possible exposures, and clinical findings of a group of patients with a similar diagnosis. Case series are non-analytic studies. They do not test a hypothesis and do not generally contain a control group. Cohort studies (Choice C) examine a group of patients with a known exposure and determine the risk for developing a given disease outcome within the exposed group compared to a control group lacking the known exposure. Cohort studies may be prospective or retrospective. Cohort studies differ from case-control studies in that patients are grouped according to exposure status in the former design and are grouped according to disease outcome status in the latter design. Cohort studies calculate relative risk (RR). An analogous cohort study to the above-described case-control study would involve comparison of a group of smokers and a group of non-smokers to evaluate the risk for developing transitional cell carcinoma. A cross-sectional study (Choice D) examines a sample population at a single point in time and describes the prevalence of a disease or risk factor. In contrast to cohort studies, which group patients according to exposure status, and case-control studies, which group patients according to disease outcome status, cross-sectional studies analyze both disease states and exposures simultaneously. A randomized control study (Choice E) is a type of prospective, interventional study design wherein patients are randomly assigned to receive a particular intervention. The intervention may be compared against placebo therapy or against standard therapy, depending on the study design and the disease being studied. When combined with blinding procedures, randomized control studies form the gold standard of medical research and generate more powerful evidence than other forms of observational or descriptive studies.

Objective: Case-control studies group patients according to disease outcome status and analyze the odds of exposure to a hazard. The appropriate statistical measure of case-control studies is the odds ratio. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

74 Exam Section 2: Item 24 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 24. A 25-year-old woman comes to the physician 2 days after noticing a mass in her right axilla. She also has a 1-week history of malaise, headaches, and night sweats. The patient says that she recently adopted a kitten and has sustained several bite and scratch marks. Her temperature is 37.8°C (100°F). Physical examination shows edema and tenderness of the right axillary lymph node. The skin over the node is erythematous, tough, and warm. There are scratches and bite marks of various ages over the upper extremities and hands. The result of an indirect fluorescent antibody test for Bartonella henselae is positive. A biopsy specimen of this lymph node is most likely to show which of the following histologic patterns? A) Diffuse neutrophil infiltration B) Granulomas containing stellate microabscesses C) Large activated lymphocytes with occasional plasma cells D) Serous inflammation with abundant intravascular fibrin and few inflammatory cells E) Vascular proliferation with fibrosis Correct Answer: B. Cat-scratch disease is caused by infection with the gram-negative coccobacillus Bartonella henselae. Infection is acquired via scratches or bites from domestic or feral cats, which may be localized or disseminated, but most commonly causes local lymphadenitis in the lymphatic drainage pattern of the scratch location. The most involved lymph nodes are the axillary and cervical lymph nodes. Histologic examination typically demonstrates necrotizing granulomas with stellate (star-shaped) microabscesses. Multinucleated giant cells may or may not be present. Incorrect Answers: A, C, D, and E. Diffuse neutrophil infiltration (Choice A) of lymph nodes is nonspecific and occurs in a wide range of infections. Neutrophils, which kill by phagocytosis, are recruited by local histocytes and lymphocytes to combat invading bacterial organisms. Large activated lymphocytes with occasional plasma cells (Choice C) describes histologic findings that may be consistent with early clonal proliferation of plasma cells in the immune response against a bacterial infection. Initially, lymphocytes enlarge and eventually transition to plasma cells, marked by a moderately sized nucleus and abundant rough endoplasmic reticulum and Golgi apparatus. Serous inflammation with abundant intravascular fibrin and few inflammatory cells (Choice D) may describe the histologic pattern seen in inflammation found within a body cavity (eg, pleura, pericardium) especially in the setting of malignancy or a hypercoagulable state. It is less consistent with common inflammatory patterns on lymph node biopsy. It may also be observed in skin blistering. Vascular proliferation with fibrosis (Choice E) may be seen for example in the retina, especially in the case of diabetic retinopathy, or in the case of angiogenesis associated with malignancy.

Objective: Cat-scratch disease is caused by infection with the gram-negative coccobacillus Bartonella henselae. Histologic examination of involved lymph nodes typically demonstrates necrotizing granulomas with stellate (star-shaped) microabscesses. Multinucleated giant cells may or may not be present. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

140 Exam Section 3: Item 40 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 40. A 65-year-old woman comes to the emergency department because of a 4-hour history of vomiting bright red blood; she also has had dizziness and nausea during this period. Her pulse is 140/min, and blood pressure is 70/30 mm Hg. Physical examination shows pale, clammy skin, diaphoresis, and decreased capillary refill time. A diagnosis of hypovolemic shock is made. Placement of a catheter in the internal jugular vein for blood transfusions is planned. Improper insertion of the catheter is most likely to result in damage to which of the following sets of underlying structures? A) Common carotid artery and pulmonary artery B) Esophagus and common carotid artery C) Esophagus and pulmonary artery D) Lung and common carotid artery E) Lung and esophagus Correct Answer: D. Placement of a central venous catheter (CVC) is frequently performed at any of three sites: the internal jugular vein, subclavian vein, and femoral vein. Indications for central venous access include volume resuscitation, emergency vascular access, and administration of vasopressors or other caustic medications. Central venous cannulation also facilitates interventions such as plasmapheresis, hemodialysis, and intracardiac pacing. The placement of an internal jugular venous catheter can be done under ultrasound guidance, or by using the anatomic landmark formed by the heads of the sternocleidomastoid approximately five centimeters superior to the clavicle. Complications related to central catheter insertion often depend on the site of insertion. At the internal jugular vein, adjacent structures include the apex of the lung and common carotid artery. latrogenic puncture of the pleura can lead to pneumothorax, with or without tension, especially in patients receiving positive pressure ventilation. The internal jugular vein commonly lies parallel and superficial to the common carotid artery; however, the common carotid artery may also be medial, lateral, or overlying the jugular vein in some cases. Inadvertent puncture of the common carotid artery near the internal jugular vein may cause a rapidly expanding hematoma, arterial dissection, or thrombosis. If unrecognized, dilation and catheter placement into the common carotid artery can lead to life-threatening hemorrhage or stroke. Incorrect Answers: A, B, C, and E. The pulmonary artery (Choices A and C) arises from the right ventricular outflow tract in the mediastinum. It is a distant structure that would be unlikely to be injured during central catheter placement unless a right-heart or pulmonary artery catheterization was attempted. The esophagus (Choices B, C, and E) would be unlikely to be injured with internal jugular catheter insertion, as it lies posterior and deep in the neck and chest, behind the trachea. It is also a midline structure, and catheter insertion in the neck is generally performed from a lateral approach.

Objective: Central venous catheter insertion is indicated for volume resuscitation, emergency vascular access, administration of vasopressors or other caustic medications, and facilitates interventions such as plasmapheresis, hemodialysis, and intracardiac pacing. A potential site of central catheter insertion, the internal jugular vein, is adjacent to the lung and common carotid artery, and complications from puncturing or injuring these structures include pneumothorax and hematoma or hemorrhage, respectively. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

12 Exam Section 1: Item 12 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 12. A 57-year-old man comes to the physician because of an intermittent cough and a 9-kg (20-lb) weight loss over the past 6 months. He has smoked 1 pack of cigarettes daily for 42 years. His serum calcium concentration is 13.3 mg/dL. An x-ray of the chest shows a 4-cm central mass in the left lung. Which of the following is the most likely diagnosis? A) Adenocarcinoma of the lung B) Large cell lymphoma C) Metastatic osteosarcoma D) Metastatic renal cell carcinoma E) Small cell carcinoma of the lung F) Squamous cell carcinoma of the lung Correct Answer: F. Squamous cell carcinoma of the lung is the second most common type of primary lung cancer after adenocarcinoma. Risk factors for all major types of lung cancer include tobacco use, secondhand smoke, asbestos, or radon exposure, and a family history of lung cancer. Features associated with squamous cell carcinoma of the lung include pulmonary cavitations, central location, and hypercalcemia because of paraneoplastic parathyroid hormone-related peptide (PTHPP) production. Histologic characteristics include polygonal cells with intercellular bridges, eosinophilic cytoplasm, keratin pearls, and extensive necrosis. Lung cancer in general typically presents with cough, unintentional weight loss, hemoptysis, chest pain, dyspnea, and hoarseness; occasionally, wheezing, focal rhonchi or hypertrophic osteoarthropathy may be noted on examination. Diagnosis is made by chest imaging and examination of a biopsy specimen. Prognosis is a function of the cancer type along with grading and staging of the disease. It is often detected once metastatic, at which point the prognosis is poor. Incorrect Answers: A, B, C, D, and E. Adenocarcinoma of the lung (Choice A) is the most common overall primary lung cancer and the most common among nonsmokers. It typically presents as a chronic consolidation in the periphery of the lung rather than centrally. It is more common in women than men. A glandular pattern is classically seen on histology with mucin-positive staining. Large cell lymphoma (Choice B) is a common form of lymphoma which may involve B or T lymphocytes and often presents with constitutional symptoms such as fatigue, unintentional weight loss, fevers, and night sweats. Histology will show characteristic abnormal enlarged lymphocytes. Physical examination findings may include diffuse lymphadenopathy and hepatosplenomegaly. Metastatic osteosarcoma (Choice C) and metastatic renal cell carcinoma (Choice D) are primary neoplasms of bone and kidney, respectively. Both types often metastasize to the lungs. Osteosarcoma causes hypercalcemia because of local destruction of bone. Renal cell carcinoma may also produce PTHPP as a paraneoplastic syndrome, also causing hypercalcemia. In the lung, metastatic cancer from other sites is more common than primary lung cancer. Metastases to the lungs typically present as multiple peripheral, rounded nodules scattered throughout both lungs. Small cell carcinoma of the lung (Choice E) is also centrally located and associated with tobacco use. It is a neoplasm of neuroendocrine cells and is associated with numerous paraneoplastic syndromes, including Cushing syndrome caused by adrenocorticotropic hormone production, syndrome of inappropriate antidiuretic hormone, Lambert-Eaton myasthenic syndrome caused by presynaptic calcium channel antibody production, and paraneoplastic myelitis, encephalitis, and subacute cerebellar degeneration. Histologic features include small dark blue tumor cells lacking nucleoli and a high nuclear to cytoplasm ratio.

Objective: Centrally located primary lung cancers include squamous cell carcinoma of the lung and small cell carcinoma of the lung. Squamous cell carcinoma is the more common subtype and is associated with hypercalcemia because of paraneoplastic PTHrP production. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

27 Exam Section 1: Item 27 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 27. A 24-year-old man comes to the physician because of fever, muscle aches, runny nose, cough, and chills for 1 day. His temperature is 39.9°C (103.8°F), pulse is 92/min, respirations are 24/min, and blood pressure is 124/78 mm Hg. Physical examination shows erythema of the throat but no pus. The lungs are clear to auscultation. Which of the following is the most likely cause of the chills in this patient? O A) An abrupt alteration in the cellular components of the blood B) An abrupt alteration in the oxygenation of the blood O C) An abrupt alteration of the hypothalamic thermoregulatory center D) Viral infection of the brain E) Viral infection of the muscles F) Viral infection of the vascular endothelial cells Correct Answer: C. Raising the body temperature set-point is a cardinal response to infection and is achieved through alteration of the hypothalamic thermoregulatory center because of signaling from pyrogenic cytokines such as interleukin-1 and interleukin-6. In addition to fever, the body uses chills (felt as rigors, which are rapid muscle contractions and relaxations) to increase the temperature through thermogenesis during muscle contraction. The patient is presenting with signs and symptoms suggestive of a viral respiratory infection, such as influenza, which would trigger release of cytokines with a resultant febrile response. When the hypothalamic set-point for body temperature rises, actual body temperature lags behind the set point. Until the actual temperature equals the set-point, the patient will experience chills, often prompting behavioral change such as insulating under a blanket or jacket, which further limits heat loss. Antipyretic medications combat the change in the hypothalamic thermoregulatory set point by limiting the degree of inflammatory mediator synthesis (eg, blocking cyclooxygenase enzymes involved in cytokine synthesis) and blunting the hypothalamic response. Incorrect Answers: A, B, D, E, and F. An abrupt alteration in the cellular components of the blood (Choice A), such as in a hemolytic anemia or a transfusion reaction, can lead to fever and chills, though this response is also mediated by pyrogenic cytokine release. While the end-effect is similar, the pathophysiology does not involve cellular components of blood, rather, the mediators are released by tissue histiocytes and local inflammatory cells. Abrupt alteration in the oxygenation of the blood (Choice B), or acute hypoxemia, can lead to respiratory distress, dyspnea, tachypnea, and use of accessory muscles to breath. It does not generally cause chills, nor does it adjust the hypothalamic thermoregulatory set point. Viral infections of the brain (Choice D), muscles (Choice E), and vascular endothelial cells (Choice F) can also present with fever and chills. This will also be mediated through the hypothalamic thermoregulatory center. In this case, the findings of fever, cough, and erythema of the posterior pharynx are more suggestive of a viral respiratory infection such as influenza. The final common pathway causing chills in infection, regardless of source, is an adjustment in the thermoregulatory set point of the hypothalamus.

Objective: Chills are involuntary, rapid muscle contractions which raise the body temperature in response to infection. The response is mediated by pyrogenic cytokine activity on the hypothalamic thermoregulatory center. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

123 Exam Section 3: Item 23 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 23. A 55-year-old woman is brought to the emergency department by police 1 hour after she was found wandering around the city park at midnight, appearing confused. She told the police that she is a prominent physician and a member of the city council. She is unwashed, malodorous, and disheveled. She is 160 cm (5 ft 3 in) tall and weighs 55 kg (121 Ib); BMI is 21 kg/m2. Her pulse is 82/min, and blood pressure is 110/82 mm Hg. Physical examination shows a broad-based gait and nystagmus. She is not oriented to person, place, or time, but she is pleasant and cooperative. Results of alcohol and drug screening are negative. An MRI of the brain is most likely to show atrophy of which of the following? A) Amygdala В) Hippоcampus C) Hypothalamus D) Mammillary bodies E) Parietal lobe Correct Answer: D. The mammillary bodies are atrophied in Wernicke-Korsakoff syndrome, a neuropsychiatric disorder commonly related to chronic alcohol use disorder and malnutrition. Patients who chronically abuse alcohol typically demonstrate a significant vitamin B1 (thiamine) deficiency as the result of poor nutritional intake and thiamine malabsorption. Over time, thiamine deficiency can result in damage and atrophy of the mammillary bodies, visible on MRI of the brain. The typical presentation of Wernicke encephalopathy is a triad of altered mental status (including disorientation), ophthalmoplegia (which can present as nystagmus), and ataxia (as indicated by this patient's broad-based gait). The symptoms of Wernicke encephalopathy are reversible, and can mimic alcohol intoxication; however, patients will have a low or undetectable serum ethanol concentration. If not immediately treated with intravenous vitamin B1 (thiamine), Wernicke encephalopathy can evolve into Wernicke-Korsakoff Syndrome, an irreversible condition that features retrograde and anterograde amnesia and the consequent tendency to confabulate, or verbalize false stories or memories, to compensate for the inability to remember. Incorrect Answers: A, B, C, and E. The amygdala (Choice A) is involved in emotional responses such as fear. Damage to the bilateral amygdala can occur in the setting of herpes encephalitis or trauma, and may present with Kluver-Bucy syndrome, characterized by disinhibited behaviors such as hypersexuality and hyperorality. The amygdala is not typically damaged in Wernicke-Korsakoff syndrome. The hippocampus (Choice B) is sometimes damaged in Wernicke-Korsakoff syndrome and may explain this patient's anterograde memory deficits. However, mammillary body atrophy is more commonly visualized on MRI of the brain in the setting of Wernicke- Korsakoff syndrome. Hippocampal atrophy is a key finding in the setting of Alzheimer disease. The hypothalamus (Choice C) is responsible for body homeostasis by secreting hormones that regulate body temperature, hunger, sleep, and other functions. The hypothalamus is not affected in Wernicke-Korsakoff syndrome. The parietal lobe (Choice E) is a large brain region that processes sensory information and maintains spatial awareness of the body. Damage to the parietal lobe can result in symptoms such as agraphia, acalculia, finger agnosia, and left-right disorientation (dominant hemisphere, known as Gerstmann syndrome) or hemineglect (nondominant hemisphere). The parietal lobe is not typically affected in Wernicke-Korsakoff syndrome.

Objective: Chronic alcohol use disorder and associated malnutrition may result in significant vitamin B1 (thiamine) deficiency, which can cause Wernicke-Korsakoff syndrome. The syndrome is characterized by symptoms such as altered mental status, ophthalmoplegia, ataxia, and retrograde and anterograde amnesia. The mammillary bodies are classically atrophied in Wernicke-Korsakoff syndrome. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

198 Exam Section 4: Item 48 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 48. A75-year-old woman with chronic obstructive pulmonary disease comes to the physician because of a 3-month history of shortness of breath while performing household chores and swelling of her feet at night. She smoked 2½ packs of cigarettes daily for 50 years until she decreased to one-half pack daily 2 years ago. She takes no medications and does not use supplemental oxygen. Her respirations are 18/min and labored. Physical examination shows cyanosis of the lips and fingertips. Crackles, rhonchi, and wheezing are heard over all lung fields bilaterally. There is a markedly prolonged expiration phase. Arterial blood gas analysis on room air is most likely to show which of the following sets of findings in this patient? pH Po2 (mm Hg) Pco, (mm Hg) HCO,- (mEq/L) A) 7.32 89 48 24 B) 7.35 65 56 30 C) 7.39 94 39 24 D) 7.41 62 43 32 E) 7.42 72 32 21 Correct Answer: B. Chronic obstructive pulmonary disease (COPD) is an obstructive lung disease characterized by decreased lung function from a combination of features of chronic bronchitis or emphysema, resulting in airflow obstruction on expiration, especially in a patient with history of smoking. The patient is presenting with COPD, characterized by shortness of breath and labored respirations along with a prolonged expiratory phase plus scattered wheezes on examination. Chronic airflow obstruction results in a primary respiratory acidosis with appropriate metabolic compensation. The arterial blood gas (ABG) will show a normal or near-normal pH, decreased PO, increased PCO, and increased HCO,. Treatment is focused on smoking cessation and supplemental oxygen to decreased mortality. Symptoms are treated with inhaled short-acting B-adrenergic agonists, long-acting B-adrenergic agonists, muscarinic antagonists, and glucocorticoids. Incorrect Answers: A, C, D, and E. Decreased pH, near-normal PO2 increased PCO, and normal HCO, (Choice A) may be seen in a primary respiratory acidosis without metabolic compensation, which occurs in acute hypoventilatory conditions such as asthma exacerbation, airway obstruction, and oversedation. Normal pH, normal PO, normal PCO, and normal HCO- (Choice C) would not be expected in chronic COPD as this set of values reflects no underlying metabolic or respiratory deficits. Normal pH, decreased PO, high-normal PCO, and increased HCO,- (Choice D) may be seen in a primary metabolic alkalosis with appropriate respiratory compensation. Among other causes, vomiting or volume depletion with aspiration may be consistent with this pattern. High-normal pH, decreased PO, decreased PCO, and decreased HCO-- (Choice E) may be seen in a primary respiratory alkalosis with metabolic compensation. Among other causes, a pulmonary embolism may be consistent with this pattern.

Objective: Chronic obstructive pulmonary disease (COPD) is an obstructive lung disease associated with tobacco use characterized by airflow obstruction and hypoventilation. Arterial blood gas (ABG) analysis typically shows a primary chronic respiratory acidosis with metabolic compensation. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

84 Exam Section 2: Item 34 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 34. A 56-year-old man with alcoholism comes to the physician because of a 2-month history of increasingly severe stomach pain and increased volumes of foul-smelling stool; he also has had a 9-kg (20-lb) weight loss during this period. He has a history of multiple visits to the emergency department because of severe abdominal pain. He has consumed one bottle of red wine daily for 5 years. His temperature is 37°C (98.6°F). Physical examination shows epigastric tenderness. His fasting serum glucose concentration is 150 mg/dL. A CT scan of the abdomen shows pancreatic calcifications. The most likely cause of this patient's current symptoms is a decrease in which of the following? A) Bile acid synthesis B) Colonic bacteria C) Duodenal pH D) Fecal elastase E) 7a-Hydroxylase activity Correct Answer: C. This patient with alcohol use disorder, steatorrhea, and chronic abdominal pain with pancreatic calcifications has chronic pancreatitis. Abdominal pain is caused by a reduced duodenal and pancreatic ductal pH, which occurs as a result of the reduced secretion of bicarbonate from the pancreas into the duodenum. Pancreatic ductal cells secrete bicarbonate to prevent pooling and early activation of digestive enzymes such as peptidases, lipase, and amylase prior to reaching the duodenum. Bicarbonate also serves to neutralize acidic gastric chyme that enters the duodenum through the pyloric valve. Impaired bicarbonate secretion can cause autodigestion of the pancreas and markedly reduced duodenal pH, which not only causes pain but also leads to chronic malabsorption observed as weight loss, vitamin deficiencies, and foul-smelling, oily diarrhea. Incorrect Answers: A, B, D, and E. A decrease in bile acid synthesis (Choice A) occurs in several rare, primarily autosomal recessive, genetic disorders where there is a mutation in enzymes required to convert cholesterol to the two primary bile acids, chenodeoxycholic acid and cholic acid. Alternatively, defects also occur in the transport of bile acids out of the liver, especially because of the obstruction of the bile ducts (eg, pancreatic cancer, cholangiocarcinoma, bile duct stricture). Symptoms include malabsorption of fat-soluble vitamins, jaundice, cirrhosis, and failure to thrive. A decrease in colonic bacteria (Choice B) occurs when normal colonic microbiota is disrupted by the administration of antibiotics or during infection with pathogenic bacteria such as Shigella species, C. jejuni, or Salmonella species. Symptoms include bloating, cramping, and diarrhea. A decrease in fecal elastase (Choice D) is a sensitive marker of exocrine pancreatic dysfunction regardless of the cause (eg, chronic pancreatitis, cystic fibrosis). A decrease of this enzyme indicates exocrine dysfunction but is not the cause of abdominal pain. 7a-hydroxylase (Choice E) is an enzyme that catalyzes the first step in the breakdown of cholesterol for the synthesis of bile acids. A decrease in this enzyme results in hypercholesterolemia, premature coronary artery disease, and gallstone formation.

Objective: Chronic pancreatitis impairs bicarbonate release from pancreatic acinar cells, predisposing to pancreatic autodigestion and abnormally acidic duodenal contents, which commonly presents as recurrent abdominal pain. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

153 Exam Section 4: Item 3 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 3. A 57-year-old man with alcoholism has a distended abdomen with shifting dullness and a fluid wave. He has caput medusae, palmar erythema, and spider angiomata. Which of the following is the most likely additional finding? O A) Digital clubbing B) Enlarged inguinal lymph nodes C) Flame-shaped retinal hemorrhages D) Gynecomastia E) Jugular venous distention Correct Answer: D. Gynecomastia is the most likely additional finding in this patient with clinical signs of cirrhosis. Functions of the liver that fail in cirrhosis include filtration and detoxification of portal venous blood, first pass medication metabolism, synthesis of coagulation factors, complement, bile acids, and albumin, biochemical metabolism of lipids and glucose, and metabolism of hormones (eg, estrogens). Manifestations of chronic liver disease can be attributed to the loss of these functions. For example, palmar erythema, spider angiomata, and gynecomastia occur because of an increased circulating concentration of estrogen. Blood that would normally be processed through the liver via the portal vein is shunted away towards peripheral tissues resulting in an increased peripheral conversion of androgens to estrogens. In men, testosterone and estradiol primarily circulate bound to sex hormone binding globulin (SHBG) and albumin. In cirrhotic patients, increased concentrations of SHBG preferentially bind testosterone over estradiol, leading to higher free (active, unbound) estradiol. In combination with greater peripheral conversion of testosterone to estradiol and reduced hepatic metabolism, increased circulating concentrations of estrogen lead to consequent feminization and neovascularization. Symptoms manifest as testicular atrophy, palmar erythema, spider angiomata, decreased libido, and erectile dysfunction. Cirrhosis in this patient has caused portal hypertension (PH), demonstrated on examination as ascites and caput medusae. Increased portal hydrostatic pressure in the setting of cirrhosis directly transmits to the spleen (splenomegaly), esophagogastric veins (varices), and superficial epigastric veins (caput medusae). Incorrect Answers: A, B, C, and E. Digital clubbing (Choice A) is a feature of chronic lung disease such as cystic fibrosis, idiopathic pulmonary fibrosis, and chronic obstructive pulmonary disease, along with congenital heart defects such as an unrepaired cyanotic heart disease, and a large patent ductus arteriosus with Eisenmenger syndrome. It is not a typical feature of cirrhosis. Enlarged inguinal lymph nodes (Choice B) may occur in infections of the lower extremities and the groin but can also occur in malignancy. If mobile, tender, and acute in onset, infection is most likely. If firm and fixed, malignancy should be investigated. Flame-shaped retinal hemorrhages (Choice C) are characteristic of systemic hypertension and are not commonly seen in patients with cirrhosis. Jugular venous distention (Choice E) is a sensitive sign for acute decompensated right heart failure; it also occurs in conditions that increase right atrial and ventricular pressure such as pulmonary hypertension or severe tricuspid regurgitation. If severe enough, right-sided heart failure can eventually cause congestive hepatopathy.

Objective: Cirrhosis presents with signs of portal hypertension such as ascites, esophageal varices, and splenomegaly, along with signs of hyperestrogenism such as palmar erythema, gynecomastia, and spider angiomata. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

187 Exam Section 4: Item 37 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 37. A 14-year-old boy is brought to the emergency department by his friend because of palpitations. His friend says the symptoms began shortly after he "snorted" cocaine. He is found to have tachycardia and increased mean arterial pressure. Which of the following mechanisms is the most likely cause of these adverse effects? A) Agonist action at ligand-gated Ca2+ channels B) Agonist action at ligand-gated Na-K+ channels C) Inhibition of cardiac M,receptor channels D) Inhibition of norepinephrine reuptake E) Inhibition of phosphodiesterase Correct Answer: D. Cocaine blocks the presynaptic reuptake of dopamine, serotonin, and norepinephrine, which increases the concentration of these substances in the synaptic cleft. Cocaine is an addictive recreational substance that typically causes euphoria, restlessness, and increased sympathetic tone (tachycardia, hypertension, pupillary dilation). Increased synaptic serotonin leads to euphoria, while increased synaptic dopamine leads to addiction (and in severe cases, hallucinations and paranoia), and increased synaptic norepinephrine can cause tachycardia and hypertension through activation of sympathetic adrenoreceptors leading to increased cardiac chronotropy and peripheral vascular vasoconstriction. Further, cocaine can act as a local anesthetic by decreasing Nat channel permeability and thereby slowing or blocking pain signal conduction. Incorrect Answers: A, B, C, and E. Agonist action at ligand-gated Ca2* channels (Choice A) results from endogenous glutamate binding to N-methyl-D-aspartate (NMDA) receptors in the brain. Endogenous glutamate agonist action at the NMDA receptor plays an important role in activity-dependent synaptic changes of learning and memory; however, NMDA agonism is not the known mechanism. Agonist action at ligand-gated Na*-K* channels (Choice B) is not a known mechanism of cocaine. A sodium-potassium pump utilizes adenosine triphosphate as energy to maintain appropriate intra- and extracellular concentrations of sodium and potassium. However, this is not cocaine's mechanism of action. Inhibition of cardiac M2-receptor channels (Choice C) would increase heart rate, as the M2-receptor is a muscarinic receptor normally under the control of the parasympathetic nervous system. Medications such as atropine inhibit cardiac M2-receptors. Cocaine does increase heart rate but does not act on the M2-receptor. Inhibition of phosphodiesterase (Choice E) leads to vasodilation and bronchodilation. Phosphodiesterase inhibition increases cyclic adenosine monophosphate (CAMP) or cyclic guanosine monophosphate (CGMP) concentrations, which relax smooth muscle. Medications for erectile dysfunction and heart failure inhibit different subtypes of phosphodiesterase, leading to increased blood flow to the penis, or decreased preload and afterload, respectively. Cocaine does not inhibit phosphodiesterase.

Objective: Cocaine inhibits the presynaptic reuptake of serotonin, dopamine, and norepinephrine. This leads to euphoria and potential addiction, with tachycardia and vasoconstriction as side effects from the adrenergic activity of norepinephrine. Previous Next Score Report Lab Values Calculator Help Pause

57 Exam Section 2: Item 7 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 7. A 60-year-old man has two-pillow orthopnea, severe dyspnea, and edema of the lower extremities. The onset of these symptoms was preceded by an episode of prolonged substernal chest pain 5 days ago. Which of the following additional findings on examination of the neck is most consistent with these findings? O A) Bifid carotid pulse B) Carotid bruit best heard at the carotid bifurcation O C) Jugular "cannon waves" D) Jugular venous pressure of 12 mm Hg E) Slow-rising, decreased-volume carotid pulse Correct Answer: D. An increased jugular venous pressure of 12 mm Hg (N=<8-10 mm Hg) would be expected in this patient presenting with classic signs and symptoms of acute congestive heart failure with orthopnea, dyspnea, and peripheral edema. Acute heart failure can be precipitated by numerous causes, including acute coronary syndrome (concerning in this patient with an episode of prolonged substernal chest pain), hypertensive crisis, valvular disease (eg, endocarditis), myocarditis, pericarditis, massive pulmonary embolism, tachyarrhythmia, complete heart block, thyrotoxicosis, anemia, and severe vitamin deficiencies (eg, thiamine deficiency leading to wet beriberi). Impaired forward flow of blood leads to congestion and increased hydrostatic pressure in the pulmonary vascular system and the systemic venous system. Physical examination findings of congestive heart failure can include increased jugular venous pressure, peripheral edema, ascites, hepatojugular reflux, pulmonary crackles, and an S3 or S4 gallop. Incorrect Answers: A, B, C, and E. A bifid carotid pulse (Choice A), also known as pulsus bisferiens or a biphasic pulse, is associated with aortic regurgitation, combined aortic regurgitation and stenosis, and less commonly hypertrophic cardiomyopathy. The arterial waveform has a double systolic peak, with an initial sharp, shortened peak followed by a decreased amplitude, broader peak. While severe aortic regurgitation can lead to acute heart failure, an increased jugular venous pressure would also be expected in this case. Carotid bruit best heard at the carotid bifurcation (Choice B) is suggestive of atherosclerotic cardiovascular disease and is a risk factor for stroke. Ischemic heart disease and atherosclerotic vascular disease commonly occur together, but also occur in isolation. Jugular "cannon waves" (Choice C), or cannon atrial waves, are large amplitude waves which can be present in certain cardiac arrhythmias (such as complete heart block and ventricular tachycardia). A normal atrial wave results from atrial contraction during diastole. Desynchronized contractions of the atria and ventricles lead to increased right atrial pressure and an exaggerated atrial wave. Slow-rising, decreased-volume carotid pulse (Choice E), or pulsus parvus et tardus (weak and late), can be present in aortic stenosis, as a result of delayed forward flow through the calcified aortic valve.

Objective: Common physical examination findings in acute congestive heart failure include increased jugular venous pressure, peripheral extremity edema, ascites, hepatojugular reflex, pulmonary crackles, and an S3 or S4 gallop. Previous Next Score Report Lab Values Calculator Help Pause

10 Exam Section 1: Item 10 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 10. A fetus exposed to an infection at 22 weeks' gestation appears normal at birth but later develops a persistent mucopurulent nasal discharge, malformed teeth, skin cracks and fissures around oral, anal, and vulvar orifices, splenomegaly, and lymphadenopathy. Which of the following best explains these findings? A) Inability of maternal rubella antibodies to cross the placental barrier O B) Increased amounts of herpes simplex virus in amniotic fluid O C) Maternal failure to generate antibodies to cytomegalovirus D) Maternally transferred spirochetes replicating in fetal tissue E) Persistence of HIV in placental tissue Correct Answer: D. Congenital syphilis is caused by the maternal transfer of Treponema pallidum spirochetes through the placenta in the second and third trimesters, which then replicate in fetal tissue. Though the infant may not demonstrate signs at birth, as the spirochetes continue to replicate, the infant develops signs and symptoms of the infection. Syphilitic rhinitis may herald the onset of symptoms and causes a mucopurulent nasal discharge. Fissures, or rhagades, around the mouth and orifices form and a maculopapular rash may occur. Dental abnormalities, including notched teeth and irregularly shaped molars, are also typically exhibited. Hepatomegaly, with or without splenomegaly and lymphadenopathy, occurs in nearly all affected infants. Deafness as a result of vestibulocochlear nerve involvement is also possible. These manifestations may present several years after birth. To prevent congenital syphilis, the mother must be treated promptly with penicillin as transmission typically begins after the first trimester. Incorrect Answers: A, B, C, and E. Inability of maternal rubella antibodies to cross the placental barrier (Choice A) would suggest a diagnosis of congenital rubella. Neonates exhibit a classic triad of cataracts, sensorineural deafness, and congenital heart disease. A dark blue, maculopapular rash reflecting extramedullary hematopoiesis may also be present. Increased amounts of herpes simplex virus in amniotic fluid (Choice B) would be seen in a congenital herpes simplex virus infection. This can cause both cutaneous herpetic lesions and systemic manifestations like meningoencephalitis. Maternal failure to generate antibodies to cytomegalovirus (Choice C) would cause hearing loss, seizures, a petechial rash, and intracranial calcifications in the neonate if cytomegalovirus infected the fetus. While splenomegaly and lymphadenopathy could be caused by cytomegalovirus infection, rhinitis, periorificial fissures, and dental malformations would not be expected. Persistence of HIV in placental tissue (Choice E) is a mechanism of vertical transmission of HIV from the mother to fetus. Manifestations include recurrent infections and chronic diarrhea. Early recognition and initiation of anti-retroviral therapy is paramount to prevent morbidity and mortality.

Objective: Congenital syphilis is caused by the transmission of Treponema pallidum in the second and third trimesters. Manifestations include syphilitic rhinitis, rhagades, dental malformations, and organomegaly, but the time of presentation after birth is variable. Previous Next Score Report Lab Values Calculator Help Pause

35 Exam Section 1: Item 35 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 35. A 1-day-old male newborn has bilious vomiting and reluctance to feed. Physical examination shows dehydration and mild abdominal distention. Abdominal x-rays show a volvulus of the small bowel between the third and fourth segments of the duodenum. Which of the following blood vessels is most likely constricted by the volvulus in this patient? O A) Celiac trunk B) Inferior mesenteric artery O C) Inferior vena cava D) Splenic vein E) Superior mesenteric artery Correct Answer: E. Superior mesenteric artery (SMA) constriction is the most likely finding in this patient with duodenal volvulus, a dangerous complication of intestinal malrotation, which occurs in many children within the first month of life. Malrotation results from incomplete rotation of the midgut during embryonal development and leaves the midgut attached to a thin stalk of mesentery at the base of the SMA instead of the normally broad-based mesenteric stalk. The cecum also remains in an abnormal position within the mid-upper abdomen instead of the right Tower quadrant and is attached to the lateral abdominal wall by fibrous bands, which cross over the top of the duodenum and can also cause constriction. This leaves the small intestine at risk for volvulus at the base of the SMA. Volvulus can lead to bilious vomiting, abdominal distention, and vascular compromise with subsequent risk for intestinal necrosis, perforation, and peritonitis. The SMA arises from the abdominal aorta inferior to the celiac trunk and passes anteriorly over the third segment of the duodenum, supplying blood not only to the duodenum but also to most of the small intestine and the proximal two-thirds of the large intestine. It is susceptible to constriction in duodenal volvulus because of its anatomical proximity and embryonal relationship to the formation of the midgut. Incorrect Answers: A, B, C, and D. Celiac trunk constrictions (Choice A) are not typical of midgut volvulus as the celiac trunk arises from the abdominal aorta superior to the SMA and does not interface with the mesentery of the midgut. The celiac trunk has three primary branches: the common hepatic artery, splenic artery, and left gastric artery, which supply their respective organs. Inferior mesenteric artery constriction (Choice B) is not typical of midgut volvulus, as the inferior mesenteric artery branches off of the anterior aspect of the abdominal aorta at the level of the third lumbar vertebrae and supplies the hindgut. It may be involved in sigmoid volvulus by virtue of its supply to this region. Inferior vena cava constriction (IVC) (Choice C) is not a feature of midgut volvulus but can occur during pregnancy as the gravid uterus expands. Compression of the IVC may also occur from a large abdominal tumor. Cessation or slowing of blood flow through the IVC is also seen in states of thrombosis. Splenic vein constriction (Choice D) can occur in the setting of acute or chronic pancreatitis as a result of the local inflammatory response from pancreatic necrosis; thrombosis of the splenic vein can result. This does not commonly occur in midgut volvulus.

Objective: Constriction of the SMA occurs in duodenal volvulus from intestinal malrotation; the SMA is embryologically related to the midgut and in turn shares a common mesentery with the duodenum, which twists during the volvulus. Previous Next Score Report Lab Values Calculator Help Pause

70 Exam Section 2: Item 20 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 20. An investigator working in a large health care system wishes to determine the attitude of physicians toward drug screening in the workplace. To collect the information, physicians in the health system are invited to respond to an e-mail survey that is distributed on the health system's network. Which of the following most closely describes this type of sampling technique? A) Convenience B) Random O C) Referral D) Stratified Correct Answer: A. An email survey is an example of convenience sampling, a non-random, non-probability-based method of sampling where the sample is simply taken from participants who are easy to contact. Examples of convenience samples include emailing persons within an organization (eg, hospital staff, office building staff), standing in a public place and recruiting participants who happen to be present, or recruiting a local group of people who belong to the same social club. In convenience sampling, the sample consists of participants who were easy to reach. It is also known as grab sampling or opportunity sampling. Advantages of this technique include the ability to recruit a large number of participants quickly, as the only inclusion criterion in a convenience sample is generally willingness to participate, and it is generally cost-effective. Disadvantages of convenience sampling include a high probability of selection bias and sampling error, as the participants may not reflect the population at-large, or even the population under study. Subgroup analysis is also generally limited, as power within subgroups may be insufficient. Incorrect Answers: B, C, and D. Random (Choice B) sampling describes probability-based choice of a set of study participants from a population at large. In a random sample, each individual participant is chosen by chance, therefore, the probability of being chosen is equal across the population at large. Random sampling aims to reduce bias and confounding and, while more costly and time-consuming, is an ideal technique to promote generalizability of study results. Referral (Choice C) sampling is a non-probability-based sampling technique wherein current study participants recommend future participants. This is also known as chain sampling or snowball sampling. It is most frequently employed when participants may be difficult to identify within the population at large (eg, those involved in illicit activities). Advantages include the ability to identify and study the population under interest when it may otherwise be occult; disadvantages include the potential for sampling bias and limited external generalizability of results. Stratified (Choice D) sampling involves subdividing a population into subgroups and sampling members of each group to generate a combined, composite sample for study. Advantages of this technique are that it permits analysis of subpopulations that may vary from a population at large. Generally, after stratifying the population, random sampling is applied within each stratum.

Objective: Convenience sampling is a non-random, non-probability-based method of sampling, where the sample under study includes participants who were easy to contact and willing to enroll. Examples include studying a classroom full of students who happen to be in the same place, or emailing persons within an organization (eg, resident physicians in a hospital). I3D Previous Next Score Report Lab Values Calculator Help Pause 田

143 Exam Section 3: Item 43 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 43. A 16-year-old boy is brought to the emergency department because of a 3-day history of episodes of sharp chest pain on the left that lasts for a few seconds. When the patient is asked to indicate the location of the pain, he points to the 4th rib on the left. The pain is not associated with change in position, and it does not radiate. There is no history of trauma. The patient appears anxious. His pulse is 90/min and regular. The left anteromedial area of the 4th rib is tender to palpation. Cardiac and pulmonary examinations show no abnormalities. Which of the following is the most likely diagnosis? A) Anxiety B) Costochondritis C) Pericarditis D) Pneumonia E) Rib fracture Correct Answer: B. This patient's acute, well-localized chest wall pain with accompanying tenderness to palpation of the affected area is consistent with costochondritis. Costochondritis is a common cause of chest wall pain and involves inflammation of the costochondral or chondrosternal junctions of the thoracic rib cage. Costochondritis is a self-limited condition and is usually idiopathic. While patients may present with symptoms that worsen during deep inspiration (as a result of chest wall expansion during inspiration), localized tenderness to palpation helps to differentiate costochondritis from other causes of pleuritic chest pain. Laboratory evaluation, x-ray, and ECG are normal in costochondritis and generally only aid the diagnosis by excluding similar presenting diagnoses (eg, pneumonia, acute coronary syndrome). Treatment for costochondritis involves non-steroidal anti-inflammatory drugs and physical therapy. Incorrect Answers: A, C, D, and E. Anxiety (Choice A) may present with chest pain and shortness of breath. The chest pain may be severe and described as pressure or as sharp pain. However, the chest is not usually tender to palpation, which is more suggestive of costochondritis. Pericarditis (Choice C) classically presents with substernal chest pain, which is often pleuritic, worse when lying down, and improves with leaning forward. It can occur secondary to inflammatory, infectious, or malignant conditions. The ribs are not usually tender to palpation. Pneumonia (Choice D) may present with chest pain, which is typically pleuritic and exacerbated by coughing. Patients with pneumonia present with signs and symptoms including fever, chills, shortness of breath, productive cough, tachypnea, and x-ray evidence of pulmonary consolidation, all of which are absent in this patient. Rib fracture (Choice E) is a common cause of acute chest pain in younger patients, who are more likely to sustain injuries from trauma (eg, car accident, sports). The injured area is tender to external palpation. Rib fractures are rarely spontaneous and are unlikely in a patient who lacks a history of trauma.

Objective: Costochondritis, inflammation of the costochondral or chondrosternal junctions, presents with pleuritic chest pain and focal tenderness to palpation, often in a younger person without risk factors for cardiac, pulmonary, or vascular causes of chest pain. Previous Next Score Report Lab Values Calculator Help Pause

64 Exam Section 2: Item 14 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 14. A 29-year-old man has a left testicular mass. A frozen section obtained during a planned orchiectomy shows a seminoma. Which of the following provides the greatest risk factor associated with this condition? A) Alcohol abuse B) Cryptorchidism C) Mumps orchitis D) Varicocele O E) Vasectomy Correct Answer: B. Cryptorchidism is the failure of one or both testes to descend into the scrotum. The fetal testes reside in the abdomen near the internal ring of the inguinal canal until early in the third trimester when they begin to migrate through the inguinal canal and into the scrotum. The path of descent is marked by the gubernaculum, a fibrous structure that regresses as the testes descend. Descent may continue through the first 6 months of life. An undescended testis may be found anywhere along this typical path and should be surgically corrected if it does not descend spontaneously. If left uncorrected for a prolonged duration, the risk for developing a germ cell tumor increases, such as a seminoma. Other testicular germ cell tumors include choriocarcinoma, teratoma, and embryonal carcinoma. Undescended testes are also associated with infertility as optimal spermatogenesis occurs at a temperature several degrees less than that of the body core. Incorrect Answers: A, C, D, and E. Alcohol abuse (Choice A) is a risk factor for liver cirrhosis which is accompanied by hypogonadism. However, it does not carry an increased risk for seminoma or other testicular tumors. Mumps orchitis (Choice C) refers to inflammation of the testes that results from an infection with the paramyxovirus that causes Mumps. It is a risk factor for infertility, especially when it occurs after puberty. A varicocele (Choice D) refers to dilated veins of the pampiniform plexus because of increased venous pressure. Varicoceles are commonly located on the left side because of the increased resistance to venous flow as the left gonadal vein drains into the left renal vein. The dilated veins result in an increased temperature of the scrotum and testicle, which increases the risk for infertility, if bilateral, caused by impaired spermatogenesis, but it does not increase the risk for a seminoma. A vasectomy (Choice E) is a surgical procedure used to induce permanent male contraception in which the vas deferens is severed to prevent sperm from entering the urethra. It is not a known risk factor for testicular malignancies.

Objective: Cryptorchidism, or failure of one or both testes to descend into the scrotum, is associated with an increased risk for testicular germ cell tumor development. Testicular germ cell tumors include seminomas, choriocarcinomas, teratomas, and embryonal carcinomas. Previous Next Score Report Lab Values Calculator Help Pause

107 Exam Section 3: Item 7 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 7. A 30-year-old man who is seropositive for HIV develops nausea, vomiting, watery diarrhea without blood, abdominal cramps, and a 2.3-kg (5-lb) weight loss over 10 days. A diagnosis is made by microscopic examination of stool. Which of the following is the most likely causal organism? A) Clostridium difficile B) Cryptosporidium species C) Enterococcus species D) Escherichia coli 0157:H7 E) Helicobacter pylori F) Mycobacterium avium-intracellulare G) Pseudomonas species H) Shigella dysenteriae I) Toxoplasma gondii O J) Vibrio cholerae Correct Answer: B. This patient with HIV infection and 10 days of watery diarrhea and abdominal cramps has been infected with Cryptosporidium, an intracellular parasite that causes a severe gastrointestinal illness in immunocompromised patients. In immunocompetent patients, it causes self-limited watery diarrhea. In patients with AIDS or severe immunosuppression, the disease can cause profound weight loss and dehydration. Diagnosis begins with stool microscopy and modified acid-fast staining, which shows red-pink oocysts. Microscopic examination of a single specimen is insensitive, and as a result, newer techniques include the use of direct immunofluorescence of antibodies against the oocyst wall or use polymerase chain reaction to detect cryptosporidial DNA, both of which have greater sensitivity than traditional microscopy. Treatment consists of highly active antiretroviral therapy (HAART), and if the condition persists, nitazoxanide can be used as an adjunct. Incorrect Answers: A, C, D, E, F, G, H, I, and J. Clostridium difficile (Choice A) classically presents with colitis when commensal colonic bacteria are eliminated by antibiotics. Mild C. difficile colitis symptoms include foul smelling, watery diarrhea with abdominal cramps. The preferred term for this organism is now Clostridioides difficile. Enterococcus species (Choice C) are found in normal colonic flora and may result in urinary or biliary tree infections, or subacute endocarditis following gastrointestinal or genitourinary invasive procedures. Escherichia coli 0157:H7 (Choice D) is a gram-negative rod associated with bloody diarrhea and hemolytic uremic syndrome, which presents in children with microangiopathic hemolytic anemia, thrombocytopenia, and acute renal failure. Helicobacter pylori (Choice E) is a gram-negative rod that infects the stomach and duodenum. It is associated with peptic ulcers. Symptoms include dyspepsia, upper gastrointestinal bleeding, and gastroesophageal reflux disease. Mycobacterium avium-intracellulare (Choice F) is a non-tuberculous mycobacterium that commonly causes chronic pulmonary infections in patients with lung disease and disseminated infections in patients with HIV/AIDS. Pseudomonas species (Choice G) are gram-negative rods that cause an array of infectious conditions such as urinary tract infections, osteomyelitis, otitis externa, folliculitis, hospital or ventilator-associated pneumonia, and bacteremia. Shigella dysenteriae (Choice H) is a gram-negative rod that can cause bloody diarrhea with systemic symptoms such as fever and myalgias. Toxoplasma gondii (Choice I) manifests often in immunocompromised hosts as pneumonia, chorioretinitis, hepatitis, myositis, and central nervous system abscesses. Diarrhea is not characteristic. Vibrio cholerae (Choice J) is a gram-negative rod that causes cholera, a severe, acute watery diarrhea resulting in substantial dehydration. It is endemic to Africa, southeast Asia, Haiti, and India.

Objective: Cryptosporidiosis presents with severe watery diarrhea in immunocompromised patients; it is diagnosed by microscopy of the stool, showing red-pink oocysts on acid-fast stain. Previous Next Score Report Lab Values Calculator Help Pause

61 Exam Section 2: Item 11 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 11. A 36-year-old man who is seropositive for HIV has had fever and shortness of breath for 2 days. An x-ray of the chest shows a diffuse interstitial infiltrate. A lung biopsy is shown. Which of the following organisms is most likely involved? A) Cryptococcus neoformans B) Cytomegalovirus C) Epstein-Barr virus D) Histoplasma capsulatum E) Influenzavirus A F) Mycobacterium avium-intracellulare G) Pneumocystis jirovecii (formerly P. carinii) Correct Answer: B. Cytomegalovirus, also known as human herpesvirus-5 (HHV-5), is an opportunistic infection commonly occurring in immunocompromised patients in the setting of solid-organ or allogeneic stem cell transplantation, severe ulcerative colitis, or HIVIAIDS. It can be transmitted through multiple modes, including sexual contact, urine, respiratory droplets, and to a fetus via the placenta. It can cause a variety of presentations, including mononucleosis in immunocompetent patients, and retinitis, esophagitis, and pneumonia in immunocompromised patients. The classic histologic findings are infected cells containing prominent intranuclear inclusion bodies as seen in this patient's lung biopsy. Incorrect Answers: A, C, D, E, F, and G. Cryptococcus neoformans (Choice A) is an encapsulated yeast and common opportunistic pathogen in immunocompromised patients. It can cause cryptococcal meningitis, encephalitis, and pulmonary cryptococcosis. Histology shows narrow-based budding and a bright red inner capsule on staining with mucicarmine. Epstein-Barr virus (Choice C) is associated with the development of B-cell lymphomas (eg, non-Hodgkin lymphoma) and oral hairy leukoplakia in HIV-positive adults with low CD4+ counts. Characteristic atypical lymphocytes will be seen on cytology. Histoplasma capsulatum (Choice D) causes the AIDS-defining illness histoplasmosis through inhalation of the organism's fungal spores. It classically presents with fever, weight loss, fatigue, cough, dyspnea, nausea, vomiting, and diarrhea. Histology will disclose oval yeast cells within macrophages. Influenzavirus A (Choice E) infection presents with respiratory, gastrointestinal, and systemic symptoms, including cough, nausea, vomiting, diarrhea, fever, chills, headache, myalgias, and arthralgias; it can be complicated by bacterial superinfection causing severe bacterial pneumonia. Mycobacterium avium-intracellulare (Choice F) are nontuberculous, gram-positive, acid-fast bacilli, which cause severe disease in patients with AIDS. The risk for infection is inversely related to CD4+ count and significantly increased when the CD4+ count is less than 50 cells/mm3. Symptoms may be nonspecific with fever, night sweats, fatigue, and weight loss, or it may present with focal lymphadenitis. Pneumocystis jirovecii (formerly P. carinii) (Choice G) is a yeast-like fungus which causes pneumocystis pneumonia in patients who are immunocompromised. Chest imaging typically show diffuse, bilateral infiltrates often prominently about the hila.

Objective: Cytomegalovirus is a common opportunistic pathogen that can cause a variety of disease, including retinitis, esophagitis, and pneumonia in immunocompromised patients. On histology, prominent intranuclear inclusion bodies within infected cells are characteristic. I3D Previous Next Score Report Lab Values Calculator Help Pause

49 Exam Section 1: Item 49 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 49. A 50-year-old man with type 2 diabetes mellitus comes to the physician for a follow-up examination. Current medications include long-acting insulin and metformin. His vital signs are within normal limits. Physical examination shows no abnormalities. His hemoglobin A1, is 7.9%. Treatment with a third drug that promotes release of endogenous insulin is initiated to improve glucose control. This drug is most likely which of the following? A) Acarbose O B) Miglitol O C) Pioglitazone D) Pramlintide E) Sitagliptin Correct Answer: E. Sitagliptin is an oral antihyperglycemic used in the control of blood glucose in type 2 diabetes mellitus. It is a competitive inhibitor of dipeptidyl peptidase-4 (DPP-4). DPP-4 is the enzyme responsible for breaking down gastrointestinal incretins, which are released during a meal and promote the secretion of insulin. By preventing the breakdown of incretin hormones by inhibiting the enzyme responsible for degrading them, their effect (increasing endogenous release of insulin) persists. Examples of incretins include glucagon- like peptide-1 (GLP-1). Sitagliptin is generally a second-line medication, used when metformin, diet, exercise, and weight loss have failed to control blood glucose appropriately (as marked by a hemoglobin A1c above 7.0%). In some cases, insulin or a sulfonylurea may be added prior to initiating a DPP-4 inhibitor. Incorrect Answers: A, B, C, and D. Acarbose (Choice A) and miglitol (Choice B) interfere with the intestinal digestion and absorption of carbohydrates by inhibiting intestinal brush border glucosidases. This delayed digestion of carbohydrates limits or delays absorption, thereby reducing hyperglycemia following a meal. They do not directly influence insulin release. Pioglitazone (Choice C) increases peripheral insulin sensitivity by activating peroxisome proliferator-activated receptor gamma. This receptor activates genes that increase peripheral insulin sensitivity. This class of medications demonstrates multiple side effects, including weight gain, peripheral edema, and heart failure; as a result, their use is limited. Pramlintide (Choice D) is an analogue of the hormone amylin, which acts by delaying gastric emptying, thereby limiting postprandial hyperglycemia by slowing absorption.

Objective: DPP-4 inhibitors prevent the early breakdown of incretins, which promote endogenous insulin release. Sitagliptin is an example of this class of inhibitor. They are generally used once metformin, diet, exercise, weight loss, or additional medications have failed to appropriately control a patient's blood glucose. Previous Next Score Report Lab Values Calculator Help Pause

165 Exam Section 4: Item 15 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 15. A 34-year-old woman comes to the burn unit where her 14-year-old son was admitted 2 weeks ago after he injured his hand while lighting firecrackers with his friends. In preparation for his discharge, the woman is taught to change the dressing and apply cream to the burn. She says that she feels revulsion when viewing the affected area and changing the dressing, but she tries to look cheerful for the sake of her son. Which of the following best describes this mother's defense mechanism? O A) Denial B) Displacement O C) Dissociation D) Isolation of affect E) Repression F) Splitting G) Suppression Correct Answer: G. Suppression is the conscious decision to expel a thought or feeling from one's mind. This patient is choosing not to think about her disgust related to the wound while changing the dressing. Suppression is thought to be a mature and adaptive defense mechanism. When any person experiences painful thoughts, realities, or feelings such as anxiety or sadness, the mind has several unconscious methods of shielding the person from conscious awareness of the thought, reality, or affect. These methods are called defense mechanisms. However, suppression differs from most other defense mechanisms in that the person is aware of the painful or acceptable feeling and consciously chooses to ignore it. Incorrect Answers: A, B, C, D, E, and F. In denial (Choice A), the mind protects itself from a painful external reality by unconsciously refusing to acknowledge it. For example, if an overweight man decides not to exercise because he does not think he is overweight, his unconscious mind may be protecting him from the painful fact that he is overweight. In displacement (Choice B), a person may transfer negative feelings from an unacceptable person to a more acceptable person. For instance, a woman cannot tolerate being angry at her spouse for cheating because of the fear that she will need to divorce him, so her unconscious mind transfers the anger to her mother who she cannot divorce and is therefore a more acceptable target. In dissociation (Choice C), the mind separates itself from external reality because the reality is too painful to handle. For instance, when encountering a reminder of a childhood sexual trauma, a woman may retreat into a state of unawareness of her surroundings and not remember what occurred while she was in this state. In isolation of affect (Choice D), the mind represses an intolerably painful affect, but the mind remains aware of associated thoughts. For example, a man states that he has no emotions related to his son's death. In repression (Choice E), the mind unconsciously protects itself from a painful internal reality (eg, a thought or feeling) by denying its existence. This is different from denial, in which the mind protects itself from a painful external reality. For instance, a mother smiles patiently when her son reads slowly because her unconscious mind denies the existence of anger at her son's reading disability, as this anger painfully conflicts with her perception of herself as a caring mother. In this example, the patient is aware of her revulsion and makes the conscious choice to override the disgust with a smile. In splitting (Choice F), the mind preserves positive emotions about one person by unconsciously transferring any negative emotions about that person to another person. For example, a man cannot bear to think of his mother as imperfect, so his mind transfers his negative emotions about her to his father, therefore vilifying his father.

Objective: Defense mechanisms protect a person from conscious awareness of painful emotions, thoughts, or realities. In suppression, however, the person is aware of the unacceptable feeling, and the person consequently chooses to place the unacceptable feeling out of mind. Previous Next Score Report Lab Values Calculator Help Pause

151 Exam Section 4: Item 1 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 1. Tissue obtained on renal biopsy from a 45-year-old woman with proteinuria is shown. These findings are most consistent with which of the following? A) History of analgesic abuse B) Hypocomplementemia C) Long-standing hyperglycemia D) Malignant hypertension O E) Sickle cell trait Correct Answer: C. Long-standing hyperglycemia, such as that in diabetes mellitus, can result in diabetic nephropathy. The pathophysiology of diabetic nephropathy involves nonenzymatic glycosylation of the glomerular basement membrane and efferent arterioles, resulting in thickening of the basement membrane along with compromise of the filtration barrier resulting in increased permeability to solutes and proteins. Thickening of the efferent arteriole initially increases glomerular filtration rate (GFR), which places the glomerulus at risk for hyperfiltration injury. Diabetic nephropathy characteristically presents with eosinophilic nodular glomerulosclerosis, also known as Kimmelstiel-Wilson lesions, on light microscopy as shown in the kidney biopsy. The combination of increased permeability and increased GFR progresses over time in patients with diabetes mellitus, initially beginning as microalbuminuria, progressing to macroalbuminuria, and eventually to end-stage kidney disease. Incorrect Answers: A, B, D, and E. History of analgesic abuse (Choice A) and sickle cell trait (Choice E) are associated with renal papillary necrosis. Renal papillary necrosis refers to the sloughing and necrosis of renal papillae, which can be triggered by infections (eg, acute pyelonephritis), diabetes mellitus, sickle cell disease, or nonsteroidal anti-inflammatory drugs (NSAIDS). It causes gross hematuria and proteinuria. NSAIDS are also associated with acute interstitial nephritis, which is caused by a hypersensitivity reaction to medications (eg, NSAIDS, diuretics, sulfonamides) and is characterized by a rash, azotemia, sterile pyuria, hematuria, and eosinophiluria. Hypocomplementemia (Choice B) is a characteristic of postinfectious glomerulonephritis, which results in low serum C3 complement as a result of C3 deposition along the glomerular basement membrane and mesangium. Postinfectious glomerulonephritis is a nephritic syndrome presenting with gross or microscopic hematuria and RBC casts on microscopy, which typically occurs two to six weeks after infection with group A beta-hemolytic streptococci. On laboratory evaluation, low serum C3 concentrations, along with positive anti-streptolysin O and anti-DNase titers are noted. On light microscopy, the glomeruli are enlarged and hypercellular. Malignant hypertension (Choice D) causes arteriolonephrosclerosis (sclerosis of renal arterioles). Hypertension causes thickening and hyalinization of vessel walls, which leads to narrowing of the lumen with resultant ischemia of the parenchyma, decreased glomerular filtration rate, and proteinuria. Untreated, it can lead to end-stage kidney disease.

Objective: Diabetic nephropathy occurs following the nonenzymatic glycosylation of the glomerular basement membrane and efferent arterioles, characteristically presenting as Kimmelstiel-Wilson lesions on light microscopy. It progresses over time in patients with diabetes mellitus, initially beginning as microalbuminuria, which can subsequently lead to macroalbuminuria and then end-stage kidney disease. II Previous Next Score Report Lab Values Calculator Help Pause

45 Exam Section 1: Item 45 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 45. A 53-year-old man comes to the physician for a health maintenance examination. He does not smoke. He is 180 cm (5 ft 11 in) tall and weighs 95 kg (210 Ib); BMI is 29 kg/m2 His blood pressure is 140/92 mm Hg. Physical examination shows no other abnormalities. Serum studies show: Cholesterol, total HDL-cholesterol LDL-cholesterol 240 mg/dL 35 mg/dL 170 mg/dL 175 mg/dL Triglycerides In addition to appropriate pharmacotherapy, it is most appropriate for this patient to decrease his dietary intake of which of the following types of fatty acids? A) cis-Unsaturated B) Monounsaturated C) Omega-3 D) Omega-6 E) Omega-9 F) trans-Unsaturated Correct Answer: F. According to the 2019 American College of Cardiology/American Heart Association (ACC/AHA) Guideline on the Primary Prevention of Cardiovascular Disease, biomarkers associated with increased atherosclerotic cardiovascular disease (ASCVD) risk include triglycerides greater than 175 mg/dL, LDL greater than 160 mg/dL, and a total cholesterol (not including HDL) greater than 190 mg/dL. Increased concentration of LDL-cholesterol and decreased concentration of HDL-cholesterol are known risk factors for the development of ASCVD. Intake of trans-unsaturated fatty acids are associated with increased LDL-cholesterol, decreased HDL-cholesterol, increased triglycerides, and increased concentration of systemic inflammation. LDL-cholesterol functions by transporting cholesterol from the liver to peripheral tissues, whereas HDL-cholesterol transports cholesterol in the reverse manner from peripheral tissues to the liver. Because of this, LDL-cholesterol is identified as pro-atherosclerotic, whereas HDL-cholesterol has a role in clearing peripheral lipids and triglycerides by way of returning them to the liver for metabolism. Management goals related to cholesterol include decreasing serum LDL-cholesterol, increasing serum HDL-cholesterol, decreasing serum triglycerides, and diminishing dietary intake of trans-fatty acids. A balanced management strategy involving diet, exercise, and lipid-decreasing therapies (eg, statins) is often required to achieve serum goals. Incorrect Answers: A, B, C, D, and E. Unsaturated fatty acids, including cis-unsaturated (Choice A) and monounsaturated (Choice B), are correlated with improving cholesterol concentration, especially when replacing unhealthy fatty acid intake. They are not shown to increase LDL-cholesterol. Omega-3 fatty acids (Choice C) can help with decreasing triglyceride concentration and have not been shown to increase LDL-cholesterol. Like omega-3 fatty acids, omega-6 (Choice D) and omega-9 (Choice E) fatty acids are unsaturated fatty acids. Consumption of these classes of lipids has not been shown to raise LDL or decreased HDL concentration to the extent of trans-fatty acid consumption, though this remains an area of active research.

