NBME 27 EDU OBJ

3 Exam Section 1: Item 3 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 3. A 55-year-old woman comes to the physician because of a 6-month history of progressive shortness of breath while climbing stairs and constant fatigue. She is 163 cm (5 ft 4 in) tall and weighs 107 kg (235 lb); BMI is 40 kg/m2. Her blood pressure is 170/110 mm Hg on three separate readings. Physical examination shows no other abnormalities. Echocardiography shows an ejection fraction of 45%. There are no regional wall motion or valvular abnormalities. An ECG shows that the sum of the magnitudes of the S wave in lead V, and the R wave in lead Vs is 45 mm (scale=10 mm/mV). Which of the following cellular changes is most likely in this patient's left ventricle compared with normal? A) Decreased sarcomeres and increased fibroblasts B) Increased sarcomeres in an array in parallel with each other C) Increased sarcomeres in an array in series with each other D) Increased sarcomeres with decreased abundance of F-actin E) Myocardial infiltration of neutrophils Correct Answer: B. Left ventricular hypertrophy (LVH) is a common disorder with numerous causes, including chronic systemic hypertension, aortic stenosis, coarctation of the aorta, chronic aortic or mitral insufficiency, ventricular septal defect, infiltrative cardiomyopathy (eg, amyloidosis, sarcoidosis, Fabry disease, hereditary hemochromatosis), or hypertrophic cardiomyopathy. Hypertension and aortic stenosis are the most common causes. In these conditions, hypertrophy develops secondary to chronic contraction against increased afterload. This stimulates growth and remodeling of cardiomyocytes, resulting in increased sarcomeres in an array in parallel with each other in order to generate greater contractile force to overcome the afterload. The condition can progress to left-sided heart failure if untreated. Diagnostic testing includes ECG; one of the criteria suggestive of LVH is if the sum of the magnitude of the S wave in lead V2 and the R wave in lead V5 is greater than 35 mm as in this patient. Echocardiography is more sensitive than ECG for the presence of LVH. Cardiac MRI is the gold standard for diagnosis, though availability, cost, and logistics limit its use as a first-line screening tool. Incorrect Answers: A, C, D, and E. Decreased sarcomeres and increased fibroblasts (Choice A) occurs in myocardial scar formation, which may result after a myocardial infarction or acute inflammatory injury. It is also seen in Fabry disease as a result of dysfunctional metabolism of sphingolipids. Increased sarcomeres in an array in series with each other (Choice C) occurs in dilated cardiomyopathy as the cardiac myofibrils lengthen. Dilated cardiomyopathy may be caused by an underlying genetic predisposition, toxin exposure (eg, chronic alcohol use, chronic cocaine use, doxorubicin chemotherapy), infections (eg, Chagas disease, viral myocarditis), or systemic disorders (eg, sarcoidosis, hemochromatosis). Increased sarcomeres with decreased abundance of F-actin (Choice D) may be seen in genetic cardiomyocyte disorders such as hypertrophic cardiomyopathy and some forms of dilated cardiomyopathy. LVH secondary to chronic hypertension is the more likely diagnosis in an older patient with obesity and hypertension on examination. Myocardial infiltration of neutrophils (Choice E) occurs in response to cardiomyocyte necrosis in acute myocardial infarction, furthering local inflammatory damage and recruiting macrophages to the site of injury.

Objective: LVH is a common complication of chronic systemic hypertension. Cardiomyocytes respond to increased afterload with cellular remodeling to increase contractile force by arranging additional sarcomeres in parallel, resulting in thickening of the myocardium. II Previous Next Score Report Lab Values Calculator Help Pause

86 Exam Section 2: Item 37 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 37. A 67-year-old man comes to the physician because of a 2-month history of unexplained weight loss. He has a 50-year history of type 1 diabetes mellitus. He underwent gallstone removal 12 years ago. He has smoked 1 pack of cigarettes daily for 45 years. He is 188 cm (6 ft 2 in) tall and weighs 120 kg (265 lb); BMI is 34 kg/m2. Physical examination shows no other abnormalities. Serum studies show a calcium concentration of 11 mg/dL. An abdominal CT scan shows a pancreatic mass. A biopsy specimen of the mass shows pancreatic adenocarcinoma. Which of the following is the strongest predisposing risk factor for pancreatic cancer in this patient? A) BMI of 34 kg/m2 B) Cigarette smoking C) History of gallstone disease O D) Hypercalcemia E) Type 1 diabetes mellitus Correct Answer: B. Cigarette smoking is the most important preventable risk factor for pancreatic cancer. Increased risk for pancreatic cancer is observed in smokers, especially heavy smokers. This excess risk slowly declines after smoking cessation. Other important risk factors for pancreatic cancer include obesity and type 2 diabetes mellitus. In addition to pancreatic cancer, cigarette smoking is an important, modifiable risk factor for numerous other cancers, including those affecting the larynx, esophagus, lung, kidney, bladder, and cervix. A hypothesized mechanism of carcinogenesis from smoking involves the promotion of malignancy through DNA mutation secondary to inhaled polycyclic aromatic hydrocarbons. Incorrect Answers: A, C, D, and E. BMI of 34 kg/m2 (Choice A) is suggestive of obesity. While obesity is associated with an increased risk for pancreatic cancer, smoking is a stronger and well-established risk factor. History of gallstone disease (Choice C) may increase the patient's risk for acute pancreatitis. A prior history of gallstone disease has not, however, been established to contribute to pancreatic cancer. Hypercalcemia (Choice D) is a rare cause of acute pancreatitis. It has not been established to contribute to pancreatic cancer. Type 1 diabetes mellitus (Choice E) has not been conclusively demonstrated to contribute to pancreatic cancer. While there is evidence of an association between type 2 diabetes mellitus and pancreatic cancer, this relationship has not been consistently demonstrated for patients with type 1 diabetes mellitus.

Objective: Cigarette smoking is the most important preventable risk factor for pancreatic cancer. The excessive risk for pancreatic cancer caused by cigarette smoking decreases slowly after smoking cessation. Previous Next Score Report Lab Values Calculator Help Pause

52 Exam Section 2: Item 3 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 3. A 48-year-old man comes to the emergency department because of a 3-day history of fever, chills, headache, and nonproductive cough. He underwent renal transplantation 5 years ago. Current medications include corticosteroids. His temperature is 40°C (104°F). Coarse rhonchi are heard over the right lower lobe on auscultation. A chest x-ray shows right lobar pneumonia with small pleural effusions. The patient is treated with an antibiotic that inhibits DNA gyrase. This drug is most likely which of the following? A) Ciprofloxacin B) Metronidazole C) Rifampin D) Tetracycline E) Vancomycin Correct Answer: A. Pneumonia commonly presents with cough, shortness of breath, pleuritic chest pain, fever, and abnormal breath sounds. It often follows an upper respiratory infection or aspiration of oropharyngeal microbes and is more common in immunocompromised persons. Corticosteroid use is a risk factor for infection with atypical organisms, such as Legionella pneumophila, which is a gram-negative rod that is transmitted primarily through aerosols from water sources (eg, air conditioning systems). The atypical pneumonia may appear as diffuse, patchy alveolar and interstitial opacities on lung imaging or as a lobar consolidation. It can also be associated with hyponatremia. First-line therapy for Legionella pneumophila includes administration of either macrolides or fluoroquinolones. Ciprofloxacin is a fluoroquinolone antibiotic that was likely prescribed for this patient. Fluoroquinolones impair DNA synthesis by the direct inhibition of DNA gyrase (prokaryotic topoisomerase II). Resistance to fluoroquinolones is frequently encountered and mediated by structural variants of DNA gyrase and production of membrane efflux pumps. Adverse effects of fluoroquinolones include gastrointestinal distress, tendonitis with risk for tendon rupture, and QT prolongation. Incorrect Answers: B, C, D, and E. Metronidazole (Choice B) is an antibacterial and antiprotozoal agent that diffuses into target organisms and alters microbial protein function, resulting in the generation of free radicals and subsequent DNA damage. It is particularly effective against anaerobes, especially infections involving the abdomen or pelvis, as well as the facultative anaerobes Helicobacter pylori and Gardnerella vaginalis. These organisms are not common causes of pneumonia. Rifampin (Choice C) is a rifamycin antibiotic that inhibits DNA-dependent RNA polymerase. Resistance is common as a result of the development of binding site mutations, especially if rifamycin monotherapy is used. Rifampin is used in combination with other antibiotics in the treatment of Mycobacterium tuberculosis infections. Tetracycline (Choice D) antibiotics act by reversibly binding to the bacterial ribosomal 30S subunit, which disrupts the elongation phase of protein synthesis. Resistance is conferred by the presence of bacterial transport pumps that decrease uptake of the drug and increase efflux. Vancomycin (Choice E) inhibits bacterial cell wall peptidoglycan formation in gram-positive bacteria by binding to D-Ala-D-Ala. It is not degraded by B-lactamases, which makes it effective in the treatment of resistant organisms such as methicillin-resistant Staphylococcus aureus and ampicillin-resistant Enterococcus species. It also has activity against Clostridium difficile when administered via the oral or rectal route.

Objective: Fluoroquinolones are effective antibiotics in the treatment of atypical pneumonia such as Legionella pneumophila. They inhibit DNA gyrase, which disrupts bacterial DNA synthesis. Resistance is common and caused by altered forms of DNA gyrase and the presence of membrane efflux pumps. Previous Next Score Report Lab Values Calculator Help Pause

95 Exam Section 2: Item 46 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 46. A 64-year-old homeless man with alcoholism is brought to the emergency department by police after he was found wandering aimlessly in a city park. He has weakness, fatigue, and hunger. He appears disheveled and cachectic. His temperature is 38.2°C (100.8°F), pulse is 110/min, respirations are 20/min, and blood pressure is 134/82 mm Hg. Physical examination shows mild scleral icterus, numerous skin abrasions, and 1+ edema of the lower extremities. He is oriented to person but not to place or time. Laboratory studies show: 6.8 g/dL 19% 111 um3 3500/mm3 Hemoglobin Hematocrit Mean corpuscular volume Leukocyte count Platelet count 95,000/mm3 A photomicrograph of a peripheral blood smear is shown. A deficiency of which of the following best accounts for the hematologic findings in this patient? A) Copper B) Folic acid C) Iron D) Vitamin B, (thiamine) E) Vitamin B, (pyridoxine) Correct Answer: B. Folic acid deficiency most likely accounts for this patient's megaloblastic anemia. Megaloblastic anemia occurs in the setting of impaired DNA synthesis, commonly related to folate or vitamin B12 (cobalamin) deficiency, and is characterized by erythrocyte macrocytosis and hypersegmented neutrophils. Laboratory evaluation shows anemia, increased mean corpuscular volume, and normal white cell and platelet indices. The presence of hypersegmented neutrophils on peripheral smear is a characteristic finding in megaloblastic anemia. Since folate and vitamin B 12 are needed for the conversion of homocysteine to methionine, deficiency is associated with increased homocysteine concentrations. However, vitamin B12 also acts as a cofactor for the conversion of methylmalonyl-COA to succinyl-CoA, and its deficiency is associated with both high methylmalonic acid and homocysteine concentrations. Common risk factors include chronic alcohol abuse, malnutrition, and high cell turnover conditions such as pregnancy and sickle cell disease. Deficiency of folate and/or B12 can be seen in patients with chronic alcohol use disorder, although folate deficiency is more common as the body maintains larger stores of vitamin B 12, thus depletion takes much longer. Incorrect Answers: A, C, D, and E. Copper deficiency (Choice A) results in sideroblastic anemia and is often seen in the setting of malabsorptive syndromes. Neutropenia is also a common finding. Iron deficiency (Choice C) anemia results in microcytosis, not megaloblastic anemia. It is common in menstruating women, those with diets low in iron, and in patients with chronic blood loss such as in colorectal carcinoma. Other cell lines are not typically affected, and treatment is with oral iron supplementation. Vitamin B, (thiamine) deficiency (Choice D) results in wet or dry beriberi. Dry beriberi is characterized by symmetrical peripheral neuropathy and wet beriberi manifests with high cardiac output heart failure. Thiamine deficiency can also result in Wernicke-Korsakoff syndrome, which classically presents as nystagmus, ophthalmoplegia, ataxia, and confusion. Vitamin B6 (pyridoxine) deficiency (Choice E) is uncommon as a result of dietary insufficiency, but when it does occur, it presents as glossitis, stomatitis, and seborrheic dermatitis. It occasionally develops as a result of chronic therapy with isoniazid for the treatment of tuberculosis and can also result in peripheral neuropathy, but this is typically prevented by prophylactic supplementation with vitamin Bg during the antituberculosis therapy regimen.

Objective: Folate deficiency results in megaloblastic anemia from impaired erythrocyte DNA synthesis. Common risk factors include chronic alcohol abuse, malnutrition, and high cell turnover conditions such as pregnancy and sickle cell disease. II Previous Next Score Report Lab Values Calculator Help Pause

82 Exam Section 2: Item 33 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 33. A 34-year-old woman comes to the office because of a 6-month history of progressive swelling of her hands and feet. Her blood pressure is 140/89 mm Hg. Physical examination shows 1+ edema of the hands and 2+ edema of the feet. Serum studies show a potassium concentration of 3.3 mEq/L, decreased aldosterone concentration and plasma renin activity, and increased cortisol:cortisone ratio. After questioning, it is found that the patient has been consuming a nutritional supplement she obtained from the Internet during the past 10 months. Serum studies obtained after the patient stops using the supplement are within the reference ranges. The supplement this patient consumed most likely inhibited the activity of which of the following enzymes? O A) A4-5a-Dehydrogenase B) 1a-Hydroxylase C) 21-Hydroxylase D) 11B-Hydroxysteroid dehydrogenase type II E) 3-Oxosteroid A4-dehydrogenase Correct Answer: D. The patient likely consumed a supplement containing licorice, the active ingredient of which is glycyrrhetinic acid, or enoxolone. This compound inhibits a variety of enzymes, including the enzyme 11B-hydroxysteroid dehydrogenase type lI, which catalyzes the conversion of cortisol to cortisone in the zona fasciculata of the adrenal gland. Inhibition of this reaction causes an increased cortisol to cortisone ratio. Cortisol demonstrates a weak mineralocorticoid effect, which becomes more pronounced as the concentration of cortisol increases. It inserts ENAC channels into the renal collecting tubule, which cause sodium and water reabsorption in exchange for potassium excretion, leading to hypertension and a mild hypokalemia. Secondary to this hypertension, the juxtaglomerular apparatus decreases its production of renin and subsequently decreases the production of angiotensin and aldosterone. This accounts for the decreased plasma renin and aldosterone concentrations seen in this patient. The enzyme also inhibits the metabolism of prostaglandins E2 and F2g to their respective inactive metabolites, which causes a downstream effect of inhibiting gastric acid secretion by the stomach. Incorrect Answers: A, B, C, and E. 4-5a-Dehydrogenase (Choice A) is a subtype of 5a-reductase and participates in the conversion of testosterone to dihydrotestosterone. This enzyme is inhibited by finasteride and does not account for the changes in glucocorticoid metabolism or renin-angiotensin-aldosterone pathway seen in this patient. 1a-Hydroxylase (Choice B) catalyzes the reaction converting 25-OH vitamin D3 to its active form, 1,25-(OH)2 vitamin D3 (calcitriol). This enzyme resides in the kidney and is not involved in glucocorticoid metabolism. 21-Hydroxylase deficiency (Choice C) is the cause of the most common form of congenital adrenal hyperplasia (CAH). This presents with salt wasting in a newborn or precocious puberty in a child, as well as virilization of infants with an XX karyotype (eg, clitoral hypertrophy, labial fusion), as a result of increased concentrations of androgens, as seen in this patient. As a result of the enzyme deficiency, concentrations of 17-hydroxyprogesterone, androstenedione, and testosterone are increased, while aldosterone and cortisol are decreased, leading to hypotension and hyperkalemia, not hypertension and hypokalemia as seen in this patient. 3-Oxosteroid A4-dehydrogenase (Choice E) is an enzyme required in the synthesis of bile acids. Inhibition of this enzyme would not cause changes in cortisol or the renin-angiotensin- aldosterone pathway.

Objective: Glycyrrhetinic acid, or enoxolone, is the active ingredient in licorice and primarily inhibits 11B-hydroxysteroid dehydrogenase type II, which is responsible for the conversion of cortisol to cortisone. With increased concentrations of cortisol, downstream mineralocorticoid effects are observed, which cause inhibition of the renin-angiotensin-aldosterone pathway and decreased serum concentrations of its associated hormones. Previous Next Score Report Lab Values Calculator Help Pause

71 Exam Section 2: Item 22 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 22. A 10-month-old female infant is brought to the physician because of a 1-day history of a rash. Her mother says that the infant also had a 3-day history of high fever and irritability, which spontaneously resolved 2 days ago. The patient is alert and playful. Her temperature is 37°C (98.6°F). Physical examination shows the findings in the photograph. Which of the following is the most likely causal agent? A) Epstein-Barr virus B) Human herpesvirus 6 C) Parvovirus B19 D) Rubeola virus O E) Varicella-zoster virus Correct Answer: B. Human herpesvirus 6 (HHV-6) is the cause of roseola infantum, or exanthem subitum. In a minority of cases, this viral syndrome may also be caused by human herpesvirus 7, enteroviruses, and adenoviruses, though HHV-6 is the most frequent cause. This viral syndrome is characterized by several days of high fevers sometimes accompanied by seizures; as the fevers subside, a characteristic, blanching, macular eruption forms on the neck and trunk, which then spreads outward to the face and extremities. The appearance of the rash indicates resolution of the viremia, so the diagnosis is made clinically rather than by viral serologies. Roseola infantum commonly affects infants but may also occur later in childhood. Other exanthems of childhood may be considered in the differential diagnosis of roseola infantum such as rubella, measles, erythema infectiosum, and enterovirus infections. The characteristic high fevers followed by a rash helps to identify roseola infantum as the etiology. Treatment of the syndrome is supportive; in most cases, it is a benign condition with a limited course. Incorrect Answers: A, C, D, and E. Epstein-Barr virus (Choice A) is a common cause of infectious mononucleosis. EBV is transmitted through respiratory secretions and saliva, causing theillness to be commonly acquired by teenagers and young adults rather than infants. Clinically, the syndrome commonly presents with fever, lymphadenopathy (typically involving the posterior cervical lymph nodes), and hepatosplenomegaly along with pharyngitis. Unless amoxicillin is administered for treatment of a misdiagnosed pharyngitis thought to be from Streptococcus pyogenes, patients do not typically develop a macular eruption as a feature of this syndrome. Parvovirus B19 (Choice C) is the cause of erythema infectiosum. This presents with pink patches on the bilateral cheeks of children giving an appearance of slapped cheeks. This facial rash may be followed by a lacy, macular eruption on the trunk and extremities. However, it typically affects school-aged children who are often otherwise asymptomatic. Rubeola virus (Choice D) is a highly contagious paramyxovirus that causes the disease measles, which is associated with an acute febrile illness and potentially severe sequelae. It is more common in children than adults. It characteristically presents with prodromal fever, cough, coryza, conjunctivitis, and a confluent maculopapular rash that starts at the head or neck and spreads to the trunk, excluding the palms and soles. In this syndrome, the rash and other symptoms occur together, unlike the exanthem of roseola infantum. Varicella-zoster virus (Choice E), also known as human herpesvirus-3, is transmitted via respiratory secretions and direct contact with infected skin lesions. It causes the disease chickenpox in immunocompetent children, and can cause encephalitis and pneumonia in immunosuppressed patients. The classic presentation of chickenpox is a pruritic, asynchronous rash with fluid-filled vesicles that crust over.

Objective: Human herpesvirus 6 is the cause of roseola infantum, or exanthem subitum, which is characterized by several days of high fevers followed by a macular eruption. Treatment of the syndrome is supportive, as it is generally a benign condition. Previous Next Score Report Lab Values Calculator Help Pause

60 Exam Section 2: Item 11 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 11. A 20-year-old man with sickle cell disease has chronic hemolytic anemia and has frequent painful crises that require transfusions. Hydroxyurea treatment significantly decreases the painful crises and transfusion requirements. Which of the following is the most likely mechanism of the beneficial effects of this treatment? A) Decreased erythrocyte cellular dehydration B) Decreased hemoglobin S concentration C) Increased affinity of hemoglobin S for oxygen D) Increased bone marrow production of erythrocytes E) Increased hemoglobin F concentration Correct Answer: E. Increased hemoglobin F concentration is the most likely mechanism by which hydroxyurea decreases the frequency of sickle cell pain crises and transfusion requirements. Hydroxyurea is a myelosuppressive drug used in a number of conditions such as essential thrombocytosis but is commonly used for the treatment of sickle cell disease. It inhibits ribonucleotide reductase, which results in the inhibition of DNA synthesis within erythrocytes leading to apoptosis. This is hypothesized to lead to recruitment of earlier erythroid stem cells that contain a higher concentration of hemoglobin F (HbF), which has a higher affinity for oxygen but does not result in sickling of the erythrocyte. Beyond inhibiting the production of erythrocytes containing hemoglobin S (HbS), hydroxyurea also inhibits other cell lines, including granulocytes and platelets. Incorrect Answers: A, B, C, and D. Decreased erythrocyte cellular dehydration (Choice A) is not the mechanism of action of hydroxyurea. Erythrocyte hydration is maintained through ion pumps with regulation of intracellular potassium, sodium, and water concentration. There exists a rare group of disorders known as hereditary hydrocytosis and xerocytosis resulting in abnormalities of erythrocyte hydration. Decreased HbS concentration (Choice B) is an indirect effect of hydroxyurea, but the primary mechanism is by increasing HbF concentrations. Increased affinity of HbS for oxygen (Choice C) is not a direct effect of hydroxyurea. Blood in patients with sickle cell disease overall has a lower oxygen-carrying capacity. HbS unto itself has the same affinity for oxygen as hemoglobin A when HbS is not polymerized, but an increased amount of 2,3-BPG in erythrocytes containing HbS alters the oxygen dissociation curve. Increased bone marrow production of erythrocytes (Choice D) is not the correct answer. Hydroxyurea is myelosuppressive and has the opposite effect.

Objective: Hydroxyurea decreases the number of sickle cell crises experienced by patients with sickle cell disease by increasing the relative concentration of HbF, thereby improving tissue oxygenation. %3D Previous Next Score Report Lab Values Calculator Help Pause

67 Exam Section 2: Item 18 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 18. A 45-year-old man is undergoing a surgical procedure. Which of the following maneuvers by the anesthesiologist would result in an increase in intracranial pressure? A) Decreasing airway pressures B) Decreasing fractional inspired oxygen (F102) C) Decreasing respiratory rate D) Increasing airway pressures E) Increasing Fio2 F) Increasing respiratory rate Correct Answer: C. Intracranial pressure (ICP) is dependent on the volumes of the brain, cerebrospinal fluid (CSF), and blood as a result of the fixed space of the skull. Brain volume can be increased with a mass or cerebral edema secondary to ischemic injury. CSF volume can be increased by increased production, decreased reabsorption, or an obstruction that prevents drainage. Blood volume can be increased in the extravascular or intravascular space. Extravascular blood accumulation occurs in the setting of intracerebral hemorrhage, epidural hematoma, subdural hematoma, and subarachnoid hemorrhage. However, the primary way an anesthesiologist or other provider can alter ICP is through alterations in cerebral blood flow (CBF) or intravascular blood volume present in the cranial vault. As a result of autoregulation, the CBF is maintained at a nearly constant level between blood pressures of 60 to 160 mm Hg. CBF, however, increases with increased metabolic demand, hypercarbia, hypoxemia, hyperthermia, and increased central venous pressure (CVP). The most marked effect on CBF is seen with hypercarbia, with an approximate increase of 1 to 2 milliliters per 100 grams per minute per millimeter of mercury in the arterial pressure of carbon dioxide (PaCO,). Therefore, decreasing the respiratory rate or the tidal volume, and thereby increasing PaCO, would result in an increase in CBF and a consequent increase in ICP. Incorrect Answer: A, B, D, E, and F. Decreasing airway pressures (Choice A) would result in a multitude of effects such as preventing barotrauma, potentially decreasing the arterial pressure of oxygen (Pao,) if the positive end- expiratory pressure (PEEP) were decreased, and lowering CVP by decreasing intrathoracic pressure. However, hypoxemia does not cause increases in ICP unless severe and the resultant decreases in CVP would decrease ICP. Decreasing fractional inspired oxygen (FIO) (Choice B) would decrease PaO, but this would be unlikely to lead to increases in ICP unless the hypoxemia was severe. Increasing airway pressures (Choice D) could potentially increase Pao, if PEEP were increased and could increase the CVP by increasing the intrathoracic pressure. However, there would have to be a significant increase in CVP to increase ICP, and hyperoxia minimally decreases CBF and ICP. Increasing Fio, (Choice E) increases Pao, which leads to minimal decreases in CBF and ICP. Hyperoxia does not cause increases in ICP. Increasing respiratory rate (Choice F) would lead to hypocapnia, which in turn would cause decreases in CBF and ICP, rather than increases. This technique can be utilized for the acute, temporizing management of increased ICP prior to definitive management with surgery, such as in the setting of an acute intraparenchymal hemorrhage with mass effect.

Objective: ICP is affected by brain volume, blood volume, and CSF. It is often iatrogenically altered by changes in the CBF, which is physiologically increased by increased metabolic demand, hypercarbia, hypoxemia, hyperthermia, and increased central venous pressure. However, the most significant effect on CBF is a result of changes in PaCO,2 which can be altered by changes in tidal volume and respiratory rate. %3D 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 H+ (nmol/L) 16. The graph shows a normal (Y) and abnormal (Z) relationship of bicarbonate to pH in the blood. Which of the following is the most likely cause of a shift from point Y to point Z in a healthy subject? 100 90 80 7O 60 50 40 30 25 20 15 44 100 80 A) Breathing a gas mixture with a high Pco2 40 60 B) Breathing a gas mixture with a low Po, 36 40 mm Hg C) Breathing air at 2 atmospheres of pressure 32- D) Ingestion of an acid 28 20 E) Metabolic alkalosis 24- 20- 16 12 8. 7.0 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 pH Correct Answer: B. The bicarbonate buffer system is an important regulator of acid-base balance in the blood. The reaction involves the conversion of carbon dioxide plus water to carbonic acid, which then dissociates into a bicarbonate ion and a hydrogen ion. To maintain a physiologic pH of 7.4 at normal body temperature, this system has a 20:1 bicarbonate to carbonic acid ratio. This relationship is determined by the Henderson-Hasselbalch equation, which is a model used to estimate the pH of a buffer solution. Blood acid-base balance in the body is coupled to the lungs and alveolar gas exchange. Breathing a gas mixture with a low Po2 stimulates hyperventilation in response to hypoxemia; this results in decreased alveolar Pco2. The alveolar-capillary interface is a thin membrane and Pco2 readily equalizes between the blood and the alveoli. The decrease in Pco2 shifts the carbonic acid buffer reaction to the left by the Le Chatelier principle, with increased conversion of bicarbonate ions and hydrogen ions into carbonic acid. The result of acute hyperventilation is an alkalotic shift in pH, causing the carbonic acid reaction to shift towards carbon dioxide and water production to account for losing carbon dioxide to the atmosphere. This process is referred to as a respiratory alkalosis. Incorrect Answers: A, C, D, and E. Breathing a gas mixture with a high Pco2 (Choice A) and breathing air at 2 atmospheres of pressure (Choice C) would result in high carbon dioxide concentrations in the alveoli and thus in the blood, shifting the carbonic acid buffer reaction to the right with increased generation of bicarbonate and hydrogen ions. This would result in a decreased blood pH from respiratory acidosis. Ingestion of an acid (Choice D) would result in an acute increase in hydrogen ions in the serum with a shift to the left in the buffer reaction. The respiratory system would compensate in an attempt to maintain a physiologic pH of 7.4. Metabolic alkalosis (Choice E) results from bicarbonate ion excess in the serum or increased hydrogen ion loss. The respiratory system would compensate in an attempt to maintain a physiologic pH of 7.4. This would be represented by a curve to the right of point Y on the graph and a point on that curve greater than 24 mmol/L on the y-axis (high bicarbonate concentration).

Objective: In healthy subjects, blood pH concentration is maintained at or near a physiologic value of 7.4 by several acid buffer systems, the most important of which is the bicarbonate-carbonic acid buffer. It is coupled to the respiratory system through the exchange of carbon dioxide between the blood and the alveoli. Changes in ventilation affect the relationship of the buffer constituents and are responsible for appropriate compensation to acid-base disturbances. Previous Next Score Report Lab Values Calculator Help Pause (7//oww)_E0OH

54 Exam Section 2: Item 5 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 5. A 24-year-old woman at 28 weeks' gestation is brought to the emergency department because of a 3-hour history of shortness of breath. She has asthma, but she currently takes no medications. Her pulse is 100/min, respirations are 32/min, and blood pressure is 120/83 mm Hg. Physical examination shows the use of accessory muscles of respiration. Diffuse inspiratory and expiratory wheezes are heard. An inhaled Bradrenergic agonist is administered. Which of the following findings is most likely in this patient after this therapy? A) Bradycardia B) Diaphoresis C) Dry mouth D) Pallor E) Tremor Correct Answer: E. Inhaled B2-adrenergic agonists, such as albuterol, formoterol, and salmeterol are used for the treatment of bronchoconstriction in asthma and in chronic obstructive pulmonary disease. Drugs in this class act preferentially on the B2-adrenergic receptor, which is found on bronchial smooth muscle, leading to bronchodilation. B2-adrenergic agonists cause a variety of side effects that are mediated by the metabolic actions of the B2- and also the B1-adrenergic receptor, and include tremor, increased activity of the sodium-potassium ATPase with subsequent hypokalemia, glycogenolysis with subsequent hyperglycemia, tachycardia, hypertension, and headache. Recognizing the effect of B-agonists on tremor forms the basis of using B-adrenergic antagonists, such as propranolol, in the treatment of benign essential tremor. Incorrect Answers: A, B, C, and D. Bradycardia (Choice A) is a side effect of some a2-adrenergic agonists, such as clonidine, which are used for the treatment of hypertension, hypertensive emergency, and opiate withdrawal. Diaphoresis (Choice B) is mediated by sympathetic stimulation from acetylcholine on muscarinic, not adrenergic receptors. This is one of few cases when the sympathetic nervous system signals via acetylcholine, whereas its effects are typically mediated by norepinephrine, epinephrine, and dopamine. Dry mouth (Choice C) is a side effect of muscarinic receptor antagonists, such as atropine, which are used for the treatment of bradycardia. Pallor (Choice D) is a potential side effect of B2-adrenergic antagonists, such as propranolol. Peripheral blood vessels dilate in response to B2-agonism. Blocking this effect promotes a net vasoconstriction.

Objective: Inhaled B2-adrenergic agonists, such as albuterol, formoterol, and salmeterol are used for the treatment of bronchoconstriction. Side effects of B2-adrenergic agonists include tremor, hypokalemia, hyperglycemia, tachycardia, hypertension, and headache. 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 34-year-old man who is seropositive for HIV has a neoplasm manifested as violaceous raised cutaneous plaques (erythematous cutaneous plaques). Examination of tissue obtained on biopsy of the neoplasm is most likely to show which of the following? A) Dense, relatively acellular fibrous tissue with few typical fibroblasts B) Large polygonal and giant cells with myofibrils C) Malignant gland-forming cells with mucin D) Poorly differentiated malignant squamous cells E) Proliferating spindle cells forming slit-like spaces filled with blood Correct Answer: E. Kaposi sarcoma is a neoplasm of vascular endothelial cells, which may be induced by human herpesvirus 8 (HHV-8) in patients with poorly controlled HIV or who are otherwise immunosuppressed. Histologically, proliferating spindle cells forming slit-like spaces filled with blood confirms the diagnosis. Kaposi sarcoma also exhibits a classic form, which occurs spontaneously in men of Eastern European descent, and an endemic form, seen in sub-Saharan Africans. Clinically, Kaposi sarcoma progresses from patch, to plaque, and then to tumor stages. In the plaque phase, as described in this patient, reddish-purple or violaceous, nonblanching plaques are seen. It commonly affects the lower extremities, face, oral mucosa, and genitalia. It may also affect the gastrointestinal and pulmonary systems. For this reason, screening for occult gastrointestinal bleeding using a fecal occult blood test and screening for pulmonic lesions using a chest x-ray is recommended. Overall, Kaposi sarcoma is a low-risk tumor that demonstrates a good prognosis, provided the underlying HIV infection is treated. Local treatments including intralesional chemotherapy, radiation therapy, and topical retinoids may help manage existing lesions, though they do not prevent new ones from forming. For those with advanced disease, systemic therapy is warranted and can include anthracyclines (doxorubicin, daunorubicin) and/or paclitaxel, though treatment regimens are broad. Incorrect Answers: A, B, C, and D. Dense, relatively acellular fibrous tissue with few typical fibroblasts (Choice A) is characteristic of a scar. Dermal collagen fibers increase as a result of scarring. While the majority of collagen fibers in healthy skin are collagen, type I, scar is initially created by collagen, type III. Large polygonal and giant cells with myofibrils (Choice B) are characteristic of a rhabdomyoma. This tumor occurs almost exclusively in children and is often located on the ventricular walls or atrioventricular walls of the heart. Malignant gland-forming cells with mucin (Choice C) are commonly seen in adenocarcinoma, regardless of the organ from which they are derived. Adenocarcinoma is derived from glandular epithelium and thus forms gland-like structures, even when malignant. Poorly differentiated malignant squamous cells (Choice D) are characteristic of poorly differentiated squamous cell carcinoma. Regardless of the organ involved, squamous cell carcinoma is derived from cells that produce keratin. When squamous cells become aberrant, they continue producing keratin but in small foci called keratin pearls, which is another feature helpful in identifying squamous cell carcinoma on histology.

Objective: Kaposi sarcoma is a low-grade neoplasm of the vascular endothelial cells, which is characterized clinically by violaceous or reddish-purple patches, plaques, or tumors. Histologic examination demonstrates slit-like vascular spaces surrounded by a proliferation of spindle cells. It is typically caused by HHV-8. Previous Next Score Report Lab Values Calculator Help Pause

78 Exam Section 2: Item 29 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 29. A 65-year-old man comes to the physician because of a 3-month history of substernal chest pain that occurs after eating. Physical examination shows no abnormalities except for systolic hypertension. Results of an exercise stress test are positive. The table shown depicts the results of an exercise stress test in 1000 patients at similar risk compared with the presence or absence of coronary artery disease (CAD). CAD Absent Present 442 208 650 80 270 522 478 1000 Positive Exercise Stress Test Negative 350 Based on these data, which of the following is the pretest probability of CAD in this patient? A) 35% B) 48% C) 52% D) 65% Correct Answer: D. Pretest probability is defined as the baseline risk for having the condition of interest, in this case coronary artery disease (CAD), in a given population. It is proportionate to the prevalence of the disease in the population. It is determined by taking the number of people in the population with the condition of interest and dividing it by the total number of people in the population. In this case, 650 individuals have coronary artery disease out of a population of 1000. 650/1000 = 0.65 = 65%. Pretest probability is important because it impacts the positive predictive value (PPV) and negative predictive value (NPV) of a test. PPV and NPV are statistical measures 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). 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. In contrast, pretest probability does not impact the sensitivity or specificity of a test. Incorrect Answers: A, B, and C. 35% (Choice A) is the rate of individuals in the population who do not have CAD. This is equal to 1-pretest probability. 48% (Choice B) is approximately the number of individuals in the population who have a negative exercise stress, and 52% (Choice C) is approximately the number of individuals in the population who have a positive exercise stress test, whether or not they have CAD.

Objective: Pretest probability is defined as the baseline risk for having the condition of interest. The pretest probability alters the PPV and NPV of a test but not its sensitivity and specificity. Previous Next Score Report Lab Values Calculator Help Pause

74 Exam Section 2: Item 25 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 25. A full-term female newborn is evaluated shortly after birth. Pregnancy and delivery were uncomplicated. During the first 5 minutes after delivery, which of the following is most likely to be increased in this newborn? A) Flow from pulmonary artery to aorta (through ductus arteriosus) B) Pulmonary artery pressure C) Pulmonary blood flow D) Right atrial pressure to left atrial pressure gradient E) Venous return when the umbilical cord is ligated Correct Answer: C. The fetal circulation undergoes important physiologic changes as the neonate transitions to extrauterine life. These changes revolve around adjusting to receiving oxygenated blood from the lungs instead of through the umbilical cord. Fetal blood flow involves three primary physiologic shunts: the ductus venosus, ductus arteriosus, and foramen ovale. Oxygenated blood from the mother flows through the umbilical vein and a portion perfuses the liver. The majority passes through the ductus venosus to enter the inferior vena cava and mix with deoxygenated blood returning from the lower portion of the fetus. As blood enters the right atrium, it is shunted through the foramen ovale into the left atrium because of high pulmonary vascular resistance resulting from hypoxic pulmonary vasoconstriction in the intrauterine environment. Blood that does pass through the right ventricle and into the pulmonary trunk is directed into the aorta via the ductus arteriosus. In the first few minutes of life, the new exposure of the lungs to an oxygen-rich environment results in a dramatic decrease in pulmonary vascular resistance and reversal of the right- to-left shunt. Pulmonary blood flow increases accordingly. Other factors that are essential in establishing gas exchange in the neonatal lungs include clearing of fluid from the alveoli (via cough and crying) and adequate surfactant production to prevent alveolar collapse. Incorrect Answers: A, B, D, and E. Flow from pulmonary artery to aorta (through ductus arteriosus) (Choice A) decreases as the pulmonary vascular resistance drops and the pressure gradient between the pulmonary and systemic circulation reverses. Pulmonary artery pressure (Choice B) decreases secondary to oxygen-induced pulmonary vascular smooth muscle relaxation. The right atrial pressure to left atrial pressure gradient (Choice D) decreases as the pulmonary vascular resistance drops and results in the cessation of flow from right-to-left through the foramen ovale. Venous return when the umbilical cord is ligated (Choice E) decreases significantly as the neonate is no longer receiving volume from the placenta.

Objective: Pulmonary blood flow increases in the first few minutes of extrauterine life secondary to reversal of the pressure gradient between the pulmonary and systemic circulation. This process is driven by oxygen-induced pulmonary vasodilation. %3D Previous Next Score Report Lab Values Calculator Help Pause

17 Exam Section 1: Item 17 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 17. A 40-year-old woman who is a construction worker comes to the physician because of pain in her shoulder that developed after she lifted a heavy beam. Physical examination shows weakness and pain only during external rotation of the arm at the shoulder. There is muscle wasting inferior to the spine of the scapula. Which of the following muscles is most likely injured? A) Infraspinatus B) Latissimus dorsi C) Subscapularis D) Supraspinatus E) Teres major Correct Answer: A. The rotator cuff is the group of muscles and tendons that act to seat the humeral head directly into the center of the glenoid during motion of the shoulder. These muscles also act, in concert with other muscles about the shoulder, to internally and externally rotate, abduct, and adduct the shoulder. These muscles include the subscapularis, teres minor, supraspinatus, and infraspinatus. The prime movements of these muscles at the glenohumeral joint include internal rotation (subscapularis), external rotation when in an adducted position (teres minor), external rotation (infraspinatus), and abduction (supraspinatus). The coupling and antagonistic action of these muscles seats the humeral head in the glenoid and provides a stable pivot upon which the larger, more powerful muscles of the shoulder can act, such as the latissimus dorsi, deltoid, and pectoralis major. A rotator cuff tear typically occurs at the insertion of the tendons of the rotator cuff on the humeral head. Commonly, this occurs in the supraspinatus tendon; however, tears can occur in any of the rotator cuff tendons or at the myotendinous junction. Risk factors for rotator cuff tears include heavy labor, repetitive overhead lifting, diabetes mellitus, and vascular disease. Chronic disease leading to tendon degeneration as well as acute or repetitive microtrauma are thought to contribute to eventual tendon failure. The classic presentation of a rotator cuff tear is with pain and weakness in one of the axes of motion corresponding with the associated rotator cuff muscle. Late manifestations of a tear include clinically appreciable atrophy of the corresponding muscle. In this case, the patient's weakness in external rotation and loss of muscle bulk inferior to the scapular spine are consistent with an injury to the infraspinatus. Incorrect Answers: B, C, D and E. An injury to the latissimus dorsi (Choice B) would demonstrate a deficit in shoulder adduction and internal rotation. Injury to the latissimus dorsi is uncommon. Injury to the subscapularis (Choice C) would demonstrate a deficit in internal rotation and is often associated with pain and tenderness at the subscapularis insertion. Subscapularis injuries are difficult to detect as the pectoralis major compensates for loss of this muscle action. Injury to the supraspinatus (Choice D) is a common presentation of a rotator cuff tear. It classically presents with pain and weakness in abduction or forward elevation of the shoulder, not external rotation. Teres major (Choice E) functions in internal rotation and adduction about the shoulder. Injuries present with pain or weakness in association with these motions but are difficult to detect because of the compensation of latissimus dorsi.

Objective: Rotator cuff injuries are common in adults. They typically present with pain and weakness in the direction of shoulder motion that corresponds with the action of the associated rotator cuff muscle. Previous Next Score Report Lab Values Calculator Help Pause

38 Exam Section 1: Item 39 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 39. A 33-year-old African American woman comes to the physician because of dry eyes for 3 weeks. Physical examination shows decreased lacrimation and uveitis. There is mild hepatosplenomegaly. A chest x-ray shows prominent bilateral hilar adenopathy. Microscopic examination of a hilar lymph node is most likely to show which of the following pathologic conditions? A) Abscess formation B) Granuloma formation C) Hodgkin disease D) Langerhans histiocytosis E) Metastatic melanoma F) Non-Hodgkin lymphoma G) Plasmacytoma H) Sinus histiocytosis Correct Answer: B. Sarcoidosis is a noncaseating granulomatous disease that often involves the lung, and classically presents with shortness of breath and cough in a younger, of characterized by the presence of noncaseating granuloma formation on histology. The clinical presentation ranges from asymptomatic, to mild shortness of breath and cough, to diffuse granulomatous infiltration of multiple organ systems. Manifestations include uveitis, lymphadenopathy, hepatosplenomegaly, hypercalcemia, and erythema nodosum. On diagnostic imaging, patients often present with bilateral hilar lymphadenopathy, with or without bilateral upper lobe predominant interstitial infiltrates depending on the stage of the disease. The diagnosis is presumed on the basis of clinical history and imaging features, although tissue sampling, most commonly from involved lymph nodes, is required to definitively diagnose a patient. Treatment typically requires steroids. Untreated chronic disease can progress to pulmonary fibrosis and restrictive lung physiology. African American female. It is Incorrect Answers: A, C, D, E, F, G, and H. Abscess formation (Choice A) occurs in the setting of an acute inflammatory response to an infectious organism, resulting in an organized collection of pus, necrotic cellular debris, and infectious organisms present on histology. Tuberculosis is commonly implicated in lymph node abscess formation. Šarcoidosis forms noncaseating granulomas without evidence of necrosis. Hodgkin disease (Choice C) is a B-cell malignancy of the lymphatic system. Patients present with constitutional symptoms such as fever, night sweats, and unintentional weight loss. Biopsy may disclose atypical cells with bilobed nuclei, a large inflammatory infiltrate with eosinophil predominance, and/or diffuse fibrosis forming bands with scattered open spaces. Langerhans histiocytosis (Choice D) is a neoplasm of dendritic cells, which may present with different clinical syndromes depending on the underlying subtype. It typically presents in childhood or the neonatal period. Disease may affect the skin, bones, or visceral organs. Pulmonary involvement may be seen later in life. Metastatic melanoma (Choice E) typically originates as a hyperpigmented skin lesion with asymmetry, irregular-appearing borders, variable coloration, a diameter greater than 6 mm, and rapid evolution in characteristics. Malignant melanoma has the ability to rapidly invade and metastasize, which carries a poor prognosis when diagnosed late. The patient's presentation is more consistent with sarcoidosis. Non-Hodgkin lymphoma (Choice F) refers to a group of neoplasms that originate from lymphatic tissue. Patients present with constitutional symptoms such as fever, night sweats, unintentional weight loss, and progressive lymphadenopathy. Hepatosplenomegaly may be present. It does not typically cause uveitis. Plasmacytoma (Choice G) is a tumor composed of abnormal, neoplastic plasma cells. They most commonly develop in bones and may result in hypercalcemia and renal failure. Sinus histiocytosis (Choice H) refers to an abundance of histiocytes on lymph node biopsy, which is a common and nonspecific finding, often seen in the setting of local inflammation or malignancy.

Objective: Sarcoidosis classically presents with shortness of breath and cough in a younger, often African American female. It is associated with uveitis, erythema nodosum, and bilateral hilar lymphadenopathy. Biopsy of affected tissue will disclose noncaseating granulomas. Previous Next Score Report Lab Values Calculator Help Pause

72 Exam Section 2: Item 23 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 23. A previously healthy 6-year-old boy is brought to the physician by his mother because of a 1-week history of an itchy rash in his armpits, both hands and feet, and groin. The itching is most intense at night and keeps him awake. His vital signs are within normal limits. Examination of the skin shows multiple erythematous papules, some with burrows and many of which are excoriated, similar to the ones shown in the photograph. Which of the following questions to the mother will be most helpful in establishing the diagnosis? A) "Do you have any pets at home?" B) "Does anyone else in the family have an itchy rash like this?" OC) "Does your child take any medications?" D) "Has there been any recent travel?" E) "Have you changed soaps or detergents lately?" Correct Answer: B. Scabies is a skin infestation by the mite Sarcoptes scabiei that is transmissible by skin-to-skin contact, as in family members or sexual partners. The skin infestation typically occurs on the sides or webs of the fingers, wrists, axillae, areolae, or genitalia and leads to intense pruritus, as in this patient. Physical examination commonly demonstrates several small, erythematous papules that are frequently excoriated from scratching and may show mite burrows. In some cases, lesions demonstrate a crusted appearance. Diagnosis is confirmed by microscopic examination disclosing mites, eggs, or feces. Scabies is managed with topical or oral antiparasitic agents such as permethrin or lindane (less frequently used). Incorrect Answers: A, C, D, and E. Skin infections transmitted by household pets (Choice A) may include cutaneous larva migrans (twisting linear skin lesions caused by burrowing of larvae), tinea corporis (pruritic, erythematous circular or oval plaques), and cellulitis. Scabies lesions appear distinct from these zoonotic skin infections. Household pets may transmit scabies to humans; however, human-to-human transmission is more common. Rashes related to medications (Choice C) include type I (urticarial lesions) and type III (urticarial or purpuric rashes) hypersensitivity reactions. Intensely pruritic papules with burrows are atypical for medication-related eruptions. Rashes related to recent travel (Choice D) include a broad differential diagnosis depending on the region of travel including infectious causes (parasites, fungi, bacteria), which typically produce nodular or ulcerative lesions. Noninfectious causes include insect bites, which also appear nodular. This patient's erythematous papules and burrows are more typical of scabies. Contact dermatitis may arise from irritation of the skin from soaps or detergents (Choice E). Skin lesions are typically focal in one or several locations and appear as mild erythema or vesicles/bullae. Lesions are typically painful, and the presence of pruritus is variable.

Objective: Scabies is a skin infestation by mites that is transmitted by skin-to-skin contact. Scabies typically presents with intensely pruritic erythematous papules of the sides or webs of the fingers, wrists, axillae, areolae, or genitalia, and burrows may be visible. Previous Next Score Report Lab Values Calculator Help Pause

79 Exam Section 2: Item 30 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 30. An investigator is studying the effects of succinylcholine and Drug X on muscle contraction in an experimental animal model. During the study, the force of contraction of abdominal muscles is recorded after the administration of succinylcholine alone. In a separate experiment, Drug X is administered concurrently with succinylcholine. Results show that the phase I neuromuscular blockade induced by succinylcholine is enhanced by concurrent administration of Drug X. However, when Drug X is administered during the succinylcholine-induced phase Il neuromuscular blockade, it is found that Drug X reverses the effects of succinylcholine. Drug X is most likely which of the following? A) Atropine B) Dantrolene C) Neostigmine D) Rocuronium E) Verapamil Correct Answer: C. Succinylcholine is a depolarizing skeletal muscle relaxant commonly used to facilitate rapid sequence tracheal intubation. It is a potent acetylcholine receptor agonist that causes depolarization of the muscle fibers, seen clinically as fasciculations, followed by muscle relaxation. It is structurally similar to acetylcholine but is not metabolized by acetylcholinesterase. Rather, after it diffuses away from the neuromuscular junction, it is hydrolyzed by pseudocholinesterase in the plasma and liver. This 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. Cholinesterase inhibitors, such as neostigmine, reverse neuromuscular blockade when administered after succinylcholine as a result of an increase in acetylcholine concentration available for receptor binding as succinylcholine concentration decreases. However, if neostigmine is administered concurrently with succinylcholine, it initially enhances paralysis provided by the depolarizing neuromuscular blocker. This enhancement is caused by the increased concentration of acetylcholine in the nerve terminal, leading to a more intense depolarization, in addition to neostigmine inhibition of pseudocholinesterase. Incorrect Answer: A, B, D, and E. Atropine (Choice A) is an anticholinergic that acts by competitively binding to muscarinic acetylcholine receptors, particularly in the heart and bronchial smooth muscle. It is most commonly used to treat symptomatic bradycardia. It does not act on nicotinic receptors in the neuromuscular junction and would not affect the potency of succinylcholine. Dantrolene (Choice B) is a ryanodine receptor antagonist, used in the treatment of malignant hyperthermia. It prevents the release of calcium from the sarcoplasmic reticulum in response to intracellular calcium, which prevents further muscle contraction and anaerobic metabolism. Rocuronium (Choice D) is a nondepolarizing neuromuscular blocker, which acts as a competitive antagonist at the acetylcholine receptors. When given concurrently with succinylcholine, it would compete for acetylcholine receptors and cause decreased depolarization and effectiveness of succinylcholine. Verapamil (Choice E) is a calcium channel blocker that slows conduction at the sinoatrial and atrioventricular nodes, decreases cardiac contractility, and decreases blood pressure. It is most commonly used to treat supraventricular arrhythmias. It does not act at the neuromuscular junction and would not be expected to cause decreased effectiveness of succinylcholine.

Objective: Succinylcholine is a nondepolarizing neuromuscular blocker that acts as an acetylcholine receptor agonist. When cholinesterase inhibitors (eg, neostigmine) are administered concurrently with succinylcholine, they initially cause increased intensity of depolarization and paralysis caused by an increase in acetylcholine concentration at the neuromuscular junction. However, when administered after succinylcholine, they increase the amount of acetylcholine available to competitively bind to the acetylcholine receptors as succinylcholine concentration decreases, leading to a reversal of paralysis. Previous Next Score Report Lab Values Calculator Help Pause

37 Exam Section 1: Item 38 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 38. A 66-year-old right-handed man has a thrombotic stroke. A CT scan of the head is shown. Which of the following is the most likely neurologic finding in this patient? A) Global aphasia B) Left hemiparesis OC) Spatial neglect D) Weakness of all muscles of facial expression on the right E) Weakness of the lower two thirds of the face on the left Correct Answer: A. This axial, noncontrast CT image shows hypodensity (darkness) of most of the left hemisphere (sparing the anterior and posterior cerebrum) and is consistent with an ischemic infarction involving the territory of the middle cerebral artery (MCA). The MCA supplies large portions of the lateral frontal, parietal, and temporal lobes. Occlusion of the MCA in the dominant hemisphere, which is frequently the left hemisphere, can lead to global aphasia caused by involvement of the Broca (inferior frontal lobe) and Wernicke (superoposterior temporal lobe) areas, along with the arcuate fasciculus, the bundle of axons that connects these two areas. Occlusion of the left MCA also leads to right-sided paralysis and sensory loss of the face and upper limb (caused by involvement of the superior and lateral precentral and postcentral gyri). Cerebrovascular accidents (CVAS, also known as strokes) occur as a result of ischemic or hemorrhagic loss of blood supply to the brain. Approximately 80%-85% of CVAS are ischemic, commonly arising from thromboembolic disease (eg, MCA occlusion from a thrombus), whereas 15%-20% of CVAS are hemorrhagic as a result of blood vessel rupture (eg, hypertension-related intraparenchymal hemorrhage from a perforating artery). Risk factors for CVAS include smoking, hypertension, diabetes, carotid or intracranial atherosclerotic disease, history of hypercoagulability, atrial fibrillation, and advanced age. Classically, CVAS manifest as a neurologic deficit related to loss of function of the affected part of the brain. Incorrect Answers: B, C, D, and E. Left-sided hemiparesis (Choice B) would be caused by right-sided lesions of the lateral corticospinal tract. Occlusion of the right MCA could lead to hemineglect, paralysis, and sensory deficits of the left-sided face and arm, and lacunar stroke affecting the internal capsule could cause total left-sided hemiparesis. Spatial neglect (Choice C) is typically caused by damage to the parietal lobe of the nondominant hemisphere, which leads to hemineglect of the contralateral side of the body. In most patients, hemineglect syndrome results from occlusion of the right MCA, which would additionally cause paralysis and sensory loss of the contralateral arm and face.Weakness of all muscles of facial expression on the right (Choice D) is associated with lower motor neuron lesions of cranial nerve VII (facial nerve) as in Bell palsy. Lesions of the anterior inferior cerebellar artery can affect the cranial nerve VII (facial) nucleus in the pons, also causing weakness of the ipsilateral side of the face. Weakness of the lower two thirds of the face on the left (Choice E) would be caused by upper motor neuron lesions of the contralateral corticobulbar tract innervating the left cranial nerve VII nucleus. Importantly, the forehead would be spared, as the cranial nerve VII nucleus that controls forehead musculature is dually innervated by corticobulbar tract fibers from bilateral precentral gyri. Because of this, the patient would be expected to symmetrically raise his eyebrows.

Objective: The MCA supplies large portions of the lateral frontal, parietal, and temporal lobes. Occlusion of the MCA in the dominant hemisphere can lead to global aphasia as a result of the involvement of Broca and Wernicke areas, along with the arcuate fasciculus, the bundle of axons that connects these two areas. MCA occlusion also presents with paralysis and sensory loss of the contralateral face and arm. %3D Previous Next Score Report Lab Values Calculator Help Pause

49 Exam Section 1: Item 50 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 50. A 75-year-old woman is brought to the physician because of a 1-day history of fever and back pain. Her temperature is 39.5°C (103.1°F), pulse is 111/min, respirations are 32/min, and blood pressure is 115/79 mm Hg. Physical examination shows left-sided costovertebral angle tenderness. Laboratory studies show a leukocyte count of 17,000/mm3 (with 9% bands) and pyuria. Urine cultures grow Escherichia coli. It is determined that the patient's fever is partially due to interleukin-6 (IL-6), which was induced by the IL-1 produced during the immune response. Which of the following best describes the role of IkB in the nuclear factor-kappa B (NF-KB) signal transduction pathway from IL-1 binding to IL-6 induction in this patient? A) Attaches to cytokine receptor B) Facilitates proteolytic cleavage of NF-KB dimers C) Phosphorylates NF-KB D) Releases NF-KB after undergoing phosphorylation E) Translocates to the nucleus after undergoing phosphorylation Correct Answer: D. NF-KB (nuclear factor kappa-light-chain-enhancer of activated B cells) comprises a family of transcription factors that induce the expression of several proinflammatory genes, promoting the production of cytokines, chemokines, and other inflammatory mediators in response to infection, as in this patient's case of pyelonephritis. The NF-KB pathway may be activated by several stimuli, including the presence of IL-1, reactive oxygen species, tumor necrosis factor-alpha, and binding of pathogen-associated molecular patterns (PAMPS) such as bacterial lipopolysaccharides to toll-like receptors on leukocytes. NF-KB is normally present in the cytoplasm bound to the inhibitor protein IKB that renders it inactive. In the signaling pathway, a ligand (such as IL-1) binding to the appropriate receptor results in activation of the IKB kinase (IKK) complex. Activated IKK phosphorylates IKB, which releases NF-KB after undergoing phosphorylation. NF-KB is then able to translocate to the nucleus and upregulate transcription and translation of proinflammatory genes. Increased synthesis of IL-6 is induced by this pathway. Incorrect Answers: A, B, C, and E. IL-1 attaches to cytokine receptors (Choice A) on inflammatory cells, which is one of the signals that initiates the NF-KB pathway. Conversely, IkB is located within the cytoplasm, bound to NF-KB. IKB does not bind a cytokine receptor. There are bacterial proteases that facilitate proteolytic cleavage of NF-KB dimers (Choice B), which act as virulence factors expressed by some strains of Escherichia coli and Chlamydia species. Cleavage of the NF-KB dimer renders it ineffective in promoting gene expression. This is not the mechanism of IkB. Phosphorylates NF-KB (Choice C) is not the role of IKKB. IKB is phosphorylated by IKK leading to release of the NF-KB dimer, which then translocates to the nucleus. Translocates to the nucleus after undergoing phosphorylation (Choice E) is incorrect as phosphorylated IKB remains in the cytoplasm.

Objective: The NF-KB signal transduction pathway is critical in producing the innate immune response to inflammatory stimuli. NF-KB exists in the cytoplasm in an inactivated state bound to IkB. Inflammatory stimuli result in the activation of IKK, leading to phosphorylation of IKB, which then releases active NF-KB. Previous Next Score Report Lab Values Calculator Help Pause

88 Exam Section 2: Item 39 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 3' 39. An investigator creates Drug Z that interferes with protein synthesis by binding to the anticodon on the human tRNA shown in the diagram. Drug Z is most likely binding in the place of which of the following codons? A A) AUG 5' A B) CCG C) GCC A D) GCU A E) GUA G A O F) UCG G G. A A, Correct Answer: A. AUG is the codon to which this TRNA anticodon would normally bind in the absence of Drug Z. RNA is a specialized RNA that recognizes a codon on mRNA molecules, which consists of three nucleotides in a specific sequence. It carries the associated amino acid for a given mRNA codon and is therefore critical in protein synthesis. There are several primary domains of the tRNA molecule, which is read in the 3'-to-5' direction. The vertical sequence at the top starting with the 3' end is called the acceptor stem, which covalently binds the specific amino acid. The loop on the right is called the T-arm, which is required for the RNA to bind to the ribosome. The D-arm lies opposite to the T-arm. The D-arm is necessary for the accurate recognition of the tRNA by the appropriate aminoacyl-tRNA synthetase. The vertically oriented loop pointing downwards is the anticodon arm, and the anticodon sequence refers to the three-base pair sequence at the bottom. In this instance, the anticodon sequence is 3'-UAC-5'. This corresponds to an mRNA sequence of 5'-AUG-3'. Incorrect Answers: B, C, D, E, and F. Choices B through F all represent mRNA codon sequences that have corresponding base pairs on the above tRNA molecule, but none of them are the anticodon for this specific 1RNA molecule. The anticodon is located at the bottom of the vertically oriented, downward-directed arm of the tRNA, which in this particular example is 3'-UAC-5'.

Objective: The anticodon of a tRNA molecule is the three-base pair sequence at the bottom of the anticodon arm that recognizes a corresponding MRNA codon. This is the location where tRNA binds to mRNA in the transfer of a specific amino acid during protein synthesis. A drug that binds to the anticodon sequence would prevent the tRNA from interacting with MRNA. Previous Next Score Report Lab Values Calculator Help Pause

31 Exam Section 1: Item 32 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 32. An autopsy is done on a 50-year-old man who died of pneumonia despite 5 days of antibiotic therapy in the intensive care unit. He had a 15-year history of alcoholism. A photograph of a sagittal section of the brain is shown. Based on this pathology, which of the following findings was most likely present on neurologic examination of the patient prior to his death? A) Dysdiadochokinesia B) Dysmetria on finger-nose testing C) Essential tremor D) Gait ataxia E) Present Romberg sign Correct Answer: D. This patient's cerebellar vermis (pictured in cross section) appears atrophic, which could present with gait ataxia. The cerebellar vermis is in the medial zone of the cerebellum (smaller in cross- sectional surface area than the cerebellar hemisphere behind the vermis). The cerebellar vermis controls the motor function of the axial and proximal limb muscles and mediates body posture and balance. Lesions of the cerebellar vermis lead to severe gait ataxia (from truncal instability), titubation (bobbing of the head or axial body), and nystagmus. The cerebellar hemispheres control ipsilateral motor functions of the limb muscles (generally distal compared with muscles controlled by the vermis). Dysfunction of the cerebellar hemispheres leads to dysdiadochokinesia, dysmetria on finger-nose testing, intention tremor, and gait ataxia (from lower extremity instability). Alcohol has direct toxic effects on the dendrites of cerebellar neurons, which may mediate cerebellar atrophy from chronic alcohol use. Cerebellar damage can persist after abstinence from drinking. Incorrect Answers: A, B, C, and E. Dysdiadochokinesia (Choice A) and dysmetria on finger-nose testing (Choice B) occur when the cerebellar hemispheres, as opposed to the cerebellar vermis, are damaged. Cerebellar hemisphere damage also causes an intention tremor and gait ataxia. Essential tremor (Choice C) is a common adult action tremor that can present as a bilateral intention tremor of the upper extremities and is commonly heritable. Essential tremor has been associated with cerebellar lesions in various locations but is not specifically correlated with lesions of the cerebellar vermis. A present Romberg sign (Choice E) signifies that a patient maintains station without wobble or sway when the eyes are open but sways or falls when the eyes are closed. A present Romberg sign reflects a proprioceptive deficit and frequently correlates to lesions of the dorsal column-medial lemniscus pathway, not cerebellar lesions.

Objective: The cerebellar vermis controls the motor function of the axial and proximal limb muscles and thus mediates body posture and balance. Lesions of the cerebellar vermis lead to severe gait ataxia (from truncal instability), titubation (bobbing of the head or axial body), and nystagmus. %3D Previous Next Score Report Lab Values Calculator Help Pause

94 Exam Section 2: Item 45 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 45. A 55-year-old woman with a history of alcohol dependence has severe epigastric pain and is vomiting large amounts of blood. She has slightly yellow skin and conjunctivae. Her abdomen is large and distended. The bleeding vessels are most likely enlarged due to increased blood flow directly supplied by which of the following veins? О А) Нерatic B) Left gastric C) Right gastro-omental D) Splenic E) Superior mesenteric Correct Answer: B. Left gastric vein pressure from portal hypertension leads to the development of esophageal varices that may rupture and lead to massive, life-threatening hematemesis. Portal hypertension (PH) is often a consequence of liver cirrhosis but may also occur in patients with schistosomiasis or portal venous thrombosis. The portal venous system describes a series of interconnected veins that drain blood from the colon, small intestines, spleen, liver, and stomach. While this blood eventually makes its way to the systemic circulation via the inferior vena cava, it must first pass through the liver. In cirrhosis, obliteration of the hepatic sinusoids through progressive fibrosis increases the resistance to blood flow through the liver, which is transmitted to the portal venous system causing PH. Anatomically, the portal vein starts at the confluence of the left gastric, splenic, and superior mesenteric veins. The left gastric vein drains the inferior esophageal veins. When engorged secondary to portal hypertension, esophageal veins can be visualized on esophagogastroduodenoscopy (EGD) as large vertical columns underneath the esophageal mucosa, representing dilated submucosal veins, and are termed esophageal varices (EVs). EVs are usually treated endoscopically by band ligation, and patients with cirrhosis must be screened for EVs, as the risk for life-threatening upper gastrointestinal bleeding is high. This patient presents with hematemesis likely from ruptured esophageal varices. Incorrect Answers: A, C, D, and E. Hepatic veins (Choice A) are not part of the portal system, as they drain blood from the liver to the inferior vena cava. Increased pressure in this system from right-sided heart failure can result in hepatic sinusoidal congestion and potential cardiac cirrhosis. Additionally, a stent placed between the hepatic vein and portal vein (TIPS procedure) can relieve portal pressure for the treatment of portal hypertension but predisposes to the development of hepatic encephalopathy. Right gastro-omental (Choice C) venous distention leads to gastric varices, which are a less common cause of catastrophic bleeding in patients with cirrhosis. The right gastro-omental vein drains into the superior mesenteric vein. Splenic vein (Choice D) distention is common in portal hypertension and can contribute to splenomegaly, but it does not lead to gastrointestinal tract bleeding as seen in this patient. Superior mesenteric vein (SMV) (Choice E) distention is characteristic of portal hypertension. The SMV drains blood from the right and left gastro-omental veins, the small intestines, and the ascending colon. Distention of the SMV causes increased venous pressure in all of these veins, although this has no relationship to EVs and would not present with hematemesis.

Objective: The inferior esophageal veins drain into the left gastric vein, which is a component of the portal venous system. In patients with portal hypertension from cirrhosis, increased pressure is transmitted to the esophageal veins leading to the development of EVs. EVs can be life-threatening when they cause massive hematemesis, thus endoscopic screening and banding are performed routinely in patients with cirrhosis. Previous Next Score Report Lab Values Calculator Help Pause

99 Exam Section 2: Item 50 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 50. A 16-year-old boy with intellectual disability is admitted to the hospital because of severe respiratory distress. Physical examination shows dislocation of the ocular lenses bilaterally, dyspnea, and tenderness and edema of the left lower extremity. Serum concentrations of methionine and homocysteine are increased, and serum cystine concentration is decreased. This patient most likely has a defect of which of the following enzymes? A) y-Cystathionase B) Cystathionine B-synthase C) Fumarylacetoacetate hydrolase D) Hepatic adenosyltransferase E) Methylmalonyl-CoA mutase Correct Answer: B. Homocystinuria is an autosomal recessive metabolic disorder most commonly caused by a defect in cystathionine B-synthase, leading to increased serum and urine concentrations of homocysteine. Additional deficits that cause homocystinuria include decreased affinity of cystathionine synthase for its cofactors and deficiency in homocysteine methyltransferase. Increased serum concentrations of methionine indicate that the defective enzyme is cystathionine B-synthase, rather than methionine synthase, another enzyme that causes homocystinuria when deficient. Patients with homocystinuria present with characteristic signs and symptoms including intellectual disability, a tall Marfanoid habitus with long limbs, pectus excavatum and pes cavus, ocular lens subluxation (with the lens typically subluxated inferomedially), and vascular disease such as thrombosis and atherosclerosis. Patients with homocystinuria are at significantly increased risk for intravascular thrombosis as illustrated by the tender, edematous left lower extremity in this case. Myocardial infarction at an early age is well-described. Homocystinuria caused by cystathionine B-synthase deficiency is treated with supplementation of cysteine, vitamins B6 (pyridoxine) and B12 (cyanocobalamin), folate, and a diet restricted in methionine. Incorrect Answers: A, C, D, and E. y-Cystathionase (Choice A) catalyzes the conversion of cystathionine to cysteine. Defects in this enzyme cause cystathioninuria, rather than homocystinuria. Fumarylacetoacetate hydrolase (Choice C) is an enzyme involved in the metabolism of phenylalanine and tyrosine. It cleaves fumarylacetoacetate, a metabolite of homogentisic acid, to acetoacetate and fumarate. Hepatic adenosyltransferase (Choice D) catalyzes the conversion of methionine to S-adenosylmethionine. Deficiency of this enzyme results in hypermethioninemia, rather than homocystinemia. Methylmalonyl-CoA mutase (Choice E) catalyzes the conversion of methylmalonyl-COA to succinyl-CoA, which is an important step in the metabolism of branched chain amino acids and fatty acids. Deficiency in this enzyme causes methylmalonic aciduria.

Objective: Homocystinuria results from deficiency in a variety of enzymes, including cystathionine B-synthase, and leads to increased serum and urine concentrations of homocysteine. Characteristic signs and symptoms include intellectual disability, Marfanoid habitus, pectus excavatum, pes cavus, inferomedial lens subluxation, venous thrombosis, and cardiovascular disease. Previous Next Score Report Lab Values Calculator Help Pause

42 Exam Section 1: Item 43 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 43. A 4-day-old male newborn who was born at home is brought to the emergency department because of respiratory distress and cyanosis. The mother reports that she found him in his crib not breathing. He began to breathe again after she picked him up. Examination shows a narrow thorax. His ears have periauricular skin tags. He also has micrognathia, glossoptosis, a mandibular cleft, and a short palate. Tracheostomy relieves his respiratory distress. Defects such as these are consistent with altered development of which of the following pharyngeal arches? A) First B) Second C) Third D) Fourth E) Sixth Correct Answer: A. Pierre Robin syndrome is a condition present at birth and is characterized by a small lower jaw (micrognathia), downward displacement or retraction of the tongue (glossoptosis), and cleft palate. The downward displacement of the tongue can lead to airway obstruction, hypoxia, difficulty feeding, and respiratory distress. The syndrome is caused by a defect in the development of the first pharyngeal arch, which forms the maxilla, mandible, masseter, pterygoids, and mylohyoid, along with cranial nerves V2 and V3. In Pierre Robin syndrome, abnormalities resulting in hypoplasia of the mandible and micrognathia lead to the posterior displacement of the tongue. Incorrect Answers: B, C, D, and E. The second pharyngeal arch (Choice B) contributes to the development of the middle ear, hyoid bone, temporal styloid process, and muscles including the facial muscles, stapedius, platysma, and stylohyoid. It gives rise to cranial nerve VII. The third pharyngeal arch (Choice C) is involved in the development of the stylopharyngeus muscle and the glossopharyngeal nerve (cranial nerve IX). The fourth pharyngeal arch (Choice D) contributes to the development of the cricothyroid muscle and muscles of the soft palate, as well as the thyroid cartilage, and it is innervated by the superior laryngeal branch of the vagus nerve. The sixth pharyngeal arch (Choice E) contributes to the development of all intrinsic muscles of the larynx except the cricothyroid muscle, the cricoid, arytenoid, corniculate, and cuneiform cartilages, and it is innervated by the recurrent laryngeal branch of the vagus nerve.

