NBME 28

18 Exam Section 1: Item 18 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 18. During a period of 36 hours, an 80-year-old woman has increasingly severe abdominal pain followed by fever, chills, tachycardia, hypotension, and, finally, shock. Blood cultures grow Escherichia coli. Her condition worsens and, despite supportive therapy and antibiotics, she dies 4 days after the onset of the illness. Which of the following is the most likely cause of the initial hypotension? A) Excessive production of nitric oxide B) Generation of hydrogen peroxide C) Hemorrhage D) Induction of endothelial adhesion molecules E) Platelet aggregation

A. Sepsis is a systemic inflammatory syndrome that results from a dysregulated and exaggerated immune response to an infection. Sepsis can be complicated by shock and multiorgan failure with a high mortality rate. Septic shock is characterized by an impaired response of the vasculature to vasoconstricting stimuli with markedly decreased systemic vascular resistance, tachycardia, increased cardiac output, oliguria, and lactic acidosis. Excessive production of nitric oxide is associated with hypotension in the setting of sepsis. Inducible nitric oxide synthetase is a nitric oxide-producing enzyme that is upregulated through tyrosine kinase activation in response to proinflammatory cytokines and binding by lipopolysaccharides. Nitric oxide activates guanylate cyclase in vascular smooth muscle resulting in muscle relaxation and vasodilation because of increased intracellular cyclic guanosine monophosphate concentration. Incorrect Answers: B, C, D, and E. Generation of hydrogen peroxide (Choice B) occurs in phagolysosomes by the enzyme superoxide dismutase, which utilizes free oxygen radicals produced by NADPH oxidase. Excess free radical production is associated with host tissue injury. Hemorrhage (Choice C) may result in shock secondary to hypovolemia and decreased oxygen carrying capacity of the blood secondary to loss of hemoglobin. This patient's abdominal pain is likely the result of an infectious enteritis, colitis, or peritonitis, and is less likely from a ruptured abdominal aortic aneurysm, which may cause shock from internal exsanguination. Induction of endothelial adhesion molecules (Choice D) is a key step in the recruitment and migration of leukocytes to sites of infection and injury. The action of nitric oxide on the vascular smooth muscle results in hypotension, whereas the expression of endothelial adhesion molecules is involved in the immune response to localized infection. Platelet aggregation (Choice E) and activation occurs in response to inflammation and helps promote the innate immune response, although it does not directly cause the patient's hypotension. Diffuse platelet aggregation may result in thrombocytopenia in the setting of sepsis. Educational Objective: Sepsis is a dysregulated systemic inflammatory syndrome that may occur in response to infection. It may progress to hypotension and septic shock. Excessive nitric oxide production causing diffuse vasodilation is one of the mechanisms of hypotension in sepsis. %3D 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 67-year-old woman comes to the physician for a health maintenance examination. Her brother and mother have a history of colon cancer. The physician recommends colonoscopy, but the patient says that she would prefer only for her stool to be tested for blood. The physician explains that testing the stool for occult blood is not appropriate in this case. The physician is most likely concerned about which of the following regarding this test? A) Low sensitivity B) Low specificity C) Potential for a false-positive result D) Uncertain negative predictive value E) Uncertain positive predictive value

A. Sensitivity is the ability of a test to detect a disease if it is present. A test is described as sensitive if it has a high likelihood of disease detection, and therefore a low likelihood of false negativity. High sensitivity is therefore useful in ruling out a disease. This is because a negative result from a high sensitivity test indicates a low likelihood that the disease is present. Because of this, high sensitivity tests are useful for screening in which proving the absence of disease and limiting the number of false negative results are of utmost importance. In the case of cancer screening, a highly sensitive test allows the clinician to be confident that a negative test means that the patient is disease-free. In contrast, a test with poor sensitivity, if negative, does not provide strong evidence or confidence that the patient does not have the disease. In the case described, the fecal occult blood test demonstrates both a low sensitivity and specificity and is of limited use to the physician and the patient for ruling out cancer. Incorrect Answers: B, C, D, and E. Low specificity (Choice B) describes a test that is subject to false-positive errors, meaning that a positive test does not have a high likelihood of disease. High specificity is required to confirm a diagnosis. An example of high specificity testing for colon cancer would be a colonoscopy with biopsy of a lesion with staining and molecular testing under microscopy. Direct visualization of malignant cells on microscopy is highly specific for a cancer diagnosis, that is, there is a low likelihood of a false cancer diagnosis. Potential for a false-positive result (Choice C) is possible with a fecal occult blood test. For example, a bleeding hemorrhoid or taking an iron supplement can cause a positive fecal occult blood test. This is unlikely, however, to cause concern by the physician in this scenario as a false positive test does not necessarily put the patient at risk for an undiagnosed colorectal cancer, although it may lead to unnecessary further diagnostic examinations. Missing an early cancer diagnosis due to poor test sensitivity could lead to early mortality and is a more pressing issue. Uncertain negative predictive value (Choice D) is not of immediate concern in this scenario. Negative predictive value is based on both the test sensitivity and the pretest probability of the patient having the disease. Although certain individuals have higher or lower risk of colon cancer based on lifestyle and genetics, pretest probability and alterations in the negative predictive value do not alter the need for a sensitive test for screening purposes. Uncertain positive predictive value (Choice E) is also not an immediate concern. Positive predictive value is based on both test specificity and the pretest probability of disease. Positive predictive value is the likelihood that a person has a disease, given a positive test. A positive predictive value is of greater importance in confirmatory testing. Educational Objective: High sensitivity tests are required for effective disease screening. High specificity tests are necessary for confirmation of the disease. Negative and positive predictive values are functions not only of sensitivity and specificity, but also of the pretest probability of the disease. %3D Previous Next Score Report Lab Values Calculator Help Pause

85 Exam Section 2: Item 35 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 35. A70-year-old man dies in a motor vehicle collision. He had been undergoing evaluation for occult blood in the stool. A photograph of a section of the transverse colon obtained at autopsy is shown. Which of the following is the most likely diagnosis? O A) Hyperplastic polyp B) Inflammatory pseudopolyp C) Juvenile polyp D) Peutz-Jeghers syndrome E) Tubular adenoma 1 cm

E. A tubular adenoma is shown in the gross picture of the transverse colon obtained during autopsy. Chronic occult gastrointestinal (GI) bleeding is a common presenting symptom of colorectal carcinoma (CRC) and for this reason, all patients with unexplained GI bleeding should undergo colonoscopy to rule out CRC. Colonic polyps present in a variety of subtypes, from non-neoplastic polyps (eg, hamartomatous, mucosal, inflammatory, hyperplastic) to potentially malignant polyps (eg, adenomatous, serrated). Adenomatous polyps may be of tubular or villous architecture on histology, with villous architecture usually having greater malignant potential, although tubular adenomas are more common. All adenomas discovered on colonoscopy should be either completely removed if less than two centimeters or biopsied. Larger adenomas must be removed in a piecemeal fashion or surgically. Histologically, these adenomas are classified as exhibiting features of high-grade or low-grade dysplasia if the lesions do not extend into the muscularis mucosa, and as invasive adenocarcinoma if they do. Incorrect Answers: A, B, C, and D. Hyperplastic polyps (Choice A) are benign polyps that do not cause an increased risk of malignancy. They are usually small, located in the rectosigmoid colon, and are not associated with any polyposis syndromes. Inflammatory pseudopolyps (Choice B) may be seen with inflammatory bowel diseases including ulcerative colitis or Crohn disease and represent normal, non-inflamed tissue surrounded by eroded mucosa and granulation tissue. They develop secondary to chronic inflammation of the colon. This patient has no reported history of underlying inflammatory bowel disease, and on the gross specimen the surrounding colonic mucosa appears normal. Juvenile polyps (Choice C) constitute benign hamartomas and are most common in children between age two and ten. Painless rectal bleeding is characteristic. In patients with more than ten polyps, consideration should be given to the presence of a familial polyposis syndrome, but isolated polyps are generally benign and confer no increased risk of malignancy. This patient's age essentially eliminates this diagnosis. Peutz-Jeghers syndrome (Choice D) is an autosomal dominant syndrome that is characterized by hamartomatous polyps in the colon and pigmented macules in the mouth, lips, hands, and genitalia. It is associated with increased risk of breast and gastrointestinal tract cancers (eg, colorectal, stomach, small bowel, pancreatic). This older patient with a solitary mass is unlikely to have Peutz-Jeghers syndrome. Educational Objective: Tubular adenomas can predispose to the development of CRC, which may present with occult, chronic lower GI bleeding. Tubular adenomas are classified by size and histologic features, with the type of intervention (eg, colonoscopic removal vs surgical) based upon these features. II Previous Next Score Report Lab Values Calculator Help Pause

88 Exam Section 2: Item 38 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 38. A 3-year-old boy is brought to the physician because of a 1-month history of pale skin. His parents are of European descent. He has no personal or family history of major medical illness. Physical examination shows pallor. Laboratory studies show: Hemoglobin Hematocrit 8 g/dL (N=11-15) 24% (N=28-45) 34% Hb/cell (N=31-36) 90 pm3 (N=77-98) Mean corpuscular hemoglobin concentration Mean corpuscular volume A photomicrograph of a peripheral blood smear is shown. Genetic testing is most likely to show which of the following findings in this patient? A) Heterozygous mutation in the ankyrin gene B) Heterozygous mutation in the a-globin gene C) Heterozygous mutation in the B-globin gene D) Homozygous mutation in the ankyrin gene E) Homozygous mutation in the a-globin gene F) Homozygous mutation in the B-globin gene

A. A heterozygous mutation in the ankyrin gene leading to hereditary spherocytosis (HS) most likely explains this patient's normocytic anemia and peripheral smear showing spherocytes and reticulocytes. HS can occur if there are mutations in one or more proteins that link the erythrocyte inner membrane skeleton to the outer lipid bilayer, and include mutations in spectrin, ankyrin, band 3, and band 4.2. These mutations are inherited in an autosomal dominant pattern, which explains why this patient would only require one mutation to exhibit the phenotype. Spherocytosis results from the progressive loss of the erythrocyte membrane over time and gives the erythrocytes a spherical shape as the surface area decreases but the volume of the cell remains unchanged. This impairs the erythrocyte's ability to conform in small vessels of the microcirculation, especially within the spleen, where spherocytes accumulate. Patients often present with splenomegaly, hemolytic anemia, and pigmented gallstones. Peripheral smear will show spherocytes and, if there is a high degree of hemolysis, reticulocytes, which appear as larger, pale blue erythrocytes. Definitive management involves splenectomy. Incorrect Answers: B, C, D, E, and F. Heterozygous mutation in the a-globin gene (Choice B) describes a-thalassemia trait (silent carrier), which presents with normal hemoglobin levels, normal hemoglobin electrophoresis, and microcytosis. There are four a-globin genes organized as pairs on chromosome 16, with one pair inherited from each parent. The severity of anemia is determined by the number and the nature of the mutations, with a greater number of deletions resulting in more severe clinical phenotypes. Heterozygous mutation in the B-globin gene (Choice C) causes B-thalassemia minor. In comparison to the four alleles of the a-globin gene, there are only two B-globin alleles. Patients are typically asymptomatic or only mildly symptomatic. Peripheral smear demonstrates microcytosis and target cells but not spherocytes. Homozygous mutation in the ankyrin gene (Choice D) is associated with the sporadic form of HS, although mutations can be inherited in an autosomal recessive manner. These are less commonly encountered than autosomal dominant mutations. Homozygous mutation in the a-globin gene (Choice E) causes a-thalassemia, which in the case of a two-allele deletion is characterized by mild microcytic anemia and hemoglobin electrophoresis with a small concentration of Hb-Barts (y4). These patients are mostly asymptomatic. There would not likely be spherocytes on peripheral smear. Homozygous mutation in the B-globin gene (Choice F) causes B-thalassemia major, which is characterized by severe, transfusion dependent, microcytic anemia and signs of extramedullary hematopoiesis such as frontal bossing. Educational Objective: Autosomal dominant mutations in ankyrin, spectrin, and band erythrocyte membrane proteins cause hereditary spherocytosis through loss of the normal connection between the inner membrane skeleton and the outer lipid bilayer. Clinical symptoms can include jaundice, splenomegaly, and hemolytic anemia. A peripheral blood smear typically exhibits spherocytes and reticulocytes. Definitive treatment is with splenectomy. %3D Previous Next Score Report Lab Values Calculator Help Pause

39 Exam Section 1: Item 39 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 39. A 67-year-old man with poorly controlled unstable angina is about to undergo coronary angiography with stent placement. Prior to the procedure, treatment is initiated with aspirin and a drug that inhibits platelet interaction with fibrinogen. This drug is most likely which of the following? A) Abciximab B) Celecoxib C) Cilostazol D) Clopidogrel O E) Dipyridamole

A. Abciximab 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. Acute coronary syndrome most commonly occurs secondary to atherosclerotic plaque rupture and thrombus formation in the coronary arteries, which occludes the vessel. Deposition of cholesterol in the endothelial walls promotes inflammatory cell migration and formation of an atherosclerotic plaque characterized by fibroblasts, smooth muscle cells, and lipid-laden macrophages that transform into foam cells. A fibrous cap forms over the plaque, which may rupture due to stress depending on its thickness and stability. This exposes subendothelial collagen and initiates thrombus formation. Antiplatelet agents are indicated in the treatment of acute coronary syndrome to reduce the risk of thrombus progression. Complications of abciximab therapy include 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. Incorrect Answers: B, C, D, and E. Celecoxib (Choice B) is a selective cyclooxygenase 2 inhibitor that demonstrates activity in inflammatory cells and the vascular endothelium. It results in the decreased synthesis and release of arachidonic acid derivatives including thromboxane A2, which stimulates platelet activation. Thromboxane is not involved in platelet binding to fibrinogen. Cilostazol (Choice C) and dipyridamole (Choice E) are phosphodiesterase IIl inhibitors, which reduce the degradation of cyclic adenosine monophosphate in platelets and vascular smooth muscle cells resulting in the inhibition of platelet aggregation and vasodilation. Cilostazol is used in the treatment of claudication. Clopidogrel (Choice D) binds to the adenosine diphosphate receptor on platelets (also called the P2Y12 receptor), which inhibits platelet aggregation. It is commonly used in the treatment of acute coronary syndrome and coronary artery stenting. Abciximab directly inhibits platelet interaction with fibrinogen via antagonism of glycoprotein Ilb/Illa. Educational Objective: Antiplatelet therapy is indicated in the setting of acute coronary syndrome to reduce thrombus progression. Abciximab directly inhibits platelet binding with fibrinogen through the blockade of glycoprotein Ilb/Illa on the platelet surface. %3D Previous Next Score Report Lab Values Calculator Help Pause

3 Exam Section 1: Item 3 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 3. Moving the forearm against resistance from palm-down to palm-up (supination) position requires the use of which of the following muscles? A) Biceps brachii B) Brachialis OC) Triceps D) Flexor carpi radialis E) Pronator teres

A. The biceps brachii muscle has two main actions, flexion of the elbow joint and supination of the forearm. The biceps brachii contains two proximal heads, with the short head attaching to the coracoid process of the scapula and the long head entering the shoulder joint and attaching to the supraglenoid tubercle. The distal biceps tendon inserts on the bicipital tuberosity of the proximal radius. Because of its orientation crossing the elbow joint, contraction of this muscle causes elbow flexion. Its eccentric insertion on the proximal radius allows for it to wind around the radius during pronation and unwind when contracted from around the proximal radius during supination. Incorrect Answers: B, C, D and E. The brachialis muscle (Choice B) originates on the anterior surface of the humerus and crosses the elbow inserting on the tuberosity of the ulna. It does not wrap around the ulna and the ulna does not rotate. Because of this, it does not contribute to supination or pronation. The triceps muscle (Choice C) serves to extend the elbow joint. Proximally, it originates from the infraglenoid tubercle of the scapula (long head), just proximal to the radial groove (lateral head), and just distal to the radial groove (medial head). Distally, it inserts on the olecranon process of the ulna. Contraction of this muscle extends the elbow and does not contribute to rotation. Flexor carpi radialis (Choice D) originates on the medial epicondyle of the humerus and inserts on the second and third metacarpal bones. This allows for flexion of the wrist. Pronator teres (Choice E) is a muscle of the proximal forearm that extends from the medial supracondylar ridge of the humerus and inserts on the lateral aspect of the radius. Contraction along this axis will promote pronation, not supination. Educational Objective: The biceps brachii muscle has two main functions. It serves to supinate the forearm through its winding mechanism around the proximal radius and bicipital tuberosity. It flexes the elbow as the muscle and tendon cross the elbow joint anteriorly. %3D Previous Next Score Report Lab Values Calculator Help Pause