Objective: Dietary intake of trans-unsaturated fatty acids has been shown to increase serum LDL-cholesterol and decrease serum HDL-cholesterol, as well as increase serum triglycerides and promote systemic inflammation, thereby increasing the risk for ASCVD. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

17 Exam Section 1: Item 17 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 17. A clinical study compares patients with HIV infection with an age-matched control group of persons without HIV infection. Men and women in both the study and control groups have a history of unprotected sexual intercourse with both male and female partners. The CD4+ T-lymphocyte counts of persons in both groups are greater than 500/mm3 (N2500). The histories of any infection are obtained from persons in both groups. A history of infection with which of the following organisms most likely increased the risk for developing HIV infection in the study group? A) Burkholderia cepacia B) Candida albicans C) Cryptococcus neoformans D) Cytomegalovirus E) Haemophilus ducreyi F) Trichophyton rubrum Correct Answer: E. Haemophilus ducreyi causes the sexually transmitted infection chancroid, characterized by development of one or more painful genital ulcers with ragged edges. Patients who contract HIV infection as a result of unprotected intercourse often do so through microabrasions. Such microabrasions occur via normal friction during intercourse and are more common in the presence of compromised, infected, nonintact, inflamed skin/genital mucosa (eg, discontinuities such as genital ulcers). Inguinal lymphadenopathy is common with chancroid. Treatment is with a single dose of azithromycin or ceftriaxone. Incorrect Answers: A, B, C, D, and F. Burkholderia cepacia (Choice A) is a gram-negative rod that causes hospital-acquired pneumonia and pulmonary infections in patients with cystic fibrosis or chronic granulomatous disease. It is not sexually transmitted. Candida albicans (Choice B) is a fungus that causes a variety of infections in both immunocompetent patients and those with HIV/AIDS or immunocompromise. Candida often causes skin infections in moist intertriginous areas such as the groin and inframammary folds but can also cause invasive infections with high morbidity and mortality, including blood stream and urinary tract infections. Candida is not sexually transmitted. Cryptococcus neoformans (Choice C) is an encapsulated fungus that most commonly causes meningitis in HIV and AIDS patients with CD4+ counts less than 200/mm3. Treatment is with liposomal amphotericin B followed by fluconazole for consolidation therapy. Cryptococcus is not acquired sexually. Cytomegalovirus (Choice D) infection can cause primary infection in immunocompetent hosts with symptoms that are indistinguishable from those of Epstein-Barr infection (mononucleosis). Patients present with fevers, myalgias, and laboratory evidence of atypical lymphocytosis, often with abnormal liver enzymes. Cytomegalovirus infection typically affects recipients of solid organ transplants who are immunosuppressed and can present with a variety of syndromes including colitis, encephalitis, pneumonia, or retinitis. It is not sexually transmitted. Trichophyton rubrum (Choice F) is a dermatophyte (skin fungus) that can cause superficial infection on any part of the skin, but most commonly affects the groin, foot, scalp, and nail. While contagious via direct contact, it is not considered a sexually transmitted infection.

Objective: Haemophilus ducreyi is a sexually transmitted infection that causes chancroid, which presents as one or more painful genital ulcers with associated inguinal lymphadenopathy. A higher number of unprotected sexual encounters increases the patient's risk for a sexually transmitted infection, and the epithelial discontinuity of the genital ulcers increases susceptibility to HIV. I3D Previous Next Score Report Lab Values Calculator Help Pause

58 Exam Section 2: Item 8 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 8. The breakdown of dipeptides and tripeptides to free amino acids takes place primarily in which of the following areas in the gastrointestinal tract? O A) Intestinal mucosa B) Lumen of the duodenum C) Lumen of the large intestine D) Lumen of the stomach E) Mouth Correct Answer: A. Small intestinal enterocytes within the brush border of the intestinal mucosa avidly absorb dipeptides and tripeptides through active transport across the luminal membrane. Once absorbed into the intestinal mucosa, dipeptides and tripeptides undergo further digestion to free amino acids within the cytoplasm. This process is mediated by cytoplasmic dipeptidases and tripeptidases. Free amino acids are secreted into the portal circulation for transport to the liver. Brush border peptidases, in addition to intracytoplasmic peptidases, are integral to this process as only small peptides can be absorbed. Incorrect Answers: B, C, D, and E. The lumen of the duodenum (Choice B) contains important proteolytic enzymes including trypsin and chymotrypsin, which aid in the digestion of complex polypeptides. While some free amino acids are produced by luminal digestion, the majority of free amino acids are produced by digestion at the intestinal mucosal brush border. The lumen of the large intestine (Choice C) is important for the absorption of water, vitamins, and any nutrients which have not been absorbed by the small intestine. The majority of free amino acids have been absorbed in the small intestine prior to reaching the large intestine. The lumen of the stomach (Choice D) contains the proteolytic enzyme pepsin. Pepsin is produced as pepsinogen by gastric chief cells and is activated by low pH within the gastric lumen. Its primary function is the digestion of complex polypeptide chains, rather than the digestion of dipeptides and tripeptides into free amino acids. The mouth (Choice E) contains digestive enzymes including salivary amylase and lingual lipase. These enzymes are important in the digestion of carbohydrates and triglycerides, respectively. They do not play a significant role in the digestion of proteins.

Objective: Dipeptides and tripeptides are produced by the activity of trypsin and chymotrypsin within the lumen of the duodenum. Free amino acids are produced upon further digestion of these small peptides within the brush border of the intestinal mucosa. Previous Next Score Report Lab Values Calculator Help Pause

99 Exam Section 2: Item 49 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 49. A 34-year-old woman comes to the office because of a 2-day history of a full-body rash. The rash is not itchy, and she has had no other symptoms. She has a staphylococcal abscess on her right buttock for which she has been taking trimethoprim- sulfamethoxazole for the past week. She last had sexual intercourse 2 years ago. She currently is menstruating. She recently adopted a cat and changed laundry detergents to a dye-free variety. She has no history of tick bites. Her temperature is 37.1°C (98.8°F), pulse is 80/min, respirations are 16/min, and blood pressure is 130/85 mm Hg. A photograph of the patient's right upper extremity is shown; there are similar findings over the trunk and remaining extremities, including the hands and feet. The mucous membranes appear intact. Examination of the buttocks shows a healed abscess; the skin has closed, and there is no surrounding cellulitis. Which of the following is the most likely cause of this patient's condition? A) Adverse effect of drug B) Cat allergy C) Contact dermatitis D) Tinea corporis E) Toxic shock syndrome Correct Answer: A. Adverse drug effects are common causes of rashes including acute urticaria, morbilliform drug eruption, and Stevens-Johnson syndrome. A morbilliform, or maculopapular, drug eruption is characterized by numerous erythematous, smooth macules and papules coalescing into patches and plaques involving the trunk and extremities, which begins 7 to 10 days after medication initiation. Features such as skin pain, sloughing, or mucosal involvement are concerning for Stevens-Johnson syndrome or toxic epidermal necrolysis. Common culprits of drug eruptions include antibiotics, antiepileptics, sulfa-containing medications including trimethoprim-sulfamethoxazole, nonsteroidal anti-inflammatory drugs, and allopurinol. Treatment includes stopping the offending agent and providing supportive care, which often includes topical corticosteroids or emollients. Incorrect Answers: B, C, D, and E. Cat allergy (Choice B) or allergy to other antigens can cause urticaria, pruritus, allergic rhinitis, or allergic conjunctivitis, but would not be expected to cause a morbilliform rash. In contrast to a morbilliform rash, urticaria is characterized by erythematous wheals which resolve within a few hours following removal of the precipitant antigen. Contact dermatitis (Choice C) is a type IV hypersensitivity reaction caused by exposure of the skin to either an irritant or allergen causing an eczematous rash to occur in the distribution of the exposure. The hands are a commonly involved site and a full-body rash is not typical. Acute contact dermatitis is characterized by small weeping vesicles with surrounding erythema. Over time, the vesicles resolve and are replaced by scaly, lichenified patches. Treatment includes identifying and discontinuing the irritant or allergen and applying topical corticosteroids. Tinea corporis (Choice D) is caused by a superficial dermatophyte infection of the skin, most often Trichophyton tonsurans. The classic appearance is a round, annular patch with a scaly border. Potassium hydroxide preparation will show hyphae. Toxic shock syndrome (Choice E) is caused by toxic shock syndrome toxin produced by Staphylococcus aureus. While toxic shock syndrome can cause a full body erythematous eruption, the patient also classically exhibits fever, vital sign instability, and signs of end-organ dysfunction related to the underlying systemic shock. Treatment of the inciting staphylococcal infection and supportive care are warranted.

Objective: Drug eruptions are a common cause of widespread rashes. Morbilliform, or maculopapular, drug eruptions typically occur 7 to 10 days after medication initiation. Most common culprits include antibiotics, sulfa-containing drugs, and antiepileptics. Previous Next Score Report Lab Values Calculator Help Pause

15 Exam Section 1: Item 15 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 15. A 46-year-old man with type 2 diabetes mellitus is brought to the emergency department by his wife 30 minutes after he developed chest pain and then lost consciousness. He has a 2-year history of exercise-induced angina pectoris that is well controlled with isosorbide dinitrate, atenolol, and diltiazem. Current medications also include metformin and glipizide. His wife is tearful and says, "He took one of his friend's Viagra pills about 4 hours ago." On arrival, he is alert. His blood pressure is 60/30 mm Hg. Physical examination shows pallor and diaphoresis. This patient is most likely having which of the following effects of a drug interaction? A) Bradycardia B) Paroxysmal supraventricular tachycardia C) Third-degree atrioventricular block D) Torsades de pointes E) Vasodilation Correct Answer: E. Sildenafil is a selective inhibitor of CGMP-specific phosphodiesterase type 5 (PDE5). Inhibition of PDE5 leads to increased concentrations of CGMP, which causes relaxation of vascular smooth muscle (vasodilation). Sildenafil is commonly used to treat erectile dysfunction and pulmonary arterial hypertension. Šide effects of sildenafil include headache, heartburn, flushing, and visual symptoms such as blurry vision, photophobia, and cyanopsia (a visualized blue tint). A notable drug interaction is observed between sildenafil and nitrate medications, such as isosorbide dinitrate. Nitrate-containing drugs are converted to nitric oxide (NO), which also causes smooth muscle relaxation. When used alongside nitrates, sildenafil can cause profound vasodilation and hypotension which manifests as lightheadedness, pallor, diaphoresis, and syncope. In rare cases, death can result if the hypotension is uncorrected. Concomitant use of nitrates is a contraindication to the administration of sildenafil. Incorrect Answers: A, B, C, and D. Bradycardia (Choice A) would be less likely, as the vasodilatory effects of sildenafil would be expected to lead to compensatory tachycardia. In this case, the patient is taking atenolol which may blunt any reflex tachycardic response. Sildenafil does not potentiate the effect of atenolol. Paroxysmal supraventricular tachycardia (PSVT) (Choice B) can manifest with acute symptoms of lightheadedness, pallor, diaphoresis, and syncope, but would more likely be accompanied by palpitations. PSVT may be because of structural defects in the cardiac electrical conduction system, such as in Wolff-Parkinson-White syndrome, or may be provoked by alcohol, nicotine, caffeine, or psychological stress. Third-degree atrioventricular block (Choice C) involves a complete blockage of the transmission of atrial electrical impulses to the ventricles, with a resultant junctional or ventricular escape rhythm. Complete dissociation of atrial and ventricular pacing can be detected on ECG. Third-degree atrioventricular block can result from myocardial infarctions or from overdose of beta- or calcium-channel blockers but is not caused by an interaction between a B-adrenergic blocking agent and sildenafil. Torsades de pointes (Choice D) is a dangerous form of polymorphic ventricular tachycardia that is often associated with a prolongation of the QT interval. Sildenafil does not lead to drug interactions that result in QT prolongation.

Objective: Dual administration of PDE5 inhibitors and nitrate containing medications may result in profound vasodilation with resulting hypotension. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

192 Exam Section 4: Item 42 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 42. A 5-year-old boy is brought to the physician by his mother because of progressive clumsiness and fatigue during the past 6 months. He says that his legs are tired. He was delivered at term after an uncomplicated pregnancy. He has met all developmental milestones, although there was some delay compared with other children his age. He is alert. He has difficulty rising from the chair; he uses his arms to push himself into a standing position. He is unable to jump with both feet together. Physical examination shows hypertrophy of the calf muscles. This patient most likely has weak hip adduction as a result of dysfunction of the muscle inserting onto the femur from which of the following locations? O A) Anterior sacrum B) lliac crest C) lliac spine D) Ischium E) Lateral ilium Correct Answer: D. This patient exemplifies the history and examination findings classically seen with Duchenne muscular dystrophy; an X-linked disease most often seen in male children. Children with Duchenne muscular dystrophy typically have an unremarkable birth history, then begin to demonstrate signs of proximal muscle weakness early in childhood. Physical examination shows difficulty rising from squatting position, calf pseudohypertrophy, and a Gower sign (using hands to "climb" up the lower extremities when rising from the floor). Proximal muscles include those of the shoulder and hip girdles, such as the adductors and flexors of the hip and thigh. The adductor muscles of the hip include adductor magnus, longus, brevis, and minimus. Adductor longus, brevis, and minimus originate from the pubic bone, whereas adductor magnus originates on the ischium and inferior pubic ramus. In this case, a large contribution of hip adduction comes from the adductor magnus, and loss of this muscle because of myocyte dystrophy results in the proximal lower limb weakness reflected in this patient. The pectineus and the gracilis also serve to adduct the hip joint. Incorrect Answers: A, B, C, and E. The anterior sacrum (Choice A) is the origin of the piriformis, the iliacus, and the coccygeus muscles, which are not primary muscles of hip adduction. The iliac crest (Choice B) serves as an origin for the gluteus maximus posteriorly. It is also the origin of the iliacus muscle, a hip flexor, on its internal aspect. The abdominal wall musculature also inserts on the iliac crest superiorly. Muscles originating from or inserting here are not involved in hip adduction. The ilium has four spines (Choice C): anterior superior, which is the origin of sartorius, anterior inferior, which is the origin of the rectus femoris, and posterior superior and inferior to which the posterior sacral ligaments and multifidus are attached. These groups are not involved in primary hip adduction. Lateral or external surface of the ilium (Choice E) is the origin of the gluteus medius and the gluteus minimus, which are involved in hip stabilization, rotation, abduction, and extension.

Objective: Duchenne muscular dystrophy, an X-linked condition, results in progressive proximal muscle weakness in male children. Symptoms include an inability to rise from a squatting position, and weakness of the lower limb girdle. The adductor muscle group originates from the ischium and inferior pubis and serves the primary role of hip adduction. Previous Next Score Report Lab Values Calculator Help Pause

110 Exam Section 3: Item 10 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment Normal 10. A7-year-old girl is admitted to the hospital because of severe anemia. Hemolytic anemia with a probable defect in glycolytic energy production is diagnosed. Results of pyruvate kinase activity in erythrocyte lysates using varying concentrations of phosphoenolpyruvate are shown. Which of the following best explains these findings? Patient A) Decrease in the rate of enzyme synthesis B) Increase in the rate of enzyme degradation C) Mutation in the enzyme that decreases V, D) Mutation in the enzyme that increases Km [Phosphoenolpyruvate] E) Presence of a noncompetitive inhibitor in cells O F) Presence of an irreversible inhibitor in cells Correct Answer: D. Pyruvate kinase (PK), the final key regulatory step in glycolysis, catalyzes the conversion of the substrate phosphoenolpyruvate (PEP) to pyruvate in a step that produces adenosine triphosphate (ATP). This step is upregulated by fructose-1,6-bisphosphate, an upstream product, and downregulated by ATP and alanine. Red blood cells, lacking nuclei and organelles in the mature form, depend on anaerobic respiration and the production of ATP from glycolysis. If a deficiency exists in ATP-generating steps in glycolysis, the red blood cell may lack sufficient ATP for its own metabolism and regulation of ion channels, leading to cellular swelling and lysis (hemolytic anemia). Enzymes are kinetically categorized by their Michaelis constant, Km, which inversely reflects their affinity for the substrate, and by Vmax, the maximum catalysis rate. In other words, as Km increases, affinity for the substrate decreases. Km is defined as the substrate concentration at which the enzyme reaches one-half of its Vmax- In this case, the graphical kinetic analysis of the patient's PK enzyme demonstrates that at one-half of Vmax, the concentration of PEP is higher than it would typically be in a normal patient. This indicates that a mutation may have occurred in the enzyme's active site that decreases its affinity for the substrate. Therefore, the catalytic ability of the enzyme is diminished, which results in a consequent hemolytic anemia from insufficient red blood cell ATP. Incorrect Answers: A, B, C, E, and F. Decrease in the rate of enzyme synthesis (Choice A), increase in the rate of enzyme degradation (Choice B), mutation in the enzyme that decreases Vmax (Choice C), and presence of an irreversible inhibitor in cells (Choice F) would demonstrate a decreased Vmax, since Vmax is directly proportional to the total concentration of active, functional enzyme. In this example, the Vmax of the patient's PK and that of normal PK appears the same. Presence of a noncompetitive inhibitor (Choice E) would exhibit a reduced Vmax and an unchanged Km- In this case, the Vmax is unchanged, and the Km is increased, indicating that if an inhibitor is present, it would likely be a reversible, competitive inhibitor.

Objective: Enzymatic mutations may alter the affinity or velocity of an enzyme for its substrate, marked by deviation from normal in Km or Vmax. A higher Km indicates decreased affinity of the enzyme active site for the substrate. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

113 Exam Section 3: Item 13 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 2500 13. A healthy 40-year-old man is participating in a study in which he is given an intravenous injection of insulin to induce hypoglycemia. The graph shows serum concentrations of catecholamines at baseline and the peak values at the nadir of serum glucose. Which of the following most likely underlies the different magnitude of the epinephrine and norepinephrine responses? 2000 A) Activation of catechol-O-methyltransferase 1500 B) Failure of monoamine oxidase to metabolize epinephrine C) Increased stability of circulating epinephrine 1000 D) Reuptake of norepinephrine into sympathetic synapses 500 E) Selective activation of medullary epinephrine release Epinephrine Norepinephrine Baseline Post-insulin Correct Answer: E. The experiment demonstrates that after hypoglycemia is induced in a healthy male, the measured serum concentrations of epinephrine and norepinephrine are markedly different at the nadir of hypoglycemia. As a physiologic stressor, hypoglycemia is known to induce sympathetic nervous system activation with a measurable release of catecholamines produced by the adrenal medulla. It has been shown in multiple experiments that the catecholamine produced may vary with the type of inciting stressor. A selective activation of adrenal medullary epinephrine release via chromaffin cells has been proposed to explain the difference. In the case of hypoglycemia, glucagon and epinephrine are the principal mediators that will restore euglycemia in the short term. Glucagon acts on hepatocytes via a cyclic adenosine monophosphate pathway activating protein kinase A, which causes the activation of glycogen phosphorylase and results in the release of glucose monomers into the serum. Epinephrine (not norepinephrine) acts on hepatocytes at a- and B-adrenergic receptors, causing the release of intracellular calcium that activates glycogen phosphorylase (a-adrenergic receptor) and protein kinase A, which subsequently also activates glycogen phosphorylase (B-adrenergic receptor). Additionally, epinephrine increases gluconeogenesis through the activation of fructose bisphosphatase-2 (B-adrenergic receptor). This downstream requirement for epinephrine as the key ligand in glycogenolysis and gluconeogenesis in response to hypoglycemia likely explains the differential production of epinephrine by the adrenal medulla during states of hypoglycemia. Incorrect Answers: A, B, C, and D. Activation of catechol-O-methyltransferase (Choice A), the initial enzyme in the metabolism of catecholamines, would result in decreased serum concentrations of each molecule, and offers no physiologic explanation as to why epinephrine is preferentially secreted. Failure of monoamine oxidase to metabolize epinephrine (Choice B) could result in persistently increased concentrations of epinephrine, though at baseline, a difference between the two neurotransmitters would be expected if one was preferentially metabolized, which was not shown in the graph. In fact, more norepinephrine existed at baseline, the opposite of what would be expected in this scenario. Increased stability of circulating epinephrine (Choice C) may result in a prolonged half-life of epinephrine, or greater concentration, however, if epinephrine were more stable, a higher concentration at the start of the experiment would also be expected. The half-life of epinephrine is between 2 and 3 minutes; as is the half-life of norepinephrine. Reuptake of norepinephrine into sympathetic synapses (Choice D) would result in a decreased overall circulating concentration of norepinephrine, whereas norepinephrine began at a higher concentration and rose to a decreased peak.

Objective: Epinephrine plays a critical role in glucose homeostasis through its action on adrenergic receptors located on the liver, the activation of which results in glycogenolysis and gluconeogenesis. The selective activation of medullary epinephrine release may reflect the critical need for epinephrine to serve this function during times of hypoglycemia Previous Next Score Report Lab Values Calculator Help Pause Plasma catecholamine (pg/mL)

56 Exam Section 2: Item 6 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 6. Which of the following curves best describes variation of follicle-stimulating hormone concentration during the menstrual cycle? |A B IE 1 14 28 Day of menstrual cycle A) B) C) D) O E) Correct Answer: B. Follicle-stimulating hormone (FSH) is produced by the pituitary in response to gonadotropin releasing hormone from the hypothalamus. Its concentration fluctuates predictably with each menstrual cycle. At the start of the follicular phase, just after menses occurs, FSH increases to support a developing ovarian follicle. Unlike the 14-day luteal phase, the length of the follicular phase is variable. As the follicle increases in size, it requires less support and FSH decreases slightly until ovulation occurs. Ovulation is accompanied by a rapid surge in FSH and luteinizing hormone (LH); this marks the transition from the follicular phase to the luteal phase. FSH then rapidly subsides and increases slowly through the luteal phase as it prepares to support another ovarian follicle at the start of the following cycle. Incorrect Answers: A, C, D, and E. LH (Choice A) remains low throughout the follicular phase until it surges and triggers ovulation. This marks the transition to the luteal phase, named for the presence of the corpus luteum. The corpus luteum is the structure that remains after a follicle ruptures and ovulation occurs. It produces estrogen and progesterone, which support the secretory endometrium and allow for implantation to occur. If implantation occurs, the corpus luteum is maintained by human chorionic gonadotropin and continues to produce progesterone. If implantation does not occur, the corpus luteum shrinks and eventually becomes a fibrous remnant called the corpus albicans. Progesterone (Choice C) is produced by the corpus luteum after ovulation and maintains the secretory endometrium in order to allow for implantation to occur. If implantation does not occur, progesterone concentrations decrease and the endometrium beings to slough, causing menses. Estrogen (Choice D) is produced by the granulosa cells of the developing follicle. As the follicle increases in size through the follicular phase, its production of estrogen increases. A second, smaller peak of estrogen occurs in the luteal phase, which stimulates the endometrium to proliferate. A hormone that fluctuates less during the overall course of the menstrual cycle is gonadotropin releasing hormone (GnRH). For example, Choice E may represent the pulsatile release of GNRH from the hypothalamus that occurs every 1 to 1 ½ hours in the follicular phase and every two hours in the luteal phase. These small pulses stimulate the pituitary to secrete FSH and LH to drive the development of the ovarian follicle and ovulation.

Objective: FSH increases during the follicular phase of the menstrual cycle in order to support the developing ovarian follicle. It surges at the time of ovulation, followed by a precipitous drop, and then subsequently slowly increases during the luteal phase, preparing to support the developing follicle of the next cycle. II Previous Next Score Report Lab Values Calculator Help Pause Concentration of hormone in blood

76 Exam Section 2: Item 26 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 26. A 42-year-old man comes to the physician for an examination prior to beginning an exercise program at a local health club. He has smoked 1 pack of cigarettes daily for 20 years. He is 183 cm (6 ft) tall and weighs 70 kg (155 lb); BMI is 21 kg/m?. His blood pressure is 130/80 mm Hg. Physical examination shows no other abnormalities. Serum studies show: 350 mg/dL 40 mg/dL 280 mg/dL 150 mg/dL Cholesterol, total HDL-cholesterol LDL-cholesterol Triglycerides Which of the following defects is the most likely cause of this patient's increased serum LDL-cholesterol concentration? A) Decreased ability to convert cholesterol to bile acids B) Decreased clearance of cholesterol by the liver C) Increased absorption of cholesterol in the gut D) Increased production of cholesterol by adipose tissue E) Increased production of cholesterol by the liver Correct Answer: B. Familial hypercholesterolemia (also known as type Il familial dyslipidemia) is a relatively common, autosomal dominant disorder of cholesterol metabolism. The disorder is caused by mutations in the LDLR, PCSK9, or APOB genes. Defects in these genes lead to decreased clearance of cholesterol by the liver and increased serum cholesterol concentrations. LDLR encodes the LDL receptor (LDLR) expressed on hepatocytes, and loss of function mutations decrease LDL uptake by the liver. Gain-of-function mutations in PCSK9 lead to decreased LDLR expression through formation of the PCSK9-LDLR complex, which is internalized by the cell and unable to be recycled to the cell surface. Apolipoprotein B-100 (ApoB-100) defects reduce the ability of the LDL receptor to bind LDL. The diagnosis is suspected when increased cholesterol concentrations are discovered on routine laboratory testing in the absence of secondary causes for hypercholesterolemia such as alcohol use, diabetes mellitus, hypothyroidism, polycystic ovary syndrome, glycogen LDLR storage diseases, nephrotic syndrome, and cholestatic liver disease. It should also be considered when patients present with severe atherosclerotic disease early in life or have a first-degree relative with history of the same. In addition to statin therapy, the patient should be counseled on lifestyle modifications to reduce the lifetime risk for premature coronary artery disease, such as the cessation of tobacco use. Incorrect Answers: A, C, D, and E. The conversion of cholesterol to bile salts by the enzyme CYP7A1 in the liver is the primary route of cholesterol elimination from the body. A decreased ability to convert cholesterol to bile acids (Choice A) can be caused by CYP7A1 deficiencies or inhibition, leading to an increased risk for cholesterol gallstones and hypercholesterolemia. Familial hypercholesterolemia is a more common disorder. Increased absorption of cholesterol in the gut (Choice C) does not lead to increased serum cholesterol concentrations in the absence of impaired liver clearance. Absorption of cholesterol occurs primarily from the enterohepatic circulation, which describes the cycle by which the liver synthesizes bile salts from cholesterol and secretes the bile salts into bile, where they are ultimately reabsorbed by enterocytes in the ileum and transported back to the liver. A minority of absorbed cholesterol comes from the diet. Increased production of cholesterol by adipose tissue (Choice D) is not a known mechanism for hypercholesterolemia. Adipocytes are the major site of free cholesterol storage. Increased production of cholesterol by the liver (Choice E) may be seen in the setting of type IV familial dyslipidemia, characterized by the overproduction of very-low density lipoprotein (VLDL). Typically, triglyceride concentrations are greater than 1000 mg/dL, although cholesterol concentrations may also be mildly increased. Hypertriglyceridemia is a risk factor for pancreatitis.

Objective: Familial hypercholesterolemia is an inherited disorder leading to increased concentrations of cholesterol in the blood and a major risk factor for premature coronary artery disease and death if untreated. The primary mechanism of serum cholesterol accumulation is decreased clearance by the liver. I3D Previous Next Score Report Lab Values Calculator Help Pause

96 Exam Section 2: Item 46 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 46. A 36-year-old woman comes to the physician for a follow-up examination 1 week after being diagnosed with severe hypertension. Her blood pressure is 180/120 mm Hg. Physical examination shows no other abnormalities. A CT scan of the abdomen shows a renal artery aneurysm. A renal angiogram obtained from a femoral approach is shown. Which of the following is the most likely cause of these findings? Augi 08 15:0 A) Atherosclerotic renal artery disease B) Calcific nephrosclerosis C) Fibromuscular dysplasia D) Polyarteritis nodosa E) Polycystic kidney disease FR. pat EAU 13 FRANE 3 Correct Answer: C. Fibromuscular dysplasia of the renal artery is the most common cause of renal artery stenosis in younger and middle-aged women. Fibromuscular dysplasia is a noninflammatory and non-atherosclerotic angiopathy of medium-sized arteries (eg, renal, carotid), that results in multifocal fibrous and muscular thickening of the arterial wall, which can lead to a bead-like appearance of the renal artery on angiography (alternating areas of dilation and stenosis). The resultant stenosis causes secondary hypertension as a result of the abnormal stimulation of the juxtaglomerular apparatus from low afferent blood flow leading to the excessive production of renin and angiotensin. Diagnosis is typically established with renal artery Doppler ultrasonography, CT scan, MR angiography, or conventional angiography. Treatment involves angioplasty or stenting of the stenosed renal artery to improve flow. ACE inhibitors can also be considered for the treatment of unilateral stenosis but can lead to acute renal failure in the setting of significant bilateral renal artery stenosis. Incorrect Answers: A, B, D, and E. Atherosclerotic renal artery disease (Choice A) is a more common cause of renal artery stenosis in older patients after years of atherosclerosis and plaque build-up in the renal arteries and would be unusual in an otherwise healthy young patient. It is a common cause of secondary hypertension in older persons with a history of atherosclerotic cardiovascular disease. Calcific nephrosclerosis (Choice B) describes the deposition of calcium salts in the renal parenchyma secondary to hypercalcemia (often caused by primary hyperparathyroidism, sarcoidosis, or hypervitaminosis D), which can cause acute kidney injury and potentially chronic renal failure. It may be seen as clusters of calcific densities in the renal parenchyma on imaging. Polyarteritis nodosa (Choice D) is a necrotizing vasculitis involving medium-sized arteries that generally affects the kidneys and visceral organs, sparing the lungs. It can cause hypertension, but also presents with other signs and symptoms including fever, headache, malaise, weight loss, and abdominal pain. Polycystic kidney disease (Choice E) is caused by genetic mutations that result in the development of multiple cysts in the kidney because of structural abnormalities of the renal tubules. Cysts may form in infancy, childhood, or adulthood, depending on the genetic mutation inherited, and eventually compress adjacent normal renal parenchyma.

Objective: Fibromuscular dysplasia of the renal artery is the most common cause of renal artery stenosis in younger and middle-aged women. It results in secondary hypertension and presents with a bead-like appearance of the renal artery on angiography. II Previous Next Score Report Lab Values Calculator Help Pause

127 Exam Section 3: Item 27 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 27. An 18-year-old man with Crohn disease is admitted to the hospital because of a 1-day history of severe abdominal pain and intermittent bloody diarrhea. His temperature is 38°C (100.4°F), pulse is 98/min, and respirations are 18/min. Physical examination shows a draining anal fistula. Treatment with broad-spectrum antibiotics and prednisone, along with intravenous hydration and parenteral nutrition, is initiated. He recovers over the next 3 weeks. In addition to resolving the infection, the most likely mechanism of action of this pharmacotherapy is suppression of which of the following? A) Antibody binding B) Complement activity C) Mast cell degranulation D) Neutrophil function E) T-lymphocyte function Correct Answer: E. T-lymphocyte function is suppressed by steroids. Glucocorticoids such as prednisone are synthetic derivatives of endogenously produced cortisol used to a treat a host of disorders including Crohn disease, various autoimmune diseases, and as a part of standard chemotherapy regimens for many lymphoid malignancies. They exert their anti-inflammatory effect through several mechanisms, including the reduction of circulating lymphocytes and by inhibiting lymphocyte migration. Prednisone binds to steroid receptors and is transported to the nucleus where it downregulates gene expression of cytokines and proinflammatory substances. When taken for prolonged periods of time, it can suppress the hypothalamic-pituitary-adrenal axis (HPA axis) leading to a hypoadrenal crisis if exogenous corticosteroids are abruptly withdrawn. Glucocorticoid analogs also have adverse long-term effects on insulin sensitivity, lipid metabolism, and bone health; Cushing syndrome and osteoporosis can result following prolonged use. Incorrect Answers: A, B, C, and D. Antibody binding suppression (Choice A) is characteristic of medications that directly target circulating antibodies. One example is omalizumab, a monoclonal IgG antibody that targets IgE and is used in the treatment of asthma and chronic idiopathic urticaria. Complement activity suppression (Choice B) is characteristic of eculizumab which binds and inhibits complement protein C5, thereby preventing formation of the membrane attack complex. It is used to treat atypical hemolytic uremic syndrome (HUS), paroxysmal nocturnal hemoglobinuria (PNH), and myasthenia gravis. Mast cell degranulation suppression (Choice C), or mast cell stabilization, is achieved by numerous medications, all of which have different mechanisms. The prototypical mast cell stabilizer is cromolyn sodium, which prevents mast cell degranulation and the release of histamine and leukotrienes. It is used in the treatment of asthma and exercise-induced bronchospasm. Neutrophil function suppression (Choice D) is typical of many medications though is often an indirect result of the inhibition of related pathways. For example, nonsteroidal anti-inflammatory drugs, by virtue of inhibiting the production of arachidonic acid derivatives such as prostaglandins and leukotrienes, and steroids, by downregulating the synthesis of immune mediators and impairing the T lymphocyte response, will both result in diminished neutrophil function. Methotrexate has similar indirect effects on neutrophils. He ver, the T lymphocyte, which mediates the inflammatory component, is the direct primary target of glucocorticoids.

Objective: Glucocorticoids such as prednisone primarily exert their anti-inflammatory properties by inhibiting the production and function of T lymphocytes. Prolonged use has myriad consequences including impaired lipid metabolism, insulin resistance, and deleterious effects on bone health. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

150 Exam Section 3: Item 50 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 50. A 19-year-old man who is a soldier in the US Army sustains a traumatic brain injury from the detonation of an improvised explosive device while on patrol in the Middle East. This patient is at increased risk for additional neuronal apoptosis caused by diffusion of which of the following substances from the cerebral capillaries to the brain parenchyma? O A) Adenosine B) Cortisol C) Glutamate D) Lactate E) Palmitate Correct Answer: C. Excitotoxicity describes the phenomenon in which neurons endure additional damage or cell death beyond an initial inciting insult such as a traumatic brain injury, stroke, spinal cord injury, neurodegenerative disease, or withdrawal from gamma-aminobutyric acid receptor agonists (eg, alcohol, benzodiazepines). In response to damage to the central nervous system, glutamate may be released rapidly into the synaptic clefts. When such concentrations persist at high levels, neuronal apoptosis ensues. Unchecked excitatory stimulus from glutamate results in excessive stimulation of the neuron, frequent firing of action potentials, and calcium influx into the cell with resultant activation of phospholipases and proteases. Calcium influx is believed to play a key role in neuronal apoptosis through the activation of intracellular enzymes including calpain and through increased mitochondrial membrane permeability. Such dysregulation can lead to the activation of caspases leading to apoptosis. Damage to subcellular organelles, membrane lipids, and DNA can also trigger the intrinsic apoptotic cascade. Excitotoxicity can occur from excess glutamate, as in this example, or from N-methyl-D-aspartic acid (NMDA). Excitotoxicity can be further exacerbated by a concomitant dysregulation of glucose; hypoglycemia in the setting of glutamate excitotoxicity can further damage the neurons and worsen the extent of neuronal loss. Incorrect Answers: A, B, D, and E. Adenosine (Choice A) is an inhibitory neurotransmitter in the brain, acting as a central nervous system depressant and limiting neuronal activation. It is not associated with excitotoxicity. Cortisol (Choice B), a glucocorticoid hormone, has multiple effects on the central nervous system (CNS). It is not a neurotransmitter and is not associated with excitotoxicity; however, it does result in limbic system activation with resultant hypervigilance and stress and has been implicated in memory loss and cognitive disorders such as Alzheimer dementia. Acutely, cortisol may improve the central nervous system response to a threat by acting to raise blood glucose, heart rate, and blood pressure; however, chronically increased concentrations of cortisol are pathologic. Lactate (Choice D) may be increased in the brain during states of acute neuronal activation. Glutamate has been shown to increase CNS lactate production, though the presence of lactate is a consequence, not a cause, of excitotoxicity. Palmitate (Choice E), a 16-carbon saturated fatty acid, is found ubiquitously throughout the body as a component of the cell membrane phospholipid bilayer and as a synthesized fatty acid stored in adipose. It has been detected in cerebrospinal fluid but is not known to cause excitotoxicity in the setting of central nervous system injury.

Objective: Glutamate and excitatory neurotransmitters are associated with excitotoxicity. Excitotoxicity causes neuronal loss via apoptosis, often in the setting of an ischemic, traumatic, or substance-induced insult to the central nervous system. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

71 Exam Section 2: Item 21 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 21. A 62-year-old man comes to the physician because of a swollen, painful left great toe that awakened him from sleep this morning. Examination of the toe shows erythema, swelling, and exquisite tenderness. Laboratory studies show hyperuricemia. Treatment with colchicine resolves the acute symptoms. Long-term therapy with allopurinol is begun. This treatment results in decreased urate biosynthesis. A diagram of purine synthesis is shown. Which of the following labeled steps is directly inhibited by allopurinol as a result of this therapy? Ribose 5-phosphate Phosphoribosylpyrophosphate B 5-Phosphoribosylamine IMP E Adenylosuccinate XMP AMP GMP Нурохаanthine G Xanthine Urate O A) B) O C) D) E) F) G) Correct Answer: G. Allopurinol, an isomer of hypoxanthine (a purine), is an inhibitor of xanthine oxidase, the enzyme responsible for the final step in metabolism of hypoxanthine and xanthine to uric acid. Excess uric acid precipitates in joints, leading to the inflammatory arthropathy, gout. Purines, adenine and guanine, are metabolized through several intermediate steps prior to the step catalyzed by xanthine oxidase, as illustrated in the diagram above. The final product is uric acid. Inhibiting the ultimate step of purine metabolism results in the accumulation of more soluble reactants and limits the generation of the offending molecule. In turn, the serum concentration of uric acid decreases, and renal excretion or alternative metabolic pathways of hypoxanthine and xanthine takes place. Reducing the serum concentration of uric acid reduces the frequency of flares of gout. Gout classically presents following consumption of a purine-rich meal, especially something rich in red meat or seafood, often in men in their fourth to seventh decades of life. As uric acid accumulates, it crystallizes and precipitates in synovial fluid. The crystals trigger an inflammatory reaction in the joint characterized by the infiltration of leukocytes with the resultant production of inflammatory cytokines and chemokines. Physical examination shows a warm, tender joint, often with a palpable effusion or overlying erythema. The first metatarsophalangeal joint is generally the most frequently affected, and arthrocentesis will disclose negatively birefringent, needle-shaped crystals on microscopy. Incorrect Answers: A, B, C, D, E, and F. Choices A and B reflect the synthesis of purines in the de novo pathway, which involves the enzymes phosphoribosylpyrophosphate (PRPP) synthetase (Choice A) and PRPP amidotransferase (Choice B). The enzymes that synthesize purines de novo are not affected by allopurinol. However, when dysregulated, excess production of PRPP can precipitate gout flares through increased forward metabolism via the LeChatelier principle resulting in an increased production of uric acid. Choice C identifies the enzyme adenylosuccinate synthetase, which catalyzes the reaction in which inosine monophosphate (IMP) reacts with aspartate in a guanosine triphosphate-dependent reaction. Choice D identifies the enzyme adenylosuccinate lyase, which converts adenylosuccinate to AMP and fumarate. Deficiencies in this enzyme are typically inherited in autosomal recessive fashion and result in psychomotor retardation and epilepsy. Choice E identifies the enzyme inosine monophosphate dehydrogenase, which catalyzes the synthesis of xanthosine monophosphate, an intermediate in the de novo synthesis of guanosine monophosphate and guanine nucleotides. Choice F identifies the enzyme guanosine monophosphate synthase, which catalyzes the formation of guanosine monophosphate from xanthosine monophosphate.

Objective: Gout classically presents with painful, swollen joints, and is most often because of the underexcretion of uric acid, which crystallizes in synovial spaces. Allopurinol inhibits xanthine oxidase, which prevents conversion of hypoxanthine and xanthine to uric acid, thereby limiting its concentration and reducing the frequency of gout attacks. II Previous Next Score Report Lab Values Calculator Help Pause

128 Exam Section 3: Item 28 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 28. A previously healthy 18-year-old woman comes to the physician because of a 3-day history of pain with urination and urinary frequency and urgency. Physical examination shows mild tenderness over the suprapubic region. A urine culture grows an organism that is lactose fermenting and spot-indole test positive. The organism is resistant to ampicillin but sensitive to ceftriaxone. Which of the following is the most likely mechanism of antimicrobial resistance exhibited by this organism? A) Alteration of the existing penicillin-binding protein B) Changes in porin C) Efflux pumps D) Elaboration of a new penicillin-binding protein E) B-Lactamase production Correct Answer: E. Urinary tract infections classically present in women with dysuria, urinary frequency, urgency, and suprapubic discomfort. They are most caused by Escherichia coli and gram-negative enteric flora. E. coli, a gram-negative bacterium, is a lactose fermenting facultative anaerobe and can convert tryptophan to indole, a reddish-purple dye (seen in this example as color change on the spot-indole test). The main modality of resistance to B-lactam antibiotics in gram-negative bacteria, such as penicillin and ampicillin, is the production of B-lactamases, also known as penicillinases. Commonly, the genes encoding B-lactamase enzymes are transferred on plasmids between bacteria. In contrast, the main modality of resistance to B-lactam antibiotics in gram-positive organisms is an altered structure of transpeptidase penicillin-binding proteins, rendering B-lactam antibiotics unable to bind and inactivate them. Incorrect Answers: A, B, C, and D. Alteration of existing penicillin-binding proteins (PBPS) (Choice A) is a mode of resistance found frequently in gram-positive bacteria such as Streptococcus pneumoniae and viridans streptococci. Incremental and generational mutations of PBPS convey progressively greater resistance to B-lactam antibiotics, such that in some cases, earlier generation antibiotics (penicillin, ampicillin) may be completely ineffective, while resistance is beginning to emerge to later generation antibiotics as well (eg, carbapenems). Porins (Choice B) are structures that allow the passive transport of hydrophilic molecules through the outer membranes of gram-negative organisms (or cell membranes in general, as a class of proteins). Serratia marcescens, Klebsiella pneumoniae, and Pseudomonas species have demonstrated B-lactam resistance via mutations in genes coding for porins. It is not the most frequent mode of resistance in E. coli. Efflux pumps (Choice C) are used by bacteria to decrease intracellular concentrations of antibiotics. Tetracycline resistance in E. coli is a classic example of this mechanism. Elaboration of a new penicillin-binding protein (Choice D) is the mechanism of resistance in methicillin resistant staphylococci, a gram-positive organism with a cell wall critical to its survival.

Objective: Gram-negative and gram-positive bacteria have different modes of acquiring antibiotic resistance that lead to differing clinical patterns in virulence. B-lactamases inactivate B-lactam antibiotics and are a common mode of resistance to the B- lactam class among gram-negative bacteria. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

22 Exam Section 1: Item 22 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 22. A 63-year-old woman comes to the physician because of vaginal bleeding for 2 weeks. Menopause occurred at the age of 49 years, and she received hormone replacement therapy for 4 years. She has osteoarthritis of the hands and knees, but she has been otherwise healthy. Pelvic examination shows a mildly enlarged uterus and a 10-cm right adnexal mass. An endometrial biopsy specimen shows simple hyperplasia without atypia. This patient most likely has which of the following types of adnexal masses? A) Dysgerminoma B) Granulosa cell tumor C) Mature cystic teratoma D) Serous cystadenocarcinoma E) Sertoli-Leydig cell tumor Correct Answer: B. Granulosa cell tumors are a type of malignant sex cord-stromal tumor. They generally occur in women in their sixth decade of life, though can occur at any age. Their presentation is marked by the effect of their functional production of hormones (eg, estrogen). Thus, it may present with precocious puberty or vaginal bleeding in younger or premenstrual women, and with postmenopausal uterine bleeding in older women. They are typically indolent and may not be detected until large or advanced. Ovarian masses, especially in women who are perimenopausal or menopausal, raise suspicion for malignancy. Younger, menstruating women frequently develop benign cysts involving unilateral or bilateral ovaries, though seldom are these cysts of neoplastic consequence. Any growth arising at or after menopause, or in women with a family history of ovarian malignancy or associated syndrome (eg, BRCA), should be evaluated for cancer. Painless vaginal bleeding must also raise suspicion for endometrial carcinoma in a post-menopausal female (the risk for which is heightened by exogenous estrogen use). This patient's adnexal mass plus normal endometrial biopsy is most consistent with a functional, hormone-producing granulosa cell tumor. On histology, granulosa cell tumors demonstrate Call- Exner bodies, which are granulosa cells arranged around eosinophilic fluid, resembling ovarian follicles. Treatment is through surgical excision, and if the tumor is early stage, prognosis is generally favorable. Incorrect Answers: A, C, D, and E. Dysgerminoma (Choice A) is a common malignant ovarian neoplasm seen in younger women and is the female equivalent of the male seminoma. It is generally rare, representing an estimated 1% of ovarian tumors. Serologically, concentrations of human chorionic gonadotropin and lactate dehydrogenase are often increased. On histology, they generally show sheets of cells with prominent nuclei and pink cytoplasm. Mature cystic teratoma (Choice C), also known as a dermoid cyst, is a benign tumor of the ovary. They are common germ-cell tumors often seen in women age 20 to 40 years. They contain tissue derived from all three germ layers and as a result, often contain differentiated structures such as hair, bone, or teeth. They may be large, and cause pain because of rapid enlargement or associated ovarian torsion. Serous cystadenocarcinoma (Choice D) is the most common ovarian malignant neoplasm. They are generally seen in women of advanced age and those having a genetic predisposition to malignancy (eg, BRCA, hereditary nonpolyposis colorectal malignancy). Frequently, they are detected at a late stage following metastatic spread or complications such as abdominal distention or bowel obstruction secondary to peritoneal carcinomatosis. Psammoma bodies are seen on histology, and serum CA-125 concentrations, while nonspecific, are often increased. Sertoli-Leydig cell tumor (Choice E) defines a rare ovarian sex cord-stromal tumor that is generally unilateral and presents in the second or third decade of life. It often produces androgens, and as a result, presenting symptoms may relate to precocious puberty or virilization.

Objective: Granulosa cell tumors are a malignant sex cord-stromal tumor in women that produces female sex hormones, often resulting in abnormal postmenopausal uterine bleeding. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

155 Exam Section 4: Item 5 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 臺E 5. A pale 62-year-old man has had increasing fatigue and indigestion over the past few months and decreasing appetite over the past few weeks. He has moderate splenomegaly but no lymphadenopathy. Laboratory studies show: 9 g/dL 27% Hemoglobin Hematocrit Leukocyte count 3100/mm3 Neutrophils Lymphocytes 75% Monocytes Platelet count 20% 5% 75,000/mm3 The lymphocytes have cytoplasmic projections and positivity for acid phosphatase even in the presence of tartrate. Which of the following is the most likely diagnosis? A) Acute lymphoblastic leukemia B) Acute myelogenous leukemia C) Chronic lymphocytic leukemia D) Chronic myelogenous leukemia E) Hairy cell leukemia F) Infectious mononucleosis Correct Answer: E. Hairy cell leukemia (HCL) is diagnosed in the setting of pancytopenia, lymphocytes with cytoplasmic projections on peripheral smear, and positive tartrate resistant acid phosphatase (TRAP). Malaise, abdominal distention, early satiety, and pain from splenomegaly are common presenting symptoms. Physical examination may show pallor or ecchymosis as a result of pancytopenia, which is generally demonstrated on laboratory evaluation. In HCL, a clonal somatic mutation of V600E leads to aberrant BRAF signaling and prolonged survival of differentiated B lymphocytes. On light microscopy, malignant cells have long, thin circumferential cytoplasmic projections that resemble hair. HČL is a relatively indolent disease, particularly when compared to aggressive lymphomas such as diffuse large B cell lymphoma or Burkitt lymphoma. HCL is diagnosed by cytochemical stains for TRAP. In normal mammalian cells, phosphatase enzymes are normally inhibited by the addition of tartrate, but the mutated cells of HCL possess a phosphatase that is resistant to inhibition by tartrate. Two methods are used to assess for the presence of TRAP: cytochemical stain or immunostains of fixed specimens. Malignant cells demonstrate red granular cytoplasmic staining. While TRAP positivity can be seen in other indolent lymphomas such as mantle-cell and splenic marginal zone lymphoma, it is relatively sensitive and specific for HCL. Treatment is reserved for symptomatic patients, as in this case, and is usually with purine analogs such as cladribine. Incorrect Answers: A, B, C, D, and F. Acute lymphoblastic leukemia (ALL) (Choice A) is primarily a malignancy of childhood but can be seen in older populations as well, often with a translocation between chromosomes 9 and 22. Cytopenias are common but the characteristic cells in ALL seen on a peripheral smear are typically blasts, not well-differentiated lymphocytes as in HCL. Acute myelogenous leukemia (Choice B) is an aggressive malignancy that primarily affects adults. It is characterized by circulating blasts of the myeloid lineage. Patients are classified into prognostic tiers based upon the presence or absence of mutations, and treatment is generally with cytarabine and daunorubicin, and is sometimes followed by consolidation therapy and stem cell transplantation. Chronic lymphocytic leukemia (Choice C) is a malignancy of differentiated B lymphocytes. Common findings include lymphocytosis, and peripheral smear classically demonstrates lymphocytes with large blue nuclei and scant cytoplasm, many of which may appear ruptured; cytoplasmic projections are not present. Treatment with ibrutinib is initiated if patients have cytopenias or disabling symptoms such as malaise and lymphadenopathy. Chronic myelogenous leukemia (Choice D) is defined by the presence of the Philadelphia chromosome, a translocation between chromosomes 9 and 22, which causes constitutive activation of ABL1 kinase. Patients demonstrate leukocytosis, with concentrations often exceeding 100,000 cells/mm3. Basophilia and eosinophilia are often present, and treatment is typically with tyrosine kinase inhibitors imatinib or dasatinib. Infectious mononucleosis (Choice F) is commonly caused by an infection with Epstein-Barr virus or cytomegalovirus. Symptoms include fatigue, fever, chills, and sore throat; examination may show splenomegaly or pharyngitis, and laboratory evaluation may show increased transaminases. Peripheral smear typically demonstrates atypical lymphocytes without cytoplasmic projections. Pancytopenia is not common.

Objective: HCL is characterized by the presence of TRAP staining by cytoplasmic or immunohistochemical analysis, along with lymphocytes containing long, thin cytoplasmic projections on blood smear. Symptoms include malaise, weight loss, and abdominal discomfort from splenomegaly. Treatment with cladribine is reserved for symptomatic patients. II Previous Next Score Report Lab Values Calculator Help Pause

175 Exam Section 4: Item 25 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 25. A7-year-old boy is brought to the physician because of a 3-day history of cramping abdominal pain, blood in his urine, and a rash on his arms, legs, and buttocks. He had pharyngitis 2 weeks ago. Physical examination shows a hemorrhagic rash involving the extensor surfaces of the upper and lower extremities and feet. There is generalized abdominal tenderness. Test of the stool for occult blood is positive. A renal biopsy specimen shows an increased mesangial matrix that stains positive for IgA on immunofluorescence. Which of the following is the most likely diagnosis? A) Diabetic glomerulosclerosis B) Focal segmental glomerulosclerosis C) Henoch-Schönlein purpura D) Minimal change disease E) Nephroblastoma (Wilms tumor) O F) Poststreptococcal glomerulonephritis G) Renal amyloidosis Correct Answer: C. Henoch-Schönlein purpura (HSP) is an acute, systemic, small-vessel vasculitis that occurs in children and typically presents with gravity-dependent purpura (eg, buttocks, legs, and feet), arthralgias, hematuria, and abdominal pain. The rash first starts briefly as erythematous papules or urticaria, which quickly progresses to petechiae and purpura, which can enlarge to ecchymoses, and usually appears on the extensor surfaces of extremities. It often occurs following a viral illness or upper respiratory tract infection. It is associated with IgA immune complex deposition in small vessels, and IgA nephropathy. When IgA deposition occurs in the renal mesangium, glomerulonephritis may ensue, causing microscopic hematuria, red cell casts, and proteinuria. Additional complications include vomiting, gastrointestinal bleeding, and intussusception. Treatment is supportive; HSP resolves in most patients. Treatment with steroids may be appropriate for children with severe symptoms or renal involvement. Incorrect Answers: A, B, D, E, F, and G. Diabetic glomerulosclerosis (Choice A) occurs following nonenzymatic glycosylation of the glomerular basement membrane and efferent arterioles, characteristically presenting as Kimmelstiel-Wilson lesions on light microscopy. It progresses over time in patients with diabetes mellitus, initially beginning as microalbuminuria, which can subsequently lead to macroalbuminuria and finally, end-stage kidney disease. Focal segmental glomerulosclerosis (FSGS) (Choice B), minimal change disease (Choice D), and renal amyloidosis (Choice G) are nephrotic syndromes. FSGS is most associated with sickle cell disease, opioid abuse, and HIV, and is characterized by segmental sclerosis of the glomeruli. Minimal change disease is the most common cause of nephrotic syndrome in children and can be idiopathic or triggered by a recent infection or immune stimulus. Amyloidosis commonly affects the kidneys in conditions that cause amyloid deposition and under polarized light, a Congo red stain shows green birefringent crystals in the mesangium. In nephrotic syndromes, patients classically present with dependent edema in the buttocks, lower back, and legs, foamy or dark-colored urine, hypoalbuminemia, and proteinuria. Complications include hypercoagulability as a result of urinary loss of antithrombin-3 and infection caused by urinary loss of gamma globulins. Nephroblastoma (Wilms tumor) (Choice E) is the most common kidney malignancy in childhood because of mutations in tumor suppressor genes WT1 or WT2. It is characterized by a large, often palpable, unilateral flank mass and hematuria. It is not associated with a rash, abdominal tenderness, or blood in the stool. Poststreptococcal glomerulonephritis (Choice F) is a nephritic syndrome characterized by hematuria, hypertension, and proteinuria following an infection with group A streptococcus.

Objective: Henoch-Schönlein purpura is an acute, systemic, small-vessel vasculitis that occurs in children and typically presents with gravity-dependent purpura (eg, buttocks, legs, and feet), arthralgias, hematuria, and abdominal pain. It is characterized by IgA immune complex deposition in small vessels and may be complicated by gastrointestinal bleeding, intussusception, and glomerulonephritis. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

5 Exam Section 1: Item 5 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 5. A 43-year-old woman comes to the physician because of vague discomfort, nausea, easy bruising, and weight loss. Her prothrombin time is increased. The most likely cause of the increased prothrombin time is damage to which of the following cells? A) Endothelial cells B) Fibroblasts C) Нерatocytes D) Macrophages E) Skeletal muscle cells F) T lymphocytes Correct Answer: C. Hepatocyte injury accounts for this patient's symptoms and coagulation abnormality. Hepatocytes filter and detoxify portal venous blood, metabolize medications, synthesize coagulation factors (including fibrinogen, prothrombin, factors V, VII, IX, X, XI, XII, proteins C and S), complement, apoproteins, triglycerides, cholesterol, bile acids, and albumin, and play critical roles in biochemical metabolism (eg, gluconeogenesis). Many manifestations of liver disease can be attributed to dysfunction of hepatocytes, which are the primary parenchymal cells within the liver. Prolonged prothrombin time (PT) is a common finding as a result of deficiency of factor VII, and in this case, the patient's easy bruising reflects this acquired coagulopathy. Hypoglycemia from impaired gluconeogenesis, nausea from increased gut transit time, jaundice from hyperbilirubinemia, and hypogonadism from altered metabolism of estrogens are all common findings in patients with chronic liver dysfunction. Malnutrition and weight loss also frequently occur. Incorrect Answers: A, B, D, E, and F. Endothelial cell damage (Choice A) results in hypercoagulability and associated vascular disease. Endothelial cells line the inner lumen of blood vessels and function to reduce clotting. Damage to endothelial cells, which occurs via multiple mechanisms (eg, hypertension, smoking, hyperlipidemia, hyperglycemia), results in local nitric oxide deficiency, fibrosis, inflammation, and calcification, and can lead to myocardial infarction, stroke, and peripheral vascular disease. Fibroblast damage (Choice B) would result in disordered synthesis of the extracellular matrix. As fibroblasts are ubiquitous and exist in nearly every organ system, local deficiencies would produce variable phenotypic effects. Damage to fibroblasts in the skin, for example, impairs the synthesis and release of collagen thereby altering the normal process of wound healing. Macrophage damage (Choice D) and dysfunction predisposes to infection. Disorders in the activation and/or chemotaxis of macrophages lead to an increased susceptibility to mycobacterial infections as a result of failed intracellular killing and granuloma formation. Skeletal muscle cell damage (Choice E) is a common finding in acquired and inherited conditions such as muscular dystrophy, mitochondrial myopathy, toxic, metabolic, and inflammatory myopathies, and rhabdomyolysis. Symptoms are determined by the mechanism of injury but result in either pain, weakness, or both affected muscles. T lymphocyte damage (Choice F), dysfunction, or absence is typical of immunodeficiency syndromes and can be genetic or acquired. HIVIAIDS affects CD4+ T lymphocytes specifically resulting in susceptibility to a wide array of opportunistic pathogens (eg, fungal, mycobacterial, viral, and bacterial). latrogenic damage to T lymphocytes such as that which occurs with cytotoxic chemotherapy also predisposes to systemic infection.

Objective: Hepatocytes are critical for the synthesis of multiple coagulation factors. Dysfunction of hepatocytes results in an increased PT/INR secondary to the impaired synthesis of factor VII and may result in a predisposition to bleeding. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

90 Exam Section 2: Item 40 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 40. A 35-year-old woman comes to the emergency department because of abdominal cramps, nausea, and vomiting for 8 hours. Her pulse is 106/min and regular, and blood pressure is 140/96 mm Hg. Physical examination shows rhinorrhea, excessive lacrimation, and diaphoresis. There is piloerection over most of the skin. The pupils are dilated, equal in size, and responsive to direct and indirect light. There is a resting tremor involving both upper extremities. During the examination, the patient is restless and yawns constantly. This patient is most likely experiencing withdrawal symptoms from which of the following substances? A) Cannabis В) Сосaine C) Diazepam D) Ethanol E) Heroin F) Phenobarbital Correct Answer: E. This patient is withdrawing from heroin, an opiate. Opiates are central nervous system (CNS) depressants. Opiate intoxication causes sedation, bradycardia and hypotension, depressed respiratory drive, and constricted pupils. Opiates are commonly used for analgesia and act on receptors within the gastrointestinal tract, reducing gut motility and thus causing constipation. Opiate withdrawal presents with restlessness, tachycardia and hypertension, pupil dilation, tremors, muscle cramping, abdominal cramping, nausea, vomiting, and diarrhea. Lacrimation and rhinorrhea can occur. Opiate withdrawal also features yawning and piloerection, which are not seen in withdrawal from other recreational substances. Withdrawal from opiates is not potentially fatal, though is generally unpleasant for patients. Treatment is supportive and includes benzodiazepines, antiemetics, and clonidine during the acute withdrawal period, ideally with transition to a partial opioid agonist (eg, buprenorphine) or long-acting agonist (eg, methadone) if the patient is at risk for relapse. Incorrect Answers: A, B, C, D, and F. Cannabis (Choice A) is a hallucinogen that causes euphoria, paranoia, hallucinations, increased appetite, conjunctival injection, and cognitive slowing. Patients withdrawing from cannabis experience irritability, anxiety, depression, insomnia, restlessness, and decreased appetite. This patient experienced lacrimation, rhinorrhea, piloerection, and yawning, which would not occur in cannabis withdrawal. Cocaine (Choice B) is a CNS stimulant that causes euphoria and increases sympathetic tone, leading to tachycardia and hypertension, pupillary dilation, and restlessness. As a result of increased synaptic dopamine, hallucinations and paranoia can occur. Therefore, withdrawal from cocaine can cause depressed mood, fatigue, and slowing of activity. Diazepam (Choice C) is a benzodiazepine medication, which is a CNS depressant with a toxidrome of slurred speech, ataxia, and altered mental status. Benzodiazepine withdrawal features anxiety, dysphoria, psychosis, and seizures. Benzodiazepine withdrawal- induced seizures are life-threatening. This patient experienced lacrimation, rhinorrhea, piloerection, and yawning, which would not occur in benzodiazepine withdrawal. Ethanol (Choice D) is a CNS depressant that, in toxic doses, can cause slurred speech, ataxia, emotional lability, and memory lapses (similar to benzodiazepine intoxication). Ethanol withdrawal causes anxiety, tremors, diaphoresis, nausea and vomiting, and hallucinations. Complicated alcohol withdrawal refers to a severe, life-threatening type of withdrawal that includes seizures or delirium tremens. This patient experienced lacrimation, rhinorrhea, piloerection, and yawning, which would not occur in alcohol withdrawal. Phenobarbital (Choice F) is a barbiturate, a CNS depressant that has similar effects to alcohol and benzodiazepines. The risk for dangerous respiratory depression is higher in barbiturates than benzodiazepines or alcohol. Withdrawal symptoms from barbiturates include life-threatening cardiovascular collapse, delirium, and seizures. This patient experienced lacrimation, rhinorrhea, piloerection, and yawning, which would not occur in barbiturate withdrawal.