Objective: In Pierre Robin syndrome, abnormalities in the development of the first pharyngeal arch result in hypoplasia of the mandible and micrognathia leading to posterior displacement of the tongue. The downward displacement of the tongue can lead to airway obstruction, hypoxia, feeding difficulty, and respiratory distress. I3D Previous Next Score Report Lab Values Calculator Help Pause

43 Exam Section 1: Item 44 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 44. A 48-year-old woman comes to the physician because of increasingly severe right flank pain during the past 3 weeks. Pelvic examination shows a normal-appearing vagina and cervix. A mass is palpated in the right pelvis during bimanual examination. Ultrasonography of the abdomen shows a mass associated with the right ovary. An MRI of the abdomen is suggestive of ovarian cancer. Which of the following is the location of the lymph nodes to which the ovaries first drain? A) Along the ovarian vessels B) Immediately lateral to the rectum C) Inside the broad ligament D) On the anterior surface of the sacrum E) On the fundus of the uterus Correct Answer: A. The para-aortic lymph nodes are a component of the lymphatic system, a network of vessels which follow a predictable pattern of drainage to lymph node beds. 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 and any malignant cells exiting the tissues. The para-aortic lymph nodes drain the ovaries, testes, uterus, and kidneys and would be the most likely first recipient of lymph from the ovaries. Lymphatic vessels from the ovaries are carried in the suspensory, or infundibulopelvic, ligament which extends laterally from the ovary to the wall of the pelvis and runs adjacent to the ovarian artery and vein. Thus, the lymphatics draining the ovaries run along the ovarian vessels, which is a common pairing in many anatomical locations of the body. After reaching the para-aortic lymph nodes, the lymph will be channeled through the cisterna chyli and thoracic duct, and ultimately drain into the left subclavian vein. This is the same final pathway for lymph from either lower extremity, the pelvis, or the left upper extremity. În contrast, 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. Incorrect Answers: B, C, D, and E. The pelvic lymph nodes include the pararectal lymph nodes, which are immediately lateral to the rectum (Choice B), and the sacral lymph nodes, which are on the anterior surface of the sacrum (Choice D). It is unlikely that the sentinel node of an ovarian malignancy will be located in either of these lymph node beds. The lymphatic vessels draining lymph from the ovaries are carried within the suspensory ligament, along with the ovarian vessels, toward the para-aortic lymph nodes, not inside the broad ligament (Choice C). The broad ligament is a continuation of the peritoneum and supports the uterus. The mesovarium is a named portion of the broad ligament that covers the ovaries but does not carry the ovarian lymphatic vessels. The fundus of the uterus (Choice E) is not the site of a lymph node bed. Lymph from the uterus, like that of the ovaries, is also carried to the para-aortic lymph nodes. The internal iliac lymph nodes may be closely located to the uterine fundus, but these drain the lower rectum to the level of the anal canal above the pectinate line, bladder, middle third of the vagina, cervix, and prostate.

Objective: Lymph from the ovary drains to the para-aortic lymph nodes via lymphatic vessels carried in the suspensory ligament alongside the ovarian artery and vein. Previous Next Score Report Lab Values Calculator Help Pause

13 Exam Section 1: Item 13 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 13. An investigator is studying a new hormone, Hormone X. To determine its cell target, a purified, radioactively labeled form of Hormone X is added to tissue cultures. Subsequent analysis shows that the radioactivity is detected only around the region of the plasma membrane. Hormone X is most similar to which of the following hormones? A) Aldosterone B) Cortisol C) Estradiol D) Prolactin O E) Thyroxine (T4) Correct Answer: D. Prolactin is a single-chain polypeptide hormone that is structurally homologous to growth hormone. It is secreted by the basophils in the anterior pituitary gland. Prolactin binds to the prolactin receptor, a member of the Class i ytokine receptor family, which spans the plasma membrane. It has an extracellular binding domain, a transmembrane domain, and an intracellular domain. Thus, a radioactive label on prolactin would concentrate around the region of the plasma membrane where its receptor is located. Prolactin binding to this receptor causes a series of downstream reactions to occur, ultimately utilizing the JAK/STAT signaling pathway. The physiologic effects of prolactin include stimulating milk production by mammary tissue, inhibition of ovulation, and inhibition of spermatogenesis. The regulation of prolactin is controlled by dopamine, which inhibits its production, and thyrotropin-releasing hormone, which stimulates its production. Increased prolactin concentrations cause dopamine production, leading to inhibition of further prolactin secretion. In this way, prolactin inhibits its own secretion. Incorrect Answers: A, B, C, and E. The receptors for aldosterone (Choice A), cortisol (Choice B), estradiol (Choice C), and thyroxine (Choice E) are nuclear receptors that contain DNA-binding domains. Nuclear receptors can initially be located in either the cytosol or nucleus. They bind to hormones that 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. A similar analysis to that described in the question would demonstrate radioactivity detected within the nucleus, not around the region of the plasma membrane. As a consequence of the time required for gene expression to occur, hormones binding nuclear receptors require more time to demonstrate action when compared with 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 require hours to days to result in physiologic effects.

Objective: Prolactin is a polypeptide hormone that binds to a transmembrane receptor along the plasma membrane. In contrast, aldosterone, cortisol, estradiol, and thyroxine bind nuclear receptors, which diffuse across the nuclear membrane and alter DNA transcription. %3D Previous Next Score Report Lab Values Calculator Help Pause

47 Exam Section 1: Item 48 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 48. A 4-year-old boy is brought to the emergency department by his mother 6 hours after she noticed that his urine was red. He is otherwise feeling well. Fifteen days ago, the patient had a sore throat, fever, and cough. His mother thought he had the flu and treated him symptomatically with rest and analgesics, and his status improved until now. His temperature is 38.5°C (101.3°F), pulse is 110/min, respirations are 22/min, and blood pressure is 100/50 mm Hg. Physical examination shows normal breath and cardiac sounds, no organomegaly, and 1+ lower extremity edema, bilaterally. Laboratory studies show: Serum Urea nitrogen Creatinine 40 mg/dL 2 mg/dL 3.6 g/dL Albumin Urine Color Blood Protein Ketones RBC RBC casts red-brown 3+ 2+ negative 30-50/hpf few Which of the following is the most likely diagnosis? A) Membranous nephropathy B) Minimal change disease C) Papillary necrosis D) Proliferative glomerulonephritis E) Tubulointerstitial nephritis Correct Answer: D. Proliferative glomerulonephritis is a general term that refers to histologic characteristics on kidney biopsy but is commonly used to describe postinfectious glomerulonephritis. In this instance, the patient is a young boy with nephritic syndrome following acute pharyngitis. This is most consistent with poststreptococcal glomerulonephritis, a condition that occurs primarily in children two to four weeks after an episode of pharyngitis with Streptococcus pyogenes, a gram-positive bacterium that also causes skin and soft-tissue infections. Postinfectious glomerulonephritis is a nephritic syndrome presenting with gross or microscopic hematuria and RBC casts in addition to variable degrees of proteinuria. This is in contrast to nephrotic syndromes, which present typically with high concentrations of proteinuria, edema, and hyperlipidemia. It is thought that S. pyogenes contains nephritogenic antigens that lead to a robust inflammatory response within the kidneys, with the deposition of immunoglobulin G and C3 diffusely throughout the mesangium and capillary walls. On laboratory evaluation, low serum C3 concentrations, positive anti-streptolysin O, and anti-DNase titers are noted. On light microscopy, the glomeruli are enlarged and hypercellular, which is termed proliferative glomerulonephritis. Treatment is supportive, as most patients experience complete recovery within the first few weeks of illness. Incorrect Answers: A, B, C, and E. Membranous nephropathy (Choice A) is a nephrotic syndrome that is characterized by proteinuria (more than 3 g/day), edema, hypoalbuminemia, and hyperlipidemia. It is the most common nephrotic syndrome in white adults and can be primary (idiopathic) or secondary to conditions such as systemic lupus erythematosus, autoimmune disease, hepatitis B or C, or medications. Untreated, it may progress to chronic renal failure. Minimal change disease (Choice B) is the most common cause of nephrotic syndrome in children and can be idiopathic or triggered by a recent infection or immune stimulus. Treatment is supportive and most children recover completely. The prognosis can be worse in adults with this disease. Papillary necrosis (Choice C), which refers to the sloughing and necrosis of renal papillae, can be triggered by infections, diabetes mellitus, sickle cell disease, or nonsteroidal anti-inflammatory drugs (NSAIDS). It presents on urinalysis with gross hematuria and proteinuria. Tubulointerstitial nephritis (Choice E) is usually caused by a hypersensitivity reaction to offending drugs (eg, NSAIDS, diuretics, and sulfonamides) and is characterized by a rash, eosinophilia, and eosinophiluria.

Objective: Proliferative glomerulonephritis, also commonly known as postinfectious glomerulonephritis, is a nephritic syndrome characterized by hematuria with RBC casts and variable amounts of proteinuria. It often occurs two to four weeks after an acute infection with Streptococcus pyogenes. Treatment is supportive and most patients recover completely. Previous Next Score Report Lab Values Calculator Help Pause

167 Exam Section 4: Item 18 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 18. A70-year-old man has angiodysplasia of the ascending colon. Which of the following is the most likely complication? A) Hemorrhage B) Infarction O C) Infection D) Malignant transformation E) Perforation Correct Answer: A. Angiodysplasia refers to disordered growth of a blood vessel. It classically occurs in the elderly and persons with a history of hereditary hemorrhagic telangiectasia. Angiodysplasia most commonly occurs in the gastrointestinal tract and is often asymptomatic, though it may cause clinically significant gastrointestinal hemorrhage. As part of aging, recurrent mechanical stress from peristalsis and colonic contractions is implicated in the pathogenesis of angiodysplasia. In hereditary hemorrhagic telangiectasia, defective proteins involved in blood vessel formation lead to vascular malformations. The lesions in angiodysplasia typically have weakened and dilated vessel walls that are prone to rupture. Lower gastrointestinal bleeding may present with melena, hematochezia, abdominal pain, weakness, fatigue, and lightheadedness. Physical examination may show tachycardia, pallor, and orthostatic hypotension in severe cases. If severe, signs and symptoms of hemorrhagic shock may be present. Treatment for significant gastrointestinal bleeding may include endoscopy, angiography and embolization, and transfusion of blood products. Incorrect Answers: B, C, D, and E. Infarction (Choice B) of the bowel results from mesenteric ischemia, which may be caused by impaired perfusion (eg, heart failure, cardiac valvular disease, hypovolemia, hemorrhagic shock) or thromboembolism (arterial occlusion in the setting of atherosclerotic plaque rupture or atrial fibrillation, hypercoagulable states, venous thrombosis). It presents with acute, severe abdominal pain and may be complicated by bowel wall necrosis, loss of immune barriers, and bowel perforation. Severe bleeding complications of angiodysplasia may result in secondary ischemic bowel injury. Infection (Choice C) of the colon, also called infectious colitis, may be caused by viral, bacterial, or parasitic organisms. Presentation depends on the causal organism. Angiodysplasia is not associated with an increased risk for infectious colitis. Malignant transformation (Choice D) may occur with colonic polyps. Colonic neoplasia often begins with the formation of a polyp, a small growth that may be flat, sessile, or pedunculated. Malignant transformation often involves mutations in APC, KRAS, and p53 that promote disordered cell growth and division. Perforation (Choice E) of the colon commonly presents with acute abdominal pain, peritonitis, fever, and free air on x-rays and CT scans. Severe bleeding complications of angiodysplasia may result in secondary ischemic bowel injury and risk for perforation.

Objective: Angiodysplasia is a common disorder found in older adults and patients with hereditary defects of blood vessel growth and formation (eg, hereditary hemorrhagic telangiectasia). The major complication is bleeding. I3D Previous Next Score Report Lab Values Calculator Help Pause

128 Exam Section 3: Item 29 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 29. An investigator is studying the selective loss of neurons of an experimental animal model of Wernicke encephalopathy. It is found that a vitamin B, (thiamine) deficiency in the animal causes downregulation of the glial high-affinity glutamate transporter before neurons are lost. Based on these findings, which of the following cells is most likely initially affected by this disease process? A) Astrocytes B) Ependymal cells C) Microglial cells OD) Oligodendrocytes E) Schwann cells Correct Answer: A. Astrocytes are a subtype of glial cell involved in glutamate homeostasis. Astrocytes support neurons by buffering the extracellular space, regulating energy metabolism, and reacting to neuronal injury. În response to damage to the central nervous system (CNS), glutamate may be released rapidly into the synaptic clefts. When such concentrations persist at high levels, neuronal apoptosis ensues. High-affinity glutamate transporters on astrocytes are the main mechanism for the removal of glutamate from synapses, thereby protecting the brain from glutamate-induced excitotoxicity. When these high-affinity glutamate transporters are downregulated, neurons may be more likely to undergo apoptosis and produce the characteristic pattern of brain atrophy in Wernicke-Korsakoff syndrome. The mammillary bodies are typically atrophied in Wernicke-Korsakoff syndrome, a neuropsychiatric disorder most commonly related to chronic alcohol abuse and malnutrition. Patients who chronically abuse alcohol can demonstrate a significant thiamine (vitamin B1) deficiency caused by poor nutritional intake and thiamine malabsorption. Incorrect Answers: B, C, D, and E. Ependymal cells (Choice B) form the permeable lining of the ventricles that separates the cerebrospinal fluid system and brain parenchyma. Though ependymal cells remove glutamate from the cerebrospinal fluid, high-affinity glutamate transporters are denser on astrocyte cell membranes given the large number of synapses that astrocytes must regulate. Microglial cells (Choice C), derived from precursor monocytes, phagocytose cellular debris and dead neurons. They play a much smaller role than astrocytes in glutamate homeostasis.Oligodendrocytes (Choice D) are glial cells that produce myelin in the CNS. They play a much smaller role than astrocytes in glutamate homeostasis. Schwann cells (Choice E) form the myelin sheath around axons in the peripheral nervous system. They are not present in the CNS and are therefore not involved in glutamate homeostasis in the brain.

Objective: Astrocytes are a subtype of glial cell that are involved in glutamate homeostasis. Using high-affinity glutamate transporters, astrocytes remove glutamate from synapses and thereby protect the brain from glutamate-induced excitotoxicity. %3D Previous Next Score Report Lab Values Calculator Help Pause

145 Exam Section 3: Item 46 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 46. An 82-year-old woman is brought to the physician because of a 1-week history of episodes of light-headedness and fainting. She has hypertension currently treated with propranolol. Her pulse is 45/min, and blood pressure is 90/50 mm Hg. Physical examination shows no other abnormalities. An ECG shows normal, regular P waves followed by normal QRS complexes. The physician decreases the dose of propranolol. Five days later, the patient's pulse is 80/min, and her symptoms have resolved. Which of the following is the most likely explanation for this patient's low pulse while taking the higher dose of propranolol? A) Decreased activation of fast sodium channels B) Decreased rate of repolarization phase C) Decreased slope of diastolic depolarization D) Increased calcium influx during late phase of slow depolarization E) Increased potassium influx during rapid depolarization phase F) Increased sodium efflux during early repolarization phase Correct Answer: C. Propranolol is a nonselective B-adrenergic blocker that is used in the treatment of hypertension. B-adrenergic blockers are also classified as class Il antiarrhythmic agents because of their ability to modulate the electrophysiology of the heart. For this indication, propranolol may also be used in the treatment of supraventricular tachycardias such as atrial fibrillation and atrial flutter. Class II agents act through the inhibition of B-adrenergic receptors that are coupled to adenylate cyclase in the sinoatrial (SA) and atrioventricular (AV) nodes. Inhibition leads to decreased concentrations of CAMP and a decreased slope of diastolic depolarization (phase 4) in pacemaker cells as depolarization in these cells relies on the spontaneous, slow, and mixed Na*/K* influx through If channels (funny current channels). The slope of phase 4 in the SA node is the main determinant of heart rate in normal sinus rhythm, as the spontaneous membrane depolarization accounts for the automaticity of pacemaker cells. Class Il agents result in slowed depolarization of phase 4 and subsequently a slower heart rate. Slowed conduction through the AV node may manifest as a prolonged PR interval on the ECG. Adverse effects of B-adrenergic blockers include bradycardia, AV block, and heart failure. Propranolol is also associated with exacerbations of Prinzmetal (vasospastic) angina. Incorrect Answers: A, B, D, E, and F. Decreased activation of fast sodium channels (Choice A) can be accomplished through the use of class IA, IB, and IC antiarrhythmics. Class IA agents include quinidine and procainamide and result in an increased action potential, increased effective refractory period (ERP), and prolonged QT interval. Class IB agents such as lidocaine and mexiletine display a tropism for ischemic or depolarized tissue and decrease the action potential and decrease the ERP. Class IC agents include flecainide and increase the ERP in the AV node. Decreased rate of repolarization phase (Choice B) occurs with the use of class I, class III, and class IV antiarrhythmic agents and manifests as a prolonged ERP before the cell is able to depolarize again. The repolarization phase (phase 3) is driven by the opening of K* channels and closing of Ca2+ channels. Class I and class III agents (eg, amiodarone, ibutilide, dofetilide, sotalol) inhibit the opening of K* channels while class IV agents (nondihydropyridine calcium channel blockers such as verapamil and diltiazem) inhibit the closing of Ca2+ channels. Increased calcium influx during late phase of slow depolarization (Choice D) would result in faster depolarization in cells of the SA and AV nodes, causing a faster heart rate.Increased potassium influx during rapid depolarization phase (Choice E), or phase 0, does not affect the heart rate. The rapid depolarization phase in pacemaker cells is driven by voltage-gated calcium channels. The rate of depolarization in phase 4 dictates the heart rate. Increased sodium efflux during early repolarization phase (Choice F), or phase 1, does not affect the heart rate. Phases 1 and 2 (early repolarization and plateau) are absent in pacemaker cells and only occur in the cardiac myocyte action potential.

Objective: B-adrenergic blockers function as class II antiarrhythmics by prolonging the slow, spontaneous depolarization phase of pacemaker cells in the sinoatrial and atrioventricular nodes. A prolonged depolarization phase results in a slower heart rate. Previous Next Score Report Lab Values Calculator Help Pause

163 Exam Section 4: Item 14 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 14. A 47-year-old man is admitted to the hospital because of acute pulmonary edema. Treatment is started with nesiritide, a drug that is structurally identical to endogenous human B-type natriuretic peptide. Which of the following is most likely to occur in this patient? A) Arterial hypertension B) Arteriolar vasoconstriction C) Decreased cardiac output D) Exaggerated orthostatic hypotension E) Increased urine flow Correct Answer: E. Brain-type natriuretic peptide (BNP) is produced primarily by cardiomyocytes of the ventricles in response to mechanical distension and neurohormonal signaling (mediated by angiotensin II, endothelin, and B-adrenergic stimulation). BNP originates as prepro-BNP, which is enzymatically cleaved to pro-BNP within the cardiomyocyte. Pro-BNP is then cleaved into BNP and NT-pro- BNP (N-terminal pro-BNP), which are released into circulation. Both BNP and NT-pro-BNP are used as sensitive biomarkers for the presence of cardiogenic pulmonary edema and heart failure in patients. BNP promotes vasodilation by acting on natriuretic peptide receptor-A (NPR-A) which stimulates guanylyl cyclase activity leading to increased intracellular concentrations of CGMP in vascular smooth muscle cells. BNP also acts on the renal vasculature, promoting natriuresis (excretion of sodium) and diuresis (excretion of water) by increasing the glomerular filtration rate and decreasing renin production from the juxtaglomerular apparatus. BNP, along with atrial natriuretic peptide (ANP), antagonizes the renin-angiotensin-aldosterone system resulting in decreased arterial blood pressure and increased removal of salt and water from the body through increased urine flow. Incorrect Answers: A, B, C, and D. Arterial hypertension (Choice A) and arteriolar vasoconstriction (Choice B) result from stimulation by angiotensin II and sympathetic nerve activity. BNP causes vasodilation and decreased arterial blood pressure. Decreased cardiac output (Choice C) is not likely to occur in this patient. The presence of acute pulmonary edema suggests a volume overloaded state, in which the end-diastolic ventricular volume (which may also be referred to as ventricular stretch/distention or preload) is increased past the point of optimal cardiac contractility per Frank-Starling mechanics. Decreasing preload by removing fluid from the body would be expected to increase cardiac output. Exaggerated orthostatic hypotension (Choice D) is not expected with use of a BNP agonist. The vasoconstrictive reflex mediated by the aortic arch and carotid sinus baroreceptors in response to postural changes remains intact.

Objective: BNP is produced by the ventricles in response to mechanical distention and stimulation by angiotensin II, endothelin, and catecholamines. It opposes the renin- angiotensin-aldosterone system by promoting vasodilation, natriuresis, diuresis, and decreased renin production. I3D Previous Next Score Report Lab Values Calculator Help Pause

192 Exam Section 4: Item 43 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 43. A 16-year-old girl with bulimia nervosa is brought to the physician by her mother because of a 4-month history of nonbloody diarrhea. Her mother is concerned that she is abusing laxatives to maintain a low body weight. The patient is 165 cm (5 ft 5 in) tall and weighs 52 kg (115 lb); BMI is 19 kg/m2. Her temperature is 37.5°C (99.5°F), pulse is 120/min, respirations are 30/min, and blood pressure is 89/61 mm Hg. Which of the following sets of serum electrolyte concentrations is most likely in this patient? Potassium Chloride Bicarbonate A) Increased increased normal B) Increased decreased decreased C) Unchanged increased decreased D) Decreased increased decreased E) Decreased normal increased Correct Answer: D. This patient experiencing chronic diarrhea from suspected laxative abuse is likely to demonstrate hypokalemia, metabolic acidosis (decreased bicarbonate), and compensatory hyperchloremia. This patient with body weight concerns, hypovolemia, and electrolyte disturbances likely has a diagnosis of bulimia nervosa (purging type). Bulimia nervosa (purging type) involves cycles of uncontrollable eating and compensatory behaviors such as vomiting, laxative overuse, or diuretic overuse that occur at least once a week over three months or more. Serum electrolyte concentrations depend on the method used to purge. In patients who abuse laxatives, hypokalemia occurs as a result of losses of potassium-rich stool. Stool is also rich in bicarbonate (to neutralize acid from gastric contacts) and organic acid anions (from ingested carbohydrates), leading to decreased bicarbonate and metabolic acidosis from chronic diarrhea. Bicarbonate is normally reclaimed in the colon, which maintains its buffering capacity. The resultant loss leads to non-anion-gap metabolic acidosis. Hypovolemia (manifested in this patient's decreased blood pressure and increased pulse) leads to increased sodium and chloride reabsorption from the renal proximal tubule, resulting in hyperchloremia. This patient's tachypnea likely represents an attempt to compensate for metabolic acidosis by breathing off carbon dioxide and thereby decreasing serum carbonic acid. Incorrect Answers: A, B, C, and E. Normal or increased potassium would be atypical because of to this patient's loss of potassium-rich stool (Choices A, B, and C). The bicarbonate concentration is rarely normal as a result of the loss of bicarbonate in the stool (Choice A). Chloride is typically increased as a result of associated hypovolemia-induced renal proximal tubule sodium and chloride reabsorption (Choice B). In rare, severe cases, hypovolemia from laxative abuse can cause contraction alkalosis (Choice E). However, these patients with metabolic alkalosis would likely also demonstrate hypochloremia and mixed acid-base deficits (related to chloride losses in stool from severe diarrhea). Metabolic acidosis is a more common presentation in the setting of chronic diarrhea.

Objective: Bulimia nervosa (purging type) can present with laxative abuse and consequent hypovolemia and electrolyte disturbances. As a result of stool losses of potassium and bicarbonate, laboratory testing typically demonstrates hypokalemia and metabolic acidosis. The associated hypovolemia can lead to the increased reabsorption of sodium and chloride in the kidney, resulting in hyperchloremia. Previous Next Score Report Lab Values Calculator Help Pause

138 Exam Section 3: Item 39 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 39. A 25-year-old woman comes to the physician because of a 3-year history of irregular menses. Menarche was at the age of 14 years. Puberty began at the age of 12 years and progressed normally. Her blood pressure is 116/62 mm Hg. Physical examination shows increased hair growth on the face and chest. Pelvic examination shows clitoromegaly and a normal-appearing uterus. Serum studies show increased concentrations of 17-hydroxyprogesterone and androstenedione. This patient most likely has a deficiency of which of the following enzyme activities? A) 11B-Hydroxylase B) 17a-Hydroxylase C) 21-Hydroxylase D) 38-Hydroxysteroid dehydrogenase E) 5a-Reductase Correct Answer: C. This patient most likely has a deficiency of 21-hydroxylase, which prevents the conversion of progesterone and 17-hydroxyprogesterone to precursors that ultimately produce aldosterone and cortisol, respectively, in the adrenal glands. In the setting of deficiency, precursors are shunted away from the production of mineralocorticoids and glucocorticoids and toward the production of androgens. Consequently, genetically female patients undergo virilization because of androgen excess and may also demonstrate salt wasting, alkalosis, hyperkalemia, and hypovolemia as a result of aldosterone deficiency. In the classic form of 21-hydroxylase deficiency, genetically female patients present with hypoaldosteronism as well as virilization during infancy, and genetically male patients present with precocious puberty in childhood. However, this patient likely has a mild, nonclassic form of the disease, resulting in the onset of symptoms during young adulthood and the absence of signs of mineralocorticoid deficiency. 21-Hydroxylase deficiency is the most common form of congenital adrenal hyperplasia (CAH). CAH refers to various adrenal enzyme deficiencies that result in decreased concentrations of cortisol with variable effects on mineralocorticoids and androgens. Because cortisol is decreased, adrenocorticotropic hormone is increased, which explains the hyperplasia present in all forms of the disease. Incorrect Answers: A, B, D, and E. 11B-Hydroxylase (Choice A) deficiency is a rare form of CAH in which the production of cortisol and aldosterone are interrupted later in the pathway in comparison to 21-hydroxylase deficiency. The result is decreased cortisol and aldosterone concentrations with an increased concentration of the intermediate mineralocorticoid, 11-deoxycorticosterone. 11-Deoxycorticosterone acts at the kidney in a manner similar to aldosterone with resultant sodium and water retention and excretion of potassium and protons. These clinical manifestations typically begin in childhood and would not be associated with increased 17-hydroxyprogesterone and androstenedione concentrations. 17a-Hydroxylase (Choice B) deficiency is a form of CAH that decreases the synthesis of glucocorticoids and androgens and therefore shunts precursors toward the production of mineralocorticoids. Consequently, hyperaldosteronism leads to hypertension and hypokalemia. Genetically male patients typically demonstrate ambiguous genitalia and undescended testes, whereas genetically female patients may not demonstrate morphological differences until puberty, when their secondary sex characteristics are underdeveloped. Concentrations of androstenedione and 17-hydroxyprogesterone would be decreased. 3B-Hydroxysteroid dehydrogenase (Choice D) deficiency leads to decreased synthesis of mineralocorticoids, glucocorticoids, and androgens. However, an intermediate in the androgen synthetic pathway, dehydroepiandrosterone (DHEA), is increased (as a result of decreased metabolism of DHEA by 3B-hydroxysteroid dehydrogenase), which can lead to virilization of genetically female infants, children, or young adults (depending on the severity of the deficiency). However, concentrations of androstenedione and 17-hydroxyprogesterone would be decreased. 5a-Reductase (Choice E) deficiency leads to abnormal external genitalia at birth and underdevelopment of secondary sex characteristics in puberty in genetic males as a result of decreased conversion of testosterone to dihydrotestosterone (DHT). 5a-Reductase deficiency affects genetic males only.

Objective: CAH results from adrenal enzyme deficiencies in the synthesis of cortisol with variable effects on mineralocorticoids and androgens. 21-Hydroxylase deficiency, the most common form, classically presents in genetic females with virilization and hypoaldosteronism during infancy. However, a nonclassic form can present in young adulthood with virilization only. Previous Next Score Report Lab Values Calculator Help Pause

125 Exam Section 3: Item 26 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 26. A 25-year-old woman, gravida 2, para 1, at 20 weeks' gestation comes to the physician for a routine prenatal visit. She has no history of any major medical illness. She says that she feels healthy, but her 2-year-old daughter recently had pneumonia. Her respirations before the pregnancy were 16/min. Her pulse is 80/min, respirations now are 22/min, and blood pressure is 100/60 mm Hg. Physical examination shows a uterine fundal height at the level of the umbilicus, consistent with a 20-week gestation. Which of the following best explains the increased respirations in this patient? A) Incipient respiratory infection B) Metabolic adjustment for fetal lung immaturity C) Occult pulmonary embolus D) Pressure on the diaphragm from the enlarging uterus E) Respiratory compensation for fetal metabolic production of CO2 Correct Answer: E. There are a variety of changes in maternal physiology during pregnancy. In the cardiovascular and respiratory system, pregnancy results in an increased cardiac output, increased oxygen demand, and increased minute ventilation. Increased minute ventilation and respirations are results of progesterone causing increased chemoreceptor CO2 sensitivity; tachypnea is the manifestation of respiratory compensation for the fetal metabolic production of CO2. Carbon dioxide produced from the fetus travels through the umbilical artery and across the placental membrane into the maternal circulation. This increased CO2 in the maternal circulation is expired into the atmosphere from the lungs by increased minute ventilation and relative tachypnea in pregnancy. Incorrect Answers: A, B, C, and D. An incipient respiratory infection (Choice A) would be unlikely in this patient without symptoms of fever or cough. The fetal lung is immature until about 34 weeks of gestation, when the fetus produces pulmonary surfactant. The maternal respiratory drive is not affected by fetal lung immaturity (Choice B). Pregnancy results in a hypercoagulable state and increases the risk for venous thromboembolism including pulmonary emboli (Choice C). Persistently increased and unnoticed tachypnea more likely relates to normal physiologic change in the absence of signs of hypoxia or the presence of thrombosis. Pressure on the diaphragm from the enlarging uterus (Choice D) can cause mechanical compression of the thoracic cavity by displacing the diaphragm cephalad and decreasing compliance of the lung. It leads to decreased functional residual capacity of the lung. However, the enlarging uterus does not affect these changes until later in pregnancy. At 20 weeks of gestation, the fundus of the uterus lies at approximately the umbilicus.

Objective: Carbon dioxide produced from the fetus travels through the umbilical artery and across the placental membrane into the maternal circulation. This increased CO2 in the maternal circulation is expired into the atmosphere from the lungs by increased minute ventilation and relative tachypnea in pregnancy. Previous Next Score Report Lab Values Calculator Help Pause

136 Exam Section 3: Item 37 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 37. A 35-year-old woman with HIV infection has a first generalized tonic-clonic seizure. An MRI shows a small, solitary, enhancing lesion in the right frontal lobe. If stereotactic biopsy of this lesion were performed, it would show intense infiltration of lymphocytes, plasma cells, and macrophages and numerous 3 x 7-mm crescent-shaped organisms with central nuclei. The lesion regresses slowly after treatment with sulfonamide and pyrimethamine. Which of the following is the most likely mode of transmission of this infection? A) Blood transfusion B) Ingestion C) Migration across the cribriform plate D) Sexual intercourse E) Small-droplet inhalation Correct Answer: B. Ingestion is the primary means by which Toxoplasma gondii enters the human body and causes disease. T. gondii is an intracellular protozoan parasite. Transmission can occur through the ingestion of oocysts from cat feces or infected meat, direct transmission between a mother and fetus, or from organ transplantation, which leads to invasion followed by dissemination of the parasite where it typically lies dormant. It is a ubiquitous organism and nearly 10% of individuals will be seropositive for T. gondii, but the parasite does not usually cause severe disease in immunocompetent individuals. In patients with HIV infection and CD4+ cell counts less than 100 cells/mm3, however, reactivation can occur with several distinct clinical syndromes of which cerebral toxoplasmosis results in the most morbidity. Cerebral toxoplasmosis typically presents with multiple ring-enhancing lesions scattered throughout the brain. Seizures, focal neurologic deficits, and systemic symptoms are common. Patients with HIV and positive IgG antibodies are typically treated empirically with sulfonamide and pyrimethamine if cerebral toxoplasmosis is suspected, but confirmatory diagnosis occurs through a brain biopsy, which will demonstrate a dense infiltrate of lymphocytes, macrophages, and plasma cells, along with the presence of the T. gondii organism. Cerebral toxoplasmosis is an AIDS-defining disease. Incorrect Answers: A, C, D, and E. Blood transfusion (Choice A) is the means by which several pathogens are acquired, including cytomegalovirus and Babesia microti. Before the era of blood screening, HIV and hepatitis C were also potentially transmitted via blood transfusions. T. gondii is not associated with blood transfusions but may be acquired during solid organ transplantation. Migration across the cribriform plate (Choice C) occurs with several pathogens. Notable examples include Naegleria fowleri, a freshwater amoeba that causes overwhelming meningoencephalitis with a mortality rate approaching 100%. It occurs in both immunocompetent as well as immunocompromised individuals. Additionally, mucormycosis is a devastating fungal infection most common in patients with diabetes mellitus that can cause severe infiltrative cerebral disease through the direct invasion of the sinuses. Sexual intercourse (Choice D) is the means by which Treponema pallidum, the spirochete that causes syphilis, is acquired. If untreated, syphilis can lie dormant for decades and eventually cause neurosyphilis. Manifestations include meningitis, meningovascular disease, tabes dorsalis, and generalized paresis. It does not typically present with a solitary enhancing lesion on MRI. Small-droplet inhalation (Choice E) describes the mechanism of transmission of Neisseria meningitidis, a primary cause of meningitis. Patients usually present with nuchal rigidity, high fever, a petechial rash, and altered mentation. Mortality is high if antibiotic therapy is delayed, and droplet precautions should be continued for at least twenty-four hours after the start of antibiotic therapy.

Objective: Cerebral toxoplasmosis is an AIDS-defining illness that occurs in patients with CD4 counts less than 100 cells/mm3 and presents with one or more enhancing lesions on MRI. Biopsy, if obtained, will demonstrate the organism amongst a dense inflammatory infiltrate. T. gondii is acquired via the ingestion of soil contaminated with cat feces or directly from infected meat. Previous Next Score Report Lab Values Calculator Help Pause

123 Exam Section 3: Item 24 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 24. A 38-year-old woman comes to the physician because of nasal congestion, cough, and shortness of breath for 5 days. She has asthma and allergic rhinitis currently treated with montelukast, an inhaled Bzadrenergic agonist, and nasal and inhaled corticosteroids. Physical examination shows an erythematous nasal mucosa. Bilateral crackles are heard at the lung bases posteriorly. Laboratory studies show peripheral blood eosinophilia. A chest x-ray shows peripheral infiltrates in the mid lungs bilaterally with a right-sided pleural effusion. A lung biopsy specimen shows necrotizing vasculitis, granulomas, and eosinophilic necrosis. Which of the following is the most likely diagnosis? A) Churg-Strauss syndrome OB) Goodpasture syndrome C) Sarcoidosis D) Takayasu arteritis E) Wegener granulomatosis Correct Answer: A. Churg-Strauss syndrome 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. Pathology shows a granulomatous, necrotizing vasculitis and eosinophilia. Nephritic syndrome is a known complication. In addition to the clinical signs and symptoms, laboratory evaluation, and biopsy, the diagnosis is often associated with the presence of antineutrophilic cytoplasmic antibodies. Treatment involves steroids and immunomodulators (eg, cyclophosphamide). Incorrect Answers: B, C, D, and E. Goodpasture syndrome (Choice B) is a type 2 hypersensitivity reaction against collagen, type IV, in the pulmonary capillaries and glomerular basement membranes, leading to hemoptysis and hematuria. Sarcoidosis (Choice C) is an immune-mediated systemic disease resulting in formation of noncaseating granulomas in a variety of involved organs, often involving the lung. Manifestations of sarcoidosis involving the kidney may include nephrolithiasis, nephrocalcinosis, and acute interstitial nephritis. Takayasu arteritis (Choice D) is a large-vessel vasculitis usually affecting females less than 40 years old 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 be associated with skin nodules and visual disturbances. Wegener granulomatosis (Choice E), also known as granulomatosis with polyangitis, is a necrotizing small-vessel vasculitis that classically presents with sinopulmonary and kidney disease. Patients classically present with epistaxis, hemoptysis, and hematuria. Nephritic syndrome is a known complication.

Objective: Churg-Strauss syndrome or eosinophilic granulomatosis with polyangiitis is a small-vessel vasculitis and may present with asthma, nasal inflammation, and lung, gastrointestinal, cardiac, or kidney involvement. Pathology shows a granulomatous, necrotizing vasculitis and eosinophilia. Previous Next Score Report Lab Values Calculator Help Pause

65 Exam Section 2: Item 16 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 16. A 25-year-old man comes to the physician because of a 6-week history of daytime fatigue, irritability, and emotional lability; he has had a 4.5-kg (10-lb) weight loss during this period. He says that he is sleep-deprived, but he is unable to fall asleep readily. He is a medical student and has recently finished an intensive care unit rotation where he often worked days and nights. He is currently on an outpatient rotation and will be starting vacation next week. Physical examination shows no abnormalities. Which of the following recommendations by the physician is most appropriate to improve this patient's sleep hygiene? A) "Have a glass of wine before bedtime." B) "Make sure you sleep late whenever you can to catch up on your sleep." C) "Stay in bed, even if you can't fall asleep." D) "Take it easy. Exercise at this point will just increase your fatigue." E) "Try getting up the same time every day, even if you didn't get much sleep." Correct Answer: E. This patient likely has a circadian rhythm sleep disorder caused by the long and irregular hours of his intensive care unit rotation and should adopt sleep hygiene habits such as a regular wake- up time. Circadian rhythm sleep disorders feature misalignment between environmental light-dark phases and an individual's internal circadian rhythm. This misalignment may derive from an extrinsic source such as shift work, which is called shift work sleep disorder. Sequelae can include fatigue, irritability, and mood lability. The diagnosis is made by history and sleep diaries. Management relies primarily on sleep hygiene. Sleep hygiene refers to sleep-promoting habits such as a regular bedtime and wake time, avoiding screen time before bedtime, avoiding coffee and other substances that may interfere with sleep, exercising regularly, getting out of bed if unable to sleep (to avoid a mental association between wakefulness and the bed), avoiding watching the clock, and maintaining a quiet and comfortable bedroom environment. By regularly waking at the same time each morning, patients can reset their circadian rhythm and take advantage of accumulated sleep pressure as a result of sleep restriction to fall asleep more quickly. Incorrect Answers: A, B, C, and D. Wine (Choice A) and other alcoholic beverages decrease the duration of rapid-eye movement and delta-wave sleep and would therefore exacerbate this patient's fatigue. Wine is not an effective long-term solution for sleep-onset insomnia. Encouraging a patient to sleep late to catch up on sleep (Choice B) would lead to a pattern of going to sleep and waking up later and may exacerbate circadian rhythm sleep disorder. By instead waking up at the same time each morning, patients can more easily fall asleep at bedtime and reestablish a normal circadian pattern. Staying in bed while awake (Choice C) would promote a mental association between the bed and wakefulness. Patients should instead arise from bed if they cannot sleep for more than 20 minutes and return to bed once they feel tired. Avoiding exercise (Choice D) would not help sleep-onset insomnia. Exercise early in the day promotes nighttime fatigue and the ability to fall asleep more quickly. Exercise should only be avoided in the evening to prevent sympathetic activation around bedtime.

Objective: Circadian rhythm sleep disorders feature misalignment between environmental light-dark phases and an individual's internal circadian rhythm. This misalignment may derive from an extrinsic source such as shift work. Management should include education about sleep hygiene. Previous Next Score Report Lab Values Calculator Help Pause

171 Exam Section 4: Item 22 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 22. A71-year-old man comes to the physician because of a 6-month history of calf pain and discomfort when he walks more than two blocks; he is unable to walk farther than three blocks. The pain quickly subsides with rest. He also has hypertension and type 2 diabetes mellitus. Physical examination shows weak dorsalis pedis pulses bilaterally and an ankle brachial index of 0.67. The physician knows of a drug that would improve this patient's symptoms and walking distance by altering platelet function and providing direct arterial vasodilation. This drug is most likely which of the following? A) Abciximab B) Aspirin C) Cilostazol OD) Clopidogrel E) Nifedipine Correct Answer: C. Leg pain that occurs following exertion and improves with rest is known as claudication and indicates the presence of peripheral arterial disease. Claudication results from increased physiologic demand of the skeletal musculature for oxygen, which cannot be met because of the fixed obstruction to flow of stenotic arteries. As demand falls during rest, the pain typically improves. Peripheral vascular disease is common among older, male smokers, with a history of hypertension, diabetes, or hyperlipidemia. The diagnosis is suggested by absent or diminished peripheral pulses, hairless legs with smooth skin that may ulcerate, and is further supported by a decreased ankle brachial index (ABI). CT, MRI, or conventional angiography can confirm the presence of arterial stenosis and atherosclerosis of the extremities. Active exercise improves collateral vascularization, which promotes improved arterial oxygenated blood flow to the skeletal musculature. Engaging in a walking program is a first-line therapy along with mitigation of risk factors. Additional therapeutic options include cilostazol and surgical revascularization (eg, bypass, stenting). Cilostazol is a phosphodiesterase III (PDE3) inhibitor, which decreases the degradation of CAMP in platelets and vascular smooth muscle cells. Increased intracellular CAMP concentrations in platelets result in increased protein kinase A (PKA) concentrations and inhibition of platelet aggregation. Increased CAMP concentrations and PKA activity in vascular smooth muscle cells lead to smooth muscle relaxation and vasodilation by inhibition of myosin light-chain kinase (MLCK), which normally stimulates contraction by phosphorylating myosin light chains. Incorrect Answers: A, B, D, and E. Abciximab (Choice A) is a monoclonal antibody that inhibits glycoprotein Ilb/Illa on the platelet surface. Glycoprotein Ilb/Illa normally binds with fibrinogen resulting in platelet aggregation and thrombus formation. Use of abciximab can result in acute thrombocytopenia and profound bleeding, and indications for use are restricted to high-risk patients with acute coronary syndrome with planned percutaneous coronary intervention within 24 hours. Abciximab does not act on vascular smooth muscle cells. Antiplatelet agents such as aspirin (Choice B) and clopidogrel (Choice D) may be used in the treatment of claudication; however, they do not directly act on vascular smooth muscle cells and do not promote vasodilation. Their use relates to their mechanism as antiplatelet agents, limiting the risk for acute thrombotic arterial occlusion. Nifedipine (Choice E) is a short-acting dihydropyridine calcium channel blocker (CCB) and a potent vasodilator. It is used in the treatment of high-altitude pulmonary edema and hypertensive emergency. Adverse effects include profound hypotension and reflex tachycardia. Nifedipine does not alter platelet function.

Objective: Claudication is a common disorder that results from stenotic arteries and impaired blood flow to the extremities during exertion. Cilostazol is a first-line therapy, in addition to a walking program and mitigation of risk factors, because of its antiplatelet and vasodilatory properties. Previous Next Score Report Lab Values Calculator Help Pause

155 Exam Section 4: Item 6 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 6. A 26-year-old woman comes to the physician because of difficulty sleeping caused by intense itching of her arms and legs for 6 days. She has a 2-year history of gluten-sensitive enteropathy. Physical examination shows groups of tense blisters over the upper extremities, buttocks, and knees. Histopathologic examination identifies the lesions as subepidermal blisters. Direct immunofluorescence testing of the basement membrane zone within the blisters is done. The presence of which of the following patterns is most likely to confirm the diagnosis? A) Bull's-eye pattern of IgE deposition B) Granular pattern of IgA deposition C) Linear pattern of C4b deposition D) Longitudinal pattern of IgG deposition E) Speckled pattern of IgM deposition Correct Answer: B. Dermatitis herpetiformis is a vesiculobullous disorder characterized by small, tense vesicles located on extensor sites (elbows, knees, and buttocks), which are extremely pruritic. Often, no vesicles will be present as they will have been intensely scratched and replaced by erosions. Dermatitis herpetiformis is caused by the deposition of immunoglobulin A (IgÀ) along the epidermal basement membrane. On histopathologic examination with H&E staining, small collections of neutrophils focally separate the epidermis from the dermis at the dermal papillae, which correspond to a vesicle clinically. Direct immunofluorescence (DIF) demonstrates a granular pattern of IgA deposition along the basement membrane at the tips of the dermal papillae. Dermatitis herpetiformis has a strong association with celiac disease, or gluten-sensitive enteropathy, and the primary treatment is the avoidance of gluten. In severe presentations, dapsone can be used to provide near immediate relief. Incorrect Answers: A, C, D, and E. Bull's eye pattern of IgE deposition (Choice A) refers to a rare DIF finding in bullous pemphigoid. Linear pattern of C4b deposition (Choice C) and longitudinal pattern of IgG deposition (Choice D) along the basement membrane are both more common DIF findings seen in bullous pemphigoid. Bullous pemphigoid most often presents in elderly men with severe pruritus and tense bullae. Compared with dermatitis herpetiformis, the blisters tend to be larger as the disease is not limited to the tips of individual dermal papillae. Bullous pemphigoid is not associated with celiac disease. Speckled pattern of IgM deposition (Choice E) is a nonspecific DIF finding that can be present in systemic lupus erythematosus, discoid lupus erythematosus, subacute cutaneous lupus erythematosus, dermatomyositis, systemic sclerosis, and epidermolysis bullosa acquisita. However, it is not a finding seen in dermatitis herpetiformis.

Objective: Clinically, dermatitis herpetiformis is characterized by severely pruritic vesicles or erosions on the extensor surfaces of extremities. DIF demonstrates a coarse, granular deposition of IgA at the dermal papillae, which confirms the diagnosis. Previous Next Score Report Lab Values Calculator Help Pause

198 Exam Section 4: Item 49 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 49. A 55-year-old man comes to the physician because of a 3-month history of intermittent pain in his right hip. He states that the pain is exacerbated by walking and resolves with rest. Examination of the right hip shows increased pain on abduction. Range of motion is limited by pain. An x-ray of the hip is shown. Which of the following types of collagen is most likely affected in this patient? A) Type I B) Type II C) Type II D) Type IV E) Type V OF) Type VI Correct Answer: B. There are a number of different types of cartilage in the human body, each with specialized function. Hyaline cartilage has a high content of collagen, type II, providing a low-friction surface for the gliding of diarthrodial joints. Loss of collagen, type II, in the hyaline cartilage of joints is a hallmark of osteoarthritis. Genetic predisposition, chronic mechanical overload (eg, obesity), and prior injury to joints all contribute to the progression of osteoarthritis. Increased IL-1 signaling and release of matrix metalloproteases in the synovial fluid contribute to the breakdown of the complex structure of articular cartilage. As breakdown progresses, cartilage becomes fissured and fibrillary, and it loses its lubrication and load dispersion properties. Subsequently, increased compressive load is borne by the subchondral bone. In turn, sclerosis, as seen on the x-ray, occurs. Synovial fluid infiltration into the subchondral bone leads to cyst formation. Correct Answers: A, C, D, E, and F. Collagen, type I (Choice A), has high tensile strength and is a main constituent of tendons and bones. Collagen, type I, mutations are implicated in osteogenesis imperfecta. Collagen, type III (Choice C), is found in high concentration in blood vessels and granulation tissue. Mutations in the gene for collagen, type III, are implicated in vascular type Ehlers-Danlos syndrome. Collagen, type IV (Choice D), is found in basement membranes and the lens of the eye. In Goodpasture syndrome, autoantibodies are formed against collagen, type IV. Mutations in the collagen, type IV, gene are associated with Alport syndrome. Collagen, type V (Choice E), is found in multiple tissues including bone and skin. Mutations in collagen, type V, are associated with the classic type of Ehlers-Danlos syndrome. Collagen, type VI (Choice F), is an important component of the extracellular matrix in skeletal muscle. It forms a microfibrillar network connecting the basement membrane and the interstitial matrix.

Objective: Collagen, type II, is an important component of articular cartilage, functioning in load dispersion and lubrication. Degradation of collagen, type II, by matrix metalloproteases is a key step in the pathophysiology of osteoarthritis. Previous Next Score Report Lab Values Calculator Help Pause

68 Exam Section 2: Item 19 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 19. An 82-year-old woman with type 2 diabetes mellitus and painful neuropathy comes to the physician for a follow-up examination. At her last visit 3 months ago, the physician prescribed a topical cream and recommended use of acetaminophen as needed for pain. The patient reports that the topical therapy helped more than she expected and did decrease her need for acetaminophen. Physical examination shows no change in her condition. The prescribed cream most likely depletes which of the following substances from sensory nerve endings? A) Arachidonic acid B) Bradykinin C) Histamine D) Serotonin E) Substance P Correct Answer: E. Substance P has likely been depleted by the use of capsaicin cream, an agonist of transient receptor potential vanilloid 1 receptor (TRPV1) on nociceptive nerves. This initially results in a painful stimulus; however, the continued use of capsaicin completely depletes these nerve fibers of substance P. As substance P release is implicated in nociception, depletion of this substance results in gradual pain relief. It has applications in the treatment of diabetic neuropathy and neuropathic pain and has been used off-label to treat cannabis hyperemesis syndrome. While this patient's neuropathy is irreversible as a result of damage to nerve fibers over many years of hyperglycemia, her pain is effectively controlled by depleting the nociceptive neurotransmitter substance P. Incorrect Answers: A, B, C, and D. Arachidonic acid (Choice A) is produced using phosphatidylinositol and this production is inhibited by exogenous corticosteroids. Additionally, arachidonic acid is required for the production of prostanoids via cyclooxygenase. Prostaglandins and leukotrienes play a role in inflammation, and this pathway is effectively inhibited by nonsteroidal anti-inflammatory drugs, which are cyclooxygenase inhibitors. Bradykinin (Choice B) is made from kininogen via the action of kallikrein, which can be blocked by kallikrein inhibitors such as ecallantide. Bradykinin is a vasodilator that is implicated in hereditary angioedema. It does not modulate pain. Histamine (Choice C) is blocked by antihistamine medications. Histamine release causes vasodilation and increased vascular permeability through its action on four separate histamine receptors; it has been implicated as a contributor to conditions such as environmental allergies. Antihistamine medications are broken into first and second generations, with first-generation medications typically resulting in off-target effects such as drowsiness. Common examples of antihistamine medications include diphenhydramine and cetirizine. Serotonin (Choice D) depletion is a commonly used strategy to treat nausea, especially when associated with chemotherapy. Serotonin signaling in the gastrointestinal tract increases gastric motility and gastric emptying. There are a variety of serotonin receptors, but medications commonly used to treat nausea mostly block the 5-HT3 receptors. Examples include ondansetron and palonosetron.

Objective: Depletion of substance P with the use of capsaicin can treat the pain often associated with peripheral diabetic neuropathy. Capsaicin acts by stimulating substance P release until it is completely depleted, which limits nociception. Previous Next Score Report Lab Values Calculator Help Pause

135 Exam Section 3: Item 36 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 36. A 20-year-old man comes to the emergency department because of a 2-day history of profuse watery diarrhea that began shortly after he returned from a trip to Central America. He appears alert. His temperature is 38.1°C (100.6°F), pulse is 120/min, respirations are 24/min, and blood pressure is 90/60 mm Hg. Physical examination shows dry mucous membranes and poor peripheral perfusion. Compared with a healthy individual, laboratory studies are most likely to show which of the following sets of findings in this patient? Serum K+ HCO, pH Pco2 A) Increased increased increased increased B) Increased increased decreased increased C) Increased decreased decreased decreased D) Decreased increased decreased increased E) Decreased decreased increased decreased F) Decreased decreased decreased decreased Correct Answer: F. Non-anion gap metabolic acidosis is identified by a pH less than 7.35 (acidemia), with decreased PC0, and HCO,. The respiratory system compensates for the acidosis by increasing minute ventilation, which eliminates carbon dioxide, an acid. This results in decreased PCO, Anion gap ([Na+]- ([CI-] + [HCO,-])) calculates unmeasured anions in serum which are contributing to acidity, whereas non-anion gap acidosis generally results from excess supply of protons or chloride, retention of protons or chloride by the kidney, or failure to secrete bicarbonate buffers. Causes of non-anion gap acidosis include infusions of normal saline, Addison disease, renal tubular acidosis, gastrointestinal losses of bicarbonate (eg, diarrhea as in this patient), spironolactone, and acetazolamide use. In the case of diarrhea or gastrointestinal disturbance, bicarbonate secreted in the small intestine is normally reclaimed in the large intestine in exchange for chloride ions. If the colon cannot reclaim bicarbonate, such as following colectomy or during diarrheal illness, the concentration of serum chloride will remain increased and serum bicarbonate decreased. In diarrheal illness, serum potassium will also generally be decreased as a result of gastrointestinal losses, and renal losses because of the effect of aldosterone and volume contraction. 1+1 Incorrect Answers: A, B, C, D, and E. The respiratory system compensates for metabolic acidosis by increasing minute ventilation, which eliminates carbon dioxide, an acid, resulting in decreased, not increased, PCO, (Choices A, B, and D). Diarrhea leads to potassium loss from the digestive tract and aldosterone and volume contraction promote renal potassium excretion resulting in hypokalemia (Choices A, B, and C). In diarrhea, the colon is unable to normally reabsorb bicarbonate, leading to decreased serum bicarbonate concentrations (Choices A, B, and D). Choice E reflects increased pH (alkalosis) and low concentrations of blood carbon dioxide and bicarbonate, consistent with primary respiratory alkalosis, as opposed to compensatory respiratory alkalosis in the setting of a primary metabolic acidosis.

Objective: Diarrhea leads to the loss of potassium and bicarbonate from the digestive tract resulting in a non-anion-gap metabolic acidosis and compensatory respiratory alkalosis. II Previous Next Score Report Lab Values Calculator Help Pause

105 Exam Section 3: Item 6 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 6. In some inherited disorders of collagen, lysyl oxidase activity is deficient. This enzyme deaminates the e-amino group of a lysyl residue in collagen and forms a reactive aldehyde. In turn, this aldehyde interacts with which of the following amino acid residues of another collagen molecule to cause cross-linking? A) Arginine B) Aspartic acid C) Cysteine D) Histidine E) Lysine Correct Answer: E. Collagen molecules are glycoproteins and are among the most ubiquitous proteins in the body. Collagen has a high tensile strength and provides much of the support of the human musculoskeletal system. During the production of collagen, tropocoilagen molecules are cross-linked at hydroxylysine residues. Collagen synthesis begins with translation of a polypeptide chain with a characteristic glycine-X-Ý repeating sequence where X and Y are often proline or lysine residues. These peptide chains undergo vitamin-C-dependent hydroxylation, which facilitates triple helix formation. This step can be disrupted in vitamin C deficiency leading to scurvy. Disulfide bonds form across the alpha helices, leading to a fully formed procollagen molecule. These procollagen molecules are then excreted by cells (fibroblasts, osteoblasts, tenocytes). Once in the extracellular space, the C-terminus and N-terminus are cleaved, forming tropocollagen and allowing molecular aggregation into fibrils. Once aggregated, cross-linking of the hydroxylysine residues occurs, strengthening the superstructure of collagen. This step is a copper-dependent process and, when pathologic, results in the impaired wound healing seen in copper deficiency. Incorrect Answers: A, B, C, and D. Arginine (Choice A) is a precursor in the synthesis of nitric oxide, which is an important second messenger particularly in the smooth muscle of blood vessels. Aspartic acid (Choice B) plays a key role in the urea cycle as a precursor for the formation of argininosuccinate. It also participates in gluconeogenesis. Cysteine (Choice C) is an amino acid that contains a sulfur atom, which makes it useful in chelating metals such as in zinc finger and cytochrome molecules. N-acetylcysteine is a derivative of cysteine, which is used as an antioxidant to treat acetaminophen overdose and cystic fibrosis. Histidine (Choice D) is an essential amino acid, meaning that it cannot be synthesized by the human body and must be taken in from an exogenous source. It is a precursor to histamine, which has functions in allergic reactions, vasodilation, and gastric acid secretion.

Objective: During collagen synthesis, tropocollagen molecules are cross-linked at the hydroxylysine residues to increase the strength of collagen fibrils. Collagen production is a multistep process that includes both vitamin-C- and copper-dependent steps. Previous Next Score Report Lab Values Calculator Help Pause

140 Exam Section 3: Item 41 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment ECG 200- 100 - 0- Left ventricle Aorta 41. A 52-year-old man comes to the emergency department 2 hours after developing shortness of breath and nearly fainting while exercising. He is currently asymptomatic. He has a 6-month history of similar episodes that are becoming increasingly severe. He also has a 2-month history of intermittent chest pain with exertion. He last experienced this pain 3 days ago. He has no history of major medical illness and takes no medications. His father has hypertension. His temperature is 36.9°C (98.4°F), pulse is 76/min, respirations are 12/min, and blood pressure is 108/88 mm Hg. Pulse oximetry on room air shows an oxygen saturation of 97%. Cardiac examination shows a systolic murmur that is heard best at the upper right sternal border. An ECG and intracardiac pressure tracings from this patient are shown. Which of the following is the most likely underlying cause of this patient's condition? A) Congenital bicuspid valve B) Degenerative sclerosis of valve leaflets C) Inadequate preload D) Increased resistance in pulmonary artery E) Stenosis of the left main coronary artery Correct Answer: A. The pressure tracing is characterized by a large peak 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 as a result of chronic inflammation with resultant calcification and fibrosis over time, with the aortic valve most commonly affected. Early-onset aortic stenosis (ages 40 to 60 years), as in this patient, can occur in the setting of a congenital bicuspid valve or chronic rheumatic heart disease. Patients may complain of fatigue, shortness of breath, cough, diminished exercise tolerance, chest pain, or syncope with exertion. Exam findings include a crescendo-decrescendo systolic murmur best heard at the upper right sternal border, and pulsus parvus et tardus (weak and delayed pulse) 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: B, C, D, and E. Degenerative sclerosis of valve leaflets (Choice B) is the most common cause of aortic stenosis. Patients typically present later in life, over the age of 65 years. Bicuspid aortic valve or chronic rheumatic heart disease are more common causes in younger, otherwise healthy adults. Inadequate preload (Choice C) results in decreased cardiac output as a result of decreased end-diastolic cardiomyocyte fiber stretch per the Frank-Starling relationship. The pressure tracing would be expected to show a decreased peak pressure during systole with no difference between the peak left ventricular and aortic pressures. Increased resistance in pulmonary artery (Choice D) may be seen in pulmonary arterial hypertension or in pulmonary embolism. Stenosis of the left main coronary artery (Choice E) may cause symptomatic coronary artery disease with progressive dyspnea, light-headedness, and exertional angina as the result of an impaired ability to match myocardial perfusion with demand. These conditions would not be expected to cause a pressure gradient between the left ventricle and aorta during systole.

Objective: Early-onset aortic stenosis is associated with a congenital bicuspid valve. Aortic stenosis can be identified on a cardiac pressure tracing by the presence of a large peak pressure difference between the left ventricle and the aorta. Previous Next Score Report Lab Values Calculator Help Pause Pressure (mm Hg)

152 Exam Section 4: Item 3 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 3. An 84-year-old woman is brought to the physician because of a 1-day history of confusion, severe fatigue, and constipation. Three days ago, she was discharged from the hospital after undergoing a total hip replacement. During the 5-day hospital stay, she received intravenous morphine therapy every 6 hours, resulting in pain relief. Her vital signs were within normal limits, and she was oriented to person, place, and time. On discharge, a morphine regimen of an oral-equivalent dose calculated using a standard intravenous-to-oral dose conversion table was prescribed. Today, her temperature is 36.1°C (97°F), pulse is 50/min, respirations are 10/min, and blood pressure is 104/64 mm Hg. Physical examination shows mildly constricted pupils that are poorly responsive to light. Which of the following age-related pharmacokinetic changes best explains this patient's response to her current dose of oral morphine? A) Decreased first-pass effect B) Decreased passive absorption C) Increased first-pass effect D) Increased passive absorption E) Increased renal elimination Correct Answer: A. A decreased first-pass effect would increase serum concentrations of morphine, leading to the signs of opioid toxicity seen in this patient. First-pass effect refers to the initial metabolism of a medication by an organ (generally the liver) such that a decreased concentration of the medication reaches the systemic circulation. Oral medications are absorbed by the gastrointestinal tract and then undergo first-pass metabolism in the liver before reaching the systemic circulation, whereas intravenous (IV) medications directly enter the circulation. Hepatic blood flow and liver size both decrease with age, signifying that medications that are heavily metabolized by the liver-such as morphine-may increase in concentration caused by a decreased first-pass effect. Oral morphine undergoes extensive first-pass metabolism in the liver, during which it is glucuronidated to form morphine-3-glucuronide and morphine-6-glucuronide (in addition to the inactive metabolite normorphine and negligible amounts of hydromorphone). Morphine has a higher affinity for the µ2 opioid receptor, which produces adverse effects, than morphine-3-glucuronide or morphine-6-glucuronide. A decreased first-pass effect would therefore lead to an increased concentration of morphine and cause miosis, sedation, respiratory depression, bradycardia, hypotension, and constipation. The treatment of morphine toxicity includes supportive care (eg, respiratory support) and naloxone, a short-acting opioid receptor antagonist. Incorrect Answers: B, C, D, and E. Decreased passive absorption (Choice B) does not typically occur with age and would result in a decreased morphine concentration, making signs of opioid toxicity unlikely. Passive absorption occurs when a medication diffuses across a cell membrane (eg, gastrointestinal epithelial cells) to reach systemic circulation and is dictated by intra- and extracellular concentrations of the medication as well as the medication's lipid solubility. Increased first-pass effect (Choice C) would lead to an increased conversion of morphine to its metabolites and therefore a decreased morphine concentration. Increased first-pass effect occurs in states of high hepatic blood flow. Increased passive absorption (Choice D) would cause an increased morphine concentration but does not typically occur with age. Passive absorption relies on the lipophilicity and concentration of the medication rather than age-related metabolic changes. Increased renal elimination (Choice E) would result in a decreased morphine concentration and would not cause signs of opioid toxicity. The glomerular filtration rate decreases with age because of renal senescence and could lead to decreased renal elimination of morphine and, consequently, an increased morphine concentration. However, a decreased glomerular filtration rate would increase both oral and IV morphine such that transitioning from IV to oral morphine would be unlikely to result in opioid toxicity.

Objective: First-pass effect refers to the initial metabolism of a medication such that a decreased concentration of the medication reaches the systemic circulation. Oral medications typically undergo a first-pass effect in the liver, whereas IV medications directly enter the systemic circulation. Hepatic blood flow and liver size both decrease in older age, signifying that medications that are heavily metabolized by the liver-such as oral morphine-may increase in concentration because of a decreased first-pass effect. %3D Previous Next Score Report Lab Values Calculator Help Pause

36 Exam Section 1: Item 37 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 37. A 51-year-old woman comes to the physician because of a 6-month history of burning abdominal pain that occurs 1 to 2 hours after eating. She sweats profusely and has light-headedness when she stands. Her blood pressure is 105/70 mm Hg while standing. Physical examination shows epigastric tenderness. A CT scan of the abdomen shows a 2-cm mass on the proximal duodenum. Gastrin released by the tumor cells stimulates which of the following labeled cells in the photomicrograph shown to release hydrogen ions? A B C E A) B) C) D) E) Correct Answer: B. This patient's findings of postprandial abdominal pain, diaphoresis, epigastric tenderness, and orthostatic hypotension in the setting of a duodenal or pancreatic lesion are suggestive of a peptic ulcer secondary to Žollinger-Ellison syndrome (ZÉS). ZES stems from a pancreatic or duodenal gastrin-secreting tumor. Gastrin is typically produced by gastric G cells and stimulates parietal cells to release protons in order to form hydrochloric acid. Parietal cells can be identified by their large size, central, round nucleus, and intensely acidophilic cytoplasm. They are found in the middle region of the gastric gland. Excessive production of acid in the setting of ZES leads to recurrent, chronic duodenal or jejunal ulcers, which can present with abdominal pain, diarrhea secondary to malabsorption, and possible hematemesis, melena, or hematochezia. Pancreatic gastrinoma and ZES may each present as a component of multiple endocrine neoplasia type 1 (MEN1). Incorrect Answers: A, C, D, and E. Choice A indicates a gastric mucous neck cell, which produces gastric mucus and is typically found near the upper region of the gastric gland. They can be identified by a pale, foamy cytoplasm and a basolateral nucleus. Choice C indicates a chief cell. Chief cells can be identified by their basophilic cytoplasm and the presence of numerous cytoplasmic vesicles, which lend the cytoplasm a granular appearance. They produce pepsin. Choice D indicates a vascular endothelial cell, which can be identified by its position lining a capillary. Choice E indicates a perivascular fibroblast, marked by its presence within stromal connective tissue and dark, oblong nucleus.

Objective: Gastrinomas are characterized by the excessive production of gastrin, which stimulates parietal cells to produce and release hydrochloric acid. Parietal cells can be identified by their large size, central, round nucleus, and intensely acidophilic cytoplasm. They are typically found in the middle region of the gastric gland. Previous Next Score Report Lab Values Calculator Help Pause

63 Exam Section 2: Item 14 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 14. A 69-year-old woman is brought to the emergency department because of confusion and light-headedness for 45 minutes. She has a 5-year history of type 2 diabetes mellitus treated with glipizide. Recently, the dose of glipizide was increased. Her vital signs are normal. Physical examination shows no abnormalities. Her blood glucose concentration is 43 mg/dL. Intravenous infusion of glucose is begun. Which of the following is the most likely explanation for hypoglycemia in this patient? A) Decreased a-glucosidase activity in the gastrointestinal tract B) Decreased hepatic production of glucose C) Increased agonist activity with insulin receptor sites D) Increased conversion of glucose to glycogen E) Increased release of insulin from the pancreas Correct Answer: E. Glipizide is a second-generation sulfonylurea. The mechanism of action of sulfonylureas involves closing a potassium channel in a pancreatic islet B-cell membrane, resulting in depolarization of the cell. Depolarization leads to opening of voltage-gated calcium channels, calcium influx, and endogenous insulin release from the pancreas. Sulfonylureas are used as treatment in type 2 diabetes mellitus, in which the pancreatic islet cells retain function and are able to release insulin, in contrast to type 1 diabetes mellitus where the islet cells are destroyed and no functional insulin is produced. Insulin induces glucose uptake by peripheral tissues, such as skeletal muscle and adipose tissue, and promotes glycogen, triglyceride, and protein synthesis. Through these mechanisms, insulin decreases the concentration of serum glucose. The patient's recently increased dose of glipizide increases the risk for inadvertent iatrogenic hypoglycemia. Hypoglycemia that results from sulfonylurea use often persists for 12-24 hours, as sulfonylureas are long-acting agents. Treatment in the short term includes administration of dextrose. The dose of sulfonylurea should also be decreased, or an alternative agent should be selected for glycemic control. Incorrect Answers: A, B, C, and D. Decreased a-glucosidase activity in the gastrointestinal tract (Choice A) is the result of administration of acarbose. Acarbose inhibits brush border enzymes in the small intestine, preventing digestion of carbohydrates and therefore limiting the total amount of dietary carbohydrate absorbed. Side effects include gastrointestinal upset (eg, diarrhea, flatulence). Decreased hepatic production of glucose (Choice B) is incorrect, as gluconeogenesis is not directly affected by sulfonylureas. Metformin acts in this manner. Increased agonist activity with insulin receptor sites (Choice C) is the mechanism of action of the exogenous administration of insulin. Glipizide and sulfonylureas stimulate the release of endogenous insulin from the pancreas. Increased conversion of glucose to glycogen (Choice D) results from insulin agonism; however, the primary mechanism for hypoglycemia caused by sulfonylurea-induced insulin release is glucose uptake by peripheral tissues.

Objective: Glipizide and sulfonylureas stimulate endogenous insulin release from the pancreas. Insulin quickly induces glucose uptake by peripheral tissues such as skeletal muscle and adipose tissue, which can increase the risk for hypoglycemia. Previous Next Score Report Lab Values Calculator Help Pause

159 Exam Section 4: Item 10 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 10. A 48-year-old man is brought to the emergency department because of a 2-day history of shortness of breath, fatigue, decreased appetite, and cough productive of blood-tinged sputum. He has a 2-month history of hypertension that is currently untreated. His temperature is 37.1°C (98.8°F), pulse is 88/min, respirations are 18/min, and blood pressure is 150/85 mm Hg. Physical examination shows no rash. A few scattered, coarse crackles are heard on auscultation. Laboratory studies show: Serum creatinine 5.1 mg/dL (1.2 mg/dL 1 month ago) Urine Blood Protein RBC RBC casts 3+ 3+ 50/hpf numerous Which of the following is the most likely diagnosis? A) Acute tubular necrosis B) Chronic renal failure C) Goodpasture syndrome D) Prerenal azotemia E) Urinary tract obstruction Correct Answer: C. The clinical presentation of hemoptysis, hematuria, hypertension, and proteinuria is consistent with a diagnosis of Goodpasture syndrome. Goodpasture syndrome is a type 2 hypersensitivity reaction that results in autoimmune antibody targeting of collagen, type IV, in the pulmonary capillaries and glomerular basement membranes, leading to hemoptysis and hematuria. Antiglomerular basement membrane antibodies in the kidney manifests as rapidly progressive, crescentic and necrotizing glomerulonephritis. This nephritic syndrome is characterized by a crescent of eosinophilic fibrin and plasma proteins on biopsy of the glomerulus adjacent to glomerular parietal cells, monocytes, and macrophages. Inflammation results in hematuria and RBC casts in urine, oliguria, and hypertension. Proteinuria is also present but to a lesser extent than in nephrotic syndrome. Antibodies targeting the alveolar basement membrane in the lungs manifest clinically as pulmonary hemorrhage with hemoptysis and shortness of breath. Incorrect Answers: A, B, D, and E. Acute tubular necrosis (Choice A) typically occurs following an ischemic or nephrotoxic insult to the kidneys, which results in necrosis of the tubular epithelium. Granular, muddy brown casts are common on urinalysis. Chronic renal failure (Choice B) can have variable causes, such as with chronic prerenal disease in the setting of heart failure or cirrhosis with decreased renal perfusion, intrinsic renal disease such as nephrosclerosis or atherosclerotic renal artery disease, chronic nephritic or nephrotic glomerular disease, or chronic postrenal obstructive disease. Chronic kidney disease is more common in older patients after years of underlying kidney injury. Prerenal azotemia (Choice D) may occur in the setting of heart failure or cirrhosis with decreased renal perfusion, or acute kidney injury secondary to insufficient renal blood flow in states of hypoperfusion. A decreased glomerular filtration rate (GFR) results. Urinary tract obstruction (Choice E) can occur at a variety of anatomic sites including the ureter (eg, nephrolithiasis, retroperitoneal fibrosis), bladder (eg, bladder malignancy), prostate (eg, benign prostatic hyperplasia, malignancy), or urethra (eg, stricture). Postrenal azotemia from impaired urinary drainage can result.