29 Exam Section 1: Item 29 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 29. A 50-year-old man comes to the emergency department because of a 2-week history of progressive shortness of breath. His pulse is 90/min, respirations are 26/min, and blood pressure is 120/80 mm Hg. Physical examination shows no other abnormalities. Laboratory studies show: Arterial Pco2 Arterial Po2 Arterial O, content Mixed venous Po2 Mixed venous 0, content 8 vol% (N=10%-16%) 30 mm Hg 96 mm Hg 12 vol% (N=17%-21%) 36 mm Hg Which of the following is the most likely explanation for these findings? A) Anemia B) Drug-induced alveolar hypoventilation C) Residence at a high altitude D) Severe regional mismatching of alveolar ventilation and pulmonary capillary perfusion E) Voluntary hyperventilation

A. The differential for dyspnea is broad and encompasses a range of disorders that involve impaired delivery of oxygen to tissue and/or reduced elimination of carbon dioxide from the body. The laboratory studies in this case indicates a reduced arterial and venous oxygen content. The oxygen content of the blood is a function of the oxygen carrying capacity (essentially the hemoglobin concentration), percent saturation of hemoglobin, and partial pressure of dissolved molecular oxygen (Po,). The equation to compute oxygen content is thus: Oxygen content = 1.34*[Hemoglobin]*(Arterial Oxygen Saturation) + 0.003*(Arterial Po,). The amount of dissolved oxygen is negligible compared to the oxygen transported by hemoglobin. The patient in this case has reduced oxygen content in the arterial and mixed venous circulation with a normal arterial Po, In the absence of a hemoglobinopathy (eg, methemoglobinemia, carboxyhemoglobinemia), the patient's oxygen saturation is expected to be normal. The most likely diagnosis is anemia with a decreased hemoglobin concentration. The arterial Pco, is decreased indicating hyperventilation, which is expected in the setting of decreased oxygen delivery to tissue. Incorrect Answers: B, C, D, and E. Drug-induced alveolar hypoventilation (Choice B) would result in an increased arterial Pco, Potential etiologies include central nervous system depressants such as opioid analgesics and benzodiazepines. Residence at a high altitude (Choice C) would be expected to result in a decreased Po, with adaptive changes that maintain an adequate oxygen carrying capacity. These changes include secondary erythrocytosis with increased hemoglobin concentration and increased levels of 2,3-bisphosphoglyceric acid, which stabilizes the deoxygenated state of hemoglobin and promotes increased oxygen release to tissue. Severe regional mismatching of alveolar ventilation and pulmonary capillary perfusion (Choice D) occurs when either ventilation or perfusion to a region of lung is impaired. An example is pulmonary embolism, in which a region of ventilated lung has obstructed blood flow. Voluntary hyperventilation (Choice E) results in a respiratory alkalosis, with a decreased Pco, Symptoms include dizziness, weakness, and syncope. The oxygen carrying capacity of the blood would not be reduced in the absence of other factors. Educational Objective: Delivery of oxygen to tissue is largely dependent on the hemoglobin concentration as it is the primary transporter of oxygen in the blood. Other contributing factors include the oxygen saturation of hemoglobin, and, less significantly, the partial pressure of dissolved molecular oxygen. II Previous Next Score Report Lab Values Calculator Help Pause

30 Exam Section 1: Item 30 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 30. An investigator is studying the effects of triiodothyronine (T3) and thyroxine (T) in hepatocytes in an experimental animal model. Which of the following best describes the action of these thyroid hormones on this target tissue? A) Both T3 and T, bind to the melanocortin 2 receptor on the cell surface B) Both T3 and T4 enter the nucleus C) T3 is converted to T in the cytosol D) Thyroid hormone receptors preferentially bind T, over T3

B. Both T3 (triiodothyronine) and T4 (thyroxine) are hormones that act on nuclear receptors, requiring them to enter the target cell to exert effects. Unlike other lipophilic hormones, thyroid hormones contain charged amino acids that prevent passive diffusion across the cellular membrane and thus enter by facilitated diffusion. Thyroid hormone transporters transport both T3 and T4 into the cell to reach their receptors. The thyroid hormone receptors are nuclear receptors that contain DNA-binding domains. Nuclear receptors can initially be in either the cytosol or nucleus. 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. Incorrect Answers: A, C, and D. Binding to the melanocortin 2 receptor on the cell surface (Choice A) does not occur with either T3 or T4. The melanocortin 2 receptor is also known as the adrenocorticotropic hormone (ACTH) receptor and is specific for ACTH. This receptor is a G protein-coupled receptor and does not actively transport ACTH inside the cell. Conversion of T3 to T4 in the cytosol (Choice C) does not occur. T4 is the less active form of thyroid hormone and is converted to T3 in target cells. Once in the cell nucleus, T3 preferentially binds the receptor with greater affinity than T4 (Choice D), although both hormones are capable of binding and activating the receptor. Educational Objective: T3 and T4 act on nuclear receptors, requiring them to enter the target cell to exert effects. Unlike other lipophilic hormones, thyroid hormones contain charged amino acids that prevent passive diffusion across the cellular membrane. Thyroid hormone transporters transport both T3 and T4 into the cell to reach their receptors. %3D 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 physician wishes to determine the proportion of newborns delivered at a local hospital who had a diagnosis of congenital heart disease within the past year. Which of the following statistical measurements best describes these data? A) Attributable risk B) Incidence C) Odds ratio OD) Prevalence E) Relative risk

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. In this case, the incidence would be the number of newborns with congenital heart disease divided by the total number of newborns in the study sample within the past year. For example, if 20 newborns are diagnosed with congenital heart disease out of a total of 1000 newborns delivered at the local hospital within the past year, the incidence would be 2%. Incorrect Answers: A, C, D, and E. Attributable risk (Choice A) is a representation of the amount of risk that is associated with a particular exposure. Formally, its definition is the incidence rate in the exposed group minus the incidence rate in the control group. For example, if over the course of a year, in a group of 100 smokers there are five myocardial infarctions (incidence rate of 5%), and in a group of 100 non-smokers there are two myocardial infarctions (incidence rate of 2%), the risk attributable to smoking would be 5% minus 2%, or 3%. Odds ratio (Choice C) is a comparison of the odds of an outcome occurring in the exposed group with the odds of that outcome occurring in a nonexposed control group. Using the example of smokers and non-smokers referenced in Choice A, the odds of myocardial infarction in the smoking group is 5/95 = 0.053. The odds of myocardial infarction in the non-smoking group is 2/98 = 0.0204. The odds ratio would simply be the odds of myocardial infarction in the smoking group divided by the odds of myocardial infarction in the non-smoking group (5/95)(2/98)) = 2.58. %3D Prevalence (Choice D) is calculated as the ratio of the number of people with a disease divided by the total number of at-risk persons in a population at a particular point in time. This is also known as point prevalence or disease frequency. For example, if a survey was conducted of a population of 1000 persons and 100 of these individuals were identified as having heart disease, the point prevalence of heart disease would be 100/1000 = 0.10, or 10%. Relative risk (Choice E) compares the risk of an outcome in one group with the risk of an outcome in another group and is often used in cohort studies. It is similar to an odds ratio and can be confused with an odds ratio. The odds ratio can be used to approximate relative risk when the disease or outcome state is rare. An example calculation of relative risk is as follows. In the described population of smokers referenced in Choices A and C, the risk of myocardial infarction in the non-smoking group is 2/100 = 0.020, or 2%. The risk of myocardial infarction in the smoking group is 5/100 = 0.05, or 5%. The relative risk is, therefore, ((5/100) /(2/100)) = 2.50, meaning that the risk of myocardial infarction in the smoking group is 2.5 times the risk in the non-smoking group. Since, in this example, myocardial infarction is rare, the small disease assumption is valid, and the odds ratio approximates the relative risk (2.58-2.50). Educational Objective: Incidence is a measure of the rate of occurrence of new disease in a population at-risk, which is distinct from prevalence, the measure of the current amount of disease burden in a population. Odds ratio is distinct from relative risk, although if disease burden is rare, the odds ratio may be used to approximate relative risk. %3D Previous Next Score Report Lab Values Calculator Help Pause

95 Exam Section 2: Item 45 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 45. A 56-year-old woman has recently diagnosed carcinoma of the breast. An x-ray of the chest shows a tumor next to the right side of the heart. An enhanced CT scan with the tumor invading the pericardium is shown. Which of the following structures is most likely involved? A) Coronary sinus B) Greater splanchnic vein C) Right phrenic nerve D) Right vagus nerve E) Thoracic duct

C. The right phrenic nerve originates from the cervical spinal cord at the levels of C3-C5, courses inferiorly with the internal jugular vein, and then crosses anterior to the subclavian artery before descending through the superior thoracic aperture to enter the thorax. In the thorax, it descends anteriorly along the right lung root and along the pericardium overlying the right atrium of the heart. It exits the thorax through the inferior vena cava hiatus to provide motor innervation to the diaphragm. The tumor as seen in this case is invading the pericardium at the level of the right atrium, which can be identified by its anterior and lateral location in the mediastinum. The right atrium abuts the right lung and makes up the right heart border. It is located posterior to and to the right of the sternum, whereas the right ventricle holds an anterior and inferior position, directly behind the sternum. Incorrect Answers: A, B, D, and E. The coronary sinus (Choice A) is located on the posterior surface of the heart and runs within the left atrioventricular groove before draining into the right atrium. It originates from the oblique vein of the left atrium and the great cardiac vein. The lesion in this case is anterior and to the right side of the heart. Greater splanchnic vein (Choice B) involvement is unlikely given the tumor location. The splanchnic venous circulation refers to drainage of the gastrointestinal tract from the lower esophagus to the upper two-thirds of the rectum, which includes the hepatic portal vein, mesenteric veins, and splenic vein. The splanchnic circulation drains into the liver inferior to the diaphragm. The right vagus nerve (Choice D) originates in the medulla and exits the cranium through the jugular foramen before passing anterior to the subclavian artery and into the thorax. The right recurrent laryngeal nerve branches off and loops underneath the right subclavian artery. The vagus nerve continues its descent through the thorax adjacent to the esophagus and enters the abdomen through the esophageal hiatus. Its location is more posterior than the lesion in the radiograph. The thoracic duct (Choice E) passes through the aortic aperture of the diaphragm and ascends adjacent to the thoracic aorta and azygos vein. It drains into the venous system near the junction of the left subclavian and left internal jugular veins. Educational Objective: The right phrenic nerve courses along the pericardium in the region of the right atrium. Infiltrative lesions in this area may result in phrenic nerve dysfunction. Previous Next Score Report Lab Values Calculator Help Pause

89 Exam Section 2: Item 39 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 39. A 24-year-old woman is brought to the emergency department 15 minutes after the sudden onset of shortness of breath following a bee sting. Her pulse is 130/min, and blood pressure is 70/30 mm Hg. Three hours later, her blood pressure returns to normal following administration of intravenous fluids, corticosteroids, antihistamines, and epinephrine. The next day, she has minimal urine output. Which of the following areas of the kidney is likely to be most affected with this patient's prolonged hypotension? O A) Glomerular epithelial cells B) Loop of Henle C) Medullary interstitium D) Mesangial cells E) Proximal tubules

E. Decreased urinary output following an episode of hypotension is consistent with a diagnosis of acute tubular necrosis (ATN). The cells of the proximal convoluted tubule, which are responsible for a large portion of ion resorption from the filtrate, are highly metabolically active and therefore especially sensitive to ischemic injury. ATN progresses through an initial oliguric phase which may last for several weeks before recovery, which, if occurring, is marked by a subsequent polyuric phase. Ischemic ATN may occur secondary to any type of shock state, in this patient, anaphylaxis lead to disseminated vasodilation and hypotension, impairing renal perfusion. Cardiogenic, obstructive, hypovolemic, and distributive shock etiologies can all result in renal hypoperfusion. ÅTN may also occur secondary to nephrotoxic exposures such as heavy metals, antibiotics, and toxic alcohols. Urinalysis typically demonstrates muddy brown casts. Incorrect Answers: A, B, C, and D. Glomerular epithelial cells (Choice A), also known as podocytes, form part of the filtration barrier of the glomerulus. Podocyte loss occurs in diabetic nephropathy and is a hallmark of minimal change disease. Podocyte loss is not characteristic of ATN. The loop of Henle (Choice B), especially the thick ascending limb, is also metabolically active and may be injured by ischemic injury. However, injury of the loop of Henle occurs less frequently than injury of the proximal convoluted tubule. The medullary interstitium (Choice C) is the site of interstitial nephritis, which frequently occurs subsequent to medication exposure and to pyelonephritis. Mesangial cells (Choice D) form the juxtaglomerular apparatus and are important in the regulation of the renin-angiotensin system. Mesangial cells are classically observed to expand and proliferate in the context of diabetic nephropathy. Educational Objective: Acute tubular necrosis frequently results from ischemic injury to the nephron and is characterized by damage to the proximal convoluted tubule. The sloughing of tubular cells leads to the formation of muddy brown casts. Acute tubular necrosis presents initially with oliguria before progressing to polyuria when the cells regenerate. %3D Previous Next Score Report Lab Values Calculator Help Pause

70 Exam Section 2: Item 20 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 20. A 45-year-old woman comes to the physician because of a 6-month history of shortness of breath with exertion and a nonproductive cough. She sometimes has difficulty swallowing and often has heartburn, especially if she lies down after a meal. She adds that her fingers have swollen, and she has had to get her wedding ring resized. Her fingers also become white and painful in cold weather or cold water. Her pulse is 75/min, respirations are 20/min, and blood pressure is 150/100 mm Hg. Physical examination shows tight, smooth facial skin without wrinkles. Additional testing is most likely to show which of the following sets of cardiovascular changes in this patient? Mean Pulmonary Artery Pressure Coronary Vascular Resistance Left Ventricular Diastolic Compliance OA) ↑ ↑ ↑ B) ↑ ↑ C) ↑ ↑ D) ↑ E) ↑ ↑ F) ↑ G) ↑ H)

E. Systemic sclerosis (scleroderma) is an autoimmune disorder characterized by collagen deposition and progressive fibrosis of the skin, soft tissues, and internal organs as well as noninflammatory vasculopathy. There are multiple phenotypes ranging from localized scleroderma to diffuse disease. Multiple organ systems may be affected, with involvement of the renal, pulmonary, gastrointestinal, and cardiovascular systems being common. Cardiac fibrosis results in decreased left ventricular diastolic compliance and increased coronary vascular resistance. Interstitial pulmonary fibrosis classically presents with dyspnea on exertion and a nonproductive cough. Patients often present with an increased mean pulmonary artery pressure secondary to chronic hypoxic vasoconstriction and left ventricular dysfunction. Gastrointestinal manifestations include esophageal dysmotility with symptoms of dysphagia and acid reflux. Vascular manifestations include Raynaud phenomenon due to cutaneous vasospasm. The skin may be taut without wrinkles on physical examination. Females are typically affected more than males. Incorrect Answers: A, B, C, D, F, G, and H. Choices A, B, C, and D are incorrect as left ventricular diastolic compliance decreases in systemic sclerosis with cardiac involvement. Compliance refers to the change in volume of a system relative to a change in pressure. Replacement of normal cardiomyocyte fibers with collagen deposition and fibrosis results in a stiff and noncompliant ventricle. Choices C, D, G, and H are incorrect as both the cardiac and pulmonary involvement in systemic sclerosis results in pulmonary hypertension with an increased mean pulmonary artery pressure. Pulmonary hypertension may result from pulmonary fibrosis and hypoxia-induced pulmonary vasoconstriction, left heart disease, and/or concentric hypertrophy of the tunica intima of the pulmonary vasculature. Similarly, choices B, D, F, and H are incorrect as coronary vascular resistance tends to increase, not decrease, in systemic sclerosis with cardiac involvement due to concentric hypertrophy of the coronary artery wall. Educational Objective: Systemic sclerosis (scleroderma) is an autoimmune disorder that may affect multiple organ systems. It is characterized by collagen deposition and progressive fibrosis of tissue as well as noninflammatory vasculopathy, which results in stiff, noncompliant organs and increased vascular resistance. %3D Previous Next Score Report Lab Values Calculator Help Pause

52 Exam Section 2: Item 2 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 2. An investigator is studying Bzadrenoreceptors in female experimental animals. During the experiment, epinephrine is injected intramuscularly into each animal, and the effects on Bradrenoreceptors are then observed. Which of the following physiologic effects is most likely to be observed in these animals? A) Increased myocardial contractility B) Internal urethral sphincter contraction C) Lipolysis D) Pilomotor contraction E) Pupillary dilation F) Uterine relaxation