Objective: Heroin withdrawal demonstrates symptoms that are common to other CNS depressants such as restlessness, tachycardia and hypertension, tremors, and nausea and vomiting. However, lacrimation, rhinorrhea, piloerection, and yawning are much more typical for heroin withdrawal than for other CNS depressants. I3D Previous Next Score Report Lab Values Calculator Help Pause

197 Exam Section 4: Item 47 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 47. An investigator is studying the thyroid hormone receptor. It is found that the receptor binds chromatin in both the presence and absence of triiodothyronine (T3). Results show that this binding inhibits transcription in the absence of the hormone and activates transcription in the presence of the hormone. It is theorized that binding of the hormone causes an allosteric conformational change that decreases affinity for complex 1 and increases affinity for complex 2. Complex 1 and complex 2 most likely contain which of the following enzyme activities? Complex 1 Complex 2 O A) Deiodinase thyroxine peroxidase B) Histone deacetylase histone acetyltransferase O C) Restriction endonuclease DNA ligase D) Tyrosine kinase protein tyrosine phosphatase E) Ubiquitin ligase SUMO ligase Correct Answer: B. Histone deacetylase and histone acetyltransferase serve to remove (deacetylase) or add (acetyltransferase) acetyl groups to histones. Histones are proteinaceous cores that are rich in lysine and arginine (positively charged amino acids). Without modification (eg, acetylation), the positively charged histones attract negatively charged DNA, which loop around them, forming a chromatin nucleosome. Condensed heterochromatin is transcriptionally inactive. Acetylation of a histone is important in gene transcription because it permits relaxation of a DNA-histone complex by reducing the molecular affinity between DNA and histone. When DNA is tightly bound to a histone, gene transcription does not occur, as binding sites for RNA polymerase II (which transcribes messenger RNA) will be occupied by the presence of the histone. When the affinity between the histone and DNA is reduced, the unpacked (euchromatin) DNA can be bound by transcriptional enzymes, promoting gene expression. In this example, affinity for Complex 1 is reduced when the hormone binds, which decreases the activity of Complex 1 (decreasing deacetylation promotes histone acetylation), and affinity for Complex 2 is increased when the hormone binds, which increases the activity of Complex 2 (promoting histone acetylation). Incorrect Answers: A, C, D, and E. Deiodinase and thyroxine peroxidase (Choice A) are hormones found in the thyroid gland that synthesize thyroid hormone. They are not involved in the transcription of genetic material. Restriction endonuclease and DNA ligase (Choice C) are enzymes related to DNA replication and modification. DNA ligase joins Okazaki fragments together by catalyzing the synthesis of the phosphodiester bond within double-stranded DNA. Restriction endonucleases catalyze the cleavage of DNA into fragments at specific sequences of nucleotides known as restriction sites. They are found in bacteria and have a role in the defense against invading viruses. Tyrosine kinase and protein tyrosine phosphatase (Choice D) are two classes of enzymes that catalyze the addition of (kinase) or removal of (phosphatase) phosphates from tyrosine residues on transmembrane or intracellular proteins. These enzymes play a role in signal transduction within the cell, often in cell growth and differentiation. Mutations of these enzymes can result in the development of neoplastic cell lines. Ubiquitin ligase and SUMO ligase (Choice E) describe two enzyme families involved in post-translational protein modification. Ubiquitin ligase catalyzes the addition of ubiquitin to intracellular proteins, which marks them for destruction by proteasomes. SUMO ligase adds Small, Ubiquitin-like Modifiers (SUMO) to intracellular proteins, which alters function but does not label the modified protein for destruction.

Objective: Histone deacetylase and histone acetyltransferase serve to remove and add acetyl groups to histones, respectively. These enzymes play a critical role in the regulation of gene transcription, as acetylation results in relaxation of the DNA- histone complex (euchromatin), which allows for gene transcription. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

102 Exam Section 3: Item 2 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 2. Patients with hyperuricemia are most likely to have a deficiency of which of the following enzymes? O A) Adenosine deaminase B) Adenine phosphoribosyltransferase C) Adenylate deaminase D) Hypoxanthine-guanine phosphoribosyltransferase E) Xanthine oxidase Correct Answer: D. Hypoxanthine-guanine phosphoribosyltransferase (HGPRT) is a key enzyme in the purine salvage pathway. HGPRT catalyzes the conversion of guanine to guanosine monophosphate and hypoxanthine to inosine monophosphate. Patients with deficient activity of HGPRT are unable to salvage guanine and hypoxanthine and develop resultant increased concentration of xanthine and uric acid. Mutations in the gene encoding HGPRT result in Lesch-Nyhan syndrome. This condition is inherited in an X-linked fashion and therefore develops primarily in males. Findings include intellectual disability, aggressive behavior, self-mutilation, gout, and dystonia. Laboratory evaluation discloses hyperuricemia. Hyperuricemia in Lesch-Nyhan syndrome is treated with xanthine oxidase inhibitors such as allopurinol or febuxostat in order to reduce the synthesis of uric acid. Incorrect Answers: A, B, C, and E. Adenosine deaminase (Choice A) is an important enzyme in purine metabolism. It catalyzes the conversion of adenosine to inosine. Adenosine deaminase deficiency results in excessive concentrations of deoxyadenosine and is a commor combined immunodeficiency (SCID). of severe Adenine phosphoribosyltransferase (APRT) (Choice B) is another important enzyme in purine metabolism. It catalyzes the conversion of adenine to adenosine monophosphate. This reaction employs phosphoribosyl pyrophosphate (PRPP) as a cofactor. Defects in APRT result in the accumulation of adenine, rather than uric acid. Adenylate deaminase (Choice C) catalyzes the conversion of adenosine monophosphate to inosine monophosphate. Defects in this enzyme result in the accumulation of adenosine monophosphate instead of uric acid. Xanthine oxidase (Choice E) catalyzes the final steps in the synthesis of uric acid by catalyzing the conversion of hypoxanthine to xanthine and xanthine to uric acid. Defects in this enzyme can lead to the accumulation of xanthine, a rare condition known as xanthinuria. More commonly, xanthine oxidase is targeted therapeutically in patients with hyperuricemia or gout. Xanthine oxidase inhibitors, such as allopurinol and febuxostat, prevent the synthesis of uric acid.

Objective: Hypoxanthine-guanine phosphoribosyltransferase (HGPRT) is a key enzyme in the purine salvage pathway and serves to recover the purines guanine and hypoxanthine by converting them to guanosine monophosphate and inosine monophosphate, respectively. Defects in HGPRT cause Lesch-Nyhan syndrome, a key feature of which is hyperuricemia. Previous Next Score Report Lab Values Calculator Help Pause

51 Exam Section 2: Item 1 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 1. Lysosomes present in hepatocytes of patients with mucolipidosis |I (I-cell disease) contain abnormal enzyme activities resulting from an inability of these cells to do which of the following? A) Glycosylate secretory proteins B) Recycle lysosomes C) Recycle mannose 6-phosphate receptors from the plasma membrane D) Synthesize lysosomal enzymes E) Synthesize mannose 6-phosphate residues on proteins Correct Answer: E. Mucolipidosis II, also known as l-cell (inclusion cell) disease, results from accumulation of oligosaccharides, lipids, and glycosaminoglycans such as heparan sulfate and dermatan sulfate within cells. I-cell disease is a type of lysosomal storage disease. It is similar in phenotype and pathophysiology to Hurler syndrome. I-cell disease results from a failure of Golgi-based post-translational modification of proteolytic enzymes that would typically be targeted to the lysosome. Proteases targeted to the lysosome are tagged with phosphate at the sixth carbon on their mannose residues, forming mannose 6-phosphate. Defective Golgi N-acetylglucosaminyl-1-phosphotransferase enzymes are unable to join phosphate onto mannose residues. This causes the synthesized proteases to be abnormally routed into vesicles for exocytosis instead of to the lysosome. In absence of these proteins within the İysosome, normal cellular debris that require lysosomal degradation accumulate within the cells and cause inclusions that can be seen on light microscopy. The resulting accumulation of such products leads to widespread cellular and organ dysfunction. Signs and symptoms of l-cell disease include failure to thrive, developmental delay, coarse facial features, restricted skeletal development, hepatosplenomegaly, cardiac structural defects, corneal clouding, and dwarfism. I-cell disease demonstrates autosomal recessive inheritance; there is no treatment for the condition other than supportive care. Incorrect Answers: A, B, C, and D. Glycosylate secretory proteins (Choice A) is a function of the endoplasmic reticulum and the Golgi apparatus. Glycosylation serves multiple purposes in protein folding, conformational stability, cell adhesion, antigen recognition, and blood grouping. I-cell disease involves the phosphorylation of a mannose residue and not the glycosylation of proteins. Inability to recycle lysosomes (Choice B) does not play a role in the pathophysiology of l-cell disease, as it is the misrouting of normal lysosomal enzymes that results in their failure to function. Recycling of mannose 6-phosphate receptors from the plasma membrane (Choice C) may play a role in tracking lysosomal hydrolases from the Golgi apparatus to the lysosome. This is not the known pathophysiology of l-cell disease, which relates to an enzymatic mutation that fails to create the mannose 6-phosphate, which does not directly affect the receptors. Synthesize lysosomal enzymes (Choice D) is incorrect as translation still occurs normally. The deficiency in l-cell disease occurs when routing the translated protein to its site of action in the lysosome.

Objective: I-cell disease is an autosomal recessive lysosomal storage disease that results from a defect in N-acetylglucosaminyl-1-phosphotransferase enzymes. This results in the failure of phosphorylation of lysosomal hydrolases, which subsequently leads to their exocytosis from the cell instead of routing to their normal site of action within the lysosome. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

40 Exam Section 1: Item 40 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 40. A 69-year-old man with hepatic disease and increased prothrombin time is scheduled to undergo a left hip replacement operation. Preoperatively, the patient receives a transfusion of 1000 mL of fresh frozen plasma. Two hours later he develops chills, abdominal cramps, dyspnea, vomiting, and diarrhea. The most likely cause of this patient's reaction is a genetic deficiency in which of the following immune mediators? A) C3 B) IgA C) Interferon gamma D) Interleukin-2 (IL-2) E) Perforin Correct Answer: B. IgA deficiency is the most likely cause of this patient's symptoms. Selective IgA deficiency describes an absence of IgA with normal concentration of IgG, and normal or increased concentration of IgM. IgA is critical for immune health and functions in two primary ways. First, IgA binds pathogen surface antigens targeting them for opsonization, complement activation, and phagocytosis. Second, IgA coats mucosal surfaces and prevents colonization, adhesion, and invasion by pathogenic microorganisms. Immunoglobulins are produced by plasma cells, which are terminally differentiated B lymphocytes; as a result, most causal deficiencies are hypothesized to be related to B lymphocyte abnormalities. Despite this, there is clinical heterogeneity in the severity of IgA deficiency. Some patients remain asymptomatic while others suffer from recurrent sinopulmonary infections. Anaphylactic reactions to blood products such as plasma (which contains donor immunoglobulin) are classic but uncommon, as not all patients with IgA deficiency have reactions to blood products. Such severe reactions are most common in patients with undetectable concentration of IgA; these reactions occur because of the presence of preformed antibodies against IgA, which in these patients is a non-native protein and thus a target of immune activation. Incorrect Answers: A, C, D, and E C3 deficiency (Choice A) leads to recurrent infections with encapsulated bacteria. Manifestations of this disease occur early in life. While transfusion of blood products, particularly platelets, does increase the risk for bacteremia, this patient's age and symptoms are more consistent with anaphylaxis from IgA deficiency. Interferon gamma deficiency (Choice C) occurs via mutations in the interferon gamma molecule or its receptor. Patients develop disseminated mycobacterial infections early in life but are not known to have anaphylactic reactions to plasma or other blood products. Interleukin-2 (IL-2) deficiency (Choice D) or mutations in the IL-2 receptor cause immunodeficiency syndromes that are like severe combined immunodeficiency (SCID) or immunodysregulation polyendocrinopathy enteropathy X-linked syndrome (IPEX). Downstream effects from the absence of IL-2 include impaired NK cell and T-lymphocyte function. Perforin deficiency (Choice E) is characteristic of childhood primary hemophagocytic lymphohistiocytosis (HLH), a syndrome of massive immune activation with fever, sepsis-like syndrome, and death if not immediately treated. Perforin creates pores in membranes of target cells, through which proapoptotic granzymes enter.

Objective: IgA deficiency can result in anaphylaxis following the transfusion of plasma and other blood products containing IgA, as patients with this disease may have existing antibodies against IgA. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

66 Exam Section 2: Item 16 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 16. In a sample of 100 individuals, the mean leukocyte count is 7500/mm3, with a standard deviation of 1000/mm3. Assuming that the leukocyte counts in this population follow a normal (gaussian) distribution, which of the following best represents the percentage of time that a randomly selected individual will have a total leukocyte count lower than 5500/mm? O A) 1% B) 2.5% O C) 5% D) 10% E) 16.5% F) 33.2% Correct Answer: B. In a normal (Gaussian) distribution, data points center around the mean value, which is equivalent to the median and mode. Standard deviation is a measure of the degree of dispersion of data points about the mean and is defined as the square root of the variance. In a normal (Gaussian) distribution, one standard deviation above and below the mean will include approximately 68% of data points, while two standard deviations above and below the mean will include approximately 95% of data points. In a population with a mean leukocyte count of 7500/mm3 and a standard deviation of 1000/mm3, individuals with a leukocyte count less than 5500/mm3 therefore fall more than two standard deviations below the mean. Of the 5% of data points found outside of the second standard deviation (95% lie within two standard deviations, therefore 5% lie beyond two standard deviations), half of these (2.5%) are found below the second standard deviation from the mean (less than 5500/mm³) and half (2.5%) are found above the second standard deviation from the mean (greater than 9500/mm3). Incorrect answers A, C, D, E, and F. 1% (Choice A), 10% (Choice D), 16.5% (Choice E), and 33.2% (Choice F) correspond to a divergence of two standard deviations from the mean value. 5% (Choice C) corresponds to the amount of data points found in total outside the range of two standard deviations above and below the mean, however, only half of these data points fall above, while the other half fall below the mean. Therefore, 2.5% of data points lie beyond two standard deviations below the mean, while the other 2.5% of data points lie beyond two standard deviations above the mean.

Objective: In a normal distribution, standard deviation describes the degree of dispersion of data points about the mean. 68% of data points lie within one standard deviation above and below the mean, and 95% of data points lie within two standard deviations above and below the mean. Previous Next Score Report Lab Values Calculator Help Pause

124 Exam Section 3: Item 24 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 24. A healthy 27-year-old woman consumes a turkey sandwich, a cup of sweetened yogurt, an apple, and three cookies for lunch. After lunch, she returns to working at a computer. Two hours later, she drinks a 16-ounce bottle of cola sweetened with high fructose corn syrup and sucrose. The activity of which of the following enzymes is most likely increased in this patient's liver? O A) Carnitine palmitoyltransferase I B) Fatty acid synthase O C) Glycogen phosphorylase D) Phosphoenolpyruvate carboxykinase E) Protein kinase A Correct Answer: B. Fatty acid synthase is upregulated in the liver during the fed state. The consumption of a large, carbohydrate-laden meal (in this patient, sweetened yogurt, apple, cookies, high fructose corn syrup, and sucrose) results in an increased concentration of blood glucose and an abundance of energy-rich substrates. In turn, the pancreas synthesizes insulin which upregulates metabolic pathways that store such energy-rich substrates. These pathways include fatty acid synthesis, glycogen synthesis, and protein synthesis. Fatty acid synthesis is upregulated by insulin and citrate and is downregulated by glucagon and palmitoyl-CoA. Fatty acid synthesis occurs in the liver, mammary glands, and adipose tissue. Two-carbon fragments are repeatedly joined together to form long-chain fatty acids, which are stored for future catabolism during states of starvation. Incorrect Answers: A, C, D, and E. Carnitine palmitoyltransferase I (Choice A) is a transmembrane protein found on the surface of mitochondria in liver, muscle, and brain, and catalyzes the bond between palmitate and carnitine, which permits the long-chain fatty acid palmitate to translocate into the mitochondrial matrix where B-oxidation (breakdown of the fatty acid) occurs. It is upregulated in states of starvation. Glycogen phosphorylase (Choice C) is upregulated in states of starvation, and results in the degradation of glycogen stores for maintenance of serum glucose concentration. It is downregulated in states of nutritional excess. Phosphoenolpyruvate carboxykinase (Choice D) is upregulated in states of starvation. It catalyzes the irreversible conversion of oxaloacetate to phosphoenolpyruvate which is a key regulatory step in gluconeogenesis. It is downregulated in states of nutritional excess. Protein kinase A (Choice E) is an intracellular protein that is activated downstream from multiple G-protein coupled receptor pathways. In the context of anabolism and catabolism, it plays a key role in the upregulation of gluconeogenesis and glycogenolysis in hepatocytes and skeletal muscle. Generally, it is downregulated in the fed state.

Objective: In the fed state, enzymes that catalyze reactions leading to the formation and storage of energy-rich molecules (eg, glycogen, fatty acids) are upregulated. The activity of fatty acid synthase is upregulated following a meal, with insulin acting as a key activating signal. Previous Next Score Report Lab Values Calculator Help Pause

53 Exam Section 2: Item 3 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 3. The solid curve represents the normal oxygen-hemoglobin dissociation curve. The dashed curve is consistent with an increase in which of the following? 100 80- A) 2,3-Bisphosphoglycerate concentration 60- B) Body temperature C) Fetal hemoglobin concentration 40 D) CO concentration 20 - O E) Pco2 20 40 60 80 100 120 PO2 (mm Hg) Correct Answer: C. Increased fetal hemoglobin concentration accounts for the dashed line in the above oxygen-dissociation curve. Fetal hemoglobin (HbF, a2-V2) has a much higher affinity for oxygen than does adult hemoglobin A (HbA, a2- B2). This is represented by a left shift on the oxygen-dissociation curve, which indicates a higher degree of oxygen saturation (% saturation, y-axis) for any given partial pressure of oxygen (PO2, x-axis). Increased HbF concentrations are seen as a compensatory response in beta-thalassemia, where a mutation in one or more of the two B-globin genes leads to reduced or absent B-globin, thereby reducing the amount of HbA in circulation (B-thalassemia minor) or eliminating it altogether (B-thalassemia major). Increased concentrations of HbF are also seen in sickle cell disease. The hemoglobin tetramer exists in two primary conformations, the taut form, which has a reduced affinity for oxygen, and the relaxed form, which has an increased affinity for oxygen. As the molecules accept oxygen from the lungs and deliver it to tissues, they undergo repeated conformational change between the taut (when unloaded) and relaxed (when loaded) form. Interactions between the monomers within the tetramer that lead to such conformational change also account for the sigmoidal shape of the hemoglobin dissociation curve, a phenomenon called cooperative binding. Fetal hemoglobin binds cooperatively, in the same manner as adult hemoglobin, therefore, while the overall affinity for oxygen is increased (which promotes fetal oxygenation by accepting oxygen from maternal hemoglobin via the placenta), the sigmoidal shape reflecting cooperative binding is maintained, as in this example. Incorrect Answers: A, B, D, and E. 2,3-Bisphosphoglycerate (2,3-BPG) concentration (Choice A) is increased during times of high metabolic activity and at increased altitude; 2,3-BPG, produced during glycolysis, binds to hemoglobin and promotes the unloading of oxygen with shifting of the curve rightward. Increased body temperature (Choice B) decreases the affinity of hemoglobin for oxygen, promoting unloading, and shifts the curve rightward. Carbon monoxide (CO) (Choice D) has a significantly higher affinity for hemoglobin than oxygen and binds to available hemoglobin. On binding, CO displaces oxygen from heme, limiting total carrying capacity, and results in the maintenance of hemoglobin in the relaxed form, with loss of cooperative binding. As a result, while the curve shifts left, it loses its sigmoidal shape, an effect not seen in the graph. Increasing Pco2 (Choice E), which results in a decreased serum pH, is characteristic of periods of metabolic activity. This acidic pH promotes oxygen unloading by reducing hemoglobin affinity for oxygen and shifts the dissociation curve to the right.

Objective: Increased concentrations of fetal hemoglobin result in a left shift of the oxygen dissociation curve with maintenance of the sigmoidal cooperative binding characteristics. I3D Previous Next Score Report Lab Values Calculator Help Pause Hemoglobin saturation (%)

97 Exam Section 2: Item 47 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 47. A 45-year-old woman with chronic hepatitis C comes to the physician because of a 3-week history of fatigue and joint pain. Her temperature is 37.5°C (99.5°F), pulse is 88/min, and blood pressure is 155/75 mm Hg. Physical examination shows the findings in the photograph and hepatosplenomegaly. Serum studies are most likely to show an increased amount of which of the following in this patient? A) Cholesterol B) Complement O C) Creatine kinase D) Cryoglobulin E) D-Dimer Correct Answer: D. Cryoglobulins are immunoglobulins (or immunoglobulin-complement complexes) that precipitate out of serum. Risk factors for excessive cryoglobulin formation include chronic viral infections such as hepatitis B and C viruses, human immunodeficiency virus (HİV), malaria, and Ebstein-Barr virus (EBV), chronic inflammatory conditions such as systemic lupus erythematosus and Sjogren syndrome, and lymphoproliferative diseases such as multiple myeloma. The deposition of the immunoglobulin-complement complexes in blood vessels results in leukocytoclastic vasculitis, which can manifest with a reddish-purple rash (purpura) and possible end-organ dysfunction. Mixed cryoglobulinemia is characterized by the triad of weakness, arthralgias, and palpable purpura. Patients may also present with peripheral neuropathy, hematuria, and hepatosplenomegaly. Incorrect Answers: A, B, C, and E. Increased serum cholesterol concentration (Choice A) can lead to yellowish cholesterol-rich skin deposits called xanthomas. Lesions that arise on the upper and lower eyelids are called xanthelasmas. These can be seen in conditions with severely increase cholesterol concentrations such as familial hypercholesterolemia. Complement concentration (Choice B), especially C3 and C4, are often decreased in the setting of autoantibody and immune-complex mediated diseases. Active flares of systemic lupus erythematosus, mixed cryoglobulinemia, antiphospholipid syndrome, Sjogren syndrome, and membranoproliferative glomerulonephritis are associated with low C3 and C4 concentrations. Creatine kinase (Choice C) is an intracellular molecule released by myocytes when damaged. Increased creatine kinase concentrations can be seen in traumatic injuries (eg, rhabdomyolysis) and inflammatory myopathies such as polymyositis and dermatomyositis. This patient's palpable purpura is not a typical feature of inflammatory myopathy and is more consistent with a diagnosis of mixed cryoglobulinemia. D-dimer (Choice E) is a fibrin degradation product and is a non-specific marker for clotting cascade activation and degradation of cross-linked fibrin molecules. It may be normal or increased in mixed cryoglobulinemia.

Objective: Increased cryoglobulin concentrations are characteristic of mixed cryoglobulinemia, which often presents with palpable purpura in association with arthralgias and peripheral neuropathy caused by immune-complex mediated vasculitis. Chronic hepatitis C virus is the most common infection associated with cryoglobulinemia. II Previous Next Score Report Lab Values Calculator Help Pause

105 Exam Section 3: Item 5 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 5. A 46-year-old woman with systemic lupus erythematosus has had chest pain and fever for 2 days. Temperature is 38.3°C (101°F), pulse is 118/min, and blood pressure is 96/68 mm Hg. Heart sounds are distant. An ECG shows low voltage. Which of the following types of pulses is most likely in this patient? O A) Alternans B) Anacrotic O C) Bigeminal D) Bisferiens E) Paradoxus Correct Answer: E. Pulsus paradoxus describes the phenomenon by which the systolic blood pressure varies during phases of the respiratory cycle. During inspiration, decreased intrathoracic pressure leads to an increase in venous return to the right side of the heart. This bows the interventricular septum into the left ventricular chamber leading to a reduction in left ventricular stroke volume compared to during expiration. This can be measured as a reduction in systolic blood pressure during inspiration relative to expiration. Normal respirophasic blood pressure variation is less than 10 mm Hg. However, in situations where expansion of the heart and pericardium is restricted, such as cardiac tamponade, the difference between the systolic blood pressure during inspiration and expiration will become further disparate (greater than 10 mm Hg) as the left ventricle is compressed during inspiration, dramatically decreasing its stroke volume. This finding is referred to as pulsus paradoxus. The patient in this case is presenting with signs and symptoms of pericarditis (chest pain, fever, tachycardia) and resultant cardiac tamponade (hypotension, distant heart sounds). Pericarditis can occur in patients with viral infection, systemic lupus erythematosus, tuberculosis, lymphoma, or as a complication of myocardial infarction. In addition to pulsus paradoxus, findings suggestive of cardiac tamponade include hypotension, jugular venous distention, distant heart sounds, and a low-voltage ECG. Incorrect Answers: A, B, C, and D. Pulsus alternans (Choice A) is an arterial waveform characterized by alternating strong and weak beats and can be observed in severe left-sided heart failure. On the first beat, the impaired left ventricle may have a reduced ejection fraction, which leads to a greater end-diastolic volume. The second beat will then have greater contractility as the myocardium starts with a greater preload per the Frank-Starling mechanism, and a greater percentage of the ventricular volume will be ejected. However, this leads to a decreased end- diastolic volume and the next beat will have a reduced ejection fraction. This phenomenon is not seen in cardiac tamponade. The anacrotic pulse (Choice B) is seen in aortic stenosis. It is characterized by a low-amplitude wave with a prolonged duration. It is also known as pulsus parvus et tardus (weak and late). A bigeminal pulse (Choice C) is a normal beat followed rapidly by a premature ventricular contraction. The second beat has a reduced amplitude because of a decreased stroke volume from the shortened diastolic filling time. Pulsus bisferiens (Choice D), also known as a biphasic pulse, is associated with aortic regurgitation, combined aortic regurgitation and stenosis, and less commonly hypertrophic cardiomyopathy. The arterial waveform has a double systolic peak, with an initial sharp, shortened peak followed by a decreased amplitude, broader peak.

Objective: Pulsus paradoxus refers to the abnormal variation of systolic blood pressure between inspiration and expiration (greater than 10 mm Hg), which is seen in conditions that restrict expansion of the heart, such as cardiac tamponade. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

81 Exam Section 2: Item 31 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 31. An investigator studying type 2 diabetes discovers an inhibitor of uncoupling protein-2 that stimulates insulin secretion from mouse pancreatic islet cells. This inhibitor most likely increases insulin secretion by which of the following actions? A) Inactivating a tyrosine kinase O B) Increasing glucokinase activity C) Increasing glucose transporter-2 (GLUT-2) expression D) Increasing the ATP:ADP ratio E) Inhibiting the target of rapamycin (TOR) protein kinase F) Stimulating the hydrolysis of phosphoinositide Correct Answer: D. Insulin secretion takes place over several steps. As glucose is transported into the islet cell by the GLUT-2 transporter, glycolysis occurs, which increases the ratio of ATP:ADP. In turn, ATP binds to the ATP-sensitive potassium channels, which close, leading to a change in the transmembrane potential. The cell depolarizes, and calcium ions accumulate within the cytoplasm, which trigger exocytosis of pre-formed insulin. Uncoupling proteins are found in mitochondria, and in this setting, permit dissipation of energy in the electron transport chain as heat, instead of shunting all protons into a gradient and all electrons down the cytochromes to produce ATP. Uncoupling proteins in this manner permit disruption of the proton gradient, uncoupling the gradient from ATP synthesis. If an uncoupling protein were therefore inhibited, as in this experiment, any dissipation of the proton gradient over such a protein would be limited, and all of the stored energy would be converted to ATP, thus, raising the ATP:ADP ratio within the cell. Incorrect Answers: A, B, C, E, and F. Inactivating a tyrosine kinase (Choice A) would limit a cell's ability to respond to insulin, as the insulin receptor contains a tyrosine kinase moiety. Increasing glucokinase activity (Choice B) and increasing glucose transporter-2 (GLUT-2) expression (Choice C) may increase insulin release, but this would occur through the mechanism of increasing the ATP:ADP ratio as glycolysis progressed. Additionally, uncoupling proteins are not directly related to glucokinase or the GLUT-2 transporter. Inhibiting the target of rapamycin (TOR) protein kinase (Choice E) is an intracellular protein that regulates cell growth, proliferation, motility, and cell cycle progression. It is not immediately related to insulin release or uncoupling proteins. Stimulating the hydrolysis of phosphoinositide (Choice F) affects and is involved with multiple cell signaling cascades, generally involving phospholipase enzymes. Downstream effects are multiple and broad but may include signaling related to cell growth or smooth muscle contraction or relaxation.

Objective: Increases in the ratio of ATP:ADP lead to islet cell depolarization by closing potassium-ATP channels. Calcium ions enter the cell, triggering the exocytosis of pre-formed insulin. Uncoupling protein prevents some of the proton gradient in a mitochondrion from being used in the synthesis of ATP. Blocking such a protein would theoretically increase the cell's ATP:ADP ratio. Previous Next Score Report Lab Values Calculator Help Pause

42 Exam Section 1: Item 42 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 42. An 18-year-old man with acute lymphoblastic leukemia comes to the physician because of a 3-day history of intermittent fever and left-sided chest pain. He is in the fifth week of induction chemotherapy consisting of asparaginase, daunorubicin, prednisone, and vincristine. His absolute neutrophil count has been less than 500/mm3 for the past month. His temperature is 39.2°C (102.6°F). Physical examination shows dullness to percussion and decreased breath sounds on the left side. A chest x-ray shows a left lower lobe infiltrate and a moderate pleural effusion on the left. A photomicrograph of a periodic acid-Schiff stain of pleural fluid obtained via thoracentesis is shown. Which of the following is the most appropriate pharmacotherapy for this patient? A) Amphotericin B B) Ertapenem O C) Ganciclovir O D) Infliximab O E) Rituximab Correct Answer: A. Amphotericin B is the most appropriate therapy for Cryptococcus neoformans, which is an encapsulated, non-dimorphic yeast and common opportunistic pathogen in immunocompromised patients. Malignancy, especially lymphoma or leukemia, HIV/AIDS, solid organ transplantation, sarcoidosis, liver cirrhosis, and long-term corticosteroid therapy are risk factors for infection. It can cause cryptococcal meningitis, encephalitis, and pulmonary cryptococcosis. Transmission is through inhalation, with risk for dissemination to the brain and meninges as well as other distal sites. Pulmonary cryptococcosis typically presents with a low-grade, intermittent fever and non-specific respiratory symptoms. Central nervous system involvement can cause headache and progressive encephalopathy. Dermatologic spread can occur with pustular, papular, nodular, or ulcerated lesions. Diagnosis is made by analysis of cerebrospinal fluid, sputum, urine, and blood. Histologic examination may show yeast cells with narrow-based budding and a bright red inner capsule on staining with mucicarmine or periodic acid-Schiff. In cryptococcal meningitis, lumbar puncture classically detects an increased opening pressure, increased cerebrospinal fluid protein, and a mononuclear cell pleocytosis. Amphotericin B is the treatment for cryptococcal meningitis or severe pulmonary cryptococcosis in an immunocompromised patient. Incorrect Answers: B, C, D, and E. Ertapenem (Choice B) is B-lactam antibiotic of the carbapenem class, which acts by inhibiting cross-linking of bacterial cell walls by penicillin-binding proteins. It has a broad range of activity against gram-positive, gram-negative, and anaerobic bacteria. Ertapenem is not used to treat fungal infections. Ganciclovir (Choice C) is a guanosine analog which preferentially inhibits viral DNA polymerase. It is used to treat infections caused by cytomegalovirus. Infliximab (Choice D) is a monoclonal antibody which targets soluble TNF-a to prevent effective binding with the TNF-a receptor. It is used in the treatment of autoimmune disorders such as Crohn disease, ulcerative colitis, rheumatoid arthritis, ankylosing spondylitis, psoriasis, and Behçet disease. Rituximab (Choice E) is a monoclonal antibody which targets CD20 found on the surface of B lymphocytes, leading to increased B lymphocyte destruction by natural killer cells. It is used to treat non-Hodgkin lymphoma, chronic lymphocytic leukemia, rheumatoid arthritis, granulomatosis with polyangiitis, idiopathic thrombocytopenic purpura, pemphigus vulgaris, and myasthenia gravis.

Objective: Infection with Cryptococcus neoformans is common in patients with immunocompromising conditions. The organism can be recognized on histology by the presence of narrow-based budding yeast cells and bright red capsules with mucicarmine or periodic acid-Schiff staining. Previous Next Score Report Lab Values Calculator Help Pause

132 Exam Section 3: Item 32 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 32. A 12-year-old boy is brought to the physician by his mother because of a persistent rash around his nose and mouth for 4 months. He also has had a decreased appetite and an unintentional 4.5-kg (10-lb) weight loss during the past 3 months. His mother states, "The rash developed shortly after he started at his new school. I wonder if he keeps getting into something. Il've tried hydrocortisone cream, but it didn't work." She adds, "And now his grades are falling, and he's not eating as much as he used to. I hope he's not depressed because of the rash." The rash began as mild redness and then became dry and started to flake. The patient says there is no pain or discomfort with the rash. Physical examination shows injection of both conjunctivae, an eczematous eruption extending inferiorly from the nasolabial fold to the upper lip, and an erythematous nasal mucosa that is friable with gold stippling. When asked, he states that his mood is fine. Which of the following is the most likely cause of this patient's condition? A) Atopic dermatitis B) Impetigo C) Inhalant abuse D) Mercury poisoning E) Wegener granulomatosis Correct Answer: C. Inhalant abuse (eg, glue, spray paint, shoe polish, toluene, nitrous oxide) is common in children and adolescents and has effects on multiple organ systems such as dermatologic (eg, perioral or perinasal dermatitis with erythro- or pyoderma), otolaryngologic (eg, nosebleeds, halitosis), ocular (eg, conjunctival injection), cardiac (eg, dysrhythmia, tachycardia), gastrointestinal (GI) (eg, nausea, anorexia), respiratory (eg, wheezing, coughing, sneezing), and neuropsychiatric (eg, headache, anosmia, mood swings, irritability, hallucinations). The inhalant can dry the skin, leading to irritant dermatitis or bacterial superinfection. Paint or glitter may be seen on the patient's face or hands, as in this patient. This patient's recent poor functioning may be related to the neuropsychiatric effects of inhalants, and his weight loss may be related to anorexia or mood changes. Severe inhalant toxicity can manifest with tachydysrhythmias, respiratory depression, tremor, ataxia, and nystagmus, and, ultimately, seizures or coma. Treatment is primarily supportive, though some inhalants have specific antidotes (eg, methylene blue for nitrites). Prevention is key; schools should monitor the use of solvent-based products and parents and children should be educated about the risks of inhalants. Incorrect Answers: A, B, D, and E. Atopic dermatitis (Choice A), or eczema, is a chronic pruritic inflammatory skin disease common in children that is suggested by a personal or family history of atopic disease (eg, asthma or allergic rhinitis). The rash is initially characterized by erythematous papules and vesicles with exudates and crusting and, with chronic scratching, becomes dry, scaly, and excoriated in a process called lichenification. Rashes associated with inhalant abuse can appear similar to atopic dermatitis. However, in this patient without an atopic history who has neuropsychiatric and GI symptoms along with gold stippling of the rash, inhalant abuse is more likely. Impetigo (Choice B) is a contagious, childhood bacterial skin infection of the face and extremities with lesions that progress from papules to vesicles and exhibit crusts that have a yellow, gold, or honey color. The rash can appear like that of inhalant abuse. However, this patient has a chronic rash with gold stippling and neuropsychiatric and GI symptoms, which are atypical of impetigo. Mercury poisoning (Choice D) can acutely cause dermatitis, stomatitis, nausea, vomiting, cough, dyspnea, and chest pain that may be related to a potentially fatal interstitial pneumonitis. Symptoms depend on the route of exposure. Chronic exposure may cause neuropsychiatric symptoms and a desquamating rash on the palms and soles, though a chronic dermatitis-like rash on the face, as in this patient, would be rare. Wegener granulomatosis (Choice E), or granulomatosis with polyangiitis, is an antineutrophil cytoplasmic autoantibody (ANCA)-associated vasculitis. The vasculitis is associated with necrotizing vasculitis in the nasopharynx, lungs, and kidneys. Rash, neuropsychiatric symptoms, and Gl symptoms, as in this patient, are atypical for this disease.

Objective: Inhalant abuse can manifest with perioral and perinasal dermatitis, conjunctival injection, unusual breath odor, nausea, anorexia, respiratory symptoms, and neuropsychiatric symptoms such as irritability and mood swings. The rash may contain paint or glitter depending on the type of inhalant. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

160 Exam Section 4: Item 10 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 10. A 62-year-old woman has an abdominal aortic aneurysm repaired. Forty-eight hours after surgery, her right distal leg is dusky and cool, and there are dark purple-to-black necrotic lesions on several toes as shown. A biopsy of one of the lesions would most likely show occluded small arteries with needle-shaped clefts. Which of the following is the most likely cause of the lesions? A) Arteriolar narrowing B) Cholesterol emboli C) Septic emboli D) Vasospasm O E) Venous thrombosis Correct Answer: B. Aortic atheroembolism, also known as cholesterol embolization syndrome, occurs secondary to embolization of atherosclerotic plaque contents (eg, cholesterol crystals) from a proximal large artery (eg, aorta) to distal small arteries and arterioles. This results in occlusion of the distal arterial vasculature with small cholesterol emboli, which induces a localized inflammatory response and end-organ damage. Dermatologic symptoms of this phenomenon are localized petechiae, livedo reticularis, and blue toe syndrome. Histology of skin lesions discloses biconvex needle-shaped clefts within the arterial lumen left by dissolved cholesterol crystals. Risk factors for cholesterol embolization syndrome include trauma and interventional procedures that increase the risk for damage and dislodgement of atherosclerotic plaques, such as aortic aneurysm repair or cardiac catheterization (eg, damage to atherosclerotic plaques of the thoracic and abdominal aorta via a femoral artery approach). Cholesterol emboli can also involve other organs, potentially resulting in kidney failure, colonic ischemia, cerebrovascular accidents, and skeletal muscle ischemia. Incorrect Answers: A, C, D, and E. Arteriolar narrowing (Choice A) occurs in arteriolosclerosis with vessel wall thickening and narrowing of the lumen, which can lead to distal ischemic injury. It is associated with chronic hypertension and diabetes mellitus, and often affects the kidneys. It does not present acutely as in this patient's case. Septic emboli (Choice C) arise from distant infectious sources such as infective endocarditis, mycotic aneurysm, or infectious thrombophlebitis. Patients typically present with fever and symptoms related to the source of the infection. In endocarditis, septic emboli can cause Janeway lesions, which are painless, macular, hemorrhagic lesions that occur on the palmar surface of the hands and feet. Vasospasm (Choice D) causes the intermittent loss of blood flow to the distal extremities in Raynaud phenomenon, resulting in sequential white, blue, and red discoloration of the affected areas of skin. Ischemic injury can result in digital ulcerations. Raynaud phenomenon may be idiopathic or associated with mixed connected tissue disease, systemic lupus erythematosus, or CREST syndrome. Venous thrombosis (Choice E) of an extremity classically presents with unilateral edema, erythema, and pain in the affected extremity. Ischemic injury can result from capillary bed congestion and impaired perfusion, which increases the risk for skin ulceration.

Objective: Interventional procedures increase the risk for atherosclerotic plaque disruption, especially during manipulation of the aorta, which can lead to cholesterol emboli. Dermatologic symptoms include localized petechiae, livedo reticularis, and blue toe syndrome, with needle-shaped clefts on histologic examination. %3D Previous Next Score Report Lab Values Calculator Help Pause

188 Exam Section 4: Item 38 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 38. A 32-year-old woman undergoes a laparoscopic cholecystectomy after ultrasonography showed a thickened gallbladder wall and gallstones. The surgeon will be unable to inspect directly which of the following organs during this procedure? A) lleum O B) Jejunum C) Pancreas D) Stomach E) Transverse colon Correct Answer: C. Laparoscopic cholecystectomy is a minimally invasive technique whereby the gallbladder is excised and removed through small port incisions of the abdominal wall. This surgical procedure takes place within the intraperitoneal space. Despite its proximity to the gailbladder, most of the pancreas is located within the retroperitoneal space and is not directly visible during laparoscopic cholecystectomy. The peritoneum is a serous membrane lining the abdominal cavity. It covers many of the major abdominal organs, including the stomach, spleen, liver, gallbladder, most of the small intestine except for the proximal duodenum, the cecum, the transverse colon, and the sigmoid colon. Retroperitoneal structures are located within the abdominal cavity but are found posterior to the peritoneum. Important retroperitoneal structures include the adrenal glands, kidneys, ureters, pancreas, proximal duodenum, the ascending and descending colon, the rectum, and the abdominal aorta and inferior vena cava. Incorrect Answers: A, B, D, and E. The ileum (Choice A) is the most distal portion of the small intestine. The ileocecal valve connects the ileum with the cecum. Both the ileum and the cecum are intraperitoneal and can be visualized during laparoscopy. The jejunum (Choice B) is the second portion of the small intestine. It is located between the duodenum and the ileum. The jejunum is intraperitoneal and can also be visualized during laparoscopy. The stomach (Choice D) is an intraperitoneal structure and is readily identifiable during laparoscopic cholecystectomy, where it will be found medial to the gallbladder and within the epigastrium and left upper quadrant of the abdominal cavity. The stomach's lesser curvature is connected to the liver by the lesser omentum, which is an important landmark during laparoscopic cholecystectomy. The transverse colon (Choice E) is an intraperitoneal segment of the large intestine, along with the cecum and sigmoid. During laparoscopic cholecystectomy, the transverse colon can be visualized anterior and sometimes inferior to the gallbladder.

Objective: Laparoscopic cholecystectomy is a minimally invasive surgical procedure that occurs within the peritoneal space. Retroperitoneal organs, including the pancreas, cannot be easily visualized during an intraperitoneal surgery. Previous Next Score Report Lab Values Calculator Help Pause

173 Exam Section 4: Item 23 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 23. A 27-year-old man is admitted to the hospital because of fever, fatigue, malaise, anemia, and lymphadenopathy. He is a marine and had been stationed in Saudi Arabia. He has a history of a lesion that developed at the site of a sand fly bite 2 weeks before his current symptoms began. Blood cultures are negative. A biopsy specimen of a skin lesion shows enlarged macrophages containing numerous intracellular organisms. Which of the following is the most likely causal organism? A) Dracunculus medinensis B) Leishmania tropica O C) Onchocerca volvulus D) Plasmodium ovale E) Trypanosoma brucei Correct Answer: B. Leishmania tropica, a member of the protozoa genus Leishmania, is the cause of Old World leishmaniasis. The organism is transmitted via the sand fly and is endemic to the Middle East, Asia, and North Africa. New World leishmaniasis, caused by Leishmania braziliensis, is endemic to Central America and South America. Leishmaniasis has both cutaneous and visceral forms. Cutaneous leishmaniasis begins with a pink papule at the site of the sand fly bite, typically on uncovered skin as the flies do not penetrate clothing. Additional lesions may develop and spread along the draining lymphatics. In Old World leishmaniasis, a characteristic hyperkeratotic eschar may form. Visceral leishmaniasis presents with a slow onset of fever, malaise, weight loss, and organomegaly over several weeks to months. In both conditions, histopathologic visualization of amastigotes within macrophages on blood smear or tissue biopsy is diagnostic. Incorrect Answers: A, C, D, and E. Dracunculus medinensis (Choice A) is a nematode found in Africa and Asia and the cause of dracunculiasis. The infection is contracted by consumption of unfiltered water containing small crustaceans infected with the larvae of D. medinensis. The larvae enter the abdominal cavity through the intestinal tract, burrow through the subcutaneous tissue, and then emerge through the skin at the site of a painful papule approximately one year later. Systemic symptoms including fever, dizziness, pruritus, or urticaria may develop prior to emersion of the nematode from the skin. Onchocerca volvulus (Choice C) is the nematode causing onchocerciasis, known as "river blindness," and is transmitted through a bite of the female blackfly. Patients exhibit ocular changes and a variety of cutaneous manifestations: subcutaneous nodules, generalized itching, skin atrophy with loss of elastin, and depigmentation on the anterior shins. The treatment of choice for onchocerciasis is ivermectin. 31 Plasmodium ovale (Choice D) is one of several species in the Plasmodium family capable of causing malaria. Patients with malaria caused by P. ovale exhibit fever, headache, anemia, and splenomegaly in a 48-hour, or tertian, cycle. Peripheral blood smear demonstrating a trophozoite ring within erythrocytes confirms the diagnosis. In P. ovale infection, red cytoplasmic granules are additionally seen in the erythrocytes. P. ovale may also survive in its hypnozoite, or dormant, form in the liver. Treatment of this organism must include primaquine as well as chloroquine in order to eradicate both the active and dormant forms. Trypanosoma brucei (Choice E) is the protozoa that causes human African trypanosomiasis, or African sleeping sickness. The parasite is transmitted through the painful bite of a tsetse fly, after which it makes its way into the blood stream. Patients exhibit lymphadenopathy, recurrent fever, and somnolence. The treatment of choice is suramin for the hematologic disease or melarsoprol for neurologic involvement.

Objective: Leishmania tropica is a cause of leishmaniasis, which has both cutaneous and visceral manifestations. Diagnosis is confirmed with histopathologic visualization of amastigotes within macrophages. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

101 Exam Section 3: Item 1 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 1. In a breast-fed infant, lipase has an integral role in the digestion of which of the following? A) Cholesterol B) Fatty acids C) Starch D) Triglycerides Correct Answer: D. Human breast milk is rich in triglycerides and fatty acids, which form an important component of the caloric intake of a breast-fed infant. Triglycerides must be digested into free fatty acids by lipase prior to absorption by the infant's intestinal enterocytes. Adults primarily employ pancreatic lipases during triglyceride digestion; however, these enzymes are comparatively inactive in infants. Infants therefore rely primarily on lingual lipase for triglyceride digestion. Incorrect Answers: A, B, and C. Cholesterol (Choice A) is also an important component of breast milk. However, cholesterol is digested via emulsification with bile salts. Once emulsified, cholesterol can be absorbed into micelles, which are taken up by intestinal enterocytes. This process does not rely on lipase. Fatty acids (Choice B) and monoglycerides are produced after the hydrolysis of triglycerides by lipase. Lipase is not involved in the further digestion of fatty acids. Starch (Choice C) is a polysaccharide derived from plants such as corn, wheat, and rice. The predominant source of carbohydrate in breast milk is lactose, rather than starch. Human breast milk also contains amylase, which aids the infant in the digestion of carbohydrates.

Objective: Lipids, including triglycerides, cholesterols, and free fatty acids form an important part of the diet of a breast-fed infant. Lipase, especially lingual lipase, plays an important role in infant digestion by hydrolyzing triglycerides into free fatty acids and monoglycerides, which can be absorbed by intestinal enterocytes. Previous Next Score Report Lab Values Calculator Help Pause

145 Exam Section 3: Item 45 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 45. A 58-year-old man comes to the physician because of a 2-week history of severe chest pain when climbing stairs. He does not have chest pain at rest. The diagnosis of angina pectoris is made. Nitroglycerin is prescribed for this patient because of which of the following mechanisms of action? A) Decreases CAMP B) Decreases CGMP C) Decreases inositol 1,4,5-trisphosphate D) Increases CAMP E) Increases CGMP F) Increases inositol 1,4,5-trisphosphate Correct Answer: E. Nitroglycerin acts within vascular smooth muscle cells by releasing nitrite ion (NO), which activates guanylate cyclase and increases CGMP leading to smooth muscle relaxation and vasodilation. Muscle contraction occurs when myosin light chains in muscle cells are phosphorylated, leading to cross-bridge formation between myosin heads and actin filaments. This reaction is performed by myosin light chain kinase (MLCK). This process is opposed by myosin light chain phosphatase (MLCP), which dephosphorylates the myosin light chains resulting in decreased smooth muscle tone (eg, smooth muscle relaxation). MLCP is upregulated by CGMP, leading to increased myosin light chain dephosphorylation, decreased vascular smooth muscle tone, and vasodilation. Nitroglycerin is a potent venodilator, which reduces cardiac preload and myocardial oxygen demand, making it a first-line therapy in the treatment of angina pectoris. Incorrect Answers: A, B, C, D, and F. Decreases CAMP (Choice A) is incorrect as this would result in increased myosin light chain kinase activity, and lead to smooth muscle contraction and associated vasoconstriction, worsening the symptoms of angina pectoris. CAMP typically inhibits the activity of myosin light chain kinase, which leads to muscle relaxation. Decreases CGMP (Choice B) is incorrect as this is the opposite action of NO on guanylate cyclase. Decreases inositol 1,4,5-triphosphate (IP3) (Choice C) is incorrect because nitroglycerin does not act on this pathway. Increases CAMP (Choice D) is incorrect as well because nitroglycerin does not impact concentrations of CAMP, rather, this is the mechanism of action of phosphodiesterase 4 inhibitors. Increases inositol 1,4,5-triphosphate (Choice F) is incorrect because this would lead to increased muscle contraction. IP3 is a signaling molecule that increases intracellular calcium release from the sarcoplasmic reticulum of muscle cells. Calcium binds to calmodulin, and the calcium-calmodulin complex activates MLCK, leading to increased muscle contraction.

Objective: Nitroglycerin is used in the treatment of angina pectoris for its vasodilatory effect, which is mediated by increased CGMP in smooth muscle cells. Previous Next Score Report Lab Values Calculator Help Pause

129 Exam Section 3: Item 29 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 29. A 46-year-old woman comes to the physician because of a 3-day history of pain and swelling of her left leg. One week ago, she underwent operative removal of several sebaceous cysts over the labia majora. Physical examination shows edema of the left lower extremity and purulent material draining from one of the operative incisions. This patient most likely has an infection that spread to which of the following groups of lymph nodes? A) External iliac B) Inguinal C) Internal iliac D) Obturator E) Sacral Correct Answer: B. The inguinal lymph nodes are a component of the lymphatic system, a network of vessels which follow a predictable pattern of drainage to lymph node beds. Inguinal nodes drain lymphatic fluid from the labia majora, vulva, scrotum, anal canal below the pectinate line, and the skin below the umbilicus with the exception of the popliteal fossa. The lymphatic network of the right side of the body above the diaphragm is drained by the right lymphatic duct which enters the right subclavian vein, while lymph from the rest of the body, including the bilateral lower extremities and pelvis, is channeled through the cisterna chyli and thoracic duct, which ultimately drains into the left subclavian vein. Lymph is generated by hydrostatic pressure in the tissues causing fluid to leak out of vascular structures and into the interstitium. It is then collected by the lymphatics along with lymphocytes exiting the tissues after being activated by antigen presenting cells, such as in the case of infection. When an infection increases the number of lymphocytes returning to the lymph nodes, the resulting congestion can cause edema of the anatomical area being drained. Incorrect Answers: A, C, D, and E. External iliac (Choice A) lymph nodes are clustered around the external iliac vessels. These nodes drain lymphatics from the glans penis, clitoris, upper portion of the vagina, and fundus of the bladder. Internal iliac (Choice C) lymph nodes drain the lower rectum to the anal canal above the pectinate line, bladder, middle third of the vagina, cervix, and prostate. Since the infection originated from the labia majora where the surgery was performed, none of these anatomical structures were involved. The obturator (Choice D) and sacral (Choice E) lymph nodes are located in the obturator fossa and sacral concavity, respectively. Both of these lymph node beds drain lymph from the internal pelvis, but not the labia majora.

Objective: Lymph from the labia majora, vulva, scrotum, and anal canal below the pectinate line drains to the inguinal lymph nodes. Previous Next Score Report Lab Values Calculator Help Pause

181 Exam Section 4: Item 31 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 31. A 14-year-old girl is brought to the physician because of a recent growth spurt of 15 cm (6 in) during the past year. She also has had increasing fatigue and palpitations during this period. Her father is 193 cm (6 ft 4 in). Her paternal aunt has a history of palpitations and severe myopia. The patient is at the 95th percentile for height and 50th percentile for weight. Physical examination shows a long, thin face. Ophthalmologic examination shows dislocated lenses. Cardiac examination shows a hyperdynamic precordium with early click and systolic murmur. Echocardiography shows an enlarged aortic root and mitral valve prolapse. Which of the following genetic mechanisms best explains the findings in this family? A) Anticipation B) Heterogeneity C) Paternal disomy D) Pleiotropy E) Polygenes Correct Answer: D. Pleiotropy describes when a single gene affects multiple, seemingly unrelated characteristics. This principle is illustrated by Marfan syndrome, an autosomal dominant condition in which a defect of collagen synthesis leads to problems in multiple organ systems. The patient in this case has several ocular, musculoskeletal, and cardiovascular abnormalities consistent with Marfan syndrome. Most patients with Marfan syndrome demonstrate a mutation in the FBN1 gene on chromosome 15, which encodes the protein fibrillin- 1, leading to structural compromise of connective tissue. Since connective tissue is ubiquitous throughout the body, defects can affect almost any organ or organ system. Patients with Marfan syndrome are often tall, have long, thin extremities, demonstrate joint hypermobility and arachnodactyly, may experience aortic incompetence and aortic aneurysms, and may present with lens dislocations. This single gene mutation affecting multiple traits is an example of pleiotropy. Incorrect Answers: A, B, C, and E. Anticipation (Choice A) is the phenomenon in which a genetic condition tends to increase in severity or present at an earlier age as it passes through multiple generations. Examples include Huntington disease, myotonic dystrophy, and fragile X syndrome. These disorders generally involve trinucleotide repeat expansions. Heterogeneity (Choice B) refers to the genetic mechanism by which different genes produce a similar phenotype. It is the reverse of pleiotropy. For example, there are over 2000 known mutations which affect the CFTR gene in cystic fibrosis, most of which lead to a similar clinical phenotype of thick, inspissated secretions, frequent pneumonia, bronchiectasis, and failure to pass meconium in infancy. Paternal disomy (Choice C) is the term describing inheritance of both copies of a chromosome from the biological father rather than one copy from each parent. Patients with Angelman syndrome have inherited both copies of chromosome 15 from the father. Angelman syndrome is characterized by short stature, intellectual disability, severe speech impairment, and ataxia. Polygenes (Choice E) describes the mechanism by which several different genes interact with each other to produce a spectrum of phenotypes. Examples include human height, skin color, eye color, and hair color.

Objective: Marfan syndrome is a genetic disorder affecting connective tissue in which mutations in one gene (FBN1) lead to a wide variety of traits such as tall and thin stature, ocular lens dislocation, mitral valve prolapse, and aneurysms. A single gene contributing to many phenotypic effects is known as pleiotropy. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

104 Exam Section 3: Item 4 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 4. The newborn of a mother with poorly controlled diabetes mellitus has a convulsion and dies shortly after birth. Which of the following morphologic changes is most likely to be seen on autopsy? A) Basophilic cell hyperplasia of the pituitary gland B) B-Cell adenoma of the pancreas C) Eosinophilic adenoma of the pituitary gland D) Hyperplasia of the adrenal zona fasciculata E) Hyperplasia of the pancreatic islet cells Correct Answer: E. Hyperplasia of the pancreatic islet cells occurs in response to hyperglycemia because of the increased production of insulin. Glucose is able to cross the placenta, while insulin is not. In poorly controlled gestational diabetes, maternal glucose crosses the placenta and leads to fetal hyperglycemia. As insulin is not able to cross the placenta, the fetal pancreatic islet cells increase their production of insulin. After delivery, the maternal source of glucose is removed but the pancreatic islet cells continue to produce the same amount of insulin. This imbalance of insulin and glucose may then lead to hypoglycemia in the newborn. Severe hypoglycemia may lead to convulsions and death. Incorrect Answers: A, B, C, and D. Basophilic cell hyperplasia of the pituitary gland (Choice A) would correspond with an increased production of follicle stimulating hormone, luteinizing hormone, thyroid stimulating hormone, and adrenocorticotropic hormone. Increase in these hormones would not lead to hypoglycemia or convulsions. B-Cell adenoma of the pancreas (Choice B) is a tumor derived from the insulin secreting cells in the pancreas. While this could be a cause of increased insulin and resultant hypoglycemia in an adult patient, it is unlikely to be the cause in a newborn. Eosinophilic adenoma of the pituitary gland (Choice C) is a benign tumor that may produce excess pituitary hormone such as growth hormone. It may lead to increased insulin-like growth factor production with eventual diabetes and hyperglycemia in the case of acromegaly. Hyperplasia of the adrenal zona fasciculata (Choice D) would lead to the increased production of cortisol. Increased cortisol causes hyperglycemia and insulin resistance, rather than hypoglycemia.

Objective: Maternal glucose crosses the placenta while insulin does not. In response to hyperglycemia, the fetal pancreatic islet cells increase the production of insulin. If the source of glucose is quickly removed, such as after delivery, the infant has a high risk for developing hypoglycemia, which can lead to convulsions and death. Previous Next Score Report Lab Values Calculator Help Pause

6 Exam Section 1: Item 6 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 6. A 63-year-old man has had this gradually enlarging lesion on his right forearm for the past 3 years. Which of the following is the most likely diagnosis? A) Actinic keratosis B) Blue nevus C) Compound nevus D) Dermatofibroma E) Halo nevus F) Hemangioma G) Malignant melanoma H) Pyogenic granuloma I) Seborrheic keratosis 1 cm J) Squamous cell carcinoma Correct Answer: G. Malignant melanoma is likely to be present when a lesion demonstrates asymmetry, irregular appearing borders, variable coloration, a diameter greater than 6 mm, and rapid evolution in characteristics. Malignant melanoma can rapidly invade and metastasize, which carries a poor prognosis when diagnosed late. Subtypes include superficial spreading, nodular, lentigo maligna, and acral lentiginous. Any lesion with features suggestive of malignant melanoma should be surgically excised with negative margins and pathologically examined for the depth of dermal invasion. Incorrect Answers: A, B, C, D, E, F, H, I, and J. Actinic keratosis (Choice A) is a premalignant lesion that may progress to squamous cell carcinoma. It typically presents as chronic rough, scaly patches of skin in areas of prolonged sun exposure (eg, face, ears, hands). Blue nevus (Choice B) is a benign proliferation of melanocytes located in the deep dermis. The lesion appears blue because of the Tyndall effect, in which dermal melanin scatters shorter wavelengths of light, including blue light. They are typically small, round, and demonstrate regular borders. Compound nevus (Choice C) is a benign proliferation of melanocytes located in both the epidermis and dermis. They are very common. Compound nevi should not display asymmetry, border irregularity, or multiple colors. Development of compound and other benign nevi should cease in the fourth to fifth decade. New or changing nevi after this time are concerning for melanoma. Dermatofibroma (Choice D) is a benign proliferation of fibroblasts induced by trauma such as a cut or bug bite. They are typically solitary, light brown in color, and dimple inward when pinched. Halo nevus (Choice E) is a benign nevus that appears as a brown macule with a halo of hypopigmentation. This nevus is caused by lymphocytic destruction of melanocytes at the edge of the nevus, which can occur in vitiligo or as an effect of melanoma elsewhere in the body. However, halo nevi do not demonstrate asymmetry, multiple colors, or increasing size and are not inherently malignant. Hemangioma (Choice F) is a benign proliferation of blood vessels that can either be present at birth or develop shortly after birth. They are bright red in appearance and do not have a pigmented component. Infantile hemangiomas can ulcerate and bleed as well as lead to disfigurement, and thus are treated with either topical timolol, propranolol, or laser surgery. Pyogenic granuloma (Choice H) is a benign proliferation of capillary-sized blood vessels. They commonly develop on the lips and fingertips during times of increased hormonal activity (eg, pregnancy). Clinically, pyogenic granulomas are pedunculated red papules with a small rim of scale; they bleed easily. Seborrheic keratosis (Choice I) is a benign proliferation of the epidermis; lesions exhibit a greasy, adherent appearance. While seborrheic keratoses are often brown, this is caused by the keratin produced by the epidermis rather than melanin. Squamous cell carcinoma (Choice J) of the skin typically presents with a non-healing ulcerative lesion with scale. It commonly occurs secondary to sun exposure, chronic draining sinus tracts, or immunosuppression.