Objective: Goodpasture syndrome is a type 2 hypersensitivity reaction that results from autoimmune antibody targeting of collagen, type IV, in the pulmonary capillaries and glomerular basement membranes, leading to hemoptysis, hematuria, and hypertension. %3D Previous Next Score Report Lab Values Calculator Help Pause

160 Exam Section 4: Item 11 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 11. A 10-year-old boy is evaluated because of malaise and dark brown urine for the past 2 days. Laboratory studies show: 7.2 g/dL 21.6% 8400/mm3 6% Hemoglobin Hematocrit Leukocyte count Reticulocyte count Platelet count 90,000/mm3 12 sec (INR=1.0) Prothrombin time Partial thromboplastin time (activated) 28 sec Serum Creatinine Fibrinogen Fibrin split products Urea nitrogen (BUN) 7.4 mg/dL 380 mg/dL (N=200-400 mg/dL) <10 (N=<10) 82 mg/dL A peripheral blood smear is shown. Which of the following is the most likely diagnosis? A) Acute lymphoblastic leukemia B) Acute poststreptococcal glomerulonephropathy C) Chronic myelogenous leukemia D) Hemolytic uremic syndrome E) Minimal change disease F) Sickle cell disease Correct Answer: D. Hemolytic uremic syndrome (HUS) most likely explains this patient's acute renal failure, thrombocytopenia, and microangiopathic hemolytic anemia (MAHA). HUS is typically seen in children and is often caused by infection with Shiga toxin-producing E. coli (0157:H7), which is usually accompanied by bloody diarrhea. There are atypical variants of HUS, which are not preceded by a diarrheal illness and are the result of mutations in the complement system (eg, mutations of C3 and CD46). Widespread microvascular thrombosis consumes platelets and predisposes circulating erythrocytes to shearing forces as a result of a disrupted vascular lumen. When erythrocytes come into contact with microvascular clots, they are damaged, resulting in schistocytes, which are apparent on the peripheral blood smear as fragmented cells. Thrombosis also occurs in the renal vasculature because of underlying endothelial damage, resulting in renal injury. Differentiating HUS from disseminated intravascular coagulation (DIC) is important and can be done by examining the prothrombin and partial thromboplastin time; these are normal in HUS and increased in DIC as a result of clotting factor consumption. Treatment of HUS is supportive, although eculizumab, a human monoclonal IgG antibody against complement protein C5 that prevents formation of the membrane attack complex, has been used in refractory cases. Antibiotics are avoided as they can result in an increased release of Shiga toxin and worsening clinical deterioration. Incorrect Answers: A, B, C, E, and F. Acute lymphoblastic leukemia (Choice A) is primarily a malignancy of childhood and is characterized by a diverse set of chromosomal mutations. Laboratory findings usually show leukocytosis with many circulating lymphoblast cells that are apparent on peripheral smear. DIC may occur during the disease course, and anemia and thrombocytopenia are also common, but MAHA is not characteristic. Acute poststreptococcal glomerulonephropathy (Choice B) is a nephritic syndrome characterized by dark urine, hypertension, and proteinuria following an infection with group A streptococcus. It does not cause MAHA. Chronic myelogenous leukemia (Choice C) is defined by the presence of the Philadelphia (Ph) chromosome, which is created by a translocation between chromosomes 9 and 22, resulting in constitutive activation of the ABL1 tyrosine kinase. Hyperleukocytosis is common but MAHA is not. Peripheral smear typically demonstrates a diverse set of immature or partially mature cells of the granulocytic cell lineage with basophilia. Minimal change disease (Choice E) is the most common cause of nephrotic syndrome in children and can be idiopathic or triggered by a recent infection or immune stimulus. It presents with significant proteinuria and peripheral edema. Renal biopsy exhibits podocyte effacement on electron microscopy. Anemia and thrombocytopenia are not typically present. Sickle cell disease (Choice F) is caused by mutations in both beta-globin alleles that result in an abnormal hemoglobin S (HbS), which polymerizes abnormally in the setting of hypoxia and causes sickling of erythrocytes. Renal findings such as renal papillary necrosis are sometimes seen in sickle cell disease presenting with gross hematuria and proteinuria. Hemolytic anemia is a component of this disease, but schistocytes are not characteristic.

Objective: HUS is most often triggered by infection with Shiga toxin-producing E. coli (0157:H7), although it may also occur as a result of genetic mutations in the complement system. It typically presents with the triad of MAHA, thrombocytopenia, and acute renal failure. Previous Next Score Report Lab Values Calculator Help Pause

134 Exam Section 3: Item 35 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 35. An 18-year-old woman comes to the physician because of frequent, loose stools since the age of 3 years. Her 42-year-old mother and her 67-year-old maternal grandfather have similar symptoms. She is 163 cm (5 ft 4 in) tall and weighs 53 kg (116 Ib); BMI is 20 kg/m2. Physical examination shows no other abnormalities. Serum studies show: Total cholesterol HDL-cholesterol LDL-cholesterol VLDL-cholesterol decreased normal decreased decreased decreased Chylomicrons Stool analysis shows an increased fat concentration. Her affected family members have had similar laboratory findings. This patient most likely has a mutation in which of the following genes? A) APOA В) AРОВ C) АРОС D) APOD E) APРОЕ Correct Answer: B. Hypobetalipoproteinemia results from defects in the function of apolipoprotein B (APOB), which is encoded by the gene APOB. Deficiency of APOB causes malabsorption of dietary fats with subsequent steatorrhea, malabsorption (including malabsorption of fat-soluble vitamins A, D, E, and K), and failure to thrive, as well as hepatic steatosis and cirrhosis. APOB plays an important role in the function of chylomicrons, very-low-density lipoprotein (VLDL), and LDL-cholesterols, serum concentrations of which are characteristically decreased. HDL alone is not dependent on APOB and has normal serum concentrations in these patients. Treatment includes high-dose supplementation with vitamin E. Incorrect Answers: A, C, D, and E. APOA (Choice A) encodes apolipoprotein A (APOA), which forms an important component of HDL-cholesterol. Defects of APOA result in hypoalphalipoproteinemia, also known as Tangier disease. Patients exhibit characteristically decreased concentrations of serum HDL. APOC (Choice C) encodes apolipoprotein C (APOC), which forms a component of chylomicrons and VLDL-cholesterol. Patients with defects in this gene present with decreased concentrations of chylomicrons and VLDL, and hepatosplenomegaly. APOD (Choice D) encodes apolipoprotein D (APOD). APOD forms a minor component of HDL-cholesterol. APOE (Choice E) encodes apolipoprotein E (APOE). APOE is expressed in all major cholesterol and lipoproteins other than LDL-cholesterol. It plays an important role in the uptake of chylomicron remnants.

Objective: Hypobetalipoproteinemia results from defects in apolipoprotein B (APOB). APOB plays an important role in the function of chylomicrons, VLDL, and LDL-cholesterols. Patients with hypobetalipoproteinemia display decreased serum concentrations of these lipoproteins and present with steatorrhea, malabsorption, fat-soluble vitamin deficiency, and failure to thrive, as well as hepatic steatosis and cirrhosis. Previous Next Score Report Lab Values Calculator Help Pause

182 Exam Section 4: Item 33 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 33. A screening program is instituted for detection of vaginal Chlamydia trachomatis infection among first-year women college students. At the initial screening, evidence of C. trachomatis infection is found in 500 of 2500 students. One year later, screening shows vaginal C. trachomatis infection in an additional 200 students. Which of the following is the annual incidence of C. trachomatis infection in this population of women students? A) 8% B) 10% C) 16% D) 20% E) 28% Correct Answer: B. Incidence is an important epidemiological measure that assesses the rate of occurrence of new disease in a population at risk. Incidence is the number of new cases expressed as a percentage of the total population at risk over a specified period of time. In this study, the total population is 2500 students. At the beginning of the specified time period, there are 500 students who screen positive for chlamydia. The 500 active infections must be subtracted from the total population to calculate the population at risk; these 500 students already have the disease and therefore are excluded from the calculation. Thus, the incidence of new chlamydia infection would be calculated as 200 new cases divided by 2000 persons at risk over the one-year period of time, or 10%. Incorrect Answers: A, C, D, and E. 8% (Choice A) can be calculated as 200 cases divided by 2500 students. This is incorrect as it uses the total population rather than the population at risk. 16% (Choice C) does not reflect the incidence or prevalence of disease. 20% (Choice D) can be calculated as the 500 initial cases at the beginning of the study period divided by 2500 students in the total population. This represents the point prevalence of chlamydia infection at the beginning of the study, not annual incidence. 28% (Choice E) can be calculated by summing the 500 initial cases with the 200 new cases and dividing this by the 2500 students in the total population. The number of cases at the beginning of a study period can be summed with the new cases to calculate total cases during a study. This number can, in theory, be used to calculate the prevalence of disease at the end of a study; however, this is not applicable to chlamydia infection, as such infection is temporary and will resolve with antibiotic treatment. Therefore, the 500 initial cases will have resolved by the end of study. Such an approach may be applicable to chronic diseases that do not resolve (eg, malignancy).

Objective: Incidence is defined as the number of new cases of a disease divided by the population at risk over a defined period of time. %3D Previous Next Score Report Lab Values Calculator Help Pause

176 Exam Section 4: Item 27 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 27. A 3-year-old girl is brought to the physician because of intermittent abdominal pain, constipation, and irritability for 1 month. She lives with her family in a 50-year-old house that is currently being renovated. Physical examination shows pallor. Laboratory studies show a normocytic, hypochromic anemia and an increased plasma concentration of 5-aminolevulinic acid (õ-ALA). Her blood lead concentration is 50 µg/dL (N<10). A decreased activity of which of the following enzymes is most likely in this patient? A) Ferrochelatase B) Porphobilinogen deaminase C) Protoporphyrinogen IX oxidase D) Uroporphyrinogen II E) Uroporphyrinogen decarboxylase Correct Answer: A. Although lead is generally no longer used in the manufacture of paint and gasoline, children can continue to be exposed to lead through ingestion of paint chips and dust in older homes or from contaminated drinking water. Children are susceptible to the toxic effects of lead because of an immature central nervous system and lead passage across the blood brain barrier. Lead inhibits the enzymes ferrochelatase and O-ALA dehydratase, resulting in decreased heme synthesis, increased serum õ-ALA, urinary coproporphyrin, and erythrocyte zinc protoporphyrin. Clinical manifestations of lead poisoning include encephalopathy, cognitive dysfunction, peripheral nephropathy, hearing loss, renal tubular dysfunction, interstitial nephritis, vomiting, abdominal pain, constipation, and anemia. Diagnosis is suspected on exposure and symptoms and confirmed with increased serum lead concentration. Management of lead poisoning depends on the blood lead concentration and presence of symptoms, and includes chelation with dimercaprol, ethylene diamine tetraacetate, or succimer. Sources of lead should be investigated and eliminated. Incorrect Answers: B, C, D, and E. Decreased activity of porphobilinogen deaminase (Choice B) leading to accumulation of porphobilinogen, õ-ALA, and coprophobilinogen is seen in acute intermittent porphyria. It is characterized by abdominal pain, polyneuropathy, psychological symptoms, and port wine-colored urine. Protoporphyrinogen IX oxidase (Choice C) catalyzes the metabolization of protoporphyrinogen IX to protoporphyrin IX in heme synthesis. Decreased activity of protoporphyrinogen IX oxidase leads to variegate porphyria, which is characterized by photosensitivity and skin blistering, along with intermittent symptoms during acute attacks such as abdominal pain, nausea, vomiting, constipation, and diarrhea. Uroporphyrinogen III (Choice D) is not an enzyme, it is an intermediate in heme synthesis that is metabolized to coproporphyrinogen III by uroporphyrinogen decarboxylase. Decreased activity of uroporphyrinogen decarboxylase (Choice E) results in porphyria cutanea tarda, which is characterized by photosensitivity and blistering of the skin.

Objective: Lead inhibits the enzymes ferrochelatase and õ-ALA dehydratase, resulting in decreased heme synthesis, increased õ-ALA, urinary coproporphyrin, and erythrocyte zinc protoporphyrin. Clinical manifestations of lead poisoning include encephalopathy, cognitive dysfunction, peripheral nephropathy, hearing loss, renal tubular dysfunction, interstitial nephritis, vomiting, abdominal pain, constipation, and anemia. Previous Next Score Report Lab Values Calculator Help Pause

194 Exam Section 4: Item 45 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 45. A 23-year-old woman with type 1 diabetes mellitus is brought to the physician for a follow-up examination. She has a 2-year history of recurrent yeast infections. Genetic analysis shows a deficiency of myeloperoxidase. Which of the following is the most likely cause of the increased susceptibility to infections in this patient? A) Decreased oxygen consumption after phagocytosis B) Defective neutrophil degranulation C) Defective production of prostaglandins D) Inability to produce hydrogen peroxide E) Inability to produce hydroxy-halide radicals Correct Answer: E. Myeloperoxidase (MPO) deficiency is an autosomal recessive immune disorder caused by mutations in the MPO gene on chromosome 17. MPO encodes the protein MPO which is contained in neutrophil granules where it catalyzes the final step in the neutrophil oxidative burst. Dysfunctional MPO results in an inability to produce hydroxy-halide radicals, namely hypochlorite from hydrogen peroxide and chloride. MPO deficiency results in impaired but not absent bacterial killing since the enzymatic products of NADPH oxidase and superoxide dismutase are also bactericidal. Patients typically present with recurrent fungal infections. Diagnostic evaluation includes flow cytometry and cytochemical stains. Patients will have an abnormal dihydrorhodamine flow cytometric test. This also occurs with NADPH oxidase deficiency in chronic granulomatous disease (CĞD); however, the two conditions can be distinguished by using a nitroblue tetrazolium reduction test. The presence of functional NADPH oxidase will decrease nitroblue, leading to a color change from yellow to blue. Patients with CGD will not demonstrate color change. Incorrect Answers: A, B, C, and D. Decreased oxygen consumption after phagocytosis (Choice A) and inability to produce hydrogen peroxide (Choice D) occurs in CGD as a result of impaired function of NADPH oxidase. NADPH oxidase normally utilizes molecular oxygen to produce the superoxide free radical, which is then converted to hydrogen peroxide by superoxide dismutase. CGD can be distinguished from MPO deficiency by using the nitroblue tetrazolium reduction test. Defective neutrophil degranulation (Choice B) is not a component of MPO deficiency. Neutrophils are activated and able to degranulate appropriately; hypochlorite production is impaired. Defective production of prostaglandins (Choice C) is most commonly iatrogenic from administration of nonsteroidal anti-inflammatory drugs, which inhibit cyclooxygenases that produce prostaglandins.

Objective: MPO deficiency is an inherited immunodeficiency syndrome characterized by the inability to produce hydroxy-halide free radicals within phagolysosomes. The disease is typically mild and may present with recurrent Candida albicans infection. %3D Previous Next Score Report Lab Values Calculator Help Pause

148 Exam Section 3: Item 49 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 25 20 15 10 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 EPP amplitude (mV) 49. An investigator is recording the amplitude of the endplate potential (EPP) in a single striated muscle fiber on stimulation of the motor nerve. The graph shows the frequency distribution for EPPS of various amplitudes for all stimuli that produce a response. Which of the following shows the correct rank order for mode, median, and mean values from smallest to largest? A) Mean < median < mode B) Mean < mode < median C) Median < mean < mode D) Median < mode < mean E) Mode < mean < median F) Mode < median < mean G) Mode = median = mean Correct Answer: F. In statistical analysis, measures of central tendency are useful in describing a dataset. Mode, median, and mean are all measures of central tendency. The mode is defined as the value that occurs most frequently within the data. In this data set the endplate potential amplitude that occurs most frequently in the data is 0.4 mV, making this number the mode. The median is defined as the center value of the data, meaning that there are the same number of values to the left and right (lesser than and greater than) of the median. If the median falls between two numbers, these numbers are summed and divided by two. The mean is defined simply as the average of all of the values included in the data set. In the histogram in this study, the mode can be identified by the greatest value on the Y-axis. The median and the mean require computation; however, the mean is more sensitive to outliers and skewed data when compared with the median. A right-skewed distribution will therefore shift the mean further to the right as compared with the median. Because of this, a right-skewed distribution will have a mean greater than the median and a left-skewed distribution will have a mean that is less than the median. When data is not normally distributed, generally speaking, the median is the preferred measure of central tendency as it is less influenced by outliers and skew. In right-skewed distributions, the mode (while potentially occurring anywhere) is often less than the median, which is less than the mean. Incorrect Answers: A, B, C, D, E, and G. Mean < median < mode (Choice A) would be seen in a left-skewed distribution with the most common value being greater than the median. Mean < mode < median (Choice B) would be seen with a left-skewed distribution with the most common value being between the mean and the median. Median < mean < mode (Choice C) would be seen in a distribution with the most common value being greater than the mean. Median < mode < mean (Choice D) describes a distribution with the most common value lying between the median and the mean. Mode < mean < median (Choice E) describes a distribution with the most common value being less than the mean. Mode = median = mean (Choice G) describes a perfectly normal, gaussian distribution. A normal distribution does not demonstrate skew and is perfectly symmetric leading to the median and mean being equivalent.

Objective: Median, mode, and mean are measures of the central tendency of data. Mode is the most common value, median is the central value of the data, and mean is the average of all values in the data set. The mean is more sensitive to outliers and skew than is the median. Previous Next Score Report Lab Values Calculator Help Pause Number of responses

156 Exam Section 4: Item 7 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 7. A 55-year-old man comes to the office for an initial visit at the insistence of his wife. He has not seen a physician for the past 15 years. He has a 1-month history of mild shortness of breath when climbing stairs. He also has had a 23-kg (50-lb) weight gain during the past 15 years. He is 175 cm (5 ft 9 in) tall and now weighs 113 kg (250 lb); BMI is 37 kg/m2. His waist circumference is 107 cm (42 in). His pulse is 95/min, respirations are 20/min, and blood pressure is 160/90 mm Hg. Pulse oximetry on room air shows an oxygen saturation of 97%. Physical examination shows no other abnormalities. Serum studies show: Glucose [fasting] Cholesterol, total HDL-cholesterol LDL-cholesterol Triglycerides 85 mg/dL 215 mg/dL 35 mg/dL 140 mg/dL 200 mg/dL The physician explains to the patient that he is at significant risk for atherosclerotic heart disease. The most appropriate therapy to decrease this patient's risk for a myocardial infarction has which of the following effects? A) Decreases serum LDL-cholesterol concentration B) Decreases serum triglyceride concentration C) Increases the serum total cholesterol/HDL-cholesterol ratio D) Increases the serum triglyceride/HDL-cholesterol ratio Correct Answer: A. Metabolic syndrome is a common disorder that is associated with an increased risk for developing cardiovascular disease, type 2 diabetes mellitus, and nonalcoholic fatty liver disease. It is characterized by hypertension, abdominal obesity, insulin resistance with hyperglycemia, increased serum triglyceride concentrations, and decreased HDL-cholesterol. The presence of three or more of these conditions is sufficient for diagnosis. First-line therapy consists of diet and activity modifications. Pharmacotherapy should be used to treat hypertension, hyperglycemia, and dyslipidemia if present. Statins are the mainstay of therapy for dyslipidemia. Statins act by inhibiting HMG-CoA reductase (which is the rate-limiting enzyme in cholesterol synthesis) and upregulating LDL receptors. This decreases serum LDL-cholesterol concentration. Statins also increase serum HDL-cholesterol and decrease triglycerides. Adverse effects are uncommon but generally involve the liver and skeletal muscle. These adverse effects may potentially manifest as elevated results on liver function testing, or statin-associated myopathy with progression to rhabdomyolysis. Incorrect Answers: B, C, and D. The use of fibrates such as gemfibrozil decreases serum triglyceride concentration (Choice B). Fibrates upregulate lipoprotein lipase and activate peroxisome proliferator-activated receptors. Statins are the first-line therapy for increased LDL-cholesterol, and concomitant use of statins and fibrates increases the risk for myopathy. Increasing the serum total cholesterol/HDL-cholesterol ratio (Choice C) and increasing the serum triglyceride/HDL-cholesterol ratio (Choice D) confer a higher risk for cardiovascular disease. Increased HDL-cholesterol concentrations are cardioprotective and associated with lower cardiovascular risk.

Objective: Metabolic syndrome is a common disorder that is associated with an increased risk for developing cardiovascular disease, type 2 diabetes mellitus, and nonalcoholic fatty liver disease. Lipid-lowering therapy with a statin is recommended for treating associated dyslipidemia. II Previous Next Score Report Lab Values Calculator Help Pause

70 Exam Section 2: Item 21 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 21. A 60-year-old man comes to the emergency department because of swelling of his right elbow that began after he struck it against the edge of a file cabinet earlier that day. On physical examination, the tip of the olecranon is obscured by a golf ball-sized fluctuant mass with well-defined borders. The mass is swollen, mildly tender, and smooth. Range of motion of the elbow is normal. Which of the following structures was most likely injured in this patient? A) Bursa B) Cartilage C) Joint capsule D) Ligament E) Synovium OF) Tendon Correct Answer: A. Bursae are thin, sac-like, fluid-filled structures that decrease friction and permit motion between layers of connective tissue. Bursae are frequently implicated in pathology and are often associated with pain and swelling, such as the subacromial bursa which can cause pain in the shoulder, the trochanteric bursa which can cause pain over the lateral aspect of the hip, and the olecranon bursa which can cause pain and swelling posterior to the elbow. When inflamed or injured, fluid can collect inside the potential space of the bursa, causing swelling that can be appreciated on examination as a fluctuant, tender, circumscribed mass. Olecranon bursitis demonstrates such findings posterior to the elbow and may be painful to the touch. If erythema and warmth are present, these findings can signify infection of the bursa (septic bursitis). Incorrect Answers: B, C, D, E, and F. Cartilage (Choice B) is a specialized connective tissue that provides both structure and flexibility. Hyaline cartilage provides the smooth surface of joints; fibrocartilage provides the flexible structure of the intervertebral discs and pubic symphysis. Joint capsule (Choice C) is the collagen-rich ligamentous tissue that surrounds joints. It provides stability to the joint as well as contains synovial fluid within the joint space. Infection inside the joint capsule is known as septic arthritis and presents with fever, significant pain with range of motion, swelling of the joint, erythema, and inability to bear weight on the joint. Ligaments (Choice D) are organized structures of collagen, type I, that provide stability to joints and define specific ranges of motion. For example, the anterior cruciate ligament of the knee provides resistance to anterior translation of the tibia on the femur and limits hyperextension of the knee. Synovium (Choice E) is the lining of the joint capsule. It secretes synovial fluid to lubricate the articular cartilage. Inflammatory cell infiltration and proliferation of the synovium is a hallmark of autoimmune joint diseases such as rheumatoid arthritis. Tendons (Choice F) are type I collagen-rich, organized structures that attach muscles to bones and transmit muscular force across joints. Chronic inflammation may lead to attenuation and eventual tendon rupture (eg, patellar tendonitis, rotator cuff tendonitis). Tendon rupture may also be precipitated by fluoroquinolone administration, particularly in older adults.

Objective: Olecranon bursitis is a common orthopedic condition that occurs secondary to trauma to or infection of the bursa. In this condition, fluid collects in the bursa, resulting in a fluctuant, tender mass. %3D Previous Next Score Report Lab Values Calculator Help Pause

146 Exam Section 3: Item 47 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 47. A 58-year-old woman comes to the physician because of a 3-month history of mild urine leakage when she washes her hands or when she hears water running. The leakage does not occur with coughing or sneezing. There is no nocturia, urinary hesitancy, or straining with urination. Her vital signs are within normal limits. Physical examination shows no abnormalities. Which of the following is the most likely cause of this patient's incontinence? A) Incomplete emptying B) Increased intra-abdominal pressure C) Polyuria D) Uninhibited bladder contractions E) Urinary tract infection Correct Answer: D. An overactive bladder is caused by detrusor muscle instability and uninhibited bladder contractions. It commonly results in urge incontinence, which presents with the urge to void followed by the leakage of urine. Treatment of urge incontinence includes pelvic floor strengthening exercises, timed voiding, techniques for distraction or relaxation, or medications. Oxybutynin is an antimuscarinic medication used to treat urge incontinence from an overactive bladder and detrusor instability. It inhibits the parasympathetic innervation of the detrusor muscle, decreasing bladder muscle tone. Incorrect Answers: A, B, C, and E. Overflow incontinence can be caused by underactive or flaccid detrusor muscles or bladder outlet obstruction, resulting in urinary retention and incomplete emptying of the bladder (Choice A), which is evidenced by an increased postvoid residual bladder volume. When the intravesical hydrostatic pressure eventually overwhelms the internal and external urinary sphincter closing pressures, the patient will demonstrate passive leakage of urine. Increased intra-abdominal pressure (Choice B) describes the pathophysiology of stress incontinence, which is triggered by an increase in intra-abdominal pressure (eg, lifting, sneezing, Valsalva). Stress incontinence results from urethral hypermobility, pelvic floor weakness, or intrinsic sphincter deficiency. Řisk factors include multiple vaginal deliveries, obesity, and prostate surgery. Polyuria (Choice C) can be caused by a variety of diseases, including syndrome of inappropriate antidiuretic hormone secretion, diabetes insipidus, and diabetes mellitus. It is characterized by the passage of a large volume of urine, rather than small amounts of urine leakage. Urinary tract infection (Choice E) can present with an acutely overactive bladder (frequency, urgency, urge incontinence). Patients generally also experience dysuria and suprapubic pain.

Objective: Overactive bladder is caused by detrusor muscle instability and uninhibited bladder contractions. It commonly results in urge incontinence, which presents with the urge to void followed shortly by the leakage of urine. Previous Next Score Report Lab Values Calculator Help Pause

124 Exam Section 3: Item 25 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 25. A 66-year-old man is brought to the emergency department 30 minutes after sustaining injuries in a motor vehicle collision. He is pronounced dead on arrival. A photograph of the left kidney and a photomicrograph of a section of the distal left ureter taken at autopsy are shown. Examination of the right kidney and ureter shows no abnormalities. Which of the following was the most likely predisposing factor in the development of the ureteral lesion in this patient? A) Alcoholism B) Arylamine exposure C) Cigarette smoking D) Radiation exposure E) Schistosomiasis OF) Vinyl chloride exposure Correct Answer: C. Cigarette smoking is a primary risk factor for the development of urothelial cell carcinoma (also known as transitional cell carcinoma), a malignancy affecting the urinary tract epithelium. It can develop in the renal collecting tubules, calyx, renal pelvis, ureter, urethra, or bladder, as all of these structures share the same embryologic origin and are comprised of the same type of epithelium. Hematuria is a common finding and is present in most patients, with ureteral obstruction occurring in about 20% of patients as well. Some affected individuals remain asymptomatic until an advanced stage of disease and therefore present with a worse prognosis. Ureteral obstruction occurs in many patients at the site of the primary tumor or in noncontiguous areas with drop metastases. This patient has developed obstruction with consequent hydronephrosis as evidenced by dilation of the renal pelvis and renal calyces with compressive atrophy of the renal cortical parenchyma on the gross specimen. Because the entire urothelial lining is exposed to the same carcinogen (eg, cigarette smoke), malignancy can develop anywhere along the collecting system. Additional carcinogens known to be associated with urothelial cancer include phenacetin, cyclophosphamide, and aniline dyes. Treatment is based on the stage of cancer but typically involves partial or complete removal of the affected kidney, ureter, and part of the bladder. Incorrect Answers: A, B, D, E, and F. Alcoholism (Choice A) predisposes to esophageal cancer and, if patients also develop cirrhosis, predisposes to the development of hepatocellular carcinoma. While it is not known to directly increase the risk for urothelial carcinoma, men who concomitantly drink and smoke are at a higher risk. Arylamine exposure (Choice B) is associated with the development of bladder cancer. It is less commonly associated with ureteral malignancy. Arylamines are found in several types of hair dyes and exposure to these chemicals is most common in beauticians and hairdressers. Arylamines require oxidation in the liver to become carcinogenic. Radiation exposure (Choice D) predisposes to cancer depending on the type of radiation and the location. Mantle radiation used in the treatment of Hodgkin lymphoma can predispose to the development of thyroid cancer. Radiation used in other malignancies can predispose to the development of leukemia, lymphoma, and myelodysplastic syndrome. Schistosomiasis (Choice E) predominantly increases the risk for urinary tract squamous cell carcinoma, which is far less common than urothelial cancer. It is typically seen in patients from endemic areas, and patients are likely to have evidence of infection with schistosomiasis. Cigarette smoking is a far more likely risk factor in patients from areas where schistosomiasis is not common (eg, United States). Vinyl chloride exposure (Choice F) would be expected in patients who work in the manufacturing industry making polyvinyl chloride, a type of plastic. Prolonged exposure predisposes to the development of hepatic angiosarcoma and lymphoma/leukemia. It is not associated with urothelial carcinoma.

Objective: Patients who smoke cigarettes, especially men, are at a higher risk than the general population for the development of urothelial carcinoma, which can present with hematuria and urinary tract obstruction or can remain asymptomatic for years until an advanced stage. Previous Next Score Report Lab Values Calculator Help Pause

40 Exam Section 1: Item 41 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 41. During an experiment, investigators use an in vitro cell culture system to study leukocyte function. As part of the study, a new drug that prevents polymerization of actin filaments is tested. Which of the following leukocyte functions is most likely to be inhibited by this new drug? A) Binding to mannose receptor B) Internalization into clathrin-coated pits C) Phagocytosis D) Pinocytosis E) Receptor-mediated endocytosis Correct Answer: C. Phagocytosis would be affected by a new drug that prevents polymerization of actin filaments. Actin provides a mechanical superstructure for the phagocyte to maintain shape, acts as a highway to shuttle enzymes and their respective substrates together to augment intracellular signaling, and aids in the endocytosis of large particles. This latter role is a crucial part of phagocytosis. When phagocytes come into contact with an opsonized pathogen at the leading edge, the phagocyte binds the Fc portion of immunoglobulin via its Fc receptor. Grouping of multiple Fc receptors results in the activation of multiple downstream signaling pathways. Some of these include signaling through the Arp2/3 complex, WASP, and Cdc42, which ultimately leads to actin polymerization and formation of the phagocytic cup that surrounds the pathogen. Actin polymerization extends the edges of the plasma membrane in a circumferential fashion around the pathogen, thereby aiding in complete endocytosis. Inhibition at any step of this process could theoretically lead to failed phagocytosis. Incorrect Answers: A, B, D, and E. Binding to mannose receptors (Choice A) on the surface of macrophages, not neutrophils, results in endocytosis of several bacterial pathogens that display mannose residues on their cell surface. Mannose receptors are recycled from the endosome back to the cellular membrane. This process is independent of actin. Internalization into clathrin-coated pits (Choice B), pinocytosis (Choice D), and receptor-mediated endocytosis (Choice E) are all descriptive terms for a similar process. Pinocytosis refers to the endocytosis of small particles from the extracellular matrix and can occur via multiple mechanisms, including via clathrin-mediated endocytosis, caveolin-mediated endocytosis, and micropinocytosis. Clathrin-mediated endocytosis is a common means of internalizing receptor-ligand pairs and involves intracellular signaling that begins with the binding of a ligand to a receptor. While there may be instances of receptor-mediated endocytosis without clathrin, these two terms are often used interchangeably.

Objective: Polymerization of actin is a crucial step in phagocytosis as it permits the extension of the cell membrane around the pathogen, allowing for its encapsulation and internalization. Drugs that interfere with this process would interfere with phagocytosis. II Previous Next Score Report Lab Values Calculator Help Pause

26 Exam Section 1: Item 26 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 26. A 55-year-old man is found to have gastroesophageal reflux disease. In addition to recommending lifestyle changes, the physician prescribes an agent that binds irreversibly to its site of action to inhibit gastric acid formation. Which of the following agents was most likely prescribed? A) Clarithromycin B) Esomeprazole C) Famotidine D) Misoprostol E) Sucralfate Correct Answer: B. Proton-pump inhibitors, such as esomeprazole, are first-line pharmacotherapy for patients presenting with gastroesophageal reflux (GERD) and associated reflux esophagitis. GERD typically presents with burning epigastric and lower chest pain, often in association with consumption of a large meal or trigger food. It is often worse with supine positioning. Acid reflux in the esophagus causes mucosal irritation and inflammation, which can present as mucosal erythema and multiple erosions (reflux esophagitis). Over time, if the GERD remains untreated, metaplasia can occur leading to Barrett esophagus, a state of premalignancy marked by intestinal metaplasia of the distal esophagus. Untreated, this can subsequently lead to the development of esophageal adenocarcinoma. Proton pump inhibitors decrease gastric acid secretion by parietal cells through irreversible inhibition of the H*/K* ATPase located on the luminal surface, which is the primary exporter of hydrogen ions into the gastric lumen. Incorrect Answers: A, C, D, and E. Clarithromycin (Choice A) is a macrolide antibiotic that may be used in the treatment of Helicobacter pylori infection. H. pylori is a spiral-shaped, gram-negative bacterium that is associated with chronic gastritis and peptic ulcer formation. Famotidine (Choice C) is a second-generation histamine type 2 (H2) receptor blocker, which is also used in the treatment of GERD. H2 receptors on gastric parietal cells are G protein-coupled receptors that stimulate increased intracellular CAMP concentrations, which leads to increased H*/K* ATPase activity. H2 receptor blockers are reversible competitive antagonists. Misoprostol (Choice D) is a prostaglandin-E, analog that may be used to prevent peptic ulcer formation in the setting of NSAID use (NSAIDS cause decreased prostaglandin concentrations as a result of COX inhibition). Prostaglandins regulate parietal cell acid production through G protein-coupled receptor inhibition of CAMP production in the cell, leading to decreased H*/K* ATPase activity. Sucralfate (Choice E) is a medication that reacts with hydrochloric acid in the stomach to form a thick viscous coating that binds to proteins on the mucosal surface and acts as a physical acid buffer.

Objective: Proton-pump inhibitors are first-line pharmacologic agents for the treatment of gastroesophageal reflux disease. The mechanism of action is direct, irreversible inhibition of the luminal H*/K* ATPase on gastric parietal cells. Previous Next Score Report Lab Values Calculator Help Pause

51 Exam Section 2: Item 2 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 2. A female newborn delivered at 26 weeks' gestation is found to have hyaline membrane disease. She is intubated and mechanically ventilated. Her vital signs are continuously monitored, and inspired oxygen is maintained at a level that will sustain an oxygen saturation between 92% and 95%. The primary goal of this treatment is protection of which of the following structures? A) Basal ganglia B) Choroid plexus C) Cochlea D) Ductus arteriosus E) Retina Correct Answer: E. Fetal retinal vascularization is not complete until term. Fetal retinal blood vessels continue to grow outward from the optic nerve and posterior pole towards the ora serrata throughout the third trimester in a process that is dependent on a chemical gradient of vascular endothelial growth factor (VEGF). Premature birth may interrupt this process, especially in the setting of high concentrations of supplemental oxygen, which are often used to treat neonatal respiratory distress syndrome. High concentrations of supplemental oxygen may result in the abolition of the VEGF chemical gradient that promotes normal retinal vascularization, and may be directly toxic to capillaries through increased formation of free radicals. Abnormal retinal vascularization leads to avascularity of the peripheral retina with subsequent abnormal neovascularization, a condition known as retinopathy of prematurity (ROP). Neovascular vessels in the peripheral retina may cause hemorrhage and induce traction that results in retinal detachment. Proper regulation of fetal supplemental oxygen administration helps to prevent the development of ROP. Neonates who are born prematurely (less than 30 weeks gestational age) or with a low birth weight (less than 1500 g) should undergo dilated examination to screen for ROP. Neonates who develop ROP can be treated with retinal laser photocoagulation to decrease the production of VEGF and prevent further complications such as retinal detachment and blindness. Incorrect Answers: A, B, C, and D. Basal ganglia (Choice A) developmental abnormalities may occur in premature infants, but their outcome is not directly affected by neonatal hyperoxygenation. Deposition of bilirubin in the brain (kernicterus) is a known pathology in premature neonates. The choroid plexus (Choice B) lies within the cerebral ventricles and produces cerebrospinal fluid. The development of the choroid plexus is not impaired by high concentrations of supplemental oxygen during the neonatal period. The cochlea (Choice C) is a component of the inner ear. Congenital hearing loss may occur in the setting of infections with cytomegalovirus, syphilis, and rubella. While congenital hearing loss is more common in premature infants, oxygenation does not play a clear role in its development. The ductus arteriosus (Choice D) connects the fetal pulmonary artery and aortic arch. Patency of the ductus arteriosus after birth results in a continuous, machine-like murmur best heard in the left second intercostal space that radiates to the clavicle.

Objective: ROP results from abnormal vascularization and neovascularization of the retina in premature infants. Complications, such as tractional retinal detachment and blindness, may be prevented by proper regulation of fetal supplemental oxygen administration and by early detection and treatment of ROP. All infants who are born prior to 30 weeks gestational age or with a birth weight below 1500 g should be screened for ROP. Previous Next Score Report Lab Values Calculator Help Pause

170 Exam Section 4: Item 21 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 21. Development of the metanephros begins with initial specification of the metanephric blastema and sprouting of the ureteric bud and culminates with nephron differentiation and formation of the glomerulus. This cascade required for development of the collecting system and the nephron is best described as a sequence of which of the following interactions? A) Autoinductive B) Dorsalizing-ventralizing C) Epithelial-mesenchymal D) Gap junction E) Reciprocal inductive Correct Answer. E. Reciprocal inductive signaling describes the development of tissues on the basis of reciprocal signaling between two different cell types or tissues, a phenomenon that is appreciated in the development of the kidney. The ureteric bud reacts to signaling from the metanephrogenic mesenchyme and leads to branching, with the subsequent development of the collecting ducts, while the mesenchyme reacts to signals from the uretic bud to cause development of the renal tubules. When these two tissues are separated in culture, they are incapable of developing completely in the absence of the other. This interaction occurs either via diffusible compounds, direct cell-to-cell contact, or through communication via gap junctions. Incorrect Answers: A, B, C, and D. Autoinductive (Choice A) signaling specifies a process in which cells or tissue are able to induce their own growth and differentiation in the absence of interaction with other cells, which is not the case in the development of the kidney. Dorsalizing-ventralizing (Choice B) refers to the development of the dorsal-ventral axis in embryology as determined by the neural tube. Signaling via release of factors such as bone morphogenetic protein, Wnt protein, transforming growth factor-B, and fibroblast growth factor help determine the dorsal-ventral axis. This is not involved in the development of the kidneys. Epithelial-mesenchymal (Choice C) transition refers to the process during development by which epithelial cells lose their polarity and adhesions and develop the ability to migrate to become mesenchymal stem cells. The reverse process of mesenchymal-epithelial transition describes the process that occurs in the development of the kidney as mesenchymal cells differentiate into epithelial cells and arrange themselves in an ordered fashion to create the kidney. Gap junctions (Choice D) are connections between the cytoplasm of two different cells that allow for the movement of small molecules in either direction. They play a role in many stages of embryologic development but do not explain the primary mechanism by which kidney development occurs.

Objective: Reciprocal inductive signaling refers to the process by which two distinct cell populations affect the development and differentiation of the other through various signaling modalities. A prime example of inductive signaling is the interaction between the ureteric bud and the mesenchyme, which work in concert to generate the mature kidney. II Previous Next Score Report Lab Values Calculator Help Pause

121 Exam Section 3: Item 22 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 22. An animal study is conducted to assess the effects of smoking on pulmonary defense and maintenance mechanisms. For 1 week, normal male rats are exposed to levels of cigarette smoke comparable to those encountered by humans who smoke cigarettes. Results of pulmonary testing are compared with baseline levels obtained the week before the smoke exposure. Which of the following sets of changes is most likely to be observed? Mucus Production and Secretion Alveolar Activity of Airway Cilia Macrophage Function A) ↑ ↑ ↑ B) ↑ ↑ C) ↑ D) ↑ ↑ E) ↑ F) Correct Answer: C. Smoking is associated with numerous detrimental effects on the respiratory system. Tobacco smoke directly disrupts the ability of the airway to clear foreign material by resulting in increased mucus production and secretion by the respiratory epithelium, decreased activity of airway cilia, and inhibition of alveolar macrophage function. The airway and lungs are constantly exposed to potentially hazardous material from the environment. Normal respiratory epithelium secretes mucus that traps pathogens, inhaled toxins, and particles in a viscous layer. The mucus is cleared by the action of beating cilia, which propel material proximally and out of the airway. Normal mucus has a low viscosity that allows it to be moved by the beating of cilia. Exposure to tobacco smoke results in increased mucus production that forms a denser layer, which is not easily cleared. It is also associated with impaired ciliary function. The inability to remove foreign particles results in prolonged exposure of the airway to antigens and promotes inflammatory changes. Mucus plugging and obstruction of the small airways occurs over time. Alveolar macrophages reside in the alveolar space and clear microbes and debris. They also phagocytose harmful environmental particles such as carbon and silica. In patients who smoke, macrophage evaluation from bronchoalveolar lavage collection shows increased numbers of atypical macrophages that contain higher than usual amounts of pigment and sequestered carbon. These macrophages are morphologically different than in nonsmokers and display impaired function. Incorrect Answers: A, B, D, E, and F. Choices A, B, and D are incorrect as the activity of airway cilia is decreased with exposure to tobacco smoke, not increased. Ciliary activity may be enhanced by both adrenergic and cholinergic signaling along with exposure to certain chemical irritants. Choices E and F are incorrect as mucus production is increased with smoking, not decreased.

Objective: Smoking disrupts the ability of the lungs to clear mucus and foreign debris by promoting increased mucus production and secretion, directly impairing the activity of airway cilia, and inhibiting alveolar macrophage function. This leads to an increased risk for developing chronic inflammatory pulmonary disorders. II Previous Next Score Report Lab Values Calculator Help Pause

165 Exam Section 4: Item 16 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 16. A2-month-old boy is brought to the physician because of decreased movement, weakness, and poor suck since birth. His sister died at the age of 2 years because of respiratory insufficiency. He is alert. Physical examination shows poor head control, absent deep tendon reflexes, and generalized hypotonia. Results of a mutation analysis show that he is homozygous for deletion of the survival motor neuron (SMN1) gene. Which of the following is the most likely mode of inheritance in this patient? A) Autosomal dominant B) Autosomal recessive C) Mitochondrial D) X-linked dominant E) X-linked recessive Correct Answer: B. Spinal muscular atrophy (SMA) is a group of autosomal recessive disorders that lead to the degeneration of motor neurons, and result in progressive hypotonia and areflexia. SMA is caused by mutations in SMN1. Patients with SMA exhibit difficulty with mobility and frequently experience respiratory distress and difficulty feeding. This patient presents with the severe, infantile-onset form of SMA, known as Werdnig-Hoffmann disease. Classically, infantile hypotonia and respiratory failure occur within the first six months of life. Treatment is primarily supportive with nutritional interventions and mechanical ventilation, although gene therapy is emerging for patients with specific SMN1 mutations. Incorrect Answers: A, C, D, and E. Autosomal dominant inheritance (Choice A) is observed in several neurologic diseases, including Huntington disease, myotonic dystrophy, tuberous sclerosis, and neurofibromatosis. Mitochondrial inheritance (Choice C) is observed in neurologic and muscular diseases such as Leber hereditary optic neuropathy, maternally inherited diabetes and deafness, myoclonic epilepsy with ragged red fibers, and MELAS (mitochondrial encephalomyopathy, lactic acidosis, and stroke-like episodes) syndrome. X-linked dominant inheritance (Choice D) is a relatively uncommon inheritance pattern that is observed in Fragile X syndrome, incontinentia pigmenti, hypophosphatemic rickets, and Rett syndrome. X-linked recessive inheritance (Choice E) is observed in glucose-6-phosphate dehydrogenase deficiency, Duchenne and Becker muscular dystrophy, and hemophilia A and B.

Objective: Spinal muscular atrophy is a group of autosomal recessive disorders that lead to the degeneration of motor neurons and result in progressive hypotonia and areflexia. SMA is caused by mutations in SMN1. The severe, infantile variant, also known as Werdnig-Hoffmann disease, presents with infantile hypotonia and respiratory failure within the first six months of life. Previous Next Score Report Lab Values Calculator Help Pause

195 Exam Section 4: Item 46 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 46. To assess antigen-presenting cell function in a clinical study, skin explants are cultured from a volunteer and analyzed for Langerhans cell maturation and differentiation. Increased expression of which of the following surface markers is most indicative of Langerhans cell activation in this subject? A) B7 B) CD3 C) Interleukin-4 (IL-4) receptor D) Membrane immunoglobulin OE) B-Microglobulin Correct Answer: A. Langerhans cells are dendritic cells that reside in the epidermis of the skin and function as antigen-presenting cells (APC). Like other APCS, their cell surface includes the protein B7, also called CD80/86. The process by which an APC, such as a Langerhans cell, activates a T ymphocyte requires two separate stimulatory signals. This is to ensure that the T lymphocyte is not erroneously activated by self-antigen, which leads to autoimmunity. The first signal of T lymphocyte activation is created when a processed antigen is presented on the major histocompatibility complex (MHC) II of a dendritic cell and matches to a T lymphocytereceptor (TLR) on a CD4+ helper T lymphocyte. The T lymphocyte is then further activated and proliferates when a second stimulatory signal is created by the binding of B7 on the surface of the Langerhans cell to CD28 on a naïve I lymphocyte. An activated Langerhans cell, which carries an antigen in its MHC |I cleft waiting to recognize a T lymphocyte, will express B7 on its surface. Once both stimulatory signals are received, the T lymphocyte is activated, produces cytokines, and is able to target infected cells. Incorrect Answers: B, C, D, and E. CD3 (Choice B) is a T lymphocyte surface marker which resides in conjunction with the TLR. It is a pan-T lymphocyte surface marker, meaning that it is present on both CD4+ and CD8+ T lymphocytes. It is not present on the surface of Langerhans cells. Interleukin-4 (IL-4) receptor (Choice C) is a cytokine receptor present on T lymphocytes, not Langerhans cells. Binding of IL-4 by a CD4+ helper T lymphocyte causes differentiation towards the Th2 phenotype. A Th2 cell secretes IL-4, IL-5, IL-6, IL-10, and IL-13 and promotes IgE production by B lymphocytes. Membrane immunoglobulin (Choice D) is a form of immunoglobulin that is bound to the cell surface of a B lymphocyterather than released into solution like soluble immunoglobulin. The membrane immunoglobulin helps the B lymphocyte to recognize its antigen more quickly on subsequent exposures and increase expression of that antibody, leading to a rapid response to a pathogen. Membrane immunoglobulin is not present on Langerhans cells. B2-Microglobulin (Choice E) makes up the short chain of MHC I. MHC I is present on the surface of all nucleated cells, including but not limited to APCS. MHC I molecules present endogenous antigens to CD8+ cytotoxic T lymphocytes by binding to the T-lymphocyte receptor (TCR) and CD8 molecules on the lymphocyte surface. MHC I, including its component B2-microglobulin, is present on the APC constantly.

Objective: T lymphocyte activation by a dendritic APC, like a Langerhans cell in the skin, requires two signals: the binding of MHC II with the TLR and the binding of B7 with CD28. The requirement of both signals decreases the activation of T lymphocytes to self-antigens and decreases the incidence of autoimmunity. %3D Previous Next Score Report Lab Values Calculator Help Pause

190 Exam Section 4: Item 41 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 41. A 45-year-old woman with chronic liver disease secondary to hepatitis C comes to the physician for a follow-up examination. Abdominal examination shows ascites and hepatosplenomegaly. Serum studies show decreased concentrations of total thyroxine (T); serum concentrations of thyroid-stimulating hormone (TSH) and free thyroxine (FT) are within the reference ranges. Which of the following best explains these laboratory results? A) Decreased T synthesis B) Decreased thyroxine-binding globulin synthesis C) Decreased TSH synthesis D) Increased thyroglobulin storage E) Increased thyroid-releasing hormone F) Increased triiodothyronine (T3) synthesis Correct Answer: B. Thyroxine-binding globulin (TBG) is synthesized by the liver and binds most available serum triiodothyronine (T) and thyroxine (T). There is a component of free T, which remains unbound and thus is able to be converted to T3 or bind to its receptors within the target cells. In the case of severe chronic liver disease, the synthetic function of the liver is impaired and TBG concentrations decrease. Without TBG, less total T, is contained in the serum though the amount of free T which is not dependent on TBG, remains within reference range. Because free T,concentrations are constant, the patient does not experience adverse metabolic effects. Free T4continues to bind its receptors on the hypothalamus and pituitary, and thus there is no perceived need by these organs to increase production of thyrotropin-releasing hormone (TRH) or TSH, and their concentrations remain constant. Decreased synthetic function of the liver also leads to decreased production of albumin and coagulation factors, which contribute to the edema, ascites, and coagulopathy seen in cirrhosis. Other functions of the liver that fail in cirrhosis include filtration and detoxification of portal venous blood, first-pass drug metabolism, biochemical metabolism of lipids and glucose, and metabolism of hormones (eg, estrogens). Incorrect Answers: A, C, D, E, and F. Decreased T4 synthesis (Choice A) occurs in cases of hypothyroidism, whether by destruction of the thyroid tissue or failure of the hypothalamus or pituitary gland to secrete regulatory hormones. Decreased thyroxine-binding globulin synthesis is a more likely explanation for the laboratory changes in this patient with chronic liver disease. Because the concentrations of free Ta remain the same even when TBG synthesis declines, there is no decrease in negative feedback at the level of the hypothalamus or pituitary. Thus, neither thyroid-releasing hormone nor TSH synthesis decrease (Choice C) or increase (Choice E). Thyroglobulin has been detected in serum of persons with chronic liver disease, suggesting that decreased, not increased, storage of it occurs (Choice D). Increased triiodothyronine (T) synthesis (Choice F) occurs as a result of either an increase in direct synthesis by the thyroid gland or an increase in peripheral conversion from T, to T, While variability in thyroid hormone concentrations can be seen in liver disease, these mechanisms do not account for the altered concentrations in this patient, which relate to decreased hepatically synthesized TBG.

Objective: TBG is synthesized in the liver and decreases in chronic liver disease. When TBG concentrations decrease, the total T4 decreases but the free T, stays the same, and no adverse metabolic effects are experienced. Previous Next Score Report Lab Values Calculator Help Pause

56 Exam Section 2: Item 7 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 7. A 3-year-old boy is brought to the emergency department by his father because of bleeding from a cut on his left arm after he fell off his tricycle 1 hour ago. Physical examination shows a 2-cm shallow laceration on the left upper extremity. The laceration has stopped bleeding and is covered with a clot. This finding results from hepatic synthesis of which of the following? A) Enzymes that increase production of platelets B) Inhibitors of naturally occurring anticoagulants C) Plasmin from plasminogen D) Protein C and protein S E) Proteins that result in the production of fibrin strands Correct Answer: E. Proteins that result in the production of fibrin strands are made by hepatocytes in the liver and aid in the formation of blood clots. Two common pathways of activating the clotting cascade are the intrinsic pathway, whereby exposure to endothelial collagen initiates clotting, and the extrinsic pathway, whereby tissue factor release from endothelial cells initiates the cascade. Regardless of the inciting event, both pathways converge on the common pathway, which begins with the activation of factor X to factor Xa. Activated factor X induces the conversion of prothrombin to thrombin, which is subsequently required for the conversion of fibrinogen to fibrin. Fibrin is an essential component of clot formation, polymerizing to form fibrin strands via the assistance of factor XIlla. This fibrin meshwork then incorporates platelets to form a mature clot over the site of injury, which stops bleeding. Fibrinogen is produced in the liver. Incorrect Answers: A, B, C, and D. Enzymes that increase production of platelets (Choice A) are not produced in the liver. Certain mutations in genes such as JAK2, MPL, and CALR can cause unregulated production of platelets in the bone marrow and are a cause of disorders such as essential thrombocytosis, polycythemia vera, and myelodysplastic syndrome, but these are not hepatically produced enzymes. Inhibitors of naturally occurring anticoagulants (Choice B) such as protein C inhibitor, a plasma serine protease inhibitor, exist naturally in the body. They aid in maintaining balance between clotting and bleeding. Plasmin from plasminogen (Choice C) describes the process of activating plasminogen to become plasmin, which subsequently degrades fibrin clots leading to clot dissolution. While plasminogen is made in the liver, its role is to degrade and prevent clot formation. Common medications that are used in clinical practice to augment this pathway include tissue plasminogen activator and streptokinase. Protein C and protein S (Choice D) are both naturally occurring anticoagulants that are produced in the liver. Activation of protein C and S attenuates the coagulation cascade and limits clot formation.

Objective: The crosslinking of fibrin strands is the terminal event in the coagulation cascade. The fibrin meshwork then incorporates platelets, which forms a mature clot resulting in hemostasis. Fibrinogen is made in the liver and is converted to fibrin via the action of thrombin. %3D Previous Next Score Report Lab Values Calculator Help Pause

197 Exam Section 4: Item 48 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 48. A 28-year-old nulligravid woman who has been unable to conceive for 2 years undergoes an endometrial biopsy. Her menses occur at regular 28-day intervals. Photomicrographs of the endometrial tissue are shown. This patient is at which of the following phases of her menstrual cycle? A) Ischemic B) Menstrual C) Proliferative D) Secretory Correct Answer: D. The menstrual cycle consists of two ovarian phases, which cause the endometrium to cycle through three endometrial phases. The phases of the ovarian cycle include the luteal phase and the follicular phase. The first half of the menstrual cycle, the follicular phase, begins with menses and varies in length. 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 for the implantation of a fertilized ovum. This creates the proliferative phase of the endometrium. 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 begins the secretory phase of the endometrium. 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. Secretory endometrium occurs during the luteal phase of the menstrual cycle, or the fourteen days prior to menstruation. It follows the proliferative phase in which the glands grow in size. An endometrial biopsy taken from the secretory phase demonstrates distended, tortuous endometrial glands, which can be appreciated on the first photomicrograph in this case. On higher power view, the plump columnar cells contain cytoplasmic vacuoles, which contain the secretions that will be released into the glands. The final phase of the endometrial cycle is menses, in which the endometrial glands outgrow their blood supply, become ischemic, and slough off into the endometrial lumen. Incorrect Answers: A, B, and C. If implantation does not occur, the corpus luteum is not maintained and the production of progesterone stops. As progesterone concentrations decrease, the spiral arteries supplying blood to the secretory endometrium atrophy. When this happens, the endometrium no longer has the blood supply necessary to support the proliferated endometrial glands and the glands become ischemic (Choice A). With persistent ischemia, the tissue begins to necrose and then slough off into the endometrial lumen, causing menstruation (Choice B). Proliferative endometrium (Choice C) occurs during the follicular phase of the menstrual cycle, in response to estrogen causing a gradual growth in the endometrial lining. A biopsy of proliferative endometrium would demonstrate smaller endometrial glands, which have not become tortuous or distended. The columnar cells of the glands would not yet contain secretory vacuoles and would be less plump.

Objective: The endometrium goes through three stages each menstrual cycle: proliferative, secretory, and menses. The secretory endometrium is characterized by dilated, tortuous endometrial glands with large intracytoplasmic vacuoles containing secretory material. Previous Next Score Report Lab Values Calculator Help Pause

137 Exam Section 3: Item 38 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 38. A feature of the host response to malaria in experimental animals is the activation of CD8+ T lymphocytes with specificity for malaria antigens. Which of the following is the best explanation for this response? A) Endocytosis of malaria parasites leads to processing of parasite proteins in endosomal vesicles and presentation by class I MHC molecules B) Invasion of the malaria parasite into the cell cytoplasm leads to processing of parasite proteins and presentation by class I MHC molecules C) The malaria genome encodes cell surface glycoproteins that mimic class I MHC molecules D) The malaria parasite incorporates host class I MHC molecules into its plasma membrane E) The plasma membrane of the malaria parasite contains a superantigen specific for CD8+ T lymphocytes Correct Answer: B. Invasion of the malaria parasite into the cell cytoplasm leads to processing of parasite proteins and presentation by class I MHC molecules that are recognized by CD8+ T lymphocytes. This activates CD8+ T lymphocytes leading to the direct killing of cells infected with malaria. Malaria is transmitted when a sporozoite is introduced into the dermis during a mosquito (Anopheles) blood meal, followed by transit of this parasite to the hepatocytes, which it infects. Parasites evolve into merozoites and are released into the circulation to infect erythrocytes. Within the infected nucleated cells, including hepatocytes, the malaria parasite invades the cell cytoplasm. Broken down malarial antigens are processed in the rough endoplasmic reticulum onto class I MHC molecules, which are presented at the cell surface. Recognition of malarial antigens on MHC class I molecules by CD8+ T cells leads to the proliferation of malaria-specific CD8+ T cells, release of interferon-gamma and tissue necrosis factor, and subsequent initiation of apoptotic pathways through expression of perforin and granzyme, resulting in the death of infected hepatocytes. It is hypothesized that CD8+ T cells may be able to recognize and kill erythrocytes that transiently express MHC molecules. Incorrect Answers: A, C, D, and E. Endocytosis of malaria parasites leads to processing of parasite proteins in endosomal vesicles and presentation by class I MHC molecules (Choice A) is an incorrect statement because parasites do not enter the cell via endocytosis; they invade the cell directly through the cell membrane and form a protective shell using the cellular membrane called a parasitophorous vacuole membrane. The malaria genome encodes cell surface glycoproteins that mimic class I MHC molecules (Choice C) is not correct. Malaria antigens are processed and expressed as foreign antigens on the surface bound to MHC class I molecules. The malaria parasite incorporates host class I MHC molecules into its plasma membrane (Choice D) is not the correct answer. While developing parasites that are replicated in the hepatocytes require synthesis of a new plasma membrane to surround new parasites, they do not routinely incorporate human MHC I molecules into their membrane. The plasma membrane of the malaria parasite contains a superantigen specific for CD8+ T lymphocytes (Choice E) is incorrect. The presence of a superantigen has been proposed since patients who are naïve to malaria will generate some degree of immune response upon exposure, but this mechanism is believed to be mediated by CD4+ T lymphocytes.

Objective: The malaria parasite enters hepatocytes by direct invasion where it replicates. The malarial antigens are processed in the rough endoplasmic reticulum onto class I MHC molecules, which are then expressed at the cell surface where CD8+ T lymphocytes recognize, activate, and proliferate, resulting in subsequent cytotoxic cell death. Previous Next Score Report Lab Values Calculator Help Pause

191 Exam Section 4: Item 42 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 42. A 66-year-old woman comes to the physician 1 day after the sudden onset of weakness of her right arm and leg. She has a 6-year history of breast cancer. On physical examination, the tongue deviates to the left when protruded. Neurologic examination shows a right pronator drift and right-sided hemiparesis of the upper and lower extremities. Sensation is intact. Injury to which of the following labeled structures in the photograph of the brain best explains this patient's hemiparesis? A C D F Right Left A) B) C) D) E) F) Correct Answer: F. This patient likely has an infarct of her left medullary pyramid (Choice F). The upper portions of the medullary pyramid carry lateral corticospinal tract fibers innervating the contralateral side of the body. 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 (inferior-most aspect of the medullary pyramids), and then descend contralaterally in the spinal cord to synapse with the contralateral lower motor neuron in the ventral horn of the spinal cord. A lesion in the left medullary pyramid at the level of the inferior olivary nuclei (white structures directly anterior and lateral to the medullary pyramids), typically leads to a right- sided (contralateral) upper motor neuron pattern of weakness involving both upper and lower extremities (eg, spastic paralysis, pronator drift, Babinski sign, and hyperreflexia). There may also be associated damage to the ipsilateral hypoglossal cranial nerve XIIl nucleus. This would manifest with deviation of the tongue toward the affected side. Absent tone on the left side results in deviation toward the left. Incorrect Answers: A, B, C, D, and E. The inferior cerebellar peduncles (Choices A and B) integrate proprioceptive information with vestibular function. Lesions of the inferior cerebellar peduncles would therefore cause postural imbalance, vertigo, and nystagmus. The medial lemnisci (Choices C and D) are white matter tracts located posteriorly to the crus cerebri. The medial lemnisci carry fibers from the dorsal column tract, which is responsible for the sensation of pressure, fine touch, and proprioception. This patient demonstrated motor rather than sensory deficits. The right medullary pyramid (Choice E) carries lateral corticospinal tract fibers innervating the contralateral side of the body. Therefore, lesions of the right medullary pyramid would lead to a left- sided upper-motor neuron pattern of weakness of both the upper and lower extremities, which is the opposite of this patient's presentation.

Objective: The medullary pyramid carries lateral corticospinal tract fibers innervating the contralateral side of the body. Lesions of the medullary pyramid lead to upper motor neuron- pattern weakness of the contralateral upper and lower extremities. Previous Next Score Report Lab Values Calculator Help Pause E.

101 Exam Section 3: Item 2 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 2. In a historical study, 4753 nulliparous women with confirmed pregnancies were randomized to either folic acid supplementation or trace element supplementation. In both the folic acid and trace element groups, 88% of the women had pregnancies ending in live birth, late fetal death, stillbirth, or termination of pregnancy after prenatal diagnosis of a defect. The following pregnancy outcomes were found: Drug Neural tube defects Folic Acid 0% 1.3% Trace Elements 0.29% 2.3% All congenital anomalies Based on these data, using the trace element group as a control group, which of the following best represents how many nulliparous women with confirmed pregnancies would have to be treated with folic acid to prevent one congenital abnormality? A) 0 B) 0.29 C) 1.3 D) 2.3 E) 80 F) 100 G) 120 H) 344 1) 4756 Correct Answer: F. Number needed to treat (NNT) is an epidemiological measure used to characterize an intervention that clarifies the effectiveness of that intervention. Formally, the NNT is the number of patients who must receive an intervention in order to prevent the occurrence of one negative outcome. For example, a cardiovascular drug which needs to be given to 100 people in order to prevent one cardiovascular event would have an NNT of 100. The NNT is a useful measure for the evaluation of an intervention as it can be intuitively understood by clinicians and the general population, whereas relative risk reduction and absolute risk reduction calculations are more abstract and difficult to translate to the number of people helped by an intervention. In this study, the rate of congenital anomalies in the folic acid group (experimental group) is 1.3%, whereas the rate of congenital anomalies in the trace elements group (control group) is 2.3%. Finding the difference between these two risks provides the absolute risk reduction (ARR) of folic acid, 1%. Thus, if 100 mothers are treated, one child would stand to benefit from the treatment. The NNT is therefore 100 persons. NNT is related to the ARR via the following equation: NNT=1/ARR. Incorrect Answers: A, B, C, D, E, G, H, and I. 0,0.29, 1.3, and 2.3 (Choices A, B, C, and D) purely reflect the percent incidence in the study of neural tube defects and all congenital anomalies in the folic acid and trace elements groups. None of these values correctly compute the NNT. 80, 120, 344, and 4756 (Choices E, G, H, and I) reflect mathematical errors or incorrect computation of the ARR or NNT.

Objective: The number needed to treat is the number of individuals who need to receive an intervention in order to prevent one negative outcome. The number needed to treat is the inverse of the absolute risk reduction. I3D Previous Next Score Report Lab Values Calculator Help Pause

130 Exam Section 3: Item 31 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 31. A 13-year-old girl is brought to the physician by her mother because of several blemishes on her face and back for 8 months. Physical examination shows scattered comedones and papules on the face and back. Treatment with a medication that decreases cohesion between epidermal cells and increases epidermal cell turnover is initiated. Which of the following drugs was most likely prescribed? A) Benzoyl peroxide B) Clindamycin C) Metronidazole D) Tetracycline E) Tretinoin Correct Answer: E. The initial lesion of acne vulgaris is a comedone, a hair follicle that has been blocked by keratin debris. The inciting event in comedone formation is hyperproliferation of the epidermis and abnormal keratinization. This is compounded by androgenic stimulation of sebaceous glands associated with the hair follicle, which are together called a pilosebaceous unit, leading to increased sebum production. This increased sebum provides a substrate for bacterial overgrowth of normal skin flora including Staphylococcus epidermidis and Propionibacterium acnes. With the accumulation of keratin debris, increased sebum production, and bacterial overgrowth, the comedone becomes inflamed and ruptures, causing a papule or cyst to form. Clinically, areas with increased sebaceous activity including the face, upper back, and chest are prone to acne. It is a common disorder that affects the vast majority of adolescents. Treatments are aimed at decreasing comedone formation (retinoids), decreasing sebum production (antiandrogens), and mitigating bacterial overgrowth (antibiotics). Tretinoin, a vitamin A analog, is used as the first-line management of mild and moderate acne. It binds the retinoic acid receptor, a transcription factor, in the nucleus, which increases the transcription of genes involved in normalizing keratinization of the follicular epithelium and increasing cell turnover. This decreases the cohesiveness of the keratinocytes and prevents the initial formation of comedones. Because it works through altering transcription, it is best used as a preventative therapy rather than to treat pre-existing acne lesions. Incorrect Answers: A, B, C, and D. Topical benzoyl peroxide (Choice A) and clindamycin (Choice B) are both commonly used to treat mild and moderate acne vulgaris. Both treat the bacterial overgrowth component of acne and should be used in combination to prevent bacterial resistance, which develops when clindamycin is used alone. These treatments neither normalize keratinization of the follicular epithelium nor increase cell turnover. Topical metronidazole (Choice C) is used to treat rosacea but not acne. In rosacea, metronidazole is used for its anti-inflammatory rather than antibacterial effects. It does not alter epidermal cell turnover, cohesion, or keratinization. Oral tetracyclines including tetracycline (Choice D), doxycycline, and minocycline may be used for moderate to severe inflammatory acne. While there is a component of bacterial overgrowth to the pathogenesis of acne, these antibiotics are used primarily for their anti-inflammatory effect. To prevent bacterial resistance, they may be used at lower doses than would be required when treating an infection. All patients prescribed oral tetracyclines should be counseled on the possibility of gastrointestinal upset, phototoxicity, and pill esophagitis. Minocycline has an additional risk for causing drug-induced lupus and pigmentary changes to the skin.

Objective: The primary lesion of acne vulgaris, the comedone, is formed when the epithelial cells of the follicle accumulate and produce excess keratin debris, which then clogs the follicle, allowing for the proliferation of bacteria and eventually follicle rupture. Tretinoin and other retinoids decrease comedone formation and are used as the first-line management of mild and moderate acne. Previous Next Score Report Lab Values Calculator Help Pause

168 Exam Section 4: Item 19 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 19. An 86-year-old man is brought to the physician because of a 3-week history of several minor wounds that are healing slowly. He lives alone, and his diet consists of toast and tea. Physical examination shows several ecchymoses and numerous red and purple macules around hair follicles. Which of the following best explains these findings? A) Decreased synthesis of coagulation factors B) Inadequate hydroxylation of procollagen C) Increased hydroxylation of lysine D) Increased rate of synthesis of collagen peptides E) Oxidative injury to mature erythrocytes Correct Answer: B. Vitamin C (ascorbic acid) is an antioxidant, facilitator of iron absorption, and coenzyme in the synthesis of collagen via prolyl hydroxylase and neurotransmitters via dopamine hydroxylase. Deficiency in vitamin C leads to inadequate hydroxylation of procollagen and presents with signs and symptoms of impaired collagen synthesis including swollen, bleeding gums, easy bruising and bleeding (eg, hemarthrosis), petechiae, impaired wound healing, perifollicular hemorrhages, and short, fragile, curly hair. The collagen and connective tissue deficiency weaken the walls of blood vessels, resulting in easy bruising and bleeding. Collagen is synthesized by fibroblasts and begins in the rough endoplasmic reticulum with translation of collagen chains, which are glycine- and proline-rich. Prolyl hydroxylase, in a reaction requiring vitamin C, hydroxylates proline and lysine residues. This step, along with glycosylation, forms alpha-chains through disulfide bridging plus hydrogen bonding. Procollagen is exocytosed, where it forms tropocollagen after removal of the terminal ends. It is cross-linked extracellularly in a reaction that requires copper. In this case, the patient's perifollicular hemorrhages and ecchymoses in the setting of a deficient diet suggest a diagnosis of scurvy caused by impaired collagen synthesis. Incorrect Answer: A, C, D, and E. Decreased synthesis of coagulation factors (Choice A) can occur with vitamin K deficiency. 2,3-Epoxide reductase is an enzyme involved in hepatic synthesis of clotting factors and involves vitamin K as a coenzyme. A deficiency in vitamin K leads to coagulopathy with increased prothrombin time and activated partial thromboplastin time. Increased hydroxylation of lysine (Choice C) and increased rate of synthesis of collagen peptides (Choice D) are the opposite of the problem in this patient. Vitamin C deficiency leads to impaired hydroxylation of procollagen and impaired collagen synthesis. Vitamin C is an antioxidant that decreases oxidative injury to mature erythrocytes (Choice E), but this would not explain the symptoms of scurvy including bruising and perifollicular hemorrhages in this patient.

Objective: Vitamin C (ascorbic acid) is an antioxidant and coenzyme in the synthesis of collagen via prolyl hydroxylase. Deficiency is common in persons having diets poor in fruits and vegetables, and results in scurvy, which presents with blood vessel fragility (easy bruising, petechiae), impaired wound healing, and disordered hair growth. 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. A 40-year-old woman at 5 months' gestation comes to the physician for amniocentesis. Results show a normal 46,XY karyotype of the fetus. Four months later, the newborn is delivered and examination shows a female phenotype with mild clitoral enlargement. Ultrasonography shows the presence of male but not female genital ducts. This newborn most likely has a mutation of the gene of which of the following factors? A) Anti-paramesonephric (müllerian) hormone B) 5a-Reductase C) SRY protein D) Steroid sulfatase E) Zinc finger transcription factor Wilms tumor 1 Correct Answer: B. 5a-Reductase is an enzyme that catalyzes the metabolism of testosterone to dihydrotestosterone (DHT). Testosterone promotes the development of the mesonephric (wolffian) duct that develops into the seminal vesicles, epididymis, vas deferens, and ejaculatory duct. DHT promotes the development of male external genitalia and the prostate from the genital tubercle and urogenital sinus. Individuals with 5a-reductase deficiency have defective conversion of testosterone to DHT, leading to decreased concentrations of DHT and impaired virilization of the male urogenital tract. Individuals with 5a-reductase deficiency therefore appear phenotypically female externally or may have ambiguous external genitalia. Individuals have normal male internal genitalia as a result of normal concentrations of testosterone. Incorrect Answers: A, C, D, and E. Anti-paramesonephric (müllerian) hormone (Choice A) or müllerian inhibiting factor (MIF) is secreted by Sertoli cells and suppresses the development of the paramesonephric (müllerian) duct that would have developed into female internal genital structures such as the fallopian tubes, uterus, and upper vagina. The SRY gene is located on the Y chromosome and is responsible for producing testis-determining factor, also known as the SRY protein (Choice C), for testes development. Following testes 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. Steroid sulfatase (Choice D) is an enzyme that catalyzes the conversion of sulfated steroid precursors to the active steroid. Examples include forming unconjugated dehydroepiandrosterone, estrone, pregnenolone, and cholesterol. Deficiency in this enzyme may congenitally manifest as X-linked ichthyosis, characterized by hyperkeratosis, rashes, and scaly skin caused by accumulation of cholesterol sulfate in the skin, resulting in impaired fluidity of the plasma membrane and adhesion between cells. Zinc finger transcription factor Wilms tumor 1 (Choice E) is a tumor suppressor gene on chromosome 11. Nephroblastoma is the most common renal malignancy in childhood caused by mutations in tumor suppressor genes WT1 or WT2. It is characterized by a large, often palpable, unilateral flank mass and hematuria.