F. Epinephrine is a direct sympathomimetic that exerts its effects through stimulation of adrenergic receptors, with a greater affinity for B-adrenergic receptors than for a-adrenergic receptors. B-adrenergic receptors have three isotypes, which are responsible for different actions of the autonomic nervous system. B, receptors are primarily responsible for heart rate and myocardial contractility, both of which are increased with the administration of epinephrine. B2 receptors cause smooth muscle dilation in the bronchi and blood vessels, as well as decreased uterine contractility. Antagonists of this receptor can also be used to decrease aqueous humor production in glaucoma, whereas agonists can be used to decrease potassium concentration in hyperkalemia by inducing cellular uptake. Additionally, B2 receptor stimulation increases insulin release and glycogenolysis. B3 receptors are less common, but stimulation leads to thermogenesis and detrusor relaxation. Incorrect Answers: A, B, C, D, and E. Increased myocardial contractility (Choice A) would be seen in these animals due to stimulation of B1 receptors. However, this investigator is examining the effects of B2 receptors. Internal urethral sphincter contraction (Choice B) is affected by a-adrenergic receptors; specifically, stimulation of a, receptors leads to increased bladder sphincter contraction. Lipolysis (Choice C) is also affected by B-adrenergic receptors, with B2 increasing lipolysis in adipose tissue, although to a lesser extent than B3 receptors. B2 receptors have a stronger effect on smooth muscle relaxation and would cause more prominent uterine relaxation than lipolysis. Pilomotor contraction (Choice D) is mediated by a, receptors. It would be unaffected by stimulation of B2 receptors. Pupillary dilation (Choice E) is mediated by a1 receptors. While it would be seen with epinephrine administration, it is not a direct effect of stimulation of B2 adrenergic receptors. Educational Objective: The stimulation of B2 adrenergic receptors causes smooth muscle dilation in the bronchi, blood vessels, and uterus, decreased plasma potassium concentration, and increased insulin release. %3D Previous Next Score Report Lab Values Calculator Help Pause

98 Exam Section 2: Item 48 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 48. Topical corticosteroid creams and phototherapy are often effective in the treatment of psoriasis. This effectiveness suggests a role for metabolism or nuclear binding of which of the following vitamins in the treatment of psoriatic lesions? A) Niacin OB) Vitamin B, (thiamine) C) Vitamin B2 (riboflavin) OD) Vitamin B6 (pyridoxine) E) Vitamin C F) Vitamin D G) Vitamin E H) Vitamin K

F. Psoriasis vulgaris is a common inflammatory skin condition caused by immune dysregulation leading to keratinocyte proliferation. The Th17 lymphocyte is central to the pathophysiology of psoriasis, which occurs secondary to the effects of some cytokines, such as interleukin-23 and tumor necrosis factor-a. Recent developments in the treatment of psoriasis have targeted these specific immune pathways with biologic medications. However, phototherapy with narrow-band ultraviolet band light and topical calcipotriene, a vitamin D analog, are frequently used in the management of psoriasis. Active vitamin D (1,25-OH2 vitamin D) binds a nuclear transcription factor, which subsequently impacts gene transcription. Vitamin D plays a role in immune modulation and thus improves the immune dysregulation seen in psoriasis. Clinically, plaque-type psoriasis vulgaris presents with thick, salmon-colored plaques on the extensor extremities with overlying silvery, white scale. Other forms of psoriasis include pustular psoriasis, psoriatic arthritis, and guttate psoriasis. Incorrect Answers: A, B, C, D, E, G, and H. Niacin (Choice A) is a component of nicotinamide adenine dinucleotide and nicotinamide adenine dinucleotide phosphate, which are used in various biochemical reduction-oxidation reactions. Niacin lowers serum cholesterol-VLDL concentration and raises serum HDL-cholesterol concentration, making it a potential treatment for dyslipidemia. It has also been used for skin cancer prevention in at-risk individuals. It is not a treatment of psoriasis. Vitamin B1 (Choice B), or thiamine, causes variable manifestations when deficient. Dry beriberi presents as peripheral neuropathy with mental status change. Wet beriberi exhibits a dilated cardiomyopathy leading to heart failure. Wernicke encephalopathy classically presents as confusion, ataxia, and ophthalmoplegia. A deficiency of thiamine does not cause psoriasis. Vitamin B2 (Choice C), or riboflavin, is another cofactor in reduction oxidation reactions in the process of glucose metabolism. A deficiency of riboflavin may cause cheilitis and corneal vascularization. It does not have a role in psoriasis pathogenesis or management. Vitamin B6 (Choice D), or pyridoxine, contributes to the synthesis of histamine, hemoglobin, and neurotransmitters including epinephrine, norepinephrine, dopamine, serotonin, and y-aminobutyric acid. Deficiency commonly presents with peripheral neuropathy, sideroblastic anemia, glossitis, and seizures (especially in isoniazid use). Deficiency may also cause dermatitis, but not psoriasis. Vitamin C (Choice E), or ascorbic acid, is a water-soluble antioxidant found in fruits and vegetables. Deficiency of this vitamin causes the constellation of symptoms known as scurvy, which is characterized by petechiae, perifollicular hemorrhage, bruising, poor wound healing, and small, curly, fragile hairs. It is not currently thought to play a role in the treatment of psoriasis. Vitamin E (Choice G) is an antioxidant that protects cells from free radical damage. It is included in many topical products intended for skin care to help mitigate the effects of ultraviolet radiation. It does not have a role in psoriasis management. Vitamin K (Choice H) plays a critical role in the synthesis of hepatic coagulation proteins; it is oxidized in the liver during the carboxylation of glutamic acid residues on coagulation factors II, VII, IX, X, and proteins C and S. It does not contribute to the pathogenesis of psoriasis. Educational Objective: Vitamin D binds to its nuclear receptor causing increased transcription of immune regulatory genes. It is used in the treatment of psoriasis, which is caused by immune dysregulation leading to keratinocyte proliferation, in the form of phototherapy and topical calcipotriene. %3D Previous Next Score Report Lab Values Calculator Help Pause

44 Exam Section 1: Item 44 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 44. A 5-year-old girl is brought to the physician because of listlessness, fatigue, and dull pain in the right upper quadrant of the abdomen. Her height and weight are below the 25th percentile. Laboratory findings indicate that the content of her B-globin chain is 15% to 20% of normal. Sequencing of the B-globin gene shows a point mutation in a sequence 3' to the coding region in which AATAAA is converted to AACAAA. Consequently, the amount of mRNA for B-globin is decreased to 10% of normal. Which of the following functions in MRNA synthesis and processing is most likely encoded by the sequence AATAAA? A) Capping with GTP B) Cleavage and polyadenylation C) Silencing of the promoter D) Splicing of the initial MRNA transcript in the nucleus E) Transport of the mRNA out of the nucleus

B. Cleavage and polyadenylation of the pre-MRNA molecule is most likely encoded by the sequence AATAAA, and a point mutation in the B-globin gene leading to an AACAAA sequence would disrupt a necessary step in pre-mRNA processing known as polyadenylation. Polyadenylation is the addition of a string of adenine nucleotides to the 3' end of the MRNA molecule that prevents degradation and permits transport out the nucleus. The DNA sequence AATAAA becomes AAUAAA on pre-MRNA. During transcription, this signal is recognized by the cleavage enzyme CPSF (cleavage and polyadenylation specificity factor), which is bound to RNA polymerase Il and signals for the polymerase to cease transcription. The pre-MRNA molecule is then cleaved at the 3' end and binding of a second enzyme, polyadenylate polymerase, catalyzes the addition of a long string of adenine nucleotides onto the end of the pre-mRNA molecule. This polyadenylate tail allows for transport out of the nucleus and into the cytoplasm where the MRNA molecule can be translated into proteins. A non-functional polyadenylation signal would result in failure to cleave pre-MRNA or failure to polyadenylate the transcript leading to early degradation within the nucleus. Incorrect Answers: A, C, D, and E. Capping with GTP (Choice A) occurs at the beginning of transcription at the 5' end of the nascent MRNA. It is recognized by ribosomes in the cytoplasm and facilitates initiation of translation. A mutation would result in the synthesis of a normal MRNA transcript but failure of translation. Silencing of the promoter (Choice C) would result in decreased expression of the B-globin gene, but the promoter sequence is not encoded by the sequence AATAAA. Mutations in the promoter region are also known to cause B-thalassemia. Splicing of the initial mRNA transcript in the nucleus (Choice D) is the process by which introns are removed and exons are connected. Splice sites exist at the end of an intron and the beginning of an exon. Splice sites are recognized by small nuclear ribonucleoproteins, which are RNA-protein complexes that combine with the pre-MRNA to form a spliceosome. Mutations in splice sites can result in the aberrant inclusion of introns or exclusion of exons. Transport of the MRNA out of the nucleus (Choice E) occurs through nuclear pores. After transcription, mRNA molecules diffuse freely through the nucleus and into the cytoplasm through these channels. While the polyadenylate tail affects overall stability of the molecule and facilitates transport, the sequence that signals polyadenylation is upstream from this process. Educational Objective: The AATAAA sequence encodes the polyadenylation signal. Mutations here would result in failure to polyadenylate the 3' end of the mRNA transcript and lead to early degradation within the nucleus prior to translation. %3D Previous Next Score Report Lab Values Calculator Help Pause

75 Exam Section 2: Item 25 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 25. A study is designed to measure the impact of exercise on the incidence of myocardial infarction. Subjects are enrolled in the study and divided into two groups based on their self-reported exercise habits. At the end of the study, subjects who reported exercising have half the incidence of myocardial infarction compared with the subjects who did not exercise. Which of the following best describes this study design? A) Case-control B) Case series C) Cohort D) Cross-sectional E) Randomized clinical trial

C. A cohort study 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. In this study, patients are observed over time and divided into groups based on exercise (exposure) and are then followed over time. Rates of myocardial infarction (outcome) are then compared between the two groups. The relative risk of myocardial infarction (a comparison of incidence) is then calculated to be twice as high in the non-exercise group. This represents a prospective cohort study design. Incorrect Answers: A, B, D, 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. 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. A case series (Choice B) 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 conducting of larger cohort studies or randomized trials with sufficient power. A cross-sectional study (Choice D) seeks to identify the prevalence of a condition at a specific point in time. In addition, the risk factor and the outcomes are measured simultaneously. A cross-sectional study does not follow patients over time. All information is collected at a single time point. A randomized clinical trial (Choice E) 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. Educational Objective: A cohort study follows patients over a period of time, either retrospectively or prospectively, and seeks to investigate the impact of exposures or risk factors on an outcome of interest. %3D 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. Aminoglycoside antibiotics are used for their synergistic action against bacteria, in combination with other agents. These antibiotics demonstrate in vitro synergy for several bacterial species when combined with which of the following classes of antibiotics? A) Fluoroquinolones B) Macrolides C) Penicillins D) Rifamycins O E) Tetracyclines

C. Aminoglycosides include gentamicin, neomycin, amikacin, tobramycin, and streptomycin. Their bactericidal function comes from the inhibition of the 30S subunit of the bacterial ribosome, which precludes protein synthesis by causing misreading of mRNA. Aminoglycosides work synergistically with penicillins, meaning that the combined effect of the two classes is stronger than the effect of either class alone. The penicillin group of antibiotics includes penicillinase-sensitive penicillins (penicillin G, ampicillin, amoxicillin), penicillinase-resistant penicillins (oxacillin, nafcillin, dicloxacillin), and anti-pseudomonal penicillins (ticarcillin, piperacillin). These all inhibit peptidoglycan cross-linking of the bacterial wall. Aminoglycosides also demonstrate synergistic activity with monobactams such as aztreonam, which also target peptidoglycan cross-linking function. However, they do require oxygen for their uptake into the bacterial cell so are ineffective against anaerobes. Bacteria may also develop resistance to this class of antibiotics due to the inactivation of the drug by bacterial transferase enzymes, which slightly modify the aminoglycoside structure through acetylation or phosphorylation. Aminoglycoside usage may be complicated by nephrotoxicity, ototoxicity, or neuromuscular damage. They should not be used during pregnancy as they are also a teratogen. Incorrect Answers: A, B, D, and E. Fluoroquinolones (Choice A) include ciprofloxacin, levofloxacin, and moxifloxacin. They inhibit prokaryotic DNA gyrase (also known as topoisomerase). Potential side effects include vascular damage, cartilage damage, tendonitis, or tendon rupture. They do not work synergistically with aminoglycosides. Macrolides (Choice B) include antibiotics such as azithromycin and erythromycin. These antibiotics also inhibit protein synthesis but instead by inhibiting the 50S subunit of the ribosome. Use of macrolides with aminoglycosides would not provide additional therapeutic benefit, particularly given that these classes of drugs have similar mechanisms of action. Rifamycins (Choice D) inhibit the RNA polymerase required to transcribe bacterial DNA. They do not lead to synergistic effects when used with aminoglycosides. Tetracyclines (Choice E) include tetracycline, doxycycline, and minocycline. Like aminoglycosides, these interfere with the 30S subunit of the ribosome. However, rather than causing misreading of MRNA, they prevent attachment of the aminoacyl-tRNA. They do not lead to a synergistic effect when used with aminoglycosides. Educational Objective: Aminoglycosides include gentamicin, neomycin, amikacin, tobramycin, and streptomycin. They inhibit the 30S subunit of the bacterial ribosome. When used with peptidoglycan linking antibiotics such as the penicillin class, they lead to a synergistic effect and result in improved bacterial killing. %3D Previous Next Score Report Lab Values Calculator Help Pause

97 Exam Section 2: Item 47 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 47. A 24-year-old woman comes to the physician for advice about contraceptive methods. She is recently married and is not interested in having children until her mid 30s. Which of the following contraceptives carries the highest risk for interference with fertility in this patient later in life? A) Cervical cap with spermicidal jelly B) Condoms and spermicidal foam C) IUD D) Oral contraceptive O E) Progestin implant

C. An IUD is a long-acting reversible contraceptive option that does not require active compliance on the part of the patient. Both copper and levonorgestrel IUDS are available. The copper IUD provides a contraceptive method without the use of hormones. Copper is spermicidal and inhibits sperm motility and the acrosomal reaction necessary for fertilization. As well, it induces local inflammatory changes of the endometrium and increases cervical mucus production, further decreasing the chance of successful sperm passage or embryonic implantation in the endometrium. Potential complications include increased menstrual flow and pain during menses. Hormonal IUDS also cause cervical mucous thickening and induce glandular atrophy of the endometrium, preventing implantation. IŪDS are a reliable form of contraception with failure rates as low as 1% or less. Depending on the type, they may be kept in place for up to ten years. The use of an IUD may slightly increase the chance of infertility later, but most women are able to conceive after removal of the device. Incorrect Answers: A, B, D, and E. Cervical cap with spermicidal jelly (Choice A) and condoms and spermicidal foam (Choice B) are both forms of a barrier contraceptive combined with a spermicide. A cervical cap or a diaphragm fits on top of the cervix and anterior wall of the vagina. It must be placed prior to intercourse and left in for six to eight hours after use. Diaphragms and condoms are not as effective at preventing pregnancy compared to an IUD but have no hormonal component and no impact on future fertility. Likewise, a condom must be placed prior to intercourse to be effective and has no impact on fertility. Oral contraceptives (Choice D), which are taken daily are a form of hormonal contraception. Hormonal contraceptives are contraindicated in patients with a history of thromboembolic disorders, liver dysfunction, and breast malignancy with estrogen or progesterone receptor positivity. They do not affect future fertility. Progestin implant (Choice E) is a long-acting hormonal contraceptive containing progestin, which is placed subcutaneously in the arm. This has the same contraindications as an oral hormonal contraceptive and leaves future fertility unaffected. Educational Objective: IUDS are one of the most reliable forms of contraception because they do not depend on patient compliance. The use of an IUD may slightly increase the chance of infertility later, but most women are able to conceive after removal of the device. Other side effects include cramping and irregular menstrual bleeding. %3D Previous Next Score Report Lab Values Calculator Help Pause

66 Exam Section 2: Item 16 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 16. A 60-year-old woman comes to the physician because of a 6-month history of pain in her hips and knees. Physical examination shows findings consistent with osteoarthritis, and the physician recommends ibuprofen. The patient refuses and asks about taking glucosamine. Which of the following responses by the physician is most appropriate? A) "Glucosamine hasn't been studied well enough for me to recommend it." B) "Glucosamine's side effects aren't listed. It may be more dangerous than we realize." C) "Ibuprofen has been proven effective for your condition." D) "What have you heard about using glucosamine to treat arthritis?" E) "Why did you come to me if you don't want to take what I recommend?" F) "You should really see a naturopathic doctor."