Objective: Melanoma should be suspected when lesions demonstrate asymmetry, border irregularity, variable coloration, diameter greater than 6mm, or changing features. II Previous Next Score Report Lab Values Calculator Help Pause

28 Exam Section 1: Item 28 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 28. A 28-year-old woman comes to the physician because of a 2-week history of generalized swelling. She has autoimmune (Hashimoto) thyroiditis and chronic hepatitis C. Her temperature is 37.1°C (98.7°F), pulse is 74/min, and blood pressure is 142/88 mm Hg. Physical examination shows 2+ peripheral pitting edema. Laboratory studies show a serum albumin concentration of 2.1 g/dL, serum cholesterol concentration of 324 mg/dL, and 3+ protein in urine. A renal biopsy specimen shows diffuse thickening of the glomerular capillary wall, granular immunofluorescent deposits of IgG along the glomerular basement membrane, and irregular subepithelial electron-dense deposits. This patient most likely has which of the following types of glomerulonephritis? A) Crescentic B) Membranous C) Mesangioproliferative D) Proliferative E) Rapidly progressive Correct Answer: B. Membranous glomerulonephritis or membranous nephropathy is a nephrotic syndrome, which is characterized by proteinuria (greater than 3 g/day), edema, hypoalbuminemia, and hypercholesterolemia. As a result of underlying damage to the glomerular filtration barrier secondary to immune complex deposition, large proteins such as albumin are filtered into the urine and excreted, causing hypoalbuminemia. To counteract the low serum oncotic pressure caused by urinary protein losses, the liver produces lipoproteins leading to hyperlipidemia. A biopsy of membranous nephropathy usually demonstrates diffuse thickening of the glomerular capillary wall, and on immunofluorescence, IgG deposition along the glomerular basement membrane. Electron microscopy also shows subepithelial deposits. Membranous nephropathy is the most common nephrotic syndrome in Caucasian adults and can be primary (idiopathic) or secondary to conditions such as systemic lupus erythematosus, autoimmune disease against phospholipase A2- receptors, chronic hepatitis infection (especially hepatitis B or C virus), or pharmaceuticals such as nonsteroidal anti-inflammatory drugs or penicillamine. Untreated, it may progress to chronic renal failure. Incorrect Answers: A, C, D, and E. Crescentic (Choice A), mesangioproliferative (Choice C), proliferative (Choice D), and rapidly progressive (Choice E) glomerulonephritis are all nephritic syndromes, which are inflammatory processes leading to hematuria and erythrocyte casts in urine, oliguria, and hypertension. They can be associated with proteinuria, but to a lesser extent than nephrotic syndromes. They do not generally cause swelling or pitting edema. Crescentic and rapidly progressive glomerulonephritis are characterized by crescents of eosinophilic fibrin and plasma proteins on biopsy of the glomerulus, which are adjacent to glomerular parietal cells, monocytes, and macrophages. This is seen histologically in the setting of Goodpasture syndrome, granulomatosis with polyangiitis, and microscopic polyangiitis. Mesangioproliferative glomerulonephritis often coexists with membranous nephropathy and is marked by subendothelial and intramembranous immune complex deposits, in contrast to subepithelial electron-dense deposits and granular IgG seen in membranous nephropathy. Proliferative nephritis is often caused by systemic lupus erythematosus, and is marked by granular immunofluorescent deposits of IgG, and irregular subepithelial electron-dense deposits, however, it also presents with hematuria and hypertension.

Objective: Membranous nephropathy is the most common nephrotic syndrome in patients of Caucasian descent and is characterized by proteinuria (greater than 3 g/day), edema, hypoalbuminemia, and hypercholesterolemia. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

25 Exam Section 1: Item 25 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 25. A 50-year-old woman comes to the physician with her husband because of a 6-month history of progressive weakness of her arms and legs. Neurologic examination shows mild spasticity, weakness, and hyperreflexia of the upper and lower extremities. A Babinski sign is present bilaterally. An MRI of the spine shows a 1.5-cm, well-defined spherical lesion indenting the cervical spinal cord on the left at the C1 level. Which of the following is the most likely diagnosis? A) Arteriovenous malformation B) Astrocytoma O C) Epidural abscess D) Herniated disc E) Meningioma Correct Answer: E. Meningioma is a common, benign central nervous system (CNS) tumor derived from the meninges that occurs along the surfaces of the brain and spinal cord. Meningiomas may present with focal neurologic deficits from compression of the underlying brain or spinal cord parenchyma or with seizures caused by peritumoral edema and other changes in the local neurochemical environment. In this patient, the C1 lesion demonstrates upper motor neuron signs in the arms and legs (spastic paralysis, hyperreflexia, and Babinski sign). MRI of the brain or spine (as demonstrated in this patient) will show a well-defined, spherical mass that appears to indent the CNS parenchyma, since the tumor derives from the meninges lining the parenchyma, not from cerebral neurons or parenchymal glial cells. Treatment includes resection and/or radiosurgery. Incorrect Answers: A, B, C, and D. Arteriovenous malformation (AVM) (Choice A) refers to an abnormal tangle of blood vessels resulting in communication between an artery and vein. AVMS can be located throughout the CNS. On MRI, AVMS appear as a collection of blood vessels within or adjacent to CNS parenchyma rather than a spherical lesion as in this patient. Astrocytomas (Choice B) are primary CNS tumors that can present with varying grades of severity, such as low-grade pilocytic astrocytomas in children or high-grade glioblastomas in adults. Less often, these tumors can occur in the spinal cord. Since astrocytomas arise from the parenchyma itself, they would not indent the parenchyma as in this patient. They often demonstrate poorly-defined margins. A spinal epidural abscess (Choice C) is a suppurative infection located between the dura mater and vertebral wall. The diagnostic triad includes fever, back pain, and focal neurological deficits. MRI of the area in concern typically shows a rim-enhancing collection. The insidious course of this patient's neurologic symptoms (versus the more acute course of a spinal epidural abscess) and absence of the cardinal symptoms of spinal epidural abscess make meningioma more likely. A herniated disc (Choice D) in the cervical spine can cause nerve root compression, which would lead to lower motor neuron weakness, sensory abnormalities, and pain. On MRI, a disc herniation would appear as a dome-like protrusion from the intervertebral disc rather than a spherical mass.

Objective: Meningiomas are benign CNS tumors arising from the meninges that may compress the brain or spinal cord and cause focal neurologic deficits as well as seizures. MRI typically demonstrates a spherical and well-circumscribed mass that indents the parenchyma. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

158 Exam Section 4: Item 8 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 8. A2-week-old boy is brought to the physician because of persistent cyanosis. Hemoglobin concentration and erythrocyte morphology are normal. An x-ray of the chest and an echocardiogram disclose no abnormalities. Which of the following is the most likely cause of the cyanosis? O A) Hemolytic anemia B) Megaloblastic anemia O C) Methemoglobinemia D) a-Thalassemia E) B-Thalassemia Correct Answer: C. Methemoglobinemia occurs when an abnormal fraction of heme iron in the blood exists in the ferric (Fe3+) form (methemoglobin). Methemoglobin cannot effectively bind oxygen, which impairs tissue oxygen delivery. Most cases of methemoglobinemia are acquired, typically caused by exposure to an oxidant stressor such as an anesthetic (eg, benzocaine), nitrate, or dapsone. Symptoms include shortness of breath, fatigue, lethargy, lightheadedness, and in severe cases, arrhythmia, seizure, and multi-system organ failure. Vital signs typically demonstrate reduced oxygen saturation, and examination may show cyanosis. Characteristically, blood appears darkened on gross visualization, although hemoglobin concentration and erythrocyte morphology are normal. Treatment includes supplemental oxygen plus the intravenous administration of methylene blue, a reducing agent that donates electrons to heme. The reduction of heme iron to the ferrous (Fe2+) form restores the ability of hemoglobin to bind, carry, and deliver oxygen to tissues. Incorrect Answers: A, B, D, and E. Hemolytic anemia (Choice A) indicates the active lysis of erythrocytes. Common causes include autoimmune hemolytic anemia, microangiopathic hemolytic anemia, infections such as malaria and babesiosis, or inborn errors of erythrocyte synthesis and metabolism such as sickle cell disease or hereditary spherocytosis. Laboratory values typically demonstrate a normocytic anemia, decreased haptoglobin, and increased indirect bilirubin. Megaloblastic anemia (Choice B) occurs in the setting of impaired DNA synthesis, most commonly related to folic acid or vitamin B12 (cobalamin) deficiency, and classically characterized by erythrocyte macrocytosis and hypersegmented neutrophils. It is most often seen in patients with malnutrition, alcohol use disorder, pernicious anemia, or Crohn disease. a-Thalassemia (Choice D) is caused by the abnormal deletion of one or more of the four a-globin genes. Patients present with a microcytic anemia, although manifest variable degrees of symptoms, which are directly related to the number of gene deletions. B-Thalassemia (Choice E) is caused by one or more mutations of the two paired B-globin chains. Mutation of one gene result in B-thalassemia minor. Mutation of both genes causes B-thalassemia major, which is characterized by severe, transfusion dependent, microcytic anemia and signs of extramedullary hematopoiesis such as frontal bossing.

Objective: Methemoglobinemia refers to the presence of heme iron in the ferric state, which has a reduced ability to deliver oxygen to peripheral tissues. Lactic acidosis and cyanosis are common manifestations. Methylene blue converts ferric iron to ferrous iron, thereby restoring the normal oxygen binding and delivery of heme. Previous Next Score Report Lab Values Calculator Help Pause

32 Exam Section 1: Item 32 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 32. A 35-year-old man comes to the physician because of a 2-year history of slowly progressive muscle weakness of his thighs. His father and paternal uncle have a disorder involving proximal muscle weakness of the lower extremities. Physical examination shows no other abnormalities. A muscle biopsy specimen shows ragged red fibers. Electron-microscopic examination of this specimen is most likely to show which of the following inclusions? A) Intranuclear filamentous B) Lysosomal C) Mitochondrial crystalline D) Paranuclear filamentous E) Sarcoplasmic reticular Correct Answer: C. Mitochondrial crystalline inclusions on electron microscopy in association with proximal muscle weakness, ragged red fibers on muscle biopsy, and a compatible family history are consistent with mitochondrial myopathy. Mitochondria are the cellular organelles that produce ATP via oxidative phosphorylation, a process that requires four enzyme complexes to produce a transmembrane potential that is ultimately used to produce ATP via ATP synthase. Mitochondrial DNA is separate from somatic nuclear DNA, and mutations in genes encoding any of these complexes can result in the failure of oxidative phosphorylation, which results in a subsequent failure to generate ATP, and consequential myopathy and neurologic symptoms. However, there are additional mutations that affect mitochondrial RNA translation, trafficking and incorporation of respiratory protein complexes, and maintenance of the inner mitochondrial membrane that can also lead to mitochondrial myopathy. As both the skeletal muscle and neurologic system require vast quantities of ATP for normal function, these are the two organ systems most commonly affected by mitochondrial DNA mutations. Musculoskeletal manifestations include muscle pain, fatigue, exercise intolerance, and increased concentrations of creatine kinase, while potential neurologic symptoms may include ophthalmoplegia, seizures, myoclonus, and peripheral neuropathy. Mitochondrial diseases are strictly inherited through the mother, as sperm do not contain mitochondria to contribute to the fertilized embryo. As a result of the failure of energy metabolism, the cells may compensate by producing additional mitochondria. On muscle biopsy, these mitochondrial aggregates, when visualized on the background of the myofibers, appear as ragged red fibers. demonstrates abnormal mitochondrial crystalline inclusions. ctron microscopy Incorrect Answers: A, B, D, and E. Intranuclear filamentous inclusions (Choice A) are seen in inclusion body myositis and autosomal recessive distal myopathy. Symptoms include the insidious onset of distal muscle weakness and dysphagia. Lysosomal inclusions (Choice B) are found in lysosomal storage diseases. Lysosomes are organelles that contain enzymes used to degrade cellular debris. Defects in this process lead to the gradual accumulation of partially degraded material, with subsequent disruption of normal cellular function. Patients present with varying manifestations based on the subtype of lysosomal storage disease, although they may also present with cardiac disease, limb-girdle weakness, and respiratory insufficiency from involvement of the diaphragmatic muscles. The presentation generally begins in childhood. Paranuclear filamentous inclusions (Choice D) are seen in rhabdoid tumors of the kidney, a rare and highly aggressive renal malignancy presenting in children that is often metastatic. Sarcoplasmic reticular inclusions (Choice E) are pathognomonic of sarcoplasmic body myopathy, a rare myopathy that exhibits early involvement of the intrinsic hand muscles. Weakness progresses to involve most other muscles and the majority of patients are wheelchair-bound within 10 years of onset.

Objective: Mitochondrial crystalline inclusions on electron microscopy and ragged red fibers on muscle biopsy in a patient with inherited proximal muscle weakness is characteristic of mitochondrial myopathy. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

75 Exam Section 2: Item 25 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 25. A 3-year-old boy is brought to the emergency department by his mother 1 hour after having a generalized tonic-clonic seizure. He has had frequent vomiting and progressive stupor during the past 24 hours. His mother says that she gave him aspirin 2 days ago because of a temperature of 39.5°C (103.1°F). Physical examination shows hepatomegaly with no jaundice. Serum studies show an increased ammonia concentration, and increased AST and ALT activities. Dysfunction of which of the following is the most likely cause of this patient's disorder? A) Glutamate transporters B) Golgi complex C) Microtubules D) Mitochondria E) Rough endoplasmic reticulum Correct Answer: D. Mitochondrial dysfunction is associated with the pathogenesis of Reye syndrome, a rare, frequently fatal encephalopathy resulting from liver dysfunction in children. Reye syndrome presents in stages one through five with early stages demonstrating nonspecific rash, lethargy, confusion, nausea and vomiting, eventually leading to stupor. In later stages, coma, cerebral edema, seizures, organ failure, hyperammonemia, and death can result. Physical examination typically shows depressed mental status or loss of consciousness depending on the stage as well as hepatomegaly in later stages. Reye syndrome classically presents in children and is rare in adults; it may be idiopathic but is associated with salicylate use (eg, aspirin) during a preceding viral infection. Influenza viruses, especially influenza B, are most frequently implicated, though varicella has also been described as a precipitant. Patients with genetic deficiencies of fatty acid oxidation (eg, medium-chain acyl-CoA dehydrogenase deficiency) appear to be at higher risk. Salicylate metabolites are proposed to disrupt B-oxidation of fatty acids within mitochondria leading to fatty liver and the progressive hepatic dysfunction that characterizes the syndrome (hepatic encephalopathy, hyperammonemia, increased serum transaminases). Prevention is key. The only generally accepted use of salicylates in children is for mucocutaneous lymph node syndrome (Kawasaki disease). Otherwise, the use of salicylates should be avoided because of the risk for Reye syndrome. Treatment is supportive, including occasional use of mannitol to decrease cerebral edema. Even with treatment, Reye syndrome is often fatal. Incorrect Answers: A, B, C, and E. Glutamate transporters (Choice A) do not have a known role in the pathophysiology of Reye syndrome. They are implicated in the pathophysiology of hepatic encephalopathy, which occurs when ammonia joins with glutamate to form glutamine in the brain. The Golgi complex (Choice B) is an organelle that transports, modifies, and packages proteins and lipids. The Golgi complex is often implicated in liver dysfunction as a result of lysosomal storage diseases (eg, l-cell disease) but does not play a known role in the pathophysiology of Reye syndrome. Microtubules (Choice C) are involved in cellular chemotaxis and transport. They also act as the intracytoplasmic structure of the cell and as motor proteins that shuttle cellular products within the cell (eg, dynein and kinesin in neurons). Dysfunction of microtubules has been implicated in neurodegenerative diseases. Microtubules are a target of chemotherapeutic and anti-inflammatory medications. Rough endoplasmic reticulum (Choice E) plays a role in the translation and modification of proteins that will generally be transported out of the cell. Dysfunction of the endoplasmic reticulum has been implicated in neurodegenerative disease, metabolic disease, and atherosclerosis.

Objective: Mitochondrial dysfunction is associated with the pathogenesis of Reye syndrome, which is a rare, frequently fatal encephalopathy resulting from liver dysfunction in children. It is most associated with salicylate use (eg, aspirin) in children during a preceding viral infection. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

69 Exam Section 2: Item 19 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 19. A 42-year-old man comes to the physician because of chronic abdominal pain for 3 months. He describes his symptom as boring epigastric pain that is occasionally exacerbated by eating. He is an investment banker and describes his job as extremely stressful. Use of an over-the-counter histaminergic (H2)-receptor antagonist has mixed results. Vital signs are normal. Physical examination and upper endoscopy show no abnormalities. Which of the following is the most likely diagnosis? O A) Gastric lymphoma B) Generalized anxiety disorder O C) Irritable bowel syndrome D) Major depressive disorder E) Nonulcer dyspepsia Correct Answer: E. Nonulcer dyspepsia is the most likely diagnosis in this patient with epigastric pain, work-related stress, and normal endoscopic evaluation. Dyspepsia is a term used to describe epigastric pain variably accompanied by multiple symptoms including nausea, bloating, or early satiety. The most common causes of dyspepsia include peptic ulcer disease (PUD) and gastroesophageal reflux disease (GERD). While guidelines vary, most patients over the age of 60, or those with symptoms concerning for a possible malignancy such as weight loss or dysphagia, should undergo esophagogastroduodenoscopy (EGD) to rule out gastric or esophageal cancer. In younger patients, or those without red-flag features, a trial of an H2-receptor antagonist or proton pump inhibitor (PPI) is considered first- line therapy. Failure to respond to appropriate medical therapy should prompt additional evaluation. Patients should undergo testing for Helicobacter pylori with stool antigen testing (if not taking a PPI) or urea breath test. Positive tests should be followed by appropriate therapy to eradicate H. pylori. Patients who have normal evaluations with persistent symptoms are diagnosed with nonulcer (functional) dyspepsia. These patients commonly have comorbid psychiatric conditions including generalized anxiety disorder or major depressive disorder, and stressful life circumstances are supportive of this diagnosis. Incorrect Answers: A, B, C, and D. Gastric lymphoma (Choice A) would be more likely to manifest with weight loss, early satiety, gastrointestinal bleeding, and systemic symptoms such as fatigue and malaise. It would be unusual for a patient with gastric lymphoma to have a normal EGD. Generalized anxiety disorder (Choice B) requires the presence of excessive worrying about numerous aspects of daily life occurring on more days than not for at least 6 months, with a consequent inability to control symptoms of anxiety, and with the presence of at least three other symptoms (irritability, muscle tension, sleep problems, poor concentration, fatigue, restlessness). The symptoms must not be attributable to another organic cause. Irritable bowel syndrome (IBS) (Choice C) is diagnosed using the Rome-IV criteria, which include the presence of abdominal pain at least once weekly that is related to defecation, a change in stool appearance, or stool frequency (at least two of three required). IBS is further subdivided into IBS with constipation (IBS-C), with diarrhea (IBS-D), or mixed (IBS-M). Major depressive disorder (Choice D) requires the presence of at least five of nine criteria (according to the DSM-V) for at least 2 weeks without an organic cause or symptoms diagnostic of other mood or psychotic disorders. The symptoms (criteria) include sadness or anhedonia (at least one of which must be present) plus feelings of guilt, energy changes, difficulty concentrating, appetite or activity changes, psychomotor retardation, sleep disturbances, or suicidality. This patient does not meet criteria for this diagnosis.

Objective: Nonulcer dyspepsia, also known as functional dyspepsia, is a cause of epigastric pain that is common in patients with other somatic symptoms and increased levels of stress. Diagnosis is made after thorough evaluation to exclude organic causes of dyspepsia. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

11 Exam Section 1: Item 11 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 11. The biological effects of thyroid hormone and retinoic acid are mediated through similar mechanisms. Which of the following properties is common to the receptors for these two compounds? A) Ability to activate phospholipase C B) DNA-binding domain C) Formation of a dimer on binding of hormone D) Interaction with a G protein E) Tyrosine kinase activity Correct Answer: B. The receptors for retinoic acid and thyroid hormone are both nuclear receptors that contain DNA-binding domains. Nuclear receptors can initially be located in either the cytosol or nucleus. They bind to hormones which are either lipophilic and therefore capable of diffusion across the outer cell membrane, or which are transported into the cell via carrier-mediated transport. Once nuclear receptors bind their respective hormones, they translocate into the nucleus, if not already there, where they act as DNA transcription factors to regulate the expression of target genes. Because of the time required for gene expression to occur, hormones binding nuclear receptors require more time to demonstrate action when compared to hormones that act on immediately available or extracellular- domain signal transduction pathways. For example, the effect of epinephrine and norepinephrine (both of which bind G protein-coupled receptors), or insulin (which binds a receptor tyrosine kinase) is demonstrated over minutes, whereas thyroid hormone and steroid hormones such as retinoic acid require hours to days to act. Incorrect Answers: A, C, D and E. Ability to activate phospholipase C (PLC) (Choice A) is a feature of the phosphatidylinositol second messenger signaling cascade. Many G protein-coupled receptors, notably those bound to a G, protein, involve this signaling pathway. PLC enzymatically cleaves the lipid molecule phosphatidylinositol 4,5-bisphosphate (PIP2) into inositol triphosphate (IP3) and diacylglycerol (DAG), both of which exert further downstream signaling effects. Nuclear receptors, including the receptors for thyroid hormone and retinoic acid, do not employ this second messenger system. Formation of a dimer on binding of hormone (Choice C) is a feature of many tyrosine kinase receptors, such as the insulin receptor. These receptors exist as monomers on the cell surface membrane until bound by hormone, at which point dimerization occurs. Once dimerized, the cytoplasmic portions of the receptor engage in autophosphorylation of specific tyrosine residues, which creates binding sites for downstream signaling effectors. Most nuclear receptors do not engage in dimerization after hormone binding. Interaction with a G protein (Choice D) is a feature of G protein-coupled receptors. G protein-coupled receptors are integral membrane proteins with a characteristic structure composed of seven transmembrane domains. Upon activation of the receptor by hormone binding, the receptor acts as a guanine nucleotide exchange factor, and exchanges guanosine-5'-triphosphate for guanosine diphosphate on an associated G protein, the activating step. The activated G protein exerts downstream effects through various second messenger systems, primarily through cyclic adenosine monophosphate (CAMP) and phosphatidylinositol pathways. Nuclear receptors do not generally interact directly with G proteins. Tyrosine kinase activity (Choice E) is a common mechanism employed by several important receptors, such as the insulin receptor and epidermal growth factor receptor. Tyrosine kinases catalyze the phosphorylation of intracellular second messenger molecules by using adenosine triphosphate (ATP). Nuclear receptors do not demonstrate tyrosine kinase activity.

Objective: Nuclear hormone receptors contain DNA-binding domains and exert their effects by serving as DNA transcription factors. Common examples of nuclear receptors include the thyroid hormone receptor and the retinoic acid receptor. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

185 Exam Section 4: Item 35 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 35. A6-month-old girl is brought to the emergency department by her parents because of vomiting for 6 hours. She has been feeding poorly for 1 day because of an upper respiratory tract infection. Her temperature is 37°C (98.6°F), pulse is 100/min, respirations are 18/min, and blood pressure is 88/58 mm Hg. Physical examination shows lethargy and a clear nasal discharge. Laboratory studies show: Serum Ammonia Citrulline Glutamine Urine orotic acid increased increased decreased increased This patient most likely has a deficiency of which of the following enzyme activities? O A) Arginase B) Argininosuccinase O C) Argininosuccinate synthetase D) Carbamoyl phosphate synthetase E) Ornithine transcarbamylase Correct Answer: E. Ornithine transcarbamylase (OTC) deficiency is an X-linked recessive condition and the most common urea cycle disorder. It results in the accumulation of nitrogenous wastes (especially ammonia) that would normally have been converted to urea. It is generally diagnosed in infancy following a metabolic stressor that increases cell turnover or protein catabolism and therefore nitrogenous waste buildup (eg, infection, hyperthyroid or hyperadrenal states, malignancy). The presenting symptoms are generally age- appropriate neurologic symptoms (eg, flaccid/listless infant, ataxic 1-year-old), suggestive of increased serum concentration of ammonia. Patients with severe deficiency are often male, and may present with poor feeding, irritability, ataxic gait, tremors, lethargy, seizures, coma, or death, though the spectrum of presentation is broad. Diagnosis involves evaluation of reactants and products within the urea cycle; serum concentrations of upstream reactants of a deficient enzyme will be increased, while serum concentrations of downstream products will be decreased. In the classic and often severe case of OTC deficiency, the serum concentration phenotype includes increased ammonia, decreased citrulline, and increased orotic acid in both serum and urine (seen this case). Orotic acid concentration increases because of accumulation and shunting of carbamoyl phosphate (a reactant of OTC), which is converted to orotic acid when present in excess. Treatment includes avoiding precipitating states, a low protein diet, supplementation of arginine and urea cycle cofactors, administration of nitrogen-scavengers, and in extreme cases, hemodialysis to clear increased serum ammonia concentration. Incorrect Answers: A, B, C, and D. Arginase (Choice A) catalyzes the final step in the urea cycle, leading to the generation of water-soluble urea and ornithine from arginine. Deficiency in arginase is rare and autosomal recessive. The disease phenotype is mild in comparison to OTC deficiency; patients generally have longer lifespan and less severe exacerbations. In this case, increased serum citrulline concentration demonstrate that the defect is upstream from arginase. Argininosuccinase (Choice B), also known as argininosuccinate lyase (ASL), precedes arginase in the urea cycle. It converts argininosuccinate into arginine and fumarate. Deficiency of this enzyme is inherited in autosomal recessive manner; as with the other urea cycle deficiencies, symptoms are primarily neurologic. In this case, increased serum citrulline concentration demonstrates that the defect is upstream from ASL. Argininosuccinate synthetase (ASS) (Choice C) catalyzes conversion of citrulline (the product of OTC) and aspartate to argininosuccinate., preceding ASL in the urea cycle. It is associated with citrullinemia; mutations are inherited in autosomal recessive pattern. As with the other urea cycle deficiencies, symptoms are primarily neurologic. In this case, increased concentrations of citrulline demonstrate that the defect is upstream from ASS. Carbamoyl phosphate synthetase (CPS) (Choice D) as related to the urea cycle, catalyzes the synthesis of carbamoyl phosphate (a reactant of OTC) from glutamine and ammonia. A deficiency of CPS would result in a low serum or urine orotic acid concentration.

Objective: OTC deficiency is an X-linked recessive condition and the most common urea cycle disorder. It results in the accumulation of nitrogenous wastes and is diagnosed based on serum concentration of urea cycle reactants and products. It generally presents in infancy and can be fatal because of the neurotoxic effects of ammonia. I3D Previous Next Score Report Lab Values Calculator Help Pause

196 Exam Section 4: Item 46 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 46. An investigator is studying the regulation of pulmonary lymphatic flow using an animal model. Catheters are implanted in the femoral vein, pulmonary artery, and main lymphatic vessel draining the lungs in anesthetized, intubated, mechanically ventilated animals. Drugs may be injected directly into the pulmonary artery catheter and inspired gas mixtures altered at the ventilator; timed collection of lymph from the catheters is used to calculate lymphatic flow. Which of the following interventions will most likely increase the flow of pulmonary lymph in these animals? A) Administration of endothelin-1 into the pulmonary artery B) Administration of phenylephrine into the pulmonary artery C) Decreasing the inspired oxygen concentration from 21% to 10% D) Increasing the inspired carbon dioxide concentration from 0.3% to 3% E) Intravenous infusion of 0.9% saline for 5 minutes F) Intravenous infusion of 20% albumin solution (20 g/100 mL saline) for 5 minutes Correct Answer: E. Lymphatic flow is dependent on interactions between capillary hydrostatic pressure, interstitial fluid hydrostatic pressure, plasma colloid oncotic pressure, and interstitial fluid oncotic pressure. If capillary hydrostatic pressure is high relative to interstitial fluid hydrostatic pressure, fluid will flow down a pressure gradient into the lymphatic system. If the opposite is true (interstitial fluid pressure is higher than capillary pressure), fluid will flow into the capillaries. If plasma fluid oncotic pressure is high relative to interstitial fluid oncotic pressure, fluid will be drawn back into the capillaries. Conversely, if interstitial fluid oncotic pressure exceeds plasma fluid oncotic pressure, fluid will move into the interstitial space, where it will be collected by the lymphatic system. To increase lymphatic flow, the investigator must increase the hydrostatic pressure inside the capillaries, decrease the hydrostatic pressure in the interstitial space, decrease the plasma oncotic pressure, or increase the interstitial fluid oncotic pressure. Of the choices, the intravenous infusion of 0.9% saline for 5 minutes is the only intervention that would result in increased lymphatic flow by increasing the capillary hydrostatic pressure. Incorrect Answers: A, B, C, D, and F. Administration of endothelin-1 (Choice A) or phenylephrine (Choice B) into the pulmonary artery, as well as decreasing the inspired oxygen concentration from 21% to 10% (Choice C), and increasing the inspired carbon dioxide concentration from 0.3% to 3% (Choice D), would all lead to vasoconstriction at the pulmonary precapillary concentration. This would result in decreased pulmonary capillary hydrostatic pressure and decreased lymphatic flow. Intravenous infusion of 20% albumin solution (20 g/100 mL saline) for 5 minutes (Choice F) would increase the oncotic pressure of the plasma, thereby pulling fluid back into the capillaries and away from the lymphatic system. This would decrease lymphatic flow.

Objective: Movement of fluid between capillaries and the lymphatic system is dependent on the hydrostatic and oncotic pressures of both the capillary and interstitial compartments. High capillary hydrostatic pressure and low plasma oncotic pressure will increase the movement of fluid out of the capillaries and into the interstitial space, which increases lymphatic flow. Conversely, high interstitial fluid hydrostatic pressure and low interstitial fluid oncotic pressure will drive fluid back into the capillaries and away from the lymphatic system. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

139 Exam Section 3: Item 39 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 39. A2-year-old boy is brought to the physician because of a 1-week history of high-grade fever and rash. His temperature is 39.6°C (103.3°F), pulse is 144/min, and blood pressure is 122/76 mm Hg. Physical examination shows marked conjunctival injection and erythema and edema of the hands and feet. A photograph of the cheeks and mouth is shown. Echocardiography shows dilation of the proximal coronary arteries. Which of the following is the most likely diagnosis? A) Acute rheumatic fever B) Ehrlichiosis C) Erythema multiforme D) Mucocutaneous lymph node syndrome (Kawasaki disease) E) Scarlet fever Correct Answer: D. Mucocutaneous lymph node syndrome (Kawasaki disease) is a multisystemic vasculitic disease in children characterized by fever for at least 5 days, bilateral nonexudative conjunctivitis, a polymorphous rash, lymphadenopathy, erythema and edema of the hands and feet, and erythema affecting the mucous membranes with dry, cracked lips. To diagnose mucocutaneous lymph node syndrome (Kawasaki disease), a fever for 5 days or more plus four of the remaining criteria are required. Laboratory evaluation may show thrombocytosis and increased erythrocyte sedimentation rate. Complications of mucocutaneous lymph node syndrome (Kawasaki disease) include coronary artery aneurysms, and if untreated, can lead to aneurysmal expansion, rupture, or myocardial infarction. Therefore, children suspected of having mucocutaneous lymph node syndrome (Kawasaki disease) require immediate evaluation with ECG and echocardiography, immediate treatment, and monitoring for progression or complications. Treatment of mucocutaneous lymph node syndrome (Kawasaki disease) disease involves high-dose aspirin and intravenous immunoglobulin. It is the only widely accepted indication for aspirin use in febrile children because of the risk for Reye syndrome. Mucocutaneous lymph node syndrome (Kawasaki disease) is one of the most common childhood vasculitides and is a common cause of pediatric heart disease. Incorrect Answers: A, B, C, and E. Acute rheumatic fever (Choice A) classically presents with migratory polyarthralgia, pancarditis (pericarditis, myocarditis, or endocarditis), subcutaneous nodules, erythema marginatum, Sydenham chorea, fever, increased erythrocyte sedimentation rate, and short PR interval on ECG following an infection from group A B-hemolytic streptococci, frequently pharyngitis. Čomplications of rheumatic fever include mitral valve inflammation leading to long-term sequelae of chronic mitral valve regurgitation or stenosis. Ehrlichiosis (Choice B) is a tick-borne illness with a wide spectrum of disease characteristics, including fever, malaise, myalgia, headache, nausea, vomiting, rash, and altered mental status. It does not generally cause edema of the hands and feet, conjunctivitis, or coronary artery aneurysms. Erythema multiforme (Choice C) is characterized by erythematous macules and papules with clearing and a dark red outer ring, often referred to as targetoid lesions. Typically, palms and soles are affected. The condition may also be associated with fever, myalgias, and arthralgias. Triggers often include medications (especially antibiotics) and infections (eg, mycoplasma, herpes simplex virus). Scarlet fever (Choice E) can present in a similar fashion to mucocutaneous lymph node syndrome (Kawasaki disease), with an erythematous tongue, rash, and desquamation of hands and feet. However, it is not characterized by conjunctivitis or dry, chapped lips. The rash of scarlet fever is characteristically a rough, papular, diffuse body rash.

Objective: Mucocutaneous lymph node syndrome (Kawasaki disease) is a multisystemic vasculitic disease in children characterized by fever for at least 5 days, bilateral nonexudative conjunctivitis, a polymorphous rash, lymphadenopathy, erythema and edema of the hands and feet, and erythema affecting the mucous membranes with dry, cracked lips. Complications of mucocutaneous lymph node syndrome (Kawasaki disease) include coronary artery aneurysms, and if untreated, can lead to aneurysmal expansion, rupture, or myocardial infarction. II Previous Next Score Report Lab Values Calculator Help Pause

138 Exam Section 3: Item 38 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 38. A 27-year-old woman comes to the physician 3 weeks after the onset of blurred vision in her right eye that has been improving during the past 5 days. She had a similar episode 1 year ago while working on a ranch in northeastern Arizona; the episode resolved in 2 weeks without treatment. She has no history of major medical illness. Ophthalmologic examination shows a right afferent pupillary defect. Visual acuity is 20/20 in the left eye and 20/200 in the right eye. Neurologic examination shows mild weakness of the lower two thirds of the right side of the face and a right pronator drift. There is mild dysmetria when she moves the left extremities. An MRI of the brain is shown; the arrows indicate multiple abnormalities. Which of the following is the most likely diagnosis? A) Cerebral abscesses B) Cerebral lymphoma O C) Coccidioidomycosis D) Multiple sclerosis O E) Neurocysticercosis Correct Answer: D. Multiple sclerosis, a prevalent immune-mediated demyelinating disease, can present initially with optic neuritis though can also demonstrate neurologic symptoms from lesions in several other central nervous system (CNS) locations. Optic neuritis typically presents with poor visual acuity and a consequent afferent pupillary defect on physical examination. According to the McDonald criteria for multiple sclerosis, patients must have evidence of demyelinating lesions that are separated in time and location. This evidence can be based on subjective symptoms and/or imaging evidence. This patient presents with multiple episodes of focal neurologic deficits, with evidence on physical examination of deficits in multiple distributions, involving her right eye, face, and extremities. On MRI, there are lesions in multiple anatomic locations. These findings fulfill the diagnostic criteria for multiple sclerosis. MRI typically shows white matter hyperintensities in the periventricular region, cortex, subcortical region, cerebellum, or spinal cord. Oligoclonal bands of immunoglobulins, indicating immune overactivation, can be found in the cerebrospinal fluid but are not necessary for diagnosis. Treatment includes corticosteroids for symptomatic management of acute flares and long-term disease- modifying treatments. Incorrect Answers: A, B, C, and E. Cerebral abscesses (Choice A) would classically present with fever, headache, and focal neurological deficits. The presence of several bilateral lesions would indicate hematogenous spread from a different primary source of infection. Further, this patient has no history of an inciting infection, and her neurologic symptoms have been episodic and enduring over a chronic time course, as opposed to the acute course expected for abscesses. Cerebral lymphoma (Choice B) is a rare variant of non-Hodgkin lymphoma that occurs commonly in immunocompromised patients. Typical presentation includes focal neurologic deficits, neuropsychiatric symptoms, and/or signs of increased intracranial pressure (eg, positional headache, nausea, vomiting). MRI typically features a solitary lesion but can show multiple periventricular white matter lesions. However, optic neuritis and episodic neurological symptoms are more typical of multiple sclerosis in this immunocompetent patient. Coccidiomycosis (Choice C) is a fungal infection endemic to the Southwestern United States that typically presents as a self-limited respiratory illness. Coccidiomycosis can disseminate, especially in immunocompromised patients, to cause coccidioidal meningitis, commonly presenting as a persistent headache. Episodic focal neurologic deficits would be atypical, and the brain lesions in this patient point to an inflammatory parenchymal process rather than meningitis. Neurocysticercosis (Choice E) is a CNS infection caused by the pork tapeworm Taenia solium that occurs most commonly in endemic rural regions. Symptoms include headache, seizures, signs of increased intracranial pressure, and, rarely, altered vision. Typical manifestations on CT scan or MRI include cystic, enhancing, and calcified lesions in the brain parenchyma, meninges, ventricles, or spinal cord. This patient possesses no risk factors for neurocysticercosis.

Objective: Multiple sclerosis is an immune-mediated demyelinating disease that presents with episodic focal neurologic symptoms that occur in different anatomic locations; optic neuritis is a common initial presentation. MRI classically shows white matter hyperintensities in periventricular locations. II Previous Next Score Report Lab Values Calculator Help Pause

166 Exam Section 4: Item 16 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 16. A 17-year-old boy comes to the physician for a follow-up examination. One year ago, he sustained a deep laceration to his right lower extremity in a motorcycle collision. He underwent operative repair of a severed neurovascular bundle below the popliteal fossa. During the next 10 months, he received regular physical therapy. Physical examination now shows no apparent atrophy of the right lower extremity. Motor and sensory functions are intact. Which of the following changes is most likely in the muscle fibers of the patient's right foot? O A) Chronic inflammation B) Degeneration C) Fibrosis D) Grouping by fiber type E) Pseudohypertrophy Correct Answer: D. Skeletal muscle is comprised of two distinct metabolic types. Type I fibers are known as slow twitch, or red muscle, caused by a high density of capillaries, mitochondria, and myoglobin, all of which permit prolonged or sustained aerobic metabolism. These fibers are specialized for long periods of use but generate decreased force. Type Il fibers are known as fast twitch and generate powerful contractions over shorter durations. Type II fibers are less rich in mitochondria and myoglobin when compared to Type I. During reinnervation of muscle after a nerve injury, grouping by fiber type is a phenomenon observed on muscle biopsy in which the normal distribution of muscle fibers (which normally includes interspersed type I and type il fibers) is lost. Instead, type I and type Il muscle fibers group together. This finding is demonstrated in neuromuscular pathology such as acute nerve transection, amyotrophic lateral sclerosis, and spinal muscular atrophy. Incorrect Answers: A, B, C, and E. Chronic inflammation (Choice A) occurs in inflammatory disorders of the muscle such as dermatomyositis, polymyositis, or inclusion body myositis. It is characterized by the infiltration of inflammatory cells (generally lymphocytes and macrophages) on histology, along with fibrosis. Degeneration (Choice B) is a nonspecific term describing the breakdown of muscle or nerve. Rather, atrophy of muscle may be seen following a nerve injury if reinnervation does not occur. Fibrosis (Choice C) occurs because of chronic muscle inflammation. It is commonly seen on histology in cases of muscular dystrophy. Fibrosis is characterized by increased production of collagen and myofibroblast differentiation. It is also demonstrated on muscle biopsy following radiation therapy. Pseudohypertrophy (Choice E) is a classic sign of Duchenne muscular dystrophy in which muscle is replaced by fatty infiltration that causes an enlarged appearance of that muscle.

Objective: Muscle pathologies are characterized by specific changes on histology. Muscle denervation and reinnervation is characterized by a pattern known as grouping by fiber type, in which type I and type Il fibers group together. Previous Next Score Report Lab Values Calculator Help Pause

48 Exam Section 1: Item 48 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 48. A 20-year-old woman comes to the physician because of a 1-year history of intermittent low-grade fever, fatigue, and joint pains. One week ago, she had a severe exacerbation of these symptoms after a trip to the beach; she also developed a red rash, even on areas that were not exposed to sun. Her temperature is 38.3°C (101°F). Physical examination shows an erythematous rash over the malar areas, trunk, and upper extremities; oral ulcerations; 1-cm axillary lymph nodes bilaterally; and mild swelling and tenderness of the joints of the hands. Serologic studies show an antinuclear antibody titer of 1:200. Further studies are most likely to show a mutation of a gene encoding which of the following proteins? A) C1q B) C-reactive protein C) Interferon gamma D) Mannose-binding lectin E) Serum amyloid A Correct Answer: A. C1q, an activation fragment of complement protein C1, is highly associated with cutaneous, renal, and neuropsychiatric manifestations of systemic lupus erythematosus (SLE). Mutations involving C1q and low serum concentration thereof have been associated with severe presentations of SLE. SLE classically presents in females with rash, oral ulcers, joint pain, fever, photosensitivity (as shown in this case, malar rash and photodermatitis even in non-exposed areas), cytopenias, and neurologic manifestations (eg, seizures, psychiatric disturbances). Complement concentration are often low in SLE, as a result of immune complex formation and deposition, leading to systemic inflammation. Laboratory evaluation often shows increased antinuclear antibody titers, which are sensitive for the diagnosis, or titers of anti-dsDNA and anti-Smith antibodies, both of which are specific for SLE. Incorrect Answers: B, C, D, and E. C-reactive protein (CRP) (Choice B) mutations are not known to be associated with SLE. CRP is an acute phase reactant produced by the liver in states of infection and inflammation, typically in response to the cytokine interleukin-6. Interferon gamma (Choice C) is a cytokine produced by T-lymphocytes, macrophages, and natural killer cells, and has antiviral and antineoplastic roles. It is also implicated in the activation of macrophages within granulomas, and the response to mycobacterial pathogens. It does not have a commonly recognized role in the pathogenesis of SLE. Mannose-binding lectin (MBL) (Choice D) is a key surface protein and opsonin within the innate immune system and serves to bind multiple carbohydrate moieties found broadly on bacteria, viruses, and parasites. Mutations of MBL result in primary immunodeficiency, with recurrent infections often beginning in childhood. There is no known pathophysiologic role in SLE, however, its concentration has been shown to be activated in acute SLE flares, making it a potential biomarker. Serum amyloid A (Choice E), like CRP, is also an acute phase reactant that is produced in inflammatory and infectious states. In pathology, excess amyloid production can result in deposition throughout tissues, a process known as amyloidosis, leading to widespread organ dysfunction including organ failure. Its concentration is also increased in rheumatoid arthritis. Mutations in serum amyloid A are not associated with SLE.

Objective: Mutations involving complement are associated with systemic lupus erythematosus (SLE). In SLE, concentration of complement is often low because of immune activation and complex deposition. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

60 Exam Section 2: Item 10 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 10. A 45-year-old man with mild hypertension uses a nasal spray for allergic rhinitis and nasal congestion. Four days later, the spray no longer relieves his symptoms and his blood pressure is increased. Which of the following best describes the action of this drug? OA) a-Adrenergic receptor agonist O B) BzAdrenergic receptor antagonist O C) Histaminergic (H,) receptor antagonist D) Muscarinic receptor agonist O E) Prostaglandin (PGE1) receptor agonist Correct Answer: A. Nasal decongestants are a1-adrenergic receptor agonists, working to reduce secretions and inflammation in the nasal mucosa. The a1-adrenergic receptor is a sympathetic G-protein-linked adrenoreceptor. It is typically located in the vascular smooth muscle, intestinal and bladder sphincter muscles, and pupillary dilator muscle. It is coupled with a Gg protein, which results in the activation of phospholipase C, and ultimately leads to increased intracellular calcium and protein kinase C activity. This results in smooth muscle contraction, which in arterioles leads to vasoconstriction. The resultant vasoconstriction, when occurring in the systemic circulation, results in an increased total peripheral resistance and mean arterial pressure. When this effect occurs in the nasal mucosa, this results in decreased mucosal perfusion, which reduces nasal secretions, assisting in the treatment of allergic rhinitis and nasal congestion. A disorder called rhinitis medicamentosa is a known adverse effect of these medications, in which excessive use leads to rebound congestion. It is generally recommended to limit the use of nasal decongestants to no more than 5 days. Incorrect Answers: B, C, D, and E. B2-adrenergic receptors antagonists (Choice B) refer to B-adrenergic blocking agents such as nadolol and propranolol, which can be used for the treatment of tachydysrhythmias, heart failure, and myocardial infarction. B2-adrenergic receptors are primarily located in blood vessels, muscle, the gastrointestinal tract, and the bronchi. They are not used in the treatment of allergic rhinitis. Histaminergic (H1) receptor antagonists (Choice C), such as diphenhydramine, are useful in the treatment of allergic rhinitis and nasal decongestion. Histamine-mediated actions include increased vascular permeability, bronchial and intestinal smooth muscle contraction, and increased airway mucous production. Side effects of antihistamines include dry mouth, sedation, confusion, nausea, vomiting, and urinary retention. They would not cause hypertension. Muscarinic receptor agonists (Choice D), such as pilocarpine, mimic the activity of the neurotransmitter acetylcholine. For example, pilocarpine is primarily used in the treatment of glaucoma to induce improved outflow of aqueous humor through action on the ciliary muscle. Adverse effects include miosis and cyclospasm. Prostaglandin (PGE1) receptor agonists (Choice E), such as latanoprost, are used in the treatment of glaucoma by reducing flow resistance through the uveoscleral pathway, which improves aqueous humor outflow. Side effects include darkening of the iris and increased eyelash growth.

Objective: Nasal decongestant sprays work through a1-adrenergic receptor activation. Sustained use can lead to rebound nasal congestion symptoms and systemic side effects such as hypertension. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

83 Exam Section 2: Item 33 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 33. An investigator is studying a tumor cell line in an experimental animal model in which the expression of class I MHC is completely inhibited. The cell line is assayed as a target for cell-mediated immune response by CD8+ T lymphocytes, CD4+ T lymphocytes, or natural killer cells. Which of the following sets of findings is most likely (+ = killing; - = no killing)? CD8+ T Lymphocytes CD4+ TLymphocytes Natural Killer Cells O A) + + + B) + + O C) + | D) | E) + O F) Correct Answer: E. Natural killer (NK) cells induce apoptosis in their target cells. When activated, NK cells synthesize perforin and granzyme which are both proapoptotic. Classically, NK cells target tumor cells and virally infected cells in response to an absence of surface MHC class I molecules. In this example, the absence of MHC class I expression triggers an activation of NK cells. NK cells are also activated by an antibody-dependent mechanism. In response to bound immunoglobulin (Ig), CD16 binds to the Fc region of the lg, which activates the NK cell. Incorrect Answers: A, B, C, D, and F. Choice A (CD8+T Lymphocyte, CD4+ T Lymphocyte, and NK cell activation) describes a broad immune response in which both cytotoxic and helper T cells are involved, which would require expression of both MHC class I and II by the target cell, along with an additional activation trigger for the NK cell. Choice B (CD8+T Lymphocyte and NK cell activation) describes an immune response in which cytotoxic T cells (requiring MHC class I) without helper T cells responded, along with an additional activation trigger for NK cells as MHC class I would be present. Choice C (CD8+ T lymphocyte activation) describes a response in which a cytotoxic T cell may activate against a virally infected, neoplastic, or graft cell expressing MHC class I without the assistance of any additional cell lines. Choice D (CD4+ T lymphocyte activation) describes the response to a cell expressing MHC class II. Helper T cells (CD4+) generally coordinate the remainder of the immune response through cytokine release by recruiting macrophages, cytotoxic T cells, plasma cells, and eosinophils. Choice F (no activation) suggests inappropriate function of NK cells, which classically induce apoptosis in cells not expressing MHC class I.

Objective: Natural killer cells use perforin and granzyme to induce apoptosis in virally infected or neoplastic cells that do not express MHC class I antigens. Previous Next Score Report Lab Values Calculator Help Pause | + | |

16 Exam Section 1: Item 16 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 16. A 22-year-old woman is admitted to the hospital because of a 10-day history of polydipsia and polyuria. She says that the urge to urinate often awakens her at night. She has been taking lithium carbonate for 2 years for bipolar disorder; her dosage was increased 6 months ago because of recurrent severe manic episodes. Her vital signs are within normal limits. Physical examination shows no abnormalities. Over the next 24 hours, urine excretion totals 6.5 L. Laboratory studies at this time show a serum sodium concentration of 148 mEq/L, serum osmolality of 315 mOsmol/kg, and urine osmolality of 75 mosmol/kg. After administration of desmopressin, urine output and osmolality do not change. Which of the following mechanisms is the most likely cause of the polyuria in this patient? A) Decreased secretion of ADH (vasopressin) B) Inhibition of CAMP-mediated processes in the collecting duct C) Osmotic diuresis due to increased amounts of glucose in the tubule D) Polydipsia-induced washout of the renal interstitial concentration gradient E) Renal medullary necrosis due to decreased papillary blood flow Correct Answer: B. Nephrogenic diabetes insipidus (NDI) is commonly acquired secondary to medications or electrolyte disturbances. The pathophysiology of NDI occurs such that the renal collecting duct becomes insensitive to antidiuretic hormone (ADH) produced by the posterior pituitary. Normally, ADH triggers the insertion of aquaporin channels into the collecting duct membrane via a CAMP-mediated pathway. By consequence, when there is inhibition of this process in the collecting duct, as in NDI, the ability of the kidney to reclaim free water is compromised. Dehydration results, as is demonstrated by this patient, along with symptoms of polydipsia (compensatory response to low serum volume or increased serum osmolarity) and polyuria or incontinence as a result of the large volume of dilute urine produced). NDI is an adverse effect of chronic lithium use. Lithium concentrates within the cells of the renal collecting duct and interferes with the signaling pathway related to ADH, leading to limited aquaporin insertion and resultant NDI. In severe cases of dehydration, vital signs may disclose tachycardia and hypotension. Physical examination discloses no abnormalities, apart from potential signs of volume depletion (eg, weak pulse, skin tenting). Laboratory studies show increased serum osmolarity, hypernatremia, and inappropriately dilute urine. Normally, urine should be concentrated in states of hyperosmolar serum, as the main physiologic function of ADH is to reclaim water, thereby maintaining serum osmolarity and sodium balance in addition to plasma volume. Desmopressin, an ADH analog, is used as a means of differentiating nephrogenic from central DI. Central DI occurs because of an absence of ADH production, not a failure of the renal collecting tubule to respond to it. If urine output and serum osmolarity decrease following desmopressin administration, a central cause of DI (eg, head trauma, hypothalamic damage, pituitary tumor) should be investigated. No change in response to desmopressin indicates that the cause of DI is nephrogenic. The treatment of NDI includes volume repletion plus thiazide diuretics. Incorrect Answers: A, C, D, and E. Decreased secretion of ADH (vasopressin) (Choice A) describes the pathophysiology of central DI. Central DI presents with dilute urine and concentrated serum as a result of a lack of secreted ADH. However, central DI responds to desmopressin, which results in the recovery of free water via aquaporin insertion in the collecting tubule. Osmotic diuresis due to increased amounts of glucose in the tubule (Choice C) describes the pathophysiology of volume depletion in hyperglycemia. The nephron can recover filtered glucose up to 180 mg/dL, but in excess, osmotic diuresis ensues. Beyond a threshold of 300 mg/dL of glucose, severe glucosuria and osmotic dehydration can result. Polydipsia-induced washout of the renal interstitial concentration gradient (Choice D) occurs in states of psychogenic polydipsia or excess intake of free water. Excess free water consumption dilutes urea and sodium; the decreased concentration of sodium and urea within the renal interstitium results in the loss of the kidney's ability to concentrate urine. The sodium and urea gradients of the interstitium drive the resorption of water via the collecting duct. This mechanism is not dependent on ADH or CAMP. Renal medullary necrosis due to decreased papillary blood flow (Choice E) can occur in states of hypotension, sickle cell disease, or nephrotoxicity.

Objective: Nephrogenic diabetes insipidus is an adverse effect of lithium carbonate. Lithium impairs the ability of the collecting tubule to reclaim free water by interfering with the pathway of antidiuretic hormone in the cells. Patients typically present with polydipsia, polyuria, hyperosmolar serum, dilute urine, and failure to respond to desmopressin. I3D Previous Next Score Report Lab Values Calculator Help Pause

152 Exam Section 4: Item 2 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 2. A 1-year-old girl has recurrent and chronic suppurative infections caused by staphylococci and some gram-negative rods. She recovers normally from viral infections. Examination of serum shows normal concentrations of complement components and immunoglobulins. Studies of leukocyte function are most likely to show defects in which of the following mechanisms? O A) Activation of macrophages B) Chemotaxis of macrophages O C) Cytotoxic activity of natural killer cells D) Ingestion of bacteria by neutrophils E) Production of superoxide anions by neutrophils Correct Answer: E. This patient presents with recurrent pyogenic infections caused by gram-positive and gram-negative bacteria but exhibits a normal immune response to viral infections and is consistent with chronic granulomatous disease (CGD). CGD results from a defect in the nicotinamide adenine dinucleotide phosphate (NADPH) oxidase complex. NADPH oxidase complex within phagocytes, such as neutrophils, uses oxygen as a substrate for the generation of oxygen free radicals (superoxide anions). Free radical oxygen species are subsequently used for the creation of hydrogen peroxide and hypochlorous acid. Activation of this pathway leads to the "respiratory burst" which results in bacterial death. Deficiency of the NADPH oxidase complex renders phagocytes incapable of neutralizing catalase-positive bacteria, which can neutralize their own hydrogen peroxide, thus leaving the host cells without a substrate necessary to complete the respiratory burst. Diagnosis is made by an abnormal dihydrorhodamine test, or a nitroblue tetrazolium reduction test. In this latter test, normal phagocytes use the action of NADPH to reduce nitroblue, which leads to a color change from yellow to blue. Patients with CGD will not demonstrate color change. Recurrent pneumonia is the most common presenting infection in patients with CGD, and the most common infecting bacteria include Staphylococcus species, Aspergillus species, Burkholderia cepacia, and Nocardia species. Patients with CGD are also at risk for fungal infections, especially for the Aspergillus species. Incorrect Answers: A, B, C, and D. Defects in the activation (Choice A) and chemotaxis (Choice B) of macrophages leads to an increased susceptibility to mycobacterial infections caused by a failure of intracellular killing of bacteria and granuloma formation. Patients are also at an increased risk for infection with Salmonella species and intracellular bacteria. Defects in the cytotoxic activity of natural killer (NK) cells (Choice C) result in recurrent viral infections and impaired defense against malignant cells. NK cells are part of the innate immune system and are essential in the response to pathogenic viruses. A defect in the ingestion of bacteria by neutrophils (Choice D) is characteristic of Chediak-Higashi syndrome. It is caused by a mutation in the LYST gene, which causes abnormal protein trafficking and vacuole formation. Patients commonly present with recurrent pyogenic bacterial infections involving the skin and respiratory tract, but typically also have physical examination findings such as fair skin, peripheral neuropathy, and ocular albinism.

Objective: Nicotinamide adenine dinucleotide phosphate (NADPH) oxidase is necessary for the formation of reactive oxygen species in phagocytes, which allows for bactericidal activity. Mutations in the proteins that make up the NADPH oxidase complex result in chronic granulomatous disease, typically characterized by recurrent pyogenic bacterial infections. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

157 Exam Section 4: Item 7 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 7. A 24-year-old woman comes to the clinic because of facial flushing and diffuse pruritus. The symptoms began shortly after she ingested a new nutritional supplement from a health food store. Examination shows erythema of the face and upper portion of the neck. The remainder of the skin shows no gross abnormalities. There is no perioral or neck edema. Findings on lung examination are normal. Which of the following nutritional supplements is the most likely cause of her symptoms? A) Ascorbic acid B) Folic acid O C) Lipoic acid D) Nicotinic acid E) Pantothenic acid Correct Answer: D. Nicotinic acid is also known as vitamin B3, nicotinamide, or niacin. It is a component of NAD* and NADP+, which are used in various biochemical reduction-oxidation reactions. Nicotinic acid decreases concentrations of very-low-density lipoprotein (VLDL) and raises concentrations of high-density lipoprotein (HDL), making it a potential treatment for dyslipidemia. However, a common side effect is facial flushing and pruritus secondary to increased prostaglandin production. These symptoms can be avoided by taking aspirin, an irreversible cyclooxygenase inhibitor that decreases prostaglandin production. An acquired deficiency of nicotinic acid leads to pellagra, characterized by diarrhea, dementia, and dermatitis. Pellagra-like symptoms can also be seen in Hartnup disease, an autosomal recessive disorder caused by decreased gastrointestinal absorption of tryptophan, the precursor to nicotinic acid. Incorrect Answers: A, B, C, and E. Ascorbic acid (Choice A), or vitamin C, is a water-soluble antioxidant found in fruits and vegetables. Deficiency of this vitamin causes the constellation of symptoms known as scurvy, which is characterized by petechiae, perifollicular hemorrhage, bruising, poor wound healing, and short, curly, breakable hairs. Folic acid (Choice B), or vitamin Bg, is converted in the body to tetrahydrofolic acid and then used as a coenzyme in the synthesis of nucleotides and nucleosides. Folate is contained in leafy vegetables and absorbed in the jejunum. Folate supplementation is recommended in pregnancy to decrease the risk for neural tube defects. It is also commonly warranted in patients with rapid cellular turnover, such as acute leukemia and sickle cell disease, to offset the consumption of folic acid in cell division. Lipoic acid (Choice C) is a cofactor of pyruvate dehydrogenase, along with vitamins B1 (thiamine), B2 (riboflavin), niacin, and B5. It may be taken as an antioxidant supplement but does not cause flushing or pruritus. Pantothenic acid (Choice E), or vitamin B5, is an essential component of coenzyme A, which serves as a cofactor for acyl transfers. Deficiency of this vitamin leads to dermatitis, alopecia, and adrenal insufficiency.

Objective: Nicotinic acid supplementation commonly causes facial flushing secondary to increased prostaglandin production. It is not a manifestation of an allergic reaction and histamine release. Aspirin can be used to avoid these symptoms by inhibiting cyclooxygenase production of prostaglandins. Previous Next Score Report Lab Values Calculator Help Pause

t disease). Increased production of IL-12 (Choice E), an inflammatory cytokine secreted by macrophages, results in increased natural killer cell activity and promotes Th, T-cell differentiation.

Objective: Osteitis deformans (

179 Exam Section 4: Item 29 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 29. During ligand-binding experiments, a new adrenergic drug displays potent binding to a-adrenergic receptors. In healthy volunteers, the drug causes a significant increase in blood pressure. However, when used in the emergency department to increase blood pressure in trauma victims, the drug decreases blood pressure. This drug is most likely from which of the following classes? O A) Competitive antagonist O B) Full agonist O C) Inverse agonist D) Noncompetitive antagonist E) Partial agonist Correct Answer: E. Partial agonists are drugs that bind to and activate a receptor at the same site as a full agonist but have decreased efficacy relative to the full agonist. In the healthy volunteers, the partial agonist binds to open a-adrenergic sites, which cause peripheral vasoconstriction, leading to increased blood pressure. In the trauma victims, endogenous a-adrenergic agonists (eg, epinephrine and norepinephrine) are produced by the patient as a physiologic response to the trauma in an attempt to maintain mean arterial pressure. The potent new drug outcompetes the endogenous adrenergic agonists for the a-adrenergic receptor sites. Önce bound, the partial agonist leads to decreased relative efficacy and activation of the a-adrenergic receptors compared to the endogenous hormones, which results in a decreased blood pressure. Incorrect Answers: A, B, C, and D. A competitive antagonist (Choice A) directly competes with an agonist for a receptor binding site, blocking its action. The competitive antagonist can be overcome by increasing the concentration of agonist to achieve full efficacy. A competitive antagonist would not cause a blood pressure increase in the healthy volunteers. A full agonist (Choice B) leads to the same maximal response as an agonist at the receptor site. It would be expected to increase blood pressure in both the healthy volunteers and the trauma victims. An inverse agonist (Choice C) binds to the same receptor site as a full agonist, but with the opposite effect of the full agonist. This contrasts with a competitive agonist, which inhibits the efficacy of the response but does not cause the reverse effect. An inverse agonist would be expected to decrease blood pressure in both the healthy volunteers and the trauma victims. A noncompetitive antagonist (Choice D) inhibits the effect of an agonist through binding at a different site. Since it does not directly compete with the agonist, the inhibitory effect cannot be overcome by increasing the concentration of the agonist. A noncompetitive antagonist would decrease the blood pressure in both the healthy volunteers and the trauma victims.