Objective: 5a-Reductase catalyzes the metabolism of testosterone to DHT. Individuals with 5a-reductase deficiency exhibit decreased concentrations of DHT and impaired virilization of the male urogenital tract. Individuals with 5a-reductase deficiency appear phenotypically female externally or may have ambiguous external genitalia, with normal male internal genitalia. II Next Score Report Lab Values Calculator Help Pause

80 Exam Section 2: Item 31 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 31. A 10-year-old girl is brought to the physician by her mother because of increased hair growth on her face and increased muscle mass during the past 3 months. Breast development is Tanner stage 1, and pubic hair development is Tanner stage 5. Physical examination shows clitorimegaly. Pelvic examination shows a normal-appearing vagina. Serum studies show: Luteinizing hormone Dehydroepiandrosterone Dehydroepiandrosterone sulfate Testosterone 0.8 mlU/mL (N=1.0-5.9) 6.2 nmol/L (N=5.2-19.8) 0.9 umol/L (N=0.9-3.4) 14 nmol/L (N=0.2-0.7) An unregulated increase in hormone production by which of the following types of cells is the most likely cause of this patient's hirsutism? A) Adrenal zona fasciculata cells B) Adrenal zona glomerulosa cells C) Ovarian follicle cells D) Ovarian Sertoli-Leydig cells E) Pituitary acidophils F) Pituitary basophils Correct Answer: D. This patient is presenting with clinical signs and symptoms of virilization, the condition in which females develop male features a a result of increased concentrations of androgens. These features include hirsutism (the presence of unwanted hair in a male distribution), increased muscle mass, pubic hair, and clitoromegaly and are confirmed by her increased testosterone concentration. Testosterone may be increased in the setting of a hormone-secreting ovarian tumor, polycystic ovarian syndrome, hypersecretion of androgens by the adrenal gland, or an adrenal tumor. An increased testosterone concentration causes negative feedback on the pituitary gland, thereby lowering the luteinizing hormone (LH) concentration. When evaluating virilization and hirsutism, it is critical to measure the concentrations of testosterone, dehydroepiandrosterone (DHEA), and dehydroepiandrosterone sulfate (DHEA-S); the latter two are made by the adrenal gland and help to narrow the differential diagnosis. In this case, because the testosterone is increased while DHEA and DHEA-S are within normal limits, the culprit is likely of an ovarian source. À Sertoli-Leydig cell tumor is a rare androgen-producing, ovarian sex-cord stromal tumor. It is a neoplastic proliferation of the cells that give rise to the specialized gonadal stroma surrounding the granulosa and theca cells in women and the Sertoli and Leydig cells in men. These proliferating cells have the capacity for differentiation into many cell types, including Sertoli-Leydig cells, which secrete testosterone. As in this case, the presence of a Sertoli-Leydig cell tumor may cause precocious puberty in boys or virilization in girls. Incorrect Answers: A, B, C, E, and F Though not in this case, virilization may also be caused by the increased adrenal secretion of androgens, which are made in the adrenal zona reticulata. The adrenal zona fasciculata (Choice A) secretes glucocorticoids, such as cortisol, which in excess cause hypertension, insulin resistance, lipodystrophy, and myopathy. The adrenal zona glomerulosa (Choice B) secretes the mineralocorticoid aldosterone, which is a key component of the renin-angiotensin-aldosterone pathway for regulating sodium and water homeostasis. These zones do not secrete androgens and would not explain the patient's virilization. Furthermore, if the adrenal gland were involved, increased concentrations of DHEA and DHEA-S would be expected along with increased testosterone. The ovarian follicle cells (Choice C) are granulosa cells and theca cells. These two cells work in tandem to secrete estrogen, not testosterone. They are under the control of follicle-stimulating hormone (FSH) and luteinizing hormone (LH). The pituitary acidophil (Choice E) is a cell type found in the anterior pituitary that secretes prolactin and growth hormone. Increased secretion of these hormones would not cause hirsutism. The pituitary basophil (Choice F) is a cell type found in the anterior pituitary. The secretory products of this cell type are FSH, LH, adrenocorticotropic releasing hormone, and thyroid-stimulating hormone. While increased production of LH could secondarily trigger the increased production of testosterone, this patient's increased testosterone and low LH concentration suggest that the hormone is coming from a source downstream to the pituitary gland, such as the ovary.

Objective: A Sertoli-Leydig cell tumor is a rare androgen-producing, ovarian sex-cord stromal tumor, which may cause precocious puberty or virilization. Hirsutism (the presence of unwanted hair in a male distribution), increased muscle mass, pubic hair, and clitoromegaly are features of virilization. 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 3-year-old girl is brought to the emergency department by her mother 1 hour after she was found with a half-empty bottle of her grandmother's diabetes medication. The mother tells the physician that the child consumed approximately 25 metformin tablets. Physical examination shows no abnormalities. This patient is at greatest risk for which of the following serum abnormalities? A) Decreased calcium concentration B) Decreased glucose concentration C) Decreased sodium concentration D) Increased AST and ALT activities E) Increased creatinine concentration F) Increased lactic acid concentration Correct Answer: F. Metformin is an oral biguanide agent used in the management of type 2 diabetes mellitus. It decreases gluconeogenesis, increases peripheral tissue glucose uptake, and decreases serum free fatty acid concentration. Because of its efficacy, tolerability, safety, and low cost, it serves as the first-line treatment for type 2 diabetes mellitus. Although it is generally safe and well-tolerated, there are a few contraindications to metformin, including impaired renal function (caused by its excretion in the urine) and active liver disease. Metformin may cause several adverse effects, including nausea, abdominal discomfort, diarrhea, and (rarely) increased lactic acid concentration resulting in lactic acidosis. Although the incidence is rare, metformin-associated lactic acidosis has a high mortality rate. Treatment is supportive, and patients with severe lactic acidosis may require hemodialysis. Metformin does not generally cause hypoglycemia. Incorrect Answers: A, B, C, D, and E. Decreased calcium concentration (Choice A) is not a common side effect of metformin. Since metformin does not stimulate insulin release, hypoglycemia and decreased glucose concentration (Choice B) are not common. Hypoglycemia can be an adverse effect in the setting of an insulin or sulfonylurea overdose. Decreased sodium concentration (Choice C) can occur in syndrome of inappropriate antidiuretic hormone secretion (SIADH) resulting in excessive free water retention and euvolemic hyponatremia. SIADH is a common side effect of many medications (eg, cyclophosphamide) but not metformin. Increased AST and ALT activities (Choice D), seen in medication-induced hepatotoxicity, can be caused by oral hypoglycemic drugs such as glitazones (eg, pioglitazone, rosiglitazone). They increase peripheral tissue insulin sensitivity and can be hepatotoxic. Increased creatinine concentration (Choice E) can be seen in acute kidney injury, for example, caused by nephrotoxic medications (eg, vancomycin). Acute interstitial nephritis can be caused by a hypersensitivity reaction to drugs including diuretics, proton pump inhibitors, sulfonamides, and nonsteroidal anti-inflammatory drugs. Metformin does not cause acute kidney injury, but metformin toxicity in patients with underlying kidney injury may increase the risk for metformin-associated lactic acidosis.

Objective: Metformin is an oral biguanide agent used in the management of type 2 diabetes mellitus. It decreases gluconeogenesis, increases peripheral tissue glucose uptake, and decreases serum free fatty acid concentration. Metformin may cause several adverse effects, including nausea, abdominal discomfort, diarrhea, and (rarely) lactic acidosis. Previous Next Score Report Lab Values Calculator Help Pause

85 Exam Section 2: Item 36 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 36. A 3-year-old boy is brought to the physician because of a 4-hour history of abdominal pain and vomiting. His mother says that she also has occasionally noticed several 20-cm, white worms in his stool during the past month. He recently immigrated to the USA from India with his family. Abdominal examination shows marked distention and tympany. Bowel sounds are high pitched. Which of the following is the most likely causal organism? A) Ancylostoma caninum B) Ascaris lumbricoides C) Enterobius vermicularis D) Strongyloides stercoralis E) Trichuris trichiura Correct Answer: B. Ascaris lumbricoides infection likely explains this patient's abdominal pain, vomiting, and presence of large white worms in the stool. Infection with Ascaris lumbricoides is called ascariasis. This worm is endemic to India and other parts of Asia, Africa, and South America, with transmission occurring via food or water contaminated with eggs, which must embryonate in the soil before becoming infectious. Ingested eggs hatch within days in the small intestine, releasing larvae that travel into the colon, followed by larval infiltration of the intestinal wall with subsequent hematogenous spread within the portal venous system to the liver. From here larvae can move to the heart and finally to the lungs where they mature over a period of two weeks, ascend the bronchi, and travel back down the esophagus after patients cough them up and swallow them. At this point, the larvae mature into adult worms that can measure up to 35 centimeters in length. Adult worms produce eggs that are released in the stool and begin the infectious cycle again. Adult worms live up to two years, and autoinfection does not occur, although reinfection is common. Clinical manifestations are related to the number of worms and the stage of the life cycle. Patients with a high volume of worms can present with bowel obstruction, which is likely the case in this patient presenting with abdominal pain and distension, vomiting, and borborygmi (high-pitched, hyperactive bowel sounds). Pulmonary manifestations include cough and eosinophilic pneumonitis. Some patients are entirely asymptomatic as a result of low worm burden. Incorrect Answers: A, C, D, and E. Ancylostoma caninum (Choice A) is a hookworm species that causes intestinal infection in dogs, not in humans. Enterobius vermicularis (Choice C) is an intestinal roundworm that causes perianal pruritus at night. It is common in children and can be diagnosed with the tape test, which involves applying clear tape to the perianal region in the morning before defecation. Eggs can be directly visualized by microscopy. Large worms are not characteristic of this infection. Strongyloides stercoralis (Choice D) is an intestinal roundworm that can cause persistent infection in humans via autoinfection. It is a common cause of peripheral eosinophilia in patients from endemic regions and can cause severe, disseminated disease with high mortality in patients with immunocompromising conditions such as HIV. Like Ascaris lumbricoides, it has both intestinal and pulmonary phases, but the mature worms are only 2-3 millimeters in length. Trichuris trichiura (Choice E) is an intestinal worm that has a similar life cycle to A. lumbricoides, but adult worms are only about 4 centimeters in length. Additionally, symptoms more commonly involve nocturnal diarrhea, hematochezia, and dysentery. Rectal prolapse may also occur.

Objective: Patients from endemic areas with large roundworms visualized in the stool and symptoms of bowel obstruction should be suspected of having ascariasis, which is caused by the intestinai roundworm, Ascaris lumbricoides. Previous Next Score Report Lab Values Calculator Help Pause

39 Exam Section 1: Item 40 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 40. A study is conducted to assess the effects of aging on cognitive and physiologic function. Participants are recruited via advertisements in the local and regional news media, and all persons who answer the advertisement are selected for study. Which of the following best describes this sampling technique? A) Case-control B) Geographic C) Population-based control D) Self-selection E) Stratified randomization Correct Answer: D. Self-selection occurs when individuals select for themselves to participate in a study. This leads to a potential bias known as self-selection bias. An example of this could be a voluntary, internet- based survey. Such a survey would be limited to persons who use the internet, and then only those who decided to participate. Therefore, non-internet users and persons who decide for whatever reason not to participate would be systematically underrepresented in the study. In this study about aging on cognitive and physiologic function, the sampling technique presented self- selects for individuals who utilize local and regional news media, and who are sufficiently cognitively intact to understand and respond to the request for participation, both potential sources of selection bias. Incorrect Answers: A, B, C, and E. A case-control study (Choice A) investigates an association between an exposure and an outcome. In this study design, a group of patients with the disease (cases) are identified by the investigators and controls are identified from the same population using a matching process. Geographic sampling (Choice B) chooses individuals from a population on the basis of geographic location. For example, sampling of the population of a city may select certain numbers of individuals from each city block. Population-based control (Choice C) refers to when the control group is selected directly from the population of interest. This is often done in case-control studies when the control group is selected directly from the same population from which the cases are selected. Stratified randomization (Choice E) is a process of randomization in which study subjects are enrolled and then stratified into groups on the basis of factors that may influence the outcome of the study. After stratification, the individuals are randomized to intervention groups within each stratum.

Objective: Self-selection sampling occurs when individuals decide for themselves to participate in a study. This leads to self-selection bias, a methodological weakness in population survey-based study designs. %3D Previous Next Score Report Lab Values Calculator Help Pause

45 Exam Section 1: Item 46 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 46. A7-year-old girl is brought to the physician by her mother because of increasingly severe left knee pain during the past 12 hours. Her temperature is 39.1°C (102.3°F), pulse is 115/min, respirations are 18/min, and blood pressure is 110/60 mm Hg. Physical examination shows a tender, swollen left knee that is unable to be flexed because of pain. Hematologic studies show a leukocyte count of 35,000/mm3 and erythrocyte sedimentation rate of 120 mm/h. Which of the following is the most likely diagnosis? A) Juvenile rheumatoid arthritis B) Lyme disease C) Rheumatic fever D) Septic arthritis E) Viral synovitis Correct Answer: D. Septic arthritis commonly occurs in children. Risk factors include trauma to the joint permitting translocation of cutaneous organisms as well as procedures causing transient bacteremia (eg, dental procedures). The most common organism found on aspiration is Staphylococcus aureus. Septic arthritis classically presents with systemic signs and symptoms of illness such as fever, chills, myalgias, arthralgias, and nausea, along with a painful, erythematous, swollen joint with limited range of motion on examination. Patients may refuse to bear weight. In children, septic arthritis commonly affects the knee and hip. It may occur concomitantly with osteomyelitis of adjacent bone. Serum laboratory values typically disclose leukocytosis, as well as increased erythrocyte sedimentation rate and C-reactive protein concentrations. Synovial fluid aspiration will demonstrate a high concentration of white blood cells and may show organisms on Gram stain with growth on cultures. This patient's presentation is highly consistent with monoarticular septic arthritis. Incorrect Answers: A, B, C, and E. Juvenile rheumatoid arthritis (Choice A) (juvenile idiopathic arthritis) occurs before the age of 16 and classically presents with joint pain and inflammation along with extra-articular findings such as uveitis, lymphadenopathy, and rash. It does not present with signs of sepsis. Lyme disease (Choice B) is transmitted by the bite of the Ixodes scapularis tick and is caused by the pathogen Borrelia burgdorferi. It classically presents with a spreading rash (erythema migrans) at the site of the tick bite. Late-stage manifestations include arthritis, heart block, and cranial nerve palsies. Rheumatic fever (Choice C) is a sequela of untreated Group A Streptococcus infection in which there is an autoimmune reaction against host proteins that have similar structure to streptococcal M protein. Its manifestations include carditis, arthritis, subcutaneous nodules, erythema marginatum, and Sydenham chorea. Viral synovitis (Choice E) (transient synovitis) occurs when a viral infection causes inflammation in the synovium of a joint. Commonly, it occurs in pediatric hip joints and may mimic septic arthritis; however, the laboratory studies generally do not show leukocytosis and will likely only show mildly increased erythrocyte sedimentation rate and C-reactive protein. Signs of sepsis would be atypical.

Objective: Septic arthritis is a common surgical emergency in the pediatric population. It presents with acute onset joint pain, erythema, swelling, refusal to bear weight, and signs and symptoms of sepsis. Previous Next Score Report Lab Values Calculator Help Pause

90 Exam Section 2: Item 41 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 41. A 45-year-old woman is brought to the emergency department because of a 6-hour history of constant right flank pain, fever, pain with urination, and blood in the urine. Antibiotic treatment was initiated 2 weeks ago for a urinary tract infection caused by Proteus mirabilis. She is writhing on the hospital bed, unable to find a comfortable position. Her temperature is 38.3°C (101°F), pulse is 110/min, respirations are 30/min, and blood pressure is 160/80 mm Hg. Abdominal examination shows no abnormalities except for tenderness of the right flank to palpation. Urinalysis shows 1+ blood and a pH of 7.5. A spiral CT scan of the abdomen shows a staghorn right renal calculus. This renal calculus is most likely composed of which of the following? A) Calcium oxalate B) Calcium phosphate C) Cystine D) Struvite E) Uric acid Correct Answer: D. Struvite is the most likely substance implicated in the formation of a staghorn renal calculus, especially in the setting of a recent urinary tract infection with Proteus mirabilis, which is a gram- negative bacterium that produces urease. Urease catalyzes the conversion of urinary urea to ammonia and carbon dioxide, which results in urine alkalinization. The solubility of struvite is decreased under alkaline conditions and precipitates as a crystal. An alkaline urine is present in this patient with a urinary pH of 7.5. Struvite crystals have an orthorhombic configuration under light microscopy. Struvite calculi are radiopaque and can be potentially identified by an x-ray or CT scan. Struvite calculi may adopt a branching morphology that molds itself to the shape of the collecting system (staghorn calculi). Given their large size and often ramified structure, it is rare for struvite calculi to pass through the urinary tract, and surgical removal is often necessary along with treatment of the underlying infection. Incorrect Answers: A, B, C, and E. Calcium oxalate (Choice A) is the most common etiology of nephrolithiasis, which is more frequently seen in patients with malabsorptive syndromes. Citric acid decreases the formation of calcium oxalate crystals by forming soluble complexes with calcium ions, and therefore, calcium oxalate calculi tend to form in the setting of hypocitraturia. Calcium oxalate crystals have an octahedral morphology under light microscopy and form radiopaque calculi. Patients can be treated with thiazide diuretics, a low-sodium diet, and supplementation with potassium citrate. Calcium phosphate (Choice B) stones are found in patients with similar risk factors as those who develop calcium oxalate stones. Most patients have consistently increased urinary pH as a result of distal renal tubular acidosis, which predisposes to the formation of calcium phosphate stones. Treatment is similar to that for calcium oxalate, although potassium citrate is usually less effective. Cystine (Choice C) precipitates as a urinary calculus in the setting of increased urinary cystine concentration in patients with a hereditary defect of the renal proximal convoluted tubule that results in decreased reabsorption of cystine, ornithine, lysine, and arginine. Cystine crystals are hexagonal and form radiolucent calculi. The urinary cyanide-nitroprusside test can be used to confirm the diagnosis. Uric acid (Choice E) calculi form most commonly in patients with hyperuricemia (eg, gout), in the setting of rapid cellular turnover caused by malignancy, or in the setting of dehydration. Uric acid crystals have a rhomboid configuration and form radiolucent calculi. Given the absence of these risk factors, the patient's history is more suggestive of a struvite calculus.

Objective: Struvite forms renal calculi in the setting of a urinary tract infection with urease-positive organisms, such as P. mirabilis. Urease catalyzes the conversion of urinary urea to ammonia and carbon dioxide, which results in urine alkalinization. The solubility of struvite is decreased under alkaline conditions and precipitates as a crystal. Struvite stones can become quite large with a branching morphology that molds to the shape of the collecting system and often require surgical removal. Previous Next Score Report Lab Values Calculator Help Pause

69 Exam Section 2: Item 20 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 20. A 5-year-old boy who had intrauterine growth restriction has continued to grow slowly. Psychomotor development is normal. His parents are of average stature. Genetic studies show that he has uniparental maternal heterodisomy for chromosome 7. Which of the following mechanisms best explains his slow growth? A) Expansion of an unstable trinucleotide repeat B) Expression of an autosomal dominant disorder C) Expression of an autosomal recessive disorder D) Genomic imprinting of growth genes E) Somatic mosaicism Correct Answer: D. Uniparental heterodisomy is the inheritance of a pair of nonidentical chromosomes from a single parent that results from an error during meiosis I. In contrast, uniparental isodisomy is the inheritance of a pair of identical chromosomes caused by errors during meiosis Il or postzygotic duplication of the chromosome. Uniparental maternal heterodisomy for chromosome 7 is associated with intrauterine growth restriction and short stature without impairment of psychomotor development. This is caused by genomic imprinting of growth genes. Genomic imprinting describes epigenetic modification in which one gene copy is silenced by methylation depending on the parent-of-origin. Genomic imprinting may be complete or partial, with the methylated allele suppressed rather than silenced. Other disorders resulting from uniparental disomy include Angelman syndrome and Prader-Willi syndrome. Incorrect Answers: A, B, C, and E. Expansion of an unstable trinucleotide repeat (Choice A) is the mechanism of disorders such as Huntington disease, myotonic dystrophy, fragile X syndrome, and Friedreich ataxia. These disorders may display anticipation, the phenomenon in which a genetic condition tends to increase in severity or present at an earlier age as it passes through multiple generations. Expression of an autosomal dominant disorder (Choice B) occurs when the disease phenotype is expressed with only a single copy of the defective allele (heterozygous) or two copies of the defective allele (homozygous). Autosomal dominant genetics follow Mendelian patterns of inheritance. Expression of an autosomal recessive disorder (Choice C) occurs when the disease phenotype is expressed only if two copies of the defective allele are present. Autosomal recessive genetics follow Mendelian patterns of inheritance. Somatic mosaicism (Choice E) occurs when a mutation alters a gene during the early mitotic divisions after fertilization, resulting in the proliferation of two or more distinct somatic cell populations with different genomes in the developing organism. McCune-Albright syndrome is an example of this phenomenon.

Objective: Uniparental disomy is an example of genomic imprinting, in which epigenetic modification to the genome occurs by selective methylation and suppression of an allele dependent on the parent-of-origin. Uniparental maternal heterodisomy of chromosome 7 is an established disorder that results in the suppression of growth genes caused by imprinting of both maternal alleles. %3D 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. A 16-year-old girl with cystic fibrosis is brought to the physician because of a 3-week history of generalized weakness, numbness and tingling of her arms and legs, and difficulty walking. She has not adhered to her medication regimen during the past 6 months. She appears alert and oriented. Her vital signs are within normal limits. Physical examination shows bilateral weakness and decreased deep tendon reflexes in the upper and lower extremities. She walks with an ataxic gait. The most likely cause of these findings is a deficiency of which of the following? A) Folic acid B) Vitamin A C) Vitamin B, (pyridoxine) D) Vitamin D E) Vitamin E Correct Answer: E. Vitamin E is an antioxidant that protects erythrocytes and 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. In contrast to vitamin B12 deficiency, patients do not have megaloblastic anemia, hypersegmented neutrophils, or increased serum methylmalonic acid concentrations. On peripheral smear, patients with vitamin E deficiency may have acanthocytosis. Incorrect Answers: A, B, C, and D. Folic acid (Choice A) is converted to tetrahydrofolic acid and used as a coenzyme in the synthesis of nucleotides and nucleosides. Folate is contained in leafy vegetables and absorbed in the jejunum. Folate deficiency is often seen in patients with malnutrition, alcoholism, and patients taking anti-folate medications (eg, phenytoin, methotrexate). Megaloblastic anemia occurs in the setting of impaired DNA synthesis. Vitamin A (Choice B) is an antioxidant necessary for differentiation of epithelial cells into specialized tissue. Deficiency is characterized by ocular manifestations including night blindness, corneal degeneration, Bitot spots on the conjunctiva, dry skin, and immunosuppression. Vitamin B6 (pyridoxine) (Choice C) deficiency limits 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 D (Choice D) deficiency commonly comes from low exposure to UV radiation and low dietary vitamin D intake. Fat malabsorption syndromes such as celiac disease or cystic fibrosis can impair absorption of vitamin D in the gastrointestinal tract. Patients will demonstrate symptoms of low bone mineral density (eg, varus bowing of the tibia, osteoporosis), or hypocalcemia (eg, seizures, muscle spasms). Hypoplastic teeth and multiple caries can result, as mineralization of dentin is impaired.

Objective: Vitamin 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 deficiency with posterior column and spinocerebellar tract demyelination. Previous Next Score Report Lab Values Calculator Help Pause

158 Exam Section 4: Item 9 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 9. A 56-year-old man comes to the physician because of a 4-day history of fatigue and malaise. He has hypertension treated with a diuretic and an ACE inhibitor and osteoarthritis and gout treated with nonsteroidal anti-inflammatory drugs. Two weeks ago, allopurinol was added to his medication regimen. His temperature is 38.1°C (100.6°F). Physical examination shows an erythematous rash over the anterior trunk. Laboratory studies show: Serum K+ Ca2+ Urea nitrogen Creatinine 5.6 mEq/L 8.5 mg/dL 55 mg/dL 2.4 mg/dL 4.9 mg/dL 8.4 mg/dL Phosphorus Uric acid Urine pH Protein WBC RBC 7.5 1+ 5-7/hpf 15-20/hpf A renal biopsy specimen shows normal glomeruli and a prominent lymphocytic/histiocytic infiltrate in the periglomerular and peritubular tissues. An occasional eosinophil and rare segmented neutrophils are also present. There is no necrosis. Which of the following is the most likely pathogenesis of the disease process in this patient? A) Immune reaction B) Ischemia secondary to prostaglandin inhibition OC) Ischemia secondary to vascular hyaline deposition D) Renal deposition of Tamm-Horsfall protein E) Vesicoureteral reflux Correct Answer: A. Immune reaction in response to the addition of an offending drug-in this case allopurinol-explains this patient's fever, erythematous rash, acute kidney injury with pyuria and hematuria, and renal biopsy showing periglomerular and peritubular lymphocytic/histiocytic infiltrate. This collection of signs and symptoms in a patient who recently started allopurinol is consistent with acute interstitial nephritis (AİN). The classic triad of fever, rash, and peripheral eosinophilia in the setting of acute kidney injury only occurs in 10% of patients. AIN can occur after exposure to numerous drug classes, but most commonly these include nonsteroidal anti-inflammatory drugs (NSAIDS), diuretics, and sulfonamides, but may also occur in the setting of systemic autoimmune disorders such as lupus or Sjögren syndrome. Unlike other forms of acute kidney injury caused by medications, AIN does not reliably occur at a set interval after medication exposure; the latent period can range from days to months. Laboratory findings often exhibit peripheral eosinophilia and urinary eosinophiluria, although both are nonspecific for AlIN. Urinalysis will also demonstrate pyuria and WBC casts along with variable proteinuria. Renal biopsy is required for definitive diagnosis and will show an interstitial infiltrate consisting primarily of T lymphocytes and monocytes, with occasional eosinophils and plasma cells. These inflammatory cells may invade the tubules to cause tubulitis. Treatment requires removal of the offending agent while providing supportive care for acute kidney injury-related electrolyte and volume derangements. Incorrect Answers: B, C, D, and E. Ischemia secondary to prostaglandin inhibition (Choice B) describes one mechanism by which NSAIDS can cause renal injury. Prostaglandins such as PGE2 and prostacyclin are vasodilatory agents that increase renal blood flow through vasodilation of the afferent arteriole. Inhibition of this action can lead to inappropriately decreased renal blood flow and ischemia which can manifest as acute tubular necrosis. Ischemia secondary to vascular hyaline deposition (Choice C) occurs in severe chronic hypertensive kidney disease. Hypertension leads to intimal thickening and luminal narrowing of the glomerular arterioles in addition to the larger renal arteries. Deposition of hyaline into the arteriolar wall further narrows the lumen. This decreases renal blood flow and results in chronic ischemia. Uncontrolled hypertension is a major contributor to chronic kidney disease. Renal deposition of Tamm-Horsfall protein (Choice D), also known as uromodulin, is associated with several rare genetic forms of kidney disease including nephronophthisis, medullary cystic kidney disease, and familial juvenile hyperuricemic nephropathy. Deposition of uromodulin does not occur in AIN. Vesicoureteral reflux (Choice E) often presents in children with postrenal azotemia and hydronephrosis evidenced by dilated ureters, renal pelves, blunted renal calyces, and/or compressed, atrophic renal parenchyma. The extent and distribution of upper urinary tract dilation and renal parenchymal atrophy depends on the severity of vesicoureteral reflux. It can predispose to pyelonephritis but does not have an association with AIN.

Objective: AIN is a clinical syndrome often characterized by acute kidney injury along with fever, eosinophilia, and rash, although this triad is present in only 10% of patients. It is caused by an inflammatory response to an offending drug or systemic autoimmune disorder, which leads to renal interstitial immune cell infiltration with consequent acute kidney injury. Treatment requires removal of the offending agent. Previous Next Score Report Lab Values Calculator Help Pause

106 Exam Section 3: Item 7 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 7. A75-year-old man has the painless lesions shown. He worked outside for 30 years maintaining the grounds of a golf course. The lesions are rough to the touch. Which of the following is the most likely diagnosis? A) Actinic keratosis B) Angioma C) Seborrheic keratosis D) Solar lentigines 1 cm E) Xanthelasma Correct Answer: A. Actinic keratosis is a premalignant lesion that may progress to squamous cell carcinoma. Clinically, lesions typically appear as light pink, ill-defined macules with a gritty texture in areas of prolonged sun exposure, such as the face, ears, and dorsal hands. Some actinic keratoses may develop a thick, hyperkeratotic, slightly yellow crust. In actinic keratosis, the basal layer of the epidermis demonstrates atypia, which may extend into the spinous layer. If atypia extends through all layers of the epidermis, it becomes squamous cell carcinoma in situ. When the atypical squamous cells penetrate the basement membrane and begin invading into the dermis, invasive squamous cell carcinoma is diagnosed. Actinic keratoses progress to squamous cell carcinoma at a slow rate of roughly 1% per year. In patients with many actinic keratoses, this risk is compounded, and treatment is warranted. Single actinic keratoses are commonly treated with cryotherapy using liquid nitrogen, while numerous actinic keratoses are treated with field therapy using either topical 5-fluorouracil or blue light photodynamic therapy. Incorrect Answers: B, C, D, and E. Angioma (Choice B) is a benign proliferation of blood vessels, which can either be present at birth, develop shortly after birth, or develop in adulthood. Cherry angiomas are bright red in appearance and are typically seen on the trunk beginning in middle age. They are not related to sun exposure and do not have any malignant potential. Seborrheic keratosis (Choice C) is a benign proliferation of the epidermis; lesions exhibit a greasy, adherent appearance. Seborrheic keratoses are often tan, brown, or black, which is caused by the keratin produced by the epidermis, rather than melanin. They increase in number with increasing patient age and do not demonstrate a predilection for sun-exposed sites. Solar lentigines (Choice D) are benign, smooth, brown macules, which develop in sun-exposed areas with increasing patient age, sometimes called "liver spots." A lentigo that is enlarging or developing multiple colors should raise suspicion for the lentigo maligna subtype of melanoma. Xanthelasma (Choice E) is a yellow-orange papule or plaque typically on the upper eyelids, which is caused by the deposition of lipid containing histiocytes, or macrophages. Xanthelasma often indicates an underlying hyperlipidemia or hypercholesterolemia. Lipid may also deposit on the tendons, forming a tendinous xanthoma, or on the cornea, forming a corneal arcus.

Objective: Actinic keratosis is a premalignant lesion, which may progress to squamous cell carcinoma and is typically seen in areas of prolonged sun exposure. Lesions clinically appear as light pink, ill-defined macules with a gritty texture. Previous Next Score Report Lab Values Calculator Help Pause

107 Exam Section 3: Item 8 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 8. A 79-year-old woman with a history of severe dementia, type 2 diabetes mellitus, and hypertension is brought to the emergency department by her son because of chest pain and agitation for 4 hours. She had smoked 2 packs of cigarettes daily until the age of 70 years, when she quit. The son is not aware of the patient's medication regimen. Her pulse is 120/min, respirations are 32/min, and blood pressure is 180/100 mm Hg. Crackles are heard in both lung bases on auscultation of the chest. Cardiac examination shows a systolic ejection murmur at the apex and a regular rhythm. An ECG shows ST-segment elevation in the anterolateral leads. A chest x-ray shows a mildly enlarged cardiac silhouette. Which of the following is the most likely diagnosis? A) Acute coronary syndrome B) Acute pericarditis C) Bilateral pneumonia D) Cerebrovascular event E) Pulmonary embolism Correct Answer: A. Acute coronary syndrome classically presents with acute-onset chest pain, often radiating to the neck, jaw, or arm, along with shortness of breath, nausea, and/or light-headedness. Risk factors include a history of smoking, diabetes, hypertension, hyperlipidemia, and/or obesity. If the infarction results in left ventricular dysfunction, patients may experience cardiogenic pulmonary edema resulting in shortness of breath, tachypnea, hypoxia, audible crackles (rales), and an S3 or S4 gallop. Diagnosis is made using clinical history, analysis of cardiac biomarkers, and an ECG that may show ST-segment elevation or depression, inverted T-waves, or Q-waves in a particular vascular distribution of the heart. Treatment requires the use of antiplatelet agents (eg, aspirin, clopidogrel) plus anticoagulants (eg, heparin), pain control, and revascularization through angioplasty, thrombolysis, or coronary artery bypass grafting. Complications include papillary muscle rupture with subsequent mitral regurgitation, dysrhythmias, cardiogenic shock, pulmonary edema, ventricular wall rupture, and postinfarction pericarditis. Incorrect Answers: B, C, D, and E. Acute pericarditis (Choice B) occurs as a result of an infectious, inflammatory, or malignant precipitant, and presents as pleuritic chest pain that is exacerbated with lying down. ECG findings include diffuse (not focal) ST elevations and PR depressions. Bilateral pneumonia (Choice C) presents with cough, shortness of breath, pleuritic chest pain, fever, and abnormal breath sounds such as rhonchi. It often follows a viral upper respiratory infection or aspiration event. The ECG is often unremarkable, and the chest x-ray can be normal early in the course but may show patchy multifocal infiltrates or regions of consolidation. Cerebrovascular event (Choice D) describes an ischemic or hemorrhagic loss of blood supply to the brain. Risk factors include smoking, hypertension, diabetes, carotid or intracranial atherosclerotic disease, history of hypercoagulability, atrial fibrillation, and age. Classically, cerebrovascular events manifest as a neurologic deficit related to the affected part of the brain, though patients may also present with nonspecific encephalopathy. ECG changes are not expected. Pulmonary embolism (Choice E) presents with acute chest pain, shortness of breath, and hypoxia, often in a patient with deep venous thrombosis or a known hypercoagulable state. Small pulmonary emboli are often undetectable on chest x-ray, and the ECG may be normal or show sinus tachycardia. Large emboli with infarction are often seen on chest x-ray by secondary signs such as wedge-shaped areas of infarction or via the Westermark sign of unilateral oligemia. The ECG in a pulmonary embolus often shows sinus tachycardia, and if there is right heart strain can also exhibit right-axis deviation, S wave in lead 1, Q wave in lead 3, and inverted T wave in lead 3 ('S1Q3T3'). ST segment elevation would not be expected.

Objective: Acute coronary syndrome with myocardial infarction results in ST-segment elevations in contiguous leads. Complications of cardiomyocyte injury and necrosis include papillary muscle rupture with subsequent valvular insufficiency, dysrhythmias, cardiogenic shock, pulmonary edema, ventricular wall rupture, and postinfarction pericarditis. Previous Next Score Report Lab Values Calculator Help Pause

110 Exam Section 3: Item 11 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 11. A 70-year-old man has coronary artery bypass grafting and aortic valve replacement. Immediately after the operation, his blood pressure is 8 0/40 mm Hg, and he requires an intra-aortic balloon pump and intravenous infusion of dopamine. After 24 hours of treatment, his blood pressure is 140/80 mm Hg. Serum creatinine concentration is 4.5 mg/dL (preoperative value 1 mg/dL). Urine output is 15 mL/h. Which of the following best explains these findings? A) Acute interstitial nephritis from dopamine B) Acute tubular necrosis from ischemia C) Atheromatous emboli dislodged by the balloon pump D) Normal postoperative renal function E) Toxic acute tubular necrosis induced by hemolysis of erythrocytes Correct Answer: B. Acute kidney injury (AKI) is defined as an acute increase in creatinine by greater than or equal to 0.3 mg/dL or an increase greater than or equal to 1.5 times the baseline creatinine, occasionally with oliguria or anuria. Patients may be asymptomatic or demonstrate symptoms of uremia (eg, fatigue, nausea, vomiting, pruritus, altered mental status), hyperkalemia (eg, peaked T-waves on ECG), and hypervolemia if the AKI is severe. The differential diagnoses for the acute change in renal function can be subdivided into prerenal, intrinsic, and postrenal causes. Prerenal causes include sepsis, hypovolemia, hypotension, and cardiogenic shock and lead to decreased filtration as a result of decreased perfusion pressure at the glomeruli. Intrinsic causes include nephritic and nephrotic syndromes, interstitial nephritis, and acute tubular necrosis, the latter of which can be caused by ischemia as a result of severe prerenal disease. Postrenal causes include obstructive pathologies, such as nephrolithiasis or benign prostatic hyperplasia. The most common cause of acute kidney injury, particularly in the setting of prolonged hypotension, is acute tubular necrosis from ischemia. Acute tubular necrosis can also be caused by nephrotoxins, such as myoglobin, hemoglobin, and ethylene glycol. The diagnosis can be made clinically and with laboratory evaluation including urinalysis, which will show muddy brown casts, and urine electrolytes, which will demonstrate an increased fractional excretion of sodium caused by the inability of the proximal tubules to reabsorb sodium. Further laboratory evaluation will also showa BUN to creatinine ratio of less than 20. Treatment is supportive, but the patient may require hemodialysis in severe cases. Incorrect Answers: A, C, D, and E. Acute interstitial nephritis (Choice A) most commonly presents as a hypersensitivity reaction following exposure to certain medications such as antibiotics, diuretics, and nonsteroidal anti- inflammatory agents. It can also occur in the setting of autoimmune diseases such as systemic lupus erythematosus. It typically presents with pyuria, predominantly eosinophilic, and a stereotypical drug rash on physical examination. It is not commonly caused by dopamine. Atheromatous emboli dislodged by the balloon pump (Choice C) could cause renal insufficiency as a result of ischemia of various blood vessels in the kidney. Hov more commonly presents one to four weeks after an inciting event, typically in association with blue toes or livedo reticularis caused by cholesterol emboli. er, atheroembolic disease Normal postoperative renal function (Choice D) may be slightly decreased from preoperative levels as a result of prerenal causes, such as hypovolemia in the setting of decreased oral intake, insensible losses, and inadequate fluid resuscitation. However, an increase of 4.5 times the patient's baseline level would be unexpected. Toxic acute tubular necrosis induced by hemolysis of erythrocytes (Choice E) could also cause acute kidney injury. It is associated with severe hemolysis (eg, ABO-incompatible blood transfusion) and presents with amber-colored urine and pigmented granular casts on urinalysis. This patient has no signs of severe hemolysis.

Objective: Acute kidney injury is defined as an acute increase in creatinine by greater than or equal to 0.3 mg/dL or 1.5 times above baseline creatinine, occasionally with oliguria or anuria. The most common cause is acute tubular necrosis, which can be caused by ischemia in the setting of prolonged hypoperfusion or toxins (eg, hemoglobin, myoglobin, ethylene glycol). Laboratory evaluation will show a BUN to creatinine ratio of less than 20, a fractional excretion of sodium greater than 2%, and muddy brown casts on urinalysis. Previous Next Score Report Lab Values Calculator Help Pause

109 Exam Section 3: Item 10 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 10. An obese 39-year-old woman with hirsutism and a long-standing history of anovulation is found to have endometrial hyperplasia and enlarged ovaries with multiple small cysts. The adipose tissue in this woman is most likely producing which of the following hormones? A) Aldosterone B) Cortisol C) Estrone OD) 17a-Hydroxyprogesterone OE) 20a-Hydroxyprogesterone F) Pregnenolone G) Progesterone Correct Answer: C. Estrone is a form of estrogen that is produced by adipose tissue, ovaries, and adrenal glands. This patient demonstrates signs of polycystic ovarian syndrome (PCOS) including hirsutism, increased body weight, irregular menses as a result of anovulation, and enlarged ovaries with numerous small cysts on ultrasonography. It is hypothesized that the initial event in PCOS is altered hypothalamic regulation of follicle-stimulating hormone (FSH) and luteinizing hormone (LH), which may be caused in part by hyperinsulinemia, leading to an increased LH:FSH ratio. The increased LH:FSH ratio increases production of testosterone by the theca interna cells of the ovary and decreases the rate of follicular maturation, leading to hyperandrogenism and formation of unruptured follicles or cysts instead of ovulation. Adipose tissue contains the enzyme aromatase, which converts the excess serum testosterone present in PCOS into estrone and further contributes to cycle irregularity. Increased circulating estrone and the high number of anovulatory cycles expose the endometrium to unopposed estrogen, which increases the risk for endometrial hyperplasia, also a finding in this patient. The unopposed estrogen can be mitigated with oral contraceptives and weight reduction to lower the risk for endometrial hyperplasia. Ovulation can be induced by clomiphene, and hirsutism can be treated with the antiandrogen medication spironolactone. Incorrect Answers: A, B, D, E, F, and G. Aldosterone (Choice A) is a mineralocorticoid produced by the zona glomerulosa of the adrenal cortex and plays an important role in the maintenance of blood volume and sodium homeostasis, primarily through the renin-angiotensin-aldosterone pathway. It is not produced by the adipose tissue and is not involved in the pathogenesis of PCOS. Cortisol (Choice B) is a glucocorticoid hormone produced by the zona fasciculata of the adrenal cortex. Its plays a role in glucose metabolism, blood pressure regulation, and stress and mood regulation. It is not produced by adipose tissue and not contributory to PCOS. 17a-Hydroxyprogesterone (Choice D) and 20a-hydroxyprogesterone (Choice E) are both related to progesterone and have weak effects on the progesterone receptor. Neither is produced by the adipose tissue. Pregnenolone (Choice F) is an intermediate in the production of most steroid hormones including progesterone, androgens, estrogens, glucocorticoids, and mineralocorticoids. The adipose tissue is not creating steroid hormones; rather, it is converting androgens to estrone via the aromatase enzyme, and thus pregnenolone is not a byproduct of adipose tissue. Progesterone (Choice G) is produced by the corpus luteum, the structure that remains after an ovarian follicle ruptures and releases an egg and maintains the secretory endometrium in order to allow implantation to occur. It is not produced by the adipose tissue. Progesterone is beneficial in PCOS as it opposes the increased estrogen and helps to decrease the risk for endometrial hyperplasia, which is one of the main reasons oral contraceptive regimens are used in the treatment of PCOS.

Objective: Adipose tissue contains the enzyme aromatase, which converts androgens to estrone. PCOS is characterized by both obesity and an increased serum concentration of androgens, providing both necessary elements for the production of estrone. Previous Next Score Report Lab Values Calculator Help Pause

169 Exam Section 4: Item 20 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 20. A 27-year-old man who works in a boiler room comes to the physician because he and his 32-year-old wife have been unable to conceive a child for the past 3 years. He had the mumps at the age of 12 years. He tells the physician that his wife was evaluated for infertility and test results were normal. He adds, "Could it be my work? I've heard that high temperatures cause infertility." He is muscular and has a deep voice. Physical examination shows abundant facial and body hair and a normal penis. His testicles are 3 cm long. Semen analysis shows azoospermia. His serum follicle-stimulating hormone and luteinizing hormone concentrations are less than 1 mlU/mL. His serum testosterone concentration is 36 nmol/L (N=10-35). Which of the following is the most likely cause of the development of azoospermia in this patient? A) Excessive testicular temperature B) Gonadotropin-releasing hormone deficiency C) Obstruction of seminal vesicles D) Use of exogenous testosterone E) Viral orchitis F) XXY karyotype (Klinefelter syndrome) Correct Answer: D. Spermatogenesis is the process by which sperm are created and takes place in the seminiferous tubules of the testes. Lining the seminiferous tubules are spermatogonia, which generate the primary spermatocytes that will ultimately become sperm. Sertoli cells are also present within the seminiferous tubules and are required to support spermatogenesis. Pulsatile gonadotropin- releasing hormone (GnRH) produced by the hypothalamus causes the pituitary to secrete follicle-stimulating hormone (FSH) and luteinizing hormone (LH). FSH acts on the Sertoli cells, leading them to produce androgen-binding protein. LH acts on the Leydig cells, located adjacent to but outside of the seminiferous tubules, and causes them to secrete testosterone. The androgen- binding protein binds to testosterone and maintains the concentration of testosterone in the seminiferous tubule required for spermatogenesis. An aberrancy in spermatogenesis causes azoospermia, or lack of sperm in the semen, which manifests as infertility. The use of exogenous testosterone causes azoospermia by exerting negative feedback on the hypothalamus and pituitary. In response to negative feedback, the hypothalamus decreases production of GNRH and the pituitary decreases production of FSH and LH. Without FSH and LH, the Leydig cells do not produce testosterone and the Sertoli cells do not produce androgen-binding protein. Thus, the environment is not optimal for spermatogenesis and consequent azoospermia occurs. Testicular atrophy occurs as a consequence of exogenous testosterone, noted on physical examination in this case. Incorrect Answers: A, B, C, E, and F. Excessive testicular temperature (Choice A) is the cause of infertility in cryptorchidism, the failure of one or both testes to descend into the scrotum, and bilateral varicoceles. Optimal spermatogenesis occurs at a temperature several degrees lower than that of the body core, which is achieved when the testes have descended into the scrotum. The dilated veins of a varicocele also cause the temperature in the scrotum to rise. GNRH deficiency (Choice B) prevents spermatogenesis by the same mechanism of decreased FSH and LH. However, in this patient an exogenous source of testosterone is a more likely explanation for his infertility given his increased serum testosterone and low FSH and LH concentrations. The seminal vesicles are two paired structures located in the pelvis posterior to the bladder, which secrete the fluid that will constitute the bulk of semen into the ejaculatory duct. Spermatogenesis occurs in the testes and by the time the secretions reach the opening of the seminal vesicles, sperm is already contained within the fluid. Thus, obstruction of the seminal vesicles (Choice C) would cause impaired production of semen but leave sperm production intact. Viral orchitis (Choice E) refers to inflammation of the testes that results from an infection with the paramyxovirus that causes mumps, which typically presents with unilateral or bilateral acute testicular pain, erythema, and tenderness. It is a risk factor for infertility as it may cause testicular atrophy, especially when it occurs in postpubertal men. However, this complication is rare, and this patient's infertility is more likely to be caused by his testosterone use. An XXY karyotype, or Klinefelter syndrome (Choice F), is another potential cause of male infertility. The XXY karyotype causes dysgenesis of the seminiferous tubules and testicular atrophy, leading to decreased spermatogenesis. However, other dysmorphic body features including a eunuchoid body shape, long extremities, and gynecomastia would be observed.

Objective: Testicular atrophy and azoospermia are induced by exogenous testosterone as a result of the inhibition of GNRH, FSH, and LH at the level of the hypothalamus and pituitary gland, respectively. Without a supportive local environment in the seminiferous tubule spermatogenesis does not occur, and infertility ensues. Previous Next Score Report Lab Values Calculator Help Pause

87 Exam Section 2: Item 38 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 38. A7-year-old boy is brought to the physician by his mother because of a 3-hour history of pain and stiffness of his neck. Two days ago, he fell off his bike and hurt his neck. Physical examination shows decreased range of motion of the neck. Neurologic examination shows no abnormalities. AP and lateral x-rays of the cervical spine show congenital fusion of the atlas to the occipital bone associated with C2-3 vertebral fusion. Flexion and extension of the neck are most likely placing additional strain on which of the following structures in this patient because of his congenital abnormalities? A) Atlantoaxial joint B) Interventricular foramen of Monroe C) Sternocleidomastoid muscles D) Vertebral artery Correct Answer: A. The spinal column is comprised of bony vertebrae, intervertebral discs made of fibrocartilage and gel-like material (nucleus pulposus), and ligaments that maintain the structure of the column. Function of the spinal column is based upon having numerous mobile segments; the facet joints and intervertebral discs between each vertebra provide for a small amount of motion in multiple directions. The summation of small motions allows for the composite range of motion of the spine, including flexion, extension, medial and lateral deviation, rotation and torsion, and combinations of these together. When segments are fused together from a congenital disorder, ankylosing spondylitis, or from surgery, no motion can occur at these segments. Because of this, the adjacent segments (facets, intervertebral discs, ligaments) will experience increased strain. This may lead to accelerated degeneration of the spinal column at such locations. The patient in this case demonstrates fusion of C1 to the occipital bone and fusion of C2 with C3. The segment with intact motion between these areas of fusion that will endure increased strain will be the C1-C2 joint, also known as the atlantoaxial joint. Incorrect Answers: B, C, and D. The interventricular foramen of Monro (Choice B) connects the lateral ventricles with the third ventricle of the brain. These structures do not see increased strain or pressure from fusion of spinal segments. Increased intracranial pressure (eg, secondary to hydrocephalus) would cause increased pressure in these regions. The sternocleidomastoid muscles (Choice C) originate at the manubrium and medial portions of the clavicle and insert at the mastoid process of the temporal bone. Spasms in these muscles can cause torticollis. These muscles do not undergo increased stress with the fusion of spinal segments. The vertebral arteries (Choice D) course through the transverse foramina of the cervical vertebrae from C6 through C1 supplying branches to the spinal cord, medulla, pons, and posterior central nervous system. Acute traumatic injury to the cervical spine, particularly fractures that involve the transverse foramina, can injure the vertebral arteries.

Objective: Fusion of a spinal motion segment leads to increased stress in the adjacent segments. Spinal motion is complex and is facilitated by motion in multiple planes across multiple segments of the vertebral column. Previous Next Score Report Lab Values Calculator Help Pause

119 Exam Section 3: Item 20 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 20. A 16-year-old girl with a medulloblastoma is scheduled to begin chemotherapy with a combination of agents, including cisplatin. Addition of ondansetron may decrease emesis by blocking which of the following receptors? OA) azAdrenergic B) Dopaminergic (D) C) Histaminergic (H) D) Muscarinic (M) E) Serotoninergic (5-HT3) Correct Answer: E. The presence of nausea and vomiting depends on a complex interaction between the central nervous system, parasympathetic autonomic nervous system, gastrointestinal system, and endocrine system. Stimuli that cause nausea and emesis can originate in the viscera, vestibular system, or chemoreceptor trigger zone. Visceral stimuli exert their action through serotonin and dopamine, whereas vestibular inputs signal nausea via histamine and acetylcholine. Finally, the chemoreceptor zone acts through alterations in serotonin and dopamine. For example, gastrointestinal dysmotility and increased tone can cause nausea through parasympathetic inputs to the central nervous system via the vagus nerve. In the setting of chemotherapy, the chemoreceptor zone responds to emetogenic chemotherapy medications via the area postrema, leading to nausea and subsequent emesis. For this reason, medications that antagonize serotoninergic (5-HT3) and dopaminergic (D2) impulses are the most efficacious for treating chemotherapy-induced nausea and vomiting. Serotoninergic antagonists include ondansetron and granisetron, while dopaminergic agents include metoclopramide and prochlorperazine. Incorrect Answers: A, B, C, and D. ɑ2-Adrenergic (Choice A) receptor antagonists can result in an increase in sympathetic tone and an increase in insulin release, with tachycardia and hypoglycemia seen clinically. Mirtazapine is a selective a2-adrenergic receptor antagonist used to treat depression, whereas phentolamine is a nonselective a-adrenergic receptor antagonist that is used to treat hypertensive emergencies, such as those caused by pheochromocytoma or cocaine use. Neither play a role in the treatment of nausea and vomiting. Dopaminergic (D1) (Choice B) receptor antagonists constrict renal vascular smooth muscle and increase the symptoms of Parkinson disease. Many antipsychotics antagonize dopaminergic (D1) receptors in addition to other dopamine receptors, whereas the commonly used antiemetics in the antidopaminergic category antagonize the D2 receptor. D1 receptor antagonists play no role in the treatment of nausea and vomiting. Histaminergic (H;) (Choice C) receptor antagonists decrease bronchoconstriction, mucus production, and vascular permeability associated with allergic states, although they are also occasionally used to treat vestibular nausea. Many cross the brain barrier and cause significant sedation, as well as anticholinergic-like effects, making them an ineffective initial choice for chemotherapy-induced nausea prophylaxis. Muscarinic (M,) (Choice D) receptor antagonists would decrease gastric acid secretions, bronchoconstriction, urinary bladder detrusor contractility, and salivary gland secretions through blockage of acetylcholine receptors. Medications in this category include ipratropium and oxybutynin, as well as some antipsychotics and antidepressants, none of which are used to treat nausea and vomiting.

Objective: Ondansetron is a serotoninergic (5-HT3) receptor antagonist that is a first-line prophylactic and treatment for chemotherapy-induced nausea and vomiting. Previous Next Score Report Lab Values Calculator Help Pause

189 Exam Section 4: Item 40 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 40. A previously healthy 54-year-old man comes to the clinic because of light-headedness for 6 hours. His symptom began after skiing at a resort at an altitude of 2743 m (9000 ft). He has been taking a carbonic anhydrase inhibitor since 2 days before arrival at the resort. His blood pressure is 110/60 mm Hg while sitting and 95/50 mm Hg while standing. Physical examination shows no other abnormalities. Which of the following is the most likely cause of his orthostatic hypotension? A) High-altitude sickness B) Hypovolemia C) Нурохіа D) Impaired sympathetic nerve activity E) Respiratory alkalosis Correct Answer: B. The patient is likely suffering from hypovolemia secondary to diuretic use, engaging in strenuous physical activity, and increased respiratory water vapor loss as a result of lower atmospheric pressure at high altitude. Hypovolemia is the result of decreased water intake, increased loss of salt and/or water, or a decrease in blood volume (eg, hemorrhage, third-spacing of fluids). Symptoms include light-headedness, dyspnea, fatigue, weakness, and thirst. Signs may include tachycardia, orthostatic hypotension (as present in this case), and decreased skin turgor when pinched beneath the clavicle. If severe, volume depletion can progress to hypovolemic shock with oliguria, cool extremities, and encephalopathy. Diuretics, such as carbonic anhydrase inhibitors, may exacerbate hypovolemia in the setting of decreased fluid intake. Carbonic anhydrase inhibitors block the conversion between CO2, H20, and H2CO3 by carbonic anhydrase in the renal proximal tubule cells. H2CO3 dissociates into HCO3 in the tubule lumen and binds with Nat to form NaHCO3. Carbonic anhydrase must function within the proximal tubule for reabsorption of HCO3. When inhibited, Na*, HCO3", and Cl accumulate in the tubule Ilumen. The increased osmotic pressure in the lumen draws water into the urine. The loss of bicarbonate also produces a systemic metabolic acidosis. This can be useful in preventing some of the symptoms of high-altitude sickness, which result from hyperventilation (caused by decreased atmospheric oxygen pressure at altitude) and the resultant respiratory alkalosis. Čarbonic anhydrase inhibitors are also used to treat glaucoma and idiopathic intracranial hypertension. Incorrect Answers: A, C, D, and E. High-altitude sickness (Choice A) results from the acute exposure of the body to increased altitude, where lower atmospheric pressure leads to decreased oxygen concentration in the alveoli. The resultant hypoxia (Choice C) and respiratory alkalosis (Choice E) lead to the signs and symptoms of the disorder. Altitude sickness is a spectrum of disease that ranges from mild acute mountain sickness (shortness of breath, headache, anorexia, nausea, vomiting, fatigue, dizziness, and difficulty sleeping) to life-threatening high-altitude pulmonary edema (HAPE) and high- altitude cerebral edema (HACE). Altitude sickness does not typically cause orthostatic hypotension. Impaired sympathetic nerve activity (Choice D) may cause orthostatic hypotension by blocking reflexive vasoconstriction in response to gravity. This effect is mediated by baroreceptors located in the carotid sinus and aortic arch, which detect changes in blood pressure. Adrenergic antagonists such as a- and ß-adrenergic blockers diminish this effect, resulting in orthostatic hypotension. The patient in this case is not taking any medications that block sympathetic nerve activity.

Objective: Orthostatic hypotension may result from intravascular volume depletion or impaired reflexive vasoconstriction in response to blood pressure changes sensed by baroreceptors in the aortic arch and carotid sinuses. High altitude and diuretic use are risk factors for developing hypovolemia. Previous Next Score Report Lab Values Calculator Help Pause

117 Exam Section 3: Item 18 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 18. A 16-year-old girl with type 1 diabetes mellitus is brought to the physician because of a 10-kg (22-lb) weight loss during the past 6 months. The patient reports that she is feeling fine, and she does not think that anything is wrong. She says that she is happy to have lost the weight, and she would like to lose more weight. She says that her diabetes has been in good control, and she is not aware of any insulin reactions. She is 165 cm (5 ft 5 in) tall and now weighs 46 kg (102 Ib); BMI is 17 kg/m2. Physical examination shows no other abnormalities. Laboratory studies show a hemoglobin A1, of 8.4%; 6 months ago, it was 5.8%. Which of the following patient behaviors most likely led to her weight loss? A) Decreasing the amount of self-administered insulin doses B) Overuse of laxatives C) Restricting calorie consumption D) Self-induced vomiting after meals E) Starting an intense aerobic exercise program Correct Answer: A. This patient with type I diabetes mellitus has experienced weight loss and a worsening of her glucose control, as demonstrated by the increase in hemoglobin A1, from 5.8% to 8.4%. Both findings may be caused by decreasing the amount of self-administered insulin. Insulin is produced by pancreatic B cells and primarily works to lower serum glucose. However, in type I diabetes, autoimmune destruction of the ß cells leads to a deficiency of insulin, which can only be corrected by exogenous insulin administration. In the absence of insulin, skeletal muscle and adipose tissue uptake of glucose is impaired, leading to glycosylation of erythrocytes and an increased hemoglobin A1. The hemoglobin A1, reflects the average blood glucose over the past three months and thus demonstrates that the patients' failure to administer appropriate doses of insulin is a chronic problem. In addition, without the successful uptake of glucose into the cells, glycogenolysis, proteolysis, and lipolysis all occur in an effort to provide a source of fuel to the tissues. Weight loss occurs as the patient consumes protein and fat reserves instead of utilizing glucose. With increased amounts of glucose in the serum and urine, the patient is also likely experiencing symptoms of increased thirst (polydipsia), increased urination (polyuria), and hunger (polyphagia). She is likely also dehydrated as a result of the osmotic diuretic effect of glucose in the urine. The patient is at a high risk for developing diabetic ketoacidosis, which is an emergent high anion-gap metabolic acidosis caused by the accumulation of ketones, the byproduct of lipolysis, in the blood. Insulin is a trophic hormone, and its absence leads to rapid catabolism. Incorrect Answers: B, C, D, and E. While the overuse of laxatives (Choice B) may cause dehydration and weight loss, it would not be responsible for hyperglycemia and subsequent increased hemoglobin A 1e Restricting calorie consumption (Choice C) causes weight loss by requiring the body to utilize glycogen and fat stores for fuel. Self-induced vomiting after meals (Choice D) is a feature of bulimia nervosa, which would have a similar effect as restricting calorie consumption. Similar to the overuse of laxatives, neither of these causes would be responsible for hyperglycemia and increased hemoglobin A1c Intense aerobic exercise (Choice E) potentiates the effects of insulin. If the patient had been exercising aggressively while still administering insulin, she would be more likely to be euglycemic or hypoglycemic, not hyperglycemic.

Objective: The absence of insulin in a patient with type I diabetes mellitus will impair skeletal muscle and adipose tissue uptake of glucose, leading to the increased glycosylation of erythrocytes. Glycogenolysis, proteolysis, and lipolysis will all occur in an effort to provide energy to tissues, leading to weight loss. Previous Next Score Report Lab Values Calculator Help Pause

75 Exam Section 2: Item 26 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 26. A 30-year-old woman comes to the physician for a health maintenance examination. She is training for a marathon and has been running up to 20 miles daily. She states that as long as she has adequate caloric intake, she feels well on long-distance runs of up to 20 miles. Physical examination shows no abnormalities. Her fasting serum glucose concentration is 60 mg/dL. After her glucose stores have been depleted, which of the following organs, in addition to the liver, is most likely to release newly produced glucose in this patient? A) Adrenal glands B) Kidney C) Pancreas D) Stomach E) Thyroid gland Correct Answer: B. The release of glucose into the serum by the kidney and liver, produced via gluconeogenesis and glycogenolysis, likely accounts for this patient's ability to maintain adequate serum glucose concentrations during periods of prolonged strenuous exercise. Depletion of glucose occurs quickly after the start of exercise, so the body must rely upon mechanisms for the continued synthesis and release of glucose into the blood to maintain appropriate glucose concentrations for tissue metabolism. Glycogen is produced during the fed state and stored primarily in the liver and skeletal muscle, but only the liver possesses the ability to release free glucose into the serum during the process of glycogenolysis as it possesses the enzyme glucose-6-phosphatase, which hydrolyzes glucose-6-phosphate to create free glucose molecules. Similarly, the kidney possesses this enzyme, which permits release of free glucose to the serum during gluconeogenesis. The liver participates in both glycogenolysis and gluconeogenesis, whereas the kidney primarily supplies glucose via gluconeogenesis. The free glucose can circulate in the serum, where it may be used by the skeletal muscle for aerobic and/or anaerobic metabolism during exercise. Incorrect Answers: A, C, D, and E. Adrenal glands (Choice A) release epinephrine in response to the stress of exercise and hypoglycemia. Epinephrine acts on hepatocytes to cause the release of intracellular calcium, which activates glycogen phosphorylase allowing for glycogenolysis in the liver. The pancreas (Choice C) releases glucagon from a cells during episodes of hypoglycemia. Glucagon acts on hepatocytes via a cyclic adenosine monophosphate pathway activating protein kinase A that activates glycogenolysis and gluconeogenesis. The stomach (Choice D) is not a site of gluconeogenesis. During periods of prolonged or intense exercise, blood flow to the stomach and the rest of the gastrointestinal tract decreases, with subsequent delayed gastric emptying and increased transit times. Digestion is slowed as a result of these changes. The thyroid gland (Choice E) does not play a role in gluconeogenesis or glycogenolysis during exercise or fasting states.

Objective: The liver and skeletal muscle are the primary areas where glycogen is stored. The liver and kidney also participate in gluconeogenesis and possess the capability to release free glucose into the serum as a result of the presence of glucose-6-phosphatase. This process helps maintain euglycemia during periods of prolonged fasting or intensive exercise. Previous Next Score Report Lab Values Calculator Help Pause

8 Exam Section 1: Item 8 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 8. A 60-year-old woman comes to the physician because of vaginal bleeding for 6 weeks. Menarche occurred at the age of 12 years, and menses had occurred at regular 30-day intervals. She had three term pregnancies by the age of 25 years. She has been receiving conjugated equine estrogen therapy since menopause 10 years ago. She has smoked 1 pack of cigarettes daily for 40 years. She drinks alcoholic beverages five times weekly. Physical examination and laboratory studies show no abnormalities. An endometrial biopsy specimen shows endometrial adenocarcinoma. Which of the following is the strongest predisposing risk factor for the biopsy findings? A) Alcohol use B) Cigarette smoking C) Early menarche D) Estrogen therapy E) Multiparity Correct Answer: D. Vaginal bleeding in a postmenopausal woman should raise suspicion for malignancy, including uterine or cervical carcinoma. Estrogen excess without adequate opposition by progestin is a strong risk factor for atypical endometrial hyperplasia and endometrial carcinoma, as this malignancy is estrogen-sensitive. This is one reason that estrogen therapy without progestin is no longer generally recommended for long-term treatment of menopausal symptoms or prevention of osteoporosis. Additionally, estrogen therapy also increases the risk for thromboembolic events in postmenopausal women. Increased exposure to estrogen can come from an exogenous supplement as in this case, or an endogenous source, such as increased peripheral production of estrogen by adipose tissue or from an increased number of lifetime menstrual cycles. Thus, risk factors for endometrial carcinoma include estrogen supplementation, obesity, early menarche or late menopause, nulliparity, advanced age, and family history. Endometrial biopsy is used to confirm the diagnosis. With biopsy and imaging, the stage of endometrial carcinoma is determined and follows TNM (tumor, nodes, metastases) staging. If the disease is confined to the uterus, surgical excision is the first-line therapy. This is generally achieved through hysterectomy with bilateral salpingo-oophorectomy. Incorrect Answers: A, B, C, and E. While alcohol use (Choice A) and cigarette smoking (Choice B) are risk factors for cancers such as hepatocellular carcinoma and squamous cell carcinoma, respectively, they have not been shown to increase the risk for endometrial carcinoma. Early menarche (Choice C) is a risk factor for endometrial carcinoma as it increases the total number of ovulatory cycles and thus endometrial estrogen exposure over a lifetime. However, this patient's use of estrogen supplementation over a 10-year period is a more direct and substantial risk factor for endometrial carcinoma. Nulliparity, not multiparity (Choice E), is a risk factor for endometrial carcinoma. There is a hiatus from estrogen surges for the course of each pregnancy as menstrual cycles are postponed. The three term pregnancies experienced by this patient would be considered protective for endometrial carcinoma.

Objective: Vaginal bleeding in a postmenopausal woman should raise suspicion for malignancy, including uterine or cervical carcinoma. Estrogen excess, whether endogenous or exogenous, is a strong risk factor for endometrial carcinoma. Previous Next Score Report Lab Values Calculator Help Pause

114 Exam Section 3: Item 15 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 15. A 50-year-old woman is admitted to the hospital for treatment of injuries sustained in a motor vehicle collision. Intubation and mechanical ventilation are initiated. The ventilator has adjustments for frequency, tidal volume, inspiratory and expiratory duration, and positive end-expiratory pressure (PEEP). The most effective way to decrease arterial Pco, in this patient is to increase which of the following? A) Expiratory duration at constant frequency B) Frequency at constant tidal volume C) Inspiratory duration at constant frequency D) Inspired O, fraction E) PEEP by 5 cm H,0 at constant frequency and tidal volume F) Size of the endotracheal tube being used Correct Answer: B. There are two main considerations when evaluating the effectiveness of the respiratory cycle in a patient on a ventilator. Oxygenation, which refers to the mechanisms of oxygen diffusion from the alveoli into the blood, and ventilation, in which carbon dioxide diffuses from the blood into the alveoli. The arterial pressure of oxygen (PaO2) is dependent on positive end-expiratory pressure (PEEP) as it prevents alveolar collapse, increasing the number of alveoli available for gas exchange and the fraction of inspired oxygen (F1o). Increases in either variable will increase Pao, The arterial pressure of carbon dioxide (PaCO2), in contrast, is dependent on minute ventilation, which is dictated by the tidal volume and respiratory rate, with increases in either variable associated with an increased exchange of carbon dioxide from the blood into the alveoli for exhalation, which results in a decreased arterial PaCO2 eing Incorrect Answers: A, C, D, E, and F. Increases in expiratory duration at constant frequency (Choice A) do not alter the minute ventilation, and therefore would not affect the PaCo, concentration. This technique does prevent incomplete expiration and breath stacking, with associated autologous PEEP, which is beneficial in patients with obstructive respiratory disorders. Increases in inspiratory duration at constant frequency (Choice C) do not alter the minute ventilation, and therefore would not affect the Pco, concentration. It can be used to decrease the peak airway pressures a patient experiences when on volume control ventilation, as it allows a longer period of time over which the volume can be delivered, decreasing the peak pressure obtained. Increases in inspired O, fraction (Choice D) and increases in PEEP by 5 cm H,0 at constant frequency and tidal volume (Choice E) affect oxygenation, not ventilation, and would be expected to increase the PaO, but have no effect on the PaCO, Increases in the size of the endotracheal tube being used (Choice F) decreases the peak airway pressure and increases the maximum inspiratory flow rate. It does not affect ventilation or oxygenation and would not be expected to cause changes in PaCO2

Objective: Ventilation, or the exchange of carbon dioxide between the blood and the alveoli, is dependent on minute ventilation (the product of respiratory rate and tidal volume), with increases in ventilation leading to a decreased arterial Pco, Oxygenation is dependent on the fraction of inspired oxygen and the positive end-expiratory pressures, with increases in either leading to increases in arterial Po, II Previous Next Score Report Lab Values Calculator Help Pause

157 Exam Section 4: Item 8 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 8. A 43-year-old man with advanced chronic renal failure comes to the physician to discuss the possibility of undergoing kidney transplantation. He also has type 2 diabetes mellitus and hepatitis C. He has been receiving treatment with hemodialysis for 1 year. Before the physician has a chance to initiate the discussion on transplantation, the patient says that he is unsure about receiving a transplant. He adds that if he were to undergo the procedure, he would rather get a donor kidney from one of his relatives than from a cadaver. Which of the following is the most appropriate next statement by the physician? A) "The chances of you dying after receiving the kidney transplant are greatest 3 to 4 years after the procedure." B) "I want to be sure that I understand your concerns about getting a transplant." C) "I want you to know that transplantation will improve your lifestyle and life expectancy as compared to hemodialysis." D) "Please reconsider using a cadaveric kidney. That way, you wouldn't have to inconvenience any relatives, and the waiting list is only 1 year." E) "That's fine, but the transplant procedure is very invasive and we need to determine if it would be safe for you to go further with this." Correct Answer: B. Physicians should elicit patient-specific concerns when discussing consequential decisions such as undergoing kidney transplantation. One important role of physicians is to understand patients' underlying values and goals and assist the patients in achieving their goals, an approach that honors autonomy. Before educating this patient, the physician should clarify the patient's underlying values and understanding of the risks and benefits of transplantation and receiving a related-donor kidney versus a cadaveric kidney. Ônce the physician appreciates the patient's values, he can tailor further discussion and education to help the patient make the decision that best aligns with his values. The physician should also assess the patient's understanding and tailor further discussion to address knowledge gaps. Incorrect Answers: A, C, D, and E. The physician should refrain from immediately educating the patient or expressing an opinion about the patient's decision (Choices A, C, D, and E) before assessing the patient's level of understanding about the decision and underlying values. Additionally, this patient voices a preformed opinion and may feel defensive if the physician immediately provides an opposing opinion, precluding an open and effective discussion.