D. When a patient asks questions about naturopathic treatments such as glucosamine, physicians should initially ask open-ended questions to explore the patient's understanding and goals. It is possible that this patient does not understand the risks and benefits of ibuprofen versus glucosamine, and the physician can tailor further discussion to address these knowledge gaps. After determining that the patient understands the risks and benefits, the physician and patient should partner to develop a treatment plan that aligns with the patient's values. This plan may ultimately involve referral to a naturopathic doctor. Using non-judgmental, open-ended questioning can improve therapeutic alliance and help the physician and patient problem solve around how to best address this patient's symptoms. Incorrect Answers: A, B, C, E, and F. Immediately educating the patient about the absence of evidence behind glucosamine or the benefits of ibuprofen (Choices A, B, and C) may preclude the collaborative formation of a treatment plan. This strategy may prevent the physician from learning about the patient's knowledge gaps and goals. Asking the patient why they came to see the physician if they are not going to take what the physician recommends (Choice E) strikes a paternalistic and judgmental tone. Starting the conversation in this manner may prevent open, effective discussion about the options. Referring the patient to a naturopathic doctor (Choice F) may be ultimately indicated if the patient maintains that she prefers glucosamine. However, the physician should first assess the patient's understanding of the risks and benefits of both treatment options so that the physician can educate the patient and collaboratively formulate a treatment plan with the patient. Educational Objective: When a patient asks questions about naturopathic treatments, physicians should initially ask open-ended questions to explore the patient's understanding and goals. The physician can tailor further discussion to address the patient's knowledge gaps and goals. The physician may ultimately refer the patient to a naturopath. %3D Previous Next Score Report Lab Values Calculator Help Pause

79 Exam Section 2: Item 29 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 29. A 55-year-old woman is brought to the emergency department after being injured in a motor vehicle collision. She has an injury of the soft tissue of the face that prevents her from drinking from a glass without spilling the contents. Several days later she still has this problem. The photograph shows her attempting to purse her lips to whistle. Which of the following nerves is most likely damaged? A) Buccal branch of the facial nerve B) Inferior alveolar branch of the mandibular division of the trigeminal nerve C) Infraorbital branch of the maxillary division of the trigeminal nerve D) Mandibular branch of the facial nerve E) Pharyngeal branches of the vagus nerve

A. Damage to the buccal branch of the facial nerve is most likely responsible for this patient's inability to purse her lips or drink from a glass. The facial nerve performs diverse afferent and efferent functions and has five main extracranial branches: the temporal, zygomatic, buccal, marginal mandibular, and cervical branches. The buccal branch of the facial nerve runs along the parotid duct and innervates the orbicularis oris (purses lips), buccinator (holds cheeks flat to aid in chewing and blowing), levator labii superioris (elevat masticate. upper lip), and anguli oris (assists in smiling). Damage to the buccal branch therefore leads to difficulty in performing tasks that require closing or pursing the lips such as drinking from a glass or whistling and may also lead to the decreased ability to Incorrect Answers: B, C, D, and E. The inferior alveolar branch of the mandibular division of the trigeminal nerve (Choice B) mediates sensation of the lower teeth and gingivae along with the lower lip and skin of the chin. The infraorbital branch of the maxillary division of the trigeminal nerve (Choice C) provides sensory innervation to the lower eyelid, lateral inferior portion of the nose, and upper lip. This patient demonstrates motor rather than sensory dysfunction. The mandibular branch of the facial nerve (Choice D) supplies the depressor muscles of the lower lip (the depressor anguli oris, depressor labii inferioris, and mentalis). Consequently, damage to the mandibular branch leads to the decreased ability to open the mouth or smile. This patient demonstrates a decreased ability to close the mouth (purse the lips). The pharyngeal branches of the vagus nerve (Choice E) innervate the pharyngeal constrictor muscles and palatine muscles, which function to initiate swallowing and move a food bolus from the pharynx to the esophagus. This patient with damage to the buccal branch of the facial nerve would demonstrate difficulties in masticating, not swallowing. Educational Objective: The buccal branch of the facial nerve innervates several perioral muscles such as the orbicularis oris and buccinator muscles. Damage to the buccal branch of the facial nerve leads to difficulty pursing the lips and masticating. %3D Previous Next Score Report Lab Values Calculator Help Pause

63 Exam Section 2: Item 13 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 13. A previously healthy 10-year-old girl is brought to the physician by her mother because of a 6-week history of headache, nausea, and difficulty walking. An MRI of the brain shows a mass in the posterior fossa that is found to be an astrocytoma. This tumor developed from cells that normally serve which of the following functions in the brain? A) Formation of myelin sheaths in the central nervous system B) Phagocytosis of recycled synaptic terminal membrane C) Production and secretion of cerebrospinal fluid D) Termination of action potentials E) Transport of hormones from the cerebral spinal fluid to capillaries F) Uptake of amino acid neurotransmitters

F. Astrocytes are a subtype of glial cell involved in the uptake of amino acid neurotransmitters such as glutamate, y-aminobutyric acid (GABA), and glycine. Astrocytes support neurons by buffering the extracellular space, regulating energy metabolism, and reacting to neuronal injury. In 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. Pilocytic astrocytomas, the most common pediatric brain tumor, are benign brain tumors typically found in the posterior fossa. Cerebellar tumors may present with dizziness and symptoms of increased intracranial pressure (due to compression of the fourth ventricle by the tumor and consequent obstructive hydrocephalus) such as headache as well as cerebellar signs (eg, gait ataxia, nystagmus, and dysmetria). Treatment centers around surgical resection. Incorrect Answers: A, B, C, D, and E. Formation of myelin sheaths in the CNS (Choice A) is mediated by oligodendrocytes, which are glial cells that produce myelin. Oligodendrogliomas can occur in children but are less common than astrocytomas and are typically located supratentorial. Phagocytosis of recycled synaptic terminal membrane (Choice B) is performed by microglial cells, which are derived from precursor monocytes. Tumors derived from microglia are rare. The production and secretion of cerebrospinal fluid (CSF) (Choice C) is accomplished by the choroid plexus, which forms the blood-CSF barrier. The choroid plexus is a complex network of capillaries lined by different cells such as the ependymal cells, which form the permeable lining of the ventricles. These periventricular cells would not form tumors in the parenchyma of the posterior fossa. Termination of action potentials (Choice D) is mediated by ion channels on neurons. Once the intracellular space becomes depolarized, sodium channels inactivate, promoting hyperpolarization of the intracellular space by potassium efflux that is unopposed by sodium influx. This drives the neuron back to its resting potential, terminating the action potential. Transport of hormones from the CSF to the capillaries (Choice E) is controlled by arachnoid villi or by endothelial cells. Arachnoid villi are small protrusions of the arachnoid mater that drain CSF into the dural venous sinuses, whereas endothelial cells line capillaries (substances that diffuse into capillaries must cross through or between endothelial cells). Proliferations of the arachnoid villi (eg, arachnoid cysts) would appear as extra-axial masses rather than masses within the posterior fossa parenchyma. Educational Objective: Astrocytes are a subtype of glial cell involved in the uptake of amino acid neurotransmitters such as glutamate, which protects neurons from excitotoxicity. Pilocytic astrocytomas are benign pediatric tumors of the posterior fossa, which may present with signs of increased intracranial pressure and cerebellar dysfunction. %3D Previous Next Score Report Lab Values Calculator Help Pause

80 Exam Section 2: Item 30 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 30. A previously healthy 6-month-old boy is brought to the physician because of a cough for 1 week. Initially he had a low-grade fever, sneezing, congestion, and runny nose. He then developed a dry intermittent cough. The parents now note that with any startle the baby chokes and gasps. He has not had any immunizations. Physical examination shows paroxysms of "machine gun" -like coughing with a forced expiratory grunt at the end of coughing. Leukocyte count is 30,000/mm3 (70% lymphocytes). Neutrophil chemotaxis and oxidative metabolism are defective due to increased activity of which of the following enzymes? OA) Adenylyl cyclase B) Myeloperoxidase C) NADPH oxidase D) Phospholipase C E) Protein kinase C

A. Episodes of severe paroxysmal coughing in the setting of a respiratory illness is concerning for pertussis. Pertussis occurs due to an infection from Bordetella pertussis and classically presents in three stages. The catarrhal stage is first, which presents with fever, rhinorrhea, and a mild cough. This progresses to the paroxysmal stage after one to two weeks, in which patients experience violent bouts of extreme coughing, at times violent enough to cause secondary traumatic injury such as rib fractures, urinary incontinence, and pneumothorax. Often, such coughing fits are followed by syncope, emesis, or apnea. The paroxysmal stage can last for weeks to months. Symptoms resolve during the convalescent stage, which lasts from one to four weeks. Pertussis can be prevented with immunization, and is treated with antibiotics, typically macrolides. Pertussis toxin is a major virulence factor that inactivates the G¡ subunits of G protein-coupled receptors in the respiratory epithelium and systemically, leading to uninhibited adenylyl cyclase activity and increased intracellular cyclic adenosine monophosphate (CAMP) concentration. In neutrophils and macrophages, this results in impaired recruitment, migration, and production of inflammatory mediators. Incorrect Answers: B, C, D, and E. Myeloperoxidase (Choice B) deficiency is a common inherited immunodeficiency syndrome characterized by the inability to produce hypochlorous acid within phagolysosomes as part of the oxidative burst. Disease is typically mild and may present with recurrent Candida albicans infection. NADPH oxidase (Choice C) deficiency is found in chronic granulomatous disease, in which the absence of an effective oxidative burst predisposes to recurrent infections with catalase-positive organisms. Phospholipase C (Choice D) cleaves membrane bound phospholipids, leading to the formation of inositol triphosphate and diacylglycerol with further downstream signaling effects, including activation of protein kinase C (Choice E). Pertussis toxin does not affect this second messenger system. Educational Objective: Pertussis toxin is a major virulence factor of B. pertussis, the causative agent of pertussis. Pertussis toxin causes increased intracellular CAMP concentration through the inactivation of G¡ subunits of G protein-coupled receptors in the respiratory epithelium and systemically, leading to uninhibited adenylyl cyclase activity. This results in impaired recruitment, migration, and production of inflammatory mediators by inflammatory cells, including neutrophils. %3D Previous Next Score Report Lab Values Calculator Help Pause

81 Exam Section 2: Item 31 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment Y 31. A 22-year-old woman comes to the office because of a 3-day history of nonproductive cough. She also has a 1-week history of fatigue, progressive shortness of breath with exertion and while lying down, and swelling of her legs and feet. She delivered a male newborn via uncomplicated vaginal delivery 1 month ago. She has no history of major medical illness and takes no medications. Her temperature is 37.7°C (99.8°F), pulse is 104/min, respirations are 20/min, and blood pressure is 126/80 mm Hg. Bilateral basilar crackles are heard. There is 1+ edema of the lower extremities bilaterally. Which of the following is the most likely diagnosis? A) Amniotic fluid embolism B) Cardiomyopathy C) Major depressive disorder D) Pneumonia E) Pulmonary embolism F) Pulmonary fibrosis

B. Heart failure presents with dyspnea on exertion, reduced exercise tolerance, paroxysmal nocturnal dyspnea, orthopnea, shortness of breath, fatigue, weight gain, and peripheral edema. Physical examination often reveals jugular venous distension, pitting peripheral edema, bibasilar rales, a displaced apical impulse, and an S3 or S4 gallop. Laboratory studies typically show increased concentration of brain natriuretic peptide, while x-ray may reveal cardiomegaly and pulmonary edema. It can result from ischemic heart disease or non-ischemic cardiomyopathy (eg, alcohol use disorder, beriberi, Chagas disease, adverse effect of chemotherapy). This patient has peripartum cardiomyopathy, which is a form of non-ischemic cardiomyopathy leading to left ventricular failure. Peripartum cardiomyopathy typically presents after 36 weeks' gestation or within 5 months of delivery and is associated with a left ventricular ejection fraction of less than 45%. Its development is multifactorial but includes oxidative stress and impaired vascular endothelial growth factor signaling, although the etiology is not completely elucidated. Treatment is similar to treatment for other forms of heart failure, with supplemental oxygen, diuretics, inotropes or vasopressors as needed. Many patients have resolution of their cardiomyopathy. Complications include formation of a left ventricular thrombus, development of atrial fibrillation or dysrhythmia, worsening cardiac function necessitating cardiac transplant, and increases in maternal morbidity and mortality. Incorrect Answers: A, C, D, E, and F. Amniotic fluid embolism (Choice A) presents with acute onset of shortness of breath, altered mental status or unconsciousness, diffuse bleeding from venipuncture sites or the vagina, hypoxia, hypotension, and signs of fetal distress during the intrapartum or immediate postpartum period. It is caused by amniotic fluid entering the maternal circulation and is associated with high levels of maternal mortality. It would not present 1 month after delivery. Major depressive disorder (Choice C) presents with depressed mood, anhedonia, insomnia, weight loss, fatigue, impairments in concentration, guilt, psychomotor slowing, and suicidal ideation. It is common in the postpartum period but does not cause shortness of breath or peripheral edema. Pneumonia (Choice D) typically presents with fever, cough productive of purulent sputum, shortness of breath, and occasionally chest pain. Hypoxia, leukocytosis, and a new infiltrate on chest x-ray are often seen. It is not associated with peripheral edema. Pulmonary embolism (Choice E) classically presents with shortness of breath, hemoptysis, tachycardia, and hypoxia in a patient with a coagulopathy, malignancy, recent surgery, or prolonged immobilization. It does not typically cause orthopnea or bibasilar crackles. Pulmonary fibrosis (Choice F) most commonly presents in older patients with exertional dyspnea and nonproductive cough, often gradual in onset and progression. Dry bibasilar crackles may be audible on physical examination, and high-resolution CT scan of the chest will reveal honeycombing and traction bronchiectasis. It is less likely in this young patient with peripheral edema and relatively acute onset of symptoms. Educational Objective: Peripartum cardiomyopathy presents after 36 weeks' gestation or within 5 months of delivery with signs of congestive heart failure such as shortness of breath, weight gain, orthopnea, paroxysmal nocturnal dyspnea, cough, and peripheral edema. It typically resolves with heart failure treatment. %3D Previous Next Score Report Lab Values Calculator Help Pause

43 Exam Section 1: Item 43 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 43. A 55-year-old man comes to the physician because of a 2-month history of increasing difficulty swallowing and regurgitation of undigested food. He also has noticed unusual rumbling sounds in his voice that he feels originate in his neck. Physical examination shows halitosis. A videofluoroscopic swallowing study shows a 4-cm, posterior midline pouch protruding between the thyropharyngeus and cricopharyngeus portions of the inferior pharyngeal constrictor muscle. These muscles are most likely innervated by which of the following nerves? A) Glossopharyngeal nerve B) Hypoglossal nerve C) Motor fibers from the vagus nerve D) Parasympathetic fibers from the vagus nerve E) Sympathetic fibers from the superior cervical ganglion

C. Motor fibers from the vagus nerve innervate the thyropharyngeus and cricopharyngeus portions of the inferior pharyngeal constrictor muscle. A Zenker diverticulum occurs as a result of an uncoordinated swallowing mechanism, which results in muscle spasms of the cricopharyngeus leading to pulsion diverticula as the wall of the hypopharynx and superior esophagus weakens. It occurs primarily in older patients. The vagus nerve innervates most of the pharynx except the stylopharyngeus muscle and possesses fibers that join with nerve fibers from the external laryngeal nerve, glossopharyngeal nerve, and sympathetic chains to form the pharyngeal plexus. The pharyngeal plexus innervates the muscles of the pharynx. Symptoms of a Zenker diverticulum include halitosis, regurgitation of undigested food, and dysphagia. Barium swallow reveals a posterior, midline outpouching above or at the level of the cricopharyngeus. Treatment includes surgical repair and resection (diverticulectomy), plus dietary modifications to include soft or liquid foods. Incorrect Answers: A, B, D, and E. The glossopharyngeal nerve (Choice A), like the vagus nerve, is a mixed sensory and motor nerve. It provides motor innervation to the stylopharyngeus muscle. It also stimulates secretion from the parotid gland. The hypoglossal nerve (Choice B) innervates both the intrinsic and extrinsic muscles of the tongue and is a purely motor nerve. Normal function helps propel a food bolus posteriorly into the pharynx to initiate the swallow reflex. Parasympathetic fibers from the vagus nerve (Choice D) innervate the gastrointestinal system down to the transverse colon as well as the heart. These fibers act to regulate peristalsis and digestion and exert control over the heart rate. The muscle fibers arising from the vagus that innervate the pharynx are not parasympathetic fibers. Sympathetic fibers from the superior cervical ganglion (Choice E) innervate the eyes, carotid body, salivary gland, and thyroid gland. Horner's syndrome results from damage to the sympathetic chain ganglion and is manifest as ipsilateral miosis, anhidrosis, and ptosis. The superior cervical ganglion does not innervate the pharyngeal muscles. Educational Objective: Motor fibers from the vagus nerve innervate all the muscles of the pharynx except the stylopharyngeus muscle. Disruption of this plexus of nerves can contribute to poor coordination of the swallow reflex leading to the development of a Zenker diverticulum. %3D 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. Which of the following is most directly responsible for concentrating testosterone in the lumen of the seminiferous tubules? A) Androgen-binding protein B) Follicle-stimulating hormone (FSH) C) FSH/gonadotropin-releasing hormone D) Inhibin E) Luteinizing hormone