Objective: Partial agonists act at the same binding site as full agonists, though with decreased efficacy. In the presence of a full agonist, competition for the binding site leads to a diminished receptor response. Previous Next Score Report Lab Values Calculator Help Pause

9 Exam Section 1: Item 9 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 9. A 17-year-old girl who is a gymnast is brought to the emergency department 30 minutes after fainting during a competition. Her mother says that the patient eats so little food that her teeth are decaying and that she has stopped menstruating. She is 165 cm (5 ft 5 in) tall and weighs 37 kg (82 lb); BMI is 14 kg/m2. Sexual development is Tanner stage 2. Her pulse is 130/min and blood pressure is 80/50 mm Hg. Physical examination shows pale, cool skin. There are scars across her knuckles that she says are cat scratches. Serum studies show a potassium concentration of 2.9 mEq/L and a chloride concentration of 92 mEq/L. Which of the following is the most likely diagnosis? A) Anorexia nervosa (binge eating/purging type) B) Anorexia nervosa (restricting type) C) Bulimia nervosa (nonpurging type) D) Chronically high metabolic demands of competition with dehydration during the meet E) Obsessive-compulsive disorder with food compulsions Correct Answer: A. Anorexia nervosa (binge eating/purging type) is characterized by an intense fear of gaining weight, decreased self-worth related to body weight, and consequent binge eating and purging behavior (eg, vomiting, laxative or diuretic misuse). Patients with anorexia nervosa often have a BMI less than 17 kg/m2. As a result of insufficient nutrition and weight loss, the hypothalamus decreases gonadotropin-releasing hormone (GNRH) secretion, which leads to amenorrhea and pubertal delay/arrest. The physical examination can show dry, scaly skin, and fine hair or hair loss. Patients with the binging/purging type of anorexia will also demonstrate tooth decay from gastric hydrochloric acid erosion of enamel and scars on the knuckles secondary to abrasions from incisors when inducing vomiting. The loss of gastric hydrochloric acid leads to hypochloremia and metabolic alkalosis. In turn, renal excretion of potassium, when faced with increased bicarbonate filtration, leads to hypokalemia. In severe cases, signs of hypovolemia such as tachycardia and hypotension may be present. Treatment of anorexia is through a combined medical and psychiatric approach and involves correcting fluid and electrolyte derangements alongside behavioral and pharmacologic therapy. Incorrect Answers: B, C, D, and E. Anorexia nervosa (restricting type) (Choice B) is characterized by an intense fear of gaining weight, decreased self-worth related to body weight, and consequent restriction of food intake (instead of binging and purging). The restricting type shares some characteristics with the binge eating/purging type, including low BMI, amenorrhea, and pubertal delaylarrest. However, tooth decay, knuckle scars, and electrolyte disturbances related to vomiting would not be expected. Bulimia nervosa (nonpurging type) (Choice C) involves cycles of uncontrollable eating and compensatory behaviors that occur at least once a week over three months or more. Unlike patients with anorexia nervosa, patients with bulimia nervosa typically have normal BMI. Rather than engaging in purging behaviors such as vomiting, patients with the nonpurging type of bulimia nervosa will fast or intensely exercise. The sequelae of vomiting demonstrated in this patient would not be expected. Chronically high metabolic demands of competition with dehydration during the meet (Choice D) may present with post-exercise tachycardia and hypotension acutely, and amenorrhea and pubertal delay/arrest and weight loss chronically. However, a BMI as low as 14 kg/m2 would be atypical, as patients would be expected to match their oral intake to metabolic demands. Additionally, athletes do not typically demonstrate physical examination findings indicating repeated vomiting. Obsessive-compulsive disorder with food compulsions (Choice E) manifests as recurrent, intrusive urges or thoughts that the patient tries to relieve or neutralize by repetitive rituals involving food (eg, counting the number of bites). In severe cases, these compulsions can lead to weight loss that causes amenorrhea. However, a ritual involving compulsive vomiting would be atypical. While the compulsions of patients with obsessive-compulsive disorder are motivated by relieving distress from obsessions, patients with eating disorders are motivated by body image, as manifested in this patient's poor oral intake combined with her vomiting behaviors. Finally, the patient evades questions about her knuckle lesions, which may indicate that the compulsions are ego-syntonic, as in eating disorders, rather than ego-dystonic, as in obsessive-compulsive disorder.

Objective: Patients with anorexia nervosa (binge eating/purging subtype) demonstrate low BMI, pubertal delay or arrest, and purging behavior leading to hypovolemia and electrolyte disturbances. In patients who repeatedly vomit to purge, physical examination findings can include tooth decay and knuckle scars. I3D Previous Next Score Report Lab Values Calculator Help Pause

190 Exam Section 4: Item 40 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 40. An 80-year-old woman is brought to the emergency department by police 30 minutes after she was found wandering in the street, confused and disoriented. She says that she has no medical conditions and takes no medications. During history taking, she becomes annoyed and agitated, saying, "Stop being so nosey! There's nothing wrong with me! I feel fine and I just want to go home." She appears frail but alert. Her vital signs are within normal limits. Physical examination shows mildly dry mucous membranes. There is no evidence of trauma. She is oriented to person but not to place or time. She is able to provide her home address. Which of the following is the most appropriate action by the physician? A) Attempt to contact the patient's family to find out more information B) Arrange for transportation to take the patient home C) Arrange for the patient to be transferred to a skilled nursing care facility D) Administer a sedative-hypnotic medication E) Admit the patient to a psychiatric inpatient facility Correct Answer: A. Attempts should be made to contact the patient's family to find out more information. This patient's significant cognitive impairment is impeding her ability to care for herself and stay safe outside the hospital. She does not appear to understand the danger of discharging home and therefore lacks decision making capacity. According to the ethical principle of beneficence, it is permissible to keep the patient in the hospital to investigate her medical status and social situation. Her family could inform the medical team whether the confusion is acute, indicating a possible delirium that should be investigated with laboratory testing and/or imaging studies, or chronic, indicating dementia that may be preventing the patient from caring for herself. If dementia is suspected, the medical team should ask the family whether they can increase their supervision of the patient outside the hospital. If the patient's family cannot increase their level of support, the patient may require hospitalization, home care, or placement within a supervised group home or skilled nursing care facility. If family cannot be reached, hospitalization will be necessary in the short term to rule out delirium, thereby ensuring patient safety while additional studies and information are obtained. Incorrect Answers: B, C, D, and E. Arranging transportation to take the patient home (Choice B) would put this confused patient at risk for physical harm, as indicated by her dry mucous membranes (sign of dehydration) and street wandering. Discharging this patient would violate the ethical principle of nonmaleficence. Arranging for the patient to be transferred to a skilled nursing facility (Choice C) may be ultimately necessary. However, the patient's family should be contacted first to determine whether the confusion is acute, representing delirium, and therefore reversible with medical workup and treatment. Additionally, it is possible that the family could provide increased supervision outside the hospital. Administering a sedative-hypnotic medication (Choice D) such as a benzodiazepine is not clinically indicated, as the patient is not physically agitated or attempting to leave. If agitation occurs, an antipsychotic medication (eg, haloperidol, quetiapine) would be less likely to cause or exacerbate delirium than a sedative-hypnotic medication. Admitting the patient to a psychiatric facility (Choice E) is not warranted, as the patient is demonstrating cognitive symptoms (eg, confusion, disorientation) rather than psychiatric symptoms (eg, mood or psychotic symptoms) that would be treated at a psychiatric facility. She likely has delirium or dementia, which would be addressed on a medical ward.

Objective: Patients with significant cognitive impairment who appear unable to stay safe outside of the hospital should not be allowed to leave, as they do not have decision-making capacity. Family should be contacted to gather more information about the patient's baseline cognitive status and/or determine whether family can increase the patient's level of supervision outside the hospital, assuming no medical reason for admission is found. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

189 Exam Section 4: Item 39 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 39. A 4-year-old girl is brought to the physician by her parents 2 days after the mother noticed streaks of blood on the child's well-formed stools. Her father has pigmented macules on his lips. A paternal aunt had breast cancer. Physical examination shows several 2- to 3-mm pigmented macules on the lips and buccal mucosa. Rectal examination shows no abnormalities. Test of the stool for occult blood is positive. Colonoscopy is done and three polyps are removed, the largest measuring 1 x 1 x 1 cm. This patient's polyps are most likely which of the following types? A) Adenomatous B) Hamartomatous C) Hyperplastic D) Inflammatory E) Lymphoid Correct Answer: B. Colonic polyps present in a variety of subtypes, from non-neoplastic polyps (eg, hamartomatous, mucosal, inflammatory, hyperplastic) to potentially malignant polyps (adenomatous, serrated). Hamartomatous polyps are characterized by disorganized growth of tissue similar to normal colonic tissue. When solitary, they are usually benign and do not have a significant risk for malignant transformation; however, when syndromic, they are associated with increased risk in gastrointestinal tract cancers. Peutz-Jeghers syndrome is an autosomal dominant syndrome in which hamartomatous polyps occur in the colon and gastrointestinal tract, and pigmented macules are seen in or around the mouth, lips, hands, and genitalia. It is associated with an increased risk for breast and gastrointestinal tract cancers (eg, colorectal, stomach, small bowel, pancreatic). Incorrect Answers: A, C, D, and E. Adenomatous polyps (Choice A) may be of tubular or villous architecture on histology, with villous architecture usually having greater malignant potential. They arise because of mutations in the APC tumor suppressor gene or the KRAS oncogene. They are associated with familial adenomatous polyposis which is an autosomal dominant genetic syndrome characterized by the presence of thousands of adenomatous polyps arising after puberty. As a result of their malignant potential, patients require prophylactic colectomy. Hyperplastic polyps (Choice C) are benign polyps that do not cause an increased risk for malignancy. They are usually small, located in the rectosigmoid colon, and are not associated with any polyposis syndromes. Inflammatory polyps (Choice D) may be seen with inflammatory bowel diseases including ulcerative colitis or Crohn disease. While they occasionally may be neoplastic, they are more often pseudopolyps, which consist of normal, non-inflamed tissue surrounded by eroded mucosa and granulation tissue. They develop secondary to chronic inflammation of the colon. Lymphoid polyps (Choice E) are uncommon lesions that are usually found near the ileocecal valve because of the presence of lymphoid tissue in the area. They are benign and are not associated with malignant transformation or any polyposis syndromes.

Objective: Peutz-Jeghers syndrome is an autosomal dominant syndrome that is characterized by hamartomatous polyps in the colon and pigmented macules in the mouth, lips, hands, and genitalia. It is associated with an increased risk for breast and gastrointestinal tract cancers (eg, colorectal, stomach, small bowel, pancreatic). Previous Next Score Report Lab Values Calculator Help Pause

172 Exam Section 4: Item 22 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 22. A 35-year-old man comes to the physician because of progressively painful erections during the past year. Examination of the penis shows abnormal curvature, shortening on erection, and a subcutaneous plaque on the dorsal surface. Which of the following is the most likely diagnosis? A) Desmoid tumor O B) Dupuytren contracture O C) Keloids D) Nodular fasciitis E) Peyronie disease Correct Answer: E. Peyronie disease is characterized by fibrosis leading to a fibrotic plaque within the tunica albuginea. On cross section, the layers of connective tissue within the penis include the superficial fascia, deep fascia, and tunica albuginea. The tunica albuginea is a fibrous sheath which surrounds the corpora cavernosa, two paired structures containing a cavernosal artery at the center surrounded by a vascular plexus, which can fill with blood to become erect. The corpus spongiosum, which contains the urethra at its center, is outside of the tunica albuginea and surrounded by the deep fascia and superficial fascia only. When the tunica albuginea is infiltrated by a thick, fibrotic plaque, it leads to shortening of the associated aspect of the penis, an abnormal curvature, and penile shortening on erection. Because the tunica albuginea does not surround the urethra, affected patients are able to maintain urinary function. Incorrect Answers: A, B, C, and D. Desmoid tumors (Choice A) are aggressive, fibrous tumors of connective tissue (sarcomas) which do not have metastatic potential but may be locally aggressive. While they can occur anywhere, the most common locations are the trunk and extremities, arising from skeletal muscle and aponeuroses in each location. The abdominal wall is a common location of desmoid tumors, which may infiltrate into the abdominal cavity and even the viscera. While the penis could be the site of a desmoid tumor, it would not be confined to the dorsal surface of the penis. Dupuytren contracture (Choice B) is caused by fibrosis of the fascia of the palm. It begins as a small nodule but over time progresses to prevent full extension of the fingers. Keloids (Choice C) are firm, skin colored nodules created by aberrant scar formation. They most commonly occur on the head, neck, and trunk secondary to trauma. While they may be disfiguring, disruption of the normal penile anatomy would be unusual. Nodular fasciitis (Choice D) is a benign tumor of fibroblast origin. It is often misdiagnosed as sarcoma, but in contrast to a malignant tumor, has a short duration of symptoms and is self-resolving.

Objective: Peyronie disease is characterized by infiltration of the tunica albuginea by a fibrous plaque causing abnormal penile curvature and shortening on erection. Previous Next Score Report Lab Values Calculator Help Pause

149 Exam Section 3: Item 49 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 49. A70-year-old woman comes to the physician because of an 8-month history of foul-smelling breath and pain before and after she swallows. Physical examination shows halitosis. Two views from a barium swallow are shown; the arrow indicates a diverticulum. This abnormality is most likely located between which of the following muscles? A) Cricopharyngeus and inferior pharyngeal constrictor B) Inferior and middle pharyngeal constrictor C) Middle and superior pharyngeal constrictor D) Stylopharyngeus and middle pharyngeal constrictor E) Superior pharyngeal constrictor and stylopharyngeus Correct Answer: A. Pharyngoesophageal (Zenker) diverticulum occurs within the hypopharynx between the cricopharyngeus muscle and inferior pharyngeal constrictor muscle. The mechanism of formation includes an uncoordinated swallowing mechanism, which results in muscle spasms of the cricopharyngeus leading to a pulsion diverticula as the wall of the hypopharynx and superior esophagus weakens. Symptoms include halitosis, regurgitation of undigested food, and dysphagia. Barium swallow shows a posterior, midline outpouching above or at the level of the cricopharyngeus. Treatment includes surgical repair and resection (diverticulectomy), plus dietary modifications to include soft or liquid foods. Incorrect Answers: B, C, D, and E. The inferior and middle pharyngeal constrictors (Choice B) form part of the posterior and lateral walls of the hypopharynx and direct food towards the esophagus during swallowing. Several important structures pass between these muscles including the superior laryngeal artery, vein, and the internal laryngeal nerve. The middle and superior pharyngeal constrictors (Choice C) form the superior part of the hypopharynx. The glossopharyngeal nerve (cranial nerve IX), the stylopharyngeus muscle, and the stylohyoid muscle pass between these muscles. The stylopharyngeus and middle pharyngeal constrictor muscles (Choice D) contribute to palatal elevation during swallowing and constriction of the middle pharynx to aid in propagation of a food bolus, respectively. The glossopharyngeal nerve travels along the lateral aspect of the stylopharyngeus to innervate the tongue. The superior pharyngeal constrictor and stylopharyngeus (Choice E) act in concert to narrow the pharyngeal lumen and elevate the palate, respectively. This assists with movement of the food bolus toward the esophagus while preventing nasal regurgitation.

Objective: Pharyngoesophageal (Zenker) diverticulum occurs between the cricopharyngeus and inferior pharyngeal constrictor muscles as a result of the incoordination of contraction between the two muscles. This results in increased pressure generation and subsequent diverticula formation in the weak connective tissue between the two muscles. ID Previous Next Score Report Lab Values Calculator Help Pause

36 Exam Section 1: Item 36 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 36. A79-year-old woman with osteoarthritis comes to the physician for an initial examination. She is otherwise healthy and lives alone. When the physician enters the examination room, he detects a slight odor suggestive of urine. It is most appropriate for the physician to approach the topic of urinary incontinence with this patient in which of the following ways? O A) "Are you having any problems that you may be too embarrassed to talk about? Would you be willing to talk about them today?" B) "I couldn't help but notice an odor when I came into the room. Are you having any difficulty holding your urine?" O C) "I understand this can be difficult to talk about, but do you find yourself also having fecal incontinence in addition to urinary incontinence?" D) "I know this can be an embarrassing topic, but people sometimes have difficulty holding their urine as they get older. Is this ever a problem for you?" E) "It is not surprising that someone your age is having problems with urinary incontinence. How many accidents do you have a day?" Correct Answer: D. Acknowledging that a topic is sensitive, normalizing the problem, and using direct yet non-presumptive language are all effective communication strategies when discussing a difficult topic such as urinary incontinence. Physicians can normalize a patient's natural tendency to feel embarrassed-or suggest that this is a normal response-and thereby invite the patient to openly discuss potentially shameful details about a physical dysfunction. Educating the patient that urinary incontinence is common in older patients can also reduce the potential embarrassment associated with the topic. Finally, asking a direct question to confirm the suspected problem can help the physician gather needed information from the patient without causing the patient shame about objective signs of urinary incontinence such as a foul odor. Incorrect Answers: A, B, C, and E. Physicians should avoid asking vague questions that do not directly assess for urinary incontinence (Choice A). This may imply that the topic is too shameful to openly discuss and discourage the patient from mentioning the issue. Before assuming a patient has a potentially embarrassing subjective concern based on a physical examination finding, physicians should ask the patient if they are having this suspected concern (Choices B, C, and E). In each of these answer choices, the physician is implying that the patient's incontinence is noticeable, which could trigger shame and defensiveness rather than an open discussion of the problem between the patient and physician. Additionally, the physician is asking invasive questions without first inviting the patient's assent to discuss the potential problem.

Objective: Physicians should acknowledge a sensitive topic, normalize the issue, and use direct, non-presumptive language to foster an effective discussion about a potentially embarrassing topic such as urinary continence. Physicians should avoid stigmatizing and presumptive language, as this could trigger defensiveness on the patient's part and prevent effective discussion of the topic. Previous Next Score Report Lab Values Calculator Help Pause

86 Exam Section 2: Item 36 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 36. A 43-year-old man comes to the emergency department because of a 1-year history of low back pain. Before the examination, the patient says, "My physician is arrogant and insensitive. He never returns my phone calls, I always have to wait forever to be seen, the tests he orders are painful and unnecessary, and he can never tell me what is causing my back pain or how to treat it." Which of the following is the most appropriate response by the emergency department physician about this patient's complaints? A) Reassure the patient that his physician's behavior is not unusual and that low back pain can be difficult to assess B) Encourage the patient to make an appointment with his physician to communicate his concerns C) Explain to the patient how he may register a complaint against his physician with the state medical licensing board D) Provide the patient with a copy of the Patients' Bill of Rights E) Telephone the patient's physician to make him aware that the patient is very dissatisfied with treatment Correct Answer: B. Encouraging the patient to make an appointment with his physician to discuss his concerns communicates to the patient that the emergency department physician believes the patient's concerns are legitimate and important to address, which will help the patient feel validated. Encouraging this direct physician-patient communication, as opposed to involving a third party, will improve interpersonal communication between the patient and his physician and increase the likelihood that the patient's specific concerns will be satisfactorily addressed. The emergency department physician does not possess all the facts, only the patient's side of the story; therefore, the emergency department physician should avoid making definitive decisions about actions that should be taken regarding the patient's concerns. Incorrect Answers: A, C, D, and E. Reassuring the patient that his physician's behavior is not unusual, and that low back pain can be difficult to assess (Choice A) would likely feel invalidating to the patient, as the patient stated that the outpatient physician does not return his phone calls among other objectively unprofessional actions. The emergency department physician does not possess all of the facts and should abstain from making a judgment about the outpatient physician's behavior. Explaining to the patient how to register a complaint against his physician with the state medical board or providing the patient with a copy of the Patients' Bill of Rights (Choices C and D) assumes wrongdoing on the behalf of the outside physician; however, the emergency department physician does not possess all of the facts about the behavior of the outside physician. Encouraging the patient to speak directly with his outside physician also increases the chance of solving the problem expeditiously. Telephoning the patient's physician to make him aware that the patient is dissatisfied with treatment (Choice E), rather than having the patient speak directly to his physician, would not solve the reported problem of poor physician-patient communication. To improve interpersonal communication and discuss specific concerns, the patient should make an appointment with the outpatient physician.

Objective: Physicians should encourage patients with concerns about a certain member of their healthcare team to discuss these concerns directly with the team member. Direct discussion will improve interpersonal communication between the patient and the team member and increase the chances that the patient will have their specific needs met. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

14 Exam Section 1: Item 14 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 14. A 34-year-old woman is brought to the emergency department by her husband because of an inability to see objects in the peripheral visual fields in both eyes for 1 day. Her temperature is 37.2°C (99°F), pulse is 68/min, respirations are 16/min, and blood pressure is 124/78 mm Hg. Automated computerized perimetry examination shows a loss of the temporal visual field in both eyes. Neurologic examination shows no other abnormalities. This patient is most likely to have which of the following additional findings? A) Amenorrhea B) Aphasia C) Double vision D) Loss of memory E) Seizures Correct Answer: A. This patient's bitemporal hemianopsia is most likely the result of optic chiasm compression from a pituitary adenoma, which is often associated with amenorrhea in women secondary to the overproduction of prolactin. The pituitary gland is in the sella turcica. A pituitary adenoma is a benign brain tumor that can present with mass effect symptoms such as headache and bitemporal hemianopsia (loss of bilateral temporal visual fields). Bitemporal hemianopsia occurs because of compression of the optic chiasm, which contains fibers from the bilateral nasal retina (containing visual information from the temporal eye fields). Pituitary adenomas can cause symptoms from hyperpituitarism or hypopituitarism, depending on whether the tumor is functionally secreting hormones. The most common functional pituitary adenoma is a prolactinoma; other types of pituitary adenomas secrete follicle-stimulating hormone (FSH), luteinizing hormone (LH), adrenocorticotropic hormone (ACTH), thyroid-stimulating hormone (TSH), or growth hormone (GH). Prolactinomas suppress gonadotropin-releasing hormone (GNRH), which results in decreased FSH and LH production and manifests as amenorrhea. Additionally, if a pituitary adenoma is nonfunctional but sufficiently large, it can interfere with normal hormone secretion by the pituitary gland, causing hypopituitarism. In this setting, decreased secretion of FSH and LH could also manifest as amenorrhea. Incorrect Answers: B, C, D, and E. Aphasia (Choice B) is commonly caused by lesions in the frontal and temporal lobes, respectively. These regions are commonly damaged in the setting of stroke, trauma, or tumors. The sella turcica is in the middle cranial fossa at the base of the brain, so pituitary adenomas would be unlikely to affect the speech centers of the cerebral cortex. Double vision (Choice C), or diplopia, may be caused by dysfunction of the extraocular muscles or refractive errors. Although large pituitary adenomas could impinge the cranial nerves that innervate the extraocular muscles (occulomotor, trochlear, and abducens nerves) as they pass through the cavernous sinuses, the optic nerve is closer to the pituitary gland and more likely to be impinged by a pituitary adenoma. This patient developed mass-related bitemporal hemianopsia, making it unlikely that these additional cranial nerves would also be affected. Loss of memory (Choice D) is unlikely to be related to the mass or hormonal effects of a pituitary adenoma. The hippocampus is a key center for memory processing, and is in the mesial temporal lobe, unlikely to be affected by a pituitary adenoma. Seizures (Choice E) are a rare sequela of a massive pituitary adenoma compressing adjacent cerebral cortex. Given that this patient's mass effect symptoms started recently and only involve the optic chiasm (bitemporal hemianopsia), a tumor this large would be unlikely. Meningioma and glioblastoma are more commonly associated with seizures caused by the location of these tumors within or adjacent to the cerebral cortex.

Objective: Pituitary adenomas can cause mass effect symptoms such as headache and bitemporal hemianopsia from compression of the adjacent optic chiasm. Additionally, they may be functional or nonfunctional in nature. Hypersecretion of hormones, most commonly prolactin, can manifest with amenorrhea and galactorrhea in females. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

95 Exam Section 2: Item 45 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 45. A plasmid has acquired a mutation that prevents it from replicating in bacteria. This mutation most likely affects which of the following regions within the plasmid? O A) Antibiotic resistance gene B) The lac 2 promoter O C) Multicloning site D) Origin of DNA synthesis E) Ribosome binding site F) Transposable genetic element Correct Answer: D. The origin of DNA synthesis, also called the origin of replication, is a sequence of DNA within a plasmid where replication is initiated and binds replicative enzymes (eg, DNA gyrase, polymerase, ligase) which permit the initiation of plasmid duplication and reproduction. A mutation at the origin of DNA synthesis would prevent replication of the plasmid. Plasmids are extrachromosomal segments of double-stranded DNA that are commonly present in bacteria; they may contain genes that confer survival advantages to the host and are usually classified by their function. Examples of plasmid classes include resistance plasmids, which confer antibiotic resistance to the host, virulence plasmids, which may convert a bacterium into a pathogen, and metabolic or degradative plasmids, which permit metabolism of nutrients or foreign substances. Another common class of plasmid is the fertility plasmid, which contains genes that permit bacterial conjugation and expression of sex pili, through which genetic material can be transferred. Plasmids must be able to replicate, or any conferred benefit will be lost. Incorrect Answers: A, B, C, E, and F. Antibiotic resistance gene (Choice A) refers to the material contained on a resistance plasmid, which confers antibiotic resistance to the host bacterium. Examples of antibiotic resistance genes include those coding for B-lactamases, which are commonly found in gram-negative rods such as Escherichia coli and Klebsiella pneumoniae. These genes are separate from the origin of DNA synthesis on a plasmid and do not directly affect bacterial DNA replication. The lac 2 promoter (Choice B) describes a region of the lac operon, a polycistronic gene that conveys the ability to digest lactose. The lac operon is commonly found in Enterobacteriaceae and permits the digestion of lactose in a glucose-poor environment. When conditions are appropriate for expression of the lac operon (glucose poor, lactose rich environment), the promoter binds RNA polymerase which begins transcription of messenger RNA (MRNA). This process is unrelated to the replication of DNA, though a promoter mutation would result in a decreased activity of gene transcription. Multicloning site (Choice C) describes a region of a plasmid that contains many restriction enzyme target sites, which allows for the acquisition or insertion of exogenous DNA. This is also known as a polylinker site, and it is typically in a location that is distinct from any expressed exons. Ribosome binding site (Choice E) describes the region of a mRNA to which ribosomes are recruited prior to initiation of translation. The site is located within the 5' untranslated region and upstream from the start codon. It is known as a Shine-Dalgarno sequence in prokaryotes, and a Kozak sequence in eukaryotes. It often includes adenine- and thymine-rich sequences (eg, TATA box). Transposable genetic element (Choice F), also known as a transposon, describes a segment of DNA that can move position within or across chromosomes. As related to plasmids, transposons have been described moving between a chromosome and plasmid; in this way, genetic material can be transferred.

Objective: Plasmids contain regions of DNA that will be transcribed for a functional purpose; they also contain regions that permit their replication (the origin of DNA synthesis) by binding replicative enzymes. A mutation in the origin of DNA synthesis region prevents the binding of enzymes, preventing replication of the plasmid. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

154 Exam Section 4: Item 4 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 4. A 50-year-old man has had progressive dyspnea on exertion over the past 6 months. He now has dyspnea at rest. He has had a persistent dry cough since having an upper respiratory tract bacterial infection 1 year ago. Examination of lung tissue obtained on biopsy shows chronic inflammation and fibrous thickening of the alveolar septa. Which of the following is the most likely diagnosis? A) Chronic bronchitis B) Diffuse alveolar damage O C) Emphysema D) Sarcoidosis E) Usual interstitial pneumonitis Correct Answer: E. Interstitial lung disease (ILD) refers to the heterogeneous group of disorders which result in pulmonary fibrosis and progressively impaired lung function. Causes include infections, occupational and environmental exposures, medication-induced pulmonary toxicity, radiation-induced lung injury, connective tissue disorders, hypersensitivity pneumonitis, and idiopathic disease. Patients classically present with progressive dyspnea and chronic nonproductive cough. Inspiratory crackles and hypertrophic osteoarthropathy may be present during physical examination. Usual interstitial pneumonitis (also called usual interstitial pneumonia) is the most common histopathologic and radiologic pattern of ILD. It is a pattern associated with idiopathic pulmonary fibrosis. Chest x-ray often shows peripheral and basilar honeycombing (clustered cystic air spaces). Histology will show interstitial fibrosis in a patchwork pattern, honeycomb changes, and fibroblast foci. Incorrect Answers: A, B, C, and D. Chronic bronchitis (Choice A) refers to long-term inflammation of the bronchi and is commonly seen in patients with tobacco use. It is defined as the presence of a productive cough on most days over a 3-month period for at least 2 consecutive years in patients who do not have another cause of cough. Histologic features include inflammatory infiltrates within the airway walls. Diffuse alveolar damage (Choice B) is a term used to describe histologic change associated with many inflammatory, infectious, toxin-mediated, and inhaled causes of acute lung injury, and is often the description associated with the broad class of causes that result in acute respiratory distress syndrome (ARDS). ARDS has many causes, for example, severe pneumonia, sepsis, inhalational injury, drowning, ventilator use, and vasculitis. Classically, on biopsy, ARDS demonstrates noncardiogenic pulmonary edema, eosinophilic alveolar infiltrates, and hyaline membranes, along with destruction of normal pneumocyte architecture. Emphysema (Choice C) describes structural changes to the lung that result in abnormal and enlarged air spaces distal to the terminal bronchioles with loss of individual alveoli. There is noted absence of fibrotic changes. Emphysema is associated with smoking histories and a1-antitrypsin deficiency. Sarcoidosis (Choice D) is a granulomatous inflammatory condition which can affect multiple organ systems, especially the lungs and lymphatic system. While sarcoidosis is an interstitial lung disease that can cause progressive pulmonary fibrosis, it is not associated with a usual interstitial pneumonitis pattern. Lung biopsy will disclose noncaseating granulomas.

Objective: Pulmonary fibrosis results from chronic inflammation of the pulmonary interstitium, classically presenting with dyspnea and a nonproductive cough. Idiopathic pulmonary fibrosis demonstrates a usual interstitial pneumonitis pattern on both imaging and histology. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

106 Exam Section 3: Item 6 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 6. A 27-year-old man has the sudden onset of fever and shaking chills 2 weeks after returning from a trip to Africa. He has mild anemia. A peripheral blood smear is shown. He is treated with chloroquine followed by primaquine. Which of the following best explains the benefit of using primaquine in this patient? A) Decreases the dose of chloroquine needed B) Decreases the duration of chloroquine needed C) Produces more rapid resolution of symptoms D) Treats chloroquine-resistant parasites E) Treats exoerythrocytic parasites Correct Answer: E. Fatigue, fever, chills, and anemia presenting in a patient with recent travel to Africa or Central/South America suggests a diagnosis of malaria. There are several different pathogenic species of malaria including Plasmodium falciparum, Plasmodium ovale, Plasmodium vivax, Plasmodium knowlesi, and Plasmodium malariae, although over 90% of malaria infections in returning U.S. travelers are secondary to P. falciparum or P. vivax. Noncyclic fevers in a traveler returning from Africa suggests infection with either P. vivax (more likely) or P. ovale. The peripheral blood smear shows a schizont form of the parasite, although any malarial stage (gametocyte, schizont, trophozoite, or ring) can be seen on the peripheral smear and is diagnostic of infection. While most serious infections are secondary to P. vivax or P. falciparum, only P. ovale and P. vivax have hypnozoite liver stages (exoerythrocytic) of the parasitic life cycle. Even after the initial parasitemia has resolved, these patients are still at risk for reactivation many weeks or months later. Chloroquine effectively eradicates circulating forms if there is no resistance but has no effect on the hypnozoites residing in the liver. For this reason, it is essential to follow chloroquine with a course of primaquine, which effectively targets the hypnozoite exoerythrocytic forms of P. vivax and P. ovale. Of note, glucose 6-phosphate dehydrogenase (G6PD) deficiency status should be confirmed prior to the administration of primaquine as patients with G6PD deficiency are at a higher risk for hemolytic anemia when given this medication. Incorrect Answers: A, B, C, and D. Decreasing the dose (Choice A) or duration (Choice B) of chloroquine or treating chloroquine-resistant parasites (Choice D) are not benefits of using primaquine. Chloroquine is given as a fixed dose of 1 g on day 1 followed by 0.5 g 6 hours later, and then 0.5 g daily for 2 days for all non-severe malaria infections unless there is a high degree of known chloroquine resistance, which is most common among P. vivax found in Papua New Guinea and Indonesia, and among P. falciparum found throughout Africa. In such instances, or in cases of severe malaria as determined by clinical and laboratory criteria, artesunate-based therapies are preferred. The addition of primaquine treats the exoerythrocytic liver stage of the parasite and does not augment the initial dose of chloroquine. Primaquine does not produce more rapid resolution of symptoms (Choice C), rather, it prevents future relapses by eliminating dormant hypnozoite forms.

Objective: Plasmodium ovale and Plasmodium vivax both have life cycles that include an exoerythrocytic liver stage (hypnozoite) of the parasite. Treatment with chloroquine alone will eliminate parasites circulating in the blood but will not affect liver hypnozoites. Addition of primaquine effectively treats these liver forms and prevents relapsing infection. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

93 Exam Section 2: Item 43 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 43. A previously healthy 43-year-old man comes to the physician because of a 1-week history of fever; vague, migratory abdominal pain; and generalized muscle pain and weakness. He has no history of a similar illness. His temperature is 38.2°C (100.8°F), pulse is 78/min, respirations are 16/min, and blood pressure is 150/94 mm Hg. Pulmonary examination shows no abnormalities. Abdominal examination shows diffuse tenderness with no organomegaly. There is generalized muscle weakness and tenderness to palpation of various muscle groups. Laboratory studies show: 15 g/dL 45% Hemoglobin Hematocrit Leukocyte count Platelet count 22,000/mm3 550,000/mm3 105 mm/h Erythrocyte sedimentation rate Urine Blood Protein 3+ 2+ Blood cultures are negative. Examination of a muscle biopsy specimen from this patient shows segmental transmural necrotizing arteritis. Which of the following is the most likely diagnosis? A) Churg-Strauss syndrome B) Giant cell arteritis C) Polyarteritis nodosa D) Takayasu arteritis E) Wegener granulomatosis Correct Answer: C. Polyarteritis nodosa is a necrotizing vasculitis involving medium-sized arteries that generally affects the kidneys and visceral organs, sparing the lungs. It is characterized by transmural inflammation of arterial walls with fibrinoid necrosis and typically presents with systemic symptoms such as fever, malaise, weight loss, fatigue, and multisystemic involvement. It commonly affects the kidneys and gastrointestinal system, causing hypertension and kidney microaneurysms, along with abdominal pain and melena, respectively. Skin lesions may include nodules, purpura, ulcers, vesicles, or bullae. When involving the peripheral nervous system, polyarteritis nodosa can cause polyneuropathy with both motor and sensory deficits. Laboratory evaluation often shows increased serum concentrations of inflammatory markers such as erythrocyte sedimentation rate and leukocytosis. Urinalysis may demonstrate hematuria and proteinuria, as seen in this case. Treatment involves suppression of the inflammation with corticosteroids and cyclophosphamide. Incorrect Answers: A, B, D, and E. Churg-Strauss syndrome (Choice A) or eosinophilic granulomatosis with polyangiitis is a small vessel vasculitis. The most common symptoms include asthma, ear, nasal, and sinus inflammation, and peripheral neuropathy. It can affect multiple organ systems and can present with clinical manifestations including arthralgia, skin lesions such as nodules or a maculopapular rash, cardiomyopathy, lung, kidney, or gastrointestinal involvement. Nephritic syndrome is a known complication. Giant cell arteritis (Choice B), or temporal arteritis, is a vasculitis that affects large vessels such as the aorta, temporal branch of the external carotid, and vertebral arteries. It is characterized by a headache, temporal tenderness, jaw claudication, and fever. Complications include monocular blindness from inflammation involving the retinal branch arteries, and therefore requires urgent administration of glucocorticoids, ophthalmologic evaluation, and temporal artery biopsy for diagnosis. Takayasu arteritis (Choice D) is a large vessel vasculitis usually affecting females less than age 40 years and is characterized by weak pulses caused by granulomatous thickening and narrowing of the aortic arch and proximal great vessels. It is also associated with systemic symptoms such as fever, night sweats, myalgias, and arthralgias. It may also be associated with skin nodules and vision disturbances. Wegener granulomatosis (Choice E), also known as granulomatosis with polyangitis, is a necrotizing small-vessel vasculitis that classically presents with sinopulmonary and renal vessel involvement. Patients classically present with epistaxis, hemoptysis, and hematuria. Nephritic syndrome is a known complication.

Objective: Polyarteritis nodosa is a necrotizing vasculitis involving medium-sized arteries that generally affects the kidneys and visceral organs, sparing the lungs. It typically presents with systemic symptoms, hypertension, kidney damage, kidney microaneurysms, abdominal pain, and melena. Treatment involves suppression of the inflammation with corticosteroids and cyclophosphamide. II Previous Next Score Report Lab Values Calculator Help Pause

4 Exam Section 1: Item 4 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 4. Assuming the positive predictive value of a test is 50% and the negative predictive value is 75%, which of the following tables are consistent with these values? Disease Disease Present Absent Present Absent A B Positive 50 50 750 250 Test Negative 250 750 50 50 Positive 250 750 50 50 Test Negative 50 50 750 250 E Cannot be determined from data given A) B) C) D) E) Correct Answer: A. Positive predictive value (PPV) and negative predictive value (NPV) are statistical measurements that define the proportion of positive test results that are true positives (positive tests in patients who have the condition) and true negatives (negative tests in patients without the condition). PPV is thus defined as (true positives [TP]) / (TP + false positives [FP]) and measures the probability of a patient having the disease given a positive test result. NPV is defined as (true negatives [TN]) / (TN + false negatives [FN]) and measures the probability of a patient being disease free given a negative test result. Both PPV and NPV vary with disease prevalence; PPV varies directly with prevalence whereas NPV varies inversely with prevalence. Therefore, the more prevalent a disease, the greater the PPV of a test used in that population and the lower the NPV of a test used in that population. Given the data above, a PPV of 50% indicates that TP / (TP + FP) = 0.5. In Choice A, the total number of positive tests (TP + FP) = 100, and the TP = 50, therefore, PPV = 50/100 = 0.5, or 50% = PPV. In Choice A, the total number of negative tests (TN + FN) = 1000, and the TN = 750, therefore, NPV = 750/1000 = 0.75, or 75% = NPV. %3D Incorrect Answers: B, C, D, and E. Choice B, PPV = TP / (TP + FP) = 750/(750 + 250) = 0.75 = 75%; NPV = TN / (TN + FN) = 50/(50 + 50) = 0.5 = 50% Choice C, PPV = TP / (TP + FP) = 250/(250 + 750) = 0.25 = 25%; NPV = TN / (TN + FN) = 50/(50 + 50) = 0.5 = 50% %3D Choice D, PPV = TP/ (TP + FP) = 50/(50 + 50) = 0.5 = 50%; NPV = TN / (TN + FN) = 250/(250 + 750) = 0.25 = 25% %3D Cannot be determined from data given (Choice E) is incorrect, as the data needed to compute positive and negative predictive values includes the numbers of true and false positive and negative cases based on the test in question.

Objective: Positive predictive value defines the proportion of positive test results that are true positive and is calculated as TP/ (TP + FP). Negative predictive value defines the proportion of negative test results that are true negative and is calculated as TN / (TN + EN). II Previous Next Score Report Lab Values Calculator Help Pause

163 Exam Section 4: Item 13 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 13. A 30-year-old woman is being evaluated for infertility. Menarche occurred at age 12 years, and she had regular menses until approximately 3 years ago. At that time, she noticed increased spacing between her menstrual periods and decreasing flow. Physical examination shows no abnormalities. Laboratory findings show normal serum testosterone and prolactin concentrations and an increased serum follicle-stimulating hormone concentration. Which of the following is the most likely cause of the amenorrhea? A) Endometriosis B) Hypothyroidism C) Pituitary tumor D) Polycystic ovarian syndrome E) Premature ovarian failure Correct Answer: E. Premature ovarian failure occurs secondary to the atresia of ovarian follicles before age 40 years. It is a relatively common cause of infertility. As the number of ovarian follicles declines, estrogen concentrations fall and the patient develops menopausal symptoms, including irregular menses with decreased flow, hot flashes, and vaginal dryness. As estrogen decreases, feedback inhibition on the hypothalamic-pituitary-ovarian axis is removed, and concentrations of gonadotropin-releasing hormone are increased. This stimulates the production of luteinizing hormone and follicle-stimulating hormone, which cause serum concentrations to rise to menopausal levels to increase the production of estrogen by the ovaries, which are unable to do so. Other causes of infertility, such as polycystic ovarian syndrome or a prolactin-secreting pituitary adenoma must be excluded. Incorrect Answers: A, B, C, and D. Endometriosis (Choice A) denotes the presence of endometrial tissue outside of the uterus, and classically presents with symptoms of dysmenorrhea, dyspareunia, and dyschezia. The gold-standard for diagnosis is a diagnostic laparoscopy, which evaluates for extrauterine endometriotic implant lesions. Hypothyroidism (Choice B) is a common cause of female infertility as a result of anovulatory cycles and increased serum prolactin concentration. Infertility secondary to hypothyroidism would be accompanied by other symptoms of hypothyroidism including fatigue, weight gain, cold intolerance, and constipation. Pituitary tumor (Choice C) is associated with gynecomastia, galactorrhea, and amenorrhea when prolactin is functionally produced by the tumor, however, serum prolactin concentration would be expected to be increased. Other signs of a pituitary tumor include features related to its mass effect in the sella turcica with loss of other pituitary hormones, or damage to the optic chiasm resulting in bitemporal hemianopsia on examination. Polycystic ovarian syndrome (PCOS) (Choice D) is a common cause of infertility, however it is usually accompanied by hirsutism and increased body weight. PCOS occurs secondary to excess androgens, and can present with polycystic ovaries on ultrasound, oligoovulation, anovulation, or clinical signs of hyperandrogenism, of which two are required for the diagnosis. PCOS increases the risk for metabolic syndrome and type 2 diabetes mellitus. If a patient with PCOS is attempting to conceive, therapy should include weight loss, clomiphene (to improve successful ovulation), or metformin.

Objective: Premature ovarian failure is defined as the early atresia of ovarian follicles before the age of forty. It is a common cause of female infertility and demonstrates postmenopausal concentrations of follicle-stimulating hormone. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

115 Exam Section 3: Item 15 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 15. The following data about cancer prevalence and mortality are collected in a small city for a calendar year: Males Females Total Population Number of persons with cancer (all types) Number of deaths from cancer (all types) Number of women with ovarian cancer Number of deaths from ovarian cancer 43,000 224 47,000 183 90,000 407 80 50 30 8 Which of the following is the period prevalence of ovarian cancer in women in this city in this calendar year? O A) 5/183 B) 5/407 C) 5/47,000 D) 5/90,000 E) 8/183 F) 8/407 G) 8/47,000 H) 8/90,000 Correct Answer: G. Prevalence, as applied to biostatistics, defines the number of cases of a disease or condition as a fraction of a population in consideration at a point-in-time. Prevalence, therefore, indicates the percentage of current cases of a condition within the population under study. Mathematically, prevalence is calculated by dividing the number of existing cases by the number of people in the population under study or at risk for the condition. In this case, the condition or disease under study is ovarian cancer, and the population in question is women. Based on the data given, prevalence therefore equals 8 (the number of females with ovarian cancer)/47,000 (the number of females in the city). It is important to note that the total prevalence of ovarian cancer would be calculated by taking the number of cases and dividing it by the total population in the city, however this population would also include males and would therefore be less statistically meaningful, as they would not be at risk for contracting ovarian cancer. Incorrect Answers: A, B, C, D, E, F, and H. 5/183 (Choice A) and 5/407 (Choice B) calculate the number of deaths from ovarian cancer within a year as a fraction of the number of females, 183, (or persons, 407) with all types of cancer within that year, respectively. They therefore represent annual mortality rates within women (or all persons) who have cancer, respectively. 5/47,000 (Choice C) and 5/90,000 (Choice D) calculate the number of deaths from ovarian cancer within a year as a fraction of the total number of women, 47,000 (or total persons, 90,000) in the population, respectively. They therefore represent annual mortality rates from ovarian cancer among women in the population, or among the total population, respectively. 8/183 (Choice E) and 8/407 (Choice F) calculate the number of women living with ovarian cancer as a fraction of the number of women with cancer, 183, (all types, 407) and the number of persons with cancer (all types), respectively. They therefore calculate the prevalence of ovarian cancer within women or persons who have cancer within the population, respectively. 8/90,000 (Choice H) calculates the number of women with ovarian cancer as a fraction of the total population; it therefore represents a total prevalence of the disease within the population (men and women), whereas the data point in question is the prevalence among women, not the total population.

Objective: Prevalence defines the number of existing cases of a disease or condition as a percentage of the number of people in the population under study who are at risk for that condition. I3D Previous Next Score Report Lab Values Calculator Help Pause

50 Exam Section 1: Item 50 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 50. A 57-year-old man is brought to the emergency department because of a 6-hour history of acute lower abdominal and flank pain. He also has a 6-week history of fatigue and generalized muscle and bone pain. He appears ill and changes positions to relieve the pain when it occurs. Physical examination shows a 0.5-cm, nontender, movable mass in the left side of the neck. Abdominal examination shows decreased bowel sounds; there is no rebound tenderness. Laboratory studies show: Serum Ca2+ Urea nitrogen Creatinine Urine RBC 12.8 mg/dL 25 mg/dL 1.5 mg/dL 3-5/hpf An x-ray of the abdomen shows a calculus in the proximal left ureter. Which of the following is the most likely underlying cause of this patient's condition? A) Parathyroid adenoma B) Parathyroid carcinoma C) Parathyroid hyperplasia D) Thyroid follicular adenoma E) Thyroid nodular hyperplasia O F) Thyroid papillary carcinoma Correct Answer: A. Primary hyperparathyroidism presents with hypercalcemia (which in turn presents with nephrolithiasis, chronic bone pain, abdominal discomfort, and psychiatric disturbances), hypophosphatemia, subperiosteal bone resorption, and kidney failure. Primary hyperparathyroidism is most caused by a parathyroid adenoma, followed by parathyroid hyperplasia, and parathyroid carcinoma. Excessive production of parathyroid hormone (PTH) results in increased serum calcium concentration from increased osteoclast activity, intestinal absorption, and renal tubular absorption. Serum phosphorus concentration is decreased secondary to increased excretion by the renal tubules. This patient demonstrates many features of primary hyperparathyroidism and resultant hypercalcemia, including abdominal pain, bone pain, increased serum calcium concentration, increased serum creatinine concentration, and a renal calculus, along with a mass in the location of the parathyroid gland. Parathyroid adenoma is estimated to comprise 90% of cases of primary hyperparathyroidism, making it the most likely diagnosis. Incorrect Answers: B, C, D, E, and F. Parathyroid carcinoma (Choice B) and parathyroid hyperplasia (Choice C) are uncommon etiologies of primary hyperparathyroidism, making up a combined less than 10% of cases. Thyroid follicular adenoma (Choice D) is a benign neoplasm of the thyroid that contains differentiated follicular cells. Thyroid nodular hyperplasia (Choice E) refers to areas of nodular, hyperplastic cells that may be scattered throughout the gland. Generally, both entities are non-functional but occasionally can produce thyroid hormone, resulting in their recharacterization as a toxic thyroid adenoma or toxic multinodular goiter. They do not generally affect serum calcium concentration or produce PTH. Thyroid papillary carcinoma (Choice F) is the most common malignant neoplasm of the thyroid gland. It typically demonstrates empty-appearing nuclei with central clearing, nuclear grooves, and psammoma bodies on histology. It is generally nonfunctional.

Objective: Primary hyperparathyroidism is most commonly caused by a parathyroid adenoma, followed by parathyroid hyperplasia, and parathyroid carcinoma. I3D Previous Next Score Report Lab Values Calculator Help Pause

162 Exam Section 4: Item 12 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 12. A 29-year-old woman has had three spontaneous abortions between the third and fourth weeks of gestation. The most likely cause is failure of her ovaries to produce sufficient quantities of which of the following hormones? O A) Estrogen B) Human chorionic gonadotropin C) Luteinizing hormone D) Oocyte maturation inhibitor E) Progesterone Correct Answer: E. Progesterone is secreted by the corpus luteum, the structure that remains after an ovarian follicle ruptures and releases an egg. Progesterone is necessary for maintaining the secretory endometrium and creating a hospitable environment for implantation. If implantation occurs, the corpus luteum is maintained by human chorionic gonadotropin and thus, progesterone continues to be secreted by the corpus luteum in order to maintain the endometrial lining and prevent menses. Following the first trimester, the placenta will make its own progesterone to maintain the pregnancy. If implantation does not occur, progesterone concentrations decrease and the endometrium beings to slough, causing menses. If implantation does occur and progesterone concentrations are not sustained, the endometrium will slough, and a spontaneous abortion will occur. Incorrect Answers: A, B, C, and D. Estrogen (Choice A) is produced by the granulosa cells of the developing follicle. As the follicle increases in size through the follicular phase, its production of estrogen increases. A second, smaller peak of estrogen occurs in the luteal phase and stimulates the endometrium to proliferate but it is not responsible for maintaining the lining as is progesterone. Human chorionic gonadotropin (Choice B) is produced by the syncytiotrophoblastic cells of the placenta in order to maintain the corpus luteum and continue its production of progesterone. Concentrations of human chorionic gonadotropin begin to rise after 3 or 4 weeks' gestation; prior to this, the corpus luteum secretes progesterone independently. Luteinizing hormone (LH) (Choice C) triggers ovulation when it surges and marks the transition between the follicular phase and luteal phase. Without LH, ovulation would not occur. LH does not play a role in maintaining the endometrial lining when implantation occurs. Oocyte maturation inhibitor (Choice D) is a factor secreted by the ovary that prevents the developing follicles from completing meiosis until the ovulatory surge of follicle stimulating hormone (FSH) and LH occurs. A deficiency in this factor would promote the premature maturation of follicles.

Objective: Progesterone is the hormone responsible for maintaining the endometrial lining when implantation occurs. Without enough progesterone, spontaneous abortion ensues as a result of endometrial sloughing. Previous Next Score Report Lab Values Calculator Help Pause

111 Exam Section 3: Item 11 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 11. An experiment is designed to assay the secretions of chondrocytes obtained from the articular cartilage of osteoarthritic joints. Which of the following compounds is most likely to be found in decreased concentration in the secretions? A) Catabolic metalloproteinase B) Interleukin-1 (IL-1) O C) Interleukin-6 (IL-6) D) Proteoglycans E) Tumor necrosis factor Correct Answer: D. Proteoglycans are extracellular matrix molecules that have a central protein core or backbone structure with covalently attached glycosaminoglycans. Proteoglycans, collagen, and water are the main constituents of human cartilage. The collagen fibers provide tensile strength while the negative charges of the sulfates on proteoglycans provide repulsive ionic forces that provide the compressive strength of cartilage at the molecular level. In osteoarthritis, the breakdown of articular cartilage caused by chronic repetitive use and inflammation results in the decreased synthesis of proteoglycans by the chondrocytes. The loss of this component of cartilage is reflected at the macroscopic level by cartilage degradation and bone-on-bone contact in osteoarthritic degenerative joint disease. On arthrocentesis, synovial fluid may show few leukocytes, but will otherwise be without evidence of acute inflammation or infection. Incorrect Answers: A, B, C, and E. Catabolic me ase (Choice A) is a type of enzyme known as a matrix metalloproteinase, which is a zinc-containing peptidase that degrades components of extracellular matrix plus cell surface proteins and receptors. They are involved in tissue remodeling, and have demonstrated roles in the pathology of cirrhosis, arthritis, and tumor transformation. In the case of arthritis, their concentration would either be unaffected, or hypothetically increased because of the upregulation in the breakdown of articular cartilage. Interleukin-1 (IL-1) (Choice B) is also known as osteoclast activating factor. IL-1 leads to an increase in RANK ligand signaling and subsequent osteoclast-mediated bone resorption. Osteoporosis is characterized by an increase in osteoclast number and activity, which is driven by IL-1. Osteoarthritis also results from chronic inflammation; IL-1 may be increased in this setting as well. Interleukin-6 (IL-6) (Choice C) is an inflammatory cytokine produced by macrophages and monocytes that triggers acute phase reactant production in the liver, for example, C-reactive protein. IL-6 is also a pyrogen. Its production in a joint would more commonly be expected in cases of rheumatoid arthritis or septic arthritis, however osteoarthritis also may demonstrate components of chronic inflammation, and concentrations of IL-6 may also be increased in this case. Tumor necrosis factor (TNF) (Choice E) is key in the pathophysiology of rheumatoid arthritis. TNF is a potent activator of macrophages and neutrophils, enhancing their cytotoxic effects and expression of endothelial adhesion molecules to promote migration into peripheral tissues. Because of this central role in rheumatoid arthritis, it is a key target of disease modifying agents such as infliximab and etanercept. Since osteoarthritis is also associated with components of chronic inflammation, TNF concentrations may also be increased in such cases.

Objective: Proteoglycans comprise the extracellular matrix and are a key element of articular cartilage. In disease states where cartilage breaks down, such as osteoarthritis, decreased proteoglycan secretion would be expected. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

171 Exam Section 4: Item 21 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 21. A 50-year-old man is admitted to the hospital after being diagnosed with a pulmonary embolus. Treatment is started with intravenous heparin. Twenty-four hours later, warfarin is added. On day 2, his partial thromboplastin time is 52 seconds (control 26 sec), and prothrombin time is 12 seconds (control 12.1 sec; INR=1). Which of the following is the best explanation for the normal prothrombin time and INR measurements in this patient? O A) Heparin-warfarin interaction B) Hereditary resistance to heparin O C) Hereditary resistance to warfarin D) Long half-life of factor II (prothrombin) E) Too low a dose of heparin F) Too low a dose of warfarin G) Undetected liver disease Correct Answer: D. Warfarin inhibits the hepatic synthesis of factor II (prothrombin), VII, IX, and X, in addition to proteins C and S. Factor Il exhibits a long half-life of nearly 60 hours. For the effect of warfarin to be shown in an increased prothrombin time (PT)INR, circulating prothrombin must first degrade. For this reason, the time from the initiation of warfarin to therapeutic anticoagulation takes several days, and patients are not adequately anticoagulated until the PT/INR has been at the goal concentration for at least 48 hours. This explains why the patient's PT/INR remains unchanged 24 hours after starting warfarin. The coagulation cascade consists of the intrinsic, extrinsic, and common pathways. The intrinsic pathway includes the activation of factors XII, XI, IX, and VIII, and activity of this pathway is measured by the partial thromboplastin time (PTT). The extrinsic pathway involves the activation of factor VII and is measured by the PT/INR; the effect of warfarin is measured here. The intrinsic and extrinsic pathways trigger the common pathway via the activation of factor X. Activated factor X converts prothrombin to thrombin, which converts fibrinogen to fibrin. Fibrin is an essential component of clot formation, polymerizing to form fibrin strands and achieve hemostasis. Incorrect Answers: A, B, C, E, F, and G. Heparin and warfarin interaction (Choice A) together will predispose the patient to bleeding, but there is otherwise no significant pharmacologic interaction between the two. Hereditary resistance to heparin (Choice B) may be because of low baseline concentrations of or mutations involving antithrombin III, high concentrations of factor VIII and fibrinogen, or accelerated clearance of heparin. Such a case would be reflected by a low or normal PT in the setting of standard heparin dosing. Hereditary resistance to warfarin (Choice C) is rare. More commonly, suboptimal anticoagulation with warfarin is a result of underdosing, use of medications that decrease drug levels, inconsistent diet (excess consumption of foods rich in vitamin K), or patient nonadherence. Too low a dose of heparin (Choice E) would be reflected by a subtherapeutic PTT. Too low a dose of warfarin (Choice F) would be reflected by a subtherapeutic PT/INR tested after 48 to 72 hours to reflect the steady-state effect on clotting factors. This patient's PT/INR remained normal because insufficient time had passed for warfarin to achieve therapeutic effect. Undetected liver disease (Choice G) has myriad effects on the clotting cascade. Patients with cirrhosis often exhibit an increased, not a normal, PT/INR because of ineffective synthesis of Vitamin K dependent clotting factors. This patient has no features to suggest liver disease (eg, encephalopathy, jaundice, ascites, varices, spider angiomata).

Objective: Prothrombin (factor II) has a long half-life. While warfarin reduces hepatic synthesis of factors II, VII, IX, and X, the anticoagulant effect does not manifest until circulating prothrombin has been depleted. This explains why the patient's PT/INR remains unchanged 24 hours after starting warfarin. I3D Previous Next Score Report Lab Values Calculator Help Pause

46 Exam Section 1: Item 46 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 46. A 1-month-old male newborn is brought to the emergency department by his mother because of a 5-day history of vomiting after feedings. His mother says the vomiting is forceful, and appears to contain only formula. She says he seems eager to feed after each episode. He appears irritable. He is at the 25th percentile for length and weight. His temperature is 37.6°C (99.7°F), pulse is 140/min, respirations are 24/min, and blood pressure is 80/40 mm Hg. Physical examination shows decreased skin turgor. Which of the following sets of serum findings (in mEq/L) is most likely in this newborn at this time? ITTT Na* (N=134-146) K* (N=3-6) CI- (N=95-115) HCO,- (N=20-28) A) 132 3.2 90 37 B) 132 3.2 100 25 C) 132 4.9 90 35 D) 135 3.5 110 14 E) 138 4.6 90 16 F) 138 4.6 110 23 Correct Answer: A. Pyloric stenosis is caused by hypertrophy of the pyloric sphincter, leading to gastric outlet obstruction and vomiting. It commonly presents with nonbloody, nonbilious emesis after feedings beginning around 3 to 6 weeks of age. It is more common in male infants. Symptoms may begin with occasional vomiting, then progress to projectile emesis after every feed, dehydration, and malnutrition as the stenosis worsens. Repeated bouts of vomiting can cause metabolic and electrolyte derangements, classically hypochloremic, hypokalemic metabolic alkalosis caused by losses of hydrochloric acid, a major component of stomach secretions. Metabolic alkalosis occurs secondary to dehydration and subsequent volume contraction plus relative excess of bicarbonate. Volume depletion also leads to the activation of the renin-angiotensin-aldosterone system, which in turn stimulates renal tubule bicarbonate reabsorption and new bicarbonate generation. Metabolic alkalosis, volume depletion, and increased concentration of aldosterone also leads to renal potassium excretion and hypokalemia. Hypokalemia also contributes to metabolic alkalosis via transcellular exchange of extracellular sodium and protons for potassium ions. Diagnosis of pyloric stenosis is typically by abdominal ultrasonography, which demonstrates thickened pyloric musculature, and treatment requires surgical pyloromyotomy. Incorrect Answers: B, C, D, E, and F. Choices B, D, and F reflect a chloride concentration within the reference ranges. Persistent vomiting results in dehydration, volume contraction, and a hypochloremic, hypokalemic metabolic acidosis. Choices C, E, and F reflect a potassium concentration within the reference ranges. Metabolic alkalosis and increased aldosterone concentration secondary to volume depletion result in a decreased serum potassium concentration. Choices D and E also reflect metabolic acidosis or respiratory alkalosis, shown by a decreased concentration of bicarbonate. These derangements are not the primary deficits observed in pyloric stenosis.