Objective: When a patient has a consequential decision to make, the physician should explore the patient's understanding of the risks and benefits and underlying values, so that further discussion can be tailored to appropriately address the patient's understanding and values. %D Previous Next Score Report Lab Values Calculator Help Pause

118 Exam Section 3: Item 19 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 19. A2-month-old boy is brought to the physician because of hypotonia and poor feeding since birth. Physical examination shows large fontanels, midface hypoplasia, hepatomegaly, and cryptorchidism. Serum studies show increased concentrations of very-long-chain fatty acids, phytanic acid, and pipecolic acid. If this patient's hepatocytes were analyzed using electron microscopy, results would most likely show the absence of which of the following organelles? A) Endoplasmic reticulum B) Lysosomes C) Mitochondria D) Nucleoli E) Peroxisomes Correct Answer: E. Zellweger syndrome is caused by a genetic mutation that leads to an absence of peroxisomes. As peroxisomes are responsible for metabolizing very long-chain fatty acids (VLCFA), phytanic acid, pipecolic acid, pristanic acid, hydrogen peroxide, and ethanol, their absence leads to an accumulation of these products. They also play an important role in the synthesis of cholesterol and bile acids, as well as cell membrane substrates. Zellweger syndrome classically presents with seizures, intellectual disability, hypotonia, hepatomegaly, kidney disease, cataracts, hearing loss, and craniofacial abnormalities. Physical examination will demonstrate absence of reflexes, and laboratory evaluation will show increased concentrations of VLCFA, phytanic acid, and pipecolic acid, as in this patient. There is no effective treatment, and life expectancy is less than six months of age. Incorrect Answers: A, B, C, and D. The endoplasmic reticulum (Choice A) is responsible for synthesizing proteins, lipids, and steroids, as well as the transport of proteins and molecules within the cell, and detoxification of toxins. While not present in erythrocytes, their absence in hepatocytes would likely lead to cell death and liver failure. It would not lead to the accumulation of VLCFA, phytanic acid, and pipecolic acid. Lysosomes (Choice B) are responsible for the degradation of a variety of substrates, including those brought into the cell through endocytosis or materials of the cell itself (eg, autophagy). Deficiencies in lysosomal enzymes lead to lysosomal storage disorders, which present broadly with neurocognitive decline, skeletal abnormalities, and craniofacial abnormalities. Mitochondria (Choice C) are responsible for producing energy for the cell in the form of adenosine triphosphate. Disorders of mitochondrial structure or function include MELAS (mitochondrial encephalomyopathy, lactic acidosis, and stroke-like episodes) syndrome and myoclonic epilepsy with ragged red fibers. Complete absence in the hepatocytes would lead to cell death and liver failure. Nucleoli (Choice D) are the sites of ribosome biosynthesis in the nucleus. Absence in hepatocytes would lead to cell death through the inability to translate cellular proteins.

Objective: Zellweger syndrome is caused by a genetic absence of peroxisomes and presents classically with seizures, intellectual disability, hypotonia, hepatomegaly, kidney disease, cataracts, hearing loss, and craniofacial abnormalities. Laboratory evaluation will demonstrate increased VLCFA, phytanic acid, and pipecolic acid. Previous Next Score Report Lab Values Calculator Help Pause

91 Exam Section 2: Item 42 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 42. A 1-year-old girl is admitted to the hospital because of failure to thrive. She has had frequent, loose stools for 6 months. She has not gained weight satisfactorily during this period. She is at the 3rd percentile for length and weight. Physical examination shows facial edema. Laboratory studies show hypoalbuminemia. Examination of the contents of the duodenum postprandially show: Trypsinogen Chymotrypsinogen Proelastase increased increased increased Lipase Amylase normal normal A deficiency of which of the following enzymes is the most likely cause of these findings? A) Aminopeptidase B) Carboxypeptidase A OC) Chymotrypsin D) Enteropeptidase E) Trypsin Correct Answer: D. This infant's findings of failure to thrive, diarrhea, edema, hypoalbuminemia, and increased concentrations of luminal proenzymes are suggestive of enteropeptidase deficiency. This rare disorder is caused by an inactivating mutation in the gene encoding enteropeptidase. In the absence of functional enteropeptidase, pancreatic trypsinogen fails to be cleaved to active trypsin, and therefore fails to catalyze the conversion of other pancreatic proenzymes, such as chymotrypsinogen and proelastase, to their active forms. The normal activity of pancreatic enzymes, including lipase and amylase, suggests that the quantity of pancreatic secretions is adequate and that the patient's defect is rather one of luminal enzymatic conversion of proenzymes to their active form. Failure to digest proteins and fats leads to osmotic and malabsorptive diarrhea and failure to thrive during infancy. Treatment is with oral replacement of enteropeptidase. Incorrect Answers: A, B, C, and E. Aminopeptidase (Choice A) is important for the digestion of dietary peptides at the brush border. However, deficiency of this enzyme would not account for the increased concentrations of several proenzymes that are observed in this patient. Carboxypeptidase A (Choice B) is a pancreatic enzyme that is important for the digestion of dietary peptides. However, deficiency of this enzyme would not account for the increased concentrations of several proenzymes that are observed in this patient. Chymotrypsin (Choice C) is produced as the proenzyme chymotrypsinogen by pancreatic acinar cells and is converted to its active form by trypsin. A deficiency of chymotrypsin would not account for the increased concentrations of other proenzymes, such as proelastase and trypsinogen, which is suggestive of deficiency of enteropeptidase. Similarly, trypsin (Choice E) is produced as a proenzyme, trypsinogen, in pancreatic acinar cells. Deficiency of trypsin and chymotrypsin are present in this patient, but only secondary to an absence of activating enteropeptidase, which is responsible for converting trypsinogen to its active form.

Objective: Enteropeptidase deficiency results in the inability to convert trypsinogen to trypsin, with resultant inability to convert numerous digestive proenzymes to their active forms. Patients present with osmotic and malabsorptive diarrhea and failure to thrive in infancy. Treatment is with oral replacement of enteropeptidase. Previous Next Score Report Lab Values Calculator Help Pause

30 Exam Section 1: Item 31 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 31. An investigator is testing the hypothesis that advanced glycosylation end products (AGES) lead to an increased synthesis of extracellular matrix by human renal mesangial cells. The cells are plated on two media: type I collagen treated extensively with glucose 6-phosphate to produce AGES (first medium), and a control medium of untreated type I collagen. To confirm this hypothesis, MRNA that encodes for which of the following is most likely to be increased in cells incubated in the first medium compared with cells in the control medium? A) Endothelial nitric oxide synthase B) Fibronectin C) NADPH oxidase D) Nuclear factor-KB E) Vascular endothelial growth factor Correct Answer: B. Fibronectin MRNA would most likely be increased in renal mesangial cells incubated in a medium containing glucose-6-phosphate, as fibronectin is a critical component of the extracellular matrix (ECM). Soluble fibronectin is secreted from cells-in this case, renal mesangial cells-into the extracellular space where it binds to cell-surface receptors such as a5B1 integrin. Binding is thought to introduce conformational changes in the fibronectin strand leading to the exposure of previously hidden binding sites. Additional fibronectin molecules bind at these sites and allow the previously soluble fibronectin molecule to become an insoluble matrix of multiple branching fibronectin units. As fibronectin is a protein that must be translated from MRNA, cells that secrete more fibronectin would be expected to have higher levels of MRNA encoding fibronectin. Incorrect Answers: A, C, D, and E. Endothelial nitric oxide synthase (Choice A) is an enzyme that synthesizes nitric oxide from L-arginine and oxygen. It is a vasoactive compound that results in the relaxation of vascular smooth muscle. It is not implicated in the formation of the ECM. NADPH oxidase (Choice C) is an enzyme important in the synthesis of hypochlorous acid and reactive oxygen species in the respiratory burst of phagocytes. It plays a role in phagocyte- mediated killing of pathogens but not in the formation of the ECM. Deficiency of NADPH oxidase in phagocytes is characteristic of chronic granulomatous disease, an immunodeficiency that results in recurrent infections with catalase-positive organisms. Nuclear factor-KB (Choice D) is a transcription factor that has broad functionality but primarily serves to increase expression of genes implicated in immune response, lymph tissue development, B-cell development, and stress response. It is activated by tumor necrosis factor-alpha, along with multiple intracellular and extracellular stimuli. Vascular endothelial growth factor (VEGF) (Choice E) is a growth factor that stimulates angiogenesis (the formation of new blood vessels). It is critical in tissue repair but is commonly implicated in tumorigenesis. There are numerous drugs that target VEGF in routine clinical use, including ranibizumab and bevacizumab.

Objective: Fibronectin is a component of the ECM. Renal mesangial cells grown in a medium that promotes the formation of the ECM would be expected to express increased levels of fibronectin mRNA. Previous Next Score Report Lab Values Calculator Help Pause

100 Exam Section 3: Item 1 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 1. A previously healthy 32-year-old woman comes to the physician because of a 3-month history of heavy menstrual bleeding, intermittent nosebleeds, and blood in her urine and stools. She says, "I've had to stop flossing because my gums bleed for an hour afterwards." Physical examination shows numerous ecchymoses and petechiae over the lower extremities. A test result for circulating antiplatelet antibodies is positive. A photomicrograph of a biopsy specimen of bone marrow is shown. Which of the following is the most likely pathogenesis of this patient's disease? A) Inability of the platelets to be released into the blood B) Megakaryocyte dysfunction C) Monosomy 7 D) Platelet destruction O E) Platelet dysfunction Correct Answer: D. Platelet destruction as a result of circulating antiplatelet antibodies accounts for this patient's presentation, which is consistent with immune thrombocytopenic purpura (ITP). Anti-Gpllb/Illa antibodies bind to platelets, which results in their opsonization. Macrophages within the spleen then phagocytose the platelet-antibody complex, resulting in thrombocytopenia. ITP is diagnosed in the setting of a low platelet count in the absence of other explanatory causes for thrombocytopenia. Bone marrow biopsy will classically show an increased number of megakaryocytes indicating adequate platelet production but increased peripheral destruction of platelets. It classically presents with petechiae and purpura, and prolonged bleeding time on laboratory analysis. Severe ITP may lead to uncontrolled hemorrhage. Acute ITP often follows an infection, and generally is self-limited. Chronic ITP is first treated with steroids (which may be required long-term), intravenous immunoglobulin, or immunomodulators. Refractory ITP that is non-responsive to such therapies is treated with splenectomy, which results in remission of thrombocytopenia in the majority of cases because of the spleen's primary role in the underlying pathophysiology. Incorrect Answers: A, B, C, and E. Inability of the platelets to be released into the blood (Choice A) is not a feature of ITP. These patients demonstrate an increased number of megakaryocytes that create and release platelets into the peripheral blood, which then are targeted by antibody-mediated destruction that results in thrombocytopenia. Megakaryocyte dysfunction (Choice B) may occur in patients with ITP but is not hypothesized to be the primary mechanism by which thrombocytopenia occurs. Megakaryocyte dysfunction also occurs in the setting of myelodysplastic syndrome (MDS), but a bone marrow biopsy in patients with MDS is abnormal. Monosomy 7 (Choice C) is characterized by childhood-onset thrombocytopenia, aplastic anemia, and development of acute myelogenous leukemia. The bone marrow in a patient with monosomy 7 would be abnormal and an increased number of megakaryocytes would not be expected. Platelet dysfunction (Choice E) occurs as a result of medications such as aspirin or clopidogrel, and in patients with chronic kidney disease and uremia. Each of these are characterized by a normal number of circulating platelets but abnormal function, which can lead to an increased risk for bleeding despite a normal platelet count.

Objective: ITP is a diagnosis of exclusion and presents with mucocutaneous bleeding and severe thrombocytopenia, often following an upper respiratory illness. It is a result of the antibody-mediated destruction of circulating platelets, and bone marrow biopsy will demonstrate increased numbers of megakaryocytes. Previous Next Score Report Lab Values Calculator Help Pause

62 Exam Section 2: Item 13 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 13. A study is conducted to assess body mass index (BMI) in a group of 100 patients with type 2 diabetes mellitus. Results show a mean (± standard error of the mean) BMI of 31 kg/m2 (+ 4 kg/m2). The 99% confidence intervals for this measurement are 20.7 to 41.3. The 95% confidence intervals are 23.3 to 38.8. Based on this information, which of the following best represents the number of individuals who have a BMI between 38.8 kg/m2 and 41.3 kg/m2? A) 2 B) 4 C) 6 D) 8 E) 10 Correct Answer: A. In a gaussian (normal) distribution without skew, data points center around the mean value. 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 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. Three standard deviations above and below the mean will include approximately 99.7% of data points. In a population with a mean BMI of 31 kg/m2 and a standard deviation of 4 kg/m2, individuals with a BMI less than 23.3 kg/m2 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 23.3 kg/m2) and half (2.5%) are found above the second standard deviation from the mean (greater than 38.8 kg/m2). The number of individuals who fall between two standard deviations higher than the mean and three standard deviations higher than the mean will be half of 4.7% (99.7% minus 95%), or roughly 2%. In a population of 100 individuals, this represents two individuals. Incorrect answers: B, C, D, and E. Four individuals (Choice B), or 4% of the population corresponds approximately to the amount of data points found in total outside the range of two standard deviations but within three standard deviations. This includes those individuals who fall both above and below the mean. Six (Choice C), eight (Choice D), and 10 (Choice E) do not correspond with a number of individuals outside the range of one, two, or three standard deviations from 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, 95% of data points lie within two standard deviations above and below the mean, and 99.7% of data points lie within three standard deviations above and below the mean. Previous Next Score Report Lab Values Calculator Help Pause

12 Exam Section 1: Item 12 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 12. A73-year-old woman has a 3-month history of increasing fatigue, weakness, and a 9.0-kg (20-lb) weight loss. Examination shows hepatosplenomegaly. An M-protein spike is observed on serum electrophoresis. Which of the following is the most likely diagnosis? A) Acute myelogenous leukemia B) B-lymphocyte neoplasia C) Chronic myelogenous leukemia D) Leukemoid reaction OE) T-lymphocyte neoplasia Correct Answer: B. B-lymphocyte neoplasia likely accounts for this patient's presentation with an M-protein spike (M-spike). While an M-spike may occur in a variety of B cell malignancies including Waldenstrom macroglobulinemia, plasma cell leukemia, smoldering myeloma, amyloidosis, or plasmacytoma, it is most commonly recognized in the setting of multiple myeloma (MM). Other signs and symptoms of MM include bone pain, anemia, hypercalcemia, renal dysfunction, and lytic bone lesions. Hepatosplenomegaly is an occasional finding. B cells are precursors to plasma cells, which secrete specific immunoglobulins of different classes. MM is a malignancy caused by the neoplastic proliferation of a single plasma cell clone, which overproduces monoclonal immunoglobulin and light or heavy chains. These clonal immunoglobulins are secreted in high numbers and appear as a monoclonal spike on protein electrophoresis in the gamma region. Diagnosis of multiple myeloma is suspected based on the presence of an M-protein spike in the presence of concerning symptoms, but is confirmed by bone marrow biopsy, which must demonstrate at least 10% clonal plasma cells. As a result of the hypersecretion of a single immunoglobulin, these patients often have a relative immunodeficiency as the production of other immunoglobulins is impaired. Incorrect Answers: A, C, D, and E. Acute myelogenous leukemia (Choice A) involves abnormal cells of the myeloid lineage, not the lymphocyte lineage, with a variety of defining mutations, each of which portends its own prognosis. Secretion of immunoglobulins is not a feature of this disease; therefore, an M-spike would not be expected on protein electrophoresis. Chronic myelogenous leukemia (Choice C) is defined by the presence of the Philadelphia (Ph) chromosome, which is created by a translocation between chromosomes 9 and 22, resulting in constitutive activation of the ABL1 tyrosine kinase. Peripheral smear will show a diverse set of immature or partially mature cells of the granulocytic cell lineage with basophilia. Hypersecretion of monoclonal immunoglobulins does not occur in this disease. Leukemoid reaction (Choice D) is a response to stress or infection that results in a significant rise in peripheral circulating leukocytes. It can sometimes be confused with acute leukemia but is differentiated by the fact that, in a leukemoid reaction, leukocytes are mature, functional, and demonstrate increased leukocyte alkaline phosphatase activity. An M-spike is not encountered during a leukemoid reaction. T-lymphocyte neoplasia (Choice E) describes a spectrum of diseases that occur as a result of mutations in T lymphocyte progenitor cells and are termed either T-cell leukemia or T-cell lymphoma. These are rare when compared to other B-cell malignancies such as B-cell lymphoma or multiple myeloma. Because T lymphocytes do not secrete immunoglobulins, an M-spike would not be expected.

Objective: An M-spike denotes a monoclonal concentration of immunoglobulins secreted from a population of malignant plasma cells. While M-spikes are encountered in several malignancies, they are most commonly seen in multiple myeloma, which is a disorder of clonal plasma cell proliferation. Plasma cells are terminally differentiated B cells and thus, this is considered a B-lymphocyte malignancy. Previous Next Score Report Lab Values Calculator Help Pause

98 Exam Section 2: Item 49 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 49. A 58-year-old man with chronic pancreatitis comes to the physician because of a 3-week history of greasy, foul-smelling stools. The stools are difficult to flush. There is no associated blood or mucus. He is 178 cm (5 ft 10 in) tall and weighs 65 kg (143 Ib); BMI is 21 kg/m2. Vital signs are normal. Physical examination shows no abnormalities; there is no jaundice. Which of the following recommendations is most appropriate for this patient? A) B vitamin supplementation B) Decreased carbohydrate intake C) Decreased protein intake D) Decreased sodium intake E) Fat-soluble vitamin supplementation F) Iron supplementation Correct Answer: E. Chronic pancreatitis is a progressive disease process caused by a response to recurrent pancreatic injury that results in pancreatic fibrosis, regeneration, and loss of acinar cells, islet cells, and ducts. In addition to chronic abdominal pain as a result of inflammation, patients may develop pancreatic insufficiency as a result of decreased pancreatic secretions. Absence of pancreatic lipase results in steatorrhea and an inability to absorb fat-soluble vitamins including vitamins A, D, E, and K. Patients may also develop diabetes mellitus as a result of decreased pancreatic insulin production. Patients with chronic pancreatitis can be treated with pancreatic enzyme repletion and with supplementation of fat-soluble vitamins. Incorrect Answers: A, B, C, D, and F. B vitamin supplementation (Choice A) may include supplementation of vitamin B1 (thiamine), B6 (pyridoxine) or B12 (cobalamin) and is useful for the treatment of various disorders such as Wernicke encephalopathy or sideroblastic anemia, or in patients being treated with isoniazid. While pancreatic insufficiency may lead to deficiency of vitamin B12 (cobalamin), it does not affect the absorption of most B vitamins, which are water-soluble. Decreased carbohydrate intake (Choice B) may be useful for patients with diabetes mellitus. While diabetes mellitus may develop in the setting of chronic pancreatitis, the patient's symptoms of steatorrhea are more concerning for the development of fat-soluble vitamin deficiencies. Decreased protein intake (Choice C) may be useful in the treatment of patients with uric acid disorders such as gout or ornithine transcarbamylase deficiency. Decreasing protein intake does not address the patient's steatorrhea and subsequent inability to absorb fat-soluble vitamins. Decreased sodium intake (Choice D) may be useful in the treatment of hypertension. Intestinal absorption of sodium is independent of pancreatic enzymes.Iron supplementation (Choice F) is useful for patients with iron deficiency anemia. Chronic pancreatitis is not a common cause of iron deficiency.

Objective: Chronic pancreatitis can result in pancreatic insufficiency as a result of chronic, progressive pancreatic inflammation, fibrosis, and loss of acinar cells and islet cells. Decreased secretion of pancreatic digestive enzymes, including lipase, results in steatorrhea and a decreased ability to absorb fat-soluble vitamins. Patients with chronic pancreatitis may benefit from supplementation of these vitamins, which include vitamins A, D, E, and K. 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 49-year-old man comes to the emergency department because of a persistent and painful erection since he took a drug for erectile dysfunction 4 hours ago. His temperature is 37°C (98.6°F), and blood pressure is 110/65 mm Hg. Physical examination shows flushing of the skin and priapism. The most likely cause of this patient's condition is a lack of input from which of the following neural pathways? A) General sensory fibers from the pudendal nerve B) Parasympathetic fibers from the pelvic plexus C) Somatic motor fibers from the pudendal nerve D) Sympathetic fibers from the prostatic plexus E) Visceral afferent fibers to the pelvic plexus Correct Answer: D. Sympathetic fibers from the prostatic plexus are involved in detumescence of the penile erectile tissue, signifying that a lack of input from these fibers can lead to priapism. From the prostatic plexus, the cavernous and pudendal nerves both carry sympathetic fibers that release norepinephrine to extinguish penile erection. Norepinephrine vasoconstricts the helicine arteries of the corpora cavernosa and thereby abolishes the parasympathetic nervous system-mediated vasodilation that produces erections. Erectile dysfunction medications such as sildenafil are phosphodiesterase inhibitors. Phosphodiesterase inhibition in the case of penile blood flow increases cyclic guanosine monophosphate (CGMP) concentrations, which relaxes the smooth muscle of the vasculature through nitric oxide signaling. This signal occurs downstream from the G protein-coupled receptors at the vascular smooth muscle membrane; therefore, erectile dysfunction medications can override sympathetic nervous system-mediated vasoconstriction. Incorrect Answers: A, B, C, and E. General sensory fibers from the pudendal nerve (Choice A) are activated by touch or pressure on the penile skin, glans, and urethra. Somatic motor fibers from the pudendal nerve (Choice C) respond to this sensory input by causing contraction of the ischiocavernosus and bulbocavernosus muscles, leading to erection. Increased pudendal nerve activity, rather than a lack of activity, promotes erections. Parasympathetic fibers from the pelvic plexus (Choice B) contribute to the cavernous nerves, which travel to the corpora cavernosa of the penis and release acetylcholine that causes vasodilation of the helicine arteries, resulting in penile erection. Increased activity of the parasympathetic fibers of the cavernous nerves promotes erections. Visceral afferent fibers to the pelvic plexus (Choice E) innervate the colon and rectum. They respond to stretch and are thought to control defecation. They are not known to play a large role in erections.

Objective: Erectile dysfunction medications include phosphodiesterase inhibitors, which directly relax the smooth muscle of the vasculature of the corpora cavernosa, leading to erection. These medications override the sympathetic input that normally vasoconstricts these blood vessels to extinguish erections. Previous Next Score Report Lab Values Calculator Help Pause

104 Exam Section 3: Item 5 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 5. A 56-year-old man comes to the physician because of a 1-week history of fever, headache, and painful sores on his penis. Two weeks ago, he had unprotected sexual intercourse with a new female partner. His temperature is 38.2°C (100.8°F), pulse is 85/min, respirations are 15/min, and blood pressure is 128/72 mm Hg. Physical examination shows inguinal lymphadenopathy. Genital examination shows shallow, vesicular lesions on the shaft of the penis. Which of the following is the most likely diagnosis? A) Chancroid B) Genital herpes C) Human papillomavirus infection D) Lymphogranuloma venereum E) Syphilis Correct Answer: B. This patient's vesicular and ulcerative lesions on the penis are likely caused by herpes simplex virus (HSV) in the form of genital herpes. HSV-1 and HSV-2 are both members of the herpesvirus family of double-stranded DNA, enveloped viruses. Genital herpes is most often caused by HSV-2 but can also be caused by HSV-1. In contrast, herpes labialis is most often caused by HSV-1 but can be caused by HSV-2. Coinfection with both strains is possible. Genital herpes is transmitted through sexual contact and perinatally. HSV may be latent in the sacral ganglia until reactivation, when it causes painful vesicles and ulcerated erosions on the genitalia with associated inguinal lymphadenopathy. Systemic manifestations are also possible, including viral meningitis and encephalitis. Treatment for herpetic infections involves drugs that inhibit viral DNA polymerase, classically by guanosine analogs such as acyclovir, valacyclovir, and famciclovir. Incorrect Answers: A, C, D, and E. Chancroid (Choice A) presents with an exudative painful genital ulcer with inguinal lymphadenopathy. It is caused by infection with the organism Haemophilus ducreyi. This patient's shallow, vesicular lesions are more classic of genital herpes. Infection with human papillomavirus (HPV) (Choice C) is the cause of genital warts (low-risk subtypes 6 and 11). Genital warts are soft, fleshy papules, not vesicles or ulcers. Infection with the high-risk subtypes of HPV, including 16, 18, 31, and 33, are potential causes of cervical cancer and Bowen disease, a premalignant lesion on the shaft of the penis. In HIV patients with low CD4+ counts, herpetic lesions can take on a verrucous, or wartlike, appearance, but this is not their typical morphology. Lymphogranuloma venereum (Choice D) is caused by infection with Chlamydia trachomatis subtypes L1, L2, and L3. It presents as small painless genital ulcers with tender inguinal lymphadenopathy, while this patient's ulcerative lesions are painful. Syphilis (Choice E), caused by an infection with Treponema pallidum, demonstrates multiple stages with varying symptoms, including primary with a painless chancre, secondary with fever, lymphadenopathy, and condylomata lata, and tertiary with tabes dorsalis, aortitis, and gummas. The chancre of primary syphilis typically presents as a painless ulcerative genital lesion, unlike the painful lesions seen in this patient.

Objective: Genital herpes infections present with painful vesicles and erosions with associated lymphadenopathy. HSV-2 followed by HSV-1 are the most common viral culprits. Previous Next Score Report Lab Values Calculator Help Pause

10 Exam Section 1: Item 10 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 10. A 28-year-old man comes to the physician for a routine health maintenance examination. He is from Jamaica. He has practiced Rastafarianism all his life and follows a diet consisting only of natural juices, fruits, and grains. He also regularly smokes marijuana as part of his religion. Physical examination shows no abnormalities. His hematocrit is 32%, and serum vitamin B12 (cobalamin) concentration is within the reference range. Which of the following best describes the erythrocytes in this patient? Color Size (in Diameter) Shape A) Normal small normal B) Normal small 10% sickle, 90% normal C) Normal large normal D) Normal large 10% sickle, 90% normal E) Pale small normal F) Pale small 10% sickle, 90% normal G) Pale large normal H) Pale large 10% sickle, 90% normal Correct Answer: E. Choice E corresponds with a diagnosis of iron deficiency anemia (IDA). Erythrocytes in the setting of IDA are pale (hypochromic) and small in size (microcytic) but are normal in shape. Iron is required for the synthesis of heme, which is a necessary component of the hemoglobin molecule and, thus, of erythrocytes. It functions to shuttle oxygen to and from peripheral tissues. In individuals who do not have an adequate intake of dietary iron in the form of heme obtained from animal meat, deficiency can develop. IDA may also develop as the result of chronic blood loss from menstruation, colorectal bleeding, or can occur as a result of malabsorption syndromes such as celiac disease and in patients who have undergone gastric bypass surgery. Erythrocytes on the peripheral blood smear are hypochromic as a result of deficient hemoglobin concentration. It is thought that erythrocytes are microcytic as a result of continuing erythrocyte division in order to reach an adequate hemoglobin concentration; because hemoglobin stores are inadequate, cell division continues beyond what would normally occur and causes the cells to be smaller than normal. Treatment in the case of poor dietary intake is with oral iron supplementation. If severe, intravenous iron can be given. Incorrect Answers: A, B, C, D, F, G, and H. Normal color, small size, and normal shape (Choice A) describes several other causes of microcytosis without hypochromia, including thalassemia and anemia of chronic disease. While thalassemia may present with target cells on peripheral smear, the majority of erythrocytes will appear as microcytic. The degree of hypochromia is dependent on the severity of decreased hemoglobin concentration and mean corpuscular volume. Normal color and small size with 10% sickle and 90% normal shape (Choice B) can be seen in patients with sickle cell disease and a concomitant cause of microcytosis without hypochromia. For example, this may occur in patients with sickle cell disease and thalassemia, anemia of chronic disease, or iron deficiency anemia, depending on the severity of hypochromia. It may also occur in the setting of hemoglobin sickle cell disease. Normal color, large size, and normal shape (Choice C) is characteristic of macrocytic anemia, which is most commonly secondary to vitamin B12 (cobalamin) or folate deficiency. This patient does not have vitamin B12 deficiency and is unlikely to have folate deficiency given his vegetarian diet. Normal color and large size with 10% sickle and 90% normal shape (Choice D) is found in sickle cell disease with a concomitant cause of macrocytosis, most commonly folate or vitamin B12 deficiency. Folate deficiency can develop rapidly in patients with sickle cell disease as a result of ongoing hemolysis and should be replaced with a daily folate supplement. Pale color and small size with 10% sickle with 90% normal shape (Choice F) is found in sickle cell disease with a concomitant cause of microcytosis with hypochromia, such as iron deficiency anemia and anemia of chronic disease. Pale color, large size, and normal shape (Choice G) defines macrocytosis with hypochromia, which may result from a combination of diseases such as vitamin B12 or folate deficiency (macrocytosis) with iron deficiency, anemia of chronic disease, or thalassemia (hypochromia). Pale color and large size with 10% sickle and 90% normal shape (Choice H) can be seen in patients with sickle cell disease who are taking hydroxyurea. Hydroxyurea increases the proportion of hemoglobin F and also results in increased mean corpuscular volume (MCV) as a result of the release of more reticulocytes, which are larger than mature erythrocytes, thus resulting in macrocytosis.

Objective: Patients who follow a strict vegetarian diet are at risk for iron deficiency anemia from inadequate dietary heme ingestion. IDA presents with hypochromic, microcytic erythrocytes on peripheral smear. Previous Next Score Report Lab Values Calculator Help Pause

61 Exam Section 2: Item 12 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 12. A3-year-old boy is brought to the physician because of developmental delay. He has a history of a cleft palate that was repaired surgically at the age of 1 year. He did not walk until the age of 18 months and did not speak any words until the age of 20 months. His height and weight are at the 5th percentile; his head circumference is below the 5th percentile. Examination shows ptosis and a smooth philtrum. Maternal use of which of the following substances during pregnancy is the most likely cause of this patient's condition? A) Alcohol B) Cocaine C) Heroin D) Marijuana E) Nicotine Correct Answer: A. Alcohol is a proven teratogen, as alcohol use during pregnancy leads to irreversible fetal neurological effects and physical malformations in a cluster of symptoms called fetal alcohol syndrome. Fetal alcohol syndrome is the leading cause of intellectual disability in the United States and would explain this patient's failure to meet developmental milestones. As a result of the failure of cell migration, fetal alcohol syndrome also presents with physical abnormalities such as microcephaly, facial abnormalities (small palpebral fissures, thin vermilion border, cleft lip/palate, smooth philtrum), limb deformities, and congenital heart disease. Though alcohol is associated with obstetric complications such as preterm birth that commonly result from the use of other substances during pregnancy, these neonatal malformations are specific to alcohol use. Incorrect Answer: B, C, D, and E. Cocaine (Choice B) and nicotine products (Choice E) cause vasoconstriction of the placental vasculature, which can lead to preterm birth, intrauterine growth restriction, low birth weight, and placental abruption. Maternal hypertension and preeclampsia can result from the use of cocaine or cigarettes. Smoking cigarettes can additionally lead to respiratory complications in the mother (eg, asthma) and poor fetal oxygen delivery. Heroin (Choice C) use during pregnancy may result in obstetric complications such as intrauterine growth restriction, placental abruption, preterm birth, fetal passage of meconium, and fetal death. Heroin withdrawal in the fetus can present as neonatal abstinence syndrome, which features irritability, a high-pitched cry, poor sleep, and uncoordinated sucking reflexes that can lead to poor feeding and longer postpartum hospital stays. Marijuana (Choice D) use during pregnancy leads to an increased risk for developmental delay, autism spectrum disorder, and mortality, though is not associated with any of the physical malformations seen in fetal alcohol syndrome. Prenatal marijuana use also increases the risk for preterm birth and a fetus that is small for gestational age.

Objective: Alcohol is a known teratogen that causes fetal alcohol syndrome when used during pregnancy. Fetal alcohol syndrome presents with intellectual disability and developmental delays, along with physical malformations such as microcephaly, facial dysmorphia, limb deformities, and congenital heart disease. %3D Previous Next Score Report Lab Values Calculator Help Pause

57 Exam Section 2: Item 8 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 8. A 3685-g (8-lb 2-oz) male newborn is delivered vaginally at term to a 26-year-old primigravid woman. Pregnancy and delivery were uncomplicated. Apgar scores are 9 and 9 at 1 and 5 minutes, respectively. Physical examination shows no abnormalities.. An antibiotic ointment is applied to his eyes to prevent ocular infections. Inhibition of which of the following is most likely to occur with this treatment? A) Bacterial cell-wall synthesis B) Bacterial protein synthesis C) DNA gyrase D) Folic acid synthesis E) Mycolic acid synthesis Correct Answer: B. Neonatal conjunctivitis is defined as conjunctival inflammation during the first 30 days of life. The three most common causal organisms are Neisseria gonorrhoeae, Chlamydia trachomatis, and herpes simplex virus. Gonococcal conjunctivitis presents within three to five days after birth and is characterized by purulent discharge. Gonococcal infections are aggressive and can rapidly progress to keratitis, corneal perforation, or endophthalmitis. Chlamydial conjunctivitis presents within five to 14 days of birth and is characterized by watery discharge. Herpetic conjunctivitis presents more than 14 days after birth. All neonates should be given topical antibiotic prophylaxis to prevent gonococcal conjunctivitis. Historically, topical silver nitrate was a common prophylactic agent that was also a frequent cause of chemical conjunctivitis, which presents within the first three days after birth. The use of silver nitrate has more recently been supplanted by the use of topical erythromycin, which binds to bacterial ribosomal 50s subunits and prevents protein synthesis. Incorrect Answers: A, C, D, and E. Inhibition of bacterial cell-wall synthesis (Choice A) is the mechanism of B-lactams such as penicillins and cephalosporins. Penicillin resistance is relatively prevalent in N. gonorrhoeae, and topical agents from this drug class are not routinely used for the prophylaxis of neonatal conjunctivitis. Inhibition of DNA gyrase (Choice C) is the mechanism of quinolone antibiotics. The side effect profile of quinolones, including arthropathy and tendinopathy, precludes their use in children and infants. Inhibition of folic acid synthesis (Choice D) is the mechanism of trimethoprim-sulfamethoxazole. Trimethoprim-sulfamethoxazole may contribute to neonatal kernicterus and is not commonly used for the prophylaxis of neonatal conjunctivitis. Mycolic acid synthesis (Choice E) is inhibited by antimycobacterial drugs such as isoniazid. Such a mechanism is employed in the prevention or treatment of tuberculosis, which is not commonly associated with neonatal conjunctivitis.

Objective: All neonates should receive topical prophylaxis against bacterial conjunctivitis caused by N. gonorrhoeae and C. trachomatis. Erythromycin is the most commonly used prophylactic agent and functions by binding to the bacterial 50s ribosomal subunit and preventing bacterial protein synthesis. Previous Next Score Report Lab Values Calculator Help Pause

131 Exam Section 3: Item 32 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 32. An investigator wants to determine if coffee drinking is associated with pancreatic cancer. Coffee drinking histories of persons with cancer were compared with those of healthy persons. Which of the following best describes this type of study? A) Case series B) Case-control study C) Cohort study D) Cross-sectional study E) Ecologic study F) Randomized clinical trial Correct Answer: B. A case-control study investigates an association between an exposure and an outcome. In this study design, a group of patients with the disease (cases) are identified. A group of patients without the disease (controls) are matched on baseline characteristics to the cases. Exposure data for the two groups is collected, and these data are compared to determine association with the outcome (disease) in question. An odds ratio may be calculated to compare exposures between groups. In this study, the cases are those persons with pancreatic cancer and the controls are those persons who are healthy and matched to the persons with pancreatic cancer. These persons are surveyed for histories of exposure, in particular, exposure to coffee drinking. Incorrect Answers: A, C, D, E, and F. A case series (Choice A) is a descriptive study design in which a number of consecutive or nonconsecutive cases of a disease and/or treatment are described in detail, with information about exposure, demographics, and comorbidities. Case series do not imply a cause-and-effect relationship. They do not test a hypothesis nor are they randomized. They are useful in characterizing the natural history of a disease or response to treatment. They are also useful in describing rare diseases, as the small population size may not permit conduction of a larger cohort or randomized trials with sufficient power. A cohort study (Choice C) identifies a group of patients and follows them over time to identify whether an exposure is associated with an outcome of interest. Cohort studies may be retrospective or prospective in design. In a prospective design, the hypothesis and analysis protocols are established prior to the start of the study period. In a retrospective design, the hypothesis or question is designed after the study time period has passed. An example of a prospective cohort study would be following a group of 1,000 smokers for a time period of 10 years and identifying the proportion of these patients who develop pancreatic cancer to identify the risk for pancreatic cancer in smokers as compared with a control group of nonsmokers. A cross-sectional study (Choice D) seeks to identify the prevalence of a condition at a particular point in time. An example of a cross-sectional study would be a single survey of a population inquiring whether patients have coronary artery disease and concurrently inquiring about activity levels and diet. Thus, the risk factor and the outcomes are measured simultaneously. The study does not follow patients over time. All information is collected at a single time point. An ecologic study (Choice E) is used to find associations between exposure and an outcome particularly when the outcome is rare. In an ecologic study, data are aggregated at the population level and therefore, conclusions about individuals cannot be drawn. A randomized clinical trial (Choice F) is an experimental study design. Patients are randomly allocated to two or more interventional arms or control arms, and these patients are followed over time to evaluate an outcome of interest. Randomized design minimizes opportunity for bias; thus, a randomized interventional study can be used to imply causation. Common examples of randomized trials include therapeutic comparisons between a new drug and the previous standard of care.

Objective: A case-control study is used to find associations between an outcome and an exposure. This study design includes identifying a group with the specified outcome (cases) and without the outcome (controls), and then comparing exposures across the groups. %3D Previous Next Score Report Lab Values Calculator Help Pause

t disease, is a disorder of osteoclast and osteoblast activity marked by the formation of disorganized, woven bone instead of lamellar bone. This disease may present with pathologic fractures, lytic or sclerotic lesions on x-rays, and cranial nerve dysfunction from compression within the neural foramina of the skull base. Concentrations of serum calcium, phosphate, and PTH will be normal, while alkaline phosphatase concentrations will be increased. Osteomalacia (Choice D) is an adult form of vitamin D deficiency that may be caused by decreased gut absorption of vitamin D, insufficient sunlight, poor diet, or chronic kidney failure. However, none of these causes account for increased serum calcium, decreased serum phosphate, and increased urine calcium.

Objective: Excessive production of PTH in primary hyperparathyroidism results in increased serum calcium from increased osteoclast activity, intestinal absorption, and renal tubular absorption. Serum phosphate concentration is decreased secondary to increased excretion by the renal tubules. Osteoporosis is a potential complication of primary hyperparathyroidism as the subperiosteal long bones undergo abnormal resorption, leaving the bones weakened and prone to fractures. Previous Next Score Report Lab Values Calculator Help Pause

177 Exam Section 4: Item 28 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 28. A 22-year-old woman comes to the physician 3 months after she noticed a painless, slowly enlarging mass on the left side of her neck. Physical examination shows a freely mobile, soft, cystic mass with a cutaneous surface opening. The physician explains that it is from incomplete fusion during embryonic development. Which of the following is the most likely location of the opening of the duct leading to the mass in this patient? A) Anterior to the sternocleidomastoid muscle B) Midline on the neck C) Postauricular D) Posterior to the parotid gland E) Submental Correct Answer: A. Branchial cleft cysts are developmental abnormalities that result from failure of fusion of the mesodermal branchial arches, with subsequent failure of obliteration of the ectodermal branchial clefts. The second through fourth branchial clefts are most commonly affected. Branchial cleft cysts appear as a cystic mass in the lateral portion of the neck. Characteristic features of branchial cleft cysts, which may aid in their diagnosis, are immobility during swallowing and a location anterior to the sternocleidomastoid muscle. A fistulous tract may or may not be present. Branchial cleft cysts may become infected. Treatment is with surgical excision. Incorrect Answers: B, C, D, and E. A location midline on the neck (Choice B) is suggestive of a thyroglossal duct cyst. Thyroglossal duct cysts are mobile during swallowing or with protrusion of the tongue. Postauricular location (Choice C) is usually suggestive of mastoiditis or lymphadenopathy. Cysts of the first branchial cleft may be located posterior or inferior to the ear and posterior to the parotid gland (Choice D), but cysts of the first branchial cleft are extremely rare. Submental location (Choice E) is usually suggestive of lymphadenopathy. Submental abscess or cellulitis may occur in the submental space, a condition known as Ludwig angina, which may acutely threaten the airway.

Objective: Branchial cleft cysts are developmental abnormalities that result from failure of fusion of the mesodermal branchial arches and failure of obliteration of the ectodermal branchial clefts. They may be identified by their location in the lateral aspect of the neck, anterior to the sternocleidomastoid muscle, and by their immobility with swallowing. Treatment is with surgical excision. %3D Previous Next Score Report Lab Values Calculator Help Pause

115 Exam Section 3: Item 16 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 16. A 47-year-old woman comes to the emergency department because of the recent onset of acute asthma-like breathing difficulty. She has had diarrhea and intermittent flushing for 4 months. An x-ray of the chest shows a well-circumscribed 2.3-cm mass in the central part of the lower lobe of the right lung. Urine 5-hydroxyindoleacetic acid concentration is increased. Which of the following is the most likely diagnosis? A) Adenocarcinoma B) Carcinoid tumor C) Malignant lymphoma D) Metastatic hepatocellular carcinoma E) Squamous cell carcinoma Correct Answer: B. Carcinoid syndrome is a rare disorder that results from the production of high concentrations of serotonin by a carcinoid tumor. Carcinoid tumors are neuroendocrine cell neoplasms that most commonly occur in the gastrointestinal tract, though they can occur as a primary lung lesion as well. Normally the liver eliminates active serotonin secreted by a primary gastrointestinal carcinoid tumor; however, if the cancer has metastasized or if the primary lesion has developed outside of the gastrointestinal tract, then physiologically active serotonin enters the systemic circulation and manifests as carcinoid syndrome. Symptoms include weight loss, flushing, diarrhea, bronchospasm, right-sided cardiac valvular disease, and abdominal pain. Physical examination may show erythema, wheezing, and a cardiac murmur dependent on which valve is involved (tricuspid valve is commonly affected). Diagnosis can be confirmed by measuring urine concentration of 5- hydroxyindoleacetic acid (5-HIAA). 5-HIAA is the primary metabolite of serotonin, and increased urine concentrations suggest increased concentrations of serotonin in the blood. Additional diagnostic information can be obtained by CT imaging and octreotide scintigraphy to localize the tumor and evaluate for metastatic disease. Treatment consists of surgical resection and medical therapy with octreotide (a somatostatin analog). Patients may suffer from carcinoid heart disease if the cardiac valves are affected, with the eventual consequent development of right heart failure. Incorrect Answers: A, C, D, and E. Adenocarcinoma (Choice A) of the lung 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. Malignant lymphoma (Choice C) often presents with constitutional symptoms such as fatigue, unintentional weight loss, fevers, and night sweats. Physical examination findings may include diffuse lymphadenopathy and hepatosplenomegaly. Malignant lymphoma may present with paraneoplastic syndromes, but they do not typically produce serotonin. Metastatic hepatocellular carcinoma (Choice D) is associated with hepatitis, cirrhosis, and chronic liver disease. It is characterized by weight loss, jaundice, hepatomegaly, and ascites. Increased alpha-fetoprotein (AFP) is seen in many hepatic malignancies. Squamous cell carcinoma (Choice E) of the lung is the second most common type of primary lung cancer after adenocarcinoma. Features associated with squamous cell carcinoma of the lung include pulmonary cavitations, central location, and hypercalcemia caused by paraneoplastic parathyroid hormone-related peptide (PTHPP) production.

Objective: Carcinoid syndrome results from increased concentrations of circulating serotonin, most commonly caused by a carcinoid tumor that has metastasized from the gastrointestinal tract (bypassing metabolism by the liver) or less commonly caused by a primary carcinoid tumor of nongastrointestinal origin. Diagnosis is made by measuring concentrations of the serotonin metabolite 5-HIAA. Previous Next Score Report Lab Values Calculator Help Pause

113 Exam Section 3: Item 14 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 14. A 45-year-old man receives the diagnosis of Achilles tendon rupture. He is a long-distance runner. He has had an upper respiratory tract infection for 3 weeks and was treated with antibiotics. Which of the following agents is the most likely cause of the Achilles tendon rupture? A) Amoxicillin B) Azithromycin C) Levofloxacin D) Rifampin E) Trimethoprim-sulfamethoxazole Correct Answer: C. Levofloxacin is a fluoroquinolone antibiotic. Its mechanism of action is to inhibit bacterial DNA gyrase, an enzyme necessary for DNA replication. It is used commonly to treat respiratory infections and urinary tract infections. A known complication of fluoroquinolone use is the weakening of tendons and subsequent rupture. This tends to occur in individuals with high tendon loading (in this case, a distance runner) and in older persons. Because of this side effect, alternative antibiotic choices are often recommended in these populations. The mechanism of fluoroquinolone-induced tendinopathy is not fully elucidated. However, it is thought that these drugs have a direct toxic effect on musculoskeletal tissues and may inhibit collagen synthesis and promote collagen degradation. Incorrect Answers: A, B, D, and E. Amoxicillin (Choice A) is a cephalosporin antibiotic that is used to treat pediatric upper respiratory tract bacterial infections. It is generally well-tolerated; however, it can cause rash when given to patients with Epstein-Barr virus (classically, when expired) and is associated with hypersensitivity and atopic reactions. Azithromycin (Choice B) is a macrolide antibiotic. Its mechanism of action involves inhibiting the 50s ribosomal subunit, which stops bacterial protein translation. It can cause gastrointestinal symptoms from increased motility, and is associated with prolongation of the QT interval. Rifampin (Choice D) is an antituberculosis drug that increases the activity of the CYP-450 enzymes in the liver. It also causes bodily fluids to exhibit an orange tinge. Trimethoprim-sulfamethoxazole (Choice E) is a folate antagonist combination antibiotic. Adverse effects include acute interstitial nephritis, hemolysis in patients with glucose-6-phosphate dehydrogenase deficiency, and teratogenesis as a result of folate antagonism in the fetus.

Objective: Fluoroquinolone antibiotics inhibit bacterial DNA gyrase enzymes leading to decreased bacterial replication. They have been shown to increase the risk for tendon rupture and tendinopathy. Previous Next Score Report Lab Values Calculator Help Pause

166 Exam Section 4: Item 17 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 17. A female newborn delivered at term is found to have a cleft palate with cleft lip bilaterally. Physical examination shows no other abnormalities. Cranial ultrasonography shows no abnormalities. Failure of fusion at which of the following sites is the primary cause of this patient's cleft lip? A) Frontonasal and eye prominences B) Frontonasal and maxillary prominences C) Mandibular process and external auditory meatus D) Mandibular prominence and secondary pharyngeal arch E) Maxillary and nasal prominences Correct Answer: E. Cleft lip and cleft palate may occur in either complete (extending through the nares) or incomplete (not involving the nares) fashion and may occur unilaterally or bilaterally. Cleft lip occurs from a failure of fusion of the maxillary and nasal prominences. Cleft palate occurs from a failure of fusion of the palatine prominences. Cleft lip and palate may occur as isolated developmental anomalies or in the setting of chromosomal abnormalities (eg, Patau syndrome or trisomy 13) or syndromes of midfacial hypoplasia (eg, Treacher Collins syndrome). Treatment is with surgical repair. Incorrect Answers: A, B, C, and D. Frontonasal and eye prominences (Choice A) contribute to the structures of the midface. The frontonasal prominence forms the medial and lateral nasal processes that form the nose and upper lip. Failure of the frontonasal prominence to fuse with the eye prominences would not lead to cleft lip and palate. Frontonasal and maxillary prominences (Choice B) contribute to the nasal structures and floors of the orbits. Failure of these structures to fuse would result in midline facial anomalies but not cleft lip. Mandibular process and external auditory meatus (Choice C) contribute to the structures of the ear. Failure of fusion of these structures would not result in cleft lip or palate.Mandibular prominence and secondary pharyngeal arch (Choice D) contribute to the structures of the jaw and neck. Failure of fusion of these structures would not result in cleft lip or palate.

Objective: Cleft lip and cleft palate may occur in either complete (extending through the nares) or incomplete (not involving the nares) fashion and may occur unilaterally or bilaterally. Cleft lip occurs from a failure of fusion of the maxillary and nasal prominences, while cleft palate occurs from a failure of fusion of the palatine prominences. %D Previous Next Score Report Lab Values Calculator Help Pause

180 Exam Section 4: Item 31 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 31. Clomiphene enhances female fertility because of which of the following characteristics? Effect on Follicle-stimulating and Luteinizing Hormone Release Activity in the Pituitary Gland A) Antiestrogenic decreases B) Antiestrogenic increases C) Estrogenic decreases D) Estrogenic increases Correct Answer: B. Clomiphene is a selective estrogen receptor modulator, a class that exhibits a mix of agonist or antagonist effects on the estrogen receptor in different tissues. Clomiphene binds the estrogen receptors in the pituitary gland. By binding the estrogen receptor, it blocks the negative feedback from circulating estrogen and thus has an antiestrogenic effect. As a result of decreased negative feedback, the pituitary gland increases the production of follicle-stimulating hormone (FSH) and luteinizing hormone (LH). FSH and LH are integral in the process of ovulation and implantation. At the start of the follicular phase, just after menses occurs, FSH increases to support a developing ovarian follicle. 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 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. LH remains low through 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. With the increased production of FSH and LH induced by clomiphene, both ovulation and implantation are facilitated. Incorrect Answers: A, C, and D. While clomiphene does have an antiestrogenic effect on the pituitary gland, this effect leads to an increase in FSH and LH rather than a decrease (Choice A). The effect of clomiphene on the pituitary gland is antiestrogenic, not estrogenic (Choices C and D). An estrogenic effect on the pituitary gland would lead to increased negative feedback and decrease the production of FSH and LH. Decreased FSH and LH would perpetuate infertility by preventing ovulation and implantation.

Objective: Clomiphene is a selective estrogen receptor modulator, which demonstrates anti-estrogenic effects on the pituitary gland. By blocking the negative feedback of estrogen on the pituitary gland, the production of FSH and LH is increased, supporting ovulation and implantation. %3D Previous Next Score Report Lab Values Calculator Help Pause

181 Exam Section 4: Item 32 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 32. A 27-year-old woman has had severe lower left quadrant abdominal pain and vomiting for 3 hours. She is weak and hypovolemic. She has a history of regular menstrual cycles and frequent pelvic infections. Her last menstrual period began 9 weeks ago and lasted 4 days. Which of the following is the most likely site of the primary pathologic process? A) Cervix B) Ovary C) Rectouterine pouch (of Douglas) D) Uterine tube E) Vermiform appendix Correct Answer: D. Pregnancy is suspected when there is a missed or delayed menstrual period. Ectopic pregnancy often presents with symptoms including lower quadrant abdominopelvic pain, vaginal bleeding, nausea, or vomiting. Ectopic pregnancy is an abnormal pregnancy in which the fertilized ovum implants in the uterine (fallopian) tube (most common), on the ovary, within the peritoneal cavity, or in any nonendometrial location. Risk factors for ectopic pregnancy include prior ectopic pregnancies, in vitro fertilization, and prior pelvic inflammatory disease or sexually transmitted infections. Rupture of an ectopic pregnancy can lead to hemoperitoneum. If undiagnosed or severe, it can lead to syncope and hemorrhagic shock. Ectopic pregnancy is nonviable. Ectopic pregnancy can be managed medically, using methotrexate for small, early pregnancies, but may require salpingectomy or evacuation with laparoscopy for larger pregnancies in cases of medical treatment failure or in any case of complication. Incorrect Answers: A, B, C, and E. An ectopic pregnancy can present on the cervix (Choice A), ovary (Choice B), peritoneal cavity, or any nonendometrial location. However, cervical or ovarian ectopic pregnancies are less common, and the majority of ectopic pregnancies implant in the uterine (fallopian) tube. The rectouterine pouch (of Douglas) (Choice C) is a potential space between the uterus and the rectum in females. In a ruptured ectopic pregnancy resulting in hemoperitoneum, blood would pool in the rectouterine pouch. It would not likely be a location of ectopic pregnancy implantation. The vermiform appendix (Choice E), when inflamed or infected in the setting of appendicitis, typically causes right lower quadrant pain, fever, nausea, and vomiting. In a patient with a delayed menstrual period and left lower quadrant abdominal pain, ectopic pregnancy is more likely.

Objective: Ectopic pregnancy is an abnormal pregnancy in which the fertilized ovum implants in the fallopian tube (most common), on the ovary, within the peritoneal cavity, or in any nonendometrial location. It typically presents with symptoms including lower quadrant abdominopelvic pain, vaginal bleeding, nausea, or vomiting and can be complicated by rupture and hemoperitoneum. Previous Next Score Report Lab Values Calculator Help Pause

183 Exam Section 4: Item 34 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 34. The photomicrograph is of a stained stool smear. Chronic infection with this organism is most likely to cause which of the following? A) Bloody diarrhea B) Constipation C) Intestinal obstruction D) Intestinal malabsorption O E) Iron deficiency 10um Correct Answer: D. The photomicrograph displays a flagellated, pear-shaped trophozoite with two large, central nuclei and a median axostyle, consistent with the appearance of Giardia lamblia. G. lamblia is a flagellated, parasitic organism that causes giardiasis. Giardiasis is usually acquired through the ingestion of cysts in contaminated water, which then mature into trophozoites in the small bowel. Parasitic overgrowth of the small bowel leads to intestinal malabsorption. Giardiasis presents with persistent diarrhea that is foul-smelling and fatty, as well as abdominal pain, bloating, cramping, and weight loss. Diagnosis is made with stool culture, microscopy, or stool antigen testing. Metronidazole is the first line treatment. Incorrect Answers: A, B, C, and E. Bloody diarrhea (Choice A) is characteristic of intestinal amoebiasis caused by Entamoeba histolytica. Amoebiasis presents with bloody diarrhea and intense abdominal pain. It can lead to colonic perforation or extrahepatic invasion with formation of amoebic cysts in the liver, lung, or brain. While the diarrhea associated with giardiasis may occasionally be bloody, this symptom is less frequently observed than intestinal malabsorption. Constipation (Choice B) and intestinal obstruction (Choice C) are not features of giardiasis. Both of these symptoms may be observed in severe infections with the parasitic roundworm Ascaris lumbricoides. Iron deficiency (Choice E) may result from chronic blood loss in association with infection with several species of worm, including the hookworms Ancylostoma duodenale and Necator americanus, the whipworm Trichuris trichiura, and the trematodes Schistosoma haematobium and S. mansoni.

Objective: Giardia lamblia is a flagellated, parasitic organism that causes giardiasis. Light microscopy discloses a pear-shaped, binucleate, flagellated trophozoite. Parasitic overgrowth of the small bowel leads to intestinal malabsorption. Previous Next Score Report Lab Values Calculator Help Pause

150 Exam Section 4: Item 1 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 1. A 60-year-old man develops pain, erythema, and swelling of the right great toe. Serum uric acid concentration is three times normal. Which of the following findings is most common in patients with this condition? A) Absence of aminotransferase B) Absence of glucose 6-phosphatase C) Absence of glutathione peroxidase D) Absence of hypoxanthine guanine phosphoribosyltransferase E) No specific enzyme or renal defect Correct Answer: E. Gout classically presents following consumption of a purine-rich meal, especially red meat or seafood, often in males 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 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. The vast majority of cases of gout and persons with hyperuricemia have no specific enzyme or renal defect. It is largely a diet and lifestyle-driven disease. Alcohol use is commonly a precipitant, as is diuretic use. The preferred treatment of a gout flare is nonsteroidal anti-inflammatory drugs (NSAIDS) or colchicine if there exists a contraindication to NSAIDS. Intra-articular corticosteroid injection may also be used in an acute flare. Chronic gout can be treated with a decrease in alcohol consumption, decrease in red meat and seafood consumption, and weight loss. Flares can also be prevented with xanthine oxidase inhibitors such as allopurinol. Incorrect Answers: A, B, C, and D. Absence of aminotransferase (Choice A) can describe many enzyme deficiencies, such as ornithine aminotransferase deficiency, in which increased concentrations of ornithine accumulate in the body leading to early-onset cataracts, night blindness, and myopia. Absence of glucose-6-phosphatase (Choice B) is the cause of glycogen storage disease type I, which presents as growth restriction, fasting hypoglycemia, and accumulation of glycogen in the liver caused by the inability to conduct glycogenolysis and gluconeogenesis. Absence of glutathione peroxidase (Choice C) leads to increased intracellular reactive oxygen species and subsequent cell membrane, DNA, protein, and organelle damage. Absence of hypoxanthine guanine phosphoribosyltransferase (Choice D) is a characteristic mutation in Lesch-Nyhan syndrome. Deficiency of this enzyme leads to increased uric acid concentrations, intellectual disability, poor muscle control, and self-injurious behavior.

Objective: Gout is a common condition resulting from high concentrations of uric acid in the plasma and tissues. Uric acid crystals precipitate when supersaturated. Commonly, this condition is caused by lifestyle-based factors without specific inborn errors of metabolism or organ dysfunction. I3D Previous Next Score Report Lab Values Calculator Help Pause

184 Exam Section 4: Item 35 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 35. A 37-year-old man with HIV infection comes to the physician for a routine examination. He began treatment with zidovudine (AZT), lamivudine (3TC), and efavirenz 8 months ago. Physical examination today shows no abnormalities. Laboratory findings from 6 months ago and today are shown in the table. 6 Months Ago 450/mm3 10,000 copies/mL Today 250/mm3 CD4+ T-lymphocyte count (N2500) Plasma HIV viral load 100,000 copies/mL The drug resistance in this patient is most likely due to which of the following steps of HIV viral replication? A) Assembly of viral particles B) Initiation complex of peptide formation C) Reverse transcription D) Synthesis of late proteins E) Uncoating of viral nucleic acid Correct Answer: C. Reverse transcription is the most likely step in HIV viral replication to experience changes resulting in drug resistance. HIV drugs are separated into several classes and each of them work via a different mechanism to prevent viral replication. Drugs of disparate mechanisms are combined to inhibit HIV replication at multiple steps, which typically prevents the development of resistance. HIV infection and replication can be broken into seven primary steps: binding, fusion, reverse transcription, integration, replication, assembly, and budding. After HIV binds to the surface of CD4 cells via interaction of its specific glycoproteins with CD4 receptors such as CCR5, the virus fuses with the cell membrane and releases its contents into the cytoplasm. HIV RNA is reverse transcribed by reverse transcriptase to create HIV DNA, followed by incorporation of this DNA into the CD4 cell genome. Once integrated, HIV hijacks the cellular machinery to produce the HIV genome and all the requisite proteins to form a new HIV viral particle. The new virus then buds off of the CD4 cell and is released to infect other lymphocytes. Drugs that block HIV reverse transcriptase are either nucleoside reverse transcription inhibitors (NRTIS) or non-nucleoside reverse transcription inhibitors (NNRTI). AZT and 3ŤC are both NRTI,, and efavirenz is a NNRTI. With regard to the steps involved in viral replication, reverse transcription of HIV RNA is highly error prone and can lead to the development of mutations. Additionally, NRTIS and NNRTIS, especially 3TC and efavirenz, have a low genetic barrier to resistance, meaning that HIV resistance to these drugs is not uncommon and can develop quickly. Once resistance is acquired, the continued use of the same medication or those in the same class can exert additional selection pressure, leading to increasingly resistant strains of HIV. Resistance rates in current clinical practice are decreasing as a result of the combination of NNRTIS and NRTIS with drugs from other classes, including integrase and protease inhibitors. Incorrect Answers: A, B, D, and E. Assembly of viral particles (Choice A) into mature viral capsules is inhibited by protease inhibitors. The HIV-specific proteins must be cleaved in order for viral particles to reach maturity. Resistance in this step is unlikely as this patient is not taking a protease inhibitor. Initiation complex of peptide formation (Choice B) and synthesis of late proteins (Choice D) are steps that occur in the nucleus of the host cell by utilizing host machinery. These processes are not completed by HIV machinery. Uncoating of viral nucleic acid (Choice E) occurs immediately upon viral entry into the host cell. This process is prevented through blocking HIV from binding to the host cell by either competitively binding CCR5 or binding to the envelope glycoprotein of HIV. This patient is not taking any fusion inhibitors to which he could develop resistance.

Objective: HIV reverse transcriptase is an error prone step during HIV viral replication, which can lead to the development of mutations that predispose to viral resistance. Drugs that block this enzyme are termed NRTIS or NNRTIS. When used alone and not in combination with drugs from other classes they can potentially induce resistance via mutation of the HIV reverse transcriptase. Previous Next Score Report Lab Values Calculator Help Pause

127 Exam Section 3: Item 28 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 28. An investigator is studying the human immune response to tumor cell antigens in malignancies. Which of the following sets of cancer types and tumor antigens is most likely to produce the highest antibody titer? Cancer Type Tumor Antigen A) B-cell lymphoma CD19 B) Breast cancer HER2/neu C) Cervical cancer human papillomavirus type 16 E6 protein D) Melanoma tyrosinase E) Prostate cancer prostatic acid phosphatase Correct Answer: C. Of the options presented, E6 protein produced by human papillomavirus (HPV) is the only nonhuman antigen and will likely produce the highest antibody titer. Immunogenicity refers to the ability of an antigen to invoke a humoral and cell-mediated immune response. Cervical cancer is a common malignancy. Risk factors for development include infection with human papillomavirus multiple sexual partners, and immunosuppression (eg, concomitant HIV infection). High-risk HPV subtypes are 16, 18, 31, and 33. HPV-16 produces a protein called E6, which is a potent p53 inhibitor. Squamous intraepithelial lesions may develop that have the potential to transform into cervical carcinoma. Cervical cancer is typically asymptomatic initially, and patients may develop vaginal bleeding, pelvic pain, and bladder outlet obstruction as a result of the enlarging mass. Advanced disease involves local invasion of the bladder, ureters, vagina, and rectum as well as distant metastases. Incorrect Answers: A, B, D, and E. Self-antigens are produced normally by the body and generally have poor immunogenicity (except in cases of autoimmune disorders). CD19 (Choice A) is a transmembrane protein located on B- lymphocytes and dendritic cells. It is a target for immunotherapy in B-cell lymphomas. HER2/neu (Choice B) is a receptor tyrosine-kinase encoded by an oncogene that is associated with breast cancer. Tyrosinase (Choice D) is an oxidase involved in the production of melanin by melanocytes. Mutations that result in its loss of function can cause albinism. Prostatic acid phosphatase (Choice E) is normally produced by the prostate. Concentrations may be increased in prostate cancer.

Objective: Immunogenicity describes the ability of an antigen to stimulate an adaptive immune response. Foreign antigens generally exhibit greater immunogenicity than self-antigens. %D Previous Next Score Report Lab Values Calculator Help Pause

187 Exam Section 4: Item 38 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 38. A 53-year-old woman is brought to the emergency department because of a 1-week history of severe chest pain and cough productive of moderate amounts of thick yellow sputum. She currently takes no medications. She is not in acute respiratory distress. Her temperature is 37.6°C (99.6°F), pulse is 95/min, respirations are 32/min, and blood pressure is 120/80 mm Hg. A chest x-ray shows consolidation of the right lower lobe of the lung. Culture of sputum grows Streptococcus pneumoniae. Treatment with antibiotics is started. This patient is most likely to sustain injury to normal pulmonary tissue by inflammatory cells releasing which of the following mediators? A) Complement proteins B) a-Defensins C) Interleukin-1 (IL-1) D) Prostaglandins E) Reactive oxygen species F) Transforming growth factor-ß Correct Answer: E. The production of reactive oxygen species is induced in the immune response to infection and is responsible for damage to invading microbes and host tissue alike. NADPH oxidase complex within phagocytes, such as neutrophils and monocytes, 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. This process is largely contained within phagolysosomes; however, a sufficiently large inflammatory response can cause significant damage to the surrounding tissue. Reactive oxygen species are unstable molecules that damage cells through disruption of DNA and RNA, oxidation of fatty acids in lipids (lipid peroxidation), oxidation of amino acids in proteins, and oxidative deactivation of functional enzymes. These mechanisms may trigger apoptosis as a result of irreversible cellular damage or necrosis, which causes further release of inflammatory cellular contents. Incorrect Answers: A, B, C, D, and F. Complement proteins (Choice A) are a group of circulating proteins synthesized by the liver that participate in both the innate immune response and the humoral (antibody-mediated) component of the adaptive immune response. The complement system may be activated via the classical pathway (antigen-antibody binding to C1), alternative pathway (C3b binding to microbial surface proteins, which is antibody-independent), and/or the lectin pathway (mannose residues on pathogen surface, which is also antibody-independent). Complement activation results in opsonization of invading microbes, direct microbial killing via formation of the membrane attack complex (MAC), and proinflammatory signaling (eg, C5a mediation of neutrophil chemotaxis). Complement proteins are not released directly by inflammatory cells. a-Defensins (Choice B) are antimicrobial peptides stored within the granules of neutrophils. They are involved in direct pathogen killing in phagolysosomes via membrane disruption and act as proinflammatory mediators when neutrophil degranulation releases them into the extracellular space. Defensins are not implicated in direct damage to host cells. Interleukin-1 (IL-1) (Choice C) is a proinflammatory cytokine secreted by macrophages, endothelial cells, dendritic cells, and B lymphocytes. It has numerous effects, including the promotion of fever, vasodilation, and stimulation of endothelial cells to express adhesion proteins for leukocyte recruitment. It is also known as osteoclast-activating factor. Prostaglandins (Choice D) are immune mediators derived from arachidonic acid that promote vasodilation in the inflammatory response. They do not directly cause cellular damage. Transforming growth factor-ß (Choice F) is an anti-inflammatory cytokine secreted by most cells in the body, which mediates downregulation of the inflammatory response as well as tissue regeneration and scar formation after injury.

Objective: Inflammation is the response to harmful stimuli such as infection, with direct microbial killing resulting from the leukocyte production of reactive oxygen species and antimicrobial peptides. Inadvertent host tissue damage can occur from exposure to reactive oxygen species, which result in structural damage to DNA, RNA, enzymes, and the cellular lipid membrane. Previous Next Score Report Lab Values Calculator Help Pause

139 Exam Section 3: Item 40 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 40. A 23-year-old woman is brought to the emergency department by her roommate because she was found unresponsive on the living room floor. Examination shows constricted pupils and stupor. Which of the following substances is most likely responsible for these effects? A) Amphetamine B) Androstenedione C) Caffeine D) Cocaine E) Heroin F) LSD G) Marijuana H) Nicotine I) Phencyclidine (PCP) Correct Answer: E. This patient is likely intoxicated with an opiate such as heroin. Opiates are central nervous system (CNS) depressants used as analgesics and recreational drugs. Opiates act throughout the CNS and peripheral nervous system and interact with several neurotransmitter systems, leading to diverse effects. Opiate intoxication causes euphoria (caused by interaction with dopamine), altered mental status, sedation (stupor if severe), bradycardia and hypotension, depressed respiratory drive, and constricted pupils. Miosis is a distinctive finding that is less common in intoxication with other CNS depressants and is caused by direct opiate receptor activity in brain areas responsible for pupillary control. Opiates also act on receptors within the enteric nervous system, decreasing gut motility and causing constipation. Incorrect Answers: A, B, C, D, F, G, H, and I. Amphetamine (Choice A) and cocaine (Choice D) are CNS stimulants that cause euphoria and increase sympathetic tone, leading to tachycardia and hypertension, pupillary dilation, and restlessness (rather than sedation). As a result of increased synaptic dopamine, hallucinations and paranoia can occur. Androstenedione (Choice B) is a steroid precursor to estradiol and testosterone. Toxicity signs may include infertility and atypical secondary sex characteristics. Severe sedation would be uncommon. Caffeine (Choice C) is a commonly used stimulant used with diverse receptor activities. In smaller doses, caffeine increases energy, alertness, and causes mild diuresis, while toxic doses of caffeine produce agitation, seizures, vasoconstriction, tachycardia, hypertension, and electrolyte disturbances. LSD (Choice F) is a potent hallucinogen known to cause auditory and visual hallucinations, depersonalization, flashbacks, and paranoia. Marijuana (Choice G) is a hallucinogen that causes euphoria, paranoia, hallucinations, cognitive slowing, conjunctival injection, and dilated pupils. Phencyclidine (PCP) (Choice I) can cause similar symptoms to LSD and is also associated with aggression, rotatory nystagmus, and ataxia. LSD, marijuana, and phencyclidine are associated with sympathetic activation rather than sedation and respiratory depression. Nicotine (Choice H) is a CNS stimulant found in tobacco. Nicotine toxicity includes agitation, seizures, heart palpitations, and vomiting. Miosis and sedation would be atypical.

Objective: Opiate intoxication causes symptoms of CNS depression including sedation, respiratory depression, bradycardia, and hypotension, along with characteristic miosis on physical examination. Sedation, depressed vital signs, and miosis are atypical of stimulant and hallucinogen toxidromes. Previous Next Score Report Lab Values Calculator Help Pause

151 Exam Section 4: Item 2 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 2. A clinical trial is conducted to compare the efficacy of a surgical procedure versus medical management of gastroesophageal reflux disease. In the study, 800 patients are randomly assigned to either the surgical procedure group or the medical management group; each group consists of 400 patients. Of the patients assigned to the surgical procedure group, 75 do not undergo the operation during the study period. Of the patients assigned to the medical management group, 50 undergo the operation during the study period. In an intention-to-treat analysis, which of the following most accurately represents how many patients would be analyzed in the surgical procedure and medical management groups, respectively? Surgical Procedure Group Medical Management Group OA) 325 350 B) 375 425 C) 400 400 D) 425 350 E) 450 325 Correct Answer: C. Intention-to-treat analysis is used to preserve randomization when attrition or crossover is introduced into the study. According to intention-to-treat analysis, all patients should be analyzed as part of the group to which they were initially randomized. Although randomization is preserved, intention-to-treat analysis leads to problems as it ignores protocol deviations, withdrawal from the study, and noncompliance. In this study, the intention-to-treat analysis ignores the fact that a proportion of the surgical group did not undergo surgery, and a proportion of the medical management group did undergo surgery. Because of this, the true effectiveness of the treatments are diluted. In this example, intention-to-treat analysis would dictate that the 400 patients initially randomized to the surgical procedure should be analyzed as the surgical procedure group without any adjustments for crossover. Similarly, the patients initially assigned to the medical management group would be analyzed as part of the medical management group. Incorrect Answers: A, B, D, and E. Choice A reflects analysis of the actual interventions of the study regarding the medical management group, and only accounts for the 325 persons originally allocated to the surgery group who actually received surgery, not the 50 additional who crossed over from medical to surgical management. This is not consistent with an intention-to-treat analysis. Choices B, D, and E do not group patients either as they were initially allocated or based on treatment received. These allocations do not adhere to intention-to-treat analysis.