A. Androgen-binding protein (ABP) is produced by the Sertoli cells of the seminiferous tubules via the regulation of follicle-stimulating hormone (FSH). Testosterone is produced by Leydig cells in the interstitium adjacent to the seminiferous tubules, the production of which is regulated by of luteinizing hormone (LH). Once released into the lumen of the seminiferous tubules, ABP facilitates spermatogenesis by binding to testosterone, allowing this otherwise lipophilic hormone to concentrate in the lumen. Normal spermatogenesis requires high local concentrations of luminal testosterone. Incorrect Answers: B, C, D, and E. Follicle-stimulating hormone (FSH) (Choice B) is produced by gonadotropic cells in the anterior pituitary. It plays an important role in spermatogenesis by stimulating Sertoli cells to produce ABP and by directly stimulating sperm development. Its role in the concentration of luminal testosterone is indirect and mediated by ABP. FSH/gonadotropin-releasing hormone (Choice C) is produced by the hypothalamus and stimulates the production and release of FSH and LH from gonadotropic cells in the anterior pituitary. Its role in the concentration of testosterone in the lumen of the seminiferous tubules is indirect and is mediated by both FSH and ABP. Inhibin (Choice D) is produced by Sertoli cells and exerts negative feedback on gonadotropic cells in the anterior pituitary to regulate the production of FSH. Luteinizing hormone (Choice E) is also produced by gonadotropic cells in the anterior pituitary. It is important for stimulating the production of testosterone but is not involved in the process of concentrating testosterone in the lumen of the seminiferous tubules. Educational Objective: Androgen-binding protein (ABP) is produced by the Sertoli cells of the seminiferous tubules under the direction of follicle-stimulating hormone. Once released into the lumen of the seminiferous tubules, ABP facilitates spermatogenesis by binding to testosterone and concentrating it in the lumen of the tubules. %3D 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 23-year-old woman has had the lesions shown in her mouth for 3 days. She has had frequent similar episodes over the past 15 years. The lesions are exacerbated by spicy, salty, and acidic food and drinks. They last approximately 1 week and resolve spontaneously. Visits to the dentist seem to trigger the development of the sores. Which of the following is the most likely diagnosis? O A) Aphthous ulcers B) Candidiasis C) Geographic tongue D) Koplik spots O E) Leukoplakia F) Lichen planus G) Psoriasis

A. Aphthous ulcers are painful, round to oval, shallow oral ulcers. They are the most common cause of mouth sores and can be idiopathic or related to underlying conditions such as lupus erythematosus (in which case they are not painful) or Behçet syndrome. They may demonstrate pathergy, which is the development of new erosions at the site of a trauma, such as after a dental procedure. When these lesions develop recurrently, a diagnosis of recurrent aphthous stomatitis is made. It is commonly seen in adolescence and young adulthood, and episodes typically decrease with increasing age. The etiology is multifactorial, but the lesions can be exacerbated by spicy, acidic, or salty foods, as in this case. Stress can also lead to exacerbations. Recurrent aphthous stomatitis may be treated by optimizing oral hygiene, avoiding exacerbating factors, and treating pain with topical anesthetics and coating agents. Incorrect Answers: B, C, D, E, F, and G. Oral candidiasis (Choice B) demonstrates thick, white plaques on the tongue or buccal mucosa, which can be scraped off with a tongue blade. It is commonly seen in immunosuppressed individuals, such as those with poorly controlled HIV infection, or patients using a steroid inhaler and altering the normal oral microbiome. Geographic tongue (Choice C) is a feature of psoriasis and often seen in children. The tongue demonstrates a maze-like pattern of white, linear patches. Ulcers are not a typical feature. Koplik spots (Choice D) are bright red macules with a bluish-white center on the buccal mucosa, which are a sign of an active measles infection. Koplik spots are accompanied by a prodromal fever, cough, coryza, conjunctivitis, and a confluent maculopapular rash that starts at the head/neck and spreads to the trunk, excluding the palms and soles. Leukoplakia (Choice E) refers to the development of white plaques in the mouth, which cannot be scraped off by a tongue blade, and are typically seen on the tongue or buccal mucosa. It may be due to an underlying Epstein-Barr virus infection and is common in patients with HIV infection or malignancy. Oral lichen planus (Choice F) is characterized by white patches with a stellate appearance on the buccal and gingival mucosa. Erosions can also occur, but the white, stellate patches will also be present, unlike in this case. Psoriasis (Choice G) may affect the oral mucosa in the form of geographic tongue. However, it is more classically characterized by thick, salmon-colored plaques with silvery-white scale on the extensor extremities. Educational Objective: Aphthous ulcers are painful, round to oval, shallow oral ulcers. They are the most common cause of mouth sores and can be exacerbated by certain foods, trauma, or emotional stress. II Previous Next Score Report Lab Values Calculator Help Pause

78 Exam Section 2: Item 28 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 28. The presence of argininosuccinate in the urine indicates a defect in the conversion of which of the following? A) Ammonia to urea B) Lysine to glutaryl CoA C) Methionine to succinyl CoA D) Phenylalanine to fumarate E) Tryptophan to indole

A. Argininosuccinate is an important intermediate within the urea cycle. It is formed from citrulline by argininosuccinate synthetase and is converted to arginine by argininosuccinate lyase (ASL). Mutations in the gene encoding ASL result in argininosuccinic aciduria, characterized by the accumulation of argininosuccinate, citrulline, and ammonia. Patients typically present in infancy with lethargy, vomiting, poor feeding, hepatomegaly, seizures, and coma. Deficiency of this enzyme is inherited in an autosomal recessive manner. Incorrect Answers: B, C, D, and E. Conversion of lysine to glutaryl CoA (Choice B) takes place during catabolism of lysine. Glutaryl CoA is subsequently metabolized by glutaryl COA dehydrogenase to crotonyl CoA. Conversion of methionine to succinyl CoA (Choice C) occurs during the catabolism of methionine via multiple intermediates, including homocysteine, cystathionine, and propionyl CoA. Succinyl CoA is an intermediate within the citric acid cycle. Conversion of phenylalanine to fumarate (Choice D) involves multiple steps and occurs via intermediates including tyrosine and homogentisic acid. Disorders in this pathway include alkaptonuria, which is caused by deficiency of the enzyme homogentisate oxidase. Conversion of tryptophan to indole (Choice E) occurs via a reaction catalyzed by tryptophanase with additional products pyruvate and ammonia. Educational Objective: Deficiency of the enzyme argininosuccinate lyase results in the autosomal recessive disease argininosuccinic aciduria, which is characterized by the accumulation of argininosuccinate, citrulline, and ammonia. Patients typically present in infancy with lethargy, vomiting, poor feeding, hepatomegaly, seizures, and coma. %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 u aVR VI V4 II aVL V2 VS m aVF V3 V6 VI II V5 27. A previously healthy 21-year-old woman comes to the office because of a 2-month history of shortness of breath and fatigue. Her most recent menstrual period was 3 months ago. Menses previously had occurred at regular 28-day intervals. She tells the physician that she thinks she may be pregnant. She takes no medications and has not seen a physician for several years. She appears healthy. She is 160 cm (5 ft 3 in) tall and weighs 54 kg (120 lb); BMI is 21 kg/m² Vital signs are within normal limits. The lungs are clear. Cardiac examination shows a normal S, a widely split S, that does not change with respiration, and a grade 3/6 holosystolic murmur that is loudest at the lower left sternal border and radiates to the upper left sternal border. ECG is shown. The most likely cause of these findings is dysfunction of which of the following structures? 1: A) Atrial septum B) Ductus arteriosus C) Interventricular septum D) Pulmonic valve E) Tricuspid valve

A. 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 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. This increased stroke volume of the right ventricle results in delayed closure of the pulmonic valve, which presents as a fixed, split S2, and low-grade physiologic ejection murmur on cardiac auscultation. The increased right heart volumes also result in a prominent right ventricular impulse on physical exam and may present an increased risk for the development of a right bundle branch block, which may be present on ECG as seen in this case. If the atrial septal defect remains uncorrected, it can result in the development of Eisenmenger syndrome secondary to prolonged pulmonary vasculature remodeling resulting in pulmonary arterial hypertension and shunt reversal leading to cyanosis. Asymptomatic atrial septal defects may become clinically significant in the setting of increased blood flow (such as during pregnancy). Incorrect Answers: B, C, D, and E. A patent ductus arteriosus (Choice B) is a persistent extracardiac conduit between the aorta and the pulmonary artery that has failed to obliterate after birth. It results in a continuous, machine-like murmur best heard in the left second intercostal space, radiating to the clavicle. Defects of the interventricular septum (Choice C) are characterized by a holosystolic murmur best heard in the left lower sternal border. They do not classically result in a fixed, split S2. Increased flow across the pulmonic valve (Choice D) due to shunting of blood from the left atrium to the right atrium results in the fixed S2 associated with an atrial septal defect. The pulmonic valve itself is typically normal. Tricuspid valve (Choice E) dysfunction results in tricuspid regurgitation, which presents as a holosystolic murmur best heard in the left lower sternal border. Educational Objective: A fixed, widely split S2 is characteristic of an atrial septal defect due to increased blood flow through the pulmonic valve. Severe defects can result in pulmonary hypertension and development of Eisenmenger syndrome over time, with reversal of the left to right shunt. II Previous Next Score Report Lab Values Calculator Help Pause

4 Exam Section 1: Item 4 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 4. A 35-year-old man is admitted to the hospital because of a 5-day history of fever and dyspnea. He underwent a bone marrow transplantation 6 months ago; the procedure was complicated by severe graft-versus-host disease. His temperature is 38°C (100.4°F), and respirations are 30/min. Scattered crackles are heard on auscultation of the chest. A chest x-ray shows patchy infiltrates. A transbronchial biopsy specimen shows findings consistent with cytomegalovirus infection. Intravenous administration of ganciclovir is begun. This drug interferes with the function of which of the following enzymes? A) DNA polymerase B) Integrase C) Reverse transcriptase D) RNA polymerase E) Thymidine kinase

A. Cytomegalovirus (CMV), also known as human herpesvirus 5 (HHV-5), is an opportunistic infection commonly occurring in immunocompromised patients with solid-organ or allogeneic bone marrow transplantation, severe ulcerative colitis, or HIV/AIDS infection. It can be transmitted through multiple modes, including sexual contact, urine, respiratory droplets, and to a fetus via the placenta. It can cause a variety of presentations, including mononucleosis in immunocompetent patients, and retinitis, esophagitis, and pneumonia in immunocompromised patients. Treatment for all human herpesvirus infections involves drugs that inhibit viral DNA polymerase, classically by guanosine analogs such as acyclovir, valacyclovir, and famciclovir. Prior to exerting their antiviral effects, most guanosine analogs must be phosphorylated by the viral enzyme thymidine kinase. They are then able to inhibit the viral DNA polymerase by terminating the nascent DNA chain during replication. These drugs are effective against herpes simplex virus and varicella zoster virus, weakly effective against Épstein-Barr virus, and not effective against CMV, which does not have the necessary thymidine kinase needed for phosphorylation. However, it does have the necessary phosphorylating enzyme to activate ganciclovir, another guanosine analog that inhibits DNA polymerase, and thus this is an effective anti-viral treatment for CMV. Incorrect Answers: B, C, D, and E. Integrase inhibitors (Choice B) prevent integration of proviral DNA into the host genome. They are a component of the highly active anti-retroviral therapy utilized in the treatment of HIV. Reverse transcriptase (Choice C) transcribes DNA from viral MRNA for incorporation into the host cell genome. This enzyme is inhibited by both nucleoside reverse transcriptase inhibitors, such as abacavir and didanosine, and non-nucleoside reverse transcriptase inhibitors, such as efavirenz and nevirapine. These medications are used to treat HIV infections. RNA polymerase (Choice D) is the enzyme that facilitates the conversion of DNA to RNA. It is the target of nucleotide and nucleoside polymerase inhibitors, which are utilized in the treatment of hepatitis C infection. Thymidine kinase (Choice E) is a phosphorylating enzyme required for activation of the guanosine analogs acyclovir, valacyclovir, and famciclovir. Development of a mutation in the viral thymidine kinase enzyme prevents drug phosphorylation and confers resistance to these medications. ČMV does not contain the thymidine kinase enzyme. Educational Objective: Ganciclovir is a guanosine analog that inhibits DNA polymerase and is used to treat CMV. Like other guanosine analogs, activation of this medication requires phosphorylation within the infected cell. %3D Previous Next Score Report Lab Values Calculator Help Pause

32 Exam Section 1: Item 32 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 32. An 18-hour-old male newborn is 61 cm (24 in) long and weighs 5443 g (12 Ib). His mother has type 1 diabetes mellitus. His serum glucose concentration is 20 mg/dL. Which of the following fetal conditions immediately prior to birth most likely precipitated the newborn's postnatal hypoglycemia? A) Decreased gluconeogenesis B) Decreased glycogen concentration C) Decreased glycogen synthetase activity D) Decreased serum insulin concentration E) Increased serum insulin-like growth factor

A. Excessive fetal exposure to insulin in utero can cause macrosomia, neonatal respiratory distress syndrome or hyaline membrane disease, and hypoglycemia in the newborn. Human placental lactogen produced by the placenta increases maternal insulin resistance. This leads to increased levels of glucose in the maternal circulation, and subsequently the fetal circulation. Fetal hyperglycemia in turn causes the fetus to produce excess insulin, as well as decrease the rate of gluconeogenesis. The insulin then acts as a growth factor, leading to macrosomia, and inhibits the production of surfactant, which leads to respiratory distress syndrome. After birth, the newborn continues to synthesize insulin but is no longer exposed to maternal blood glucose levels, which leads to hypoglycemia. Treatment is supportive and includes administration of supplemental glucose while neonatal endocrine regulation self-corrects. Incorrect Answers: B, C, D, and E. Decreased glycogen concentration (Choice B) would lead to fasting hypoglycemia as there would be deficient glycogenolysis to maintain adequate blood glucose concentration. This patient would likely have increased glycogen concentration due to its prolonged exposure to insulin in utero. Decreased glycogen synthetase activity (Choice C) would lead to fasting hypoglycemia due to inadequate glycogen synthesis and insufficient stores. However, glycogen synthetase is stimulated by insulin, which is increased in this newborn. Decreased serum insulin concentration (Choice D) would lead to hyperglycemia; this patient has increased insulin concentration due to prolonged exposure to maternal hyperglycemia with resultant islet cell hypertrophy. Increased serum insulin-like growth factor (Choice E) plays a role in the development of long bones and muscle mass, rather than blood glucose levels. It is increased in gigantism and acromegaly. Educational Objective: Maternal diabetes exposes a fetus to high blood glucose concentration, in response to which the fetus increases production of insulin and decreases gluconeogenesis. Following birth, this can lead to postnatal hypoglycemia as the newborn is no longer exposed to the increased blood glucose concentration in the maternal circulation. %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 16. A 20-year-old woman comes to the emergency department 30 minutes after slipping on ice and extending her hand to break her fall. Palpation of the anatomic snuff-box produces pain. A wrist x-ray is most likely to show a fracture of which of the following carpal bones? A) Scaphoid B) Lunate C) Triquetrum D) Pisiform E) Trapezium F) Trapezoid G) Capitate H) Hamate

A. Falling onto an outstretched hand can lead to traumatic injuries such as distal radius fractures, elbow dislocations, or fractures of the carpal bones of the wrist. A common pattern of injury with this mechanism is a scaphoid fracture. The scaphoid bone is part of the lateral column of the wrist and supports force transmission from the hand to the lateral aspect of the radius. Fractures of this bone typically present with lateral wrist pain and tenderness in the anatomic snuff-box, which is the dorsal extensor pollicis longus and abductor pollicis longus. The scaphoid has a blood supply that proceeds from distal to proximal. Displacement of a fracture of this bone may lead to decreased blood supply of the proximal fragment, leading to avascular necrosis and debilitating wrist pain and deformity. Because of this, it is important to identify and appropriately treat scaphoid fractures. pression between Incorrect Answers: B, C, D, E, F, G, and H. Fractures of the lunate (Choice B) are an uncommon injury. More commonly, with high energy injuries to the wrist, the capitate and the remaining carpal bones may dislocate from the concave surface of the lunate. This is known as a perilunate dislocation and may lead to acute compression of the median nerve. The triquetrum (Choice C) is a carpal bone located in the ulnar aspect of the wrist. Fractures in this location are uncommon. If fractured, it would present with tenderness along the ulnar aspect of the wrist. The pisiform (Choice D) is a carpal bone in the ulnar aspect of the wrist. It is a pea-shaped sesamoid bone that can be mistaken for a fracture fragment. Fractures of the pisiform are uncommon. The trapezium (Choice E) is the bone at the base of the thumb that can also be palpated in the floor of the anatomic snuff-box. Fracture of the trapezium is uncommon and typically presents with pain after trauma to the thumb. The trapezoid (Choice F) is a wedge-shaped carpal bone just proximal to the second metacarpal. Fracture of the trapezoid is an uncommon carpal injury. The capitate (Choice G) is a larger carpal bone in the center of the wrist. It articulates with the lunate and is dislocated in a perilunate dislocation. The hamate (Choice H) is a carpal bone in the ulnar aspect of the wrist that has a process along its volar surface, referred to as the hook of the hamate, that acts as a covering for the finger flexor tendons of the ulnar digits. Fracture of the hook of the hamate can occur while placing a high load through the wrist while holding a handle (eg, baseball bat, golf club, sledgehammer). This injury can lead to compression of the ulnar nerve in Guyon canal and irritation or impingement of the adjacent tendons. Educational Objective: Fracture of the scaphoid presents with radial wrist pain and tenderness in the anatomic snuff-box, typically following a fall on an outstretched hand. Identification and appropriate treatment of this injury is important as the retrograde blood supply to this bone places it at risk for avascular necrosis of the proximal fragment. %3D Previous Next Score Report Lab Values Calculator Help Pause