Objective: Pyloric stenosis is caused by hypertrophy of the pyloric sphincter, resulting in gastric outlet obstruction, which typically first presents at around 3 to 6 weeks of life. It is characterized by repeated vomiting after feeds, leading to dehydration and a hypochloremic, hypokalemic metabolic alkalosis. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

72 Exam Section 2: Item 22 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 22. A70-year-old man comes to the physician with his wife because of a 6-month history of abnormal behavior during sleep. His wife says, "He suddenly punches at the air and yells as if he were acting out his dreams, but he doesn't wake up." As a result of his agitation, he has fallen on the floor several times. Neurologic examination shows no abnormalities. Which of the following additional findings is most likely in this patient? A) Absent muscle atonia during REM sleep B) Anterior cingulate mass lesion O C) Evidence of major depressive disorder with psychotic features D) Left temporal lobe spikes on EEG E) Nocturnal hypoglycemia Correct Answer: A. Absent muscle atonia during REM sleep is abnormal and a sign of REM sleep behavior disorder (RBD). In typical REM sleep, several neural circuits terminate on spinal cord motor neurons to cause sleep atonia. When this muscle atonia is lost, patients can violently act out their dreams. They typically remember the dreams. REM sleep behavior disorder typically occurs in adult males and can be idiopathic or related to underlying alpha-synuclein degeneration (eg, Parkinson disease, multiple system atrophy, or Lewy body dementia). In the absence of neurologic abnormalities on physical examination, this patient's REM sleep behavior disorder is likely idiopathic and therefore does not warrant further workup for a neurodegenerative disorder. Polysomnography showing a lack of atonia during REM sleep confirms the diagnosis. Treatment includes creating a safe sleep environment and, if the behavior is severe, initiating melatonin or clonazepam. Patients with idiopathic REM sleep behavior disorder should be closely monitored because of the high conversion rate to neurodegenerative disease. Incorrect Answers: B, C, D, and E. An anterior cingulate mass lesion (Choice B) would disrupt functions of the cingulate gyrus, which include decision making, social interactions, empathic responses, and autonomic motor efferent signals. Spontaneous arm movements and vocalizations as in this patient are atypical of an anterior cingulate mass lesion. Evidence of major depressive disorder with psychotic features (Choice C) would include weeks of depressed mood, sleep changes, decreased appetite, decreased energy, and psychotic symptoms such as delusions of guilt. This patient's strange behavior during the nighttime only without consistent mood or psychotic symptoms is inconsistent with major depressive disorder with psychotic features. Left temporal lobe spikes on EEG (Choice D) would suggest a seizure disorder with a seizure focus in the left temporal lobe. A focal left temporal lobe seizure would not generally produce motor or verbal symptoms. If the temporal lobe seizure generalized secondarily, motor and verbal symptoms would be rhythmic and uncoordinated, as opposed to this patient's coordinated acting out of his dreams. Nocturnal hypoglycemia (Choice E) can cause seizures. Motor and verbal symptoms would be rhythmic and uncoordinated, as opposed to this patient's coordinated acting out of his dreams.

Objective: REM sleep behavior disorder features absent muscle atonia during REM sleep. Patients typically act out their dreams in a coordinated, though sometimes violent, manner. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

178 Exam Section 4: Item 28 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 28. A 25-year-old woman comes to the physician because of a 3-month history of difficulty falling and remaining asleep. She says that she has been under stress and anxious during this period. Physical examination shows no abnormalities. Nocturnal polysomnography shows normal sleep architecture, including normal volumes and latencies of all non-REM and REM sleep stages. There is no evidence of sleep-disordered breathing or other sleep-fragmenting disorder. Compared with this patient's non- REM stage 3 to 4 (deep) sleep, which of the following is most likely to be increased in her REM sleep? A) Cerebral blood flow B) Delta wave EEG activity C) Intercostal muscle activity D) Serum growth hormone concentrations E) Thermoregulation Correct Answer: A. Cerebral blood flow is typically increased in REM sleep compared to non-REM stage 3 to 4 sleep (deep sleep). Cerebral blood flow during REM sleep is like that occurring during wakefulness, whereas the cerebral blood flow during stages 3 and 4 is significantly reduced compared to the awake state. Cerebral blood flow reflects total synaptic activity. Increased cerebral blood flow therefore reflects the dream state specific to REM sleep. This blood flow is especially increased in the visual association areas, which is thought to allow for complex visual processing during dreams. This patient likely has psychophysiological insomnia, or insomnia related to anxiety, which demonstrates normal sleep architecture. Incorrect Answers: B, C, D, and E. Delta wave EEG activity (Choice B), waves of the lowest frequency and highest amplitude on polysomnography, are typical of stages 3 and 4 (also known as "slow-wave sleep"). REM sleep is characterized by beta waves. Intercostal muscle activity (Choice C) is decreased in REM sleep, which may be related to atonia of the intercostal muscles. Respiratory parameters such as tidal volume demonstrate increased variability during REM sleep compared to other stages. Serum growth hormone concentrations (Choice D) increase the most steeply during the deep sleep stages. This variation in growth hormone concentration is related to altered hypothalamic activity during the different stages of sleep. Thermoregulation (Choice E) diminishes during REM sleep as a result of reduced temperature sensitivity of the hypothalamic preoptic nucleus during this stage. In contrast, non-REM sleep is associated with vasodilation, leading to the cooling of body temperature. One postulated function of deep, non-REM sleep is to cool the body and brain temperature and therefore conserve energy.

Objective: REM sleep is characterized by increased total brain synaptic activity. Therefore, cerebral blood flow increases during REM sleep. Previous Next Score Report Lab Values Calculator Help Pause

24 Exam Section 1: Item 24 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 24. A 52-year-old woman with breast cancer comes to the physician for a follow-up examination. She had a 4-week course of radiation treatment 6 months ago. Her respirations are 26/min. Physical examination shows no recurrence of the cancer. A CT scan of the chest shows bilateral patches of atelectasis in the upper lung fields. The atelectasis in this patient most likely developed because of which of the following primary pathophysiologic processes? O A) Compression B) Consolidation C) Contraction D) Obstruction E) Resorption Correct Answer: C. Atelectasis describes the collapse and deflation of pulmonary alveoli, resulting in impaired gas exchange. There are a variety of causes of atelectasis. In this case, this patient's findings of tachypnea and patchy atelectasis on CT scan following radiotherapy are likely sequelae of radiation pneumonitis. Radiation pneumonitis typically presents in temporal association with radiotherapy with signs and symptoms including malaise, dyspnea, nonproductive cough, and pleuritic chest pain, with examination showing tachypnea, pulmonary crackles, or a pleural rub. lonizing radiation damages pulmonary tissues through the generation of free radicals and DNA strand breaks, leading to a local response characterized by inflammation, deposition of hyaline membranes, and activation of fibroblasts. The late stages of radiation pneumonitis are characterized by fibrosis and contraction of lung tissue. Fibrotic pulmonary tissue displays increased elastic recoil and reduced compliance. These properties render the affected lung vulnerable to reduced alveolar expansion, accumulation of pulmonary secretions, and resultant atelectasis. Some chemotherapeutic agents such as doxorubicin and bleomycin act as radiosensitizing agents and their concurrent use is a risk factor for radiation pneumonitis. Treatment for radiation pneumonitis is primarily supportive; pulmonary physiotherapy with lung-expanding exercises such as incentive spirometry can reduce the degree of atelectasis. Incorrect Answers: A, B, D, and E. Compression atelectasis (Choice A) can result from the intrapleural accumulation of fluid or air, such as in the setting of a pleural effusion, hemothorax, or pneumothorax. Affected areas of the lung can develop atelectasis as a result of direct compression of alveoli, and when occurring, such areas of atelectasis are often directly adjacent to the compressive lesion. Compression atelectasis is unlikely since the patient's CT scan did not show a source of compression. Consolidation (Choice B) occurs when the alveoli become filled with material. This contrasts with atelectasis, where the alveoli are collapsed. Consolidation commonly occurs in the setting of pneumonia, pulmonary edema, pulmonary hemorrhage, or aspiration. Consolidation can be readily identified by chest x-ray or CT scan. Obstruction (Choice D) of pulmonary airways is a common cause of atelectasis and may be caused by bronchial tumors, mucus plugging, or inhaled foreign body. The absence of local alveolar expansion as a result of airway obstruction leads to distal airway collapse and atelectasis following resorption (Choice E) of residual alveolar air. While radiation pneumonitis may lead to airway collapse, this occurs primarily through fibrosis and contraction of the pulmonary tissue.

Objective: Radiation pneumonitis is a potential sequela of radiotherapy for the treatment of malignancy. Fibrosis and contraction of lung tissue are characteristic of the later stages of radiation pneumonitis and can produce alveolar collapse and impaired gas exchange, a phenomenon known as atelectasis. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

156 Exam Section 4: Item 6 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment Please Wait I 1 Marker alleles 1,2 1,1 II 1 2 3 Marker alleles 1,2 1,1 1,1 Affected female Affected male Pregnancy, sex unspecified Unaffected female I Unaffected male 6. A 25-year-old woman who is at 12 weeks' gestation asks for prenatal counseling. Her husband has a clinically variable autosomal dominant disorder with 100% penetrance that is evident within the first year of life and for which linkage analysis is available. They have two unaffected daughters 3 and 5 years old. A pedigree is shown. A single polymorphic marker has a recombination fraction of 0.1 (10%) and is informative. Alleles 1 and 2 for this marker are indicated in the pedigree. Which of the following is the risk that this fetus will have this disorder? A) 1% B) 10% C) 50% D) 90% E) 99% Correct Answer: D. The chance that the fetus will have the disorder is 90%. The genotype of the three unaffected individuals in the pedigree compared to that of the affected individual indicates that allele 2 is the polymorphic marker associated with the disease trait. Allele 2 is not the disease trait itself but is geographically localized near the disease trait on the chromosome. Recombination takes place in prophase I of meiosis and is the process by which two chromatids exchange a small portion of genetic material and separate, or unlink, two traits previously present on the same chromatid. The likelihood that the DNA will recombine between the two sites and separate allele 2 and the disease trait is 0.1 (10%), the recombination fraction. The likelihood that it will not recombine is 0.9 (90%), one minus the recombination fraction. Thus, there is a 90% likelihood that the fetus will inherit the disease trait along with allele 2. Because the disease has 100% penetrance, an individual with the genotype of disease will demonstrate the phenotype of the disease in all cases. However, because it is clinically variable, not all individuals with the disease will have the same manifestations. Incorrect Answers: A, B, C, and E. 1% (Choice A) and 99% (Choice E) would be the risks of not inheriting the disease (1%) or inheriting the disease (99%) if the recombination fraction was 0.01 (1%) and the fetus inherited allele 2. 10% (Choice B) is the risk for a recombination event that unlinks the informative marker from the disease trait. This is the likelihood that the fetus will not have the disorder because it has inherited allele 2. 50% (Choice C) is the likelihood of a fetus inheriting a disease trait from a heterozygous parent. In an autosomal dominant disorder with complete penetrance, this corresponds to a 50% likelihood of the fetus having the disease. However, because the marker alleles of the fetus are known in this case, the likelihood of disease is now based solely on the likelihood that recombination does not occur.

Objective: Recombination takes place in prophase I of meiosis and is the process by which two alleles closely located on the same chromatid can be separated. If the genotype at only one of these sites is known, linkage analysis allows for the prediction of the likelihood of inheriting a given allele at the other site. %3D Previous Next Score Report Lab Values Calculator Help Pause

144 Exam Section 3: Item 44 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 44. A married couple and their 12-year-old son come to the emergency department because of a 2-hour history of nausea, vomiting, and diarrhea. They all ate lunch at a company picnic 6 hours ago. The physician suspects staphylococcal food poisoning and notifies the local health department. The health department contacts other picnic attendees to collect data on which foods they consumed at the event. Based on the investigation, ice cream is suspected as the source of the toxin. Among those who ate the ice cream, 96 became ill and 24 did not experience symptoms. Among those who did not eat the ice cream, 3 became ill and 27 did not experience symptoms. Which of the following is the relative risk for illness in those who ate the ice cream compared with those who did not? O A) 0.125 B) 0.7 C) 8 D) 32 E) 36 Correct Answer: C. Relative risk (RR) describes the difference in likelihood of the occurrence of a disease outcome between two groups of patients with or without a particular exposure. Calculations of relative risk are commonly performed in cohort studies. RR is calculated by dividing the fraction of patients with a positive exposure and who developed disease (a) amongst all patients who were exposed which includes those exposed who did not develop disease (b), (a + b), by the fraction of patients with a negative exposure and who developed disease (c) amongst all patients who were not exposed which includes those who did not develop disease (d), (c + d). RR thus equals (a / (a + b) / (c / (c + d)). In this case, the RR is calculated as follows: RR = (96 / (96 + 24)) / (3/ (3 + 27)) = 0.80 /0.10 = 8. RR values greater than 1.0 indicate an increased risk for developing disease in association with the exposure, whereas values less than 1.0 indicate a reduced risk for developing disease, and RR equal to 1.0 indicates that the disease outcome and the exposure are not related. Incorrect Answers: A, B, D, and E. 0.125 (Choice A), calculates an inverse of the relative risk. 0.7 (Choice B), calculates the attributable risk (AR). AR describes the portion of the overall risk for developing disease which can be attributed to the exposure as compared to the risk that exists without exposure. AR is defined as AR = (a / (a + b)) - (c / (c + d)). In this scenario, AR = (96 / (96 + 24)) - (3 / (3 + 27)) = 0.8 - 0.1 = 0.7. 32 (Choice D), calculates the ratio of the ill patients who ate ice cream to the ill patients who did not eat ice cream, 96:3. Calculations of relative risk must consider the entire sample of patients, not only those who developed illness. 36 (Choice E), calculates the odds ratio (OR). OR is defined as (a x d) / (b x c). OR is commonly employed in case-control studies. In this case, OR = (96 x 27) / (24 x 3) = 2592 / 72 = 36. %3D

Objective: Relative risk (RR) describes the risk for developing disease in a group with an exposure divided by the risk for developing disease in a group without the exposure. RR is calculated as (a / (a + b)) / (c / (c + d)). Previous Next Score Report Lab Values Calculator Help Pause

20 Exam Section 1: Item 20 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 20. An 85-year-old man comes to the physician because of a 4-month history of headaches, shortness of breath, and leg swelling. He has a 15-year history of well-controlled hypertension. His pulse is 80/min, respirations are 18/min, and blood pressure is 210/110 mm Hg. Crackles are heard at the lung bases. There is a bruit in the left flank and edema in both lower extremities. Urinalysis shows: Specific gravity Blood Protein RBC 1.015 (N=1.001-1.038) trace 30 mg/24 h 6/hpf 0/hpf negative WBC Bacteria He dies the next day. Which of the following findings is most likely on autopsy? A) Chronic interstitial nephritis B) Ostial stenosis of one renal artery C) Scattered nodular aneurysms in the renal arteries D) Solitary left kidney Correct Answer: B. Renal artery stenosis is a cause of secondary hypertension because of the abnormal stimulation of the juxtaglomerular apparatus from low afferent blood flow leading to excessive production of renin and angiotensin. Stenosis of the renal artery leads to decreased afferent blood flow and ischemic nephropathy of the affected kidney, which can result in progressive kidney atrophy. There are two main causes of renal artery stenosis, fibromuscular dysplasia and atherosclerosis. In younger patients, fibromuscular dysplasia is the most common cause, while older patients develop renal artery stenosis more commonly from atherosclerosis. On physical examination, a bruit may be auscultated on the side of the renal artery stenosis. Urinalysis is typically within reference ranges. Diagnosis is established with renal artery Doppler ultrasonography or MR angiography. Treatment involves angioplasty or stenting of the stenosed renal artery to improve flow. ACE inhibitors can be considered for unilateral stenosis but can lead to acute renal failure in the setting of bilateral renal artery stenosis. Incorrect Answers: A, C, and D. Chronic interstitial nephritis (Choice A) is a chronic inflammatory disease of the kidneys that can originate from a broad range of causes, most commonly from medication reactions (eg, antibiotics, analgesics), and systemic inflammatory disorders such as systemic lupus erythematosus, Sjogren syndrome, or sarcoidosis. It is often asymptomatic, presenting with gradual renal function decline. Urinalysis may disclose hematuria, sterile pyuria, or mild to moderate proteinuria. A lymphoplasmacytic infiltrate is present on histology. Scattered nodular aneurysms in the renal arteries (Choice C) are found in fibromuscular dysplasia. While fibromuscular dysplasia is a potential cause of renal artery stenosis in young patients, atherosclerosis of the renal artery is the most likely cause in this older patient. Solitary left kidney (Choice D) is a common congenital abnormality in which only a single kidney is formed at birth secondary to renal agenesis of the contralateral kidney. Alternatively, both kidneys may form but only one is functional (renal dysplasia), or a kidney may be lost to trauma or surgical resection (eg, kidney lacerations, malignancy, or donation). In general, people with a solitary kidney demonstrate normal overall renal function but are more susceptible to develop chronic kidney disease.

Objective: Renal artery stenosis is characterized by refractory hypertension and progressive renal atrophy. Fibromuscular dysplasia is the most likely cause in younger patients, while atherosclerosis is more common in older patients. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

7 Exam Section 1: Item 7 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 7. A 66-year-old man has a surgical excision of a brain mass; a photomicrograph of excised tissue is shown. He has smoked 1 pack of cigarettes daily for the past 50 years. His blood pressure is 130/90 mm Hg. Laboratory studies show polycythemia, hypercalcemia, and microscopic hematuria. Immunostaining of the tissue is positive for epithelial membrane antigen and negative for carcinoembryonic antigen. Which of the following is the most likely site of the primary neoplasm? O A) Colon B) Kidney C) Liver D) Lung E) Lymph node O F) Prostate G) Testis Correct Answer: B. Renal cell carcinoma (RCC) is an adenocarcinoma of tubular epithelial cells. RCC is the most common primary malignancy of the kidney and most commonly occurs in older, male smokers. It can present with gross or microscopic hematuria, flank pain, weight loss, or fever. Laboratory analysis may show polycythemia or hypercalcemia as a result of associated paraneoplastic syndrome production of erythropoietin or parathyroid hormone-related peptide. Hypercalcemia can also result from bony metastasis. Diagnosis of RCC typically occurs with contrast-enhanced CT scan or MRI and is confirmed by biopsy at the time of nephrectomy. Biopsy of RCC is uniquely characterized by polygonal clear cells, because of the accumulation of lipid and carbohydrate content in the cells. It spreads hematogenously and commonly presents as a metastatic neoplasm. The brain and lung are frequent sites of metastasis. Incorrect Answers: A, C, D, E, F, and G. Colon (Choice A) and pancreatic cancers often produce carcinoembryonic antigen (CEA) as a nonspecific marker. This patient's tumor was negative for CEA, making colon cancer less likely. Liver (Choice C) cancer is most often hepatocellular carcinoma and is associated with chronic hepatitis B and cirrhosis. It is characterized by weight loss, jaundice, hepatomegaly, and ascites. a-fetoprotein (AFP) is frequently increased. Lung (Choice D) cancer is a known consequence of smoking, however lung cancer generally presents nonspecifically with weight loss or symptoms related to the site of metastasis, or with cough, hemoptysis, wheezing, recurrent pneumonia, or bronchial obstruction in severe disease. Lymph node (Choice E) malignancies (eg, lymphoma) most frequently present with fever, night sweats, and weight loss, potentially with palpable lymphadenopathy. Prostate (Choice F) cancer is diagnosed by a biopsy of the prostate; increased prostate-specific antigen (PSA) is used as a tumor marker. Late stages of disease may metastasize to bone, especially the lumbar vertebrae, causing back pain. Testicular cancer (Choice G) most often occurs in younger men as an isolated testicular mass; it is generally marked by an increased concentration of AFP or human chorionic gonadotropin. Metastasis to the brain is uncommon.

Objective: Renal cell carcinoma often presents in older men with a history of cigarette smoking, and the brain is a potential site of metastasis. Laboratory analysis may show hematuria along with hypercalcemia or polycythemia secondary to paraneoplastic syndromes. Histology typically demonstrates polygonal clear cells related to the accumulation of lipid and carbohydrate content in the cells. ID Previous Next Score Report Lab Values Calculator Help Pause

68 Exam Section 2: Item 18 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 18. A 58-year-old woman is brought to the emergency department because of a 2-hour history of shortness of breath and chest pain that radiates to her back between the shoulder blades. Her respirations are 28/min. Physical examination shows diaphoresis. An ECG shows no abnormalities. Coronary angiography shows occlusion of the marginal branch of the left anterior descending coronary artery. A percutaneous coronary revascularization is done. Subsequent to the stent's placement, her serum concentrations of myocardial creatine kinase (CK-MB) and troponin I are increased. Which of the following mechanisms best explains these laboratory findings? O A) Cell shrinkage B) Formation of apoptotic bodies C) Liquefactive necrosis of the myocardium D) Membrane lipid peroxidation E) Protease inactivation by cytoplasmic free calcium ions Correct Answer: D. This patient presents with a cardiac reperfusion injury subsequent to percutaneous coronary revascularization. Reperfusion injury can occur in any organ following restoration of blood flow after a period of ischemia. Restoration of perfusion leads to a rapid increase in the production of free radicals. In turn, free radicals damage cells through membrane lipid peroxidation, oxidative damage to intracellular proteins, and DNA strand breaks. Reperfusion injury typically occurs within hours of revascularization. The damage to the membranes and intracellular proteins results in cellular necrosis and lysis. Grossly, this is described as a red infarct. Serum studies will demonstrate increased concentrations of normally intracellular myocardial proteins, including troponin I and CK-MB. Incorrect Answers: A, B, C, and E. Cell shrinkage (Choice A) and formation of apoptotic bodies (Choice B) are seen in apoptosis or programmed cell death. It is not an inflammatory process, and intracytoplasmic contents are generally not released to interstitial spaces and serum. In contrast, myocardial infarction produces coagulative necrosis most pronounced between 1 and 3 days after infarction. Histologic features of coagulative necrosis include the loss of myocardial nuclei and cytoplasmic hypereosinophilia. Liquefactive necrosis (Choice C) is most commonly associated with bacterial or fungal abscesses and brain infarcts. It is not generally seen following myocardial infarction, which is characterized by coagulative necrosis. Altered membrane permeability as a result of oxidative damage leads to changes in calcium flux and a rise in intracellular calcium. However, this intracellular calcium flux leads to activation, rather than inactivation (Choice E) of intracellular calcium-dependent proteases such as calpain. Once activated by calcium, such proteases function to potentiate myocyte injury during ischemia and reperfusion.

Objective: Reperfusion injury occurs when tissue oxygenation is restored following a period ischemia. This occurs because of the generation of free oxygen radicals and peroxides which result in lipid peroxidation, protein modification, and DNA strand breaks. Previous Next Score Report Lab Values Calculator Help Pause

13 Exam Section 1: Item 13 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 13. A 45-year-old woman has had progressive fatigue and mild shortness of breath with exertion during the past 3 months. She thinks she had rheumatic fever at the age of 10 years. A grade 2/6, rumbling, diastolic murmur is heard at the fifth left intercostal space in the midclavicular line. A lateral x-ray of the chest shows mild posterior displacement of the esophagus. Which of the following cardiovascular structures is most likely responsible for the esophageal displacement? O A) Descending aorta B) Left atrium C) Left ventricle D) Right atrium E) Right ventricle Correct Answer: B. In the typical anatomic orientation, the left atrium is the most posterior part of the heart and lies anterior to the esophagus in the mediastinum. In this patient, the left atrium is most likely enlarged secondary to mitral valve stenosis in the setting of prior rheumatic fever. Rheumatic fever, a complication of untreated streptococcal pharyngitis, can lead to endocardial valve inflammation and damage. The mitral valve is most affected. Acute valve inflammation typically causes valvular insufficiency (regurgitation), while late sequelae can result in fibrosis of the valve and stenosis. A rumbling, diastolic murmur best heard at the fifth left intercostal space is characteristic of mitral stenosis. Left atrial enlargement, which can displace the esophagus posteriorly, occurs secondary to impaired left atrial emptying in the setting of the chronic mitral valve disease. Incorrect Answers: A, C, D, and E. The descending aorta (Choice A) is located lateral to the esophagus and is not typically affected by rheumatic fever. The left ventricle (Choice C) is located inferior and lateral to the left atrium. Left ventricular hypertrophy may occur in the setting of aortic stenosis, which can also be a chronic sequela of rheumatic fever, although this would be unlikely to displace the esophagus posteriorly. The right atrium (Choice D) commonly enlarges in the setting of tricuspid regurgitation. The tricuspid valve is less commonly affected than the mitral or aortic valves in rheumatic fever. As well, enlargement of the right atrium would not displace the esophagus posteriorly. The right ventricle (Choice E) is the most anterior chamber of the heart. Right ventricular hypertrophy may occur in the setting of pulmonic stenosis, which is seen in tetralogy of Fallot. As well, right ventricular dilation is a potential finding in the setting of right heart strain from a large pulmonary embolus. Neither of these are related to prior rheumatic fever and would not displace the esophagus posteriorly.

Objective: Rheumatic fever commonly affects the mitral valve, potentially leading to mitral stenosis later in life and subsequent left atrial enlargement. The left atrium and esophagus have a close anatomic relationship and left atrial enlargement can displace the esophagus posteriorly. Previous Next Score Report Lab Values Calculator Help Pause

135 Exam Section 3: Item 35 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 35. A 40-year-old woman comes to the physician because of pain of her hands, wrists, and knees for 2 weeks. Physical examination shows swelling, tenderness, and warmth of the proximal interphalangeal, wrist, and knee joints. Laboratory studies show an increased erythrocyte sedimentation rate and normochromic, normocytic anemia. A test result for rheumatoid factor is positive. Which of the following sets of findings in synovial fluid obtained from the knee joint is most likely in this patient? Complement (C3) Segmented Neutrophils Interleukin-1 (IL-1) Tumor Necrosis Factor O A) ↑ B) ↑ ↑ C) ↑ ↑ D) ↑ ↑ ↑ E) ↑ O F) ↑ Correct Answer: D. Rheumatoid arthritis is an autoimmune disease that causes extensive inflammation in synovial tissues. The clinical course is characterized by joint destruction, periarticular bone loss, and synovial hypertrophy, along with extra-articular manifestations such as rheumatoid nodules and skin rash. Patients often present with symmetric patterns of arthritis particularly in the wrists and hands in middle-aged women. Serologic studies generally demonstrate a positive rheumatoid factor, and anti-cyclic citrullinated peptide antibodies. Rheumatoid arthritis is associated with the HLA-DR4 genotype. During an immune response, there are several changes in the homeostasis of cytokines and chemokines. In rheumatoid arthritis, both the classic and alternative pathways of complement are activated, along with a broad cellular and humoral immune response in the synovial space. Autoantibodies and immune complexes contribute to complement activation, the consumption of which decreases the serum and synovial concentrations of it. Infiltration of immune cells into the synovial space including macrophages, neutrophils, and T lymphocytes, results in the production of cytokines and acute phase reactants within the joint space such as interleukins and tumor necrosis factor (TNF). TNF is key in the pathophysiology of rheumatoid arthritis. TNF is a potent activator of macrophages and neutrophils, enhancing their cytotoxic effects and expression of endothelial adhesion molecules to promote migration into peripheral tissues. Because of this central role in rheumatoid arthritis, it is a key target of disease modifying agents such as infliximab and etanercept. Interleukin-1 (IL-1) is a potent immune regulator, also known as osteoclast activating factor; it promotes bone reabsorption and cartilage destruction in rheumatoid arthritis. Incorrect Answers: A, B, C, E, and F. In rheumatoid arthritis, complement concentrations in the synovial space would decrease because of consumption, making Choices A, B, and C incorrect. As a result of the inflammatory response, concentrations of TNF within the synovium would be increased, making Choices B, E and F incorrect. Neutrophils would be expected within the joint space, making choices B, C, and F incorrect. The production of IL-1 would be increased, making Choices A, C, and E incorrect.

Objective: Rheumatoid arthritis is a common autoimmune disease characterized by joint destruction, pain, and upregulation of immunomodulatory cytokines. In rheumatoid arthritis and similar inflammatory states, inflammatory cells, interleukins, and tumor necrosis factors are increased. In this setting, complement activation results in the consumption of complement proteins, resulting in decreased concentrations within the synovial fluid. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

131 Exam Section 3: Item 31 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 31. A 35-year-old man is brought to the physician by staff of the group home where he resides because of worsening behavior for 2 weeks. During the interview, he tells the physician that he believes the CIA is spying on him through his television set. He then reports hearing voices in the hall outside the examination room and states that the CIA now plans to kill him. He appears disheveled with unkempt hair and poor hygiene. He has difficulty answering the physician's questions because he seems to be listening to internal stimuli. Mental status examination of the patient is most likely to show which of the following findings? A) Flattened affect B) Inability to state his name C) Inability to write his name D) Lack of orientation to place or time E) Long-term memory impairment Correct Answer: A. Patients with chronic psychotic disorders such as schizophrenia or schizoaffective disorder frequently have flattened affect, or a decreased range of affect. According to the DSM-5, patients with schizophrenia have two clusters of symptoms: positive symptoms (addition of mental dysfunction including delusions, hallucinations, and disorganized thinking and behavior) and negative symptoms (absence of certain mental functions such as affect, volition, and speech). Given that the patient lives in a group home, he likely has a disorder that chronically impairs his function, which makes a chronic, primary psychotic disorder such as schizophrenia more likely than a substance-induced psychotic disorder (which would likely present with positive symptoms but not negative symptoms). This patient demonstrates positive symptoms, such as delusions and disorganized behavior, and appears to also exhibit negative symptoms, such as avolition (lack of ability to perform tasks), which has likely resulted in the patient's poor hygiene and grooming. Given this likely negative symptomatology, he is also likely to demonstrate a flattened affect on mental status examination. Incorrect Answers: B, C, D, and E. Inability to state his name (Choice B) and lack of orientation to place or time (Choice D) would suggest the patient's basic cognitive functions are compromised. In patients with chronic psychotic disorders, while cognition is often impaired, basic cognitive functions such as orientation to name, place, and time are likely to be intact. Dementia or delirium are more likely to influence orientation. Inability to write his name (Choice C) would indicate impairment of the patient's gross or fine motor coordination such as apraxia. This would more commonly be seen in neurological disorders such as stroke, Huntington disease, or Parkinson disease. Long-term memory impairment (Choice E) is a less common manifestation of primary psychotic disorders than positive or negative symptoms of psychosis. While patients with schizophrenia can develop cognitive dysfunction that affects long-term memory, negative symptoms such as flattened affect are more common.

Objective: Schizophrenia is associated with both positive and negative psychotic symptoms. Negative symptoms of psychosis include lack of affect, avolition, apathy, and alogia. This can manifest as poor hygiene and a flattened affect on mental status examination. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

65 Exam Section 2: Item 15 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 15. A 45-year-old man is brought to the physician by his parents because he is reclusive, has "weird ideas," and still wants to live at home. He is dressed in several layers of clothes. During the interview, he is cooperative but he has poor eye contact and seems uninterested in the interview. He says, "I believe some people can read other people's minds. I carry these crystals to help me realize self-actualization." He has no delusions or hallucinations. This patient most likely has which of the following personality disorders? A) Borderline B) Dependent C) Histrionic D) Narcissistic E) Schizoid F) Schizotypal Correct Answer: F. Schizotypal personality disorder is one of the cluster A personality disorders, the odd or eccentric cluster. The disorder is characterized by odd behavior and thinking and a constricted affect. Patients with schizotypal personality disorder may have strange, overly metaphorical or magical thinking, though they fluctuate in the degree of their conviction about these beliefs (which differentiates these beliefs from the fixed delusions of schizophrenia). The strange beliefs may also manifest as an eccentric appearance as in this patient. Patients may experience somatosensory illusions or other abnormal perceptions. Patients with schizotypal personality disorder may appear to lack interest in cultivating relationships and, because of an inability to interpret others' motivations, may be deeply distrustful and anxious around others. Personality disorders are persistently maladaptive ways of relating to the self and to society that typically appear by early adulthood. Incorrect Answers: A, B, C, D, and E. Borderline personality disorder (Choice A) is a cluster B personality disorder, which is the dramatic or emotional cluster. Likely caused by a combination of genetics and emotional invalidation during childhood and maturation, patients with borderline personality disorder unconsciously learn to make impulsive and dramatic gestures, including self-harm, to obtain emotional fulfilment from others. Their ability to regulate their emotions is impaired; therefore, they experience rapidly changing affective states and tumultuous relationships, as opposed to this patient, who presented with a constricted and aloof affect. They do not demonstrate the strange beliefs or behavior of patients with schizotypal personality disorder. Dependent personality disorder (Choice B) is one of the cluster C personality disorders, the anxious or fearful cluster. Patients with dependent personality disorder have an excessive need to be cared for by others that manifests as severe separation anxiety and clinging behavior. This patient, however, is disinterested in relating to other people. Histrionic personality disorder (Choice C), a cluster B personality disorder, is characterized by theatrical, superficial expressions of emotion that unconsciously serve to garner attention from others to fulfill emotional needs. These patients may dress in a seductive way for the same purpose. Patients with histrionic personality disorder are emotionally excitable and overly concerned with others' reactions, as opposed to this patient's indifference. Narcissistic personality disorder (Choice D), a cluster B personality disorder, is characterized by fragile self-esteem and compensatory arrogant, self-aggrandizing behavior to gain approval, sometimes at others' expense. For example, a patient with narcissistic personality disorder may berate their physician for making a small mistake, which unconsciously inflates the patient's ego. Like the other cluster B personality disorders, people with narcissistic personality disorder are overly sensitive in social settings, in contrast to this patient's social apathy. Schizoid personality disorder (Choice E), a cluster A personality disorder, manifests as extreme social detachment and cold, restricted affect. Like patients with schizotypal personality disorder, these patients tend to be reclusive. However, patients with schizoid personality disorder typically do not demonstrate the odd beliefs and behavior or perceptual disturbances of patients with schizotypal personality disorder.

Objective: Schizotypal personality disorder manifests as an enduring pattern of strange or eccentric beliefs and behavior, social detachment or suspiciousness, and abnormal perceptual experiences such as illusions. Social indifference distinguishes schizotypal disorder from cluster B and C disorders, and eccentric beliefs and behavior differentiate the disorder from the other cluster A disorders. I3D Previous Next Score Report Lab Values Calculator Help Pause

200 Exam Section 4: Item 50 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 50. An investigator is studying the mechanisms of nerve regeneration after injury. A repetition injury is induced in a peripheral nerve in an experimental animal. It is found that sensory and motor axons are lost by Wallerian degeneration beyond the site of the injury. Further study is most likely to show which of the following cell structures acting as a path-finding guide for the regeneration of the damaged axons? A) Axon hillock B) Dendritic spine O C) Myelin sheath D) Nissl substance E) Terminal bouton Correct Answer: C. The myelin sheath is formed by Schwann cells in the peripheral nervous system. Schwann cells are arranged longitudinally along the length of an axon, each cell forming the myelin sheath. This sheath encircles and insulates axons, thereby increasing the conduction velocity of neural signals. When an axon is injured or severed, the axon and associated myelin distal to the site of injury are degraded in a process called Wallerian degeneration. Despite this degradation, the Schwann cells remain in the same orientation and position. The Schwann cells recruit macrophages to clear the cellular debris and secrete growth factors that promote axonal regeneration via a neuronal growth cone. This growth cone responds to these growth factors as it grows along the line of Schwann cells. In this way, Schwann cells and the myelin sheath they form guide axonal regeneration to the axon's appropriate destination. Incorrect Answers: A, B, D, and E. The axon hillock (Choice A) is the part of the cell body closest to the axon. The axon hillock sums excitatory and inhibitory inputs and can generate action potentials because of its high concentration of voltage-gated sodium channels. It is not known to play a role in axonal regeneration. Dendritic spines (Choice B) are small membranous protrusions from dendrites that serve as the postsynaptic membrane, binding neurotransmitter released from the presynaptic membrane. Dendrites and their spines are not involved in axonal regeneration. Nissl substance (Choice D), or Nissl body, is a neuronal organelle containing rough endoplasmic reticulum that produces neurotransmitters. In response to axonal injury, the Nissl substance disperses throughout the cell body, indicating disruption of normal neurotransmitter synthesis, as the neuron must now synthesize materials for axonal growth instead. The dysfunctional Nissl substance is not involved in guiding axonal regeneration. The terminal bouton (Choice E) is the swelling at the end of an axon that serves as the presynaptic membrane, releasing neurotransmitters to bind to neurotransmitter receptors on the postsynaptic neuronal membrane. The terminal bouton, generally distal to the site of axonal injury, plays no role in axonal regeneration.

Objective: Schwann cells form the myelin sheath around axons. After Wallerian degeneration of the axon, Schwann cells and the myelin sheath form a pathway for the axon to regenerate. Previous Next Score Report Lab Values Calculator Help Pause

176 Exam Section 4: Item 26 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 26. A 33-year-old nulligravid woman comes to the physician with her 35-year-old husband because she has not been able to conceive for 18 months. Previous evaluation of the husband showed no abnormalities. Physical and pelvic examinations show no abnormalities. Examination of an endometrial biopsy specimen shows secretory glands and a decidualized stroma. Which of the following hormonal changes underlies the maturation of this patient's endometrium? O A) Decreased follicle-stimulating hormone from the pituitary gland O B) Decreased luteinizing hormone from the pituitary gland C) Decreased production of testosterone from the adrenal cortical cells D) Increased estradiol from the follicle E) Increased B-human chorionic gonadotropin from the ovarian theca cells F) Increased progesterone from the corpus luteum Correct Answer: F. Secretory endometrium with decidualized stroma occurs during the luteal phase of the menstrual cycle, or the 14 days prior to menstruation. The first half of the menstrual cycle, the follicular phase, which varies in length, begins with menses. During menses, follicle-stimulating and luteinizing hormone (FSH and LH, respectively) concentrations increase and stimulate the developing follicle. The follicle produces estrogen, which leads to the proliferation of the endometrium in preparation of implantation of a fertilized ovum. As estrogen rises, a surge occurs, which in turn stimulates a surge in LH that causes ovulation. Immediately following ovulation, the luteal phase begins as the corpus luteum forms. The corpus luteum secretes progesterone to maintain the endometrial lining, which becomes secretory in histologic character. However, if no implantation occurs, the corpus luteum degrades to the corpus albicans, and estrogen and progesterone concentrations decrease, causing menstruation and minor increases in FSH and LH. Incorrect Answers: A, B, C, D, and E. Decreased follicle-stimulating hormone from the pituitary gland (Choice A) and decreased luteinizing hormone from the pituitary gland (Choice B) occurs following ovulation and the LH surge, which is proceeded by secretory changes in the endometrium. However, it is the increased progesterone from the corpus luteum, not decreased FSH or LH, that is responsible for the change. Decreased production of testosterone from the adrenal cortical cells (Choice C) would not adversely influence the character of the endometrium, as its controlling hormones during the menstrual cycle are primarily estrogen and progesterone. Increased estradiol from the follicle (Choice D) is responsible for the initial development of proliferative endometrium, but progesterone is responsible for the maturation of the secretory and decidual endometrium. Increased B-human chorionic gonadotropin from the ovarian theca cells (Choice E) is incorrect as B-human chorionic gonadotropin is secreted early by the syncytiotrophoblast of the implanting embryo, thereby supporting the corpus luteum.

Objective: Secretory endometrium with decidualized stroma occurs during the luteal phase of the menstrual cycle, the creation of which is governed by progesterone from the corpus luteum. Previous Next Score Report Lab Values Calculator Help Pause

47 Exam Section 1: Item 47 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 47. A 33-year-old woman with HIV infection is brought to the emergency department 30 minutes after she had a generalized tonic-clonic seizure. She also has a 2-month history of daily headaches. Physical examination shows no signs of meningismus. Muscle strength is 3/5 in the left upper extremity and 5/5 in the right upper extremity. Her CD4+ T-lymphocyte count is 22/mm3 (N2500), and plasma HIV viral load is 50,000 copies/mL. A CT scan of the head shows a 3-cm lesion in the right cerebral cortex. Serologic studies show a positive IgG antibody titer to Toxoplasma gondii. Treatment with pyrimethamine and sulfadiazine is initiated. During the next 2 weeks, she has three additional seizures. Two weeks after starting antibiotic therapy, a CT scan of the head shows that the lesion has increased to 3.5 cm. Which of the following is the most likely cause of this mass? A) Bacterial abscess B) Cerebral toxoplasmosis C) Glioblastoma D) Metastatic disease E) Non-Hodgkin lymphoma Correct Answer: E. This patient with HIV/AIDS and a solitary expanding mass in the cerebral cortex most likely has primary central nervous system (CNS) lymphoma, a variety of non-Hodgkin lymphoma and an AIDS-defining malignancy. Primary CNS lymphoma classically presents with seizures, lethargy, subacute memory loss, and headache. Physical examination may disclose neurologic deficits stemming from structural disruption as a result of the location of the mass. Risk factors include a low CD4 count (often less than 50 cells/mm3) and a high HIV viral load. Clinical suspicion for CNS lymphoma should be highest in patients with uncontrolled HIV and a newly discovered brain mass. CNS lymphoma commonly involves the deep structures of the brain (eg, basal ganglia), cerebral white matter, corpus callosum, and periventricular areas. Diagnosis is made by imaging, biopsy, and cerebrospinal fluid analysis. The differential diagnosis for a space-occupying brain lesion in a patient with HIV/AIDS includes cerebral toxoplasmosis, brain abscesses, malignant primary brain tumors, and metastatic disease. MRI of the brain is helpful in differentiating between these diseases. Lumbar puncture is frequently performed as part of the evaluation for suspected primary CNS lymphoma unless there is a contraindication such as impending brain herniation. Treatment includes methotrexate and highly active antiretroviral therapy. Rituximab, a monoclonal antibody against CD20, can be used as an adjunct. Incorrect Answers: A, B, C, and D. Bacterial abscess (Choice A) of the brain commonly presents with fever, headache, and a focal neurologic deficit depending on the location of the abscess. These are most commonly associated with infections from viridans streptococcus and Staphylococcus aureus. In patients with HIV/AIDS, pathogens such as Salmonella species, Aspergillus species, Nocardia species, and Listeria monocytogenes should be considered. Cerebral abscesses may occur from hematogenous seeding resulting in multiple abscesses, or secondary to contiguous spread from otitis media, mastoiditis, or sinusitis. Cerebral toxoplasmosis (Choice B) typically presents with multiple ring-enhancing lesions scattered throughout the brain. In this patient with a cerebral mass and a positive IgG antibody to toxoplasmosis, empiric antimicrobial therapy is reasonable. However, given the aggressive nature of primary CNS lymphoma, frequent clinical and x-ray reassessments are required. Expansion of this patient's mass despite treatment for toxoplasmosis suggests an alternative diagnosis. Glioblastoma (Choice C) is a high-grade, aggressive primary brain astrocytoma that has a poor prognosis. Imaging features include an expansile mass crossing the corpus callosum with surrounding vasogenic edema. Seizures are a common presenting symptom, as are headaches and focal neurologic deficits. While glioblastomas may occur in patients with AIDS, primary CNS lymphoma is more common. Metastatic disease (Choice D) is typically associated with multiple brain lesions, often occurring at the gray-white matter junction. Breast cancer, lung cancer, renal cell carcinoma, and melanoma frequently metastasize to the brain. This patient demonstrates no signs or symptoms of a distant, extracranial primary malignancy.

Objective: Solitary brain lesions in patients with HIVIAIDS, particularly when associated with a low CD4+ count and a high viral load, should raise suspicion for primary CNS Iymphoma. Alternative diagnoses include a bacterial abscess, cerebral toxoplasmosis, primary brain tumors, and metastatic lesions. I3D Previous Next Score Report Lab Values Calculator Help Pause

146 Exam Section 3: Item 46 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 46. A 45-year-old man comes to the physician because of a 2-week history of pain of his left testis. Six weeks ago, he had a vasectomy. Physical examination shows a bead-like mass of the left testis. Excision of the mass is done. Microscopic examination of the mass shows sheets of extravasated spermatozoa engulfed by histiocytes, many with prominent enlarged nuclei. This patient's lesion is most likely the result of which of the following processes? O A) Adaptation B) Autoimmunity O C) Hemorrhage D) Infection E) Neoplasia Correct Answer: B A vasectomy is a surgical procedure used to induce permanent male contraception in which the vas deferens is ligated to prevent sperm from entering the urethra during sexual intercourse. In this case, the procedure was complicated by a leak of spermatozoa into the surrounding tissue which elicited a granulomatous response characterized by histiocytes and multinucleated giant cells. In normal male genitalia anatomy, the Sertoli cells lining the seminiferous tubules create a blood-testis barrier through tight junctions, which allows the seminiferous tubule to be an immune-protected site. The spermatozoa developing within the seminiferous tubule contain genetic material that may be foreign to the body as a result of recombination during meiosis. Thus, when this barrier is intact, it prevents potential novel antigens from leaving the seminiferous tubule and encountering the host immune system. If this occurs, the spermatozoa can elicit an immune response that may include the production of autoantibodies against the gametes. In the case of a vasectomy, this barrier is broken. Spermatozoa leak out of the vas deferens and are identified by the immune system as a foreign body, triggering a granulomatous reaction and autoantibody production. Incorrect Answers: A, C, D, and E. Adaptation (Choice A) may occur as a response to tissue injury in order to preserve the function of the tissue. In the setting of a vasectomy, this could be demonstrated by recanalization of a sealed vas deferens. In this case, however, the reaction is because of an immune response to gametes considered foreign rather than an attempt to preserve the function of the vas deferens. Hemorrhage (Choice C) is a potential complication of any surgical procedure but would not demonstrate histiocytes on histology. Instead, layered erythrocytes and thrombin would be expected. Infection (Choice D) is also a potential complication of a vasectomy but would be expected to occur earlier in the post-operative course. Microscopic examination would demonstrate neutrophils and bacterial organisms. Histiocytes may be present, but spermatozoa would likely be absent. Neoplasia (Choice E) should be considered in the evaluation of any testicular mass. Testicular tumors include germ cell tumors and non-germ cell tumors, such as those caused by the proliferation of Sertoli cells or Leydig cells. However, sheets of spermatozoa engulfed by histiocytes is not a feature of any neoplastic process.

Objective: Spermatozoa may leak into the surrounding interstitium after a vasectomy. Without the benefit of the normal blood-testis barrier, the gametes may elicit an immune foreign body granulomatous reaction and production of autoantibodies. Previous Next Score Report Lab Values Calculator Help Pause

39 Exam Section 1: Item 39 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 39. An investigator is designing a study to compare a new behavioral program for attention-deficit/hyperactivity disorder (ADHD) with the standard behavioral modification program for this disorder. Because ADHD is more common among boys, girls and boys are randomized into the two treatment groups separately. Which of the following types of treatment allocation is most likely being used in this study? O A) Alternation B) Open O C) Outcome-adapted D) Simple random assignment E) Stratification Correct Answer: E. Stratification describes the allocation of subjects or the analysis of results within a trial by a factor other than the treatment given. It involves subdividing a study population into homogeneous subgroups then analyzing each subgroup in comparison to the study population or other subgroups. Advantages of this technique are that it permits analysis of subpopulations that may vary from a population at large. In this case, the researchers speculate that since ADHD is more common among boys, there may be a meaningful difference when comparing the new and standard treatment programs applied to boys as applied to girls, whereas that meaningful difference may not be detected if the combined population is analyzed. Stratification is one method of reducing confounding variables. In this case, the study population has been stratified based on gender between the new and standard interventions. Incorrect Answers: A, B, C, and D. Alternation (Choice A) describes assigning each participant in a study to a treatment or control group on an alternate basis and is a type of a non-randomized controlled trial. Participants who agree to participate in such as trial are assigned one-by-one in a time- series fashion to the treatment group, then the control group, then the treatment group, then to the control group, alternating in assignment until target enrollment has been met. Open (Choice B) studies, or open trials, are types of trials where information about the treatment or control allocation is not withheld from those participating (study subjects or researchers). In contrast, blind trials limit one or both parties from knowing which intervention is being administered. Open trials may introduce observer bias, though sometimes may be bioethically appropriate. There is no information provided in this case to suggest whether the study is being conducted in an open or blind fashion. Outcome-adapted (Choice C) allocation describes a change in allocation method over time in a controlled trial based on results of either the treatment or control arm. Initial allocation in such a study is generally random or one-to-one between treatment and control groups at the outset, and as data shows which arm has better outcomes, participants are generally allocated to the arm that is yielding a better outcome more frequently. The study described has not yet reported any outcome data, thus, an outcome-adapted allocation cannot yet occur. Simple random assignment (Choice D) describes probability-based allocation of participants to treatment or control arms. In this example, a non-probability-based allocation has occurred with stratification of participants based on gender.

Objective: Stratification describes allocating subjects or results by a factor that is not necessarily the treatment administered; groups may be stratified or subdivided based on age, gender, profession, ethnicity, or other demographic features. It is one method to control for confounding variables. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

116 Exam Section 3: Item 16 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 16. A 34-year-old man is brought to the emergency department semiconscious and combative. In addition to sedation, a short-acting neuromuscular blocking agent is administered for intubation to prevent aspiration. Within a few seconds after administration of the drug, he has transient muscle fasciculations in his face; he develops generalized paralysis within 1 minute. Forty-five minutes after completion of the procedure, he is still paralyzed. A genetic abnormality of which of the following enzymes is the most likely cause of his unusually slow recovery from paralysis? A) Angiotensin-converting enzyme B) Choline O-acetyltransferase C) Monoamine oxidase O D) Phenylethanolamine N-methyltransferase E) Pseudocholinesterase F) Tyrosine hydroxylase Correct Answer: E. Succinylcholine is a depolarizing skeletal muscle relaxant commonly used to facilitate rapid sequence tracheal intubation. The generally short duration of action is caused by its rapid metabolism by pseudocholinesterase. Succinylcholine is a potent acetylcholine receptor agonist that causes depolarization of the muscle fibers, seen clinically as fasciculations, followed by muscle relaxation. It is structurally like acetylcholine but is not metabolized by acetylcholinesterase. Its alternative metabolism causes it to persist in the synaptic cleft for a longer duration, leading to prolonged depolarization and prevention of further skeletal muscle activation by acetylcholine. As serum concentration falls, it diffuses away from the neuromuscular junction. It is then hydrolyzed by pseudocholinesterase in the plasma and liver. Therefore, a genetic abnormality of pseudocholinesterase would prevent succinylcholine from being metabolized, which would lead to an abnormally slow recovery from neuromuscular blockade. Incorrect Answers: A, B, C, D, and F. Angiotensin-converting enzyme (ACE) (Choice A) is an enzyme present primarily in the lungs that plays an important role in the renin-angiotensin-aldosterone system, which is responsible for regulating blood pressure via vascular tone and sodium and fluid balance. ACE converts angiotensin I to angiotensin II, which causes vasoconstriction and the promotion of aldosterone release. It has no role in neuromuscular blockade. Choline O-acetyltransferase (Choice B) is an enzyme involved in the synthesis of acetylcholine. It transfers an acetyl group from acetyl-CoA to choline, forming acetylcholine. A deficiency of this enzyme would thus cause muscle weakness because of a decrease in acetylcholine synthesis. However, it would not prolong the action of succinylcholine. Monoamine oxidase (MAO) (Choice C) acts to degrade a variety of neurotransmitters, including serotonin, norepinephrine, epinephrine, and dopamine. A genetic abnormality in this enzyme would be expected to cause increased concentrations of these neurotransmitters, leading to hypertension and tachycardia. MAO does not play a role in the neuromuscular junction. Phenylethanolamine N-methyltransferase (Choice D) is an enzyme that converts norepinephrine to epinephrine by transferring a methyl group from S-adenosyl-L-methionine (SAM) to norepinephrine. It does not play a role in neuromuscular blockade. Tyrosine hydroxylase (Choice F) is the rate limiting step of catecholamine synthesis and catalyzes the conversion of tyrosine into dihydroxyphenylalanine (DOPA). An abnormality in this enzyme would not be expected to prolong neuromuscular blockade.

Objective: Succinylcholine is a depolarizing neuromuscular blocker. It causes depolarization, seen clinically as fasciculations, followed by skeletal muscle relaxation. It is degraded by pseudocholinesterase, and its action is prolonged by a deficiency of this enzyme. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

88 Exam Section 2: Item 38 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 38. A 63-year-old man with alcoholism is diagnosed with hepatic encephalopathy. Treatment with lactulose therapy via nasogastric tube is begun. The effectiveness of this intervention requires which of the following mechanisms? A) Absorbed lactulose binding alcohol in the blood, decreasing blood alcohol concentrations O B) Lactulose binding of ingested alcohol within the colonic lumen for excretion C) Metabolism to glucose and galactose, increasing serum glucose concentrations D) Osmotic diarrhea, flushing out ingested alcohol from the gut E) Trapping of ammonia in the colon by acidic metabolites of lactulose Correct Answer: E. The trapping of ammonia in the colon by acidic metabolites of lactulose is the mechanism of action of lactulose, which prevents the further uptake of ammonia, thereby permitting urea cycle clearance of existing serum ammonia and improving the symptoms of hepatic encephalopathy. Degradation of lactulose by colonic bacteria generates hydrogen ions that create an acidic environment within the gut. The abundance of extra hydrogen ions converts ammonia (NH3) to ammonium (NH,*), which cannot freely be absorbed by the intestines and diffuse into the bloodstream, causing it to be excreted in the feces. Lactulose is given prophylactically to many patients with cirrhosis to prevent the development of hepatic encephalopathy and is titrated to a goal number of bowel movements daily. While the measurement of serum ammonia is commonly performed, concentrations do not reliably correlate with the degree of encephalopathy. Incorrect Answers: A, B, C, and D. Absorbed lactulose does not bind alcohol in the blood (Choice A). Lactulose is an osmotically active compound with minimal systemic absorption. Similarly, lactulose does not bind to ingested alcohol within the colonic lumen for excretion (Choice B); rather, it exerts its effects by preventing the absorption of ammonia from the gut through the production of acidic metabolites. Lactulose is not metabolized to glucose and galactose (Choice C). Breakdown of lactose (not lactulose) via the action of the enzyme lactase leads to the production of glucose and galactose. Galactose is subsequently converted to glucose in the liver. Osmotic diarrhea (Choice D) is a known adverse effect of lactulose, but this does not result in the flushing out of ingested alcohol from the gut. Alcohol is absorbed rapidly in the stomach and small bowel, and there is no meaningful interaction between alcohol and lactulose. Additionally, hepatic encephalopathy often occurs separate from alcohol intoxication.

Objective: Systemic ammonia absorption is prevented by the administration of lactulose, which is degraded by gut bacteria to create an acidic environment that facilitates the conversion of ammonia to ammonium, trapping it in the lumen of the bowel for subsequent excretion in the stool. Previous Next Score Report Lab Values Calculator Help Pause

147 Exam Section 3: Item 47 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 47. A2-year-old boy is brought to the physician because of chronic bacterial respiratory infections since birth. He is currently asymptomatic. Physical examination shows no abnormalities. T- and B-lymphocyte counts and serum antibody concentrations are within the reference ranges. Natural killer cell count and function are normal. Analysis of cellular expression of human leukocyte antigen by flow cytometry shows absence of class I MHC-expressing cells. A diagnosis of bare lymphocyte syndrome, type I, is made. This patient most likely has mutations in the genes encoding which of the following? A) Adenosine deaminase O B) Fas ligand (CD178) C) Interleukin-2 (IL-2) receptor a chain (CD25) D) Peptide transporter (TAP) Correct Answer: D. Peptide transporter (TAP) gene mutation accounts for this patient's diagnosis of bare lymphocyte syndrome, type . There are two TAP genes (1 and 2) that bind in the endoplasmic reticulum (ER) to form the TAP complex. This protein normally shuttles antigenic peptides obtained from foreign pathogens into the ER where the antigens are complexed with MHĊ-I before being shuttled back to the surface of the cell for antigen presentation. This antigen-MHC-I complex is recognized by CD8+ T lymphocytes, which upon recognition of foreign peptides, kill the presenting cell. Absence of TAP results in the failure to present foreign peptides to CD8+ T lymphocytes via MHC-I and thereby results in susceptibility to a wide range of infections. Pulmonary bacterial and skin infections are most common. Incorrect Answers: A, B, and C. Adenosine deaminase (ADA) mutations (Choice A) result in the accumulation of a purine derivative called deoxyadenosine, which in large quantities inhibits the action of ribonuclease reductase and prevents DNA synthesis in T and B lymphocytes. Additionally, deoxyadenosine is toxic to lymphocytes. Patients often exhibit severe combined immunodeficiency and present with failure to thrive and opportunistic infections. Fas ligand (CD178) mutations (Choice B) cause autoimmune lymphoproliferative syndrome. The Fas ligand is a member of the tumor necrosis factor family and plays a role in the induction of lymphocyte apoptosis. Mutations result in the proliferation of lymphocytes because of the failure of apoptosis, often resulting in the development of generalized lymphadenopathy, splenomegaly, and hepatomegaly from lymphocyte expansion. Interleukin-2 (IL-2) receptor a chain (CD25) mutations (Choice C) result in lymphopenia with abnormal T lymphocyte development and normal B lymphocyte development. Concomitant infiltration of the gastrointestinal tract with abnormal lymphocytes is also seen.

Objective: TAP mutations prevent the presentation of foreign antigens on MHC-I to CD8+ T lymphocytes. This limits the normal immune response resulting in susceptibility to multiple bacterial infections, most commonly pulmonary and skin infections. Patients demonstrate normal numbers of lymphocytes. Previous Next Score Report Lab Values Calculator Help Pause

108 Exam Section 3: Item 8 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 8. A 25-year-old woman, gravida 2, para 2, comes to the physician because of a painful lump in the right vulva. Her last menstrual period was 1 week ago. She does not recall any recent trauma and has had no prior vaginal infections. Examination shows an acutely tender, 2.5-cm mass that is located near the vaginal orifice posteriorly and distends the labia majora. It appears to point vaginally. Which of the following is the most likely diagnosis? O A) Bartholin abscess B) Inclusion cyst O C) Skene gland occlusion D) Vulvar carcinoma E) Vulvar hematoma Correct Answer: A. The Bartholin glands, or greater vestibular glands, are a pair of glands on either side of the posterior vaginal orifice that secrete a lubricating fluid through two small ducts into the vestibule. When one of these glands becomes blocked, a cyst may occur. If bacteria collect behind the blockage, an abscess may result, as in this case. This is most seen in women of reproductive age. The infection is generally polymicrobial, but anaerobes such as Bacteroides fragilis, Clostridium perfringens, Peptostreptococcus species, and Fusobacterium species are key pathogens. While a Bartholin abscess may be caused by Chlamydia trachomatis or Neisseria gonorrhoeae, it is not considered an exclusively sexually transmitted infection, although Neisseria is commonly cultured. When an abscess develops, the gland enlarges in size and becomes fluctuant, erythematous, and indurated. It may cause vulvar pain with walking, sitting, or sexual intercourse. Incorrect Answers: B, C, D, and E. An inclusion cyst (Choice B) is an epidermal derived cyst that is not associated with a secretory gland. It most commonly occurs on the labia majora when affecting the genitals. An inflamed inclusion cyst becomes enlarged, erythematous, and painful but true infection is uncommon. Skene gland occlusion (Choice C) causes a cyst to form adjacent to the urethral meatus, where the Skene gland enters the vestibule. This is typically asymptomatic, but if large may cause urethral obstruction, difficulty urinating, incontinence, or pain. Vulvar carcinoma (Choice D) is a neoplasm of the vulva derived from the squamous epithelium, which can be primary or secondary to the direct extension of cervical carcinoma. Associated risk factors include human papillomavirus infection and smoking. Vulvar carcinoma classically presents in a postmenopausal woman. Vulvar hematoma (Choice E) is a collection of blood in the soft tissue of the vulva, which is typically induced by trauma. It may occur anywhere on the vulva and can be painful.

Objective: The Bartholin glands are located on both sides of the vaginal orifice posteriorly with ducts oriented superiorly that drain into the vestibule. Infection of an occluded Bartholin gland causes a Bartholin abscess to form. Previous Next Score Report Lab Values Calculator Help Pause

33 Exam Section 1: Item 33 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 33. A 32-year-old man is being evaluated for increased jugular venous pressure, a systolic murmur, hepatomegaly, ascites, and 2+ pitting edema of the ankles. He is an intravenous drug user. Which of the following is the most likely diagnosis? A) Aortic regurgitation O B) Aortic stenosis O C) Chronic anemia D) Mitral regurgitation E) Mitral stenosis F) Tricuspid regurgitation Correct Answer: F. Intravenous drug use is a risk factor for infective endocarditis and valvular dysfunction caused by the nonsterile injection of material into the venous system. Though left-sided infective endocarditis is more common in general, the tricuspid valve is most commonly involved in the setting of intravenous drug use. Tricuspid valve endocarditis is typically associated with Staphylococcus aureus, Pseudomonas aeruginosa, and Candida organisms. The formation of vegetations on the valve and local inflammatory damage can lead to severe tricuspid regurgitation, which presents as a holosystolic murmur best heard at the left lower sternal border. Severe tricuspid regurgitation is a risk factor for the development of right-sided heart failure, which is characterized by an increased jugular venous pressure, hepatomegaly, ascites, and peripheral extremity edema. Incorrect Answers: A, B, C, D, and E. Aortic regurgitation (Choice A) presents with an early diastolic decrescendo murmur best heard at the right second intercostal space. It is commonly associated with endocarditis, acute rheumatic fever, and aortic root dilation. In this case, the peripheral volume overload, systolic murmur, and history of intravenous drug use are more suggestive of tricuspid regurgitation. Aortic stenosis (Choice B) presents with a crescendo-decrescendo systolic murmur, best heard at the upper right sternal border and radiates to the carotid arteries. It classically occurs secondary to age-related fibrotic and calcific changes of the valve but can occur earlier in life in cases of bicuspid aortic valve or chronic rheumatic disease. Chronic anemia (Choice C), when severe, can lead to a syndrome similar to high-output heart failure, as cardiac output is increased to compensate for decreased oxygen-carrying capacity of the blood. A systolic flow murmur may be appreciated at the left upper sternal border. Mitral regurgitation (Choice D) presents with a holosystolic murmur best heard at the left fourth or fifth intercostal space along the midclavicular line and radiates to the left axilla. It is commonly associated with mitral valve prolapse or prior myocardial infarction. While the mitral valve is often affected in infective endocarditis, intravenous substances more commonly involves the tricuspid valve. Mitral stenosis (Choice E) is classically heard as an opening snap followed by a diastolic rumble, loudest over the cardiac apex and radiates to the axilla. If severe enough, it can result in left atrial enlargement, cardiogenic pulmonary edema, and arrhythmias such as atrial fibrillation and flutter.