Objective: Intention-to-treat analysis groups patients as initially randomized regardless of the actual occurrence of treatment within the study. This analysis strategy preserves randomization but may dilute the true effects of the intervention. %3D Previous Next Score Report Lab Values Calculator Help Pause

129 Exam Section 3: Item 30 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 30. A 35-year-old man comes to the physician because of a tremor for 6 weeks. He has a 15-year history of schizophrenia treated with haloperidol. Physical examination shows bradykinesia, tremor, and cogwheel rigidity. Haloperidol is discontinued, and treatment with aripiprazole is started because it has a decreased likelihood of causing the same adverse effects. Which of the following types of interaction with the dopamine receptor best explains this decreased likelihood? A) Agonism B) Antagonism C) Inverse agonism D) Irreversible antagonism E) Partial agonism F) Reversible antagonism Correct Answer: E. Unlike most antipsychotics, aripiprazole partially agonizes the dopamine (D2) receptor. Most antipsychotics are reversible antagonists of the D2 receptor, which block dopamine signaling nonspecifically across different brain areas. Blocking dopamine signaling in the limbic system decreases psychotic symptoms, while blocking dopamine signaling in the basal ganglia leads to medication-induced parkinsonism including bradykinesia, tremor, and cogwheel rigidity. Aripiprazole is an atypical (or second-generation) antipsychotic medication that is a partial D2 agonist, producing a pharmacodynamic response that is weaker than a full agonist but stronger than an antagonist. As a result of less attenuation of dopamine signaling in the basal ganglia, aripiprazole is less likely to lead to the drug-induced parkinsonism that is demonstrated in this patient. Incorrect Answers: A, B, C, D, and F. Agonism (Choice A) of the D2 receptor would increase dopamine signaling and therefore relieve parkinsonism. Dopamine agonists such as pramipexole are utilized to treat Parkinson disease. However, increasing dopamine signaling in the limbic system can lead to psychotic symptoms, so dopamine agonists should not be utilized in patients with psychotic disorders. Antagonism (Choice B), irreversible antagonism (Choice D), and reversible antagonism (Choice F) describe when an agent blocks the native signaling ligand from binding to its receptor, preventing downstream signaling. A reversible antagonist eventually dissociates from the receptor, leading to a transient pharmacodynamic effect, whereas an irreversible antagonist cannot be outcompeted for the receptor and may permanently bind and inactivate the receptor. Most antipsychotic medications such as haloperidol reversibly antagonize D2 receptors, blocking dopamine's downstream effects nonspecifically across different brain regions. D2 receptor blockade in the basal ganglia leads to medication-induced parkinsonism. Because antipsychotics are reversible antagonists, discontinuing an antipsychotic typically results in resolution of parkinsonism. Inverse agonism (Choice C) of the D2 receptor refers to a ligand binding to the D2 receptor and actively decreasing the baseline level of dopamine signaling, resulting in a pharmacodynamic effect opposite to dopamine's effects. This mechanism of action would likely lead to intolerable parkinsonism and acute dystonia, and is therefore not a mechanism of action of any antipsychotic medications.

Objective: Most antipsychotic medications antagonize dopamine receptors and therefore decrease psychotic symptoms while also leading to medication-induced parkinsonism. Aripiprazole uniquely partially agonizes dopamine receptors, which decreases psychotic symptoms without profound effects on motor function. Previous Next Score Report Lab Values Calculator Help Pause

153 Exam Section 4: Item 4 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 4. A 34-year-old man with AIDS comes to the physician because of muscle aches and reddish brown urine for 12 hours. He began treatment with trimethoprim-sulfamethoxazole 1 day ago as prophylaxis against Pneumocystis jirovecii (formerly P. carinii) infection. A sample of blood after centrifugation shows red serum. Which of the following is the most likely explanation for these changes? A) Allergic interstitial nephritis results in hematuria B) Enzyme deficiency permits hemoglobin oxidation C) Precipitation of sulfamethoxazole leads to cystitis D) Trimethoprim decreases erythrocyte permeability of hemoglobin E) Trimethoprim imparts a red color to body fluids Correct Answer: B. Enzyme deficiency of gluose-6-phosphate dehydrogenase (G6PD) permits hemoglobin oxidation and increased oxidative stress leading to hemolysis. G6PD is an enzyme in erythrocytes that makes NADPH from NADP*. An inadequate supply of NADPH leads to decreased concentrations of glutathione, which is a critical scavenger of reactive oxygen species within erythrocytes. Increased reactive oxygen species lead to cell membrane damage and precipitation of hemoglobin as it becomes oxidized and denatured. Membrane damage results in hemolysis, while precipitated hemoglobin presents as Heinz bodies in erythrocytes on a peripheral blood smear. Damage to erythrocytes leads to hemolysis with the subsequent development of unconjugated hyperbilirubinemia from saturation of hemoglobin conjugation processes in the liver and dark reddish-brown urine from the release of cell-free hemoglobin. Hemolytic reactions can be precipitated by certain medications, of which sulfa medications (eg, trimethoprim-sulfamethoxazole) are primary offenders. Incorrect Answers: A, C, D, and E. Allergic interstitial nephritis (AIN) results in hematuria (Choice A); however, hemolysis in G6PD deficiency results in the release of cell-free hemoglobin, indicating the patient's dark urine is a result of hemoglobinuria, not hematuria. Additionally, AIN typically results in pyuria, eosinophiluria, and acute kidney injury and is commonly precipitated by medications such as proton pump inhibitors, nonsteroidal anti-inflammatory medications, and classes of antibiotics including penicillins and sulfonamides. Precipitation of sulfamethoxazole leads to cystitis (Choice C) in patients receiving high doses of sulfamethoxazole who have inadequate urine output and acidic urine. Sulfamethoxazole crystals will precipitate when urine pH is below seven and can cause both acute kidney injury and cystitis. This can be prevented by alkalinization of the urine using sodium bicarbonate to achieve a urinary pH greater than seven. This would not account for red serum as seen in this patient. Trimethoprim decreases erythrocyte permeability of hemoglobin (Choice D) is not a characteristic of this drug as it has no direct effect on erythrocyte permeability. Trimethoprim imparts a red color to body fluids (Choice E) is not correct as this most commonly occurs in patients taking rifampin for the treatment of tuberculosis.

Objective: NADPH is produced from NADP* by the enzyme G6PD and is required for the decrease of glutathione, which helps scavenge reactive oxygen species in erythrocytes. The absence of G6PD results in increased oxidative stress that can lead to hemolysis when patients are exposed to certain medications such as trimethoprim-sulfamethoxazole and other sulfa drugs. Previous Next Score Report Lab Values Calculator Help Pause

154 Exam Section 4: Item 5 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 5. A 1000-g (2-lb 3-oz) male newborn is delivered at 28 weeks' gestation to a 25-year-old primigravid woman. Apgar scores are 6 and 8 at 1 and 5 minutes, respectively. Three days after oral feeding is started, he has abdominal distention, and he passes soft, bloody stools. Which of the following is the most likely cause of the bloody stools in this patient? A) Allergic gastroenteritis B) Atrophic gastritis C) Meckel diverticulum D) Milk allergy E) Necrotizing enterocolitis F) Thrombosis of the inferior mesenteric artery Correct Answer: E. Necrotizing enterocolitis (NEC) is classically associated with premature infancy and is characterized by inflammation and necrosis of the bowel wall. While the pathogenesis is unknown, it is suspected to be related to an immature intestinal tract and immune system, leading to severe intestinal inflammation, necrosis, and perforation. Signs and symptoms of NEC include abdominal distention, abdominal tenderness, vomiting, bloody diarrhea, lethargy, and feeding intolerance. It is treated with intravenous antibiotics and bowel rest, and if complicated by perforation, may require surgical resection. X-rays or CT imaging may disclose pneumatosis intestinalis, which refers to air within the thickened bowel wall. Incorrect Answers: A, B, C, D, and F. Allergic gastroenteritis (Choice A) can be caused by a hypersensitivity reaction to certain foods or allergens. It is rare, and typically presents later in childhood with nausea, vomiting, abdominal pain, and occasionally diarrhea. Peripheral eosinophilia is common, as are eosinophilic infiltrates in the bowel wall. Food protein-induced enterocolitis syndrome is a similar entity but is a non- İgE-mediated atopic reaction to food protein that presents with abdominal cramping, nausea, vomiting, and diarrhea. Atrophic gastritis (Choice B) is a form of chronic gastritis that leads to mucosal atrophy and loss of gastric glands that secrete gastric acid and pepsin, commonly leading to iron and vitamin B12 (cobalamin) deficiency. Meckel diverticulum (Choice C) results from persistence of the vitelline duct and can demonstrate ectopic, functional gastric, pancreatic or, rarely, endometrial tissue. It can present with acute or chronic hematochezia or melena. It is often painless, and typically presents in childhood, before two years of age. In children, it may serve as a potential lead-point for intussusception. Milk allergy (Choice D) is a common food allergy in children occurring when cow's milk is introduced into the diet. It can present with typical allergic reaction symptoms such as urticaria, atopic dermatitis, vomiting, nonbloody diarrhea, or life-threatening anaphylaxis. In a newborn infant who has not yet been exposed to cow's milk, milk allergy would be unlikely. Thrombosis of the inferior mesenteric artery (Choice F) can present with acute, severe abdominal pain, nausea, vomiting, and bloody diarrhea, often in a patient with atrial fibrillation, splanchnic atherosclerosis, or a hypercoagulable state. It is less likely than necrotizing enterocolitis in a premature neonate.

Objective: NEC is classically associated with premature infancy and is characterized by inflammation and necrosis of the bowel wall. It typically presents with abdominal distention, abdominal tenderness, vomiting, bloody diarrhea, lethargy, and feeding intolerance. II Previous Next Score Report Lab Values Calculator Help Pause

149 Exam Section 3: Item 50 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment A B Class I MHC negative variant tumor Sarcoma tumor 12- 10- 6- 4- 103 104 105 106 107 103 104 105 106 107 Cell dose inoculated Cell dose inoculated 50. The graphs depict experiments in which cells of a murine sarcoma were injected subcutaneously into a histocompatible mouse, and the diameter of tumors in the recipient mouse was measured 10 days later. Graph A shows the results obtained with sarcoma cells. Graph B shows the results obtained with a variant of the tumor that fails to express class I MHC molecules. The difference in growth of the variant tumor is best explained by the action of which of the following cell types in the tumor cell recipient? A) CD4+ T lymphocyte B) CD8+ T lymphocyte C) Eosinophil O D) Macrophage E) Natural killer cell O F) Neutrophil Correct Answer: E. Natural killer (NK) cells are a primary line of defense against tumorigenesis. Classically, NK cells target tumor cells in response to an absence of surface MHC class I molecules. In this example, the wild-type sarcoma tumor cells demonstrate growth that is based on the inoculation dose, while the sarcoma cells lacking MHC class I expression initially exhibit no growth at the lower inoculation concentrations. The likely mechanism of this finding is the recognition of these MHC class I negative sarcoma cells by NK cells, which would result in NK cell activation and release of perforin and granzyme, proapoptotic agents that induce tumor cell death. However, at higher inoculation concentrations, the number of tumor cells may overwhelm the capacity of NK cells to combat tumor proliferation, resulting in marked growth of the tumor. Incorrect Answers: A, B, C, D, and F. CD4+ T lymphocytes (Choice A) respond to cells expressing MHC class Il molecules. They recruit macrophages, cytotoxic T cells, plasma cells, and eosinophils. CD8+ T lymphocytes (Choice B) are not likely responsible for the disparities in tumor growth between the two different tumor cell lines as CD8+ T cells induce apoptosis in cells that express MHC Class I. They would be more likely to induce tumor death in the wild-type sarcoma cells, which express MHC class I. Eosinophils (Choice C) express both protumorigenic molecules and antitumorigenic molecules (eg, IL-18, TNF-a). They would not respond to an absence of MHC-I on the surface of sarcoma cells and would not account for this observed disparity in growth. Macrophages (Choice D) are recruited to the site of tumor growth by growth factors secreted by tumor cells. When recruited, they are thought to favor angiogenesis and tumor growth. As such, they would not be responsible for the initial control of the variant tumor in this example. Neutrophils (Choice F) are abundant cells of the innate immune system that are critical in the response to invading bacterial pathogens. In the presence of tumor cells, however, neutrophil infiltration leads to further propagation of inflammation and has been shown to promote tumorigenesis and metastasis.

Objective: NK cells recognize and attack tumor cells that do not express MHC-I, leading to tumor apoptosis through the release of perforin and granzyme. However, at higher inoculations, the number of tumor cells may overwhelm the capacity of NK cells to combat tumor proliferation, resulting in marked growth of the tumor. Previous Next Score Report Lab Values Calculator Help Pause Tumor diameter (mm)

141 Exam Section 3: Item 42 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 42. A 22-year-old woman is brought to the emergency department 30 minutes after stepping on a piece of glass while jogging on the beach. Physical examination shows a 1-cm laceration of the right foot with a deeply embedded glass shard. The shard is removed, and the wound is cleaned and sutured. After 3 days, the pain at the injury site has subsided, but the patient has numbness of the surrounding area within 4 mm of the sutures. Three weeks later, sensation is normal. Which of the following cell types most likely facilitated the reinnervation of the skin of this patient's foot? A) Astrocytes B) Basal keratinocytes C) Fibroblasts D) Myoepithelial cells E) Satellite cells F) Schwann cells Correct Answer: F. Schwann cells surround peripheral neurons with layers of myelin to increase the speed of action potential conduction. In addition, they are essential in nerve repair and axon regrowth following nerve injury. During injury to a peripheral nerve, the severed distal axon degenerates in a process known as Wallerian degeneration. In turn, the debris from the damaged axon is phagocytosed by macrophages and Schwann cells. Following removal of the damaged axon, the Schwann cells maintain their position along the previous tract of the axon. The proximal segment of the nerve uses this maintained tract to extend a regenerated axon toward the denervated distal tissue along the pathway maintained by the Schwann cells. In comparison to wound healing and the regenerative processes of epithelial tissues, nerve regrowth is generally slower. The rate of axonal regrowth has classically been described at 2-3 cm per month. Incorrect Answers: A, B, C, D, and E. Astrocytes (Choice A) are the most abundant glial cell in the central nervous system. They provide support and structure to the brain and spinal cord, maintain the chemical and neurotransmitter microenvironment, and store small amounts of glycogen. When an injury occurs in the central nervous system (eg, ischemic stroke), glial cells proliferate, forming a scar. Basal keratinocytes (Choice B) are skin cells that reside in the stratum basale of the epidermis. They proliferate and differentiate from deep to superficial, regenerating the epidermis following injury. Fibroblasts (Choice C) are spindle-shaped cells that are found ubiquitously in connective tissue. Their main function is to synthesize extracellular matrix (eg, collagen, proteoglycans). In wound healing, they are often responsible for scar tissue formation. Myoepithelial cells (Choice D) are specialized cells that contain actin and myosin; contraction of these cells results in expulsion of secreted contents from glands. Tissues such as sweat, lacrimal, and mammary glands all contain myoepithelial cells. Satellite cells (Choice E) are muscle cells that are located between the myofibrillar and basement membranes. They demonstrate regenerative properties and can give rise to other satellite cells or skeletal muscle cells.

Objective: Schwann cells play a key role in the regrowth of injured peripheral nerves. They provide a conduit for axonal regrowth toward the target tissue and remove debris from the regrowth pathway. II Previous Next Score Report Lab Values Calculator Help Pause

111 Exam Section 3: Item 12 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 12. During a study of digestion, a healthy 25-year-old woman volunteers to eat a large meal consisting of turkey, dressing, sweet potatoes, mashed potatoes, peas, and rolls. She then has pecan pie and coffee. Pancreatic secretions show an immediate response in this subject with the secretion of fluid and pancreatic proenzymes, including trypsinogen. Which of the following substances is the most likely cause of trypsinogen activation in this woman's intestine? A) Acetylcholine B) Cholecystokinin C) Enterokinase D) Histamine E) Secretin Correct Answer: C. Trypsinogen is a proenzyme that serves as an important enzyme in the digestion of proteins. Once activated, it is known as trypsin, and functions as a peptidase cleaving dietary polypeptides into small fragments. Trypsinogen is produced by pancreatic acinar cells and is secreted as an inactive proenzyme. Trypsinogen is subsequently cleaved by enterokinase, also known as enteropeptidase, in the duodenal brush border. Ónce activated, trypsin catalyzes the proteolytic cleavage of other digestive enzymes, as well as of trypsinogen, and creates a positive feedback loop. Incorrect Answers: A, B, D, and E. Acetylcholine (Choice A) is produced by parasympathetic branches of the vagus nerve (cranial nerve X). It stimulates M3 receptors on gastric parietal cells to promote gastric acid secretion. Cholecystokinin (Choice B) is produced by duodenal and jejunal l-cells and functions to promote the release of pancreatic secretions, as well as the relaxation of the sphincter of Oddi, contraction of the gallbladder, and delay of further gastric emptying. Histamine (Choice D) is produced by gastric enterochromaffin-like cells and stimulates the production of gastric acid by parietal cells. Histamine antagonists are therapeutically used in peptic ulcer disease and gastrointestinal reflux disease to decrease acid production by blocking histamine receptors. Secretin (Choice E) is produced by duodenal S cells. It promotes the release of bicarbonate-rich pancreatic secretions and bile, and inhibits gastric acid production.

Objective: The inactive proenzyme trypsinogen is cleaved to the active enzyme trypsin by enterokinase, which is found in the duodenal brush border. Trypsin is important for the digestion of protein polypeptides. It also cleaves and activates other digestive proenzymes. Previous Next Score Report Lab Values Calculator Help Pause

178 Exam Section 4: Item 29 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 29. A 35-year-old woman undergoes flexible nasopharyngoscopy because of chronic episodes of bacterial sinusitis. Purulent discharge from the right sphenoethmoidal recess is noted during the procedure. Which of the following best describes the location of this structure? A) Anterior to the nasolacrimal duct B) Distal to the vestibule C) Inferior to the hiatus semilunaris D) Posterior to the middle concha E) Proximal to the fusion of the hard and soft palates F) Superior to the superior concha Correct Answer: F. The sphenoethmoidal recess is the site of drainage of the sphenoid sinus and the posterior ethmoid air cells. It is found in the superior aspect of the nasal cavity both superior and posterior to the superior nasal concha. The sphenoethmoidal recess can be used to visualize drainage from the sphenoid sinus, as in this case. Additionally, the sphenoid sinus may be used as a surgical approach to the pituitary gland, as in transsphenoidal hypophysectomy. Incorrect Answers: A, B, C, D, and E. The region anterior to the nasolacrimal duct (Choice A) is the site of maxillary bone. The nasolacrimal duct originates from the inferior aspect of the lacrimal sac fossa, which is bound anteriorly by the anterior lacrimal crest, a component of the maxillary bone, and posteriorly by the posterior lacrimal crest, a component of the lacrimal bone. The nasolacrimal duct then travels posteromedially to its opening in the inferior nasal meatus. The region distal to the vestibule (Choice B) is the site of the middle ear and the tympanic membrane. The region inferior to the hiatus semilunaris (Choice C) is the site of the inferior nasal concha. The semilunar hiatus is located in the middle nasal meatus and is the site of drainage of the frontal, ethmoidal, and maxillary sinuses. The region posterior to the middle concha (Choice D) is the site of the nasopharynx. The region proximal to the fusion of the hard and soft palates (Choice E) is the site of the soft palate, which lies beneath the floor of the nasal cavity.

Objective: The sphenoethmoidal recess is the site of drainage of the sphenoid sinus and the posterior ethmoid air cells. It is found in the superior aspect of the nasal cavity, superior and posterior to the superior nasal concha. Previous Next Score Report Lab Values Calculator Help Pause

173 Exam Section 4: Item 24 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 24. A 20-year-old man comes to the physician because of a 2-month history of recurrent headaches. Funduscopic examination shows papilledema and a small retinal lesion consistent with a retinal hemangioblastoma. An MRI of the brain shows a lesion in the cerebellar midline. A CT scan of the abdomen shows cystic lesions in the pancreas. Which of the following is the most likely diagnosis? A) Neurofibromatosis 1 B) Neurofibromatosis 2 C) Sturge-Weber syndrome D) Tuberous sclerosis E) von Hippel-Lindau disease Correct Answer: E. This patient with a retinal hemangioblastoma, cerebellar lesion, and cystic lesions of the pancreas likely has von Hippel-Lindau disease. Von Hippel-Lindau disease is a heritable neurocutaneous disorder that presents with multiple benign and malignant tumors. Many tumors originate from the vasculature, including hemangioblastomas of the retina and cerebellum and angiomatosis of the skin and mucosa. Additional tumors include renal clear cell carcinomas, pancreatic masses, and pheochromocytomas. Diagnosis occurs by genetic testing that shows the pathogenic variant of the VHL gene, which is typically inherited in an autosomal dominant fashion. Management involves surgical excision of the tumors. Incorrect Answers: A, B, C, and D. Neurofibromatosis 1 (Choice A) is a heritable neurocutaneous disorder that presents with cutaneous neurofibromas (benign neoplasms derived from neural crest cells), cafe-au-lait spots, pigmented iris hamartomas (Lisch nodules), optic gliomas, and pheochromocytomas. This autosomal dominant disorder results from a de novo or inherited mutated NF1 tumor suppressor gene. Neurofibromatosis 2 (Choice B) is a heritable neurocutaneous disorder that presents with cutaneous neurofibromas, bilateral vestibular schwannomas, schwannomas of the cranial nerves or in the spinal cord, intracranial meningiomas, and neuropathy. Neurofibromatosis 2 results from a mutated NF2 gene, causing insufficient merlin (also known as schwannomin) protein, which normally acts as a tumor suppressor. Sturge-Weber syndrome (Choice C) is a neurocutaneous disorder that presents with proliferation of capillary-size blood vessels, seen on examination as nevus flammeus (a benign birthmark also known as a port-wine stain), angiomatosis of the leptomeninges (leading to seizures, intellectual disability, focal neurologic deficits), and episcleral hemangioma (leading to early-onset glaucoma). The causal activating mutation of the GNAQ gene is typically sporadic rather than inherited. Tuberous sclerosis (Choice D) is an autosomal dominant neurocutaneous disorder characterized by benign tumors in several organ systems. Hamartomas can form in the skin (leading to shagreen patches and hypomelanotic macules known as ash leaf spots) or brain (leading to epilepsy and intellectual disability), along with cardiac rhabdomyomas (leading to mitral regurgitation), and renal angiomyolipomas. Though these tumors are benign, patients are at risk for developing malignancies in the soft tissues, brain, or kidney.

Objective: Von Hippel-Lindau disease is an autosomal dominant neurocutaneous disorder that presents with multiple benign and malignant tumors. Many tumors originate from the vasculature, including hemangioblastomas of the retina and central nervous system, and angiomatosis of the skin and mucosa. Tumors of the kidney and pancreas also occur. Previous Next Score Report Lab Values Calculator Help Pause

103 Exam Section 3: Item 4 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 4. A 30-year-old woman has excessive gingival bleeding 8 hours after a tooth extraction. She has taken 400 mg of acetaminophen every 3 hours since the procedure. Laboratory studies show: Platelet count Platelet morphology Platelet aggregation Partial thromboplastin time Bleeding time Factor VIII (antihemophilic factor) coagulant activity moderately suppressed Ristocetin cofactor activity normal normal normal by collagen stimulation assay normal moderately increased moderately suppressed Which of the following is the most likely diagnosis? A) End-stage hepatic disease B) Hemophilia A C) Hemophilia B D) Vitamin K deficiency E) von Willebrand disease Correct Answer: E. Von Willebrand disease accounts for this patient's mucocutaneous bleeding in the setting of a normal platelet count with moderately low concentrations of Factor VIII, suppression of ristocetin cofactor activity, and increased bleeding time. Von Willebrand disease is one of the most common hereditary bleeding disorders and is caused by quantitative or qualitative abnormalities of von Willebrand factor (vWF), which binds platelets and subendothelial collagen in primary hemostasis. Impaired platelet adherence to the subendothelial lining, typically mediated by vWF, leads to a prolonged bleeding time. Von Willebrand factor also transports factor VIII in plasma, which degrades rapidly when unbound. Factor VIII is a critical component of the intrinsic coagulation pathway, and decreased concentrations can occasionally lead to a prolonged partial thromboplastin time (PTT), although it can also be normal as in this patient. Von Willebrand disease is inherited in an autosomal dominant pattern. It can present with epistaxis, gingival bleeding, petechiae, easy bruising, and menorrhagia. Additional diagnostic testing can be performed with the ristocetin cofactor assay, which requires functional vWF for platelet aggregation to occur. Treatment depends on the site and severity of bleeding and may include desmopressin (which promotes release of additional vWF from endothelial cells) or vWF concentrates. Incorrect Answers: A, B, C, and D. End-stage hepatic disease (Choice A) often results in mild to moderate thrombocytopenia and increases in the PTT and prothrombin time (PT) as a result of the decreased hepatic synthesis of clotting factors. Abnormal PT and PTT do not always correlate with an increased risk for bleeding, and patients with end-stage hepatic disease may in fact be prone to both bleeding and clotting. Factor VII concentrations are often increased in chronic liver disease. Hemophilia A (Choice B) and Hemophilia B (Choice C), X-linked coagulopathies, arise as a result of a genetic deficiency in factor VIII or IX synthesis or activity, respectively. Patients often present with easy bruising and hemorrhage involving muscles and joints. Hemophilia does not cause platelet dysfunction and would not cause an increased bleeding time on laboratory evaluation. Vitamin K deficiency (Choice D) results in the impaired synthesis of factors II, VII, IX, and X. Deficiency commonly results in an increased INR but has no effect on platelet number or function.

Objective: Von Willebrand disease is a common inherited cause of mucocutaneous bleeding and results from a deficiency in vWF concentrations or function. It is reflected by increased bleeding time, decreased factor VIII activity, and suppressed ristocetin cofactor activity. Treatment depends on the site and severity of bleeding and may include desmopressin or vWF concentrates. 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. The incidence of Haemophilus influenzae type b meningitis in children below age 2 years has decreased since the introduction of the current vaccination for this infection. This vaccine is composed of which of the following? A) Anti-idiotype B) Attenuated bacterial strain OC) Conjugated capsular polysaccharide D) Killed whole cell E) Outer membrane complex F) Peptide G) Recombinant cell protein H) Toxoid Correct Answer: C. The Haemophilus influenzae type b vaccination uses conjugated capsular polysaccharide to provoke an adaptive immune response. Vaccination is a form of active immunity designed to prevent disease by stimulating the immune system to develop and maintain the ability to respond to a foreign antigen. The maintenance of immunity is referred to as immune memory. Vaccinations may contain the whole infectious organism in a killed or a live-attenuated state that is unable to produce virulent disease. They may also be fractional, containing a specific component of a pathogen that the immune system can recognize. Fractional vaccines may be developed from a particular protein or polysaccharide associated with the organism. Bacterial vaccinations commonly employ a capsular or cell-wall polysaccharide, or a denatured toxin that is produced by the bacterium (referred to as a toxoid). Examples of capsular polysaccharide-based vaccines include vaccinations against Haemophilus influenzae type b, Neisseria meningitidis, and Streptococcus pneumoniae. Examples of toxoid-based vaccines include those against Corynebacterium diphtheriae, Clostridium tetani, and Bordetella pertussis. The presence of the polysaccharide alone is usually insufficient to provoke a lasting immune response as the polysaccharide is not usually well- presented to T cells on major histocompatibility complexes (MHC). Conjugation of the pathogen-specific polysaccharide with another antigen that is known to be highly immunogenic (presented on MHC) leads to a robust B- and T-cell response, with lasting immune memory. Toxoids are commonly used as conjugates. Incorrect Answers: A, B, D, E, F, G, and H. Anti-idiotype (Choice A) antibodies are those that bind to the variable region of another antibody. Attenuated bacterial strain (Choice B) is a form of vaccine in which the live organism has been rendered nonvirulent. The oral typhoid vaccine is an example of a live-attenuated bacterial vaccine. Killed whole cell (Choice D) vaccines utilize killed, inactive bacterial cells to stimulate the immune system. The Vi polysaccharide intramuscular typhoid vaccine is an example of a killed whole- cell bacterial vaccine. Outer membrane complex (Choice E) is a target for vaccine development. The outer membrane complex is composed of surface antigens that can elicit an immune response in vitro. The current Haemophilus influenzae type b vaccine utilizes a capsular polysaccharide. Peptide (Choice F) vaccines are typically synthetic derivates that mimic pathogen proteins and can be selected for high immunogenicity. There are currently no approved peptide vaccines for humans. Recombinant cell protein (Choice G) vaccines are developed by inserting the DNA sequence of the target antigen into a cluster of host cells, which are then proliferated, harvested, and purified to obtain a concentrate of the immunogenic protein. One form of the annual influenza vaccine is produced through recombinant DNA techniques. Toxoid (Choice H) vaccines are composed of exotoxins that have been chemically modified to remove their virulence. They stimulate the host production of antibodies that target the pathogenic bacteria's virulent toxin, disabling its function.

Objective: Capsular polysaccharides are immunogenic antigens that are utilized in vaccine development for immunity against Haemophilus influenzae type b, Neisseria meningitidis, and Streptococcus pneumoniae. The immunogenic response can be enhanced by conjugating the polysaccharide component with a toxoid. Previous Next Score Report Lab Values Calculator Help Pause

76 Exam Section 2: Item 27 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 27. A 10-year-old girl is brought to the physician by her parents 30 minutes after she sustained a red ant bite to her right foot. The parents report that within 15 minutes, redness and swelling occurred at the site of the bite. Physical examination shows edema and erythema at the wound site. The action of which of the following chemical mediators on postcapillary venules is the most likely cause of these findings? OA) Bradykinin В) СЗЬ C) Interleukin-1 D) Phospholipase C E) Thromboxane A2 F) Tumor necrosis factor Correct Answer: A. The body responds to harmful stimuli such as infectious agents, mechanical disruption, and chemical irritants with a localized inflammatory response, which is mediated by the innate immune system. The cardinal signs of inflammation are rubor (redness or erythema), calor (warmth), tumor (swelling or edema), dolor (pain), and functio laesa (loss of function). Erythema, warmth, and edema are caused by localized vasodilation of the postcapillary venules, resulting in increased blood flow to the affected area, which occurs within minutes of an insult. The primary inflammatory mediators of vasodilation are histamine, prostaglandin-E2, and bradykinin. Prostaglandin-E2 and bradykinin are also involved in sensitizing nociceptive nerve endings resulting in the increased pain of inflammation. Incorrect Answers: B, C, D, E, and F. C3b (Choice B) is a core component of the complement cascade and is formed by the cleavage of C3. The C3b fragment may then participate in one of three of pathways. It may bind with factor B to form C3 convertase and amplify C3 cleavage in a positive feedback loop. It may bind with C4b and C2b fragments to form C5 convertase. Or, it may opsonize the microbe by binding directly to the pathogen surface. It is not a vasoactive mediator. Interleukin-1 (Choice C) is a proinflammatory cytokine secreted by macrophages, endothelial cells, dendritic cells, and B lymphocytes. It has numerous effects, including the promotion of fever, vasodilation, and stimulation of endothelial cells to express adhesion proteins for leukocyte recruitment. Interleukin-1 is produced in the inflammatory response after the recruitment and migration of initial innate immune cells. Phospholipase C (Choice D) 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 ultimately leading to increased intracellular calcium and protein kinase C activity. In vascular smooth muscle cells, this results in increased smooth muscle contraction, which leads to vasoconstriction. Thromboxane A2 (Choice E) is an eicosanoid derivative of arachidonic acid that stimulates platelet activation and aggregation. It is also a vasoconstrictor. Tumor necrosis factor (Choice F) is a potent activator of macrophages and neutrophils, enhancing their cytotoxic effects and expression of endothelial adhesion molecules to promote migration into peripheral tissues.

Objective: Erythema, warmth, and edema are cardinal signs of inflammation, which result from localized vasodilation. Histamine, prostaglandin-E2, and bradykinin are the primary inflammatory mediators that promote vasodilation in the initial innate immune response. Previous Next Score Report Lab Values Calculator Help Pause

133 Exam Section 3: Item 34 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 34. A 41-year-old man comes to the physician because of a 6-week history of numbness of his right hand. He works in construction. Sensation to pinprick is decreased in the thumb and over a portion of the anterior forearm. The physician suspects that there is compression of the C5 and C6 spinal nerves as they exit the spinal canal. Weakness of which of the following movements on the right is most likely to confirm the diagnosis? A) Abduction of the upper extremity B) Adduction of the index finger C) Elevation of the shoulder D) Extension of the index finger E) Flexion of the wrist Correct Answer: A. The C5 and C6 cervical spinal nerve roots innervate many of the muscles of shoulder abduction. These muscles include supraspinatus, infraspinatus, and teres minor (muscles of the rotator cuff) and the deltoid, which is the strongest and largest shoulder abductor. Compression of these nerve roots at the neural foramen will lead to weakness in abduction of the upper extremity. The C5 and C6 nerve roots also innervate the biceps brachii and brachialis muscles. Thus, this injury would also cause weakness in flexion of the elbow. The C5 dermatome is located on the lateral aspect of the shoulder and upper arm. The C6 dermatome is located on the lateral aspect of the forearm and extends to the lateral two digits of the hand. Injury to these nerve roots will cause decreased sensation, pain, and/or paresthesia in these distributions. Incorrect Answers: B, C, D, and E. Adduction of the index finger (Choice B) is controlled mainly by the palmar interosseous muscle between the first and second metacarpals. This muscle is innervated primarily by the T1 nerve root with variable contributions from C8 via the deep branch of the ulnar nerve. Elevation of the shoulder (Choice C) is controlled by the accessory cranial nerve (cranial nerve XI) via the superior portion of the trapezius muscle. C5 and C6 nerve roots do not contribute to the innervation of this muscle. Extension of the index finger (Choice D) is controlled by the extensor indicis proprius and extensor digitorum muscles. These muscles are innervated by the posterior interosseous nerve, which is a branch of the radial nerve. The cervical nerve root contributions to this motor function are primarily C7 and C8. Flexion of the wrist (Choice E) is controlled by the flexor carpi radialis and flexor carpi ulnaris muscles, which are innervated by both the median and ulnar nerve with contributions from C6, C7, and C8 cervical nerve roots. Muscles of finger flexion also contribute to flexion of the wrist as they cross the radiocarpal joint.

Objective: C5 and C6 cervical nerve roots contribute the primary innervation to the deltoid muscle and the muscles of the rotator cuff. Compression of these nerve roots at the neural foramen will result in weakness of upper extremity abduction. Previous Next Score Report Lab Values Calculator Help Pause

142 Exam Section 3: Item 43 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 43. During an experiment, a strain of knockout mouse is developed that has an isolated deficiency of carbonic anhydrase in erythrocytes. No other type of cell or tissue is deficient in this enzyme. Compared with normal mice, venous blood in the knockout mouse will most likely have a higher concentration of which of the following substances? A) Ca2+ B) CI- O C) HCO3- D) K+ E) Mg2+ F) Na+ Correct Answer: B. Chloride (Cr) will be present in greater concentrations in the venous blood of mice with erythrocytes that are deficient in carbonic anhydrase, an enzyme utilized in erythrocytes to convert carbon dioxide (CO2) and water (H2O) to bicarbonate (HCO3) and hydrogen ions (H*). H* binds to hemoglobin and modulates the unloading and loading of oxygen while HCO3 is shuttled out of the erythrocyte in exchange for chloride via a chloride/bicarbonate exchange pump located on the surface of the erythrocyte. A deficiency in carbonic anhydrase would lead to lower concentrations of intracellular bicarbonate, which would result in an increased influx of bicarbonate into the cell in exchange for chloride efflux out of the cell. This would consequently result in higher concentrations of chloride in the serum. Incorrect Answers: A, C, D, E, and F. Ca2+ (Choice A) is utilized in a wide variety of reactions as an intracellular signaling molecule but is not involved in the action of carbonic anhydrase. Common physiologic pathways that utilize calcium signaling include skeletal muscle contraction and cardiac myocyte contraction. HCO3 (Choice C) concentrations will be lower in the serum if there is a deficiency of carbonic anhydrase, since there would be an increased influx of bicarbonate into the cell to maintain intracellular acid-base homeostasis. K* (Choice D) concentrations in the serum are tightly regulated. During periods of acidosis, potassium is shuttled out of cells in exchange for hydrogen ions, and during periods of alkalemia, potassium is shuttled into the cells in exchange for hydrogen ions. Potassium does not play a role in the action of carbonic anhydrase. Mg2+ (Choice E) and Na* (Choice F) concentrations of the venous blood would not be affected by an isolated deficiency of carbonic anhydrase in the red blood cell, as these enzymes play no role in the homeostasis of magnesium or sodium ions in the intracellular and extracellular compartments.

Objective: Carbonic anhydrase is an enzyme present in erythrocytes that catalyzes the formation of bicarbonate from water and carbon dioxide. A deficiency would result in decreased intracellular bicarbonate concentrations, which would stimulate the increased influx of bicarbonate into the cell in exchange for chloride efflux out of the cell. This would consequently result in higher concentrations of chloride in the serum. Previous Next Score Report Lab Values Calculator Help Pause

144 Exam Section 3: Item 45 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 45. An investigator is studying a respiratory epithelial cell line. The cell line is infected with a newly discovered coronavirus. Which of the following steps is essential for the reproduction of this new virus? A) Formation of double-stranded DNA B) Integration into host DNA C) Lysis of the host cell D) Replication in the nucleus E) Translation by the host ribosome Correct Answer: E. Translation by the host ribosome is necessary for the reproduction of this new coronavirus in the respiratory epithelium. The coronavirus family describes a set of enveloped RNA viruses of great clinical significance. Some examples include the virus that causes severe acute respiratory syndrome (SAŘS), middle eastern respiratory syndrome (MERS), and coronavirus disease 2019 (COVID-19). The replication cycle is unique and starts with binding to a host receptor (variable), followed by viral entry via direct fusion or via endocytosis in a vesicle, with subsequent release of viral genomic RNA into the cytoplasm. This RNA is translated into a large protein which, through subsequent proteolysis, generates a viral RNA-dependent RNA polymerase. This RNA polymerase then creates an antisense negative-strand RNA, followed by the synthesis of mRÑA from this template. These mRNA strands then utilize host ribosomes to synthesize viral structural proteins. The virus is assembled within vesicles, which then fuse with the host cell membrane to release mature virus into the extracellular space. Thus, utilization of the host ribosome is an essential step in the reproduction of the new coronavirus. Incorrect Answers: A, B, C, and D. Formation of double-stranded DNA (Choice A) does not occur in the reproduction cycle of coronavirus. Examples of viruses with double-stranded DNA genomes that must be replicated are adenovirus, herpesviruses, and poxviruses. Integration into host DNA (Choice B) is characteristic of the HIV virus, a retrovirus containing two single-strand RNA fragments, which use reverse transcriptase to generate HIV DNA that is integrated into the host genome. Lysis of the host cell (Choice C) is a characteristic of many viruses. As a class, these are referred to as lytic viruses and lysis of the cell is required for release of viral particles. This is not a feature of coronavirus. Replication in the nucleus (Choice D) is a feature of many viruses, most commonly the DNA viruses such as adenovirus, herpesviruses, and poxviruses. Coronavirus replication occurs in the cytosol, not in the nucleus.

Objective: Coronavirus is an RNA virus that utilizes host ribosomes to synthesize essential viral proteins in the cytoplasm before the assembly of mature virus particles in vesicles. %3D Previous Next Score Report Lab Values Calculator Help Pause

108 Exam Section 3: Item 9 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 9. The initial therapeutic effect of hydrochlorothiazide in patients with hypertension is a result of decreased sodium reabsorption in which of the following parts of the nephron? A) Ascending limb of the loop of Henle OB) Descending limb of the loop of Henle C) Collecting duct D) Distal tubule E) Proximal tubule Correct Answer: D. The distal convoluted tubule (DCT) is the main site for sodium and chloride reabsorption in the nephron, allowing for production of dilute, hypotonic filtrate. Hydrochlorothiazide (HCTZ) and other thiazide diuretics (eg, chlorthalidone, metolazone) inhibit sodium and chloride reabsorption in the DCT by inhibiting the action of the sodium-chloride cotransporter (NCC) channel. These drugs result in urinary losses of sodium and potassium with associated water loss. HCTZ has the opposite effect on calcium; calcium ion loss is decreased by thiazide diuretics. Thiazide diuretics are a first-line therapy for hypertension. Incorrect Answers: A, B, C, and E. The ascending limb of the loop of Henle (Choice A) is the site of action of loop diuretics (eg, furosemide, torsemide, bumetanide, ethacrynic acid), which act on the sodium-potassium-chloride cotransporter to decrease solute reabsorption and induce an associated water loss.The descending limb of the loop of Henle (Choice B) passively reabsorbs water and is impermeable to sodium. It is not a significant site of diuretic action. The collecting duct (Choice C) is the terminal segment of the nephron and reabsorbs sodium and water while secreting potassium and protons, primarily through epithelial sodium channels and aquaporins. The collecting duct is the site of action of potassium-sparing diuretics (eg, spironolactone, eplerenone, triamterene, amiloride). The proximal tubule (Choice E) is the site for reabsorption of the majority of electrolytes (bicarbonate, sodium, chloride, phosphate, and potassium), water, uric acid, glucose, and amino acids from the filtrate. The proximal tubule is the site of action of carbonic anhydrase inhibitors (eg, acetazolamide) and osmotic diuretics (mannitol).

Objective: Hydrochlorothiazide (HCTZ) and other thiazide diuretics (eg, chlorthalidone, metolazone) inhibit sodium and chloride reabsorption in the DCT by inhibiting the action of the sodium-chloride cotransporter (NCC) channel. These drugs result in urinary losses of sodium and potassium with associated water loss. %D Previous Next Score Report Lab Values Calculator Help Pause

147 Exam Section 3: Item 48 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 48. Norepinephrine is the primary neurotransmitter for which of the following nerves? O A) Postganglionic sympathetic nerves in sweat glands B) Postganglionic sympathetic nerves in the prostate O C) Preganglionic sympathetic nerves D) Sympathetic nerves innervating the adrenal gland Correct Answer: B. Postganglionic sympathetic nerves release norepinephrine to prostatic smooth muscle during ejaculation, which leads to smooth muscle contraction and emission of ejaculate. The sympathetic nervous system functions to arouse the body during stress, involving increased heart rate and blood pressure, glycogenolysis, and inhibition of gastrointestinal peristalsis. The cell bodies of first- order sympathetic nerves (also known as preganglionic sympathetic nerves) reside in the intermediolateral column (also known as the lateral horn) of the spinal cord. The preganglionic fibers exit the spinal cord and enter the paravertebral sympathetic trunk, which contains the sympathetic ganglia (a ganglion is a collection of neuronal cell bodies). In the sympathetic ganglia, preganglionic sympathetic nerves release acetylcholine onto postganglionic sympathetic nerve cell bodies. These postganglionic nerves then travel to and innervate the smooth muscle of the viscera (including the prostate). Most postganglionic fibers release norepinephrine onto their target organs. The sweat glands and adrenal gland are notable exceptions to this sympathetic pathway. Incorrect Answers: A, C, and D. Postganglionic sympathetic nerves in sweat glands (Choice A) release acetylcholine. These postganglionic nerves also produce vasoactive intestinal peptide (VIP), which regulates sweat secretion. Preganglionic sympathetic nerves (Choice C) release acetylcholine. Preganglionic parasympathetic nerves similarly release acetylcholine. Sympathetic nerves innervating the adrenal gland (Choice D) are preganglionic sympathetic nerves that release acetylcholine directly onto the chromaffin cells of the adrenal medulla. These chromaffin cells then release norepinephrine and epinephrine into the systemic circulation.

Objective: In the sympathetic nervous system, preganglionic sympathetic nerves release acetylcholine onto postganglionic sympathetic nerves located in the paravertebral sympathetic trunk. The postganglionic sympathetic nerves typically release norepinephrine to stimulate the smooth muscle of the viscera and blood vessels. The adrenal gland and sweat glands are part of the sympathetic nervous system but are instead innervated by acetylcholine-releasing neurons. %D Previous Next Score Report Lab Values Calculator Help Pause

188 Exam Section 4: Item 39 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 39. The neoplasm indicated by X in the MRI of the head is most likely to first cause which of the following symptoms? A) Dilated pupil B) Diplopia C) Dry cornea D) Ptosis E) Visual field deficit Anterior Posterior Correct Answer: E. Benign adenomas are the most common neoplasms identified in the pituitary gland. The pituitary gland rests in the bony sella turcica, just inferior to the optic nerve and chiasm. The medial fibers that supply the nasal retina within each optic nerve decussate at the optic chiasm, which is located immediately superior to the pituitary gland. Mass lesions of the pituitary can therefore compress the optic chiasm and produce a visual field deficit. Most patients will present with a bitemporal hemianopsia in such cases, with loss of the lateral field of view bilaterally. Incorrect Answers: A, B, C, and D. Dilated pupil (Choice A) is a feature of oculomotor nerve palsy. Nerve fibers that control the pupillary light response travel to the eye on the surface of the oculomotor nerve and may be compressed by mass lesions, which most frequently occurs in the setting of aneurysms of the posterior communicating artery. Diplopia (Choice B) may occur in the setting of palsy of either the oculomotor nerve, trochlear nerve, or abducens nerve, because of entrapment of extraocular muscles, or as a result of refractive error and ocular surface dryness or tear film abnormalities. Compression of the optic nerve or the optic chiasm does not present with diplopia. Dry cornea (Choice C) is not a common result of optic nerve compression. Corneal dryness may be observed in Sjögren syndrome or secondary to lagophthalmos and corneal exposure in the setting of facial nerve palsy, also known as Bell palsy. Ptosis (Choice D) may be observed as a feature of oculomotor nerve palsy, Horner syndrome, or myasthenia gravis.

Objective: Pituitary mass lesions, such as a pituitary adenoma, may lead to compression of the optic chiasm and produce visual field defects. Bitemporal hemianopsia is the most frequently observed visual field defect as a result of chiasmal compression. Previous Next Score Report Lab Values Calculator Help Pause

174 Exam Section 4: Item 25 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 25. A 65-year-old man develops his third episode of pneumonia during the past 8 months. The pneumonias have been localized to the right middle lobe of the lungs. Which of the following studies is most likely to establish the cause of these recurrent pneumonias? A) Pulmonary function tests B) Ventilation-perfusion lung scans C) Bronchoscopy D) Pleural biopsy OE) Pulmonary arteriography Correct Answer: C. Recurrent pneumonia in the same location of the lung is concerning for obstructive pneumonia secondary to a central obstructing lesion of the airway. The presence of a lesion (eg, tumor) results in the impaired clearance of mucus, microbes, and foreign debris, which is a risk factor for the development of clinically significant infection of the alveolar space and lung parenchyma. This may be caused by extrinsic compression of the airway or an intrinsic space-occupying lesion in the lumen. The patient should be evaluated initially with noninvasive chest imaging. Recurrent right middle lobe pneumonia suggests that the right middle lobar bronchus is the site of obstruction. The next diagnostic study to establish the underlying cause of the recurrent pneumonias should be bronchoscopy, which will allow for the direct visualization of the airway and the cause of the obstruction. If an obstructing lesion is identified, biopsy tissue samples can be obtained for histologic evaluation. Potential etiologies include primary lung malignancy, metastatic disease, mucus plug, or an aspirated foreign object. Treatment includes antibiotics for the pneumonia and management of the underlying lesion. Incorrect Answers: A, B, D, and E. Pulmonary function tests (Choice A) include spirometry, full-body plethysmography, single-breath diffusing capacity, arterial blood gas analysis, and pulse oximetry. They are useful in the diagnosis of obstructive and restrictive lung diseases. Analysis of the flow-volume loops obtained through spirometry may also suggest the presence of an upper airway obstruction in the trachea or larynx. Ventilation-perfusion lung scans (Choice B) utilize scintigraphy to evaluate regions of lung for ventilation and perfusion defects. It can useful in evaluating for a pulmonary embolism if a contrasted CT angiogram of the pulmonary arteries is contraindicated (eg, contrast allergy, pregnancy). Pleural biopsy (Choice D) may be indicated for evaluation of a pleural effusion of unclear etiology, pleural thickening, or a pleural mass. Pleural biopsy is typically obtained percutaneously. An obstructing lesion in a central airway is better evaluated by bronchoscopy. Pulmonary arteriography (Choice E) refers to the invasive catheter angiographic evaluation of the pulmonary arteries. This is no longer necessary for the diagnosis of pulmonary embolism given advances in ventilation-perfusion scans and CT pulmonary angiography, but may still be performed during pulmonary embolectomy.

Objective: Recurrent pneumonia in the same pulmonary lobe raises concern for a central obstructing lesion of a lobar bronchus. Bronchoscopy is the preferred study for evaluating the underlying cause and if a mass is discovered, obtaining biopsy samples for histologic evaluation to establish a definitive diagnosis. Previous Next Score Report Lab Values Calculator Help Pause

161 Exam Section 4: Item 12 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 12. A 25-year-old man sustains perforation of the large bowel in the area of the cecum during a motor vehicle collision. An exploratory operation is planned to resect the affected bowel. To prevent excessive intraoperative hemorrhage in this patient, it is most appropriate to ligate the terminal branch of which of the following arteries? A) Celiac B) Inferior mesenteric C) Inferior rectal D) Left colic E) Middle colic F) Superior mesenteric Correct Answer: F. The superior mesenteric artery (SMA) supplies arterial blood to the distal duodenum, jejunum, and ileum, as well as the large bowel including the cecum, ascending colon, and proximal two- thirds of the transverse colon. The arterial blood supply to the cecum is supplied by the ileocolic artery, a branch of the SMĀ. The ileocolic artery can be ligated during exploratory abdominal surgery and repair of perforating cecal injuries to prevent excessive intraoperative hemorrhage. Incorrect Answer: A, B, C, D, and E. The celiac artery (Choice A) is the first major gastrointestinal branch of the abdominal aorta and supplies much of the foregut, including the stomach, liver, spleen, and proximal duodenum. It does not provide arterial blood supply to the large bowel. The inferior mesenteric artery (Choice B) is the third major gastrointestinal branch of the abdominal aorta and supplies the large bowel from the distal third of the transverse colon to the upper portion of the rectum via the left colic, sigmoidal, and superior rectal arteries. It does not provide direct arterial blood to the cecum. The inferior rectal artery (Choice C) is a branch of the pudendal artery and supplies the distal portion of the rectum and anus, where it forms anastomoses with the superior and middle rectal arteries, which are themselves branches of the inferior mesenteric artery and internal iliac artery, respectively. The inferior rectal artery does not provide blood supply to the cecum. The left colic artery (Choice D) is a branch of the inferior mesenteric artery. It divides into the marginal artery that supplies much of the descending colon distal to the splenic flexure. It forms anastomoses near the splenic flexure with branches of the middle colic artery (Choice E), which is a branch of the SMA that supplies much of the transverse colon. Neither of these arteries provide direct arterial blood to the cecum.

Objective: The cecum derives its arterial blood supply from the ileocolic artery, which is a branch of the SMA. The ileocolic artery may be ligated during surgical repair of cecal trauma to prevent excessive intraoperative bleeding. Previous Next Score Report Lab Values Calculator Help Pause

112 Exam Section 3: Item 13 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 13. A 45-year-old man with Li-Fraumeni syndrome agrees to participate in an investigational study of tumor suppressor gene disorders. A mutation is found in his p53 tumor suppressor gene that converts an arginine to a proline at amino acid position 248 (R248P). This change alters hydrogen bonding between the mutant p53 tumor suppressor gene protein and the thymidine oxygen and adenine ring nitrogen of DNA. Which of the following is the most likely result of this mutation on the transcription of genes that inhibit cell division and contain the consensus sequence TATA (A/T) (A/T) at -30 bp 5' to the transcription start site? A) Decreased binding of DNA polymerase B) Decreased binding of RNA polymerase C) Increased binding of DNA polymerase D) Increased binding of RNA polymerase E) Increased binding of trans factors Correct Answer: B. The consensus sequence TATA lies upstream from the transcription start site and functions as a portion of the promoter, where RNA polymerase and transcription factors bind to DNA. Also known as the TATA box, this sequence is a noncoding regulatory sequence that governs transcription of coding sequences. Similarly, DNA contains noncoding sequences such as enhancer and silencer regions, noncoding sequences that augment or inhibit transcription of regulated genes, respectively. Mutations at the TATA sequence within the promoter result in decreased gene transcription of whatever gene may be associated with that particular promoter because of the decreased binding of RNA polymerase to the promoter. Regulatory proteins (eg, p53) also bind these regions. p53, a tumor suppressor, limits oncogene transcription and promotes transcription of genes that inhibit cell division. In this case, the transcription of genes that inhibit cell division would be limited, as the main function of p53 is to control the cell cycle, prevent cell growth and division, and initiate apoptosis. Absent or dysfunctional mutations of the p53 tumor suppressor gene are a common finding in many solid tumors as the function of p53 is lost and cell growth and division are permitted unchecked. Li-Fraumeni syndrome, inherited in an autosomal dominant manner, results from non- or subfunctional p53. The syndrome is characterized by multiple malignancies at an early age. This inherited loss of p53 predisposes the patient to neoplasia as additional oncogenic mutations accumulate. Incorrect Answers: A, C, D, and E. Decreased binding of DNA polymerase (Choice A) or increased binding of DNA polymerase (Choice C) would not affect the transcription of RNA and resultant gene expression. DNA polymerase is involved in DNA replication, not transcription. Mutations in the promoter sequence that limit the binding affinity of the TATA box would decrease binding, not increase binding of RNA polymerase (Choice D) and trans factors (Choice E).

Objective: The consensus sequence TATA lies upstream from the transcription start site and functions as a binding site for RNA polymerase and regulatory proteins. Mutations at the TATA sequence or the promoter result in a decreased level of gene transcription as a result of the inability of RNA polymerase to bind to the promoter. Previous Next Score Report Lab Values Calculator Help Pause

186 Exam Section 4: Item 37 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 37. A 25-year-old woman is brought to the emergency department 4 hours after she fell on an icy sidewalk and landed on her right arm. Examination of the right upper extremity shows decreased strength with elbow flexion. Sensation to pinprick is decreased over the lateral aspect of the forearm. A CT scan-style cross section of the right upper extremity is shown in the figure; the arrow indicates the injured nerve. This patient most likely sustained injury to a nerve supplied by which of the following nerve roots? A) C4-6 B) C5-7 O C) C5-T1 D) C6-8 O E) C8-T1 Correct Answer: B. The musculocutaneous nerve arises from the lateral cord of the brachial plexus and receives contributions from the C5, C6, and C7 cervical nerve roots. This nerve travels between the biceps brachii and the brachialis muscles providing motor innervation to both of these, as depicted in the diagram. The musculocutaneous nerve also supplies sensory innervation to the lateral aspect of the forearm via its terminal branch, the lateral antebrachial cutaneous nerve. In this patient, decreased lateral forearm sensation and decreased elbow flexion strength are consistent with a musculocutaneous nerve injury in the C5-7 distribution. Incorrect Answers: A, C, D, and E. The axillary nerve consists of fibers from the C5 and C6 nerve roots predominantly, however in some cases it includes fibers from C4-C6 (Choice A). This nerve provides motor innervation to the deltoid muscles well as teres minor, and sensory innervation to the posterior part of the upper arm. The radial nerve arises from the posterior cord of the brachial plexus and receives contributions from C5, C6, C7, C8, and T1 (Choice C). It provides sensory innervation to the lateral upper arm as well as the posterior forearm and posterior hand. It provides motor innervation to the extensors of the elbow as well as the extensors of the wrist and digits. The thoracodorsal nerve receives contributions from C6, C7, and C8 cervical nerve roots (Choice D) and innervates the latissimus dorsi. It arises from the posterior cord of the brachial plexus. The ulnar nerve arises from the medial cord of the brachial plexus and receives contributions from the C8 and T1 cervical nerve roots (Choice E). It provides sensory innervation to the medial border of the forearm as well as the ulnar digits. It provides motor innervation to the majority of the intrinsic muscles of the hand.

Objective: The musculocutaneous nerve arises from the lateral cord of the brachial plexus and consists of fibers from the C5-C7 cervical nerve roots. It travels between the biceps brachii and brachialis muscles and contributes to elbow flexion and sensation of the lateral forearm. Previous Next Score Report Lab Values Calculator Help Pause

126 Exam Section 3: Item 27 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 20- 27. The graph shows the plasma concentration of an antibiotic as a function of time after an intravenous bolus. If the antibiotic is administered continuously by intravenous infusion, approximately how many hours will it take for the plasma concentration of the drug to first achieve almost 97% of its steady-state concentration? 17.5- 15- 12.5- A) 3 10- B) 6 7.5- C) 9 5- D) 15 2.5- E) 24 3 6. 9. Time (h) 12 Correct Answer: D. A steady-state concentration of a drug occurs when the rate of elimination of a drug is equal to the rate of drug administration. Because of this, the time to steady-state concentration is dependent on the rate of elimination. Rate of elimination can be quantified by the half-life of the drug, or the time required for the plasma concentration of the drug to decrease by one half. This antibiotic has an initial concentration of 20 mcg/mL. This concentration is decreased to half, or 10 mcg/mL, in three hours, making the half-life of this drug three hours. The equation that defines the plasma concentration of a drug with respect to its half-life and steady-state concentration is C=Cs[1-(½)"], where C is the concentration, Css is the steady-state concentration, and n is the number of half- lives. This equation can be restated as C/C= 1-(½)", allowing for the simple calculation of the percentage of steady-state concentration achieved given n half-lives. By this equation, one half-life results in a serum concentration of 50% of the steady-state concentration, and two half-lives result in a serum concentration of 75% of the steady-state concentration. Similarly, five half-lives are required to achieve 97% of the steady-state concentration (C/C= 1-[½]°), making the time required 15 hours (3 hours x 5 half-lives). %3D %3D ss Incorrect Answers: A, B, C, and E. 3 hours (Choice A) would lead to 50% of steady-state concentration by CICss= 1-(2)". 6 hours (Choice B) would lead to 75% of steady-state concentration by C/Ces= 1-()2. %3D 9 hours (Choice C) would lead to 87.5% of steady-state concentration by C/Cs 1-(2)3. %3D 's, 24 hours (Choice E) would lead to 99.6% of steady-state concentration by C/Css= 1-(½)8.

Objective: Time to steady-state concentration of a drug is a function of the half-life of the drug. The equation for calculating the percent of steady-state concentration is C=Css[1-(½)^], where C is the concentration, Css is the steady-state concentration, and n is the number of half-lives. Previous Next Score Report Lab Values Calculator Help Pause [Plasma drug] (µg/mL)

81 Exam Section 2: Item 32 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 32. A 57-year-old man comes to the physician because of increasingly severe joint pain and weakness during the past 2 months. Physical examination shows pallor. Laboratory studies show: 8 g/dL 25% Hemoglobin Hematocrit Leukocyte count Myeloid cells (immature and mature) Platelet count 150,000/mm3 90% 50,000/mm3 present Philadelphia (Ph') chromosome A chimeric growth-promoting protein with which of the following activities is most likely encoded in this patient due to the presence of this chromosome? A) CAMP B) Dihydrofolate reductase C) MAP kinase D) Platelet-derived growth factor E) Tyrosine kinase Correct Answer: E. This patient carries a diagnosis of chronic myelogenous leukemia (CML), which is defined by the Philadelphia chromosome (Ph). The Ph chromosome is created by a translocation between chromosomes 9 and 22, leading to constitutive activation of the ABL1 tyrosine kinase. Activation of the ABL1 tyrosine kinase leads to subsequent activation of the JAK/STAT and Ras/MAPK/ERK pathways, with the downstream consequences of increased cellular proliferation, loss of normal checkpoint inhibition, and resistance to apoptosis. While the Ph chromosome is required for the diagnosis of CML, it can also be found in acute lymphoblastic leukemia (ALL) and acute myeloid leukemia (AML). Typical laboratory findings in CML include hyperleukocytosis, which is present in this case, with an increase in nearly all of the myeloid cell lines. Basophilia and eosinophilia are characteristic. Leukostasis, a condition in which hyperleukocytosis leads to end organ damage from occlusion of capillaries by malignant, nondistensible cells, is uncommon in CML because malignant cells are well differentiated and pliable. This is in contrast to conditions such as AML and ALL in which circulating blasts, which are large and not particularly pliable, easily lodge in capillaries to cause pulmonary and neurologic complications. Treatment of CML is primarily with tyrosine kinase inhibitors such as imatinib or desatinib. They are remarkably effective and often result in complete molecular response within several months of initiation. Incorrect Answers: A, B, C, and D. CAMP (Choice A) is a ubiquitous signaling molecule critical in a variety of cellular signaling pathways. Mutations in pathways that utilize CAMP are found in many types of cancer, including breast, liver, skin, and lung cancer. CAMP is not necessarily implicated in the pathogenesis of CML. Dihydrofolate reductase (DHFR) (Choice B) is an enzyme that converts dihydrofolate into tetrahydrofolate, which is required for the synthesis of purines for DNA synthesis. Methotrexate is a medication used in the treatment of various disorders, most notably rheumatoid arthritis and B-cell lymphomas such as Burkitt lymphoma, which competitively inhibits DHFR. DHER is not implicated in the pathogenesis of CML. MAP kinase (Choice C) is an enzyme involved in a signaling cascade leading to increased cellular growth and proliferation. It includes the proteins Ras, Raf, MEK, and ERK. One common mutation in this pathway involves BRAF, which is commonly mutated in melanoma and is the target of several therapies. Platelet-derived growth factor (Choice D) stimulates the growth and proliferation of multiple cell types; overexpression has been implicated in breast cancer, gliomas, and certain leukemias. It is not known to play a role in the development of CML.

Objective: CML is a neoplasm of the myeloid cell line that is defined by the presence of the Ph chromosome resulting in constitutive activation of the ABL1 tyrosine kinase. Typical findings include hyperleukocytosis with a diverse array of myeloid lineage cells present on the peripheral smear. Previous Next Score Report Lab Values Calculator Help Pause

46 Exam Section 1: Item 47 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 47. Agroup of men between the ages of 25 and 30 years and a group of men between the ages of 75 and 80 years were deprived of water for 24 hours. Renal function measurements for both groups are shown: 25-30 Years Urine specific gravity Urine osmolality (mOsmol/kg) 1.030 1100 75-80 Years 1.020 850 Which of the following changes associated with aging is the most likely cause of the difference in urine concentrating ability between the two groups? A) Decreased renal blood flow B) Decreased renal tubule responsiveness to ADH (vasopressin) OC) Decreased single nephron filtration rate D) Increased release of atrial natriuretic peptide E) Increased sodium transport in the thick ascending loop of Henle Correct Answer: B. Decreased renal tubule responsiveness to ADH (vasopressin) most likely explains the disparities in urine concentrating abilities between younger and older men. ADH is a peptide hormone made in the hypothalamus and released from the posterior pituitary in response to decreased effective circulating volume or states of hyperosmolarity. It acts on vascular smooth muscle via the V, receptor to cause vasoconstriction, thereby increasing mean blood pressure. It also acts on the principal cells of the renal collecting duct to increase the expression of aquaporin channels on the luminal surface of these cells. Increased expression of aquaporin channels leads to a greater ability to resorb water from the lumen of the collecting duct to regulate serum osmolality. This free water resorption consequently increases the urine osmolality and specific gravity. In the setting of water deprivation, serum osmolality rises, which signals the hypothalamic osmoreceptors to secrete increased amounts of ADH. This subsequently results in increased free water absorption from the collecting tubule to maintain homeostatic serum osmolality. On the basis of the data provided, this ability to concentrate urine through the action of ADH during water deprivation appears to wane with age. ADH can also be secreted in response to myriad other stimuli including pain and nausea and can cause an inappropriate resorption of water leading to a condition known as the syndrome of inappropriate ADH release. Incorrect Answers: A, C, D, and E. Decreased renal blood flow (Choice A) occurs in states of hypovolemia or hypotension, each of which have many potential causes. This stimulates the release of renin from the juxtaglomerular apparatus of the kidney, which is necessary for the formation of angiotensin I from angiotensinogen. This is converted to angiotensin II by an angiotensin-converting enzyme, which acts on vascular smooth muscle to increase systemic vascular resistance and on the renal tubules to increase sodium resorption. Decreased single nephron filtration rate (Choice C) correlates with decreased overall glomerular filtration rate (GFR), which decreases with age, but this should not result in changes to urine osmolality unless the decline is severe. A severe decline in GFR as occurs with acute oliguric or anuric kidney injury can result in defects of collecting tubule concentrating capabilities, but this is from injury to tubular cells, not from the action of ADH. Increased release of atrial natriuretic peptide (Choice D) occurs in patients with increased circulating volume as a result of conditions such as heart failure. It induces water diuresis and natriuresis instead of water resorption. Increased sodium transport in the thick ascending loop of Henle (Choice E) would result in increased absorption of water in addition to the sodium and would have minimal effect on the urine osmolality, which is primarily regulated by the absorption of free water.

Objective: The concentrating ability of the nephron in response to ADH decreases with time, which is reflected as a lower urine osmolality in older men as compared to younger men in the setting of water deprivation. Previous Next Score Report Lab Values Calculator Help Pause

33 Exam Section 1: Item 34 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 34. A 12-year-old girl is brought to the physician by her parents because of a 2-day history of dizziness and headache. She is intelligent and does well in school. Her mother had a cerebellar tumor removed during childhood and is concerned that her daughter might have a similar condition. A normal reaction to which of the following physical examination maneuvers is most likely to rule out a cerebellar tumor in this patient? A) Babinski sign B) Doll's eye (oculocephalic) maneuver C) Glabellar tap D) Rapid alternating movements E) Tinel sign Correct Answer: D. The inability to perform rapid alternating movements, also known as dysdiadochokinesia, indicates cerebellar dysfunction. Rapid alternating movements involve quickly alternating between clapping the palms and the dorsum of one hand. The cerebellum mediates motor learning and calculates movement course changes in response to sensory input. Therefore, cerebellar dysfunction results in a decreased ability to alternate between opposite movements as in the rapid alternating movements physical examination maneuver. This patient may have a pilocytic astrocytoma or medulloblastoma, two of the most common cerebellar tumors in children. Cerebellar tumors may present with dizziness and symptoms of increased intracranial pressure (as a result of compression of the fourth ventricle by the tumor and consequent obstructive hydrocephalus) such as headache as well as gait ataxia and nystagmus (for midline tumors) or dysmetria on finger-to-nose testing and dysdiadochokinesia (for lateral tumors). Incorrect Answers: A, B, C, and E. The Babinski sign (Choice A), or extensor plantar response, is an upper motor neuron sign that indicates damage to the corticospinal tract, a descending motor tract located in the central nervous system. The Babinski sign is present when the lateral part of the plantar surface of the foot is firmly stroked and the hallux extends (moves rostrally) while the other toes abduct. The cerebellum does not contain the corticospinal tract, so cerebellar dysfunction would not cause the Babinski sign. The doll's eye (oculocephalic) maneuver (Choice B) refers to the brainstem reflex that occurs in normal adults when the examiner moves the patient's head and the eyes move in the opposite direction to hold the patient's gaze on an object. In a negative oculocephalic reflex, the patient's eyes remain midline, which indicates severe brainstem dysfunction. The glabellar tap or glabellar reflex (Choice C) refers to blinking in response to repetitive tapping on the forehead. The glabellar reflex is a primitive reflex that is related to frontal lobe immaturity in infancy and also occurs in advanced dementia as a frontal release sign. Tinel sign (Choice E) refers to pain or paresthesias after an examiner repetitively percusses the median nerve as it passes through the wrist. A positive Tinel sign is suggestive of carpal tunnel syndrome.

Objective: The inability to perform rapid alternating movements indicates cerebellar dysfunction. The cerebellum mediates motor learning and calculates movement course changes in response to sensory input. Previous Next Score Report Lab Values Calculator Help Pause

77 Exam Section 2: Item 28 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 72% 100% Superior vena cava Pulmonary veins Right atrium 84% 95% Left atrium Right ventricle 84% 95% Left ventricle Pulmonary artery Aorta 84% 95% 28. A 10-year-old girl is referred to a cardiologist because of a 6-year history of a murmur. She is asymptomatic. Physical examination shows a grade 2/6, systolic ejection murmur heard over the left sternal border. An ECG shows right axis deviation and right ventricular hypertrophy. She undergoes cardiac catheterization. Oxygen saturation findings are shown in the block diagram. This patient most likely has which of the following congenital heart defects? A) Abnormal pulmonary venous return B) Atrial septal defect C) Membranous ventricular septal defect D) Tetralogy of Fallot Correct Answer: B. Atrial septal defect is a common congenital malformation of the interatrial septum. The most common type is an ostium secundum defect, although ostium primum defects are commonly associated with trisomy 21. The atrial septal defect results in a left-to-right shunt with abnormal flow of blood from the left atrium to the right atrium, resulting in relative volume overload of the right atrium and ventricle. The presence of a channel between the left and right atrium leads to mixing of oxygenated blood from the left atrium with returning venous blood from the superior vena cava, resulting in a higher oxygen saturation in the right atrium than expected (as seen in the patient's cardiac catheterization findings). Atrial septal defects are also associated with a fixed, split S2 as a result of increased blood flow across the pulmonic valve, and a low-grade ejection murmur on cardiac auscultation. The increased right heart volumes also result in a prominent right ventricular impulse on physical examination and may present an increased risk for the development of a right bundle branch block. Chest x-rays characteristically demonstrate increased caliber of the main pulmonary artery and increased pulmonary vascular markings. If the atrial septal defect remains uncorrected, it can result in the development of Eisenmenger syndrome secondary to prolonged pulmonary vasculature remodeling with consequent pulmonary arterial hypertension and shunt reversal leading to cyanosis. Incorrect Answers: A, C, D, and E. Abnormal pulmonary venous return (Choice A) occurs in total or partial anomalous pulmonary venous return, a rare congenital malformation in which one or more of the four pulmonary veins do not insert appropriately into the left atrium and instead drain into the right atrium. Increased right atrial and decreased left atrial oxygen saturation would be expected. Total anomalous return is incompatible with life unless an atrial septal defect or patent ductus arteriosus is present in order to provide oxygenated blood to the systemic circulation. Membranous ventricular septal defect (Choice C) and muscular ventricular septal defect (Choice D) present with a holosystolic murmur best heard in the left lower sternal border. The oxygen saturation would not vary between the superior vena cava and the right atrium, and a higher oxygen concentration would be expected in the right ventricle as a result of mixing with oxygen-rich blood from the left ventricle. Tetralogy of Fallot (Choice E) consists of pulmonary infundibular stenosis, overriding aorta, ventricular septal defect, and right ventricular hypertrophy, resulting in a right-to-left shunt with cyanosis first noted in infancy. The murmur heard with tetralogy of Fallot relates to the underlying ventricular septal defect, characterized by a holosystolic murmur best heard in the left lower sternal border, and to the pulmonic stenosis, characterized by a systolic crescendo-decrescendo murmur in the pulmonic auscultation area.