49 Exam Section 1: Item 49 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 49. A 41-year-old woman is evaluated because of increasingly severe headaches for 6 weeks. Her blood pressure is 160/100 mm Hg while standing and supine. A bruit is heard over the left costovertebral angle. Urinalysis shows no abnormalities. An angiogram of the left renal artery shows alternating areas of stenosis and aneurysmal dilatation ("string of beads" sign). Which of the following conditions of the renal artery is the most likely diagnosis? A) Fibromuscular dysplasia B) Hyaline arteriolosclerosis C) Intimal fibroplasia D) Periarterial fibroplasia E) Perimedial hyperplasia

A. Fibromuscular dysplasia is a non-inflammatory and non-atherosclerotic angiopathy of medium-sized arteries (eg, renal, carotid), that results in multifocal fibrous and muscular thickening of the arterial wall. The resultant stenosis causes secondary hypertension due to abnormal stimulation of the juxtaglomerular apparatus from low afferent blood flow leading to excessive production of renin and angiotensin. It can lead to severe hypertension in an otherwise healthy, young patient with no medical comorbidities or laboratory abnormalities. Physical examination findings often include signs of left ventricular hypertrophy such as S4 gallop and a renal artery bruit auscultated lateral to the umbilicus. Angiography may reveal a bead-like appearance of the renal artery. ACE inhibitors can be considered for unilateral stenosis but can lead to acute renal failure in the setting of significant bilateral renal artery stenosis. Incorrect Answers: B, C, D, and E. Hyaline arteriosclerosis (Choice B) refers to thickening of arteriolar walls with hyaline deposits seen on hematoxylin and eosin staining that occurs over time secondary to chronic hypertension. Narrow arterioles in the kidneys lead to decreased blood flow and glomerular filtration rate, resulting in increased activation of the renin-angiotensin-aldosterone system and resultant hypertension. Intimal fibroplasia (Choice C) is an uncommon subtype of fibromuscular dysplasia present in less than 10% of cases. Periarterial fibroplasia (Choice D) is a rare subtype of fibromuscular dysplasia and perimedial hyperplasia (Choice E) is also uncommon. The most common subtype of fibromuscular dysplasia is medial fibroplasia. Educational Objective: Fibromuscular dysplasia is a non-inflammatory and non-atherosclerotic angiopathy of medium-sized arteries that can lead to renal artery stenosis and secondary hypertension due to the excessive production of renin and angiotensin. It commonly presents in a young patient with no medical comorbidities or abnormal laboratory findings. %3D Previous Next Score Report Lab Values Calculator Help Pause

58 Exam Section 2: Item 8 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 8. A patient with a 6-month history of heat intolerance, fatigue, episodes of tachycardia, and weight loss has a diffusely enlarged thyroid gland. Serum concentrations of triiodothyronine (T) and thyroxine (T) are increased; thyroid-stimulating hormone is decreased. Which of the following is the most likely diagnosis? A) Autoimmune thyroid hyperplasia B) Pituitary neoplasm C) Surreptitious ingestion of T4 D) Thyroid neoplasm

A. Graves disease is the most common cause of hyperthyroidism and is an autoimmune thyroid hyperplasia. It is caused by an autoantibody that stimulates the thyroid-stimulating hormone receptors (TSHR) on thyroid follicular cells. Excessive stimulation of TSHR leads to follicular thyroid hyperplasia and a diffusely enlarged thyroid. Serum laboratory evaluation typically reveals increased concentrations of T3 and T4, decreased thyroid-stimulating hormone (TSH), and the presence of TSHR stimulating antibodies. Graves disease presents with symptoms of hyperthyroidism, including heat intolerance, weight loss, tremor, hyperreflexia, warm, moist skin, and pretibial myxedema. Some patients with Graves disease also develop thyroid ophthalmopathy, which can cause diplopia, proptosis, and restrictive strabismus. Incorrect Answers: B, C, and D. Pituitary neoplasm (Choice B) may lead to hyperthyroidism if the pituitary neoplasm, most commonly an adenoma, secretes excessive TSH. Serum concentrations of TSH would be markedly increased, rather than decreased. TSH-secreting adenomas are rare and constitute only a small portion of patients with hyperthyroidism. Surreptitious ingestion of T4 (Choice C), also known as factitious thyrotoxicosis, may result from deliberate ingestion of thyroid hormone, such as levothyroxine. Factitious thyrotoxicosis can be distinguished from other causes of hyperthyroidism, such as Graves disease, by the absence of other suggestive clinical findings, including goiter, thyroid swelling, orbital involvement, or pretibial myxedema. Thyroid neoplasm (Choice D), such as papillary or medullary thyroid carcinoma, rarely leads to hyperthyroidism. Most thyroid carcinomas present as a cold nodule on radioactive iodine uptake scan, which can distinguish these lesions from hyperfunctioning thyroid nodules. Educational Objective: Graves disease is caused by an autoantibody that stimulates the TSHR on thyroid follicular cells. Excessive stimulation of TSHR leads to follicular thyroid hyperplasia and a diffusely enlarged thyroid. Serum laboratory evaluation typically reveals increased concentrations of T3 and T4, decreased TSH, and the presence of TSHR stimulating antibodies. %3D Previous Next Score Report Lab Values Calculator Help Pause

6 Exam Section 1: Item 6 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 6. A 14-year-old girl is brought to the physician by her parents because of a 1-month history of a rash that appears with sun exposure. Her parents tell the physician that she has been eating little food. Physical examination shows a pruritic rash on the exposed areas of the body. Her serum tryptophan concentration is decreased. Urine studies show increased excretion of amino acids, predominantly alanine, isoleucine, leucine, phenylalanine, tryptophan, and valine. Production of which of the following vitamins is most likely impaired in this patient? A) Niacin B) Vitamin B, (thiamine) C) Vitamin B2 (riboflavin) D) Vitamin B5 (pantothenic acid) E) Vitamin C

A. Hartnup disease is an autosomal recessive disorder involving a defect in a kidney and intestinal neutral amino acid transporter protein. This defect leads to aminoaciduria and a decreased absorption of neutral amino acids from the gastrointestinal tract, resulting in deficiencies of neutral amino acids. Neutral amino acids include tryptophan, phenylalanine, glycine, alanine, valine, isoleucine, leucine, methionine, and proline. Tryptophan is converted to niacin, so a deficiency in tryptophan can result in niacin deficiency. Niacin deficiency is characterized by rash, glossitis, diarrhea, and neuropsychological disturbances such as dementia and hallucinations. Incorrect Answers: B, C, D, and E. Vitamin B1 (thiamine) (Choice B) is a cofactor for several enzymes in glucose metabolism and adenosine triphosphate production, including pyruvate dehydrogenase and a-ketoglutarate dehydrogenase. Deficiency is characterized by Wernicke encephalopathy, a triad of confusion, ophthalmoplegia, and ataxia. Wernicke encephalopathy is theoretically reversible with administration of high-dose thiamine; if untreated, it can progress to Korsakoff syndrome which is characterized by dementia, confabulation, hallucinations, and psychosis. Vitamin B2 (riboflavin) (Choice C) deficiency is characterized by inflammation and cracking of skin around the lips, mouth, and tongue. It is not associated with aminoaciduria or Hartnup disease. Vitamin B5 (pantothenic acid) (Choice D) deficiency is characterized by dermatitis, enteritis, alopecia, and adrenal insufficiency. It is not associated with aminoaciduria or Hartnup disease. Vitamin C (Choice E) is found in fruits and vegetables and is necessary for collagen synthesis, iron absorption, immune function, and conversion of dopamine to norepinephrine. Deficiency causes scurvy, which is characterized by swollen gums, bruising and poor wound healing, petechiae, perifollicular and subperiosteal hemorrhages, and short, fragile, curly hair. Educational Objective: Hartnup disease is an autosomal recessive disorder involving a defect in a kidney and intestinal neutral amino acid transporter, leading to deficiencies in neutral amino acids such as tryptophan. Tryptophan is converted to niacin, so a deficiency in tryptophan can result in niacin deficiency. Niacin deficiency is characterized by rash, glossitis, diarrhea, and neuropsychological disturbances such as dementia and hallucinations. %3D Previous Next Score Report Lab Values Calculator Help Pause

60 Exam Section 2: Item 10 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 10. A 62-year-old woman develops difficulty breathing. Pulmonary function tests before and after bronchodilator therapy show no changes. Predicted and patient values are: Test FVC (L) FEV, (L) FEVFVC Total lung capacity (L) Residual volume (L) Predicted Patient 5.0 4.0 0.8 6.0 1.6 4.0 2.4 0.6 7.2 2.7 Which of the following is the most likely explanation for these findings? Airway Resistance Lung Compliance A) ↑ ↑ B) ↑ normal C) Normal D) ↑ E)

A. Increased airway resistance and increased lung compliance are features of chronic obstructive pulmonary disease (COPD). Lung compliance is the ability of the lung to stretch and expand; increased compliance means that the lung has a greater ability to expand, but it does not imply that the lung has equal capability to contract. Airway resistance is the resistance to airflow through the bronchi and bronchioles. Spirometry is a useful tool for differentiating various forms of lung disease. Patients can usually be categorized as obstructive, restrictive, or mixed patterns. Patients with obstructive lung disease have an FEV/EVC ratio less than 0.7 and an FEV, less than 80% of predicted values while patients with restrictive lung disease have an FEV/FVC ratio greater than 0.8. Patients with COPD, which is most commonly caused by prolonged smoking or exposure to secondhand smoke, experience dynamic collapse of the bronchi during expiration leading to outflow obstruction and air trapping. Outflow obstruction leads to increased airway resistance and is reflected by a reduced FEV, The lung parenchyma in patients with COPD lacks elasticity and the normal recoil of the lung is impaired, which is reflected as increased compliance. This, in conjunction with air trapping, is usually demonstrated by an increased residual lung volume (RV) and total lung capacity (TLC). The severity of outflow obstruction is graded according to the FEV, with lower readings consistent with severe disease. Response to bronchodilators is important in distinguishing COPD from asthma and from asthma/COPD overlap syndromes. In asthma, improvement in outflow obstruction as measured by an increase in FEV, by 12% and 200-mL absolute volume is expected. In COPD, no response to bronchodilators is expected, which is the case in this patient. Incorrect Answers: B, C, D, and E. Increased airway resistance and normal lung compliance (Choice B) is seen in patients with asthma. Patients with asthma will have a reduced FEV/FVC but the FEV, will respond to bronchodilators, which represents a reversible defect. As patients do not have parenchymal disease, their lung compliance is normal. Normal airway resistance and decreased compliance (Choice C) is seen in restrictive lung disease, which can occur as a result of conditions such as idiopathic pulmonary fibrosis or neuromuscular disorders such as muscular dystrophy. Decreased airway resistance and increased lung compliance (Choice D) may occur in the setting of treated COPD or chronic bronchitis. The lung compliance is increased as a result of loss of alveolar septa and destruction of alveoli, and if therapy to reduce bronchial inflammation such as through suctioning and corticosteroids is effective, airway resistance falls to permit ventilation. Decreased airway resistance and decreased lung compliance (Choice E) can be seen in normal patients at maximum inspiration. In normal lungs, increasing the volume of the lungs expands the airways and decreases airway resistance while the compliance of the lungs decreases at maximum inspiration. Educational Objective: COPD results in loss of the normal elastance of the lung with a consequent increase in compliance. During expiration, dynamic airway collapse manifests as a reduced FEV, while chronic air trapping and loss of elastance are reflected as increased RV and TLC. 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. A 25-year-old man comes to the physician because of a 3-day history of pain and swelling of his right leg. He has no history of major medical illness or recent trauma. Examination of the right lower extremity shows edema and tenderness. Duplex ultrasonography of the right lower extremity shows a thrombus extending into the superficial femoral vein. Further studies show protein C deficiency. Inactivation of which of the following coagulation factors is most likely as a result of this deficiency in this patient? A) Factors V (proaccelerin) and VIII (antihemophilic factor) B) Factors V (proaccelerin) and IX (plasma thromboplastin component) C) Factors V (proaccelerin) and XI (plasma thromboplastin antecedent) D) Factors VII (antihemophilic factor) and IX (plasma thromboplastin component) E) Factors VIII (antihemophilic factor) and XI (plasma thromboplastin antecedent) F) Factors IX (plasma thromboplastin component) and XI (plasma thromboplastin antecedent)

A. Protein C inactivates factors V and VIII; deficiency of protein C leads to a hypercoagulable state predisposing to the development of both venous and arterial thrombi. The coagulation cascade consists of the intrinsic, extrinsic, and common pathways. The intrinsic pathway consists of sequential activation of factors XII, XI, IV, and VIII, and activity of this pathway is measured by the partial thromboplastin time. The extrinsic pathway involves the activation of factor VII and is measured by the prothrombin time or the INR. Both the intrinsic and extrinsic pathways can trigger the common pathway via activation of factor X to Xa. Activated factor X (factor Xa) has several functions, but one is to modify factor V and allow it to form a prothrombinase complex with factor Xa. This complex helps to form a fibrin clot. Factor Xa is also necessary for the conversion of prothrombin to thrombin. Thrombin binds to thrombomodulin on the surface of endothelial cells inducing a conformational change that allows it to activate protein C, while protein C localizes to the endothelium by binding to the endothelial protein C receptor (EPCR). Activated protein C binds to the surface of activated platelets and degrades factors Va and VIlla, which is the means by which it exerts its anticoagulant effects. Incorrect Answers: B, C, D, E, and F. Factors V and IX (Choice B) and Factors V and XI (Choice C) are incorrect as factors IX and XI are not under inhibitory control by protein C. Activation of factor IX by Xla is an important step in the coagulation cascade and deficiency of factor IX is associated with hemophilia. Factor XI activates factor IX in the intrinsic pathway. Factors VIII and IX (Choice D), when activated, both take part in the intrinsic clotting cascade. While factor VIII is under inhibitory control by protein C, factor IX is not. Factors VIII and XI (Choice E) are incorrect as factor VIII is under inhibitory control by protein C but factor XI is not. Factors IX and XI (Choice F) are both activated in the intrinsic clotting cascade and serve to accelerate clotting. They are not influenced directly by protein C. Educational Objective: Protein C is a natural anticoagulant that is activated by thrombin after thrombin binds to the endothelial surface. Activated protein C binds to the surfaces of activated platelets and degrades factors Va and VIlla, thereby exerting negative feedback on the clotting cascade. %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 200- 100- M M 25 27 1 3 5 7 9 11 13 15 17 19 21 23 25 2713 5 Cycle (day) M = menstruation 5. The graph shows changes in serum estradiol concentration during a normal menstrual cycle. Which of the following ovarian cells is primarily responsible for the aromatization of androstenedione to estradiol at the time indicated by the arrow? A) Granulosa B) Luteal C) Stromal D) Theca externa E) Theca interna

A. The first half of the menstrual cycle, the follicular phase, which varies in length, begins with menses. During menses, follicle-stimulating hormone (FSH) and luteinizing hormone (LH) concentrations increase and stimulate the developing follicle. Androstenedione is converted to estrone and estradiol via aromatase in the granulosa cells of the follicle. The estrogen then secreted from the granulosa cell is responsible for follicle growth and endometrial proliferation. 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. However, if no implantation occurs, the corpus luteum degrades to the corpus albicans, and estrogen and progesterone levels decrease, causing menstruation and minor increases in FSH and LH. Incorrect Answers: B, C, D, and E. Luteal cells (Choice B) are present in the corpus luteum and are derived from the granulosa cells of the pre-ovulatory follicle. They secrete progesterone and estrogen. However, they do not develop until after ovulation (14 days prior to menstruation). Stromal cells (Choice C) are the connective tissue and supporting cells of the ovary. They do not secrete estradiol. Theca externa cells (Choice D) are the cells that form the outer layer of a developing follicle. The theca externa is primarily loose connective tissue and therefore the cells are generally fibroblasts, macrophages, and smooth muscle; these cells do not secrete hormones. Theca interna cells (Choice E) are cells of the follicle that are responsible for generating androstenedione from cholesterol, after which the androstenedione is transported to the granulosa cell to be converted to estradiol. Educational Objective: Granulosa cells in the developing follicle are responsible for converting androgens received from the theca interna cells into estradiol via aromatase. Previous Next Score Report Lab Values Calculator Help Pause