Objective: Tricuspid valve endocarditis is typically associated with intravenous drug use. Damage to the valve can lead to tricuspid regurgitation, which is a risk factor for the development of right-sided heart failure. Previous Next Score Report Lab Values Calculator Help Pause

62 Exam Section 2: Item 12 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 12. A study is conducted on patients with asthma to compare a new asthma treatment with a placebo. After 12 weeks of treatment, the mean (+ standard error of the mean) scores are 2.8 ± 0.1 in the treated group (n = 103) and 3.1 ± 0.1 in the placebo group (n = 100). The p-value is 0.035. Which of the following best describes the meaning of the p-value? A) 3.5% of the treated patients had asthma scores lower (better) than the mean score of the controls B) 96.5% of the treated patients had asthma scores lower (better) than the mean score of the controls C) The drug worked in 3.5% of the patients D) The drug worked in 96.5% of the patients E) If the drug really does not work, there is a 3.5% chance of finding an average difference of 0.3 or greater F) If the drug really does not work, there is a 96.5% chance of finding an average difference of 0.3 or greater Correct Answer: E. When conducting any experiment, two hypotheses are created. The null hypothesis, or Ho, states that there is no difference between the two groups being compared, in this case the new asthma treatment with the placebo. The alternative hypothesis, or H1, suggests that there is a difference between the two groups being compared. The likelihood of erroneously rejecting the null hypothesis (concluding that the alternative hypothesis is true) is called a type I error. The acceptable rate of a type I error is a percentage called the alpha level (a) and is set by the investigators. When the alpha level is converted to a decimal, it can be compared to the P-value of the experiment. An alpha level of 5% or 0.05 indicates that if the null hypothesis is true, there is 5% chance that the investigators will erroneously reject it based on the study results. In this case, the investigators found a difference in the mean scores of the two groups of 0.3 with a P-value of 0.035. If the drug is no different than placebo and the null hypothesis is correct, there is a 3.5% chance of finding an average difference of 0.3 or greater under this circumstance. In other words, by chance and sampling error, these results might occur 3.5% of the time. A P-value of 0.05 or less is generally considered statistically significant. Incorrect Answers: A, B, C, D, and F. The P-value does not convey any information about the percentage of patients with improvement in their asthma (Choice A and B). It is a measure of statistical significance rather than a measure of the study outcome. The P-value indicates the probability that a type I error has been made. It does not provide any information on the proportion of subjects with a certain outcome (Choices C and D). The P-value indicates the probability of a null hypothesis being incorrectly rejected. There is a 96.5% chance (Choice F) of finding an average difference of 0.3 or greater if the alternative hypothesis is correct and the drug works.

Objective: The P-value is the probability that a type I error has been committed and the null hypothesis has been erroneously rejected. A P-value of 0.05 or less is generally considered statistically significant. Previous Next Score Report Lab Values Calculator Help Pause

103 Exam Section 3: Item 3 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 3. A 38-year-old pregnant woman had genetic amniocentesis that showed a 46,XX karyotype in the cultured amniotic fluid cells. A newborn was delivered with normal male external genitalia, testes, and genital ducts. Which of the following is the most likely explanation? O A) Adrenogenital syndrome B) Androgen insensitivity syndrome O C) Fragile X syndrome D) Maternal exposure to oral progestins E) SRY gene translocation Correct Answer: E. The SRY gene is located on the Y chromosome and is responsible for producing testis-determining factor, which in turn drives male gonadal differentiation and development. In development, two testicular cell types, Sertoli cells and Leydig cells, play key roles in genital embryology and phenotypic sex determination. Sertoli cells are responsible for secreting Mullerian inhibitory factor (MIF) which suppresses the development of the paramesonephric (Mullerian) duct that would have developed into female internal genital structures such as the fallopian tubes, uterus, and upper vagina. Leydig cells secrete testosterone, which promotes the development of the mesonephric (Wolffian) duct that develops into the seminal vesicles, epididymis, vas deferens, and ejaculatory duct. Testosterone is also converted to dihydrotestosterone (DHT) by 5a-reductase, which promotes the development of the male external genitalia and prostate from the genital tubercle and urogenital sinus. SRY gene translocation can occur during recombination in which the SRY gene on the Y chromosome becomes part of the X chromosome, leading to an XX embryo developing male characteristics. Incorrect Answers: A, B, C, and D. Adrenogenital syndrome (Choice A) is also known as congenital adrenal hyperplasia (CAH) and is caused by congenital enzyme deficiencies in the adrenal cortex that lead to abnormal concentrations of mineralocorticoids, glucocorticoids, and androgens. For example, a deficiency in 17a-hydroxylase causes increased concentration of mineralocorticoids (eg, aldosterone) and decreased concentration of cortisol and androgens, resulting in ambiguous genitalia and undescended testes in XY chromosome infants. Infants having two X chromosomes lack secondary sexual development later in life, observed as delayed puberty, amenorrhea, and failure to develop breasts and pubic hair. Androgen insensitivity syndrome (Choice B) is caused by a defect in the androgen receptor complex, resulting in a genetically XY male appearing phenotypically female (female external genitalia, scant pubic and axillary hair, absent uterus and fallopian tubes, and a rudimentary vagina). Despite having testes and testosterone, the inadequate androgen receptor prevents hormone binding, and therefore, genetic expression that would have resulted in a phenotypic male never occurs. Testes develop normally, and patients demonstrate increased concentration of testosterone, estrogen, and luteinizing hormone. Fragile X syndrome (Choice C) is caused by a trinucleotide repeat in the FMR1 gene resulting in DNA methylation and decreased expression. It is characterized by postpubertal macroorchidism, large jaw and face, large everted ears, and congenital heart defects, and is the second most common cause of genetic intellectual disability after Down syndrome. Maternal exposure to oral progestins (Choice D) has been associated with hypospadias, an anatomic abnormality in which the opening of the penile urethra lies on the ventral surface of the penis. In most cases, it is a solitary defect, however, it can be associated with an inguinal hernia or cryptorchidism.

Objective: The SRY gene is located on the Y chromosome and is responsible for producing testis-determining factor, which results in male gonadal differentiation. In testis development, hormones secreted by Sertoli cells (MIF) and Leydig cells (testosterone and DHT) promote the development of male internal and external genitalia and suppress the development of female structures. SRY gene translocation can occur during recombination in which the SRY gene on the Y chromosome becomes part of the X chromosome, leading to an XX embryo developing male characteristics. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

18 Exam Section 1: Item 18 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 18. A77-year-old woman dies in the hospital after a long illness. Her vertebral column, obtained at autopsy, is shown in the photograph. The process shown is most likely associated with an increase in which of the following? A) Calcium B) Estrogen C) Interleukin-1 (IL-1) D) Monoclonal immunoglobulin O E) Vitamin D Correct Answer: C. Osteoporosis is a common condition that is characterized by the progressive loss of bone mineral density leading to decreased bone strength. This decreased bone mineral density often leads to fragility fractures, which can greatly decrease mobility and increase the risk of death in elderly individuals. The disease classically affects postmenopausal women with inflammatory disorders, and individuals with metabolic or endocrine disorders such as hypercortisolism or hyperparathyroidism. Osteoporosis can also be induced by long-term treatment with medications that cause an increase in bone resorption such as corticosteroids. The diagnosis is made using a DEXA scan via calculation of a T-Score. A T-score of -2.5 or less, that is a bone density measurement that is less than 2.5 standard deviations below the mean, is diagnostic of osteoporosis. T-scores between -1.0 and -2.5 are defined as osteopenia. Interleukin-1 (IL-1) is also known as osteoclast activating factor. IL-1 leads to an increase in RANK ligand signaling and subsequent osteoclast-mediated bone resorption. Osteoporosis is thereby characterized by an increase in osteoclast number and activity, which is driven by IL-1. Incorrect Answers: A, B, D, and E. Serum concentrations of calcium (Choice A) and phosphorus are typically normal; electrolyte balance is generally maintained in osteoporosis. In some cases of secondary osteoporosis (eg, hyperparathyroidism), increased serum calcium concentration may be detected. However, this variety of osteoporosis is much less common then postmenopausal or senile osteoporosis, neither of which present with calcium dysregulation. Estrogen (Choice B) is a key hormone in the pathophysiology of osteoporosis in postmenopausal females. Prior to menopause, estrogen stimulates osteoblasts and decreases osteoclast activity. The decrease in estrogen seen with menopause leads to an imbalance of greater osteoclast activity. In turn, osteoporosis is associated with a decrease in estrogen. Estrogen supplementation, or administration of selective estrogen receptor modulators, can treat or prevent osteoporosis in select patients. Monoclonal immunoglobulins (Choice D) are seen in B-lymphocyte proliferative disorders such as multiple myeloma. Clonal expansion of plasma cells leads to the overproduction of genetically and structurally identical nonfunctional immunoglobulins. This expansion of plasma cells leads to an increase in the production of RANK ligand and thereby an increase in the number and activity of osteoclasts. The gross section of vertebral bone shown does not demonstrate the typical lytic lesions of multiple myeloma. Vitamin D (Choice E) is a steroid hormone involved in calcium and phosphorus homeostasis and bone mineral density maintenance. In general, low vitamin D is associated with decreased bone mineral density, as opposed to increased vitamin D concentrations.

Objective: The balance of osteoblast and osteoclast interactions is key in the maintenance of bone mineral density. Interleukin-1, also called osteoclast activating factor, can result in bone mineral density loss when increased. ID Previous Next Score Report Lab Values Calculator Help Pause

148 Exam Section 3: Item 48 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 48. A 30-year-old woman and her 35-year-old husband come to the physician for genetic counseling prior to conception. They have a 2-year-old daughter with an isolated ventricular septal defect. The wife's sister had a heart defect, for which she underwent operative repair. Previous evaluation of the couple, including echocardiography, showed no abnormalities. Which of the following best represents this couple's risk for having another child with a congenital heart defect? O A) 0% B) 3% O C) 25% D) 50% E) 100% Correct Answer: B. Inheritance of disease is based on autosomal, mitochondrial, and X-linked genetics of the parents, genes distributed into the gametes, de novo mutations acquired during embryonic cell division and differentiation, penetrance and expressivity of the condition, mosaicism, and epigenetic modifications of expression. In this case, an isolated ventricular septal defect (VSD) is present in one of the offspring of two phenotypically normal parents. VSDS are among the most common congenital heart malformations, and they may occur in isolation or in association with complex structural abnormalities. The genetic cause of VSDS is heterogeneous and includes known associations with well-defined syndromes such as Down, DĪGeorge, or Holt-Oram syndrome. It may also occur in isolation and via nonspecific, random mutations and individual point mutations that are de novo and do not follow a particular Mendelian inheritance pattern. The baseline frequency of muscular-type VSDS in newborns has been estimated at 2% to 3% in populations of otherwise genotypically and phenotypically normal parents. Regarding inheritance, the exact frequency cannot be calculated with certainty for this particular couple as the patterns are complex and multifactorial. Additionally, the parents are phenotypically normal with no evidence in the history to suggest that the parents are harboring a particular mutation; therefore, the risk most likely approximates the baseline VSD risk across the population. Incorrect Answers: A, C, D, and E. 0% (Choice A) is incorrect, as with every inheritable disease, there is always a risk for de novo mutations, regardless of the genetic composition of the parents. 25% (Choice C) describes the inheritance of an autosomal recessive condition in which both parents are carriers. It also assumes that the condition is completely expressed and completely penetrant. 50% (Choice D) describes the inheritance of an autosomal dominant condition with one parent demonstrating heterozygosity for the disease, or an X-linked recessive condition to sons of carrier mothers. 100% (Choice E) describes the inheritance of an autosomal dominant condition with a single or both parents having a homozygous genotype, or an autosomal recessive condition in which both parents demonstrate a homozygous genotype.

Objective: The baseline population VSD incidence is approximately 2% to 3%, with inheritance patterns that do not follow conventional Mendelian patterns apart from those associated with known genetic syndromes. Previous Next Score Report Lab Values Calculator Help Pause

31 Exam Section 1: Item 31 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 31. A 62-year-old man comes to the physician for a follow-up examination. Two weeks ago, he was discharged from the hospital after sustaining an acute myocardial infarction. He smokes 2 packs of cigarettes and drinks four 12-oz beers daily. His diet mostly consists of cured meats and fast food. He does not exercise. The patient tells the physician, "I know that I need to make some changes in how I live so that my heart can be healthier. I just don't have the willpower to quit smoking and drinking and all that stuff right now." Which of the following best describes this patient's stage of behavioral change? A) Precontemplation B) Contemplation C) Preparation D) Action E) Maintenance Correct Answer: B. In this scenario, the patient is aware of the potential consequences of his unhealthy habits and is considering reversing the habits. Although he has identified that his behavior is detrimental, he is indecisive about whether he is ready to act, which is consistent with the contemplation stage. The stages of behavioral change are used to define a patient's readiness to change a health-related behavior such as substance use, diet, or exercise habits. In sequential order, the stages of behavioral change are precontemplation, contemplation, preparation, action, maintenance, and termination. Physicians aim to move patients through these stages over time with an interview technique called motivational interviewing. Motivational interviewing involves using open-ended, non-judgmental questions to help the patient explore their reasons for wanting to change or maintain the habit. Incorrect Answers: A, C, D, and E. Patients in the precontemplation stage (Choice A) are not interested in changing their habit and may not see their detrimental habit as a problem. This patient believes his habits are creating problems for his health and is considering change, which means he has surpassed the precontemplation stage. Patients in the preparation stage (Choice C) have committed to making a change and are ready to discuss strategies and resources to help them make the change. For example, a patient may research information about a low-carbohydrate diet to improve his eating habits. This patient expresses knowledge of his detrimental behavior and its impact on his health but is not ready to act. In the action stage (Choice D), patients start to take steps to change their behavior. For example, a patient may begin a low-carbohydrate diet to improve his unhealthy eating habits. After making the change, the patient must maintain the change (Choice E) and strategize how to cope with temptation to return to the old habit. For example, a patient may join a support group to assist with maintaining their improved behavior choice.

Objective: The contemplation stage of behavioral change refers to ambivalence about changing a behavior, in which the patient expresses knowledge of the consequences of their detrimental behavior but is not yet prepared to make plans to change. Motivational interviewing is a technique used by physicians to promote a patient's readiness for change. Previous Next Score Report Lab Values Calculator Help Pause

80 Exam Section 2: Item 30 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 30. An 82-year-old woman is admitted to the hospital for treatment of a small right pleural effusion. Physical examination shows dullness to percussion at the level of the 7th rib and below on the right. A thoracentesis is done. The patient is sitting up, supported by leaning against a bedside table. After preparing the skin and delivering a local anesthetic, into which of the following locations on the right is it most appropriate to insert the catheter? O A) Above the 5th rib in the midscapular line B) Above the 5th rib just to the right of the sternum C) Above the 7th rib in the paravertebral region D) Above the 9th rib in the midscapular line E) Below the 5th rib in the midaxillary line F) Below the 5th rib in the midclavicular line G) Below the 9th rib in the midscapular line Correct Answer: D. Pleural effusions occur as a result of multiple underlying conditions including infection such as pneumonia, malignancy such as lung cancer or mesothelioma, and increased hydrostatic pressure such as left-sided heart failure. They are clinically suspected on examination in the setting of dullness to percussion and diminished breath sounds, and can be visualized on imaging including chest x-ray, CT scan, or ultrasonography. A diagnostic thoracentesis should be performed whenever the cause of the effusion is unknown. A therapeutic thoracentesis can be used to relieve clinical symptoms. The preferred site for the procedure is on the affected side, along the midaxillary line if the patient is supine or the posterior midscapular line if the patient is upright. Bedside ultrasonography can be used to identify a suitable fluid pocket. If ultrasonography is not available, the preferred location is two rib spaces below the most superior point of dullness to percussion. The catheter should be inserted along the top of the inferior rib to avoid damaging the intercostal neurovascular bundle which runs along the inferior margin of the superior rib. Caution must also be used to avoid damage to the liver and spleen, which can rise during exhalation to the level of the 5th rib on the right and the 9th rib on the left, respectively. For the patient in this case, the correct location would be on the right side, above the 9th rib in the midscapular line. Any fluid obtained is then analyzed for cell count, organisms, pH, glucose, total protein, and enzyme concentrations. Incorrect Answers: A, B, C, E, F, and G. Inserting the catheter above the 5th rib in the midscapular line (Choice A) or above the 5th rib just to the right of the sternum (Choice B) are both incorrect as the effusion extends only to the level of the 7th rib and the catheter would likely injure the lung parenchyma. This can result in pneumothorax, or less commonly pulmonary hemorrhage or hemothorax, all of which would require emergent management. Placement above the 7th rib in the paravertebral region (Choice C) is incorrect because the costophrenic recess is located more inferior along this line, and this location risks injury to the lung parenchyma. Inserting the catheter below the 5th rib in the midaxillary line (Choice E), below the 5th rib in the midclavicular line (Choice F), or below the 9th rib in the midscapular line (Choice G) would be incorrect as these approaches would risk damage to the intercostal neurovascular bundle.

Objective: The correct positioning for a thoracentesis is above the 9th rib on the affected side, along the midaxillary line if the patient is supine or the posterior midscapular line if the patient is upright, which avoids damage to the lung parenchyma and adjacent structures. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

180 Exam Section 4: Item 30 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 30. A 70-year-old man dies of coronary artery disease. He had a cerebral infarction 8 years ago. A photograph of a specimen of the brain stem is shown. Which of the following neurologic deficits was most likely present in this patient after his cerebral infarction? Left Right A) Dysmetria on the left B) Dysmetria on the right C) Loss of joint position sense in the left extremities D) Loss of joint position sense in the right extremities E) Spastic hemiparesis on the left F) Spastic hemiparesis on the right Correct Answer: F. The infarction occurred in the left crus cerebri, which is illustrated in the gross section as severely atrophied tissue in comparison to the normal crus cerebri on the right. Myelin stains black; the absence of stain on the left side suggests a neurodegenerative process involving the region at the anterior-most portion of this inferior section of the midbrain. The crus cerebri carries lateral corticospinal tract fibers. Upper motor neurons of the lateral corticospinal tract originate in the primary motor cortex (precentral gyrus), descend ipsilaterally through the internal capsule and midbrain, decussate in the caudal medulla, descend in the contralateral spinal cord, and synapse with lower motor neurons in the contralateral anterior horn, which innervates a muscle of the contralateral extremity. Therefore, a lesion in the left crus cerebri at the midbrain level would lead to a right-sided upper motor neuron pattern of weakness. Incorrect Answers: A, B, C, D, and E. Dysmetria on the left (Choice A) or right (Choice B) would result from lesions of the superior cerebellar peduncle. The decussation of the superior cerebellar peduncle is a white matter tract pictured in the gross section crossing anterior to the left medial longitudinal fasciculus and trochlear nucleus (approximately halfway between the anterior and posterior extremes). The superior cerebellar peduncles carry efferent fibers of the cerebellothalamic tract from the deep cerebellar nuclei to the motor nuclei of the red nuclei and thalamus, causing dysmetria when disrupted. Lesions rostral to the pictured decussation would cause contralateral cerebellar signs, whereas lesions caudal to the decussation would lead to ipsilateral cerebellar signs. Loss of joint position sense in the left extremities (Choice C) or right extremities (Choice D) would be caused by a lesion to the medial lemniscus, a white matter tract located in the lateral aspect of the above section immediately posterior to the crus cerebri bilaterally. The medial lemniscus carries fibers from the dorsal column tract, which is responsible for the sensation of pressure, fine touch, and proprioception (joint position sensation). The dorsal column-medial lemniscus tract originates from the sensory nerve ending and then synapses with the second-order neuron at the ipsilateral nucleus cuneatus or gracilis, ascending ipsilaterally, decussating in the medulla, synapsing at the thalamus, and terminating in the primary sensory cortex. Lesions inferior to the medulla will produce ipsilateral proprioception deficits, whereas lesions superior to the medulla lead to contralateral proprioception deficits. Spastic hemiparesis on the left (Choice E) would be caused by a lesion of the right crus cerebri as a result of the decussation of lateral corticospinal tract in the caudal medulla. Spastic hemiparesis on the left could be caused by any corticospinal lesion rostral to the caudal medulla.

Objective: The crus cerebri carries fibers from the lateral corticospinal tract. Lesions to the crus cerebri lead to contralateral upper motor neuron weakness because of the caudal decussation of the lateral corticospinal tract in the medulla. II Previous Next Score Report Lab Values Calculator Help Pause

125 Exam Section 3: Item 25 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 25. During a study of familial hypercholesterolemia, hepatocytes from an affected patient are extracted and analyzed. Results show that the LDL receptors are evenly distributed on the surface of the cells. The cells also bind LDL-cholesterol with normal affinity but cannot internalize it. The cells are shown to internalize transferrin. Based on these findings, which of the following is the most likely location of the causal mutation in this patient? O A) Apo B O B) Apo E O C) Clathrin heavy chain D) Clathrin light chain E) Cytoplasmic domain of the LDL receptor F) Extracellular domain of the LDL receptor Correct Answer: E. Mutation of the cytoplasmic domain of the LDL receptor on this patient's hepatocytes explains why LDL is able to bind to its receptor but endocytosis fails. The LDL receptor is a transmembrane receptor that resides in clathrin coated pits along the cell membrane. When Apo B, an embedded component of LDL particles, binds to the LDL receptor, the entire receptor-ligand complex is internalized (endocytosis). The acidic environment of the endosome leads to dissociation of LDL from its receptor, followed by fusion with the lysosome where LDL and Apo B are degraded into their respective components to form amino acids, hydrolyzed fatty acids, and cholesterol esters. The LDL receptor is then recycled back to the cell membrane. Signaling for endocytosis occurs via the cytoplasmic domain of the LDL receptor so defects in this portion would affect endocytosis. Absence or dysfunction of the LDL receptor results in familial hypercholesterolemia. Incorrect Answers: A, B, C, D, and F. Apo B mutations (Choice A) would render LDL incapable of binding to the LDL receptor but would not prevent internalization. Apo B mutations are another cause of familial hypercholesterolemia. Apo E mutations (Choice B) would have no effect on LDL binding and endocytosis as Apo E is not a normal component of the circulating LDL particle. While it does bind to the LDL receptor, it is associated with chylomicrons, VLDL, IDL, and HDL. Clathrin heavy chain (Choice C) and clathrin light chain (Choice D) mutations would result in widespread dysfunction of endocytosis affecting more than just the internalization of the LDL receptor. Endocytosis of transferrin occurs via clathrin-mediated mechanisms, so the normal trafficking of transferrin in this patient excludes clathrin mutations as the causal defect. Mutations of the extracellular domain of the LDL receptor (Choice F) would inhibit binding of LDL particles to the LDL receptor.

Objective: The cytoplasmic domain of the LDL receptor is responsible for signaling that leads to endocytosis, so mutations at this location allow for normal binding of LDL but an inability to transport it across the membrane. Mutations that cause dysfunction or absence of the LDL receptor result in familial hypercholesterolemia. Previous Next Score Report Lab Values Calculator Help Pause

82 Exam Section 2: Item 32 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 32. A 43-year-old woman, gravida 2, para 1, at 10 weeks' gestation comes to the physician for an initial prenatal visit. Physical examination shows a uterus consistent in size with a 10-week gestation. This patient is at increased risk for a child with Down syndrome if she has which of the following sets of ultrasound and serum findings? Fetal Ultrasound Nuchal Maternal Serum Pregnancy-associated Plasma Protein Human Chorionic Translucency Gonadotropin O A) ↑ B) ↑ ↑ O C) ↑ O D) ↑ E) ↑ F) Correct Answer: B. Increased nuchal translucency, decreased pregnancy-associated plasma protein, and increased human chorionic gonadotropin are the components of the first trimester combined test that indicate an increased risk for a child to have Down syndrome. When all three elements are present, the test has a detection rate of 90% and a false positive rate of 11%. Like any screening test, a positive result should be followed by a definitive, diagnostic test, such as chorionic villus sampling or amniocentesis. Down syndrome is caused by a trisomy of chromosome 21. In most cases, it is because of a meiotic nondisjunction, which occurs when two chromatids fail to separate, and a single, haploid gamete contains two copies of chromosome 21 instead of one. The incidence of meiotic nondisjunction increases with increasing maternal age: 1 in 25 women older than age 45 will have such an event occur. In less than 5% of cases, it is because of an unbalanced translocation called a Robertsonian translocation, which typically occurs between chromosomes 14 and 21. Incorrect Answers: A, C, D, E, and F. Decreased, rather than increased (Choice A), maternal serum pregnancy-associated plasma protein is associated with Down syndrome. Human chorionic gonadotropin is produced by the syncytiotrophoblastic cells of the placenta in order to maintain the corpus luteum and continue its production of progesterone. This is critical to maintaining the pregnancy through the first eight to ten weeks when the placenta is not yet producing its own progesterone. Human chorionic gonadotropin is increased, not decreased (Choice C), in Down syndrome. Increased human chorionic gonadotropin is also seen in the setting of multiple gestations and hydatidiform mole. A decreased fetal nuchal translucency is not associated with Down syndrome (Choices D, E, and F). The nuchal fold is a translucent area on the posterior fetal neck seen with ultrasonography; the average thickness is increased in Down syndrome.

Objective: The first trimester combined test suggests a fetal diagnosis of Down syndrome when there is increased fetal ultrasound nuchal translucency, decreased maternal serum pregnancy-associated plasma protein, and increased human chorionic gonadotropin. It is followed by definitive, diagnostic testing such as chorionic villus sampling or amniocentesis. Previous Next Score Report Lab Values Calculator Help Pause

182 Exam Section 4: Item 32 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 32. A 46-year-old man receives the diagnosis of squamous cell carcinoma of the esophagus. A barium swallow is shown. Esophagectomy at the region indicated by the arrows is most likely to involve ligation of arterial branches of which of the following vessels? A) Aorta B) Internal thoracic arteries C) Pulmonary arteries D) Superior mesenteric artery E) Thyrocervical trunk Correct Answer: A. The esophagus has a complex, segmental blood supply derived from multiple vessels based on position from superior to inferior. The superior (cervical) esophagus is supplied by arterial branches from the inferior thyroid arteries. Branches from the bronchial arteries supply the superior and middle thoracic esophagus, with the inferior thoracic esophagus (as denoted by the arrows in the barium swallow above) being supplied by esophageal branches directly from the aorta. Below the diaphragm, the esophagus is supplied by the left gastric artery and short gastric branches. A substantial amount of variability in vascular supply is present across patients, and occasionally, contributing branches from intercostal arteries can be observed. During surgery, vessels must be ligated to prevent exsanguination. Incorrect Answers: B, C, D, and E. The internal thoracic arteries (Choice B) originate from the subclavian artery bilaterally and course inferior along the anterior, medial body wall on the posterior aspect of the ribcage. They supply the anterior chest wall and breasts and are often harvested for use in cardiac bypass grafting. The pulmonary arteries (Choice C) originate from the pulmonary trunk at the right ventricle and carry deoxygenated blood to the lungs. Ligation of a pulmonary artery would be required in a pneumonectomy. The superior mesenteric artery (Choice D) supplies the distal duodenum, jejunum, ileum, and proximal one-half to two-thirds of the large bowel. Branches of this artery would require ligation during bowel resection. The thyrocervical trunk (Choice E) is a branch of the subclavian artery; one of its branches is the inferior thyroid artery, which supplies the superior, not inferior, esophagus.

Objective: The inferior thoracic esophagus is arterially supplied by branches directly from the aorta that must be ligated during an esophagectomy. Below the diaphragm, the esophagus is supplied by the left gastric artery and short gastric branches. II Previous Next Score Report Lab Values Calculator Help Pause

119 Exam Section 3: Item 19 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 19. A 68-year-old man comes to the emergency department in a mountain resort town with his wife because of a 1-month history of shortness of breath and headaches. One month earlier, he retired and moved from his home at sea level to the resort town which is 9000 feet above sea level. His wife is asymptomatic. His respirations are 16/min. Laboratory studies show an erythrocyte count of 4 million/mm3. This patient most likely has a history of disease of which of the following organs? O A) Heart B) Kidney O C) Liver D) Lung O E) Thyroid gland Correct Answer: B. High altitudes exhibit decreased atmospheric oxygen (PO2) when compared to sea level. In the short term, the body compensates for the resultant hypoxemia by increasing ventilation, which in turn may cause respiratory alkalosis and compensatory metabolic acidosis. In the long term, the body compensates for a decreased PO2 by increasing the hemoglobin concentration, which in turn increases oxygen carrying capacity, and by shifting the oxygen-hemoglobin dissociation curve to the right via increased 2,3- bisphosphoglycerate production. Erythropoietin (EPO) is the principal molecule involved in increasing the production of red blood cells. EPO is released in response to hypoxia by renal interstitial cells, which then stimulates red blood cell progenitor proliferation in the bone marrow. However, in chronic kidney disease, the kidney's ability to create erythropoietin is reduced, and the consequent anemia results in a decreased oxygen carrying capacity of blood and difficulty in acclimatizing to higher altitudes. Altitude sickness symptoms result, which include shortness of breath, headache, anorexia, nausea, vomiting, fatigue, dizziness, and difficulty sleeping. Incorrect Answers: A, C, D, and E. Heart failure (Choice A) is known to cause shortness of breath, however it classically presents with symptoms of volume overload such as orthopnea, dyspnea on exertion, jugular venous distention, and pitting edema. When left-sided, symptoms include shortness of breath, and examination findings include rales on auscultation of the lungs, and an S3 or S4 gallop on cardiac auscultation. Liver disease (Choice C) also causes volume overload characterized by ascites but would also be accompanied by jaundice, scleral icterus, and (in extremis) hepatic encephalopathy. Lung disease (Choice D) such as chronic obstructive pulmonary disease may present with shortness of breath, but is generally also accompanied by cough, increased sputum production, and wheezing caused by the destruction of lung parenchyma and/or the obstruction of small airways. It is not associated with anemia by itself, in fact, chronic hypoxia can result in erythrocytosis as a result of renal compensation. Thyroid gland disease (Choice E) has many manifestations depending on the relative derangement in thyroid hormone. Hyperthyroid states classically present with heat intolerance, weight loss, warm, flushed skin, hair loss, loose stools, or oligo- or amenorrhea. Hypothyroid states classically present with cold intolerance, weight gain, dry skin, constipation, and menorrhagia. Neither set of symptoms clearly explains this patient's shortness of breath at altitude.

Objective: The kidneys synthesize EPO in states of hypoxemia, which promotes the synthesis of new red blood cells to increase oxygen carrying capacity. Chronic kidney disease leads to anemia via the reduced synthesis of EPO. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

126 Exam Section 3: Item 26 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 26. A 28-year-old man comes to the physician because of a 2-week history of light-headedness, mild chest pain, palpitations, and shortness of breath after exercising. He has no history of major medical illnesses. His temperature is 36.9°C (98.4°F), pulse is 72/min, respirations are 17/min, and blood pressure is 120/82 mm Hg. Cardiac examination shows a bounding point of maximal impulse that is shifted to the left. Echocardiography shows cardiomegaly and marked thickening of the ventricular septum. It is most appropriate for the initial therapy to increase which of the following in this patient? A) Contractility B) Diastolic filling C) Heart rate D) Peripheral resistance E) Pulmonary perfusion pressure Correct Answer: B. The patient's presentation and diagnostic studies are consistent with hypertrophic obstructive cardiomyopathy, which classically presents with dyspnea, chest pain, syncope, or sudden cardiac death in an exercising young athlete with a potential family history of a similar event. Physical examination findings include a systolic murmur loudest in the lower left sternal border, which is quieter with increased preload and afterload, and a laterally displaced point of maximal impulse. It is secondary to hypertrophy of the interventricular septum, which can be seen on echocardiography. Hypertrophic obstructive cardiomyopathy, if severe, can result in left ventricular outflow tract obstruction, especially during instances of decreased preload or afterload, such as exercise. In addition to left ventricular outflow tract obstruction, the walls of the left ventricle are stiff and noncompliant, which impairs filling of the chamber during diastole. During exercise or other stress events, the associated tachycardia results in decreased diastolic filling time, which further decreases the left ventricular preload, worsens the left ventricular outflow tract obstruction, and results in decreased cardiac output. The reduced cardiac output can result in syncope during exercise, and possible ventricular arrhythmias caused by cardiac ischemia, which may lead to sudden death. Medications that increase diastolic filling by reducing the heart rate, such as B-adrenergic blocking agent or non-dihydropyridine calcium channel blockers, decrease the degree of left ventricular outflow tract obstruction and improve cardiac output. Incorrect Answers: A, C, D, and E. Increasing contractility (Choice A) can worsen symptoms of hypertrophic cardiomyopathy by increasing the degree of left ventricular outflow tract obstruction as the chamber contracts down with greater force. Medications that decrease contractility (negative inotropes) are instead used for treatment. Increasing the heart rate (Choice C) would worsen the symptoms of hypertrophic cardiomyopathy because of shorter diastolic filling times, which would decrease left ventricular preload and result in increased left ventricular outflow obstruction. Increasing peripheral resistance (Choice D) leads to increased afterload, which is favored in hypertrophic cardiomyopathy as it helps reduce the degree of left ventricular outflow tract obstruction by keeping the left ventricular chamber stented open. While avoiding unnecessary afterload reduction is important in the treatment of hypertrophic cardiomyopathy, pharmacologically increasing afterload is not commonly employed as it can lead to increased heart strain. Increasing pulmonary perfusion pressure (Choice E) would reflect increasing cardiac output from the right side of the heart or increasing pulmonary vascular resistance (both of which would increase pulmonary artery pressure) and potentially distend the chamber of the right ventricle. Septal bowing could occur, which would further impair left ventricular filling and worsen the symptoms of hypertrophic cardiomyopathy.

Objective: The left ventricular septum is asymmetrically thickened in hypertrophic obstructive cardiomyopathy, leading to outflow tract obstruction and diastolic dysfunction. Medications that decrease the heart rate improve symptoms by allowing more time for diastolic filling, which increases left ventricular preload and decreases the degree of left ventricular outflow tract obstruction. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

174 Exam Section 4: Item 24 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 24. A study is conducted to examine the effects of polychemotherapy on patients with lung carcinoma. Bronchoalveolar lavage samples taken before and after treatment with methotrexate, doxorubicin, cyclophosphamide, and lomustine are examined for changes in pulmonary surfactant. In order to access a lingular bronchus for lavage, the bronchoscope must pass through which of the following bronchi? A) Left lower lobe B) Left upper lobe O C) Right lower lobe D) Right middle lobe E) Right upper lobe Correct Answer: B. The gross lobar anatomy of the lung contains three lobes of the right lung (upper, middle, and lower) and two lobes of the left lung (upper and lower). The lingula is a distinct projection of the upper lobe of the left lung and is homologous to the middle lobe of the right lung. It is comprised of superior and inferior lingular segments. To access the lingular bronchus, a bronchoscope will need to pass from the trachea to the left mainstem bronchus and into the left upper lobe bronchus. Incorrect Answers: A, C, D, and E. The left lower lobe (Choice A) and right lower lobe (Choice C) are divided into five bronchopulmonary segments, including superior, anterior-basal, medial-basal, lateral-basal, and posterior-basal segments. The lingula is a distinct projection of the left upper lobe, not the lower lobes. The right middle lobe (Choice D) is comprised of medial and lateral bronchopulmonary segments. It is separated from the right lower lobe inferiorly and posteriorly by the oblique fissure and from the right upper lobe superiorly by the horizontal fissure. The right upper lobe (Choice E) is divided into three segments, the apical, posterior, and anterior segments. The lingula is a distinct projection of the left, not right upper lobe.

Objective: The lingular bronchus is a distinct projection of the left upper lobe, which can be accessed by bronchoscopy through the left upper lobe bronchus. Previous Next Score Report Lab Values Calculator Help Pause

121 Exam Section 3: Item 21 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 21. A 69-year-old woman comes to the physician because she has had weakness of her left leg since awakening that morning. Physical examination shows weakness of the extremity. A Babinski sign is present on the left. Sensory testing shows decreased somatic sensation in the left foot, agraphesthesia on the plantar surfaces of the toes, and decreased position sense in the toes. An MRI of the brain shows an edematous area in the cerebral cortex of the right hemisphere. The most likely cause of this condition is a lesion located at which of the following labeled areas in the photographs of normal brains shown? A B E C G F H. A) B) C) D) E) F) G) H) I) O J) Correct Answer: G. Strokes occur because of ischemic or hemorrhagic loss of blood supply to the brain. Approximately 80% to 85% of strokes are ischemic, arising from thromboembolic disease (eg, middle cerebral artery (MCA) occlusion from a thrombus), whereas 15% to 20% of strokes are hemorrhagic, caused by blood vessel rupture (eg, hypertension-related intraparenchymal hemorrhage). Risk factors for stroke include smoking, hypertension, diabetes, carotid or intracranial atherosclerotic disease, history of hypercoagulability, atrial fibrillation, and age. Classically, strokes manifest as a neurologic deficit related to the affected part of the brain. This patient is experiencing weakness and numbness, loss of proprioception, and signs of upper motor neuron damage (Babinski sign) involving the left lower extremity. These symptoms correlate with the precentral and postcentral gyri of the right cerebral hemisphere, which control motor and sensory function, respectively. The cortical neurons that control motor and sensory function involving the left lower extremity are located in the medial regions of the precentral and postcentral gyri. In the gross specimen shown, this area is indicated most closely by label "G", which reflects the medial aspect of the posterior frontal lobe and anterior parietal lobe, referred to as the paracentral lobule. This area is generally supplied by a branch of the ipsilateral anterior cerebral artery (ACA), the likely location of the acute occlusion or hemorrhage. Incorrect Answers: A, B, C, D, E, F, H, and I. Choices A and B reflect the lateral aspect of the anterior parietal (postcentral gyrus) and posterior frontal (precentral gyrus) lobes, respectively. These areas of the cerebral cortex control the sensory function (primary somatosensory cortex, postcentral gyrus) and motor function (primary motor cortex, precentral gyrus) of the contralateral face, upper extremity, and trunk. Strokes involving this area present with contralateral facial and upper extremity loss of sensation and weakness, respectively. The MCA supplies this territory. Choice C identifies the prefrontal cortex. This area is associated with multiple functions including learning, reasoning, problem solving, emotion, behavioral control, memory, self-regulation, and personality. In a stroke, lesions affecting this area may result in behavioral dysregulation and the development of psychiatric symptoms (eg, post-stroke depression). The prefrontal cortex is a watershed area supplied by both the ACA and MCA. Choice D identifies the posterior, superior temporal gyrus, an area that in the dominant hemisphere makes up one portion of Wernicke area, a brain region involved in the understanding of language. In a stroke, lesions involving this area can result in Wernicke aphasia. In the nondominant hemisphere, it is involved in discriminating the pitch of sound. This area is supplied by the MCA. Choice E identifies the supramarginal gyrus and the inferior lateral parietal lobe, which function in reading, word meaning, language, and interpretation of spoken sounds. This area is supplied by the MCA. Choice F identifies the area anterior to the premotor and supplemental motor cortices. The frontal eye fields are found here, which control voluntary eye movements. Strokes affecting this area may result in tonic deviation of the eyes toward the side of the lesion. Choice H identifies the medial parietal cortex posterior to the somatosensory cortex. This area plays a role in spatial and visual coordination and orientation. Choice I identifies the occipital lobe, whose primary function relates to reception and integration of visual information.

Objective: The lower extremity is controlled by the medial aspects of the precentral (motor) and postcentral (sensory) gyri in the region referred to as the paracentral lobule. Strokes affecting this region result in the loss of motor and sensory function of the contralateral lower extremity. II Previous Next Score Report Lab Values Calculator Help Pause

134 Exam Section 3: Item 34 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 34. A75-year-old man is found to have an 8-cm mass in the head of the pancreas that is encasing and markedly narrowing the inferior vena cava. Blood from this patient's lower extremities is most likely returning to the superior vena cava from which of the following veins? A) Celiac and paraspinal veins B) Lumbar, azygos, and hemiazygos veins C) Right and left renal veins and suprarenal veins D) Superior and inferior mesenteric veins E) Superior mesenteric and hepatic veins Correct Answer: B. The lumbar, azygos, and hemiazygos veins provide an alternate route for blood from the inferior body to return to the right atrium when the inferior vena cava is blocked. Similarly, if the superior vena cava is occluded, the azygos vein can provide a potential route for blood from the superior body to return to the right atrium. The azygos vein runs inferosuperiorly along the right side of the vertebral column, toward the superior vena cava. The azygos vein is formed by ascending lumbar veins when they join the subcostal veins; additional tributaries include the hemiazygos vein, posterior right intercostal veins, and bronchial veins. The hemiazygos vein (and accessory hemiazygos vein) travels superiorly in the inferior thoracic region on the left side of the vertebral column and drain the left body wall and posterior intercostal spaces. In this patient, blood that would have normally drained through the inferior vena cava cannot do so as a result of constriction from the pancreatic mass. The lumbar, azygos, and hemiazygos veins provide an alternate pathway for venous drainage of the lower extremities to the superior vena cava. Incorrect Answers: A, C, D, and E. Celiac and paraspinal veins (Choice A) is incorrect, as there is no formally named celiac vein. Paraspinal veins drain the vertebrae and spinal cord depending on their position related to the vertebrae and dura, and do not provide an anastomosis to drain the inferior body. Right and left renal veins and suprarenal veins (Choice C) drain the kidneys and adrenal glands, and generally do so directly into the inferior vena cava without direct anastomosis to the azygos system. In general, the left suprarenal vein drains into the left renal vein, whereas the right suprarenal vein drains into the inferior vena cava. Superior and inferior mesenteric veins (Choice D) drain the small and large bowel into the portal vein, and do not provide an anastomosis with the caval venous system. Superior mesenteric and hepatic veins (Choice E) drain the small and large bowel, and the liver. The superior mesenteric veins drain into the portal vein, while the hepatic veins drain directly into the inferior vena cava, just prior to its insertion into the right atrium.

Objective: The lumbar, azygos, and hemiazygos veins provide an alternate route for blood from the lower extremities to return to the superior vena cava in the setting of an inferior vena cava obstruction. Previous Next Score Report Lab Values Calculator Help Pause

193 Exam Section 4: Item 43 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 43. During an experiment, Drug X is administered to treat trematode and cestode infestations. Results show that Drug X increases the permeability of the cell membranes to calcium, causing paralysis, dislodgement, and death of the parasite. Drug X most closely resembles which of the following? A) Albendazole O B) Bithionol O C) Diethylcarbamazine D) Niclosamide E) Praziquantel Correct Answer: E. Praziquantel is an anti-helminthic drug used for treating infections caused by parasitic worms. Commonly treated infections include schistosomiasis, cysticercosis, hydatid cysts, and clonorchiasis. The mechanism of praziquantel relates to altered parasite calcium homeostasis via increased membrane permeability. Resultant effects of calcium influx into the cytosol include paralysis, dislodgement of the parasite position within host organs, and death of the parasite. The mechanism of Drug X in question, also used to treat trematode and cestode infections, matches this mechanism. Incorrect Answers: A, B, C, and D. Albendazole (Choice A) is an anti-helminthic drug that binds to B-tubulin, which inhibits microtubule polymerization, decreases parasite glucose utilization, and decreases energy production. Blocking microtubule polymerization inhibits cell motility and intracellular transport. It is used for treating Echinococcus granulosus, Taenia solium, and Enterobius vermicularis, among other parasitic worms. Bithionol (Choice B) is an anti-helminthic agent that is no longer in widespread use because of severe photosensitivity as a side effect. Its mechanism relates to the inhibition of adenylyl cyclase. Diethylcarbamazine (Choice C) is a microfilaricidal medication that is used to treat filariasis, Loa loa, Wuchereria bancrofti, and Brugia malayi. Its mechanism involves the inhibition of arachidonic acid metabolism. Niclosamide (Choice D) inhibits glucose uptake and oxidative phosphorylation by reducing electric potential across the inner mitochondrial membrane. It can be used for the treatment of Taenia infections. It is less effective against tissue cestodes such as Echinococcus or Cysticercosis.

Objective: The mechanism of praziquantel relates to altered parasite calcium homeostasis through increasing membrane permeability. Calcium influx into the cytosol causes paralysis, dislodgement of the parasite position within host organs, and death of the parasite. Previous Next Score Report Lab Values Calculator Help Pause

137 Exam Section 3: Item 37 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 37. A 56-year-old woman is brought to the emergency department because of progressive shortness of breath during the past month. Two weeks ago, she underwent coronary artery bypass grafting. Her respirations are 26/min. There is dullness to percussion at the left base of the chest. A chest x-ray shows substantial fluid in the left pleural cavity. Thoracentesis yields 1200 mL of pale, milky fluid that is identified as chyle. Which of the following procedures during the operation most likely produced the iatrogenic chylothorax? A) Dissection of the internal thoracic artery B) Midsternal thoracotomy C) Placement of the central line via the left internal jugular vein D) Placement of the epicardial pacemaker lead E) Placement of the pulmonary artery catheter via the right subclavian vein Correct Answer: C. The iatrogenic chylothorax is most likely a complication of placement of the central line via the left internal jugular vein as a result of direct injury to the thoracic duct. Damage to or blockage of the thoracic duct or its major lymphatic tributaries is the most common cause of chylothorax, which is accumulation of chyle (milky fluid consisting of fat droplets and lymph) in the pleural space. The thoracic duct is the largest lymphatic vessel in the body, originating in the abdomen at a confluence of lymphatics called the cisterna chyli. It passes through the aortic aperture of the diaphragm and ascends adjacent to the thoracic aorta and azygos vein. It ultimately drains into the venous system near the junction of the left subclavian and left internal jugular veins. The proximity to the vein is a risk factor for direct injury from a central venous cannulation attempt or extrinsic compression and blockage from local venous extravasation. Malignancy, especially lymphoma, chronic lymphoid leukemia, lung cancer, esophageal cancer, and metastatic carcinoma, is the most common cause of nontraumatic chylothorax. Patients classically present with dyspnea, chest heaviness, fatigue, and weight loss. Diagnosis is made by pleural fluid analysis by thoracentesis. The classic appearance is thick, milky white fluid, with increased concentrations of triglycerides and cholesterol. Incorrect Answers: A, B, D, and E. Dissection of the internal thoracic artery (Choice A) is performed to harvest the vessel for use in the coronary artery bypass grafting procedure. The internal thoracic artery originates from the subclavian artery and courses inferior along the anterior body wall on the posterior aspect of the ribcage. It is unlikely to injure the thoracic duct given its distant anatomic relationship. Complications of a midsternal thoracotomy (Choice B) include sternal wound infection and mediastinitis. Injury to the thoracic duct and chylothorax would not result from this procedure given the duct's course along the posterior body wall. Placement of the epicardial pacemaker lead (Choice D) can be complicated by infection, ventricular arrhythmias, perforation of cardiac chambers, hemopericardium, and cardiac tamponade. It would be unlikely to injure the thoracic duct and cause chylothorax. Placement of the pulmonary artery catheter via the right subclavian vein (Choice E) can cause a chylothorax from injury to the right lymphatic duct (which drains lymph from the right thoracic duct, the right upper extremity, and the right side of the head and neck). The right lymphatic duct has several anatomic variants with drainage into the right internal jugular vein or the right subclavian vein. However, this would be expected to result in a right-sided chylothorax, not a left-sided lesion.

Objective: The most common cause of chylothorax is injury to the thoracic duct. The thoracic duct can be injured during central venous cannulation because of its proximity to the left internal jugular and subclavian veins. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

85 Exam Section 2: Item 35 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 35. In an experiment on ion transport in epithelial cells isolated from various regions of the nephron, hydrochlorothiazide inhibits Na+entry into one group of cells. These cells were most likely obtained from which of the following regions? O A) Cortical collecting tubule B) Distal convoluted tubule O C) Proximal tubule D) Thick ascending limb of the loop of Henle E) Thin descending limb of the loop of Henle Correct Answer: B. The distal convoluted tubule (DCT) is the main site for Na* and Cl reabsorption in the nephron, resulting in a more dilute and hypotonic filtrate. Hydrochlorothiazide (HCTZ) and other thiazide diuretics (eg, chlorthalidone, metolazone) inhibit NaCI reabsorption in the distal convoluted tubule by inhibiting the action of the sodium-chloride cotransporter (NCC) channel. This class of medications reduces the diluting capacity of the nephron and results in urinary losses of sodium and potassium. HCTZ has the opposite effect on calcium; calcium ion loss is reduced by thiazide diuretics. Thiazide diuretics are a first-line therapy for hypertension. Incorrect Answers: A, C, D, and E. The cortical collecting tubule (Choice A) is the terminal segment of the nephron and is the site of reabsorption of Na* and secretion of K* and H*, It is regulated by aldosterone and antidiuretic hormone. The V, -receptors in the collecting tubule are regulated by ADH and promote the uptake of free water via aquaporins. The proximal tubule (Choice C) is the closest to the glomerulus. It is the site for reabsorption of most electrolytes (HCO3, Na*, Cr, PO,3, K*), water, uric acid, glucose, and amino acids from the initial glomerular filtrate. Acetazolamide acts at the proximal tubule by inhibiting carbonic anhydrase. The thick ascending limb of the loop of Henle (Choice D) is impermeable to water and reabsorbs Na*, K*, and Clr. Loop diuretics (eg, furosemide, torsemide, bumetanide, ethacrynic acid) act at the thick ascending limb of the loop to reduce solute reabsorption. The thin descending limb of the loop of Henle (Choice E) passively reabsorbs water and is impermeable to Na*. It is not a significant site of diuretic action.

Objective: The nephron is composed of a glomerulus, proximal convoluted tubule, loop of Henle, distal convoluted tubule, and collecting duct. HCTZ acts at the distal convoluted tubule to inhibit sodium and chloride reabsorption and promote diuresis. Previous Next Score Report Lab Values Calculator Help Pause

19 Exam Section 1: Item 19 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 19. A 16-year-old boy is brought to the physician because of a 3-day history of difficulty walking because his right foot drops when he lifts it. He is a member of a wrestling team at his high school. Physical examination shows weakness of the right ankle dorsiflexor muscles. The right ankle evertor muscles have full power. Sensation to pinprick is decreased between the great and second toes of the right foot. Sensation over the rest of the foot is normal. Which of the following nerves is most likely damaged in this patient? A) Common fibular (peroneal) B) Deep fibular (peroneal) C) Sciatic D) Superficial fibular (peroneal) E) Tibial Correct Answer: B. Peripheral nerves, particularly in areas that are not protected by surrounding soft tissue are vulnerable to injury. Peripheral nerve injuries can occur as a result of repetitive motion, sporting activities with strenuous motion or physical contact, compression during positioning for a long period of time, or traumatic disruption (eg, laceration). The nerves of the lower extremity are commonly implicated in injuries and the affected nerve can often be identified by the pattern of weakness and sensory loss. The deep fibular or peroneal nerve provides sensory innervation to the first dorsal webspace and motor innervation to the tibialis anterior muscle, which is the primary dorsiflexor of the ankle joint. This nerve also innervates the extensor digitorum longus and brevis muscles, which serve to extend the four lesser digits of the foot and dorsiflex the ankle, extensor hallucis longus and brevis muscles, which extend the great toe and dorsiflex the ankle, and the tibialis posterior muscle, which supports the arch of the foot and inverts the foot. Injury to the deep fibular nerve will cause a loss of ankle dorsiflexion as seen in this patient. It will also result in greater relative strength of eversion because of the loss of tibialis posterior function which is the primary inverter of the foot. Incorrect Answers: A, C, D, and E. The common fibular (peroneal) nerve (Choice A) gives rise to both the deep fibular and superficial fibular nerves. Injury to this nerve would result in loss of ankle dorsiflexion as well as loss of ankle eversion. The alteration in sensation would include the first dorsal webspace as well as the dorsum of the foot. The sciatic nerve (Choice C) is a large nerve that originates from the lumbosacral plexus and supplies the hamstrings as well as the innervation to the entire lower extremity below the knee. It can be compressed via the piriformis muscle and can cause neuropathic pain of the posterior, lateral, and distal lower extremity. Injury to this nerve can mimic a spinal cord nerve root injury. The superficial fibular (peroneal) nerve (Choice D) supplies the peroneus longus and brevis muscles. These muscles are the main evertors of the foot. Injury to this nerve would cause weakness in eversion as well as a decrease in sensation or paresthesias over the dorsolateral aspect of the foot. The tibial nerve (Choice E) supplies motor innervation to the posterior compartment of the lower leg and the intrinsic muscles of the foot. It supplies sensation to the plantar surface of the foot. An injury to the tibial nerve would compromise strength of toe flexion and ankle plantar flexion. It is an infrequent site of injury, as it is protected by the overlying gastrocnemius muscle.

Objective: The peripheral nerves of the leg are often sites of compression injury or entrapment causing sensory loss and motor weakness. The superficial fibular nerve everts the foot, while the deep fibular nerve dorsiflexes the foot, and the tibial nerve plantarflexes the foot. Weakness in these movements can indicate which nerve is injured. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

164 Exam Section 4: Item 14 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment Injection 71 Injection Injection Injection Injection Injection 00 4. 2 4 10 Treatment week Active drug+ Saline Saline 14. A study is conducted to assess the effectiveness of injections of lidocaine into "trigger points" of pain symptoms in patients with fibromyalgia. Fifty patients are randomly assigned to biweekly trigger-point injections with 0.9% saline only or 0.9% saline plus lidocaine. The graph shows daily self-reported pain scores in relation to the injections. Which of the following is the most likely explanation for these findings? A) Placebo effect B) Regression to the mean C) Selection bias D) Type Il error E) Uncontrolled confounding Correct Answer: A. To reduce the risk for procedural or observer-expectancy bias in a study, the use of a placebo (a substance or treatment without known therapeutic value such as an inert medication or sham surgery) is common. A placebo limits the potential attribution of benefit to a belief in the therapy itself by making it seem that all parties in the study are receiving the therapy. The placebo effect, a phenomenon in which a beneficial effect appears to arise from the administered placebo, cannot be attributed in any way to the placebo's properties. It therefore arises as a result of a patient belief that the placebo must be effective. In this example, the patients in the study receive a placebo administered in identical fashion (0.9% saline) or the medication under investigation (lidocaine administered in 0.9% saline). The data reflects that patient-reported pain scores trend in a similar manner regardless of whether the placebo or active drug was administered. In some instances, the placebo (0.9% saline) was reported to reduce pain even more than the injected lidocaine (Treatment weeks 1 and 8 to 10). There does not appear to be any statistically significant benefit of lidocaine compared to placebo based on this graph, and the overall trend of pain control was improved with each administration, an effect that cannot be attributed to known properties of 0.9% saline, therefore arising purely from belief in the placebo. Incorrect Answers: B, C, D, and E. Regression to the mean (Choice B) describes a statistical phenomenon in which extreme or outlier samples skew early interpretation of a population mean. With repeated samples, fewer outlier data points occur as the data more closely approaches the actual mean. Regression to the mean describes the phenomenon of future samples more closely approximating the true mean. Selection bias (Choice C) describes a fundamental difference between studied groups, often occurring from non-random assignment of subjects to each group. Examples of selection biases include the Berkson bias, when a study population is selected from a local hospital and is less healthy at baseline than the population at large, and the non-response bias, where subjects volunteering to participate in a study differ from the nonparticipants in a meaningful way. Ensuring correct choice of the study groups is key in reducing selection bias. Type Il error (Choice D) occurs when a study demonstrates no difference between the null and alternative hypotheses, when in fact, a difference exists. In this study, the null hypothesis would suggest there is no benefit from injecting lidocaine to treat fibromyalgia, while the alternative hypothesis would suggest that at least some benefit exists. Type Il error and can be reduced by increasing the power of the study through a larger sample size and increasing precision in measurements. It is possible that a Type Il error exists in this study, though no evidence is presented to suggest it has occurred. Uncontrolled confounding (Choice E) can bias a study when a factor is related to both the exposure and the outcome but is not related to a causal pathway (eg, there are more sick people in the presence of doctors, therefore doctors must be making people sick). Matching characteristics in control and experimental groups, plus randomization can help eliminate confounding, both of which appear to have been employed in this study.

Objective: The placebo effect is a phenomenon in which a beneficial effect appears to arise from the administered placebo that cannot be attributed to the placebo's properties. It arises because of patient belief that the placebo must be effective. Previous Next Score Report Lab Values Calculator Help Pause Pain score

194 Exam Section 4: Item 44 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 44. Investigators are evaluating oxygen consumption by different segments of the renal tubules in experimental animals. Arterial oxygen saturation is decreased in blood delivered to isolated perfused pig kidneys. Results show that as oxygen delivery declines, urine output decreases. A biopsy specimen of the ischemic model is obtained. Tubular cell death is most likely to be observed in which of the following areas? A) Cortical collecting duct B) Distal convoluted tubule O C) Loop of Henle D) Papilla E) Proximal convoluted tubule Correct Answer: E. Proximal convoluted tubule (PCT) cell death is most likely to be observed as a result of decreased oxygen delivery to the kidney, which varies directly with changes in blood flow through the afferent arteriole. Increased perfusion leads to increased glomerular filtration rate (GFR), while decreased perfusion leads to decreased GFR. GFR decreases in the setting of hypovolemia, hypotension, and with use of certain medications such as nonsteroidal anti-inflammatory drugs (NSAIDS). The cells lining the PCT are highly metabolically active and they perform the majority of electrolyte resorption (an energy-dependent process). As such, these are more sensitive to reduced oxygen delivery when compared to other parts of the nephron. Additionally, blood flow to the cells of the PCT is more frequently compromised in comparison to other segments of the nephron since the PCT is in the renal cortex where blood supply is delivered via the distal, small vessels of the vasa recta. The renal cortex and outer medulla, by virtue of this distant blood supply, is considered a watershed area and is highly sensitive to changes in perfusion pressure. Inadequate oxygen delivery compromises PCT cell generation of adenosine triphosphate (ATP), without which they cannot regulate their own osmolarity and metabolic requirements. In turn, effacement and destruction of the brush border ensues, which compromises the ability of the PCT to resorb bicarbonate, sodium (65% to 80% of total reabsorption), chloride, phosphorous, potassium, and free water, and limits the ability to excrete nitrogenous wastes. If low blood flow states are not rapidly reversed, irreversible cell death will occur in the PCT. This is known as ischemic acute tubular necrosis (ATN). Incorrect Answers: A, B, C, and D. Cortical collecting duct (CCD) cell death (Choice A) may occur in severe acute tubular necrosis, however these cells are less vulnerable to decreased perfusion pressure as their blood supply is more robust. Additionally, they are less metabolically active than the cells of the PCT. Similarly, cell death of the distal convoluted tubule (Choice B) is also not typical of ATN as the blood supply to these cells is also more robust, and they have a decreased metabolic demand as compared to the PCT. Loop of Henle cell death (Choice C) can occur in the setting of ischemia, especially in the thick ascending limb which lies in the outer medulla, but injury to the thin ascending loop of Henle is uncommon. Renal papilla cell death (Choice D) can occur with renal ischemia but is more commonly seen in the setting of NSAID use, diabetes mellitus, or sickle cell disease.