Objective: Atrial septal defect is a common congenital malformation of the interatrial septum that results in a left-to-right shunt between the atria. Mixing of oxygenated blood from the left atrium with deoxygenated blood returning from the superior vena cava results in higher oxygen concentration in the right atrium and ventricle than otherwise expected. 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 60-year-old man comes to the physician because of a 1-month history of progressive shortness of breath with exertion. Examination of the chest shows: Breath sounds Percussion note Tactile fremitus Adventitious sounds Right Lung Base decreased dullness decreased Left Lung Base normal normal normal crackles none Which of the following is the most likely diagnosis? A) Asthmatic bronchitis O B) Bronchiectasis C) Bullous emphysema D) Chronic bronchitis E) Lobar pneumonia F) Pleural effusion G) Pneumothorax H) Pulmonary embolism Correct Answer: F. Unilateral decreased breath sounds, dullness to percussion, and decreased tactile fremitus are most suggestive of a pleural effusion. Fluid occupying the space between the lung and the body wall dampens transmission of sound waves and mechanical vibrations (in contrast to consolidated lung, which presents with decreased breath sounds as a result of decreased air movement into the region of consolidation and increased tactile fremitus as dense, fluid-filled lung tissue transmits mechanical vibrations better than ventilated lung). Unilateral crackles (rales) are nonspecific. 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 can be visualized on imaging including chest x-ray, CT scan, or ultrasound. A diagnostic thoracentesis should be performed whenever the etiology of the effusion is unknown. A therapeutic thoracentesis can be used to relieve clinical symptoms. Incorrect Answers: A, B, C, D, E, G, and H. Asthmatic bronchitis (Choice A) is caused by short-term inflammation of the bronchi and is commonly provoked by environmental triggers such as pollen, molds, smoke, cold, and exertion. Auscultation typically discloses wheezes. Bronchiectasis (Choice B) is abnormal dilatation of the bronchi that may result from recurrent necrotizing infections and chronic inflammation. Risk factors for bronchiectasis are cystic fibrosis, Kartagener syndrome, and allergic bronchopulmonary aspergillosis. Bullous emphysema (Choice C) describes structural changes to the lung which 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 and a1-antitrypsin deficiency. Auscultation typically discloses decreased breath sounds or expiratory wheezing. Chronic bronchitis (Choice D) refers to chronic 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 three-month period for at least two consecutive years in patients who do not have another etiology of cough. Wheezing may be appreciated on auscultation. Lobar pneumonia (Choice E) 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. Examination of the affected lobe shows decreased breath sounds and increased tactile fremitus as the consolidated lung is adjacent to the body wall. Pneumothorax (Choice G) presents with chest pain and tachypnea with diminished breath sounds and hyperresonance to percussion on the affected side. Pulmonary embolism (Choice H) presents with acute chest pain, shortness of breath, and hypoxia, often in a patient with deep venous thrombosis or a known hypercoagulable state. Small pulmonary emboli are often undetectable on chest x-ray, and the ECG may be normal or show sinus tachycardia. Examination is typically unremarkable.

Objective: Auscultation and percussion of the lungs can be used to differentiate between common pulmonary disorders. Decreased or absent breath sounds are suggestive of pleural effusion, consolidated lung, or pneumothorax. Dullness to percussion is present in pleural effusion and consolidated lung, while hyperresonance may be appreciated in pneumothorax. Decreased tactile fremitus is present in pleural effusion and pneumothorax, while increased tactile fremitus is suggestive of consolidated lung. Previous Next Score Report Lab Values Calculator Help Pause

53 Exam Section 2: Item 4 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 4. A 20-year-old woman is brought to the emergency department by her roommate 30 minutes after she ingested a large quantity of acetaminophen tablets during a suicide attempt. The physician asks her why she tried to kill herself. She replies tearfully, "My boyfriend told me that he doesn't want to see me again, and he won't return any of my phone calls. I loved him more than I've loved anyone else in my entire life. I was going to marry him! Now I hate his guts for what he's done to me. I just wanted to die." On further questioning, the physician learns that she had only two dates with this man. She tells the physician, "I just can't bear being alone. But I can tell that you understand. You're the only doctor who's ever understood how I feel." This patient most likely has which of the following types of personality disorders? A) Borderline B) Dependent C) Histrionic D) Narcissistic E) Obsessive-compulsive Correct Answer: A. Borderline personality disorder (BPD) is a cluster B personality disorder, the emotional or dramatic cluster, that features an unstable sense of self and tumultuous relationships. Likely caused by a combination of genetic polymorphisms in serotonin and dopamine receptors and emotional invalidation during childhood, patients with BPD unconsciously learn to make impulsive and dramatic gestures, including self-harm, to obtain attention and emotional fulfilment from others. Chronic invalidation also leads to poor self-esteem and the consequent reliance on others for self-esteem needs, resulting in an intense fear of abandonment and severe distress when abandonment happens. As a result of inadequate emotional attunement from caregivers, patients with BPD do not learn to label, understand, or regulate their emotions. Their negative emotional experiences are so intensely painful that people with BPD must separate these experiences from their consciousness. This splitting leads people with BPD to black-and-white thinking that includes seeing people as all good and others as all bad, as in this patient. Incorrect Answers: B, C, D, and E. Dependent personality disorder (Choice B) is a cluster C personality disorder, which is the anxious cluster. Dependent personality disorder presents with an excessive need to be cared for by others that manifests as severe separation anxiety and clinging behavior. Though this patient does illustrate clinging behavior, she also illustrates difficulty with emotional regulation, black-and- white thinking, and suicidal gestures that are more typical of BPD. 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 self-esteem needs. These patients may dress in a seductive way for the same purpose. Patients with histrionic personality disorder do not typically experience emotions as intensely as BPD patients or make suicidal gestures. Narcissistic personality disorder (Choice D) is a cluster B personality disorder that is characterized by fragile self-esteem and compensatory arrogant, self-aggrandizing behavior to gain admiration, sometimes at others' expense. Patients with BPD do not seek admiration from others, instead alternating between seeking emotional closeness and emotional distance. Patients with narcissistic personality do not typically experience the intense emotions or fear of abandonment experienced by patients with BPD. Obsessive-compulsive personality disorder (Choice E) is a cluster C personality disorder that features a pattern of inflexible orderliness and perfectionism as well as an intense fear of making mistakes. These patients typically present as emotionally constricted as opposed to the emotional tumult of patients with BPD.

Objective: Borderline personality disorder is a cluster B personality disorder that features an unstable sense of self, emotional dysregulation, and tumultuous relationships. To unconsciously fulfill their emotional needs, patients with borderline personality disorder make suicidal gestures, desperately avoid abandonment, and either idealize or devalue other people. Previous Next Score Report Lab Values Calculator Help Pause

58 Exam Section 2: Item 9 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 9. A70-year-old man comes to the office because of a 6-day history of abdominal pain. He has no history of major medical illness and takes no medications. His vital signs are within normal limits. Physical examination shows epigastric tenderness. A CT scan of the abdomen shows a mass in the liver. As part of a study, electron microscopy of a biopsy specimen of the mass is done. Results show nuclear chromatin margination under the nuclear membrane, intact cell membranes, and cells that are smaller in size than normal cells in the tissue. The mechanism causing the cell size anomaly in this patient was most likely triggered by the release of which of the following substances into the cytoplasm? A) BAX protein B) BCL2 protein C) Cytochrome c OD) Procaspase-8 E) Procaspase-9 Correct Answer: C. Apoptosis may be triggered through the intrinsic pathway by cellular damage such as radiation, oxidative damage, ischemia, or toxin exposure, or through the extrinsic pathway by the activity of perforin, granzyme, tumor necrosis factor, or the interaction between Fas and Fas ligand. Natural killer cells induce apoptosis in cells that no longer express major histocompatibility complex class I, the loss of which is a marker of malignant transformation. Apoptosis initiated in this manner occurs via the extrinsic pathway and involves caspases 8, 10, and 3. Both pathways converge upon the release of proapoptotic factors from the mitochondria. Regulatory proteins, including BAX and BAK, interact with the mitochondrial outer membrane and induce permeability that leads to the release of cytochrome c into the cytoplasm. Once in the cytoplasm, cytochrome c activates caspase enzymes. Cellular features of apoptosis include cell shrinkage caused by degradation of the cytoskeleton, nuclear pyknosis (condensation of nuclear chromatin), karyorrhexis (fragmentation of the nucleus), and blebbing of the cellular membrane. Incorrect Answers: A, B, D, and E. BAX protein (Choice A) is found in the cytoplasm under normal conditions. Upon initiation of apoptosis, BAX inserts itself into the outer mitochondrial membrane, allowing for the release of cytochrome c. BCL2 protein (Choice B) is found in the cytoplasm under normal conditions, where it is bound to the outer mitochondrial membrane. BCL2 inhibits the action of proapoptotic factors such as BAX and BÁK, and promotes cell survival. Overexpression of BCL2 is a feature of malignancies such as B-cell lymphoma, from which BCL2 protein derives its name. Procaspase-8 (Choice D) and procaspase-9 (Choice E) are found in the cytoplasm and are inactive precursors to caspase enzymes. Caspases exist as inactive proenzymes that must be activated to potentiate apoptosis.

Objective: Both the intrinsic and extrinsic apoptotic pathways involve the release of cytochrome c from the mitochondria. Once released into the cytoplasm, cytochrome c activates caspases that potentiate apoptosis. Cellular features of apoptosis include cell shrinkage, nuclear pyknosis, karyorrhexis, and blebbing of the cellular membrane. Previous Next Score Report Lab Values Calculator Help Pause

64 Exam Section 2: Item 15 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 15. A 44-year-old man comes to the physician for a follow-up examination 1 month after receiving a cadaveric kidney transplant. He says he feels well. Current medications include mycophenolate and prednisone. Physical examination shows an arteriovenous fistula in the left upper extremity and a well-healing surgical wound over the right lower quadrant of the abdomen. Serum studies show a urea nitrogen concentration of 44 mg/dL and creatinine concentration of 2.6 mg/dL, compared with 21 mg/dL and 1.8 mg/dL, respectively, 2 weeks ago. Examination of a biopsy specimen of the transplanted kidney shows an inflammatory infiltrate affecting the small arterioles. These changes are most likely caused by which of the following mechanisms? A) Donor CD4+T lymphocytes recognize class II MHC molecules on host dendritic cells B) Donor CD4+ T lymphocytes recognize class II MHC molecules on host lymphocytes C) Donor CD8+ T lymphocytes recognize class I MHC molecules on host dendritic cells D) Host CD4+ T lymphocytes recognize class II MHC molecules on renal allograft cells E) Host CD8+ T lymphocytes recognize class I MHC molecules on renal allograft cells Correct Answer: E. Host CD8+ T lymphocytes recognize class I MHC molecules on allograft cells. This leads to acute T-lymphocyte-mediated transplant rejection (TCMR), typified by an asymptomatic rise in creatinine and biopsy showing an inflammatory infiltrate affecting the small arterioles. This most commonly occurs within the first six months after transplantation. TCMR is caused by host CD8+ T lymphocytes reacting to donor MHC I complexes present on cells within the glomeruli, tubules, interstitium, and blood vessels of the allograft. The recognition of the foreign MHC I complex induces a robust immune response against the allograft. Biopsy is required for the evaluation of graft dysfunction because it can distinguish between acute rejection and other causes of renal injury, including acute tubular necrosis, interstitial nephritis, or viral infections (eg, cytomegalovirus or BK virus). Primary histologic changes include renal interstitial infiltration with lymphocytic cells, in addition to the obliteration of the tubular basement membrane. Treatment is based on the degree of rejection and allograft dysfunction but typically includes steroids and lymphocyte- depleting agents. Incorrect Answers: A, B, C, and D. Donor CD4+ T lymphocytes recognize class II MHC molecules on host dendritic cells (Choice A) describes one pathologic mechanism in acute graft versus host disease (GVHD). Host dendritic cells present host antigens on their surface continuously via MHC II molecules, and these are recognized by donor CD4+ T cells, which induce a robust inflammatory response resulting in GVHD. These CD4+ cells help recruit donor CD8+ T cells, which recognize class I MHC molecules on host dendritic cells (Choice C) and augment the immune response against host tissues. Common manifestations of acute GVHD include rash, hepatitis, and diarrhea. Donor CD4+ T lymphocytes recognize class II MHC molecules on host lymphocytes (Choice B) in the graft-versus-leukemia effect following hematopoietic stem cell transplantation. Leukemic cells express MHC II on their surface that contain leukemic antigens, which are recognized by the donor CD4+ cells. Host CD4+ T lymphocytes recognize class II MHC molecules on renal allograft cells (Choice D) describes the mechanism of chronic allograft dysfunction. T-lymphocyte activation leads to cytokine production along with humoral and cellular hypersensitivity reactions (type Il and IV, respectively). It occurs over months to years and is typified by vascular arteriosclerosis and smooth muscle proliferation with parenchymal fibrosis and atrophy.

Objective: Acute renal allograft rejection is primarily mediated by CD8+ T lymphocytes, which recognize class I MHC molecules present on the surface donor allograft cells. As these are foreign antigens, this leads to a robust immune response that can cause dysfunction of the allograft and must be treated with increased immunosuppressive therapy. Previous Next Score Report Lab Values Calculator Help Pause

34 Exam Section 1: Item 35 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 35. A 40-year-old man comes to the physician for a routine examination. He has smoked 2 packs of cigarettes daily for 20 years. Which of the following initial statements by the physician is likely to be most effective in encouraging this patient to stop smoking? A) "Do you have any relatives who died of lung cancer?" B) "Let me show you a picture of what lungs look like after a lifetime of smoking." C) "Please tell me how I can help you to stop smoking." D) "Smoking causes lung cancer and many other illnesses." E) "What thoughts have you had about quitting smoking?" F) "Why do you continue to smoke when it is so unhealthy?" Correct Answer: E. Asking an open-ended question about the patient's thoughts about quitting smoking is the best way to engage and motivate the patient and can also assess the patient's readiness for behavioral change. The stages of behavioral change are used to define a patient's readiness to change a health-related behavior such as smoking. Physicians aim to move patients through these stages over time with an interview technique called motivational interviewing. For instance, physicians can utilize motivational interviewing to move a patient who is precontemplative, or not interested in changing, to the contemplative stage, or indecisive about changing. Motivational interviewing involves using open-ended, nonjudgmental questions to help the patient explore their reasons for wanting to change or maintain the habit. Motivational interviewing is a collaborative, nondirective process that capitalizes on the patient's autonomy rather than striking a paternalistic tone. Physicians should refrain from providing advice or education initially, as this advice is unlikely to be effective (especially for patients who are in the precontemplative stage). Incorrect Answers: A, B, C, D, and F. Educating the patient about lung cancer or other health concerns related to smoking (Choices A, B, D, and F) strikes a tone that is too directive and may make this patient feel defensive about his smoking habit. Questioning why the patient continues to smoke despite the health risks (Choice F) would likely make him feel judged. An effective discussion is more likely to result from an open- ended question about the patient's thoughts about his habit. Education about the health risks of smoking may be appropriate later in the discussion after the physician has established that the patient has surpassed the precontemplative stage and allowed the patient to explore his own motivations. Encouraging the patient to share how the physician can help him quit smoking (Choice C) assumes that the patient may be interested in quitting. However, the physician has not yet established that the patient is interested in changing. This directive approach is unlikely to increase the patient's motivation to change and would therefore decrease the patient's chance of successfully quitting.

Objective: Asking open-ended questions to explore a patient's thoughts about a health-related habit engages and motivates the patient to change the habit. Motivational interviewing is a technique that uses this strategy and avoids providing advice or education until establishing that the patient is considering a change. 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. A 51-year-old man has a cardiac output of 5 L/min and a right atrial pressure of 0 mm Hg at rest. The solid curve in the graph shows the relationship between right atrial pressure and cardiac output in a normal heart. Which of the following lettered points best represents the change in this relationship 5 seconds after a large arteriovenous fistula opens? 15- 10 AO PE Boo¯OF 0- -4 -2 0 2 4 6. 8 10 Right atrial pressure (mm Hg) A) B) C) D) E) F) Correct Answer: D. The sudden surge of arterial blood into venous return that results from a large arteriovenous fistula (AVF) opening would result in increased right atrial pressure and increased cardiac output caused by the increase in preload. The AVF results in an abnormal communication between an artery and vein, leading to low-resistance, high-volume flow of blood between the artery and vein. An AVF decreases systemic vascular resistance through the creation of a shunt to the high-capacitance, low-pressure venous system. The increase in preload causes a greater distension in the cardiomyocyte fibers at the end of diastole, which results in increased cardiac contractility per the Frank-Starling relationship. This is represented by point D on the cardiac function curve in this patient with a normal heart that can respond to changes in preload and generate the increased contractile force. Patients with an AVF present with a palpable thrill, diminished distal pulses, and a continuous machine-like murmur at the site of the fistula. Over time, this can lead to the development of high-output cardiac failure. The size of the AVF is most predictive of the risk for developing heart failure, as a larger AVF results in higher flow between the artery and vein, requiring the heart to compensate with greater output to supply the remainder of the systemic circulation. Incorrect Answers: A, B, C, E, and F. Point B on the cardiac function curve represents the cardiac output relative to a decrease in preload from the patient's heart at rest. This may result from positive intrathoracic pressure during expiration (which decreases venous return to the right atrium), hypovolemia, or exposure to a venous vasodilator. Point A represents a shift upward from this state of decreased preload onto a new curve as a result of enhanced cardiac performance, which can be caused by exposure to exogeneous or endogenous inotropes (eg, epinephrine, norepinephrine), tachycardia, or a decrease in afterload. Points C and F indicate a shift downward to a decreased cardiac performance curve, which may be seen with negative inotropes, bradycardia, or heart failure (in which the cardiomyocyte fibers cannot produce the same force as a normal heart). Point E represents the baseline of this patient's heart at rest.

Objective: Arteriovenous fistulas result in a large flux of blood from the high-pressure arterial system into the low-resistance venous system. In a normal heart, the increased preload results in greater cardiac output per Frank-Starling mechanics. High-output heart failure may develop over time. Previous Next Score Report Lab Values Calculator Help Pause Cardiac output (L/min)

29 Exam Section 1: Item 30 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 1.0- 30. A study is conducted to measure survival following a diagnosis of lung, colon, or bladder cancer. Data from a statewide cancer registry are used. The graph shows survival plotted against the number of years after diagnosis for each cancer. Which of the following is the most accurate interpretation of these data concerning mortality? 0.9- 0.8- 0.7- 0.6- 0.5- A) 1-Year mortality after a diagnosis of colon cancer is the same as 1-year mortality after a diagnosis of lung cancer B) 3-Year mortality after a diagnosis of lung cancer is lower than 3-year mortality after a diagnosis of bladder cancer C) 3-Year mortality is the same after a diagnosis of lung, bladder, and colon cancer 0.4- D) 5-Year mortality after a diagnosis of bladder cancer is lower than 5-year mortality after a diagnosis of colon 0.3- cancer 0.2- E) 5-Year mortality after a diagnosis of lung cancer is lower than 5-year mortality after a diagnosis of colon cancer 0.1- 1 3 4 Years after diagnosis Bladder Colon Lung Correct Answer: D. Survival curves are a useful tool in evaluating the natural history of a disease or the efficacy of treatment. The survival curve in this study plots the number of patients still alive (Y-axis) against time (X-axis). A perfect treatment, in which all patients survived the disease, would be reflected by a horizontal line across the top of the graph, demonstrating a surviving proportion of 1.0 during the entire study duration. Thus, the higher the value on the Y-axis of a survival curve, the greater the chances of survival in that patient group. If patients were to die at a constant rate after the diagnosis of the disease, this would be reflected by a straight line with a constant downward slope. In most cases, however, cancer demonstrates a steep initial decline, followed by a plateau on survival curve analysis. This is because initial disease severity and the success or failure of treatment typically occurs early. After successful treatment, survival rates become more stable for a longer duration of time. This phenomenon is exemplified in the lung cancer survival curve in this study. The bladder cancer group has a proportion surviving of 0.90 at 5 years, meaning that 90% of patients are still alive at 5 years. The colon cancer group has a proportion surviving of 0.65 at 5 years, meaning that 65% of patients are alive at 5 years. This means that the diagnosis of bladder cancer has a lower 5-year mortality rate than does colon cancer. Incorrect Answers: A, B, C, and E. One-year mortality after a diagnosis of colon cancer is approximately 15%, whereas 1-year mortality after lung cancer is approximately 55% on the basis of the survival curves. Therefore, the 1- year mortality rates of these diseases are not the same (Choice A). Three-year mortality after a diagnosis of lung cancer is 70%, whereas the 3-year mortality of bladder cancer is approximately 8%. Thus, 3-year mortality after a diagnosis of lung cancer is not lower than 3-year mortality after a diagnosis of bladder cancer (Choice B). Three-year mortality of bladder cancer is less than 10%, while 3-year mortality for colon cancer is approximately 30%, and lung cancer mortality at 3 years is approximately 70%. Thus, 3-year mortality is not the same for these diseases (Choice C). Five-year mortality after a diagnosis of lung cancer is significantly higher than that of colon cancer (Choice E).

Objective: Survival curves are a useful depiction of survival rates in cancer and other diseases. High values on the Y-axis and curves with gradual slope reflect longer survival, while curves with more abrupt downward slope reflect higher and earlier mortality. II Previous Next Score Report Lab Values Calculator Help Pause Proportion surviving

48 Exam Section 1: Item 49 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 49. A9-year-old girl is brought to the clinic by her mother because of areas of hair loss on her scalp that the mother first noticed 1 week ago. The patient otherwise has been well. There is no personal or family history of serious illness. She receives no medications. The patient transferred from her neighborhood elementary school to a private school for gifted children 3 months ago. The mother says that no one in the house has used new shampoos, soaps, or detergents recently. The patient's growth and development have been normal. The child appears healthy. Vital signs are within normal limits. Examination of the scalp shows two irregularly shaped, quarter-sized areas of alopecia on the right side of the head. Short, broken hairs are noted within these areas, but there is no scaling or erythema. Which of the following is the most likely diagnosis? A) Alopecia areata B) Hypothyroidism C) Menkes disease D) Tinea capitis E) Trichotillomania (hair-pulling disorder) Correct Answer: E. Focal, irregular patches of hair loss with broken hairs is consistent with trichotillomania. Trichotillomania is a psychiatric impulse control disorder that presents with repetitive hair pulling resulting in noticeable hair loss and several unsuccessful attempts to stop hair pulling. Patients also commonly experience tension prior to hair pulling and relief after hair pulling. Stressors can exacerbate hair pulling behavior. Physical examination typically demonstrates bizarrely shaped patches of hair loss on the scalp or loss of eyelashes or eyebrows without significant underlying skin changes. Broken hairs are usually different lengths. Trichotillomania is managed with habit reversal therapy, which helps patients identify triggers for hair pulling and develop alternative ways to resolve the tension that precedes hair pulling behavior. Incorrect Answers: A, B, C, and D. Alopecia areata (Choice A) is a chronic, immune-mediated disorder of hair loss that can present with diffuse or focal hair loss. Areas may demonstrate smooth or irregular borders. Hairs are typically narrower proximally than distally and are prone to breaking. The broken hairs are usually the same length, while in trichotillomania, the broken hairs are different lengths. In this child with a known stressor, trichotillomania is more likely. Hypothyroidism (Choice B) commonly presents with diffuse alopecia instead of the focal alopecia of trichotillomania. Thyroid hormone helps maintain hair follicles. Menkes disease (Choice C) is a congenital X-linked disorder that causes a defect in intestinal copper absorption, leading to copper deficiency. The disease presents with progressive neurologic symptoms, hypopigmentation of the skin, bony abnormalities, and kinked hair instead of alopecia. Further, most children with Menkes disease have a life expectancy of less than four years. Tinea capitis (Choice D) is a fungal infection of the scalp common in children. The most common presentation is scaly patches with alopecia and patches of alopecia with black dots that represent broken hairs. This patient did not demonstrate scaly patches or black dots on the scalp.

Objective: Trichotillomania is a psychiatric disorder that presents with repetitive hair pulling resulting in noticeable hair loss. Patients typically present with irregularly-shaped patches of hair loss with broken hairs of different lengths. Previous Next Score Report Lab Values Calculator Help Pause

11 Exam Section 1: Item 11 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 11. A 22-year-old man is brought to the emergency department 30 minutes after collapsing at home. He has a 3-year history of intravenous drug use. He appears confused. His temperature is 39.5°C (103.1°F). Physical examination shows subungual hemorrhages of the fingers and jugular venous distention. A holosystolic murmur is heard at the lower left sternal border. Blood cultures are positive for Staphylococcus aureus. A chest x-ray shows multifocal airspace opacities. The most likely cause of these findings is infection of which of the following labeled cardiac valves in the CT scans of the chest? -B C A A) B) C) D) Correct Answer: C. Intravenous drug use is a risk factor for infective endocarditis and valvular dysfunction as a result of 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 in the left lower sternal border. Severe tricuspid regurgitation is a risk factor for the development of right-sided heart failure, which is characterized by increased jugular venous pressure, hepatomegaly, ascites, and peripheral extremity edema. Other complications include septic pulmonary emboli, which present as multifocal peripheral cavitary lung lesions. The tricuspid valve can be identified on CT imaging by its location between the right ventricle and right atrium. The right ventricle is the most anterior and inferior part of the heart, as seen in the axial slices of the patient's CT scan. Incorrect Answers: A, B, and D. Choice A represents the aortic valve, which separates the left ventricle from the ascending thoracic aorta. It is located superior to the left ventricle, anterior to the left atrium, and posterior to the pulmonic valve. It may be, along with the mitral valve, more commonly affected in cases of endocarditis that are unrelated to intravenous drug abuse. Systemic septic emboli are common. Choice B represents the pulmonic valve, which separates the right ventricle from the pulmonary trunk. It is anterior to the aorta. It is the least common valve affected by endocarditis. Choice D represents the mitral valve, which separates the left atrium from the left ventricle. The left atrium is the most posterior structure in the heart. It is more commonly affected in cases of endocarditis that are unrelated to intravenous drug abuse.

Objective: Tricuspid valve endocarditis is typically associated with intravenous drug use. Complications include tricuspid regurgitation and septic pulmonary emboli. 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 19-year-old woman comes to the physician because of a 3-day history of fever, shortness of breath, cough productive of blood-tinged sputum, rash on her legs, and pain of her groin. She lives on a small farm in rural New Mexico and says that all three of her cats have been infested with fleas. She is in respiratory distress. Her temperature is 39°C (102.2°F), pulse is 104/min, respirations are 28/min, and blood pressure is 100/60 mm Hg. Physical examination shows a 1-cm ulcer on the right lower extremity and numerous erythematous macular lesions over both lower extremities. There is a fluctuant 3 x 5-cm mass in the femoral lymph node chain. A chest x-ray shows bilateral infiltrates. A Gram stain of purulent material aspirated from the femoral mass shows numerous neutrophils and few gram-negative bacilli. Which of the following is the most likely causal organism? A) Coxiella burnetii B) Ehrlichia chaffeensis C) Francisella tularensis D) Streptococcus pneumoniae E) Yersinia pestis Correct Answer: E. Yersinia pestis is the causal organism responsible for this patient's sepsis and ulceroglandular disease, which is consistent with the bubonic plague. Yersinia pestis is a gram-negative bacillus that is most often transmitted by infected fleas, but can also be acquired via direct handling of infected animals (eg, cats), consumption, or aerosolized droplets from other infected humans. This patient's right lower extremity ulcer likely represents the site of an initial flea bite, and the femoral lymph node mass is known as a buboe, which is a large inflammatory infiltrate containing Y. pestis in a lymph node that drains the infected leg. Pneumonia, which this patient also demonstrates, can either occur as a primary syndrome or can develop as a complication (termed secondary pneumonic plague) of disseminated disease and bacteremia. Disseminated intravascular coagulation can occur with disseminated disease, and meningitis is an additional complication. Leukocytosis and thrombocytopenia are common laboratory findings. Treatment is primarily with aminoglycosides although doxycycline is a valid alternative therapy. Incorrect Answers: A, B, C, and D. Coxiella burnetii (Choice A) is an intracellular gram-negative bacterium that causes Q fever. Exposure occurs through contact with contaminated livestock. It is found throughout the United States. Acute infection presents with high fevers, myalgias, malaise, and gastrointestinal symptoms; ulceroglandular disease is not a feature. Chronic Q fever can result in endocarditis. Ehrlichia chaffeensis (Choice B) is an intracellular bacterium that causes ehrlichiosis, which is characterized by fevers, myalgias, headache, and occasionally neurologic symptoms. It is endemic in the southeast, mid-Atlantic, and central United States, but the vector is a tick, not a flea. Common laboratory findings include thrombocytopenia and morulae within neutrophils. Francisella tularensis (Choice C) is a fastidious gram-negative bacterium that causes tularemia, an ulceroglandular disease acquired through contact with infected animals (often rabbits) or through inhalation of the bacteria. Common symptoms include a local eschar at the site of entry and lymphadenopathy proximal to the site of entry, but pneumonia may also occur. While tularemia might initially be suspected in this patient, she has risk factors and biopsy findings that are most consistent with Y. pestis infection. Streptococcus pneumoniae (Choice D) is a gram-positive coccus that causes pneumonia, otitis media, sinusitis, and meningitis. It does not typically cause ulceroglandular disease and biopsy would show gram-positive organisms, not gram-negative organisms.

Objective: Yersinia pestis is the causal organism of the bubonic plague, with disease manifestations ranging from local ulceroglandular disease to sepsis, meningitis, and pneumonia. It is acquired via contact with infected animals, through flea bites, or through inhalation of droplets from infected humans. Treatment is with aminoglycosides or doxycycline. Previous Next Score Report Lab Values Calculator Help Pause

66 Exam Section 2: Item 17 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 17. A 15-month-old boy is brought to the physician by his mother because of a 9-month history of recurrent bacterial infections. The patient has not had recurrent viral infections. He is at the 10th percentile for length and weight. Physical examination shows multiple areas of honey yellow, crusted lesions over the lower extremities. A Gram stain of one of the lesions shows many gram-positive cocci in clusters but no leukocytes. Laboratory studies show no abnormalities except for a leukocyte count of 30,000/mm3. This patient most likely has a rare autosomal recessive disease leading to a lack of CD18 expression. The leukocytes would be deficient in which of the following characteristics? A) Cytokine production B) Helper T-cell function C) Immunoglobulin gene rearrangement D) Killing of intracellular bacteria E) Migration Correct Answer: E. Leukocyte adhesion deficiency (LAD) results from a defect in the attachment of leukocytes to the vascular endothelium, which consequently results in the impaired recruitment and migration to sites of extravascular inflammation or infection. LAD type 1 is caused by a dysfunctional LFA-1 integrin (CD18) protein on the leukocyte surface, which does not allow for normal attachment of the cell to the vascular endothelium. It is typically characterized by recurrent bacterial infections, impaired wound healing, a delayed detachment of the umbilical cord after birth, and lack of leukocytes at sites of infection with an absence of pus. The actions of leukocyte phagocytosis and bacterial killing are otherwise unimpaired. Laboratory studies in patients with LAD will show increased leukocyte concentrations in the blood. Incorrect Answers: A, B, C, and D. Cytokine production (Choice A) is not impaired in leukocyte adhesion deficiency. Cytokines are signaling proteins that modulate cellular response to inflammation. Proinflammatory cytokines include IL-1, 6, 8, 12, and 18, interferons, and tumor necrosis factor. Anti-inflammatory cytokines include İL-4, 10, 11, and 13. Helper T-cell function (Choice B) remains intact in LAD. Helper T lymphocytes are an essential component of the adaptive immune response to infection. They activate CD8+ effector T lymphocytes, promote humoral immunity, activate macrophages and neutrophils, and recruit eosinophils. Impaired helper T-lymphocyte function is the underlying mechanism that allows opportunistic infections to occur in HIV infection and AIDS. Immunoglobulin gene rearrangement (Choice C), also called V(D)J recombination, is the mechanism by which lymphocytes are able to generate the enormous range of the adaptive immune response to foreign antigens. Deficiency results in severe combined or common variable immunodeficiency. Killing of intracellular bacteria (Choice D) remains intact in LAD. Disorders such as chronic granulomatous disease, Chediak-Higashi syndrome, and myeloperoxidase deficiency are characterized by deficient killing of intracellular bacteria.

Objective: LAD is a group of disorders characterized by impaired leukocyte adhesion to the vascular endothelium. Leukocytes retain the ability to phagocytose and eliminate foreign pathogens, but are unable to migrate to sites of infection or inflammation in the extravascular space. II Previous Next Score Report Lab Values Calculator Help Pause

7 Exam Section 1: Item 7 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 7. A 24-year-old man with AIDS comes to the physician because of a 1-month history of abdominal cramps and watery diarrhea. These symptoms have become increasingly severe during the past 5 days. His pulse is 95/min. Physical examination shows dry mucous membranes and decreased skin turgor. A biopsy specimen of the colon is shown in the photomicrograph. Which of the following is the most likely causal organism? A) Candida albicans B) Cryptococcus neoformans C) Cryptosporidium parvum OD) Cytomegalovirus E) Giardia lamblia OF) Mycobacterium avium complex Correct Answer: C. The photomicrograph demonstrates small, round oocysts on the colonic epithelial surface, consistent with cryptosporidiosis, an AIDS-defining illness. Cryptosporidium is a genus of protozoal pathogens that cause cryptosporidiosis, a diarrheal illness that is mild and self-limited in immunocompetent hosts but which is severe and often life-threatening in immunocompromised hosts. Signs and symptoms include fever, weight loss, symptoms of dehydration and orthostasis, severe watery diarrhea, cramping abdominal pain, nausea, and vomiting. Immunocompromised patients may develop disseminated infection with involvement of the liver and lungs. Diagnosis is made with stool antigen testing or with detection of luminal oocysts by acid-fast staining. Treatment is supportive, with repletion of fluids and electrolytes, as well as antiparasitic therapy. Nitazoxanide is the treatment of choice in immunocompetent patients, whereas immunocompromised patients require antiretroviral therapy to improve their immune status and CD4+ cell counts, as well as combination regimens that involve nitazoxanide and other agents, such as azithromycin or paromomycin. Incorrect Answers: A, B, D, E, and F. Candida albicans (Choice A) can produce oral thrush and esophagitis in immunocompromised patients. Candida esophagitis is an AlIDS-defining illness. Staining with Gomori methenamine silver or periodic acid-Schiff shows pseudohyphae and budding yeasts. Cryptococcus neoformans (Choice B) is a potential cause of meningitis in AIDS patients. Staining with India ink or mucicarmine shows an encapsulated yeast with narrow-based budding. Cytomegalovirus (CMV) (Choice D) is a herpesvirus (human herpesvirus-5) that causes esophagitis, pneumonia, and retinitis in AIDS patients. Light microscopy may show intracellular inclusion bodies known as Cowdry bodies. Diagnosis is made with enzyme-linked immunosorbent assay (ELISA) to detect anti-CMV immunoglobulin and with polymerase-chain reaction amplification of viral DNA. Giardia lamblia (Choice E) is a flagellated, parasitic organism that results in giardiasis, which also presents with watery diarrhea. Light microscopy shows a flagellated, pear-shaped trophozoite with two large, central nuclei, and a medial axostyle. While giardiasis may also demonstrate cysts on light microscopy, this patient's history of AIDS and lack of history of exposure to unfiltered water (such as while hiking or camping) are more suggestive of cryptosporidiosis. Mycobacterium avium complex (MAC) (Choice F) is a mycobacterial pathogen that causes atypical pneumonia and gastrointestinal infection in AIDS patients. Light microscopy shows acid-fast, periodic acid-Schiff positive bacilli. Patients with AIDS are given routine MAC prophylaxis with azithromycin.

Objective: Cryptosporidium parvum causes cryptosporidiosis, a severe, often life-threatening illness in AIDS patients that is characterized by fever, watery diarrhea, and cramping abdominal pain. Light microscopy shows intraluminal oocysts on acid-fast stain. II Previous Next Score Report Lab Values Calculator Help Pause

83 Exam Section 2: Item 34 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 34. During a clinical study, a researcher finds that as children approach puberty, the consumption of milk in the diet is partially replaced by the consumption of soft drinks. This results in a decrease in calcium intake. This dietary change is most likely to have which of the following effects on bone mineralization or metabolism? A) Decreased parathyroid hormone secretion because of increased phosphorus intake B) Decreased peak bone accretion during adolescence C) Hypophosphatasia D) Rickets resulting from calcium deficiency E) Rickets resulting from vitamin D deficiency Correct Answer: B. Bone tissue is constantly remodeling throughout human life. During childhood and adolescence, bone mineral density increases until the early third decade of life in females and the late third decade in males. According to Wolff's law, bone density will increase in response to stress applied. In turn, weightbearing exercise or physically strenuous work will increase bone mineral density. In addition, availability of calcium and vitamin D are required for increased bone mineral density. Estrogen and testosterone are key regulators of ossification and bone mineral density as well; because of this, bone mineral density significantly increases during puberty. Deficits in any of these factors during the first three decades of life will lead to decreased peak bone mineral density. After peak bone mineral density is reached, increasing bone mineralization seldom occurs. The normal aging process biases bone remodeling toward greater osteoclast activity and less osteoblast activity, which leads to progressive losses of bone mineral density over time. In this study, decreased calcium intake will lead to decreased mineralization of bone, leading to a decreased peak bone accretion during adolescence. Incorrect Answers: A, C, D, and E. Parathyroid hormone increases the secretion of phosphorus at the kidney. Increased phosphorus intake, reflected as serum hyperphosphatemia, leads to increased (not decreased) concentrations of parathyroid hormone secretion through normal feedback mechanisms (Choice A). Hypophosphatasia (Choice C) is a rare inherited genetic disorder caused by mutations in the ALPL gene, which codes for alkaline phosphatase. Deficiency of alkaline phosphatase leads to the accumulation of pyrophosphate and decreased serum concentrations of phosphate ion, which results in impaired bone mineralization. Rickets resulting from calcium deficiency (Choice D) is rare; although decreased calcium intake will lead to decreased bone mineralization over a long period of time, slight decreases in calcium intake as described in this study will initially result in lower overall bone mineral density without signs and symptoms of rickets. Rickets resulting from vitamin D deficiency (Choice E) classically presents with bowing of long bones, especially the tibia. It typically presents in younger children. A decrease in one source of calcium in an adolescent will lead to a subclinical decrease in bone mineral density rather than overt malformation and growth plate abnormalities as seen in rickets.

Objective: Decreased calcium or vitamin D intake during adolescence and the time leading up to peak bone mineral density will lead to subclinical deficits in bone mineral accretion. This will lead to lower peak bone mineral density and increase the risk for osteoporosis later in life. Previous Next Score Report Lab Values Calculator Help Pause

172 Exam Section 4: Item 23 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 23. In an isolated population, a single mutation allele of the cystic fibrosis transmembrane regulator (CETR) gene is present at a frequency of 0.1. The normal CFTR allele and the mutation allele are in Hardy-Weinberg equilibrium in this population. Which of the following is the approximate frequency of heterozygosity for the mutation in this population? A) 0 B) 0.01 C) 0.05 D) 0.2 E) 0.5 F) 0.8 G) 1.0 Correct Answer: D. 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 + q2 = 1. In this case, the frequency of the recessive allele is 0.1 (10%). Using the equation p + q = 1, the frequency of the dominant allele is 0.9 (90%). The frequency of heterozygosity is equal to 2pq, or 2*0.1*0.9 = 0.18, or approximately 0.2 (20%). In order for Hardy-Weinberg equilibrium to be met, there must be a single mutation allele with no new allele formation by mutation, gene duplication, or gene deletion. The population must be large and isolated without entrance of individuals carrying alternative alleles. There must be random mating and neither gene allele may confer a survival advantage. While these assumptions are rarely met in clinical practice, Hardy-Weinberg equilibrium provides a useful estimate of the frequency of homozygosity and heterozygosity at the gene locus in a population. Incorrect Answers: A, B, C, E, F, and G. A heterozygosity frequency of 0 (Choice A) or 1 (Choice G) would indicate that all individuals in the population are homozygous for the same allele. As the mutation is present in the population at a frequency of 0.1, there are individuals who carry the recessive allele, so the rate of heterozygosity must fall between 0 and 1. A frequency of 0.01 (Choice B) corresponds with individuals in the population who are homozygous for the recessive allele. 0.8 (Choice F) is the frequency of individuals in the population who are homozygous for the dominant allele. Half of the frequency of the recessive allele is 0.05 (Choice C), but this not relevant to the frequency of heterozygosity in the population. A frequency of 0.5 (Choice E), or 50%, corresponds with the likelihood that a heterozygous parent will pass on the recessive allele to his or her offspring.

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 + q2 = 1. In a population with a recessive allele frequency of 0.1, the frequency of heterozygosity is roughly 0.2. Previous Next Score Report Lab Values Calculator Help Pause

122 Exam Section 3: Item 23 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 23. A 70-year-old man comes to the physician because of difficulty sleeping since his wife died 8 months ago. He states that he sometimes wakes up an hour earlier than usual without being able to go back to sleep. He also often cries when he thinks about his wife. He tells the physician that he once briefly followed a woman down the street who resembled her. The patient continues to enjoy visits with his grandchildren and remains an active member of his bowling team. He does not have any other depressive symptoms or suicidal ideation. Physical examination shows no abnormalities. Which of the following is the most appropriate initial action by the physician? A) Schedule regular appointments to monitor the patient B) Obtain neuropsychological testing C) Order a sleep study D) Prescribe antidepressant therapy E) Prescribe diphenhydramine Correct Answer: A. In this patient who is likely experiencing an uncomplicated grief reaction, the physician should schedule regular appointments to monitor the patient for suicidality and the development of psychiatric disorders such as major depressive disorder. Acute, uncomplicated grief presents in diverse ways and can include sadness, shock, anguish, guilt, regret, difficulty concentrating, decreased interest in usual activities, preoccupation with thoughts and memories of the deceased, and neurovegetative symptoms such as sleep, appetite, and energy disturbances. In some patients, these symptoms can endure for months. However, in uncomplicated grief reactions, the criteria for major depressive disorder are not met. This patient is not experiencing anhedonia and appears to be functioning in his activities, which makes uncomplicated grief much more likely than major depressive disorder. Management includes frequent reassessment, encouraging reconnection with family, and, in severe cases, referral for grief counseling. Acute grief ultimately becomes integrated grief, in which thoughts of the deceased continue to engender sadness but acute symptoms resolve. Incorrect Answers: B, C, D, and E. Obtaining neuropsychological testing (Choice B) would be unnecessary in this patient with uncomplicated grief and without significant cognitive impairment. In patients with cognitive impairment, neuropsychological testing identifies specific cognitive domains that are affected and is typically used to elucidate the cause of cognitive impairment when unclear from the clinical history. Ordering a sleep study (Choice C) would be unnecessary, as this patient's early morning awakening can likely be attributed to uncomplicated grief or typical aging-related sleep changes. Prescribing antidepressant therapy (Choice D) would be inappropriate in this patient without a diagnosis of major depressive disorder. In older patients with major depressive disorder, antidepressants should be prescribed with caution because of interactions with other medications and potential cognitive side effects. Prescribing diphenhydramine (Choice E) would be relatively contraindicated in this older patient because of the high risk for cognitive impairment from its antihistaminergic and anticholinergic activity (according to the Beers Criteria). Further, this patient lacks a psychiatric indication for diphenhydramine such as episodic anxiety or insomnia, off-label uses that are not advised in older patients.

Objective: For patients with uncomplicated grief reactions, physicians should schedule regular appointments to monitor the patient for suicidal ideation and psychiatric disorders such as major depressive disorder. Uncomplicated grief can include sadness, guilt, difficulty concentrating, preoccupation with thoughts and memories of the deceased, and disturbed neurovegetative symptoms such as sleep, appetite, and energy changes without meeting the criteria for major depressive disorder. Previous Next Score Report Lab Values Calculator Help Pause

179 Exam Section 4: Item 30 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 30. A9-year-old girl is brought to the emergency department by her father because of severe pain in her right shoulder after a fall 1 hour ago. Physical examination shows pain on movement of the right shoulder and a tender right clavicle. An X-ray of the shoulder shows a nondisplaced fracture of the right clavicle. Which of the following structures of the clavicle will most likely assist in producing new bone to heal this fracture? A) Epiphysis B) Haversian canal C) Lamella D) Periosteum E) Volkmann canal Correct Answer: D. When a fracture occurs, stem cells from the periosteum and endosteum are recruited to the site of injury to generate a healing response. There are two main types of bone healing. Primary bone healing occurs when bones are rigidly fixed together, with remodeling occurring via the haversian canal system. More frequently, secondary bone healing occurs when there is significant motion at the fracture site. In this case, stem cells from the periosteum and endosteum form a fibrocartilaginous callus, which immobilizes the fracture area and adheres the ends of the bone together. Over time, this flexible union will ossify and remodel into organized bone. Bone healing follows an organized and staged pattern. Initially, a hematoma forms and inflammatory cells infiltrate the area. At this stage, there is a significant release of cytokines that includes platelet-derived growth factor, transforming growth factor-ß, and interleukins. Fibroblasts and osteoblasts differentiate and migrate into the area, and the fibrocartilaginous fracture callus is formed. Endochondral ossification occurs, converting this fibrocartilage to woven bone. Over time, the woven bone is remodeled to form cortical and trabecular bone via concerted osteoblastic and osteoclastic activity. Incorrect Answers: A, B, C, and E. The epiphysis (Choice A) is the region of bone between the physis (growth plate) and the joint surface. This region has a broadened surface area to support the load of the joint along with a thick subchondral plate of cortical bone. The haversian canal (Choice B) is the central canal of an osteon, which is the basic functional unit of compact bone tissue. The haversian canal contains the blood supply to a unit of bone. A lamella (Choice C) is a layer of osteoid that surrounds the central canal. The lamella provides strength and support around the vascular supply. The Volkmann canals (Choice E) interconnect the haversian canals and the periosteum.

Objective: Fracture healing occurs in an organized fashion that begins with hematoma formation, inflammation, and growth factor release. This leads to the formation of a fibrocartilaginous fracture callus, which is followed by ossification and remodeling. Periosteal cells and endosteal cells are key in this repair process. %3D Previous Next Score Report Lab Values Calculator Help Pause

196 Exam Section 4: Item 47 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 47. A 4-year-old boy is brought to the physician by his parents because of developmental delays and speech difficulty. He has two maternal uncles with intellectual disability. His height and weight are at the 30th percentile. Physical examination shows a large, long head with a prominent jaw, wide forehead, and large ears. There is a high arched palate, and hyperextensible fingers and joints. His speech is difficult to understand, he has a 15-word vocabulary, and he uses single words only. Which of the following is the most likely diagnosis? A) Down syndrome B) Fragile X syndrome C) Gonadal dysgenesis 45,X (Turner syndrome) D) Hurler syndrome E) Trisomy 18 Correct Answer: B. This patient's constellation of findings, including intellectual disability, an elongated face with a prominent jaw, a high-arched palate, large ears, and hyperextensible joints is highly suggestive of Fragile X syndrome. Patients may also present with macroorchidism after reaching puberty. Fragile X syndrome is the most common cause of inherited intellectual disability and is inherited in an X-linked dominant fashion. It is caused by a CGG-trinucleotide repeat expansion within the FMR1 gene. Patients with Fragile X syndrome are at an increased risk for mitral valve prolapse and educational difficulties. They are also often diagnosed with autism. As with all trinucleotide repeat expansions, genetic anticipation is seen, wherein future generations have increased severity and/or earlier onset of disease. Incorrect Answers: A, C, D, and E. Down syndrome (Choice A) results from trisomy of chromosome 21 and presents with characteristic physical features including broad, flat, facial features with prominent epicanthal folds and a single palmar crease. Patients with Down syndrome are at increased risk for Alzheimer dementia, acute lymphoblastic and acute myeloid leukemia, cardiac septal defects, duodenal atresia, and Hirschsprung disease. Gonadal dysgenesis 45,X (Turner syndrome) (Choice C) results from monosomy of the X chromosome and presents with characteristic physical features including short stature, a wide webbed neck, and a broad chest with widely spaced nipples. Patients with Turner syndrome may also present with bicuspid aortic valve, aortic coarctation, or a fused kidney. Hurler syndrome (Choice D) is a mucopolysaccharidosis that results from deficiency of the enzyme a-L-iduronidase, with subsequent lysosomal accumulation of glycosaminoglycans (eg, heparin sulfate and dermatan sulfate). It is inherited in an autosomal recessive fashion and presents with characteristic physical features including coarse facial features with a large, elongated face, widely spaced orbits, corneal opacification, and hepatosplenomegaly. Trisomy 18 (Choice E), also known as Edwards Syndrome, presents with characteristic physical features including intellectual disability, a prominent occiput, low-set ears, micrognathia, clenched, overlapping fingers, and feet with prominent calcanei and convexly rounded soles.

Objective: Fragile X syndrome is an X-linked dominant cause of inherited intellectual disability that presents as a result of a CGG-trinucleotide repeat expansion in the FMR1 gene. It presents with clinical features including intellectual disability, an elongated face with a prominent jaw, a high-arched palate, large ears, hyperextensible joints, and postpubertal macroorchidism. Previous Next Score Report Lab Values Calculator Help Pause

143 Exam Section 3: Item 44 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 44. A 57-year-old man with a 4-year history of type 2 diabetes mellitus has increased skin pigmentation, moderate ascites, and esophagogastric varices. Tissue obtained from biopsy of the liver is shown. Which of the following substances is most likely to be markedly increased in hepatocytes? A) a-Antitrypsin B) Copper C) Glycogen D) Iron E) Melanin Correct Answer: D. Hemochromatosis may be acquired or inherited secondary to mutations in the HFE gene, leading to abnormally increased intestinal absorption of iron. This results in accumulation of iron in the body, increased serum iron, and increased ferritin. Iron can accumulate in several organs, including the liver, pancreas, skin, heart, and joints. As a result of increased free radical generation and oxidative damage, hemochromatosis in this capacity can manifest with failure of these affected organs. It typically presents after decades of iron accumulation with liver failure manifested as cirrhosis and portal hypertension, diabetes mellitus, arthritis secondary to calcium pyrophosphate deposition, cardiomyopathy with resultant symptoms of heart failure, darkening of the skin, and gonadal atrophy. Hemochromatosis, when acquired, often results in the setting of transfusion-dependent anemias such as thalassemia. Diagnostic studies may include liver biopsy, which commonly demonstrates excess iron deposition in hepatocytes on Prussian blue stain as shown in the photomicrograph. Treatment for inherited hemochromatosis involves serial phlebotomy. Incorrect Answers: A, B, C, and E. a-Antitrypsin (Choice A) deficiency presents as early panacinar emphysema and cirrhosis. It is a hereditary disease resulting from the misfolding of or a mutated nonfunctional a1-antitrypsin, an antiprotease. The misfolded protein accumulates in the liver causing hepatocyte damage and is unable to prevent protease-induced damage to the pulmonary parenchyma. Hepatolenticular degeneration (Wilson disease) results from impaired metabolism of copper (Choice B); patients often present with liver dysfunction and neuropsychological impairment (eg, parkinsonism, ataxia, tremors, dystonia, dementia, hallucinations, and/or personality changes). Glycogen (Choice C) can accumulate in glycogen storage diseases (eg, von Gierke disease). Von Gierke disease is secondary to glucose-6-phosphatase deficiency and is characterized by excess glycogen in the liver, fasting hypoglycemia, increased concentrations of triglycerides and uric acid, and hepatomegaly. Melanin (Choice E) is a dark brown pigment found primarily in cutaneous melanocytes. It absorbs ultraviolet radiation. It is not stored in hepatocytes and has no association with hemochromatosis.

Objective: Hemochromatosis presents with liver failure, diabetes mellitus, arthritis, heart failure, darkening of the skin, and gonadal atrophy secondary to excess total body iron. Diagnostic studies may include liver biopsy, which commonly demonstrates excess iron deposition in hepatocytes on Prussian blue stain. %3D Previous Next Score Report Lab Values Calculator Help Pause

162 Exam Section 4: Item 13 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment | 13. A 40-year-old woman comes to the physician for genetic counseling because of a family history of muscle weakness. She says that she has not sought treatment because her symptoms are not severe, but her children have profound muscle weakness and fatigue. Sequence analysis of mitochondrial DNA shows a mutation in 50% of the mother's mitochondrial DNA and in 100% of each of her children's mitochondrial DNA. A pedigree is shown. Which of the following genetic principles best explains the findings in this family? 50% A) Anticipation II B) Genetic drift 2 3 100% 100% 100% 100% 1 4 C) Heteroplasmy OD) Imprinting O Unaffected female Unaffected male E) Pseudodominant inheritance Affected female Affected male Correct Answer: C. 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. 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. Typically, all offspring and affected females show signs of disease, but the disease may be variably expressed as a result of heteroplasmy. Heteroplasmy describes the presence of both normal and mutated mitochondrial DNA, such as in the mother. The presence of both mutated and normal mitochondrial DNA results in variable expression of mitochondrially inherited diseases. Since the mother has 50% mutated mitochondrial DNA, her disease severity is decreased compared with her children. Incorrect Answers: A, B, D, and E. Anticipation (Choice A) describes a genetic inheritance phenomenon in which future generations are affected at a younger age or with more severe symptoms. The mother's decreased symptom severity is caused by mutation in 50% of mitochondrial DNA in contrast to her children's mutations in 100% of mitochondrial DNA. Genetic drift (Choice B) describes the phenomenon of a change in frequency of an allele in a population over generations due to chance. Small population sizes with small numbers of particular alleles are more subject to genetic drift. Imprinting (Choice D) describes the inactivation of an allele by methylation. With one allele inactive or imprinted, mutations at the other, active allele result in diseases such as Prader-Willi syndrome or Angelman syndrome. Imprinting is an epigenetic phenomenon that results in the genes inherited from one parent being expressed while those from the other parent are silenced. Pseudodominant inheritance (Choice E) refers to the inheritance of a recessive trait in a dominant pattern, such as in X-linked recessive inheritance in which males express the recessive disease whereas females need to be homozygous to express the disease.

Objective: Mitochondrial diseases are strictly inherited through the mother, as sperm do not contain mitochondria to contribute to the fertilized embryo. Heteroplasmy describes the presence of both normal and mutated mitochondrial DNA, which results in variable expression in mitochondrially inherited diseases. %3D Previous Next Score Report Lab Values Calculator Help Pause %3D

175 Exam Section 4: Item 26 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 26. An 83-year-old man from Puerto Rico is admitted to the hospital because of abdominal pain and blood in his stool for 1 month. He has had an 11-kg (25-lb) weight loss during this period. The patient speaks little English and is accompanied by his children and extended family. The physician wishes to evaluate the patient for possible colon cancer, including ordering a colonoscopy. The patient's family requests that he not be told any bad news and that they be allowed to make the decisions about his treatment. Which of the following is the most appropriate next step by the physician? A) Have the family present when the diagnosis is shared with the patient, and allow them to participate in treatment planning B) Inform the family that HIPAA requires that the patient receive this information and make his own decisions regarding his care C) Refer the matter to the hospital's legal counsel to have a power of attorney appointed to a family member D) Respect the wishes of the family and do not share the diagnosis or treatment plan with the patient E) Use a Spanish-speaking interpreter to determine how much the patient wishes to know about diagnosis and treatment Correct Answer: E. Physicians should elicit patients' preferences about how much information they prefer to receive about consequential diagnoses. When a patient's family expresses an opinion about a patient's care, including the delivery of a diagnosis, the physician should take this opinion under advisement but needs to ultimately honor the patient's autonomy. To fully understand this patient's perspective, the physician should use a professional Spanish-speaking interpreter. Professional interpreters can provide high-quality translations that maintain patient safety. After gathering information about the patient's preferences about the delivery of a diagnosis, the physician and patient can collaboratively decide who to include in treatment decisions. Incorrect Answers: A, B, C, and D. Sharing the diagnosis with the patient with family present (Choice A) presumes that the patient wants to know all of the information about his diagnosis and desires his family to be present. The first priority of this physician should be to clarify the patient's perspective and preferences regarding his care. Though the patient should make his own care decisions, HIPAA law is not related to this right to autonomy (Choice B). HIPAA law instead prevents patient's health information from being disclosed to other parties without the patient's explicit consent. According to HIPAA, the physician should ask the patient-using a Spanish interpreter-whether he wants to involve his family in treatment decisions. Respecting the family's wishes or appointing power of attorney to a family member without asking the patient (Choices C and D) would violate this patient's autonomy. Patients should be assumed to have decisional capacity, and physicians should therefore make every effort to elicit and respect the patient's wishes.

Objective: Physicians should elicit patients' preferences about their medical care, including how much information they wish to receive about consequential diagnoses. If family members want to be involved in treatment decisions, the physician should first get the patient's consent to involve the family. In patients who speak other languages, professional interpreters should be utilized to understand the patient's preferences. Previous Next Score Report Lab Values Calculator Help Pause

132 Exam Section 3: Item 33 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 33. An investigator studying the energy requirements for acid secretion in gastric mucosa cells observes that the energy needed to secrete 1.0 nmol of H* is 7 x 10-9 kcal. Given this result, how many ATP molecules must be hydrolyzed during the secretion of one H+ ion? A) 1 B) 2 C) 3 D) 6 E) 38 Correct Answer: A. Adenosine triphosphate (ATP) is used in metabolic processes for intracellular energy transfer. 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 synthesize ATP via ATP synthase. Hydrolysis of ATP into either ÁDP and one inorganic phosphate molecule or AMP and two inorganic phosphate molecules generates energy, which is then used to power other enzymatic reactions. The HK*ATPase is the enzyme used to generate gastric acid by secreting a single hydrogen ion into the gastric lumen in exchange for a single potassium ion. This reaction takes place in the parietal cells, the acid secreting cells of the stomach, which are found primarily in the gastric body. With each exchange of hydrogen and potassium, one ATP is used. Therefore, there is a one-to-one ratio of H+ ions secreted to ATP hydrolyzed. Inhibition of this H+K+ATPase is the mechanism by which proton-pump inhibitors (PPIS) exert their effect on gastric acid production. PPI irreversibly inhibit the H+K+ATPase in gastric parietal cells and thereby limit the secretion of hydrogen ions, which results in an increased gastric pH. Incorrect Answers: B, C, D, and E. Reduction of a single molecule of FADH, leads to the generation of two (Choice B) molecules of ATP. In contrast, reduction of a single molecule of NADH leads to generation of three (Choice C) molecules of ATP. Only one ATP molecule is required to secrete a single H+ into the gastric lumen, not six (Choice D). Taking into account the ATP generated both directly and indirectly (through reduction of NADH and FADH,), the theoretical maximum number of ATP generated by metabolizing a single molecule of glucose is 38 (Choice E). This is not the number of ATP required to secrete a single H+ ion into the gastric lumen.

Objective: Secretion of H+ ions into the gastric lumen is performed by the H*,K ATPase in parietal cells. For each enzymatic reaction, the energy from one molecule of ATP is utilized to secrete one molecule of H*. %3D Previous Next Score Report Lab Values Calculator Help Pause

116 Exam Section 3: Item 17 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment A B 17. A 16-year-old boy comes to the physician because of wrist pain for 1 week. X-rays of the upper extremities from 4 years ago (Figure A) and today (Figure B) are shown. The x-rays are read as normal. Which of the following hormones most likely mediated the change observed in the x-rays? A) Aldosterone B) Cortisol C) Parathyroid hormone D) Testosterone O E) Thyroxine (T4) Correct Answer: D. The physis is the cartilaginous plate that lies near the end of long bones, between the epiphysis and metaphysis, that provides axial growth to the skeleton. On the x-rays, it is the area of radiolucency at the distal radius in a child that in the adult appears as an ossified radiopaque stripe. It is divided into different zones. The zone of resting cartilage displays limited activity. The zone of proliferation is where active mitosis of chondrocytes occurs, contributing to axial growth as chondrocytes multiply. The zone of hypertrophy is where chondrocytes produce collagen and mineralization begins. In the zone of calcification, apoptosis of chondrocytes occurs and blood vessels infiltrate. The zone of ossification is where osteoblasts form osteoid and contribute to mature bone mineralization. During puberty, hormones such as growth hormone and insulin-like growth factor-1 stimulate long bone growth at the physis. Testosterone and estrogen also serve to increase physeal axial growth. Aromatization of testosterone to estrogen eventually leads to closure of the physis in later stages of puberty. Estrogens signal for the ossification of the cartilaginous physis. Such ossification progresses until the physis is completely ossified, leaving a sclerotic area of bone known as a physeal scar. When this occurs, axial growth of a long bone is no longer possible. Incorrect Answers: A, B, C, and E. Aldosterone (Choice A) is a steroid hormone that acts on mineralocorticoid receptors at the collecting duct of the nephron leading to increased sodium and water resorption, and excretion of potassium and hydrogen ions. Cortisol (Choice B) is a glucocorticoid that has multiple effects throughout the body including decreasing inflammation, increasing adrenergic tone, and increasing blood glucose concentrations. It also increases osteoclast activity and decreases bone mineral density. Parathyroid hormone (Choice C) is a peptide hormone that regulates calcium homeostasis by increasing osteoclast activity and bone resorption. It is synthesized by the parathyroid glands primarily in response to hypocalcemia. Thyroxine (T4) (Choice E) is the main hormone produced in the thyroid gland. It acts on receptors in the peripheral tissues to regulate basal metabolic rate, adrenergic tone, and bone remodeling.

Objective: Testosterone and estrogen regulate the rate of physeal growth, ossification, and closure. Closure occurs in adolescence and is marked by ossification of the radiolucent physis on x-rays. II Previous Next Score Report Lab Values Calculator Help Pause

120 Exam Section 3: Item 21 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 21. A 47-year-old man comes to the physician because of a 2-day history of yellow eyes and skin. He has undergone a total colectomy and reanastomosis at the age of 20 years because of the presence of numerous colon polyps, and he has a family history of colon cancer. Physical examination shows scleral icterus, generalized jaundice, and a nontender mass that is palpable in the right upper quadrant of the abdomen. Upper endoscopy shows four duodenal polyps; a 4-cm mass is seen arising from one of them. A CT scan of the abdomen shows extension of the duodenal mass into the head of the pancreas. Histopathologic examination of the resected mass shows an adenocarcinoma arising from a duodenal adenoma. Molecular analysis of the adenocarcinoma is most likely to show which of the following? A) Biallelic loss of function of the APC gene B) Instability of microsatellite regions of the genome C) Lack of expression of a mismatch repair gene D) Methylation of the MLH1 promoter region E) Presence of the V600E mutation of the BRAF gene Correct Answer: A. Familial adenomatous polyposis (FAP) is an autosomal dominant genetic syndrome resulting from biallelic loss of function of the APC gene. It is characterized by the presence of thousands of adenomatous polyps arising after puberty. Colonic neoplasia often begins with the formation of a polyp that undergoes malignant transformation. Adenomatous polyps, in general, typically occur as a result of chromosomal instability; most precursor mutations involve the APC gene, a tumor suppressor, which results in increased cellular proliferation and increased risk for malignancy. FAP results from inactivating mutations of the APC gene on chromosome 5. Cancer development follows the two-hit hypothesis, in which a second mutation in the other APC allele leads to malignant transformation. Patients with FAP are also at increased risk for tumor development from p53 and KRAS mutations caused by the preceding loss of function of APC. Because of its malignant potential, patients require prophylactic colectomy. In addition to colorectal cancer, FAP is associated with duodenal adenocarcinoma, hepatoblastoma, thyroid cancer, and pancreatic cancer. Obstruction of the pancreatic duct leads to cholestatic liver injury, jaundice, and scleral icterus. Incorrect Answers: B, C, D, and E. Lynch syndrome is a similar autosomal dominant disorder with a high risk for developing colorectal cancer. The underlying pathogenesis of Lynch syndrome is instability of microsatellite regions of the genome (Choice B), which results from the lack of expression of a mismatch repair gene (Choice C) secondary to methylation of the MLH1 promoter region (Choice D). Mismatch repair genes are important in correcting DNA replication errors. Dysfunctional mismatch repair proteins increase the risk for spontaneous mutations that may lead to cancer. Lynch syndrome is also associated with increased risk for endometrial, ovarian, and skin cancers. Presence of the V600E mutation of the BRAF gene (Choice E) occurs in several different cancers, notably malignant melanoma. The BRAF gene (oncogene) encodes a protein involved in the regulation of cell growth and proliferation.

Objective: The APC gene is mutated in FAP, resulting in the development of numerous adenomatous polyps that substantially increase the risk for colorectal carcinoma. The APC gene is located on chromosome 5 and encodes a tumor suppressor protein. FAP is also associated with duodenal adenocarcinoma, hepatoblastoma, thyroid cancer, and pancreatic cancer. Previous Next Score Report Lab Values Calculator Help Pause

50 Exam Section 2: Item 1 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 1. A 52-year-old man comes to the physician because of inability to achieve an erection 6 weeks after undergoing operative removal of a tumor on the distal portion of the rectum. During the excision, the tumor was found to be extensively attached to the posterior aspect of the prostate. Damage to which of the following nerves is the most likely cause of this patient's symptoms? A) Gray rami from the sympathetic chain B) Pelvic splanchnics C) Pudendal D) Superior hypogastric E) Ventral rami of S2-4 Correct Answer: B. Damage to the pelvic splanchnic nerves is the most likely cause of this patient's erectile dysfunction. The pelvic splanchnic nerves are efferent parasympathetic preganglionic nerves that arise from the intermediolateral column of spinal cord levels S2 to S4. These nerves innervate the smooth muscle of the distal anterior rectum, located immediately posterior to the prostate, explaining this patient's pelvic splanchnic nerve damage. The pelvic splanchnic nerves then synapse with the postganglionic parasympathetic cavernous nerve in the inferior hypogastric plexus. The cavernous nerve travels to the corpora cavernosa of the penis and releases acetylcholine to promote vasodilation of the helicine arteries, resulting in penile erection. Damage to the pelvic splanchnic nerves results in impaired parasympathetic innervation to the erectile tissue of the penis, which manifests as erectile dysfunction. Incorrect Answers: A, C, D, and E. The gray rami from the sympathetic chain (Choice A) are located immediately dorsal to the paravertebral sympathetic chain. The gray rami are located too posteriorly to be affected by this patient's tumor excision. As well, sympathetic innervation plays a role in detumescence. The pudendal nerve (Choice C) carries sensory fibers that originate from the penile skin, glans, and urethra as well as motor neurons innervating the ischiocavernosus and bulbocavernosus muscles, leading to the contraction of these muscles and thereby mediating the sensorimotor aspect of erections. Though damage to the pudendal nerve would lead to erectile dysfunction, the pudendal nerve is located near the posterior aspect of the rectum and is therefore less likely to be damaged by surgical manipulation of the periprostatic region. The superior hypogastric nerves (Choice D) are located just anterior to the bifurcation of the abdominal aorta, which is significantly superior to the distal rectum. These nerves provide sympathetic control of the emission of semen into the posterior urethra during ejaculation but do not mediate erection. The ventral rami of S2-4 (Choice E) contain somatic sensory and motor neurons of the S2-4 spinal nerves, innervating the genital area. The ventral rami from each sacral spinal cord level travel inferomedially to unite in the sacral plexus, which is located along the posterolateral surface of the pelvic cavity. Nerves at this location would not be affected by periprostatic procedures.

Objective: The pelvic splanchnic nerves are efferent parasympathetic preganglionic nerves that synapse with the cavernous nerves. Activation of the cavernous nerves leads to vasodilation in the corpora cavernosa and, consequently, penile erection. The pelvic splanchnic nerves are vulnerable to damage during periprostatic surgeries, which results in erectile dysfunction. Previous Next Score Report Lab Values Calculator Help Pause

193 Exam Section 4: Item 44 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 44. A 24-year-old woman is brought to the emergency department 30 minutes after sustaining abdominal trauma in a motor vehicle collision. She is hypotensive. Rupture of the spleen is suspected. Several attempts to insert a catheter in the left subclavian vein for a blood transfusion are unsuccessful, and she develops worsening dyspnea. Breath sounds are decreased in the left lung. A chest x-ray is shown. Which of the following structures is most likely indicated by the arrows? A) Parietal pleura B) Phrenic nerve C) Pulmonary artery D) Thoracic duct E) Visceral pleura Correct Answer: E. 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 caustic medications. Complications related to central catheter insertion often depend on the site of insertion. The subclavian vein lies adjacent to the subclavian artery and the apex of the lung. The lung is covered by two layers: the outer parietal pleural and the inner visceral pleura. latrogenic puncture of the pleura by a needle or catheter from attempted subclavian catheter insertion can introduce air into the pleural space between the visceral and parietal pleura either from the external body wall or from injury to the visceral pleura resulting in air escape with each breath. Air in the intrapleural space, a pneumothorax, separates the visceral and parietal pleura, decreases the compliance of the lung, and prevents complete expansion, leading to eventual collapse and respiratory distress. Incorrect Answers: A, B, C, and D. The parietal pleura (Choice A) is separated from the visceral pleura in the setting of a pneumothorax, but the parietal pleura is the outer membrane and is attached to the inner surface of the thoracic cavity. Injury to the phrenic nerve (Choice B) can cause respiratory distress because of diaphragmatic paralysis or dysfunction, but this is a rare complication of central venous catheter insertion. It lies next to the heart in the mediastinum, and an x-ray may show elevation of the affected unilateral hemidiaphragm. The pulmonary arteries (Choice C) arise from the right ventricular outflow tract in the mediastinum. They are distant structures that would be unlikely to be injured during central catheter placement unless a right-heart or pulmonary artery catheterization was attempted. The thoracic duct (Choice D) is the largest lymphatic vessel in the body, originating in the abdomen at a confluence of lymphatics called the cisterna chyli. It ascends adjacent to the thoracic aorta and azygos vein and drains into the venous system near the junction of the left subclavian and left internal jugular veins. The close proximity to the vein is a risk factor for direct injury from a central venous cannulation attempt. Damage to or blockage of the thoracic duct or its major lymphatic tributaries is a common cause of chylothorax, which is accumulation of chyle (milky fluid consisting of fat droplets and lymph) in the pleural space.