64 Exam Section 2: Item 14 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 14. A 26-year-old woman comes to the physician because of a 1-week history of rectal pain that is made more severe by defecation, and occasional blood on the toilet tissue after a bowel movement. She says that her stools appear normal and that she has not had any trauma. She has a history of chronic constipation. Visual rectal examination shows the findings in the photograph. Which of the following is the most likely diagnosis? A) Anal fissure B) Bowen carcinoma C) Condyloma acuminatum D) Perianal abscess E) Prolapsed internal hemorrhoid

A. This patient's examination demonstrates an anal fissure at the posterior midline position, which explains the patient's rectal pain and scant hematochezia. Anal fissures are a common cause of severe rectal pain with defecation and occur in patients with risk factors such as chronic constipation, those who partake in receptive anal intercourse, or after vaginal birth, although diarrhea can predispose to fissures. Most fissures occur at the posterior midline with the remainder typically occurring in the anterior midline. A fissure starts with an initial tear in the anoderm with exposure of part of the anal sphincter that subsequently leads to muscle spasm with compromise of local blood flow. This can delay healing of the fissure. Treatment is with sitz baths, fiber supplements, topical lidocaine jelly, and topical nitroglycerin. Fissures that persist for greater than two months or those that do not occur along the midline can indicate the presence of underlying disease such as Crohn disease, and these patients should receive further evaluation. Incorrect Answers: B, C, D, and E. Bowen carcinoma (Choice B), also called squamous carcinoma in situ, describes a squamous carcinoma of the epithelium that primarily affects areas with heavy sun exposure. It presents with a slow growing scaly patch or plaque. It would be unlikely to present in the perianal area or in such a young patient. Condyloma acuminatum (Choice C) are anogenital warts and are caused by infection with human papillomavirus. Lesions may spontaneously resolve or gradually increase in number and enlarge, so treatment is usually offered with imiquimod. There is no evidence of a wart-like lesion in this patient's perianal area. Perianal abscess (Choice D) presents as a painful, palpable, fluctuant mass in the perianal region, occasionally with surrounding erythema. Abscesses can involve the anal sphincter, injury to which can cause fecal incontinence. While this patient has pain with defecation, her examination reveals a fissure. Prolapsed internal hemorrhoids (Choice E) appear as a red or purple mass protruding from the anus during defecation. Internal hemorrhoids are generally painless, unlike external hemorrhoids. If severe, hemorrhoids may remain prolapsed following a bowel movement requiring manual reduction. Educational Objective: Anal fissures are a common cause of rectal pain and scant hematochezia with defecation. Risk factors include pregnancy, chronic constipation, and low fiber diets. They generally appear in the posterior midline. Treatment is with sitz baths and topical lidocaine or nitroglycerin. Previous Next Score Report Lab Values Calculator Help Pause

61 Exam Section 2: Item 11 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 11. A3-year-old girl is brought to the physician because of a 2-week history of diarrhea. Her temperature is 37.6°C (99.8°F), pulse is 70/min, respirations are 18/min, and blood pressure is 110/70 mm Hg. Physical examination shows generalized lymphadenopathy. A CT scan of the chest and abdomen shows enlarged lymph nodes in the mesentery and para-aortic region. Examination of a lymph node biopsy specimen shows marked proliferation of histiocytes and numerous segmented neutrophils. Granulomata are absent, and special stains show numerous acid-fast bacilli, which are subsequently identified as Mycobacterium avium-intracellulare. Serum studies show normal concentrations of IgA, IgG, IgM, B lymphocytes, T lymphocytes, and CD4+ and CD8+ Tlymphocytes. This patient most likely has defective function or expression of which of the following proteins? A) Class I MHC molecules B) Interferon-gamma receptor C) Interleukin-2 (IL-2) receptor D) Leukocyte function-associated antigen-1 E) NADPH oxidase

B. Certain intracellular pathogens can avoid detection by the immune system by replicating within macrophages. To kill intracellular microbes, macrophages rely on complex signaling between the innate and adaptive immune systems. Initial macrophage activation occurs with binding of pathogen-associated molecular patterns to tol-like receptors. This stimulates release of proinflammatory cytokines including interleukin-12 (IL-12), which stimulates T lymphocytes to produce interferon-gamma (IFN-y). IFN-y then binds to the IFN-y receptor on macrophages, which triggers a JAK kinase signaling cascade that allows macrophages to begin eliminating intracellular pathogens. This cycle is referred to as the IL-12/IFN-y axis. IL-12 also induces helper and cytotoxic T-lymphocyte differentiation, as well as activation of natural killer cells. Deficiencies in the IFN-y receptor typically present in early childhood with severe mycobacterial and/or salmonellal infections. Patients are unable to form granulomas due to the impaired stimulation of macrophages. Incorrect Answers: A, C, D, and E. Class I MHC molecules (Choice A) are found on all nucleated cells and platelets. They present fragments of antigens generated by degradation of proteins in the cytosol to T lymphocytes. Foreign antigens will stimulate cytotoxic T lymphocytes to initiate apoptosis in the presenting cell. Interleukin-2 (IL-2) receptors (Choice C) are primarily found on T lymphocytes. Activation stimulates helper T lymphocytes, cytotoxic T lymphocytes, and regulatory T lymphocytes. They are not involved in the IL-12/IFN-y axis, and deficiency is not associated with mycobacterial infections. Leukocyte function-associated antigen-1 (LFA-1) (Choice D) is an integrin involved in leukocyte adhesion to the vascular endothelium during inflammatory recruitment and migration. Mutation or absence results in leukocyte adhesion deficiency type I. NADPH oxidase (Choice E) deficiency is found in chronic granulomatous disease, in which granulomas form but the absence of an effective oxidative burst impairs microbial killing. NADPH oxidase is required in the formation of free radicals. Educational Objective: The IL-12/IEN-y signaling axis is required for macrophages to be able to kill intracellular pathogens. Deficiency in the IFN-y receptor increases the risk of severe mycobacterial and salmonellal infections early in life. %3D Previous Next Score Report Lab Values Calculator Help Pause

41 Exam Section 1: Item 41 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 41. The specimen shown is from a 65-year-old man. Which of the following is the most likely diagnosis? A) Acute leukemia B) Colonic carcinoma C) Hepatic cell carcinoma D) Leiomyosarcoma E) Liposarcoma 1 cm

B. Colonic carcinoma, also known as colorectal carcinoma (CRC), is the most likely diagnosis. The gross specimen in the photograph is a cross-section of the liver than demonstrates innumerable metastases obliterating the normal liver parenchyma. Of the various malignancies that metastasize to the liver, CRC is one of the most common. Other cancers that metastasize to the liver include those of organs that drain into the enterohepatic venous system such as the pancreas and the gallbladder, although lung and breast cancers also have a tendency to metastasize to the liver. Metastases can be differentiated from primary tumors of the liver such as hepatocellular carcinoma by the appearance both on imaging and gross examination in the case of autopsy or resection. Primary tumors tend to be solitary, or multiple but with a dominant mass, while metastases tend to be multifocal and of varying size. Incorrect Answers: A, C, D, and E. Acute leukemia (Choice A) is a malignancy of white blood cell progenitor lines within the bone marrow. Clones are commonly found in the bone marrow, blood, or lymphatic system. Occasionally, certain varieties of leukemia such as acute monomyelocytic leukemia can present with infiltration of leukemic cells in the skin (leukemia cutis) or in the gums, but infiltration of the liver does not typically occur. Hepatic cell carcinoma (Choice C) typically presents as a dominant, singular mass, or as a dominant mass with a small number of additional foci. It is most common in patients who have chronic hepatitis B or cirrhosis, and additional findings on gross examination of the liver would include a nodular contour consistent with cirrhosis. Leiomyosarcoma (Choice D) is a cancer of smooth muscle cells and is much less common than CRC. While liver metastases from retroperitoneal leiomyosarcoma occur, these generally result in one to three metastatic lesions rather than the diffuse infiltration evident in the gross photograph. Liposarcoma (Choice E) is a cancer of adipose tissue. It is exceedingly rare compared to CRC and rarely metastasizes to the liver. It would be unlikely to cause the lesions seen in the photograph. Educational Objective: CRC is a common malignancy that metastasizes to the liver. As compared to primary malignancies of the liver, which generally appear as solitary lesions, metastases from CRC are often spread diffusely throughout the liver and appear as innumerable masses on imaging and gross examination. %3D Previous Next Score Report Lab Values Calculator Help Pause

65 Exam Section 2: Item 15 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 15. A moderately obese 55-year-old man is brought to the emergency department because of a 10-hour history of severe chest pain. He has a 5-year history of exercise-induced angina. His pulse is 109/min, respirations are 15/min, and blood pressure is 132/92 mm Hg. Physical examination shows diaphoresis. A blood sample obtained 2 hours after admission shows increased serum activity of creatine kinase MB. Which of the following is the most likely cause of this laboratory finding? A) Increased Golgi complex activity B) Increased permeability of the plasma membrane C) Mitochondrial swelling D) Nuclear lysis E) Proliferation of the endoplasmic reticulum

B. Creatine kinase MB (CKMB) is a cardiac enzyme that is detectable within six hours of myocardial ischemia and early coagulation necrosis. Increased plasma membrane permeability is a feature of early coagulative necrosis and allows CKMB and troponin to be released from cardiac myocytes. Other features of irreversible cellular damage include pyknosis (chromatin condensation), karyorrhexis (nuclear fragmentation), and cytoplasmic swelling. Ischemic injury manifests within 24 hours as waviness of myocardial fibers. Leakage of intracellular proteins and the onset of necrosis prompts the infiltration of neutrophils. CKMB concentration return to its baseline at approximately 48 hours after acute infarction and is useful for the detection of re-infarction after this time. In contrast, troponin concentration remains increased for 7 to 10 days after myocardial infarction. Incorrect Answers: A, C, D, and E. While increased Golgi complex activity (Choice A) results in additional secretion of proteins at the cell surface through exocytosis, this is not the mechanism of CKMB release. CKMB is not contained with a secretory vesicle. Rather, it leaks from the cytoplasm through damaged and permeable cellular membranes without the aid of vesicles. Mitochondrial swelling (Choice C) is caused by oxidative stress in myocardial ischemia and may be reversible. If severe enough to cause mitochondrial rupture, pro-apoptotic mediators, such as cytochrome c, are released and lead to irreversible cell death. CKMB is not stored within mitochondria. Nuclear lysis (Choice D) is also a feature of coagulative necrosis within the context of myocardial ischemia, but does not directly result in release of CKMB, since CKMB is not stored in the nucleus. Proliferation of the endoplasmic reticulum (Choice E) may occur in the cellular ischemia due to altered folding of cellular proteins, which accumulate in the endoplasmic reticulum. This process triggers the unfolded protein response, which seeks to restore normal function by degrading unfolded or misfolded proteins and by downregulating protein synthesis. If unsuccessful, the unfolded protein response induces apoptosis. Educational Objective: CKMB is a cardiac enzyme that is detectable within six hours of myocardial ischemia. Increased plasma membrane permeability is a feature of early coagulative necrosis and allows CKMB and troponin to be released from cardiac myocytes. CKMB concentration return to its baseline at approximately 48 hours after acute infarction and is usefulfor the detection of re-infarction. %3D 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. A 16-year-old boy is brought to the physician because of a 3-month history of shortness of breath while playing sports. He has no shortness of breath at rest. He says, "Whenever I run around I cough, so I don't want to be on the basketball team anymore." He takes no medications and has no known allergies. There is a family history of hypertension and asthma. He is 165 cm (5 ft 5 in) tall and weighs 68 kg (150 lb); BMI is 25 kg/m2. His respirations are 12/min, and blood pressure is 115/75 mm Hg. Cardiac examination shows no abnormalities except for a midsystolic click at the apex. The lungs are clear to auscultation of the chest. Which of the following best explains this patient's symptoms? A) Deconditioning B) Exercise-induced asthma C) Malingering D) Mitral valve prolapse E) Thyroid disease

B. Exercise-induced asthma most likely accounts for this patient's exertional dyspnea. Asthma is characterized by reversible obstruction of the bronchi secondary to hyperreactivity and airway inflammation. Patients present with episodes of wheezing, dry cough, and dyspnea occurring during or shortly after exercise, relieved after rest or the use of bronchodilators. Physical examination during an exacerbation often reveals tachycardia, tachypnea, diffuse wheezes (or rhonchi), and prolonged expiration relative to inspiration. Decreased tactile fremitus may be noted due to air trapping which decreases lung density (leading to reduced transmission of vibrations through the lung parenchyma to the body wall). Treatment is usually with a short acting bronchodilator (SABA) immediately before exercise, although in patients with concomitant asthma not related to exercise, treatment is directed by the severity of underlying asthma. If patients do not tolerate SABAS, montelukast is an alternative option. Incorrect Answers: A, C, D, and E. Deconditioning (Choice A) could cause dyspnea, or the subjective experience of shortness of breath, but it should not cause a cough. It would also be unlikely in a young person with a normal BMI who has previously participated in sports without difficulty. Malingering (Choice C) is defined by falsification of symptoms to obtain a secondary gain. This patient's symptoms have another possible explanation, exercise-induced asthma, and there is no clear secondary gain that he might obtain. Mitral valve prolapse (MVP) (Choice D) is less likely to explain this patient's symptoms than exercise-induced asthma. The mid-systolic click of MVP is often, but not always, followed by a systolic murmur of mitral regurgitation (MR) when symptomatic and causing cardiogenic pulmonary edema. Patients with congenital MVP often have physical findings including scoliosis, pectus excavatum, and low BMI. Symptoms of MVP (if symptomatic) are more likely to include chest pain, palpitations, and lightheadedness in addition to dyspnea. Thyroid disease (Choice E) is unlikely in this patient, as other associated findings should be present. Dyspnea can result from hyperthyroidism or thyroid storm, but these conditions generally present with diaphoresis, weight loss, exophthalmos, and tremor. Thyroid disease is also less common in male patients of this age group. Educational Objective: Exercise-induced asthma presents with dyspnea, cough, and/or wheezing that begins during or shortly after exercise. It is usually treated with SABAS although montelukast is an alternative therapy. %3D 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 25-year-old woman comes to the physician after her blood pressure was found to be 180/105 mm Hg at a health fair. She takes no medications. There is no family history of hypertension. Her last menstrual period was 1 week ago. Her blood pressure today is 180/110 mm Hg. Bilateral abdominal bruits are heard. Treatment with an angiotensin-converting enzyme (ACE) inhibitor will most likely have which of the following acute effects on this patient's renal function? A) Decreased concentrating ability secondary to renal angioedema B) Decreased glomerular filtration rate secondary to dilation of efferent arterioles C) Decreased renal blood flow secondary to dilation of afferent arterioles D) Increased concentrating ability secondary to a change in permeability of the collecting duct E) Interstitial nephritis secondary to allergic drug reaction

B. Fibromuscular dysplasia of the renal artery is the most common cause of renal artery stenosis in younger and middle-aged women. Fibromuscular dysplasia is a non-inflammatory and non-atherosclerotic angiopathy of medium-sized arteries (eg, renal, carotid) that results in multifocal fibrous and muscular thickening of the arterial wall, which can lead to stenosis. Renal artery stenosis is a cause of secondary hypertension due to abnormal stimulation of the juxtaglomerular apparatus from low afferent blood flow leading to excessive production of renin and angiotensin. The reduced afferent blood flow can result in progressive renal atrophy. Secondary hypertension should be considered in new-onset or treatment-resistant hypertension, or in younger, otherwise healthy patients. ACE inhibitors are a first-line treatment for hypertension as they block the conversion of angiotensin I to angiotensin II, which has direct vasoconstrictive effects as well as promotes salt and water retention. ACĒ inhibitors may result in a transient acute decrease in glomerular filtration rate (GFR) secondary to dilation of efferent arterioles. This effect is more pronounced in patients with renal artery stenosis, as the baseline reduced afferent blood flow leaves the nephron dependent on efferent arteriole vasoconstriction (mediated by angiotensin II) to maintain adequate filtration pressure across the glomerulus. ACE inhibitors block this effect. Incorrect Answers: A, C, D, and E. Decreased concentrating ability secondary to renal angioedema (Choice A) is incorrect. ACE inhibitors are associated with angioedema as an adverse effect due to increased bradykinin levels, which may result in swelling of the face, lips, tongue, upper airway, and gastrointestinal tract. Decreased renal blood flow secondary to dilation of afferent arterioles (Choice C) is incorrect as dilation of the afferent arterioles would result in increased renal blood flow. Increased concentrating ability secondary to a change in permeability of the collecting duct (Choice D) occurs with anti-diuretic hormone (ADH) release from the pituitary or exogenous administration of ADH analogs. ADH results in increased aquaporin expression on the luminal surface of collecting duct cells which increases the membrane permeability to water. Interstitial nephritis secondary to allergic drug reaction (Choice E) is a possibility with many medications, but is commonly associated with sulfa-based diuretics, nonsteroidal anti-inflammatory medications, antibiotics, proton pump inhibitors, and rifampin. Patients classically present with fever, hematuria, eosinophiluria, rash, and arthralgias. Educational Objective: ACE inhibitors should be used with caution in patients with renal artery stenosis, as reduced efferent arteriole vasoconstriction may result in a decreased GFR. %3D Previous Next Score Report Lab Values Calculator Help Pause