Objective: The proximal convoluted tubule is the most susceptible to decreased afferent blood flow and oxygen delivery compared to the other parts of the nephron because it is highly metabolically active and is situated in a watershed zone of the renal cortex. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

199 Exam Section 4: Item 49 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment Coronal Ахial 49. A 78-year-old man who is a Vietnam War veteran comes to the physician because of a 3-month history of low back pain that has become increasingly severe during the past 2 weeks, and has caused difficulty walking because of pain radiating down his left leg. He has a 3-week history of night sweats and an intermittent low-grade fever. He also has had a 9-kg (20-lb) weight loss during the past 3 months. He appears ill. He is 173 cm (5 ft 8 in) tall and weighs 55 kg (121 Ib); BMI is 18 kg/m2 His temperature is 38.1°C (100.6°F), and pulse is 85/min. Physical examination shows pallor, right convex lumbar scoliosis, and tenderness over the L2-3 area. He has difficulty climbing onto the examination table. Coronal and axial views of a CT scan of the abdomen and pelvis with contrast are shown. Which of the following active motions of the left hip is likely to be the most painful for this patient? A) Abduction B) Adduction C) Extension D) External rotation E) Flexion F) Internal rotation Correct Answer. E. The patient presents with subacute low-grade fever, night sweats, back pain, and weight loss, with difficulty ambulating which should raise suspicion for an inflammatory, malignant, or infectious process. The CT scan shows a loculated, enhancing, heterogeneous mass within the left psoas major muscle, consistent in this setting with a psoas abscess. The psoas originates from the lateral margins of the lumbar vertebrae, courses anterior to the ilium within the iliac fossa and inserts on the lesser trochanter of the femur. It is innervated via the lumbar plexus and has the prime action of flexion at the hip, which would be limited in the case of a psoas abscess. Incorrect Answers: A, B, C, D, and F. Abduction (Choice A) of the hip is controlled by muscles within the gluteal compartment, and includes the gluteus medius, minimus, and tensor fascia latae. These muscles can be visualized on the CT scan lateral to the ilium. Adduction (Choice B) of the hip is controlled by muscles within the adductor compartment of the thigh, and includes adductor magnus, longus, brevis, and minimus. Extension (Choice C) of the hip is controlled by the gluteus maximus, posterior head of the adductor magnus, and the hamstrings (semitendinosus, semimembranosus). External (lateral) rotation (Choice D) of the hip is controlled by the lateral rotator group, which includes the piriformis, gemellus superior and inferior, obturator externus and internus, and quadratus femoris. Internal (medial) rotation (Choice F) of the hip is controlled by multiple muscles, including the tensor fascia lata, gluteus minimus, anterior gluteus medius, and adductor longus and brevis.

Objective: The psoas muscle originates from the lateral margins of the lumbar vertebrae, courses anterior to the ilium within the iliac fossa and inserts on the lesser trochanter of the femur. It is innervated via the lumbar plexus and has the prime action of flexion at the hip, which would be limited in the case of a psoas abscess. II Previous Next Score Report Lab Values Calculator Help Pause

30 Exam Section 1: Item 30 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 30. A 26-year-old man comes to the physician because of a 3-week history of numbness of his penis and difficulty achieving and maintaining an erection. He is a competitive bicyclist who rides at least 20 to 30 miles daily. Physical examination shows no abnormalities. This patient's symptoms are most likely caused by increased pressure on the branches of which of the following nerves? A) Genitofemoral O B) llioinguinal O C) Inferior gluteal D) Pudendal E) Sacral splanchnic Correct Answer: D. The pudendal nerve supplies the perineum and genitourinary organs. It is the primary sensory afferent nerve from the external genitalia (via the dorsal nerve of the penis), the skin of the anus, and the perineum. In this case, the patient reports numbness of the penis and difficulty achieving an erection, a consequence of damage to the afferent component of the pathway. The pudendal nerve courses within the bony pelvis, exiting at the greater sciatic foramen, crossing the sacrospinous ligaments, and reentering the pelvis through the lesser sciatic foramen, all potential sites of compression. From here, it courses in the pudendal canal where the pudendal vessels join it. It divides into branches within the canal. The branches include the dorsal nerve of the penis or clitoris, the inferior rectal nerve, and the perineal nerve. Peripheral nerves, particularly in areas that are not protected by surrounding soft tissue are vulnerable to injury from strenuous, repetitive motion, physical contact, compression during positioning for a long period of time, or traumatic disruption. In this case, the patient's bicycle saddle has likely caused a compression injury of the pudendal nerve or its branches, resulting in the loss of sensory afferent signals from the genitalia with resultant numbness and difficulty maintaining an erection. Incorrect Answers: A, B, C, and E. The genitofemoral nerve (Choice A) supplies sensation to the anterior scrotum in males, and the anterior thigh. It generally does not innervate the penis. It is responsible for both the sensory and motor portions of the cremasteric reflex in men. The ilioinguinal nerve (Choice B) supplies sensation to the anterior scrotum along with the genitofemoral nerve and is also involved in the cremasteric reflex arc. It supplies the base of the penis but generally not the mid or distal shaft, or glans. The inferior gluteal nerve (Choice C) is primarily a motor neuron that supplies the gluteus maximus. It is not involved in the sensory or motor components of erection. Sacral splanchnic nerves (Choice E) originate deep within the pelvis adjacent to the sacrum. They are protected from compression injury and would be unlikely to be injured in this patient. They control the autonomic aspects of urinary, gastrointestinal, and sexual organs within the pelvis.

Objective: The pudendal nerve supplies the perineum and genitourinary organs. It is the primary sensory afferent nerve from the external genitalia (via the dorsal nerve of the penis), the skin of the anus, and the perineum. It is vulnerable to compression as it courses around the bony pelvis. Previous Next Score Report Lab Values Calculator Help Pause

91 Exam Section 2: Item 41 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 41. A 55-year-old woman comes to the physician because of a 3-day history of persistent right shoulder pain. She began a weight-training program 6 weeks ago. She says that the pain intensified after she increased the amount of weight that she had been lifting above her head. She rates the pain as 8 on a 10-point scale. Examination of the right shoulder shows point tenderness just lateral to the acromion, over the humeral head. Passive motion of the shoulder is full. Pain is reproduced with resisted abduction of the shoulder when the shoulder is abducted 90 degrees and the arm is giving the "thumbs down" sign. Sensation is intact over the right upper extremity. Which of the following tendons is most likely injured in this patient? A) Deltoid B) Infraspinatus C) Subscapularis D) Supraspinatus E) Teres major O F) Teres minor Correct Answer: D. The supraspinatus tendon is commonly injured or impinged during overhead motion, especially in persons unaccustomed to such motion. It is one of four tendons associated with the rotator cuff muscle group and is involved in glenohumeral joint stabilization as with the remainder of this muscle group. It is the continuation of the supraspinatus muscle, and it courses beneath the acromion process and superior to the humeral head to insert on the greater tuberosity of the humerus. The muscle originates in the supraspinous fossa of the scapula. As the tendon courses inferior to the acromion, it can become impinged or inflamed. As the primary role of the supraspinatus is the initiation of shoulder abduction (first 15 degrees), patients may report pain with abduction or inability to abduct the shoulder, and the pain may be worsened with maneuvers that exacerbate tendon impingement, such as resisted abduction beyond 90 degrees with internal (medial) rotation of the extremity. Diagnosis is generally based on the clinical examination and association of symptoms with maneuvers known to provoke symptoms; treatment involves rest, nonsteroidal anti-inflammatory drugs, and physical therapy with range-of-motion exercises. Other muscles involved with the rotator cuff are the infraspinatus, teres minor, and subscapularis. Incorrect Answers: A, B, C, E, and F. The deltoid (Choice A) is a large muscle over the superior and lateral shoulder that is the prime mover of arm abduction. Injury to the deltoid or its tendon would likely limit arm abduction over a wide range of degrees, as opposed to when impingement of the supraspinatus occurs, which affects the initiation of abduction (first 15 degrees). The infraspinatus (Choice B) is a rotator cuff muscle that originates in the infraspinous fossa of the scapula and inserts on the greater tubercle of the humerus; its main action is external (lateral) rotation of the arm. Limitation in this function, or pain elicited on passive internal rotation or resisted external rotation would suggest an injury to the infraspinatus muscle or tendon. The subscapularis (Choice C) is a rotator cuff muscle that originates in the subscapular fossa and inserts on the lesser tubercle of the humerus, with the prime action of internal (medial) rotation and adduction of the humerus. Limitation in this function, or pain elicited on passive external rotation or resisted internal rotation would suggest an injury to the subscapularis muscle or tendon. The teres major (Choice E) is an internal (medial) rotator and adductor of the arm. It is not considered a component of the rotator cuff. Injury to this muscle is marked by loss of the prime function (difficulty internally rotating or adducting the arm against resistance), and pain is felt inferior to the shoulder. The teres minor (Choice F) is a rotator cuff muscle that originates from the lateral scapula and inserts on the greater tubercle of the humerus. It is involved in external (lateral) rotation of the arm. Limitation in this function, or pain elicited on passive internal rotation or resisted external rotation would suggest an injury to the infraspinatus muscle or tendon.

Objective: The supraspinatus tendon is commonly injured or impinged during overhead motion. Abduction with internal rotation of the arm can produce pain with impingement of the injured tendon. I3D Previous Next Score Report Lab Values Calculator Help Pause

92 Exam Section 2: Item 42 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 42. A 35-year-old woman comes to the physician in the spring because of a 2-week history of tearing and itching of her eyes. She has a 10-year history of similar symptoms during this time of year. She attributes her condition to an allergy to pollen and grasses. Her symptoms improve when she takes an over-the-counter antihistamine. The antihistamine therapy most likely affects which of the following aspects of the host response in this patient? O A) Chemotaxis O B) IgE-mast cell binding O C) Leukocyte adhesion D) Phagocytosis E) Vascular permeability Correct Answer: E. Increased vascular permeability, driven by histamine release from IgE-crosslinked mast cells in the presence of an allergen, causes the symptoms of seasonal allergies. These include rhinitis, sneezing, and itching of the eyes, nose, and palate. Histamine is stored in preformed granules within mast cells. Exposure to an allergen leads to the crosslinking of IgE antibodies on the surface of presensitized mast cells, which results in rapid degranulation with release of histamine and inflammatory cytokines. Pollens, grasses, and outdoor molds are common causes of seasonal allergies. Antihistamines, either first-generation histamine antagonists such as diphenhydramine, or the less-sedating second-generation agents such as loratadine, are a mainstay of therapy. Incorrect Answers: A, B, C, and D. Chemotaxis (Choice A) is the movement of cells in response to a chemical stimulus. While other inflammatory mediators may promote chemotaxis, such as interleukins, histamine does not. While IgE-mast cell binding (Choice B) is a key step in the host response to an allergen, it is not affected by antihistamines. Anti-IgE therapies that may prevent binding to a mast cell include omalizumab, a monoclonal antibody used in the management of asthma. Leukocyte adhesion (Choice C) to the blood vessel wall is a key step in the host response to an allergen or infection, which is mediated by integrins. The immunodeficiency syndrome leukocyte adhesion deficiency type 1 is characterized by a defect in the LFA-1 integrin. Leukocyte adhesion is not affected by antihistamine therapy. Phagocytosis (Choice D) is one of the primary mechanisms of the innate immune system in response to antigens. It is involved in the initial allergen presentation and formation of IgE antibodies but is not affected by antihistamines.

Objective: The symptoms of seasonal allergies are secondary to the vasodilation and increased capillary permeability caused by histamine. Previous Next Score Report Lab Values Calculator Help Pause

133 Exam Section 3: Item 33 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 33. A male newborn is delivered at term with a right-sided aortic arch. Physical examination shows full cheeks, low-set ears, and a small chin. Serum studies show a decreased calcium concentration. Which of the following is the most likely site of the malformations in this patient? O A) First and second branchial arches B) First and second branchial grooves C) First and second branchial pouches D) Third and fourth branchial arches E) Third and fourth branchial grooves F) Third and fourth branchial pouches Correct Answer: F. DiGeorge syndrome, a 22q11 deletion syndrome, presents as a result of failure of the third and fourth branchial pouches to develop. The third branchial pouch gives rise to the thymus and inferior parathyroid glands, and the fourth branchial pouch to the superior parathyroid glands. Člassically, it presents with immunodeficiency caused by failure of thymus development (thymic aplasia), hypocalcemia from failure of parathyroid gland development, structural cardiac defects, and craniofacial malformations. Common physical examination findings include a cleft palate and low-set ears; however, the expressivity varies and combinations of abnormal palate, facial, and cardiac malformations are observed. The immunodeficiency results from aplasia of the thymus, the site of maturation of T lymphocytes, which results in an increased susceptibility to predominantly viral and fungal infections. The definitive diagnosis requires fluorescence in-situ hybridization to detect the missing DNA segment with a fluorescent probe. Treatment is supportive, as there is no cure for the condition. Incorrect Answers: A, B, C, D, and E. The first and second branchial arches (Choice A) give rise to the mandible, malleus, maxilla, stapes, styloid process of the skull, lesser horn of the hyoid, and stylohyoid ligament, along with the muscles of mastication, anterior digastric, tensor tympani, tensor veli palatini, stapedius, stylohyoid, platysma, and posterior digastric. They also give rise to the maxillary, mandibular, and facial nerves. Treacher Collins syndrome and congenital pharyngocutaneous fistula are associated with abnormal development of the first and second branchial arches. The first and second branchial grooves (Choice B), also known as clefts, give rise to the external acoustic meatus (first cleft) and the cervical sinus (second cleft in combination with the third and fourth branchial grooves [Choice E]), which do not typically persist into adulthood. When the second through fourth branchial grooves abnormally persist, a branchial cleft cyst is often the result. The first and second branchial pouches (Choice C) give rise to the middle ear cavity, mastoid air cells, and palatine tonsils. The third and fourth branchial arches (Choice D) give rise to the greater horn of the hyoid, cricoid, arytenoids, corniculate, cuneiform, and thyroid, muscles such as the stylopharyngeus and pharyngeal constrictors, and the glossopharyngeal and vagus nerves.

Objective: The third branchial pouch gives rise to the thymus and inferior parathyroid glands, and the fourth branchial pouch to the superior parathyroid glands. Abnormalities involving development of these pouches typically stem from a deletion of chromosome 22q11, which presents as DiGeorge syndrome or velocardiofacial syndrome. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

141 Exam Section 3: Item 41 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 41. A57-year-old woman comes to the emergency department because of a 7-week history of fever, night sweats, and nonproductive cough; she also has had a 6.8-kg (15-lb) weight loss during this period. She is currently receiving pharmacotherapy for Crohn disease but does not remember the name of the medication. Her temperature is 38.2°C (100.8°F), pulse is 90/min, respirations are 23/min, and blood pressure is 130/90 mm Hg. Amphoric breath sounds are heard over the right upper lung field. A chest x-ray is shown. This patient is most likely being treated with a monoclonal antibody targeted against which of the following? A) CD3 B) CD20 C) Complement protein C5 O D) IgE E) Tumor necrosis factor-a Correct Answer: E. Monoclonal antibody therapy targeted against tumor necrosis factor-a (TNF-a) increases a patient's risk for active mycobacterial infection. This patient presents with fevers, weight loss, and cough, in association with a cavitary right apical lung lesion on chest x- ray, which are consistent with a reactivated Mycobacterium tuberculosis (MTB) infection. Monoclonal antibodies against TNF-a that are specifically used to treat Crohn disease include infliximab, adalimumab, and certolizumab. These medications are known to increase the risk for reactivating latent MTB because of their deleterious effect on granuloma formation and maintenance. All patients who are considered for treatment with these agents must undergo screening for latent MTB infection. If positive, treatment for latent MTB with nine months of isoniazid is indicated. Incorrect Answers: A, B, C, and D. CD3 (Choice A) is a primary surface cell marker on T lymphocytes. A monoclonal antibody against this protein was first marketed as muromonab and was used to treat acute cellular rejection in kidney, cardiac, and liver transplant recipients. Use of this medication did not predispose to reactivation of MTB but was associated with cytokine release syndrome. It is no longer in use. CD20 (Choice B) is a surface marker found on B lymphocytes. The most commonly used monoclonal antibody targeting this site is rituximab. It is currently used as a primary component of chemotherapy regimens to treat a variety of B-cell lymphomas. It is also used to treat vasculitis such as granulomatosis with polyangiitis, microscopic polyangiitis, neurologic disorders such as myasthenia gravis and autoimmune encephalitis, and refractory immune thrombocytopenia. It is well tolerated, but patients must be tested for hepatitis B as the reactivation of hepatitis B is a known side effect. Eculizumab is a monoclonal antibody that binds to and inhibits complement protein C5 (Choice C) and thereby prevents formation of the membrane attack complex. It is used to treat atypical hemolytic uremic syndrome (HUS), paroxysmal nocturnal hemoglobinuria (PNH), and myasthenia gravis. Like patients with terminal complement deficiencies, patients taking eculizumab are predisposed to infection with Neisseria meningitidis. Omalizumab (Choice D) is a monoclonal antibody against IgE used to treat patients with asthma and increased IgE concentrations, as well as those with refractory chronic idiopathic urticaria. It can increase the risk for helminthic infection.

Objective: The use of monoclonal antibody therapy against TNF-a is associated with an increased risk for reactivated latent MTB. Because of this, all prospective candidates for this therapy should be screened for latent MTB and accordingly treated if positive. II Previous Next Score Report Lab Values Calculator Help Pause

41 Exam Section 1: Item 41 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 41. A 55-year-old woman is scheduled to undergo transvaginal hysterectomy and oophorectomy for dysfunctional uterine bleeding. During the procedure, the uterus must be separated from all surrounding pelvic structures. Identification and incision of which of the following structures that attaches to the cervical region and extends posteriorly is most appropriate in this patient? O A) Mesometrium O B) Mesosalpinx C) Mesovarium D) Ovarian ligament E) Round ligament of the uterus F) Uterosacral ligament Correct Answer: F. The uterosacral ligament is one of the structures that supports the uterus and holds it in place in the pelvis. It is a paired structure that connects the uterus to the sacrum at the level of the cervix. They may also be called the rectouterine ligaments or sacrocervical ligaments. The uterus is also supported by the broad ligament, which attaches to the superior aspect of the uterus and is a layer of peritoneum. The pubocervical ligaments and cardinal ligaments join the uterosacral ligament in supporting the middle portion of the uterus. The inferior portion of the uterus receives further support from the muscular components of the pelvic floor such as the levator ani. Each of these connections must be severed in order to successfully remove the uterus during a hysterectomy. Incorrect Answers: A, B, C, D, and E. The mesometrium (Choice A), mesosalpinx (Choice B), and mesovarium (Choice C) are named portions of the broad ligament. While these structures are all contiguous, the mesometrium covers the uterus and creates most of the broad ligament, the mesosalpinx covers the fallopian tubes, and the mesovarium covers the ovaries. The ovarian ligament (Choice D) is a fibrous structure that extends medially from the ovary to the uterus. In contrast, the suspensory ligament of the ovary extends laterally from the ovary to the wall of the pelvis and contains the ovarian artery and vein. The round ligament of the uterus (Choice E) is a remnant of the gubernaculum, an embryonic structure that assists in the descent of the gonads. The ligament begins at the cornua of the uterus where the fallopian tubes insert, passes through the inguinal canal, and terminates in the labia majora.

Objective: The uterosacral ligament is a paired structure that extends bilaterally from the cervical region to the sacrum, helping to support the uterus in the pelvis. Previous Next Score Report Lab Values Calculator Help Pause

23 Exam Section 1: Item 23 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 23. During a study on exercise, a 45-year-old woman runs on a treadmill for 30 minutes. An increase in which of the following is most likely to change gastrocnemius muscle blood flow in this woman? A) Interstitial adenosine concentration B) Interstitial amino acid concentration C) Interstitial glucose concentration D) Interstitial oxygen tension E) Parasympathetic stimulation F) Sympathetic stimulation Correct Answer: A. During aerobic exercise, the cellular consumption of adenosine triphosphate (ATP) increases. Myosin motor proteins dephosphorylate ATP and subsequently use the energy released to generate muscle contraction. Adenosine will become diphosphate, monophosphate, or completely without phosphate as a result of this reaction, and it must then be reconverted to ATP via the Krebs cycle and oxidative phosphorylation in the mitochondrion. Byproducts of energy metabolism regulate regional blood flow, such as carbon dioxide and organic acids. Additionaly, as the interstitial adenosine concentration increases, blood vessels dilate in response. This is known as exercise hyperemia, and it occurs through mediation from local regulators. Încreased acidity, adenosine, hypoxia, nitric oxide, and prostaglandins are all important mediators of exercise-induced vasodilation. Adenosine is also a vasodilator in coronary vascular smooth muscle as well as in skeletal muscle. Adenosine modulates vasodilation by binding to the adenosine receptor and increasing cyclic adenosine monophosphate (CAMP). This leads to inhibition of myosin light chain kinase and vascular smooth muscle relaxation. Incorrect Answers: B, C, D, E, and F. Interstitial amino acid concentration (Choice B) decreases during exercise as protein and amino acids are oxidized as an energy source. Interstitial glucose concentration (Choice C) falls during exercise as it is used in glycolysis, the Krebs cycle, and oxidative phosphorylation to supply the active muscle with ATP. Glucose is stored in muscle as glycogen. Glucose itself is not a key regulator of vasodilation. Interstitial oxygen tension (Choice D), when decreased, may trigger vasodilation in skeletal muscle, but when increased, vasoconstriction would be expected. Parasympathetic stimulation (Choice E) leads to increased blood flow to splanchnic circulation, thereby shunting blood away from the peripheral tissues and toward the abdominal viscera. Parasympathetic stimulation occurs during times of minimal activity. During exercise, sympathetic stimulation would occur. Sympathetic stimulation (Choice F) can lead to both vasoconstriction and vasodilation. Vasodilation occurs through B2-adrenoreceptors and vasoconstriction through a1-adrenoreceptors via G-protein coupled receptors. Sympathetic stimulation is not focal, and a better explanation for the increase in blood supply to the gastrocnemius comes from local autoregulation through the production of local mediators.

Objective: There are multiple mediators of local muscle blood flow. These include adenosine, lactate, hydrogen ions, and the partial pressures of oxygen and carbon dioxide. These autoregulatory mediators serve to increase skeletal muscle blood flow during exercise. Previous Next Score Report Lab Values Calculator Help Pause

87 Exam Section 2: Item 37 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 37. A 5-month-old boy is brought to the physician by his parents because of a 1-week history of weakness. His parents are first cousins. Physical examination shows pallor, jaundice, and hepatosplenomegaly. His hemoglobin concentration is 6 g/dL, hematocrit is 19%, and serum bilirubin concentration is 4.2 mg/dL (mostly indirect). A photomicrograph of a peripheral blood smear is shown. Hemoglobin (Hb) electrophoresis shows HbF, no HbA, and an increased concentration of HbA, Which of the following is the most likely molecular cause of this patient's condition? A) Deletion of one or more a-globin genes B) Deletion of one or more B-globin genes C) Mutation of the a-globin gene inherited from each parent D) Mutation of the a-globin gene inherited from one parent E) Mutation of the B-globin gene inherited from each parent O F) Mutation of the B-globin gene inherited from one parent Correct Answer: E. This 5-month-old boy with anemia, indirect hyperbilirubinemia, and a peripheral blood smear demonstrating microcytosis and target cells has B-thalassemia major. There are two B-globin genes, and mutations of both lead to an absence of functional B-subunits and an inability to synthesize normal adult HbA (a2-ß2). Hemoglobin electrophoresis will demonstrate an absence of HbA and increased concentrations of HBA2 (a2-02) and HbF (a2-V2), as seen in this patient. The subsequent imbalance in hemoglobin concentrations leads to ineffective erythropoiesis, and microcytic sometimes hemolytic anemia. Patients with B-thalassemia major are transfusion dependent and have physical examination findings consistent with extramedullary hematopoiesis, such as frontal bossing. Incorrect Answers: A, B, C, D, and F. Deletion of one a-globin chain (Choice A) causes alpha-thalassemia trait (silent carrier), which presents with normal hemoglobin concentrations, normal hemoglobin electrophoresis, and microcytosis. There are four a-globin genes organized as pairs on chromosome 16, with one pair inherited from each parent. The severity of anemia is determined by the number and the nature of the mutations, with a greater number of deletions resulting in more severe phenotypes. Deletion of one or more B-globin genes (Choice B) would cause either B-thalassemia minor or major depending on the number affected; heterozygotes are generally asymptomatic apart from mild anemia and increased HBA2 on electrophoresis, whereas homozygotes with absence of all B-globin would present with severe, transfusion-dependent anemia. Deletion of B-globin genes is rare in comparison to mutations, which are often point mutations located within splice sites or regulatory sequences. Mutation of one a-globin chain from each parent (Choice C) or mutation of both a-globin chains from one parent (Choice D) cause alpha-thalassemia, which in the case of a two-allele deletion is characterized by mild microcytic anemia and hemoglobin electrophoresis with a small concentration of Hb-Barts (Y4). These patients are generally asymptomatic. Mutation of the B-globin gene inherited from one parent (Choice F) causes B-thalassemia minor. This is primarily an asymptomatic disease with mild microcytic anemia. Hemoglobin electrophoresis will show a small amount of HBA2 (~4%) and HbF (-5%) but near normal concentrations of HbA (-90%).

Objective: There are two B-globin genes, with one inherited from each parent. Mutations or deletions of one copy produces a mild disease called B-thalassemia minor while deletion or mutation of both genes causes B-thalassemia major, which is characterized by severe anemia from ineffective erythropoiesis and hemolysis. Point mutations affecting splice sites and promoter sequences are the most common cause. Previous Next Score Report Lab Values Calculator Help Pause

52 Exam Section 2: Item 2 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 2. A woman who weighs 70 kg (154 Ib) is participating in a research study. Her hematocrit is 40%, and her plasma volume is 3 L. The subject's total blood volume (in L) is closest to which of the following? A) 4 B) 5 C) 6 D) 7 E) 8 Correct Answer: B. Blood is a composite of both red blood cells and plasma. Plasma describes the acellular component of blood, though contains soluble proteins such as clotting factors. The percentage of total blood volume that is pure red blood cells is known as the hematocrit. From this information, the total blood volume can be calculated by the following equation: Plasma volume (L) = Total Blood Volume (L) x (100% - hematocrit). In this example, the patient's plasma volume is three liters. Therefore, 3L = Total Blood Volume x (100% - 40%) = Total Blood Volume x 60%. Dividing 3L by 60% solves for the Total Blood Volume, which is 5L. An alternative method for estimating total blood volume in an adult is to multiply the patient's weight in kg by the estimating factor 70 mL/kg, reflected in the equation Total Blood Volume (L) = weight (kg) x 70 mL/kg. In this case, 70 kg x 70 mL/kg = 4,900 mL, or 4.9L, roughly equal to 5L. Incorrect Answers: A, C, D, and E. Choices A, B, C, and E do not mathematically reflect total blood volume in this patient, and instead reflect mathematical or assumption-based errors.

Objective: Total blood volume can be calculated by the following equation: Plasma volume (L) = Total Blood Volume (L) × (100% - hematocrit). Total blood volume in a healthy adult without obesity can also be grossly estimated by Total Blood Volume (L) = weight (kg) x 70 mL/kg. Previous Next Score Report Lab Values Calculator Help Pause

98 Exam Section 2: Item 48 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 48. A 39-year-old woman with type 1 diabetes mellitus comes to the physician because of episodes of fainting during the past 3 weeks. The patient's insulin monitoring has shown that the episodes occur when she becomes hypoglycemic. In the past, her hypoglycemic episodes were characterized by increased sweating, heart rate, and shaking of her hands, but now she loses consciousness without warning. Current medications are insulin, lisinopril, and simvastatin. Her pulse is 72/min, and blood pressure is 128/70 mm Hg. Physical examination shows no abnormalities. It is most appropriate for this patient to receive an injection of a drug with which of the following mechanisms of action during these episodes? A) Decreased intestinal absorption of glucose B) Enhanced metabolism of insulin C) Promotion of peripheral insulin uptake D) Promotion of skeletal muscle glucose release E) Stimulation of hepatic glucose production Correct Answer: E. This patient presents with symptoms of hypoglycemia, which are generally neuroglycopenic (agitation, headache, blurry vision, diplopia, tremor, loss of consciousness), autonomic (diaphoresis, tremoring, nausea, vomiting, cramping, tachycardia, anxiety), and nonspecific (fatigue, weakness, lethargy), often exacerbated by periods of fasting or exercise. In extreme cases, hypoglycemia can present with seizures, coma, arrhythmias, and death. The use of insulin, especially if the timing is distant from the oral consumption of carbohydrates, can result in precipitous drops in blood sugar with the resultant occurrence of any combination of neuroglycopenic, autonomic, or nonspecific symptoms. The treatment of hypoglycemia during such an episode should involve the immediate administration of oral or intravenous glucose, or intramuscular glucagon, which causes the stimulation of hepatic glucose production via gluconeogenesis and glycogenolysis, in turn stabilizing the patient's blood sugar. Incorrect Answers: A, B, C, and D. Decreased intestinal absorption of glucose (Choice A) would cause, not alleviate, hypoglycemia. Some antihyperglycemic medications such as acarbose and miglitol interfere with the intestinal digestion and absorption of carbohydrates but would not be appropriate to use in a case of hypoglycemia. Enhanced metabolism of insulin (Choice B) and promotion of peripheral insulin uptake (Choice C) would theoretically limit the effect of insulin and limit hypoglycemic episodes. Insulin is metabolized primarily by cells in the liver and kidney following endocytosis after binding to its receptor. It is metabolized by insulin-degrading enzyme, an intracellular protease. This mechanism is not the target of widespread antihypoglycemic therapies. Promotion of skeletal muscle glucose release (Choice D) is not feasible as skeletal muscle lacks the glucose 6-phosphatase enzyme necessary to synthesize and release glucose. Once glucose is phosphorylated within a myocyte, it can only be used within the metabolic pathways of the cell in which it resides.

Objective: Treatment of hypoglycemia should involve the immediate administration of oral or intravenous glucose, or intramuscular glucagon, which causes the stimulation of hepatic glucose production via gluconeogenesis and glycogenolysis, in turn stabilizing the blood sugar concentrations. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

130 Exam Section 3: Item 30 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 30. A 57-year-old woman comes to the physician 3 days after she found a lump in her left breast on self-examination. Physical examination shows a 1.3-cm mass in the left breast. Microscopic examination of a biopsy specimen of the mass shows carcinoma. It is most appropriate for the physician to consider which of the following when grading the carcinoma? O A) Distant metastasis B) Maximum tumor size C) Number of mitoses per unit area D) Peritumoral lymphocytic response E) Presence of extracellular mucin F) Regional lymph node metastasis Correct Answer: C. Tumors are classified by grade and stage. Grading describes the cellular and histologic features of the tumor. Low grade tumors remain well-differentiated and have a better prognosis than high grade tumors, which are poorly differentiated or undifferentiated. Grading systems differ by the type of tumor being described. A high number of mitoses per unit area is evidence of rapid cell growth and division and is a common feature of high-grade tumors. In general, the stage of a tumor is more prognostically significant than grade. Staging describes the size and extent of a primary tumor, the degree of involvement of regional lymph nodes, and the presence of distant metastases. It is usually standardized using the "TNM" (Tumor, Nodes, Metastasis) staging system. Incorrect Answers: A, B, D, E, and F. Distant metastasis (Choice A) and regional lymph node metastasis (Choice F) are the most prognostically important elements of tumor staging. The presence of a metastasis is not a feature of tumor grading. Maximum tumor size (Choice B) is an element of tumor staging but does not play a role in tumor grading. Peritumoral lymphocytic response (Choice D) occurs when the immune system mounts a local inflammatory response against the tumor. While this may be prognostically significant in some malignancies, including breast carcinoma, the presence or absence of a lymphocytic response does not generally play a role in cancer staging or grading. Presence of extracellular mucin (Choice E) is a feature of ductal breast carcinoma. However, the presence of this finding does not play a role in breast cancer grading.

Objective: Tumor grading describes the cellular and histologic features of a tumor. The number of mitoses per unit area, when increased, reflects active cell growth and division within a tumor. Previous Next Score Report Lab Values Calculator Help Pause

34 Exam Section 1: Item 34 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 34. A 30-year-old woman, gravida 2, para 1, who is Rh-negative delivers a male stillborn at 24 weeks' gestation. She had no prenatal care. The stillborn is Rh-positive. Autopsy findings are consistent with erythroblastosis fetalis. The mechanism of this reaction is most similar to that found in which of the following conditions? A) Graft-versus-host disease B) Graves disease O C) Peanut allergy D) Poison ivy dermatitis E) Tuberculin reaction Correct Answer: B. Graves disease is an autoimmune hyperthyroid condition caused by thyroid-stimulating antibodies that activate the TSH receptor. It is an example of type Il hypersensitivity. Type II hypersensitivity is characterized by antibodies binding to cell-surface antigens, leading to cellular destruction or manifestation of clinical disease. Erythroblastosis fetalis is also an example of type II hypersensitivity. In a primigravid Rh(D)-negative mother with a Rh(D)-positive fetus, antibodies to Rh(D) antigen form during delivery or any exposure of maternal antigen-presenting cells to fetal red blood cells (eg, trauma, vaginal bleeding). When the mother becomes pregnant for a second time with an Rh(D)-positive fetus, the antibodies to Rh(D)-antigen cross the placenta and bind Rh(D)-antigen on the fetal erythrocytes, leading to cell destruction and hemolytic anemia that can result in fetal demise. This process, called erythroblastosis fetalis or hemolytic disease of the newborn, can be stopped by administering anti-D-immunoglobulin to Rh(D)-negative women during the third trimester and again after delivery. This shields the antigen on the fetal erythrocytes and prevents the maternal antigen-presenting cells from initiating antibody development. Incorrect Answers: A, C, D, and E. Graft-versus-host disease (Choice A) and poison ivy dermatitis (Choice D) are both examples of type IV hypersensitivity. Type IV, or delayed-type, hypersensitivity is characterized by a cell-mediated response, which involves the maturation of antigen-specific CD4+ or CD8+ lymphocytes to a specific antigen. When the antigen is encountered, CD4+ cells release cytokines leading to inflammation and macrophage activation, while CD8+ cells directly kill cells expressing the antigen. This type of hypersensitivity takes several days to manifest. Peanut allergy (Choice C) is an example of type I hypersensitivity. In this type, preformed IgE on the surface of mast cells and basophils is crosslinked by an antigen. This leads to immediate degranulation and release of histamine. Tuberculin reaction (Choice E) is a test of type IV hypersensitivity to tuberculin, an antigen made by Mycobacterium tuberculosis. If the patient has been exposed to the bacteria, they will have developed cell-mediated immunity to tuberculin, thus, injecting a small amount in the skin will cause an indurated nodule to form, indicating the presence of an immune response.

Objective: Type II hypersensitivity is characterized by the formation of antibodies to cell surface antigens. Both erythroblastosis fetalis and Graves disease are examples of type II hypersensitivity. Previous Next Score Report Lab Values Calculator Help Pause

44 Exam Section 1: Item 44 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 44. A 2-month-old boy is brought to the physician by his mother for a well-child examination. The mother tells the physician that her son has been gazing at her more directly. He also indicates delight with gestures when he sees her and shows signs of distress when he cannot see her. Which of the following best describes this patient's recent behavior? O A) Attachment B) Enmeshment O C) Manipulation D) Merging O E) Separation anxiety Correct Answer: A. This infant appears to be forming a secure attachment with his mother, as indicated by his eye contact (occurring at the expected age of 2 months) and positive emotional response. Infants younger than approximately 7 months typically cry when their attachment figure is absent because they have not yet developed a sense of object permanence, or the understanding that their attachment figure's absence is only temporary. Most infants will form a secure attachment to their caregiver if the caregiver is regularly providing physical and emotional care. The quality of this attachment predicts a child's social and emotional health later in life. Incorrect Answers: B, C, D, and E. Enmeshment and merging (Choices B and D) refer to pathologically poor interpersonal boundaries between two people or a family and can signify incomplete separation-individuation. An incestuous mother-son relationship is an extreme example of enmeshment. Separation-individuation is the process by which children become aware that they are individuals and begin to exert autonomy in their environment. This patient is illustrating normal attachment behavior and is too young to have started the separation-individuation process, which typically occurs later in the first year of life to early childhood. Manipulation (Choice C) refers to exerting influence over others' behaviors using deceptive methods. Manipulation tactics typically start in early childhood and manifest in behaviors such as faking illness to stay home from school. This infant is too young to manipulate his caregivers. Separation anxiety (Choice E) is an indication of secure attachment that typically develops at age nine to twelve months in which infants cling to their caregiver, are severely distressed when the caregiver leaves, and are afraid of unfamiliar people. At this age, infants have developed a sense of object permanence, or the knowledge that their caregiver continues to exist even when the caregiver is out of sight. Infants typically learn to cry in these situations because crying will encourage their caregiver to return sooner than later.

Objective: Typically developing infants who form secure attachments with their caregiver will begin to make eye contact at age 2 months and illustrate delight when their caregiver is present and distress when their caregiver is absent. Previous Next Score Report Lab Values Calculator Help Pause

77 Exam Section 2: Item 27 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 27. A 19-year-old man is admitted to the hospital following a motor vehicle collision. Physical examination shows a penetrating wound to the right cerebral cortex with complete paralysis of the left lower extremity, fracture of the right mid humerus with severing of the radial nerve, and a fracture of the right tibia. Treatment includes cast immobilization of the right upper extremity and right knee and ankle. After 10 weeks, the casts are scheduled to be removed from the right upper and lower extremities. At this point, the deep tendon reflex is most likely to be strongest in which of the following locations in this patient? A) Left Achilles tendon B) Left biceps tendon C) Right brachioradialis tendon D) Right patellar tendon Correct Answer: A. Deep tendon reflexes are caused by a motor neuron reflex arc in which a sensory afferent neuron synapses directly with a lower motor efferent neuron to yield an abrupt motor response to a stimulus. Upper motor neurons modulate these reflexes but are not required for the reflex to occur. Cerebral cortex motor neurons aid in controlling these reflexes as they descend to and synapse on the lower motor neuron cell body in the anterior horn of the spinal cord. Alterations in the reflex arc can be categorized by abnormal strength or response of the reflex. Hyporeflexia is an absent or diminished response and is characteristic of injury to the neurons which make up the reflex arc. Hyperreflexia is an increase in the strength of response and results from injury to the cerebral cortex or upper motor neuron, resulting in the loss of modulation of the lower motor neurons. Hyperreflexia results from a loss of the inhibitory function of the descending neuron. The patient had a traumatic upper motor neuron injury to the left cerebral hemisphere motor cortex, which controls the left lower extremity. His left lower extremity reflexes will be increased as the reflex arc will lack the descending motor neuron inhibition. Thus, his Achilles tendon reflex will be strongest and may present as hyperreflexia or even clonus. It should be noted that many muscle groups will demonstrate atrophy from his prolonged course of bed rest, casting, as well as lack of voluntary control of the extremity. Incorrect Answers: B, C, and D. The left biceps tendon (Choice B) reflex will be relatively normal as the patient has had no direct peripheral nerve injury or upper motor neuron injury that affects this extremity. A traumatic injury to the right cerebral cortex, if large enough, could affect both the upper and lower extremities. In this case, the patient demonstrates an affected lower extremity only. The right brachioradialis tendon (Choice C) reflex is innervated by the radial nerve, which was severed by the right midshaft humerus fracture. This is a common injury pattern as the radial nerve abuts the humerus as it courses within the spiral groove. A complete transection of a nerve will lead to a loss of neurons required for any associated reflex arc, demonstrated by hypo- or areflexia. The right patellar tendon (Choice D) reflex arc remains intact, however because of 10 weeks of cast immobilization, the muscles of the right leg will be significantly atrophied leading to an overall decreased muscle strength and bulk, which may diminish the strength of the reflex.

Objective: Upper motor neuron injuries lead to hyperreflexia and/or spasticity as the descending motor neurons no longer provide inhibitory inputs to the lower motor neurons of the reflex arcs. Lower sensory or motor neuron lesions interrupt the reflex arc, thereby causing hyporeflexia. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

169 Exam Section 4: Item 19 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 19. A4-year-old boy undergoes radiographic imaging studies of the urinary tract after having two bacterial urinary infections during the past year. His left kidney is found to be abnormally small and nonfunctional. The right kidney appears normal. He undergoes a left nephrectomy and ureterectomy. A photograph of the resected specimen is shown. Which of the following disorders most likely accounts for the pathologic changes in this kidney? A) Arteriolonephrosclerosis B) Hypospadias C) Panhypopituitarism D) Renal papillary necrosis E) Ureteral obstruction Correct Answer: E. Congenital urinary tract abnormalities include, but are not limited to, unilateral renal agenesis, fused kidneys, ureteral stricture and stenosis, duplex collecting system, posterior urethral valves, bladder agenesis, bladder exstrophy, hypo- or epispadias, and urethral strictures. A duplex collecting system is one of the most common and is caused by a bifurcation of the ureteric bud resulting in two ureters or a Y-shaped ureter. It is often associated with a division of the kidney into two parts, an upper and lower lobe. Regardless of the cause of this malformation, ureteral obstruction and vesicoureteral reflux are common complications, both of which increase the risk for urinary tract infections, as demonstrated in this patient's history. Chronic vesicoureteral reflux also causes hydroureter and hydronephrosis, with compression atrophy (reflux nephropathy) of the renal parenchyma, which results in a nonfunctional kidney, as seen in this patient. Incorrect Answers: A, B, C, and D. Arteriolonephrosclerosis (Choice A) is characterized by the sclerosis of renal arterioles secondary to long-standing hypertension. Hypertension causes thickening and hyalinization of vessel walls, which leads to narrowing of the lumen with resultant ischemia of the parenchyma, decreased glomerular filtration rate, and proteinuria. Untreated, it can lead to end-stage kidney disease. Hypospadias (Choice B) is an anatomic abnormality in which the opening of the penile urethra lies on the ventral surface of the penis. It is a congenital abnormality that occurs secondary to the failure of urethral folds to fuse and is a common birth defect found in up to 0.5% of all male children. In most cases, it is a solitary defect, however it can be associated with an inguinal hernia or cryptorchidism. It is more common than an abnormal opening on the dorsal surface of the penis (epispadias). Panhypopituitarism (Choice C) describes the absence of all pituitary hormones, including growth hormone, adrenocorticotropic hormone, thyroid stimulating hormone, luteinizing hormone, follicle stimulating hormone, antidiuretic hormone, oxytocin, and prolactin. This may be because of a congenital or acquired cause, such as radiation, severe head trauma, infections, or tumors. enal papillary necrosis (Choice D) refers to the necrosis and sloughing of renal papillae, which can be triggered by infections (eg, acute pyelonephritis), diabetes mellitus, sickle cell disease, or nonsteroidal anti-inflammatory medications. It presents with gross hematuria and proteinuria.

Objective: Urinary tract malformations, such as duplex collecting system, ureteral stricture, ureteropelvic junction stenosis, or posterior urethral valves are common, and often lead to complications involving vesicoureteral reflux and ureteral obstruction. This increases the patient's risk for urinary tract infections and eventually may lead to ipsilateral reflux nephropathy. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

184 Exam Section 4: Item 34 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 34. A 72-year-old man with multiple myeloma agrees to participate in an investigational chemotherapy clinical trial. Prior to enrollment in the study, peripheral blood B and T lymphocytes are isolated and undergo Southern blot analysis of lymphoid cell-specific genes. Using a single B-lymphocyte J-region probe, analysis of the B-lymphocyte DNA shows a 1.5-kb band. Analysis of the T-lymphocyte DNA using the same J-region probe shows a 6-kb band. The 6-kb band most likely signifies which of the following? A) Constant region gene rearrangement B) D to J gene rearrangement C) V to D gene rearrangement D) V to J gene rearrangement E) Unrearranged immunoglobulin gene Correct Answer: E. Unrearranged immunoglobulin gene is most likely signified by the 6-kb band. V(D)J recombination is a process that occurs in both B and T lymphocytes and generates an array of diverse immunoglobulins (Ig) and T-cell receptors, respectively. Complete Igs are made up of a heavy and light chain, each with a variable (V) and constant (C) region that are encoded by chromosomes 14 (Ig heavy), 22 (lambda light chain), and 2 (kappa light chain). The heavy chain, before V(D)J rearrangement, contains multiple C and V genes, in addition to dozens of Diversity (D) and Joining (J) genes. The light chains are structured similarly but lack the D genes. Rearrangement refers to the process by which one gene from each of these categories is selected randomly for expression, while the remainder of the V, D, J, and C genes are deleted from the genome. Each B lymphocyte will end the process of recombination possessing its own unique V(D)J rearrangement, thereby leading to an entirely novel Ig. This diversity of Igs allows for a robust immune system capable of recognizing many pathogens. The same process occurs in T lymphocytes, but as these cells bind and recognize antigens and do not secrete Igs, their rearrangement results in the creation of a novel T-cell receptor that recognizes a specific antigen. T-cell receptors are composed of an alpha and beta chain, with the alpha segment containing multiple V, D, and J genes and the beta segment containing multiple V and J segments. Through a similar process as that described above for Igs, a gene from each of these segments is randomly selected to create a unique T-cell receptor. This V(D)J rearrangement occurs in the bone marrow for B cells and in the thymus for T cells, with T cells undergoing an additional step whereby they are exposed to self-antigens. T cells that have receptors that bind to self-antigens are eliminated to prevent the formation of autoimmunity (negative selection). The circulating B lymphocyte in this question demonstrates a single 1.5-kb band while the T lymphocyte shows a 6-kb band for the same J-region probe. This implies that the B-lymphocyte has already undergone V(D)J rearrangement in the bone marrow and thus only expresses a single J gene whereas the T lymphocyte still possesses multiple genes within the J region. Thus, this T-lymphocyte has not undergone V(D)J rearrangement in the thymus yet and continues to harbor multiple genes in the J region, which is a surrogate for an unrearranged Ig chain. Incorrect Answers: A, B, C, and D. Constant region gene rearrangement (Choice A), D to J gene rearrangement (Choice B), V to D gene rearrangement (Choice C), and V to J rearrangement (Choice D) occur sequentially once a B or T lymphocyte starts the process of V(D)J rearrangement. Because circulating B-lymphocytes have already undergone rearrangement before leaving the bone marrow, a J-region-specific probe should demonstrate only a single J-region gene, and such a probe would not be helpful in identifying a particular step of V(D)J rearrangement. Because T lymphocytes undergo rearrangement in the thymus, circulating cells are either completely rearranged or completely native and unrearranged. Thus, any J-region probe that binds to circulating T cells would show either a fully rearranged T lymphocyte, demonstrated by a 1.5-kb band, or a completely unrearranged T lymphocyte as is the case here with a 6-kb band.

Objective: V(D)J rearrangement is a process that occurs in B lymphocytes in the bone marrow, and in T lymphocytes in the thymus, resulting in the generation of completely novel Igs and T-lymphocyte receptors, respectively. Because T lymphocytes circulate as either completely rearranged or unrearranged cells, application of a region-specific probe (eg, J-region) should demonstrate the presence of a single J gene (1.5 kb) or multiple J-genes (6 kb). The presence of a 6 kb band signifies the presence of a T- lymphocyte with an unrearranged lg gene. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

161 Exam Section 4: Item 11 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 11. A 72-year-old man comes to the physician because of a 1-month history of discoloration of his legs. He lives alone and his diet consists mostly of oatmeal and prepared meats. Physical examination shows ecchymoses of the lower extremities, splinter hemorrhages of the nailbeds, perifollicular purpura, and corkscrew hairs. A deficiency of which of the following vitamins is most likely responsible for these findings? O A) A B) B, (thiamine) О с) В, (ругidoxine) D) C O E) E Correct Answer: D. Vitamin C is an antioxidant that is involved in tissue repair and collagen synthesis through its role in the hydroxylation of proline and lysine. It is also required for the enzymatic conversion of dopamine to norepinephrine and iron absorption. It is an essential nutrient found in fruits (especially citrus) and vegetables. Deficiency of vitamin C results in scurvy, which is characterized by swollen gums, bruising, poor wound healing, petechiae, perifollicular and subperiosteal hemorrhages, and short, fragile, curly hair. Immunodeficiency and anemia may also be present. Causes of vitamin C deficiency include an insufficient intake of vitamin C, such as with selective eating habits (eg, diets consisting mostly of refined grains or meats, or with limited fruits or vegetables), alcohol use disorder, or intestinal malabsorption. It is reversible, and treatment is with supplemental oral or parenteral vitamin C. Incorrect Answers: A, B, C, and E. Vitamin A (Choice A) is an antioxidant necessary for the differentiation of epithelial cells into specialized tissue. Deficiency is characterized by ocular manifestations including night blindness, corneal degeneration, dry skin, immunosuppression, and Bitot spots on the conjunctiva. Vitamin B, (thiamine) (Choice B) is a cofactor for several enzymes in glucose metabolism and adenosine triphosphate production, including pyruvate dehydrogenase and a-ketoglutarate dehydrogenase. Deficiency is characterized by Wernicke encephalopathy, a triad of confusion, ophthalmoplegia, and ataxia. Wernicke encephalopathy is theoretically reversible with the administration of high-dose thiamine; if untreated, it can progress to Korsakoff syndrome, which is characterized by dementia, confabulation, hallucinations, and psychosis. Vitamin B6 (pyridoxine) (Choice C) deficiency limits the synthesis of histamine, hemoglobin, and neurotransmitters including epinephrine, norepinephrine, dopamine, serotonin, and GABA. Deficiency commonly presents with peripheral neuropathy, dermatitis, sideroblastic anemia, glossitis, and seizures (especially in the setting of isoniazid use). Vitamin E deficiency (Choice E) is an antioxidant that protects cells from free radical damage. Deficiency may present with hemolytic anemia and generalized muscle weakness. It can have a similar presentation to vitamin B12 (cobalamin) deficiency with posterior column and spinocerebellar tract demyelination.

Objective: Vitamin C is an essential antioxidant nutrient that is found in fruits and vegetables, and is necessary for collagen synthesis, iron absorption, immune function, and the conversion of dopamine to norepinephrine. Deficiency causes scurvy, characterized by swollen gums, bruising, poor wound healing, petechiae, perifollicular and subperiosteal hemorrhages, and short, fragile, curly hair. Previous Next Score Report Lab Values Calculator Help Pause

3 Exam Section 1: Item 3 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 3. A9-year-old boy has had alopecia and hypocalcemia since birth. His serum 1,25-dihydroxycholecalciferol concentration is within the reference range, serum 24,25-dihydroxycholecalciferol is undetectable. Serum parathyroid hormone concentration is above the upper limit of the laboratory assay and needs to be reassayed at a dilution. The results are pending. Which of the following is the most likely cause of his condition? O A) Loss-of-function mutation in the calcium-sensing receptor B) Multiple endocrine neoplasia type I C) Mutations inactivating the vitamin D 24-hydroxylase gene D) Mutations inactivating the vitamin D receptor Correct Answer: D. Vitamin D plays a role in serum calcium and phosphate homeostasis by promoting the intestinal absorption of calcium and phosphate. Parathyroid hormone (PTH) also has a role in calcium and phosphate regulation by stimulating osteoclastic bone reabsorption and distal convoluted tubular calcium reabsorption and phosphate excretion in the kidney. Vitamin D deficiency can be caused by malabsorption in the intestinal tract, malnutrition or insufficient dietary intake, and decreased sun exposure. Decreased concentration of vitamin D results in decreased intestinal calcium absorption and hypocalcemia, sensed by the parathyroid gland via calcium-sensing receptors, which leads to an increase in secretion of PTH to normalize serum calcium concentration. Increased PTH causes bone resorption, liberating calcium and phosphorus, otherwise stored in bone. In this patient, 1,25-dihydroxycholecalciferol, the active form of vitamin D, is within reference range. However, the vitamin D receptor must also exist and function appropriately for the hormone to exert its effect. Vitamin D binds a nuclear receptor, exerting its effect by gene transcription. In this case, a normal concentration of Vitamin D with persistent hypocalcemia suggests an inactivating mutation in the receptor itself, or a mutation in a regulatory sequence that prevents transcription of the genes coding for the receptor. In turn, the patient experiences persistent hypocalcemia despite normal vitamin D concentration, as the receptor mutation prevents the action of vitamin D in promoting the intestinal absorption of calcium. Incorrect Answers: A, B, and C. Loss-of-function mutation in the calcium-sensing receptor (Choice A) in the parathyroid gland would prevent the parathyroid gland from sensing low concentrations of calcium and reacting by secreting PTH. This would cause hypoparathyroidism, not hyperparathyroidism as seen in the patient. Multiple endocrine neoplasia type I (Choice B) is characterized by often functional pituitary, pancreatic, and parathyroid tumors. Pituitary tumors may secrete prolactin or growth hormone, pancreatic tumors may secrete insulin, glucagon, or vasoactive intestinal peptide, and parathyroid adenomas may secrete PTH. A functionally active parathyroid adenoma would cause an increase in PTH, and because of the role of PTH in calcium reabsorption and bone resorption, hypercalcemia would result. Mutations inactivating the vitamin D 24-hydroxylase gene (Choice C) would result in decreased concentration of 24,25-dihydroxycholecalciferol, an inactive metabolite. Vitamin D would persist in its active form as a result, which may indirectly result in hypercalcemia.

Objective: Vitamin D plays a role in serum calcium and phosphate homeostasis by promoting the intestinal absorption of calcium and phosphate. Vitamin D deficiency typically leads to decreased intestinal calcium absorption, hypocalcemia, increased PTH secretion, and increased PTH-mediated bone resorption. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

63 Exam Section 2: Item 13 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 13. A7-day-old female newborn is brought to the physician because blood has been oozing from the umbilical area for more than 24 hours. She was born at home, has had no previous medical treatment, and has been breast-fed since birth. Which of the following biochemical reactions is most likely defective in this patient? O A) Cross-linking of collagen and elastin B) Hydroxylation of proline C) Post-translational carboxylation of glutamate residues D) Synthesis of thymidylate Correct Answer: C. Vitamin K plays a critical role in the synthesis of hepatic coagulation proteins; it is oxidized in the liver during the post-translational carboxylation of glutamate residues on coagulation factors II, VII, IX, X, and proteins C and S. It is a co-factor for vitamin K epoxide reductase, which reduces vitamin K following its involvement in glutamate carboxylation. Vitamin K is often deficient in the neonatal period, as the sterile gastrointestinal tract of the neonate lacks the commensal bacteria that normally act as a source of vitamin K in the adult; additionally, the neonatal diet tends to be vitamin K poor, especially those who are breast-fed as it has a low concentration in breast milk. Newborns are often administered vitamin K at the time of birth to prevent bleeding in the neonatal period as a result of this natural deficiency. In this case, the home-born, breast-fed neonate is likely deficient in vitamin K, impairing the synthesis of coagulation proteins and therefore impairing hemostasis. Incorrect Answers: A, B, and D. Cross-linking of collagen and elastin (Choice A) generally involves lysine residues and oxidation or hydrolysis reactions, often using copper as a cofactor. Deficiencies in this process result in connective tissue diseases such as Ehlers-Danlos syndrome and Menkes disease. Ehlers-Danlos syndrome can present with easy bruising, but also presents with characteristic hyperextensible skin and hypermobile joints; it is less likely than vitamin K deficiency in a home-born, breast-fed neonate as a cause of bleeding. Hydroxylation of proline (Choice B) occurs in the rough endoplasmic reticulum during collagen synthesis and requires vitamin C as a cofactor. Deficiency of vitamin C causes scurvy, which presents with fragile hair and easy bruising. It is less likely than vitamin K deficiency to explain this patient's bleeding. Synthesis of thymidylate (Choice D) is required for DNA synthesis and requires folate as a cofactor. Folate deficiency during pregnancy increases the risk for fetal neural tube defects, and folate deficiency following birth may result in impaired DNA synthesis often marked by megaloblastic anemia. Folate is a water-soluble vitamin and deficiency at this stage would be unlikely as it is naturally found in breast milk.

Objective: Vitamin K deficiency is common in breast-fed, homeborn neonates since they lack commensal bacteria that would otherwise synthesize it in the gut. An absence of vitamin K leads to coagulopathy and bleeding since factors II, VII, IX, and X require vitamin k for the post-translational carboxylation of glutamate residues during synthesis. Previous Next Score Report Lab Values Calculator Help Pause

195 Exam Section 4: Item 45 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 45. A 33-year-old man comes to the physician for a physical examination prior to starting a new job. He feels well and recently became engaged to be married. Physical examination shows a 1.5-cm fixed, nontender mass on the right side of the neck. During the examination, he appears anxious and says, "You seem to be spending a lot of time examining my neck. Is there something wrong? Do I have cancer?" After informing the patient about the discovery of the suspected mass, which of the following is the most appropriate initial statement by the physician? A) "I know that you're anxious about this, so I think it would be a good idea for your fiancée to come in with you when you return for a biopsy next week." B) "It's understandable that you're worried, so l'l walk you through each step. We'll need to schedule a time as soon as possible to get a sample of the mass for further examination." C) "The only way to know if the mass in your neck is malignant or benign is to do a biopsy. Please take this referral to the front desk so that they can schedule the procedure and a follow-up visit with me." D) "We'll need to do a biopsy of the mass to determine if it is malignant or benign. Until we have the results of the biopsy, I don't think you should worry about it." E) "Whatever this mass turns out to be, it'll be easier to face because you'll have the support of your fiancée." Correct Answer: B. Physicians should address a potentially concerning physical examination finding by explaining the next diagnostic steps in simple language that is delivered in small pieces. By validating the patient's understandable anxiety and then walking him through the next steps, the physician can help the patient refocus on actions that can be taken to investigate the finding rather than the uncertainty of the future. This strategy to decrease the patient's anxiety can reduce the patient's distress and assist him in processing the information he is about to receive. The physician should support the patient throughout the diagnostic process by being available for questions and addressing the patient's emotions. Incorrect Answers: A, C, D, and E. Physicians should take responsibility for supporting patients through anxiety-provoking health problems rather than immediately implying that the patient should look elsewhere for support (Choices A and E). Further, the next diagnostic steps and the reasons for these steps should be thoroughly explained to the patient. Immediately mentioning the possibility that the mass may be malignant (Choices C and D) may overwhelm this patient who is already anxious. Discussing the possible etiologies of the mass using simple language and small pieces of information may be appropriate later in the conversation. Additionally, the uncertainty of this situation would likely be stressful for any patient, so the physician should normalize this understandable reaction rather than telling the patient that he should refrain from worrying. As well, the physician should avoid dismissing the patient prematurely, which leaves no opportunity for the patient to process the information or ask questions.

Objective: When discussing potentially concerning physical examination findings, physicians should validate patients' anxiety and explain the next steps in simple language that is delivered in small pieces. Physicians should take responsibility for supporting patients throughout the process. I3D Previous Next Score Report Lab Values Calculator Help Pause 田

55 Exam Section 2: Item 5 of 50 Național Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 5. A 45-year-old man has shortness of breath. He has never smoked cigarettes. There is no evidence of cardiac disease. X-rays of the chest show bullous air spaces in the lower lobes. The pathophysiologic basis of this process is most likely to be a deficiency of which of the following? O A) Ceruloplasmin O B) Epidermal growth factor O C) Ferritin D) Plasminogen E) Protease inhibitor Correct Answer: E. a1-Antitrypsin is a protease inhibitor that regulates neutrophil elastase. Neutrophil elastase is a serine protease secreted by neutrophils and macrophages as part of a localized inflammatory response. a1-Antitrypsin deficiency is an inherited disorder caused by mutations in the SERPINA1 gene, which encodes a1-antitrypsin. Reduced concentration of effective protein result in host tissue damage from neutrophil elastase, especially in the lungs and liver. Pulmonary manifestations classically include panlobular emphysema, which results in the development of bullous air spaces predominantly in the lower lobes. The diagnosis should be suspected in patients presenting with evidence of chronic lung disease with emphysematous changes and no history of tobacco use. Incorrect Answers: A, B, C, and D. Ceruloplasmin (Choice A) is a ferroxidase enzyme and the major transporter of copper in the blood. Decreased ceruloplasmin concentration is found in Wilson disease, in which excess copper concentration in the body lead to characteristic neurologic and hepatic dysfunction. Kayser-Fleischer rings may be seen around the edge of the cornea. Óther conditions associated with decreased ceruloplasmin concentration are Menkes disease and aceruloplasminemia. Epidermal growth factor (Choice B) is a protein which stimulates cell growth and differentiation by binding to the epidermal growth factor receptor (EGFR). EGFR signaling is a common pathway involved in several types of cancer, including colorectal, lung, and breast cancer. Ferritin (Choice C) is the major site of iron storage in the body. Deficiency is associated with iron deficiency anemia. Plasminogen (Choice D) is the zymogen of plasmin, which is a serine protease responsible for the degradation of many plasma proteins, especially fibrin clots. Deficiency may cause thrombosis but is not associated with emphysema.

Objective: a1-Antitrypsin deficiency should be suspected in patients who present with emphysematous lung changes in the absence of a smoking history. a1-Antitrypsin is a protease inhibitor responsible for regulating neutrophil elastase. Previous Next Score Report Lab Values Calculator Help Pause