Objective: The subclavian vein lies adjacent to the subclavian artery and the apex of the lung. The lung is covered by two layers: the outer parietal pleura and the inner visceral pleura. latrogenic puncture of the parietal pleura by a needle or catheter from attempted subclavian catheter insertion can introduce air into the pleural space between the visceral and parietal pleura, leading to a pneumothorax. Previous Next Score Report Lab Values Calculator Help Pause

185 Exam Section 4: Item 36 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 36. A 12-year-old boy is brought to the physician by his mother because of swelling, itching, and redness of his back that began 10 minutes after he was lying in the grass shirtless. Physical examination shows linear edematous eruptions on the back. A secondary mediator of this patient's hypersensitivity reaction is produced by mast cells de novo in response to the allergen. After release of histamine, mast cells will begin de novo synthesis of which of the following substances? A) C-reactive protein B) C3b component of complement C) Interferon alpha D) Interleukin-1 (IL-1) E) Prostaglandin Correct Answer: E. The patient is displaying signs and symptoms of a type I hypersensitivity reaction to an environmental allergen. The type I hypersensitivity reaction is composed of two phases, the immediate and the late phase. In the immediate phase, preformed IgE on the surface of mast cells and basophils is crosslinked by an antigen leading to their immediate degranulation and release of histamine, which results in vasodilation, vascular pooling, and increased vascular permeability. The late phase typically occurs hours after the initial insult and is mediated by the release of prostaglandins and leukotrienes synthesized by the mast cells. Prostaglandins are derived from arachidonic acid via cyclooxygenase (COX) enzymes. Once released, they modulate cell behavior via interaction with G protein-coupled receptors and activation of intracellular signaling pathways. Type I hypersensitivity reactions may present with symptoms ranging from localized pruritis, edema, and erythema to systemic, life-threatening anaphylaxis. Other examples include allergic asthma, atopic rhinitis, and atopic dermatitis. Skin testing can be performed to identify specific allergens. Incorrect Answers: A, B, C, and D. C-reactive protein (Choice A), also referred to as CRP, is an acute phase reactant synthesized by the liver that increases in response to inflammation. Expression of CRP is induced by IL-6. CRP is not synthesized by mast cells. C3b component of complement (Choice B) is a core component of the complement cascade and is formed by the cleavage of C3. The C3b fragment may then participate in one of three of pathways. It may bind with factor B to form C3 convertase and amplify C3 cleavage in a positive feedback loop. It may bind with C4b and C2b fragments to form C5 convertase, or it may opsonize the microbe by binding directly to the pathogen surface. Complement is not synthesized by mast cells. Interferon alpha (Choice C) is a glycoprotein produced by cells that are infected with a virus. Interferon alpha upregulates MHC expression for recognition and binding with CD8+ T cells, and downregulates protein synthesis to impair viral replication. Interleukin-1 (IL-1) (Choice D) is a proinflammatory cytokine secreted by macrophages, endothelial cells, dendritic cells, and B lymphocytes. It has numerous effects, including the promotion of fever, vasodilation, and stimulation of endothelial cells to express adhesion proteins for leukocyte recruitment. It is also known as osteoclast-activating factor.

Objective: Type I hypersensitivity reactions are characterized by an initial histamine release from presensitized mast cells, followed by the synthesis and release of prostaglandins, leukotrienes, and inflammatory mediators. Previous Next Score Report Lab Values Calculator Help Pause

102 Exam Section 3: Item 3 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 3. A 13-year-old girl is brought to the physician by her mother because of a 5-month history of behavioral problems. The mother states that her daughter alternates from being sad and socially withdrawn to being extremely angry and aggressive. After further evaluation, a diagnosis of bipolar disorder is made. Treatment with valproic acid, which inhibits histone deacetylase, is started. This drug is most likely to affect which of the following processes in this patient? A) MRNA splicing B) Polyadenylation C) Post-translational processing D) Transcription E) Translation Correct Answer: D. Valproate, a mood stabilizer used in bipolar disorder, increases DNA transcription by inhibiting histone deacetylase. Histones are basic proteins with many positively charged amino acid residues, which attract negatively charged phosphate groups on DNA, condensing and packing the DNA into complexes called chromatin. As welÍ, histones regulate the transcription of DNA through the epigenetic mechanisms of methylation and acetylation. Acetylation of histones decreases the affinity between DNA and histones, making it accessible for transcription. Deacetylation conversely decreases transcription. By inhibiting deacetylation, valproate increases the transcription of certain genes including the transcription factor activating protein-1 (AP-1), ultimately increasing the expression of brain-derived neurotrophic factor (BDNF) and other neuroprotective factors. Valproate is therefore postulated to exert long-lasting effects on brain plasticity. Incorrect Answers: A, B, C, and E. MRNA splicing (Choice A) describes the process by which introns are removed from the nascent mRNA strand, promoting a continuous sequence of exons only. This process involves the spliceosome complex rather than histones. Polyadenylation (Choice B) of the 3' end of MRNA, created by poly-A polymerase, assists in protecting the mRNA as it courses from the nucleus into the cytoplasm. There are no current medications that target the polyadenylation process. Post-translational processing (Choice C) encompasses many reversible or irreversible modifications made after MRNA is translated to protein, which diversify the structure and function of proteins. Phosphorylation, which regulates enzyme activity, is one common example. Translation (Choice E) is the process by which a polypeptide is synthesized from an MRNA sequence and occurs via ribosomes in the cytosol or rough endoplasmic reticulum. Ribosomes are often the target of antibiotics rather than valproate.

Objective: Valproate, a mood stabilizer used in bipolar disorder, increases transcription by inhibiting histone deacetylase. Acetylation is an epigenetic change that increases DNA accessibility for transcription. Previous Next Score Report Lab Values Calculator Help Pause

41 Exam Section 1: Item 42 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 42. A 1-year-old girl is brought to the physician by her parents because of developmental delay. Physical examination shows a coarse facial appearance, clouded corneas in both eyes, hepatosplenomegaly, stiff joints, and developmental delay. Urine studies show an increased concentration of dermatan sulfate and heparan sulfate. Which of the following enzymes is most likely impaired in this patient? A) a-L-Fucosidase B) B-Galactosidase C) Glucose 6-phosphatase D) B-Hexosaminidase A E) a-lduronidase F) Neuraminidase G) Steroid sulfatase Correct Answer: E. Mucopolysaccharidoses are caused by an enzyme deficiency that prevents the degradation of glycosaminoglycans. The buildup of atypical substrates leads to varying levels of cellular dysfunction on the basis of the enzyme affected. The diagnosis is not typically clinically apparent at birth, as the abnormal deposition of metabolites has not had time to occur. The two primary mucopolysaccharidoses are Hurler syndrome and Hunter syndrome. Hurler syndrome and Hunter syndrome are caused by deficiencies in a-iduronidase and iduronate-2-sulfatase, respectively, with both resulting in the deposition of heparan sulfate and dermatan sulfate. Hurler Syndrome is an autosomal recessive disorder and presents in the first year of life with coarse facial features, corneal clouding, hepatosplenomegaly, joint contractures, intellectual disability, and developmental delay, as seen in this patient, as well as skeletal abnormalities, heart failure, hydrocephalus, hearing loss, and recurrent respiratory tract infections. Hunter syndrome is an X-linked recessive disorder and presents with milder features of Hurler syndrome with the absence of corneal clouding, as well as aggressive behavior and pearly papular skin lesions. Diagnosis is made with urinary glycosaminoglycans and an analysis of enzyme activity; treatment is largely supportive. Average lifespan is five years for Hurler syndrome and 13 to 20 years for Hunter syndrome. Incorrect Answers: A, B, C, D, F, and G. a-l-Fucosidase (Choice A) deficiency leads to a rare glycoproteinosis called fucosidosis secondary to the accumulation of glycoproteins, glycolipids, and oligosaccharides. Fucosidosis is an autosomal recessive disorder that presents with coarse facial features, hepatosplenomegaly, intellectual disability, seizures, skeletal abnormalities, alterations in perspiration, and neurologic decline. It does not cause corneal clouding. B-Galactosidase (Choice B) deficiency leads to mucopolysaccharidosis IV type B. It exhibits a milder presentation than Hunter or Hurler syndrome, with skeletal abnormalities and clouded corneas, as well as heart disease. It does not typically cause intellectual disability. Glucose 6-phosphatase (Choice C) deficiency, or Von Gierke disease (glycogen storage disease, type I) prevents the liver from releasing glucose into the serum from glycogenolysis and gluconeogenesis. Patients present with hepatomegaly, increased glycogen in the liver, and severe fasting hypoglycemia. It does not cause corneal clouding. B-Hexosaminidase A (Choice D) deficiency leads to a lysosomal storage disorder (sphingolipidosis) that is clinically indistinguishable from Tay-Sachs disease, and presents with neurological decline, a cherry-red spot on the macula, blindness, deafness, and developmental delay. It does not cause corneal clouding. Neuraminidase (Choice F) deficiency causes sialidosis (mucolipidosis I), which presents with myoclonus, gait abnormalities, cherry-red macules, coarse facial features, skeletal malformations, and mild intellectual disability. It does not present with corneal clouding. Steroid sulfatase (Choice G) deficiency leads to X-linked ichthyosis, or hyperkeratosis of the skin with scaling. It can occasionally present with corneal opacification, attention deficit hyperactivity disorder, and cryptorchidism. It does not present with developmental delay or hepatosplenomegaly.

Objective: Hurler syndrome is an autosomal recessive mucopolysaccharidosis caused by a deficiency in a-iduronidase, leading to the deposition of heparan and dermatan sulfates. It presents in the first year of life with coarse facial features, corneal clouding, hepatosplenomegaly, joint contractures, intellectual disability, and developmental delay. %3D Previous Next Score Report Lab Values Calculator Help Pause

27 Exam Section 1: Item 27 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 27. A 56-year-old woman with type 2 diabetes mellitus is being treated with pioglitazone to improve her sensitivity to insulin and with gemfibrozil to decrease her serum triglyceride concentration. These drugs both initiate their therapeutic effects by binding to which of the following receptor groups? A) B-Adrenergic receptors B) Farnesoid X receptors C) Insulin-like growth factor receptors D) Peroxisome proliferator-activated receptors E) Toll-like receptors Correct Answer: D. Type 2 diabetes mellitus is a common disorder in resource-rich nations. Initial treatment consists of diet and activity modification, weight loss, and metformin monotherapy. If these measures fail to achieve adequate blood glucose control, additional oral antihyperglycemics may be considered. Common options for second-line pharmacotherapy include sodium-glucose co-transporter 2 (SGLT2) inhibitors, glucagon-like peptide-1 (GLP-1) analogs, dipeptidyl peptidase-4 (DPP-4) inhibitors, thiazolidinediones, and sulfonylureas. Pioglitazone (a thiazolidinedione) achieves its effect by binding to peroxisome proliferator-activated receptors (PPARS), which are expressed in most cell types including adipose cells, pancreatic cells, myocytes, and hepatocytes. PPARS are nuclear receptors that act as transcription factors to regulate metabolism. Activation results in increased cell sensitivity to insulin and adiponectin. Insulin has multiple effects, including increased glycogen, triglyceride, and protein synthesis and storage, as well as glucose uptake by adipose tissue and myocytes. Adiponectin is secreted by adipocytes and helps regulate fatty acid and glucose metabolism. Gemfibrozil is a fibrate lipid-lowering agent that also activates PPARS and upregulates lipoprotein lipase, resulting in decreased serum LDL, increased HDL, and decreased triglyceride concentrations. Incorrect Answers: A, B, C, and E. B-Adrenergic receptors (Choice A) are ubiquitous in the body. B-Adrenergic receptor activation in adipocytes results in increased lipolysis, in pancreatic cells results in increased insulin release, and in hepatocytes stimulates gluconeogenesis and glycogenolysis. Farnesoid X receptors (Choice B) are nuclear receptors expressed in hepatocytes and intestinal cells. Activation results in inhibition of cholesterol 7-a-hydroxylase which is responsible for the synthesis of bile acids. Thiazolidinediones and fibrates do not have activity at these receptors. Insulin-like growth factor receptors (Choice C) are stimulated by insulin-like growth factor (IGF-1). IGF-1 is produced by the liver in response to growth hormone from the anterior pituitary. Binding to IGF receptors promotes cell growth. IGF-1 is structurally similar to insulin but is not a target for antihyperglycemics or lipid-lowering drugs. Toll-like receptors (Choice E) are involved in neutrophil activation as part of the innate immune response to infection. Binding of pathogen-associated molecular patterns to toll-like receptors on the neutrophil surface leads to NF-KB activation and production of proinflammatory mediators.

Objective: Peroxisome proliferator-activated receptors are transcription factors that regulate glucose and lipid metabolism. They are the site of action of thiazolidinedione oral antihyperglycemics as well as lipid-lowering fibrates. Previous Next Score Report Lab Values Calculator Help Pause

35 Exam Section 1: Item 36 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 36. A 46-year-old man comes to the office because of a 1-year history of difficulty reading small print. He has no other history of vision difficulty. He has hypertension treated with hydrochlorothiazide. Pulse is 72/min, and blood pressure is 142/86 mm Hg; other vital signs are within normal limits. The pupils are 2 mm, equally round, and reactive to light and accommodation. Ocular movements are intact. On wall chart testing, visual acuity is 20/20 bilaterally. On handheld chart testing, he can read the 20/20 line only if he holds the card more than 14 inches away. Disc margins are sharp; there are no hemorrhages, exudates, or arteriolar narrowing. No other abnormalities are noted. Which of the following is the most likely diagnosis? O A) Hyperopia B) Hypertensive retinopathy C) Муорia D) Nuclear sclerosis E) Normal age-related changes Correct Answer: E. The Helmholtz theory of accommodation holds that contraction of the ciliary muscle relaxes tension of the zonular fibers upon the lens, allowing the lens to become rounder, thereby increasing its dioptric power and allowing for focusing on objects at close distances. The accommodative ability of the lens is dependent upon its ability to distend its shape. Age-related hardening of the lens and loss of accommodative ability is known as presbyopia and is a normal age-related change. Patients present with blurry vision at near-distance and notice that objects must be held at further working distances to be seen clearly. Normal pupillary reflexes to accommodation are not affected by presbyopia. Treatment is with reading glasses with positive dioptric power. Incorrect Answers: A, B, C, and D. Hyperopia (Choice A) occurs when the focal point of the eye is found at a point behind the retina and may result from short axial length or diminished converging power of the cornea and lens. Patients present with blurry vision both at a distance and when near. The condition usually presents in childhood. Treatment is with corrective lenses with a positive dioptric power. Hypertensive retinopathy (Choice B) is often asymptomatic but may cause blurry vision. Fundus examination discloses optic disc edema, retinal nerve fiber layer infarcts, retinal edema, and retinal arteriolar attenuation. While this patient has hypertension, the absence of characteristic fundus findings makes hypertensive retinopathy an unlikely diagnosis. Myopia (Choice C) occurs when the focal point of the eye is found at a point anterior to the retina and may result from long axial length or increased converging power of the cornea and lens. Patients present with blurry vision at distance. The condition usually presents in childhood. Pathologic forms of myopia may be progressive and put the patient at increased risk for retinal tears, retinal detachment, and glaucoma. Treatment is with corrective lenses with a negative dioptric power. Nuclear sclerosis (Choice D) refers to age-related opacification of the lens, also known as cataracts. Lens fibers are generated in the superficial equatorial regions of the lens and migrate into the center of the lens nucleus as they age, leading to age-related increases in lens thickness and opacification. Patients with cataracts experience slowly progressive blurry vision both at a distance and when near, as well as glare, loss of contrast sensitivity, and difficulty seeing at night. Treatment is with surgical removal of the lens and implantation of a synthetic lens.

Objective: Presbyopia is a normal age-related change that involves loss of the accommodative power of the crystalline lens. Patients experience loss of accommodation and blurry vision at near-distance, and notice that objects must be held at further working distances to be seen clearly. Treatment is with reading glasses with a positive dioptric power. Previous Next Score Report Lab Values Calculator Help Pause

44 Exam Section 1: Item 45 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 45. A 30-year-old man is brought to the physician by his wife because of a 1-day history of irrational behavior and severe abdominal pain. He began treatment with trimethoprim- sulfamethoxazole for a urinary tract infection 3 days ago. His urine turned a light burgundy color during a similar episode 1 year ago. His mother and his maternal grandfather have had similar symptoms. He appears anxious and restless. His pulse is 96/min. Physical examination shows diaphoresis. Serum studies show increased concentrations of 5-aminolevulinic acid (0-ALA) and porphobilinogen. The physician suspects that a mutant allele is causing decreased activity of an enzyme involved in heme biosynthesis. This enzyme is most likely which of the following? A) õ-ALA dehydratase B) 0-ALA synthase C) Ferrochelatase D) Porphobilinogen deaminase O E) Uroporphyrinogen decarboxylase Correct Answer: D. This patient's presenting signs and symptoms, including psychosis, abdominal pain, and burgundy-colored urine following medication exposure is consistent with a diagnosis of acute intermittent porphyria (AIP). AIP is an autosomal dominant disorder caused by inactivating mutations of the enzyme porphobilinogen deaminase, resulting in deficient synthesis of heme and accumulation of porphobilinogen. Porphobilinogen has a neurotoxic effect, leading to symptoms of abdominal pain, psychosis, and peripheral polyneuropathy. In contrast to other porphyria disorders, cutaneous photosensitivity and blistering is not observed in patients with AlP. Acute exacerbations of AIP are commonly precipitated by metabolic stressors such as starvation, alcohol ingestion, or ingestion of medications. Commonly offending medications include sulfonamides (including trimethoprim-sulfamethoxazole), barbiturates, oral contraceptives, and rifampin. Treatment includes discontinuation of the offending agent as well as intravenous infusion of glucose and heme, which inhibit the upstream enzyme ALA synthase, thereby inhibiting synthesis of porphobilinogen. Incorrect Answers: A, B, C, and E. Ö-ALA dehydratase (Choice A) catalyzes the conversion of ō-aminolevulinic acid to porphobilinogen. Defects in this enzyme cause a rare form of porphyria that may resemble AIP but which would not display increased serum concentrations of porphobilinogen. Ō-ALA synthase (Choice B) catalyzes the conversion of succinyl-CoA and glycine to õ-aminolevulinic acid. Defects in this enzyme result in X-linked sideroblastic anemia. Ferrochelatase (Choice C) catalyzes insertion of ferrous iron into protoporphyrin to form heme. Ferrochelatase is inhibited by lead poisoning. Defects in ferrochelatase can cause erythropoietic protoporphyria. However, photosensitivity is usually observed in this condition. Uroporphyrinogen decarboxylase (Choice E) catalyzes the conversion of uroporphyrinogen IIl to coproporphyrinogen II. Defects in this enzyme cause porphyria cutanea tarda. Porphyria cutanea tarda is the most common of the porphyrias but is characterized by cutaneous photosensitivity and blistering.

Objective: Acute intermittent porphyria is an autosomal dominant disorder caused by inactivating mutations of the enzyme porphobilinogen deaminase, resulting in deficient synthesis of heme and accumulation of porphobilinogen. Patients typically present after an acute metabolic stress with psychosis, abdominal pain, and burgundy-colored urine. In contrast to other porphyria disorders, cutaneous photosensitivity and blistering is not a feature of acute intermittent porphyria. Previous Next Score Report Lab Values Calculator Help Pause

73 Exam Section 2: Item 24 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 24. Which of the following abnormalities is most likely to result from occlusion of the vessel indicated by the arrow in the vertebral angiogram shown? A) Contralateral facial weakness B) Contralateral visual field deficit C) Ipsilateral anesthesia of the face D) Ipsilateral ataxia of extremity movements E) Ipsilateral palsy of lateral gaze Correct Answer: B. The vessel identified in the vertebral angiogram represents the posterior cerebral artery (PCA), the occlusion of which would result in infarction of the ipsilateral occipital lobe and a contralateral homonymous hemianopsia with macular sparing. The PCA originates near the rostral end of the basilar artery and contributes to the circle of Willis. The PCA forms anastomoses with the internal carotid arteries via the posterior communicating arteries. The PCA supplies much of the occipital lobes as well as the inferomedial temporal lobe, the thalamus, and parts of the brainstem. Visual symptoms are a prominent feature of a PCA territory infarction and include cortical blindness, contralateral homonymous hemianopsia with macular sparing, visual agnosia, and prosopagnosia. Incorrect Answers: A, C, D, and E. Contralateral facial weakness (Choice A) and ipsilateral anesthesia of the face (Choice C) are features of infarctions involving the primary motor cortex and primary somatosensory cortex, respectively, which are supplied by the middle cerebral artery. Ipsilateral ataxia of extremity movements (Choice D) can result from infarctions of either the anterior inferior cerebellar artery or the posterior inferior cerebellar artery. In the latter case, ipsilateral cerebellar findings, such as ataxia, form a core feature of lateral medullary syndrome, also known as Wallenberg syndrome. Ipsilateral palsy of lateral gaze (Choice E) is a feature of palsy of the abducens nerve, also known as the sixth cranial nerve. While sixth cranial nerve palsies may originate as a result of nuclear or fascicular ischemic injury in the brainstem, peripheral nerve injury is more common. Most of the blood supply to the abducens nucleus is derived from branches of the basilar artery.

Objective: The posterior cerebral artery originates from the basilar artery and supplies much of the occipital lobe. Infarctions of the posterior cerebral artery territory commonly lead to a contralateral homonymous hemianopsia with macular sparing. Previous Next Score Report Lab Values Calculator Help Pause

59 Exam Section 2: Item 10 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 10. A 35-year-old woman comes to the physician because of fever and left flank pain for 2 days. She has had several similar episodes during the past 4 years, all of which resolved with antibiotic therapy. Her temperature is 38.6°C (101.5°F). Physical examination shows exquisite left costovertebral angle tenderness. Urinalysis shows small numbers of WBCS and WBC casts, and numerous triple phosphate crystals. A course of antibiotic therapy does not resolve the pain and the patient undergoes resection of the left kidney. A photograph of the patient's resected kidney is shown. Which of the following was the most likely causal organism in this patient's disease process? A) Enterobacter cloacae B) Enterococcus faecalis C) Escherichia coli D) Klebsiella pneumoniae E) Proteus mirabilis F) Pseudomonas aeruginosa 1 ст Correct Answer: E. Proteus mirabilis is the most likely causal organism. The gross photograph shows a large staghorn calculus occupying the near-entirety of the renal pelvis and branching into multiple calyces. Staghorn calculi are made of struvite, also known as triple phosphate or magnesium ammonium phosphate. Bacteria such as P. mirabilis or K. pneumoniae are urease-producing organisms. Urease catalyzes the conversion of urinary urea to ammonia and carbon dioxide, which results in urine alkalinization. The solubility of struvite is decreased under alkaline conditions and precipitates as a crystal. An alkaline urine is present in this patient with a urinary pH of 7.5. Struvite crystals have an orthorhombic configuration under light microscopy. Struvite calculi are radiopaque and can be potentially identified by an x-ray or CT scan. Struvite calculi may adopt a branching morphology that molds itself to the shape of the collecting system (staghorn calculi). Given their large size and often ramified structure, it is rare for struvite calculi to pass through the urinary tract, and surgical removal is often necessary along with treatment of the underlying infection. The presence of calculi in the collecting system predisposes to recurrent episodes of pyelonephritis. Treatment is usually with a third-generation cephalosporin, although indole-positive P. mirabilis may have inducible resistance to B-lactams mediated by the AmpC gene. Incorrect Answers: A, B, C, D, and F. Enterobacter cloacae (Choice A) is an AmpC inducible gram-negative rod and intrinsically resistant to B-lactam antibiotics. It can cause urinary tract infections, pneumonia, bacteremia, and rarely meningitis. Treatment is based on resistance detected by standard microbiologic techniques. E. cloacae is not known to predispose to struvite stones although it can produce urease, albeit at a much lower rate than P. mirabilis. Enterococcus faecalis (Choice B) is a gram-positive bacterium that causes urinary tract infections, endocarditis, bacteremia, and biliary tract infections such as cholangitis and cholecystitis. It is often susceptible to ampicillin but occasionally vancomycin is required. It does not produce urease and does not predispose to the formation of struvite stones. Escherichia coli (Choice C) is a gram-negative rod and is commonly implicated in urinary tract infections. It also causes a wide variety of other infections including infectious diarrhea, bacteremia, hospital-acquired or ventilator-acquired pneumonia, and infections of the biliary tree. It does not produce urease. Klebsiella pneumoniae (Choice D) is a gram-negative rod that causes many hospital-acquired infections such as hospital-acquired pneumonia. Like P. mirabilis, it can produce urease, but at a much lower concentration, and does not predispose to the formation of struvite stones as much as P. mirabilis. Pseudomonas aeruginosa (Choice F) is a gram-negative rod that is resistant to many varieties of antibiotics. It is commonly cultured in diabetic foot infections, the lungs of patients with cystic fibrosis, and in the urine. It causes a wide variety of infections of the skin, bone, lungs, urinary tract, gastrointestinal system, and occasionally the nervous system. It does not produce urease.

Objective: Patients with urinary tract infections secondary to Proteus mirabilis, including pyelonephritis, are at risk for developing struvite stones as P. mirabilis produces urease, an enzyme that breaks down urea into ammonia and carbon dioxide. At higher urinary pH, struvite (triple phosphate crystals) precipitate and form large, branching, staghorn calculi. Surgical removal is often necessary. Previous Next Score Report Lab Values Calculator Help Pause

89 Exam Section 2: Item 40 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 40. A 70-year-old woman comes to the physician for advice after her husband of 45 years suddenly dies of cardiac arrest. She wants to donate $1,000,000 to a small city's department of public health to decrease the number of sudden cardiac deaths. She asks the physician to recommend the most effective way to put this money to use. She says, "I want to see the results of my donation over the next 1 to 2 years." The patient is most likely to accomplish her goal if she contributes to which of the following? A) Creation of an antismoking campaign B) Creation of a citywide aerobic exercise program C) Fortification of the water supply with aspirin D) Placement of external automatic defibrillators in public spaces throughout the city E) Provision of free HMG-COA reductase inhibitor drugs to everyone in town over the age of 50 years F) Provision of free relaxation and stress-management programs for all adults Correct Answer: D. Cardiac arrests in public places (nonhealthcare settings) affect thousands of persons in the United States every year. The effectiveness of defibrillation is greatest at the earliest possible time after cardiac arrest. Easy to use, battery-powered automatic external defibrillators (AEDS) placed in public areas have been shown to be highly effective in preventing sudden cardiac death from ventricular fibrillation or ventricular tachycardia. Thus, if the individual wishes to support an effective public health intervention to decrease the incidence of sudden cardiac death over the next 1 to 2 years, placement of AEDS in public places throughout the city should be considered a viable option. This treatment is on-target with respect to the proximate cause of sudden cardiac death. Incorrect Answers: A, B, C, E, and F. Creation of an antismoking campaign (Choice A) may be an effective public health intervention; however, the effects of smoking occur over a long period of time and the impact of such an intervention would likely not be measurable for many years. Further, there are many factors that influence smoking such as advertising campaigns promoting smoking and the addictive properties of nicotine. Smoking is an indirect cause of sudden cardiac death, leading to it through increasing the risk for atherosclerotic cardiovascular disease and structural and ischemic heart disease. Creation of a citywide aerobic exercise program (Choice B) would provide numerous health benefits to the community. Exercise has been shown to decrease the risk for cardiac disease, obesity, and depression. However, the effects of this intervention would be difficult to measure in the short term. Fortification of the water supply with aspirin (Choice C) is inappropriate, as aspirin has been shown to have benefits in patients at high risk for acute coronary syndrome, not in the general population. Additionally, many persons in whom salicylates are contraindicated, including children, would be exposed. Provision of free HMG-COA reductase inhibitor drugs to everyone in town over the age of 50 years (Choice E) is not an optimal intervention, as these drugs lower cardiac risk for the subset of patients with high concentrations of total cholesterol, low-density lipoprotein, and increased risk for atherosclerotic cardiovascular disease, not necessarily all persons over age 50. Provision of free relaxation and stress-management programs for all adults (Choice F) may be beneficial, but the effects would be difficult to measure.

Objective: AED placement in public areas such as shopping centers, airports, and train stations is an effective intervention that decreases sudden cardiac death as a result of the decreased time from event to initiation of treatment. %3D Previous Next Score Report Lab Values Calculator Help Pause

5 Exam Section 1: Item 5 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 5. A 26-year-old woman is brought to the emergency department by her husband after she ingested an overdose of narcotics. She is arousable only with deep painful stimuli. Her pulse is 110/min, respirations are 10/min, and blood pressure is 110/70 mm Hg. Cerebral blood flow is increased. Which of the following is the most likely cause of the increase in cerebral blood flow? A) Decreased blood pressure B) Decreased hydrogen ion concentration in the blood C) Decreased oxygen concentration in the blood D) Increased carbon dioxide concentration in the blood E) Increased heart rate Correct Answer: D. The intracranial compartment is fixed, surrounded by a rigid skull containing brain parenchyma, cerebrospinal fluid, and intravascular blood. Serum carbon dioxide concentration directly increases or decreases cerebral blood flow (CBF). Hypocapnia induces cerebral vasoconstriction, whereas hypercapnia induces cerebral vasodilation, increasing the amount of CBF and the volume of intracranial blood. In a patient with decreased respirations secondary to narcotic-induced respiratory depression, decreased air exchange and ventilation will lead to an increased amount of carbon dioxide in the blood. The resulting hypercapnia leads to vasodilation and increased CBF. Incorrect Answers: A, B, C, and E. Decreased blood pressure (Choice A) would not increase CBF. Mean arterial pressure (MAP) can have an effect on CBF, but as a result of cerebral autoregulation, CBF is maintained at a constant state over a wide range of MAP. Small changes or decreases in blood pressure have no effect on CBF; only blood pressure below or above the autoregulatory range will affect CBF. Decreased hydrogen ion concentration in the blood (Choice B) is the opposite of what would occur with opioid-induced respiratory depression, which leads to hypercapnia and increased hydrogen ion concentration. Decreased oxygen concentration in the blood (Choice C) results from hypoventilation, but CBF is more directly affected by concentrations of carbon dioxide. Increased heart rate (Choice E), just like blood pressure and MAP, has little effect on CBF provided that MAP remains within the autoregulatory range. Extreme bradycardia or tachycardia leading to severe hypotension could decrease CBF.

Objective: As a result of cerebral autoregulation, changes in MAP have minimal impact on CBF. Serum carbon dioxide concentration elicits a reversible effect on CBF. Hypocapnia induces cerebral vasoconstriction, whereas hypercapnia induces cerebral vasodilation, increasing the amount of CBF. I3D Previous Next Score Report Lab Values Calculator Help Pause

92 Exam Section 2: Item 43 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 43. A 64-year-old woman comes to the physician because of a 6-month history of heartburn and difficulty swallowing solid food and liquid. She has not had nausea or vomiting. She also has a 2-year history of pain and swelling of her wrists and hands. She has not seen a physician in 5 years. She takes no medications. She does not smoke. Physical examination shows no wrinkles on the face and neck. Her vital signs are within normal limits. There is synovial thickening of the wrists bilaterally. A photograph of the left hand is shown. Abdominal examination shows no abnormalities. Which of the following sets of findings is most likely on esophageal manometry? Esophageal Peristalsis Lower Esophageal Sphincter Tone A) Increased increased B) Increased decreased C) Normal increased D) Normal decreased E) Decreased increased F) Decreased decreased Correct Answer: F. Localized scleroderma, also called CREST syndrome, is characterized by calcinosis cutis, Raynaud phenomenon, esophageal dysmotility, sclerosis, and telangiectasias. Esophageal dysmotility, evidenced by heartburn and dysphagia, along with tightening of the skin on the face, around the mouth, and on the fingers is suggestive of this diagnosis. The esophageal dysmotility is caused by an infiltration of the esophageal muscular layer with fibrous tissue, which substantially decreases normal peristaltic activity. Infiltration of the lower esophageal sphincter by fibrosis prevents it from contracting and generating adequate tone to prevent reflux of gastric contents into the esophagus. Esophageal manometry measures the pressure generated along the length of the esophagus and lower esophageal sphincter and can be utilized in the evaluation of suspected esophageal dysmotility disorders such as CREST syndrome, achalasia, and diffuse esophageal spasm. Incorrect Answers: A, B, C, D, and E. Esophageal peristalsis relies on contraction of the muscular layer of the esophageal wall. Decreased, not increased (Choices A and B), contraction of this muscle layer characterizes CREST syndrome. Though it does not generate successful peristalsis, diffuse esophageal spasm is a condition in which the muscular layer of the esophagus is overactive. These spasms cause severe dysphagia and chest pain, often acutely in the setting of swallowing. Lower esophageal sphincter tone is increased (Choices C and E) in achalasia, a condition in which there is impaired relaxation of the lower esophageal sphincter. It manifests as dysphagia, odynophagia, weight loss, halitosis, and regurgitation of undigested food. It is diagnosed through a barium swallow plus esophageal manometry. The other findings of CREST syndrome are not present. Normal peristalsis and decreased lower esophageal sphincter tone (Choice D) characterizes gastroesophageal reflux disease (GERD). Severe GERD presents with cough, sour or metallic taste, chest discomfort, and nausea. It is often worse when supine or following the ingestion of trigger foods. It would not be expected to cause difficulty swallowing solids and liquids or the other signs of CREST syndrome.

Objective: CREST syndrome is characterized by calcinosis cutis, Raynaud phenomenon, esophageal dysmotility, sclerosis, and telangiectasias. The esophageal dysmotility is caused by fibrosis of the esophageal wall muscular layer, which causes substantially decreased peristalsis and lower esophageal sphincter tone. I3D Previous Next Score Report Lab Values Calculator Help Pause

93 Exam Section 2: Item 44 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 44. A 34-year-old woman is brought to the emergency department 30 minutes after being involved in a motor vehicle collision. Sensation to pain is decreased over the left lower extremity. Which of the following labeled regions on the diagram of the spinal cord shown is most likely damaged in this patient? BAJ H. C E F Right Left A) B) C) D) E) F) G) H) |) J) Correct Answer: D. The lateral spinothalamic tract carries sensory nerve fibers for pain and temperature. Sensory neurons for pain and temperature synapse in the ipsilateral gray matter in the spinal cord, then the second order neuron decussates in the anterior white commissure of the spinal cord and ascends contralaterally in the lateral spinothalamic tract to synapse in the thalamus. As a result of the immediate decussation in the spinal cord, sensations of pain and temperature from the left side of the body ascend in the lateral spinothalamic tract on the contralateral right side of the spinal cord. Incorrect Answers: A, B, C, E, F, G, H, I, and J. The dorsal column (Choices A, B, I and J) holds ascending sensory nerves transmitting pressure, vibration, proprioception, and fine touch sensation. These sensory neurons ascend ipsilaterally in the dorsal column, so fibers from the right side of the body ascend on the right side of the spinal cord (Choices A and B), whereas fibers from the left side of the body ascend on the left side of the spinal cord (Choices I and J). Sensory nerves from the lower body ascend in the fasciculus gracilis in the medial aspect of the dorsal column (Choices A and J) whereas sensory nerves from the upper body ascend in the fasciculus cuneatus (Choices B and I). Upper motor neurons originating from the motor cortex in the brain descend and decussate at the caudal medulla and descend contralaterally in the lateral corticospinal tract (Choices C and H). Therefore, the lateral corticospinal tract on the right of the spinal cord (Choice C) carries nerves that control the right side of the body, and the left corticospinal tract (Choice H) controls the voluntary movement of the left side of the body. However, these signals originated in the contralateral primary motor cortex of the brain. The anterior corticospinal tract (Choices E and F) carries the remaining descending corticospinal fibers from the ipsilateral primary motor cortex that did not decussate at the level of the medullary pyramids. A majority of descending motor fibers decussate in the medullary pyramids to form the lateral corticospinal tract. The motor fibers in the anterior corticospinal tract decussate through the anterior spinal commissure at the relevant level to synapse with ventral horn motor neurons. The left lateral spinothalamic tract (Choice G) carries the sensory nerves for pain and temperature for the right side of the body. This patient's deficit involves the contralateral side.

Objective: The lateral spinothalamic tract carries ascending sensory fibers for pain and temperature. As a result of the immediate decussation of these sensory nerves in the spinal cord, sensations of pain and temperature from the left side of the body ascend in the lateral spinothalamic tract on the contralateral right side of the spinal cord. II Previous Next Score Report Lab Values Calculator Help Pause

84 Exam Section 2: Item 35 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 35. A 1-year-old boy is admitted to the hospital for antimicrobial treatment of perianal pain and erythema. He has a history of recurrent infections, including an admission to the hospital at the age of 2 weeks because of high-grade fever and swelling and redness around the umbilical cord stump. Physical examination shows perianal erythema and tenderness but no fluctuance. Laboratory studies show: Leukocyte count Segmented neutrophils Bands 74,000/mm3 (N=5000-19,500) 67% 9% Eosinophils Basophils Lymphocytes Monocytes 1% 1% 20% 2% This patient most likely has a genetic deficiency of which of the following immune system components? A) C9 B) IgG C) B2 Integrin D) Interferon alpha E) Tumor necrosis factor-a Correct Answer: C. Leukocyte adhesion deficiency (LAD) results from a defect in the attachment of leukocytes to the vascular endothelium, which consequently results in the impaired recruitment and migration to sites of extravascular inflammation or infection. LAD type 1 is caused by dysfunctional LFA-1 integrin (CD18) protein on the leukocyte surface, which does not allow for the normal attachment of leukocytes to the vascular endothelium. Specifically, it is the absence of the B, integrin subunit that leads to impaired LFA-1 function. LAD is typically characterized by recurrent bacterial infections, impaired wound healing, a delayed detachment of the umbilical cord after birth, and complete lack of leukocytes at sites of infections with no pus. The actions of leukocyte phagocytosis and bacterial killing are not impaired. Laboratory studies in patients with LAD will show increased absolute leukocyte counts in the blood since the leukocytes are unable to maintain their normal marginated position outside of the blood stream against the vessel walls. Incorrect Answers: A, B, D, and E. C9 (Choice A) is a component of the complement system that polymerizes to form the membrane attack complex (along with C5b, C6, C7, and C8) when activated. Pore formation by C9 causes membrane disruption and death of the targeted cell. C9 deficiency is associated with recurrent Neisseria infections. Immunoglobulin G (IgG) (Choice B) deficiency may be present in B- and T-lymphocyte disorders such as X-linked agammaglobulinemia, common variable immunodeficiency, ataxia- telangiectasia, hyper-IgM syndrome, and Wiskott-Áldrich syndrome. Patients are susceptible to recurrent infections. It is not associated with increased absolute leukocyte counts. Interferon alpha (Choice D) is a glycoprotein produced by cells that are infected with a virus. Interferon alpha upregulates MHC expression for recognition and binding by CD8+ T cells and downregulates protein synthesis to impair viral replication. Deficiency may result in severe progressive or fulminant viral disease. Tumor necrosis factor-a (TNF-a) (Choice E) deficiency is most often iatrogenic secondary to the use of anti-TNF-a monoclonal antibodies in the treatment of autoimmune disorders such as rheumatoid arthritis, psoriasis, ankylosing spondylitis, inflammatory bowel disease, and hidradenitis suppurativa. Deficiency results in increased risk for infections, lymphoma, and reactivation of latent tuberculosis.

Objective: LAD is a group of disorders characterized by impaired leukocyte adhesion to the vascular endothelium in response to infection. Leukocytes retain the ability to phagocytose and eliminate foreign pathogens, but they are unable to migrate to sites of infection or inflammation in the extravascular space. Previous Next Score Report Lab Values Calculator Help Pause

96 Exam Section 2: Item 47 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 47. A 35-year-old woman with leiomyomata uteri undergoes an experimental treatment that involves instillation of an embolizing agent directly into the vessels that feed the leiomyomata. During this procedure, in order to reach the involved vessels, which of the following is the most direct course of the catheter after entering the femoral artery? A) Aorta gonadal artery uterine artery B) External iliac artery - internal iliac artery → uterine artery C) External iliac artery superior gluteal artery uterine artery D) Internal iliac artery - inferior vesical artery → uterine artery E) Internal iliac artery superior vesical artery → uterine artery Correct Answer: B. The main vessels supplying blood to the uterus are the uterine arteries, which have smaller subdivisions that supply the different portions of the uterus, cervix, and vagina. The uterine artery is a branch of the internal iliac artery, and travels along the broad ligament anterior to the ureter before dividing into the ascending and descending branches. The ascending branch joins the ovarian artery and further divides in the myometrium into the arcuate artery, radial artery, spiral artery, and basal artery. The descending branch is the major blood supply to the cervix and vagina. The internal iliac artery is a branch of the common iliac artery, which also gives rise to the external iliac artery. The femoral artery, in turn, is a continuation of the external iliac artery as it courses below the inguinal ligament. Therefore, to reach the patient's leiomyomata and uterine arteries in this case, the most direct course to travel is from the femoral artery to the external iliac artery and then ascend proximally until the common iliac artery bifurcation. Further advancements should occur down the internal iliac artery and, subsequently, the uterine artery. Incorrect Answer: A, C, D, and E. Entering the femoral artery and proceeding to the aorta to descend along the gonadal artery into the uterine artery (Choice A) is not feasible. The origin of the gonadal arteries occur from the upper abdominal aorta, near the level of the second lumbar vertebra. The gonadal artery, termed the ovarian artery in women, descends in the abdomen to supply arterial blood to the ovary. While the ovarian artery demonstrates an anastomotic connection with the uterine artery, this would not permit the passage of the embolizing agent to treat the patient's leiomyomata. The superior gluteal artery (Choice C) is a branch of the internal iliac artery that supplies blood to the posterior pelvis and gluteal musculature. It does not anastomose with the uterine artery, and therefore would not allow access to the blood supply of the uterine leiomyomata. The inferior vesical artery (Choice D) is a branch of the internal iliac artery that supplies blood to the bladder, as well as the prostate and seminal vesicles in males. It does not anastomose with the uterine artery, and access to the leiomyomata blood supply would not be obtained. The superior vesical artery (Choice E) is a branch of the internal iliac artery that supplies blood to the bladder, as well as the vas deferens and testicles in males. It does not anastomose with the uterine artery, and access to the leiomyomata blood supply would not be obtained.

Objective: Leiomyomata of the uterus receive their blood supply from the uterine arteries, which are direct branches of the internal iliac artery. The uterine arteries are frequently accessed for embolization in the setting of ongoing vaginal bleeding caused by uterine fibroids. The most direct course of action for embolization is to travel from the femoral artery to the external iliac artery, ascend to the common iliac artery bifurcation, and then descend through the internal iliac artery and uterine artery. Previous Next Score Report Lab Values Calculator Help Pause

18 Exam Section 1: Item 18 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 18. A 56-year-old man with a history of alcohol dependence comes to the emergency department because of fever, night sweats, and a cough productive of bloody sputum. Laboratory studies of his sputum show numerous acid-fast rods arranged in small arrays. Which of the following components of the causal organism may be directly cytotoxic to macrophages? A) Cord factor B) Cytoplasmic membrane C) Endotoxin D) Polysaccharide capsule E) Teichoic acid Correct Answer: A. Cord factor is a glycolipid found as part of the cell wall of Mycobacterium tuberculosis (MTB) and is responsible for much of its virulence. Ingestion of MTB by phagocytes should lead to the intracellular killing of the organism, but the presence of cord factor prevents fusion of the vesicles containing MTB with the lysosome, thereby preventing intracellular killing. Additionally, cord factor is directly cytotoxic to macrophages. It also plays a role in the organization of MTB in various tissues, and this ability to organize MTB into straight, closely grouped arrays gives cord factor its name. Finaly, cord factor is thought to be responsible for inducing secretion of many of the cytokines that are important in granuloma formation. The virulence of MTB is complex and includes the presence of cell wall lipids, cytoplasmic transport proteins, and certain enzymes, but cord factor is known to be one of the most important virulence factors associated with bacterial growth and survival. Incorrect Answers: B, C, D, and E. The cytoplasmic membrane (Choice B) is a lipid bilayer containing various cellular proteins that encapsulates the cytoplasm. It houses several transporter proteins, including ESX and TAT proteins, involved in iron and zinc transport, but none of these contribute as much to the virulence of MTB as cord factor. Endotoxin (Choice C) production is not a feature of MTB but is commonly seen in infections with gram-negative bacteria, which contain lipopolysaccharide (LPS) in their cell wall. LPS is highly immunogenic and is involved in the pathophysiology of many of the symptoms of septic shock (eg, rigors, fevers, vasodilation). Polysaccharide capsule (Choice D) is not directly cytotoxic to macrophages, although it does play an important role in the interaction of MTB with phagocytes. There are four primary layers of the MTB cell envelope: the plasma membrane, peptidoglycan-arabinogalactan complex (AGP), outer membrane linked to AGP by mycolic acids, and the polysaccharide capsule, which is the outermost layer. Teichoic acid (Choice E), a common component of gram-positive bacterial cell walls, is not found in the cell wall of MTB.

Objective: MTB derives much of its virulence from a host of mechanisms; cord factor, which is a glycolipid on the surface of the cell wall, protects the bacteria from phagocytic killing and is directly cytotoxic to macrophages. %3D Previous Next Score Report Lab Values Calculator Help Pause

22 Exam Section 1: Item 22 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 22. A31-year-old man who works as a chef comes to the emergency department 30 minutes after accidentally cutting his right thumb with a knife. Examination of the right hand shows a 3-cm-long, 0.6-cm-deep laceration of the thumb. The wound is cleaned, sutured, and heals normally, leaving a faint scar by 1 year after the injury. The remodeling of the scar that occurred during the first year after injury is primarily mediated by which of the following? A) Chemokines B) Interleukin-1 (IL-1) C) Matrix metalloproteinases D) Transforming growth factor-ß E) Tumor necrosis factor Correct Answer: C. Matrix metalloproteinases (MMPS) are enzymes that lyse proteins using a chelated metal ion cofactor. MMPS are necessary for the healing of wounds as well as for the remodeling of bone and cartilage. Wound healing occurs via a staged process. In early wound healing, platelet aggregation and platelet plug formation occur to achieve hemostasis. In the following one to two days, neutrophils and macrophages infiltrate the area and release growth factors and cytokines that stimulate fibroblast proliferation. Around day three, granulation tissue forms as collagen is deposited into the area and angiogenesis begins to occur. During this time, wound edges contract from the action of myofibroblasts, further aiding in healing. Epidermal cells migrate across the newly deposited collagen matrix to reconstitute normal skin appearance. In the following weeks and months, scar formation and remodeling occurs via metalloproteinase-mediated collagen breakdown. Incorrect Answers: A, B, D, and E. Chemokines (Choice A) are cell signaling proteins that specifically promote cellular chemotaxis. An example of a chemokine is IL-8, which induces chemotaxis of neutrophils and granulocytes. Interleukin-1 (IL-1) (Choice B) is a pyrogen secreted by macrophages and monocytes; it also promotes vasodilation, activates osteoclasts, and stimulates the proliferation of granulocytes. IL-1 dysregulation has been implicated in cartilage damage in osteoarthritis. Transforming growth factor-ß (TGF-B) (Choice D) is a cytokine that is secreted by multiple cells including macrophages. TGF-ß has multiple functions that include inhibiting B-cell proliferation and Th17 cells. Tumor necrosis factor (Choice E) is secreted by activated macrophages, and it is key in the formation of granulomas and immune protection against mycobacterial infections. It is increased in multiple autoimmune diseases, and its activity is inhibited by drugs such as infliximab and adalimumab.

Objective: Matrix metalloproteinases are implicated in the breakdown of extracellular matrix. They are essential for the remodeling of collagen that occurs during wound healing. They are also implicated in the breakdown of bone and cartilage tissues. Previous Next Score Report Lab Values Calculator Help Pause

14 Exam Section 1: Item 14 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 14. In the hypothetical distribution of laboratory values for a particular laboratory test in "normal" and "diseased" populations, respectively, which of the following lettered decision points will produce the maximum sensitivity for the detection of the disease? Normal Number of Diseased subjects tested D Range of values for a particular test- A) B) C) D) E) Correct Answer: A. Any population of patients will include those who have the disease and those who do not, and overlap may exist between these groups when considering testing with a range of possible values. For example, one diagnostic criteria for type 2 diabetes mellitus is two fasting serum blood glucose concentrations greater than 126 mg/dL. However, not all patients with fasting serum blood glucose greater than 126 mg/dL will truly have diabetes mellitus. These patients may receive positive tests but do not have the underlying condition (false-positive result). Therefore, the set points for diagnostic tests will result in some true positives (patients who have the disease and who the tests identifies as having the disease), true negatives (patients who do not have the disease who the test identifies as not having the disease), false positives (patients who do not have the disease who the test identifies as having the disease), and false negatives (patients who have the disease who the test identifies as not having the disease). The set point will determine the fraction of patients who are true- or false-positive and true- or false-negative and should be chosen to optimize the sensitivity and specificity of the test. If the graph were being used to determine the sensitivity and specificity of the fasting serum blood glucose test in diagnosing diabetes mellitus, Point A would result in all persons with type 2 diabetes mellitus being identified as positive. This would mean the number of false negatives would be zero, and the number of true positives would be high. Moving the cut point from point À to point B (raising the threshold for a positive test result) would cause fewer people with the disease to have a positive test, therefore increasing the number of false negatives and decreasing the number of false positives. Sensitivity is defined as the proportion of all people with the disease who test positive with the disease. The set point at A would yield a sensitivity of 100%. As the set point is moved to the right, the sensitivity will decrease. At E, the sensitivity would be 0%, meaning that none of the people with the disease would test positive. Incorrect Answers: B, C, D, and E. Compared with point A, a cutoff at point B (Choice B) would yield a higher number of false negatives. Individuals with the disease would test negative, which would decrease sensitivity. However, it would also improve specificity, the proportion of people without the disease who test negative, because the rate of false positives would decrease. At point C (Choice C), sensitivity decreases even further. At this point, there are zero false positives. All persons who have the disease test positive, yielding a specificity of 100%. At point D (Choice D), only approximately half of the diseased individuals test positive. This yields a sensitivity of roughly 50%. At point E (Choice E), every individual would test negative. The sensitivity is 0%.

Objective: Moving the cut point higher for a test decreases sensitivity and increases specificity by increasing the number of false negatives and decreasing the number of false positives. In this case, point A represents 100% sensitivity, the cut-off point at which all people who have the disease test positive. II Previous Next Score Report Lab Values Calculator Help Pause -E-

25 Exam Section 1: Item 25 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 25. A 70-year-old woman with osteoporosis comes to the physician for her scheduled injection of ibandronate. She is accompanied by her 75-year-old husband. Physical examination shows no abnormalities. The husband asks the physician, "Why don't I have osteoporosis? For the past 50 years we have done everything together, including eating the same foods and exercising. I even used to smoke and drink. Yet she has osteoporosis, and I don't." Which of the following is the most appropriate response? A) "Drinking alcohol when you were younger was protective against osteoporosis." B) "Exercising is good for both of you, but it is more protective for men." C) "Men start out with a higher peak bone mass compared with women." D) "Your estrogen concentration is lower than your wife's." E) "Your testosterone concentration is probably low." 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 even lead to 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 dual energy x-ray absorptiometry via calculation of a T-Score. AT-score of -2.5 or lower, 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 and -2.5 are defined as osteopenia.Osteoporosis tends to affect females at a higher rate than males. There are a number of reasons for this. Males generally develop higher peak bone mass during the first three decades of life. Given the chronic and generally linear rate of bone mineral loss, males will therefore tend to retain higher bone mineral density for a longer period of time given the higher bone mineral density starting point than females. Moreover, as estrogen output declines during menopause, bone mineral density loss occurs more rapidly during this phase of life. As both testosterone and estrogen stimulate osteoblasts and increase or maintain bone mineral density, the loss of estrogen leads to an accelerated loss of bone mass in women. Incorrect Answers: A, B, D, and E. Alcohol consumption (Choice A) has a negative effect on bone mineral density; however, this effect is mostly seen with heavy drinking over many years. Alcohol is a diuretic and can cause loss of electrolytes including calcium. Exercise (Choice B) is protective against osteoporosis for both men and women. Exercises that include cyclic loading of the skeleton promote remodeling and an increase of bone mineral density in response to stress (Wolff's law). Estrogen (Choice D) is a steroid hormone that increases bone mineral density. A low estrogen concentration would lead to less bone mineral density. Low testosterone concentration (Choice E) would lead to a decrease in bone mineral density, as testosterone increases bone mineral density. Men who have hypogonadism are at a higher risk for osteoporosis, although osteoporosis is overall much less common in men.

Objective: Osteoporosis is a progressive loss of bone mineral density that occurs with aging and can lead to fragility fractures. It is more common in females as bone mineral density decreases more rapidly after menopause, and females begin with lower peak bone mass as compared with males. 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. A 33-year-old woman comes to the physician because of a 2-week history of fatigue, difficulty swallowing, and paresthesias of her lips and fingers. She also has had episodes of painful wrist spasms during this period. She has Graves disease and underwent a surgical procedure to remove part of her thyroid gland 3 weeks ago. Her only medication is low-dose propranolol. Vital signs are normal. Physical examination shows no abnormalities except for a well-healing incision on the neck. An ECG shows a prolonged QT interval. Which of the following is the most likely cause of this patient's new symptoms? A) Adverse effect of propranolol OB) Hyperparathyroidism C) Hyperthyroidism OD) Hypoparathyroidism O E) Hypothyroidism F) Laryngeal nerve injury Correct Answer: D. Parathyroid hormone (PTH) is a peptide hormone that increases osteoclast activity and bone resorption in order to maintain calcium homeostasis. It is produced by the four parathyroid glands, which are located on the posterior aspect of the thyroid gland. PTH is decreased in hypoparathyroidism, which may be primary or caused by a PTH-independent increase in serum calcium. Primary hypoparathyroidism can occur following the unintentional resection of the parathyroid glands during thyroidectomy, as in this case, but may also be a result of autoimmune destruction of the parathyroid glands. As a consequence of surgical removal of the parathyroid glands, patients will develop symptoms of hypocalcemia including perioral paresthesia, carpopedal spasm, laryngospasm, and prolonged QT interval. The Chvostek sign, in which tapping of the facial nerve leads to spasm of the facial muscles, and Trousseau sign, in which an inflated blood pressure cuff occluding the brachial artery causes carpopedal spasm, are two physical examination signs that may indicate hypocalcemia. While this patient demonstrates signs of hypocalcemia, PTH may also be decreased when increased concentrations of calcium exert negative feedback on the parathyroid gland. For example, in the case of excessive calcium intake or hypervitaminosis D, calcium will inhibit the production of PTH, causing the hormone concentration to fall. Lytic osseous primary or metastatic cancer, in which calcium is often released from bones that are infiltrated with malignancy, is another potential cause of hypercalcemia and decreased PTH concentrations. Incorrect Answers: A, B, C, E, and F. Major side effects of B-adrenergic blockers, including propranolol (Choice A), are decreased heart rate and contractility, bronchospasm, and hypoglycemia. This patient's symptoms of hypocalcemia postoperatively are better explained by hypoparathyroidism secondary to inadvertent parathyroidectomy. Hyperparathyroidism (Choice B) leads to bony resorption through the increased production of PTH. Primary hyperparathyroidism results from a functionally active parathyroid adenoma, hyperplasia, or carcinoma, and is characterized by increased serum calcium and decreased serum phosphate concentrations. Secondary hyperparathyroidism is commonly caused by chronic hyperphosphatemia in the setting of end-stage renal disease. It is characterized by decreased calcium and increased phosphate concentrations. Hyperthyroidism (Choice C) classically presents with heat intolerance, weight loss, flushed skin, hair loss, loose stools, or oligo- or amenorrhea. Hypothyroidism (Choice E) classically presents with cold intolerance, weight gain, dry skin, constipation, and menorrhagia. The patient's symptoms are better explained by hypocalcemia induced by hypoparathyroidism than by an aberrancy in thyroid function. Laryngeal nerve injury (Choice F) is a potential complication of thyroidectomy. If the recurrent laryngeal nerve is damaged, paralysis of the vocal cord presenting with hoarseness will result. This may also be accompanied by dysphonia and/or dyspnea.

Objective: PTH is decreased in hypoparathyroidism, which may be caused by the inadvertent removal of one or more parathyroid glands during thyroidectomy. Symptoms of hypocalcemia, including perioral paresthesia, carpopedal spasm, laryngospasm, prolonged QT interval, Chvostek sign, and Trousseau sign, develop postoperatively Previous Next Score Report Lab Values Calculator Help Pause

32 Exam Section 1: Item 33 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 33. A healthy 25-year-old man participates in a study of muscle function. The electrophysiologic observations made on a muscle biopsy specimen are shown. Via iontophoresis, 1 µM acetylcholine (ACh) was applied to the muscle surface. Extracellular Ca2+ concentration was decreased to prevent end-plate potentials from acting as a suprathreshold for muscle action potentials. epp Amplitude (in mV) 10 mepp Amplitude (in mV) 1 Response (in mV) to ACh ( 1 μΜ) 1 Normal muscle epp=end-plate potential; mepp= miniature epp Based on these findings, which of the following electrophysiologic characteristics is expected in a muscle biopsy specimen from a patient with acute botulism? epp Amplitude (in mV) mepp Amplitude (in mV) Response (in mV) to ACh (1 μΜ) A) 1 0.5 0.5 B) 1 1 C) 2 0.2 0.2 D) 0.2 0.1 E) 15 1 1 Correct Answer: B. Skeletal muscle activity is controlled by neurons which synapse on muscles at the neuromuscular junction. At this junction, the action potential from the neuron signals for the presynaptic release of acetylcholine leading to endplate potentials on the myocyte membrane. These endplate potentials then stimulate the dihydropyridine receptors and subsequently the ryanodine receptors which cause the release of calcium from the sarcoplasmic reticulum. This calcium then activates the binding of myosin and actin leading to muscle contraction. Endplate potentials are the depolarization events that occur at the neuromuscular junction which lead to muscle fiber depolarization and subsequent contraction. Miniature endplate potentials are the small depolarization events that occur from the release of a single vesicle of acetylcholine into the neuromuscular junction. These potentials are not sufficient for reaching the action potential threshold on the muscle cell membrane. These miniature potentials occur sporadically when no depolarization signal is given to the neuromuscular junction as there is a small amount of baseline vesicle release. Given that the extracellular calcium concentration was significantly decreased from physiologic values in this experiment, less acetylcholine will be released from the axon terminal when an action potential occurs as this is a calcium-dependent process. Miniature endplate potentials are caused by ligand-gated nonspecific cation channels and are more dependent on the concentrations of potassium and sodium as well as calcium. Thus, miniature endplate potentials will not vary significantly based on calcium concentration. Similarly, the response to 1 µM acetylcholine placed in solution will be relatively unchanged. The major change as a result of botulinum toxin will be failure of the release of acetylcholine by the motor neuron, leading to a primarily decreased endplate potential. Incorrect Answers: A, C, D, and E. The magnitude of miniature endplate potentials and the response to 1 µM acetylcholine would not decrease significantly with decreased extracellular calcium concentration as this is dependent on the quantity of acetylcholine released in a single vesicle and on the sodium concentration (Choices A, C, and D). Botulism prevents release of presynaptic acetylcholine which would decrease, not increase, the magnitude of the endplate potential (Choice E).

Objective: Calcium concentration affects the release of vesicles at the neuromuscular junction, thereby affecting the amount of acetylcholine released and the overall magnitude of endplate potentials. Botulism prevents the release of presynaptic acetylcholine, decreasing the endplate potential while not affecting the miniature endplate potential or response to exogenous acetylcholine. Previous Next Score Report Lab Values Calculator Help Pause 1,

97 Exam Section 2: Item 48 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 48. A 37-year-old woman is brought to the emergency department after her husband found her unconscious. Her temperature is 36°C (96.8°F), pulse is 128/min, and blood pressure is 70/40 mm Hg. Physical examination shows cool, pale extremities, jugular venous distention, faint peripheral pulses, and crackles over the bottom two thirds of both lung fields. Heart sounds are normal, and there are no murmurs. She withdraws to painful stimuli in all four extremities. This patient is most likely experiencing which of the following types of shock? A) Anaphylactic B) Cardiogenic C) Hypovolemic D) Neurogenic E) Septic Correct Answer: B. Cardiogenic shock can occur in the setting of acute myocardial infarction, severe valvular dysfunction, severe heart failure, heart block, or arrhythmia. Patients may experience cardiogenic pulmonary edema resulting in shortness of breath, tachypnea, hypoxia, and audible rales on examination. Physical examination demonstrates hypotension, cool and pale extremities, and jugular venous distention. Pulmonary artery catheterization typically demonstrates decreased cardiac index with increased pulmonary capillary wedge pressure if there is left-sided heart dysfunction and increased systemic vascular resistance (SVR). In cardiogenic shock, dysfunction of cardiac myocytes results in decreased chronotropy, lusitropy, or inotropy. Decreased cardiac output and increased pulmonary capillary wedge pressure result from such dysfunction. Systemic vascular resistance is increased as a physiologic attempt to maintain mean arterial pressure in the setting of decreased cardiac output and peripheral perfusion. Incorrect Answers: A, C, D, and E. Anaphylactic (Choice A) and septic (Choice E) shock are forms of distributive shock, which are characterized by a primary decrease in SVR. This decrease in SVR is secondary to IgE-mediated histamine and cytokine release in anaphylactic shock, and cytokine or toxin-mediated endothelial dysfunction in septic shock. Cardiac output increases to compensate for low SVR in the early stages of distributive shock. If the inciting state is not appropriately treated, further decompensation can occur resulting in progressive hypotension and insufficient perfusion with end-organ damage. Pulmonary artery catheterization would demonstrate an increased cardiac index and decreased SVR in cases of distributive shock. Hypovolemic (Choice C) shock occurs secondary to a significant decrease in intravascular volume, commonly related to hemorrhage, dehydration, or gastrointestinal volume losses, which results in decreased cardiac preload and therefore stroke volume and cardiac output. In turn, there is arteriolar vasoconstriction to increase ŠVR in an attempt to maintain mean arterial pressure. If progressive and severe, compensation fails, which leads to the development of hypotension and inadequate organ perfusion. Neurogenic (Choice D) shock occurs in the setting of a catastrophic central nervous system injury resulting in diminished sympathetic output to the heart and vasculature. This results in hypotension secondary to decreased myocardial contractility and arteriolar vasodilation with associated bradycardia. Pulmonary artery catheterization would demonstrate decreased cardiac index and SVR.

Objective: Cardiogenic shock can occur in the setting of acute myocardial infarction, severe valvular dysfunction, severe heart failure, heart block, or arrhythmia. Signs and symptoms include hypotension, cool and pale extremities, jugular venous distention, and pulmonary edema. 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. A 55-year-old man with long-standing type 2 diabetes mellitus comes to the physician because of a 2-day history of fever, chills, nausea, and swelling of his left leg. His temperature is 38.9°C (102°F), and pulse is 110/min. Examination of the left lower extremity shows a warm, tender, erythematous, blanching rash. A photograph of the left lower extremity is shown. A photomicrograph of a Gram stain of a blood culture is shown. Which of the following is the most likely causal organism? A) Clostridium perfringens B) Enterococcus faecalis C) Haemophilus influenzae D) Pasteurella multocida E) Pseudomonas aeruginosa F) Staphylococcus aureus G) Streptococcus pyogenes (group A) Correct Answer: G. Cellulitis presents with cutaneous erythema, warmth, and induration, often following inoculation from an injury such as an abrasion. Systemic symptoms, including fever, tachycardia, and leukocytosis are often present. Most commonly, it is acute and results from a bacterial infection of the skin. The two most common pathogens to cause cellulitis are Staphylococcus aureus and Streptococcus pyogenes (group A). S. pyogenes is a gram-positive bacterium that forms chains. It is part of the B-hemolytic group of Streptococcus species and is bacitracin sensitive. Clinically, the cellulitis caused by S. aureus is typically purulent while that caused by S. pyogenes is not. In addition to cellulitis, infection with S. pyogenes can result in various diseases such as pharyngitis, scarlet fever, necrotizing fasciitis, glomerulonephritis, and rheumatic fever. Incorrect Answers: A, B, C, D, E, and F. Clostridium perfringens (Choice A) is an anaerobic gram-positive bacillus. It often infects an open wound and causes necrotizing fasciitis. Clostridium perfringens secretes alpha toxin, a phospholipase that degrades cell membranes to cause gas gangrene. Enterococcus faecalis (Choice B) belongs to Group D Streptococcus, which are gamma hemolytic. Gamma-hemolytic species are identified by their ability to grow in bile and 6.5% sodium chloride solution. Clinically, Enterococcus faecalis is a common cause of nosocomial urinary tract infections. Haemophilus influenzae (Choice C) is an encapsulated gram-negative bacterium that can result in various mucosal infections such as conjunctivitis, otitis media, bronchitis, and pneumonia. It was the leading cause of epiglottitis in children prior to widespread immunization. It is an uncommon cause of cutaneous infections. Pasteurella multocida (Choice D) is a gram-negative coccobacillus, which is frequently the culprit pathogen in infections secondary to cat, and to a lesser extent dog, bites. Pseudomonas aeruginosa (Choice E) is a gram-negative rod. Its cutaneous manifestations include folliculitis, infections of the nail unit, otitis externa maligna, and ecthyma gangrenosum. Pseudomonas species may cause cellulitis, classically as a sequela of a puncture wound of the foot in a diabetic patient. Staphylococcus aureus (Choice F) is a gram-positive coccus. It forms clusters rather than chains of cocci. S. aureus is a common cause of cellulitis, though it usually causes purulence. It also causes impetigo, erysipelas, and folliculitis.

Objective: Cellulitis is characterized by cutaneous erythema, warmth, and induration. The two most common pathogens are Staphylococcus aureus and Streptococcus pyogenes (group A), which can be differentiated clinically by the presence or absence of purulence. S. pyogenes is a gram-positive bacterium, which forms chains, is B-hemolytic, and is bacitracin- sensitive. %3D Previous Next Score Report Lab Values Calculator Help Pause

55 Exam Section 2: Item 6 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 6. A 54-year-old woman comes to the physician because of a 3-month history of fatigue and a persistent cough productive of colorless mucoid sputum. She also has had a 14-kg (30-lb) weight loss during this period. She has smoked 2 packs of cigarettes daily for 25 years. She is 168 cm (5 ft 6 in) tall and now weighs 52 kg (115 lb); BMI is 19 kg/m2. Percussion of the chest shows hyperresonance. Her serum calcium concentration is 12.8 mg/dL. A chest x-ray shows a mass in the right middle lobe of the lung near the hilum. Examination of aspirate from the mass shows malignant cells. The neoplasm in this patient is most likely which of the following? A) Adenocarcinoma B) Giant cell carcinoma C) Large cell undifferentiated carcinoma D) Small cell carcinoma E) Squamous cell carcinoma Correct Answer: E. 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 as a result of paraneoplastic parathyroid hormone-related peptide (PTHPP) production. Histologic characteristics include polygonal cells with intercellular bridges, eosinophilic cytoplasm, keratin pearls, and necrosis. Lung cancer typically presents with cough, unintentional weight loss, hemoptysis, chest pain, dyspnea, and hoarseness when symptomatic; occasionally, wheezing, focal rhonchi, or hypertrophic osteoarthropathy may be noted on examination. Diagnosis is made by chest imaging and biopsy. Prognosis is a function of the cancer type along with grading and staging of the disease. It is often detected late in stage, once metastatic, at which point the prognosis is poor. Incorrect Answers: A, B, C, and D. Adenocarcinoma (Choice A) of the lung is the most common overall primary lung cancer and the most common among nonsmokers. It typically presents 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. Giant cell carcinoma (Choice B) is a rare carcinoma of the lung that contains pleomorphic giant, multinucleated cells on histology. It most commonly involves the upper lobes and the lung periphery. Large cell undifferentiated carcinoma (Choice C) is a rare, poorly differentiated lung neoplasm that displays large atypical cells on histology. It most commonly involves the lung periphery. Small cell carcinoma (Choice D) 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, syndrome of inappropriate antidiuretic hormone, Lambert-Eaton myasthenic syndrome caused by presynaptic calcium channel antibody production, paraneoplastic limbic encephalitis, and subacute cerebellar degeneration. Histologic features include small, dark blue tumor cells lacking nucleoli with 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 caused by PTHRP production. 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. A 48-year-old man with renal artery stenosis undergoes stent placement. Femoral access is used to place the stent. After entrance into the aorta, the guide wire should be advanced superiorly just beyond which of the following structures to approach the right renal artery in this patient? A) Celiac trunk B) Internal iliac artery C) Superior mesenteric artery D) Testicular artery Correct Answer: D. The abdominal aorta has numerous branches that arise to supply the entirety of the gastrointestinal system, spleen, kidneys, spine, and testicles before dividing into the left and right iliac arteries. The testicular arteries are paired and arise just inferior to the takeoff of the renal arteries at the level of the second lumbar vertebrae, before they pass into the retroperitoneum and inferiorly through the deep inguinal ring to supply the testes. In this patient, the guidewire should be advanced just superior to the testicular artery orifices, placing it at the level of the renal arteries at approximately L1-2, to access the right renal artery. Incorrect Answers: A, B, and C. Celiac trunk (Choice A) gives rise to the common hepatic artery, left gastric artery, and splenic artery at the level of T12. It supplies blood to the liver, stomach, proximal duodenum, pancreas, inferior esophagus, and spleen. It lies above the level of the renal arteries. Internal iliac arteries (Choice B) are paired vessels that arise at the level of L5-S1 and are branches of the common iliac arteries. The internal iliac arteries supply blood to the buttocks and the pelvic organs, including the bladder and inferior part of the ureters, but do not supply blood to the kidneys. They lie below the level of the renal arteries. Superior mesenteric artery (Choice C) is a branch of the abdominal aorta that arises just inferior to the celiac trunk and supplies blood to the duodenum, jejunum, ileum, and proximal two-thirds of the colon. It arises just above the level of the renal arteries.

Objective: To access the renal arteries from the femoral artery approach, the catheter should be advanced to the area just superior to the takeoff of the testicular artery. %D Previous Next Score Report Lab Values Calculator Help Pause