45 Exam Section 1: Item 45 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 45. A 76-year-old man comes to the physician for a follow-up examination. He has hypertension treated with a B-adrenergic antagonist. He lives on a farm in central California and says he has always distilled his own liquor. Before retiring 10 years ago, he worked in a hat factory and subsequently in a textile factory. He has smoked 2 packs of cigarettes daily for the past 55 years. He tells the physician that he has had several episodes of painful swelling of his right great toe. Physical examination shows several lesions consistent with gouty tophi over the elbows bilaterally. Laboratory studies show: Hemoglobin A10 5.6% Serum Glucose 93 mg/dL 3.2 mg/dL 7.9 mg/dL Creatinine Uric acid The most likely cause of this patient's condition is which of the following? A) Cigarette smoking B) Drinking home-distilled liquor C) Farming in central California D) Working in a hat factory E) Working in a textile factory

B. Gout is a chronic recurring inflammatory arthropathy in which deposition of uric acid crystals in both large and small joints, most commonly the first metatarsophalangeal joint, leads to pain and inflammation of the joint. Deposition of uric acid crystals may occur in the surrounding tissues (tophi). Increased intake of foods that are rich in purines such as seafood and red meats have been shown to increase the risk of gout flares, as they result in increased production of uric acid. Other causes of increased uric acid levels include diuretic use, tumor lysis syndrome, alcohol consumption, and obesity. It is hypothesized that alcohol and its metabolites compete with uric acid transporters in the kidney leading to decreased uric acid excretion. Thus, the ingestion of home-distilled liquor may result in decreased uric acid excretion leading to gout flares. Incorrect Answers: A, C, D, and E. Cigarette smoking (Choice A) has been linked to numerous pathologies including lung cancer, osteoporosis, bladder cancer, and chronic obstructive pulmonary disease. It has not been directly associated with gout flares and alteration of uric acid metabolism or excretion. Farming in central California (Choice C) is not associated with any specific pathology. In general, farming work may lead to increased sun exposure and risk of cutaneous malignancy, osteoarthritis linked to heavy manual labor, traumatic injury from farm equipment, and hypersensitivity pneumonitis related to the inhalation of organic materials. Working in a hat factory (Choice D) is classically associated with exposure to mercury used in the making of felt hats. This can lead to a neurological disorder known as erethism mercurialis (mad hatter disease) characterized by personality changes, memory loss, depression, apathy, delirium, neuropathy, gastrointestinal distress, and anemia. Working in a textile factory (Choice E) is associated with the inhalation of cotton particulates that leads to lung scarring and eventual respiratory failure (byssinosis). Educational Objective: Gout is a recurrent inflammatory arthropathy of one or more joints that is caused by increased serum levels of uric acid depositing in the synovium resulting in synovial inflammation. The increased intake of purines (eg, red meat, seafood) contributes to increased blood levels of uric acid. Alcohol metabolites are thought to compete with uric acid for excretion in the kidney, leading to gouty flares with ingestion. %3D Previous Next Score Report Lab Values Calculator Help Pause

77 Exam Section 2: Item 27 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 27. A 50-year-old man comes to the physician because of a 2-month history of pain of his wrists, changes in skin color, and progressive fatigue. His brother has type 2 diabetes mellitus and cirrhosis. Physical examination shows bronze-colored skin, tenderness of the metacarpophalangeal joints in both hands, and hepatosplenomegaly. Serum studies show: AST ALT Ferritin 100 U/L 110 U/L 1200 ng/mL Total iron-binding capacity 200 ug/dL (N=250-400) Transferrin saturation 80% (N=20-50) Analysis of a liver biopsy specimen shows a markedly increased iron concentration and cirrhosis. Which of the following is the most likely cause of the findings in this patient? A) Increased erythropoietin action B) Increased intestinal iron absorption C) Increased oral iron intake D) Decreased erythropoiesis E) Decreased iron excretion F) Decreased serum transferrin concentration

B. Hemochromatosis may be acquired or inherited secondary to mutations in the HFE gene, leading to abnormally increased intestinal iron absorption. 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. Due to increased free radical generation and oxidative damage, hemochromatosis can manifest with failure of the affected organs. It typically presents after decades of iron accumulation with liver failure manifest 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/or gonadal atrophy. Hemochromatosis, when acquired, often occurs in the setting of transfusion-dependent anemias such as thalassemia. Diagnostic studies may include liver biopsy, which commonly demonstrates excess iron deposition seen in hepatocytes on Prussian blue stain. Treatment involves serial phlebotomy. Incorrect Answers: A, C, D, E, and F. Increased erythropoietin action (Choice A) results in erythrocytosis and can be associated with certain cancers including renal cell carcinoma as a paraneoplastic syndrome. Erythropoietin is produced in the interstitial cells of the kidney and stimulates erythrocyte production in the bone marrow. It is triggered by hypoxia to increase production of red blood cells. Polycythemia can lead to itching and erythromelalgia but does not involve increased iron levels and iron deposition in tissues. Increased oral iron intake (Choice C) and decreased iron excretion (Choice E) are not the pathologic mechanisms in hereditary hemochromatosis. Normal iron sensing regulates the amount of intestinal absorption of iron. Defects in iron sensing in the HFE gene leads to hemochromatosis. Decreased erythropoiesis (Choice D) is seen in chronic kidney disease and end-stage renal disease, which results in anemia. In severe renal disease, erythropoietin may need to be supplemented. Decreased serum transferrin concentration (Choice F) is not a cause of hemochromatosis. Transferrin functions to transport and sequester iron to various tissues, and transferrin is saturated with iron in hemochromatosis. Increased intestinal absorption of iron in hemochromatosis leads to increased serum iron, ferritin, and transferrin saturation, as well as decreased total iron binding capacity. Educational Objective: Hemochromatosis may be acquired or inherited secondary to mutations in the HFE gene, leading to abnormally increased intestinal iron absorption. Hemochromatosis presents with liver failure, diabetes mellitus, arthritis, heart failure, darkening of the skin, and/or gonadal atrophy secondary to excess total body iron. I3D 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 53-year-old man has had progressive difficulty swallowing for the past 3 months. He has a 10-year history of heartburn with esophageal regurgitation of gastric contents. Tissue obtained on biopsy of the lower third of the esophagus is shown. Which of the following best describes the nature of this lesion? A) Basal zone hyperplasia of submucosal glands B) Intestinal metaplasia of squamous epithelium C) Malignant transformation of epithelium into squamous carcinoma OD) Squamous metaplasia of submucosal glands

B. Intestinal metaplasia of squamous epithelium in the esophagus, also known as Barrett esophagus, can be a consequence of prolonged gastroesophageal reflux disease (GERD), which occurs when acidic gastric contents reflux backward through the lower esophageal sphincter into the esophagus. The mucosa of the esophagus is comprised of squamous epithelium and does not traditionally encounter such an acidic environment. Constant exposure to acidic intraluminal contents induces a change in cell type from squamous epithelium to the columnar glandular epithelium found in the intestines as an adaptive response. These metaplastic cells will exhibit a brush border and goblet cells. Metaplasia can eventually lead to dysplasia, which is premalignant. Patients with confirmed Barrett esophagus should be evaluated at regular intervals determined by the presence and/or grade of dysplasia. Treatment involves ablation of the dysplastic cells via endoscopy and management of the underlying GERD with a proton pump inhibitor, dietary modification, and smoking cessation. Incorrect Answers: A, C, and D. Basal zone hyperplasia of submucosal glands (Choice A) is not the pathologic change observed in Barrett esophagus, although submucosal gland secretions do neutralize acidic luminal contents. They also lubricate the esophagus which allows for the food bolus to pass. Malignant transformation of epithelium into squamous carcinoma (Choice C) occurs with esophageal squamous carcinoma, which is more common in patients who consume alcohol and smoke cigarettes. Barrett esophagus primarily predisposes to adenocarcinoma, not to squamous carcinoma. Squamous metaplasia of submucosal glands (Choice D) is also associated with the development of esophageal adenocarcinoma. Submucosal glands contain progenitor cells that may play a role in the pathogenesis of dysplasia as these progenitor cells serve as a source of potentially dysplastic or neoplastic cells, however, the pathophysiology of Barrett esophagus involves intestinal metaplasia. Educational Objective: Barrett esophagus develops in individuals with chronic GERD and is histologically characterized by intestinal metaplasia whereby the normal squamous epithelium is replaced by columnar epithelium. Over time, dysplasia can develop, predisposing to esophageal adenocarcinoma. II Previous Next Score Report Lab Values Calculator Help Pause

73 Exam Section 2: Item 23 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 23. A case-control study is conducted to determine if obesity is a risk factor for gastroesophageal reflux disease (GERD). A questionnaire is used to select subjects with severe symptoms of GERD and subjects with no symptoms. A BMI is calculated for each subject. The results (in kg/m2) are shown: BMI<25 300 700 25<BMI<30 900 900 30<BMI<35 200 200 BMI>35 50 Subjects with GERD symptoms Subjects with no symptoms 30 Which of the following represents the odds ratio for GERD symptoms in subjects with BMIS greater than 35 compared with subjects with BMIS less than 25? A) (50×30) / (300x700) B) (50x700) / (30x300) OC) [50/(30+50)] / [300/(300+700)] OD) [50/(50+300)] / [30/(30+700)] O E) [50/(50+200+900+300)] / [30/(30+200+900+700)]

B. Odds ratio is a comparison of the odds of an outcome occurring in the exposed group with the odds of that outcome occurring in a nonexposed comparison group. It is calculated as the odds of the outcome of interest in the exposed group divided by the odds of the outcome of interest in the nonexposed group. In this case-control study, the odds ratio would be calculated as follows. The odds of gastroesophageal reflux disease (GERD) in patients with a BMI > 35 would be 50 (number of subjects with GERD symptoms) divided by 30 (number of subjects with no symptoms) (50/30 = 1.667) indicating that a person in this category is more likely than not to have symptoms of GERD. The odds of having GERD in a patient with a BMI < 25 would be 300 divided by 700 (300/700 = 0.429). Čalculating the ratio between these two odds (odds ratio) would be: (50/30) / (300/700) = 3.889. This equation can be rearranged to (50×700) / (30x300). Incorrect Answers: A, C, D, and E. (50x30)/ (300x700) (Choice A) is an erroneous expression of the true odds ratio (50/30) / (300/700)), which incorrectly substitutes multiplication for division. [50/(30+50)] / [300/(300+700)] (Choice C) is a calculation of a relative risk ratio. This is an inappropriate calculation as a case-control study has a defined number of outcomes of interest (eg, GERD) as these individual cases were selected during the study design. Because the number of GERD cases has been predetermined, calculating the risk of disease is inappropriate. [50/(50+300)] / [30/(30+700)] (Choice D) is also an erroneous calculation. It does not calculate relative risk or odds ratio. [50/(50+200+900+300)] / [30/(30+200+900+700)] (Choice E) calculates the number of patients with a BMI > 35 in the GERD group divided by all patients with GERD symptoms (risk of BMI > 35 given GERD symptoms). Also calculated is the number of patients with a BMI > 35 with no symptoms divided by the total number of patients without symptoms (risk of BMI >35 given no GERD symptoms). These two numbers are then divided to form a ratio (0.0344/0.0164 2.098), meaning that patients with GERD are twice as likely to have a BMI > 35. Although this calculation is technically reasonable, it is not the focus of the study and is an expanded ratio. Educational Objective: Odds ratio is calculated commonly in case-control studies to evaluate the likelihood of exposure to a certain risk factor given a disease or non-disease state. Relative risk ratios are not calculated in case-control studies as the disease or outcome rate (cases) has been predetermined by the investigator. %3D Previous Next Score Report Lab Values Calculator Help Pause

48 Exam Section 1: Item 48 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 48. Based on the graph of p-aminohippurate (PAH) concentration versus PAH secretion, which of the following is lower at point Y than at point X? Y 80 A) Glomerular filtration rate B) PAH clearance C) PAH excretion rate X 40 D) PAH filtered load E) Renal blood flow 20 40 60 80 100 Plasma p-aminohippurate (PAH) concentration mg/dL

B. PAH clearance is lower at point Y than at point X. Renal clearance is defined as the volume of plasma that is completely cleared of a given substance in a specified unit of time. Increased clearance of PAH would mean that more plasma is cleared of PAH per unit of time, while decreased clearance would indicate the opposite. The pictured graph indicates that as plasma PAH concentrations increase, the amount of PAH cleared by the kidneys into the urine increases in a linear fashion until an inflection point where the slope of the graph becomes zero. This inflection point indicates the plasma PAH concentration at which the ability of the kidneys to clear higher concentrations of PAH reaches its maximum. Beyond this concentration, the kidneys are unable to clear any additional PAH. At point Y, there is a higher plasma concentration of PAH when compared to X. While the absolute secretion of PAH at point Y exceeds that at point X, the clearance is lower because the kidneys are unable to increase their rate of clearance despite an increasingly high concentration of PAH. Incorrect Answers: A, C, D, and E. Glomerular filtration rate (GFR) (Choice A) describes the volume of renal blood flow through the glomerular apparatus per unit of time. It is highly regulated by changes in the size of the afferent and efferent arterioles but is also affected by the state of the glomerular apparatus, which is affected in patients with chronic kidney disease. The GFR would equal the PAH clearance if PAH were not actively excreted. PAH excretion rate (Choice C) is not correct as the excretion rate of PAH is higher at point Y than at point X. PAH filtered load (Choice D) is higher at point Y than at point X. The filtered load is dependent upon the concentration of PAH in the plasma, with increased plasma concentrations correlating with an increased filtered load. The filtered load limit is ultimately determined by the GFR. Renal blood flow (Choice E) cannot be interpreted from this graph. Educational Objective: Renal clearance is defined as the volume of plasma that is completely cleared of a given substance in a specified unit of time. Clearance often reaches an inflection point beyond which increasing plasma concentrations result in no additional urinary clearance of the substance, which is demonstrated in the pictured graph. While the excretion rate and filtered load of PAH are higher at point Y as compared to point X, the PAH clearance is lower. %3D Previous Next Score Report Lab Values Calculator Help Pause PAH secreted mg/min

82 Exam Section 2: Item 32 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 32. A 63-year-old man is scheduled to undergo coronary artery bypass grafting with a portion of the great saphenous vein. An incision to remove a portion of the vein should begin in which of the following locations? A) Along the lateral surface of the leg B) Along the medial side of the ankle joint OC) Along the plantar surface of the foot D) Anterior to the knee joint E) Posterior to the hip joint

B. The great saphenous vein is the longest vein in the body and runs subcutaneously along the medial aspect of the leg. It starts just proximal to the arch of the foot and passes anterior to the medial malleolus of the ankle. On persons with lower BMI, it can be palpated at this location. It courses proximally along the medial aspect of the knee and continues along the medial aspect of the thigh until it joins the common femoral vein in the region of the femoral triangle in the groin. This vein is often used for bypass grafting by cardiothoracic and vascular surgeons, and to remove a portion, the incision should begin along the medial side of the ankle joint. Its subcutaneous location makes it easy to harvest and redundancy of the venous outflow of the leg (eg, tibial and popliteal veins) makes it a non-essential vein for drainage of the lower extremity. Incorrect Answers: A, C, D, and E. The lateral surface of the lower leg (Choice A) is drained by the small saphenous vein, which bridges off from the femoral vein at the level of the popliteal fossa. This vein is less commonly used as a bypass graft. The plantar surface of the foot (Choice C) has a venous network that includes the medial plantar vein, the lateral plantar vein, and the deep plantar venous arch, which connects the medial and lateral veins. This vascular network is not used for grafting because the veins are short and surgery to the plantar surface of the foot can cause long-term pain due to scarring on a weightbearing surface. Anterior to the knee joint (Choice D) are the genicular veins. These are small veins that drain the area around the knee joint into the popliteal vein. They are short and are not used for bypass grafting. Posterior to the hip joint (Choice E) are the superior and inferior gluteal veins. These veins drain the gluteal musculature posterior to the hip and drain into the internal iliac vein. Educational Objective: The great saphenous vein is the longest vein in the body. It runs along the medial aspect of the lower extremity. It can be easily identified in some patients along the anterior aspect of the medial malleolus of the ankle. It not essential for venous drainage of the lower extremity, making it optimal for bypass grafting. %3D Previous Next Score Report Lab Values Calculator Help Pause

69 Exam Section 2: Item 19 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment I II 6. 1 4 II 36 1 1 IV 2|5 2 Affected male Affected female Unaffected male O Unaffected female Affected male, deceased Ø Unaffected female, deceased 19. An investigator is studying a large family with many members who are affected by a disorder caused by a fully penetrant autosomal dominant inherited gene mutation. A pedigree is shown. Most affected members also have a rare allele at a locus thought to be closely linked to the disease locus. A father (individual III-3) and his daughter (individual IV-3) have the disorder, but they have the wild-type allele at the linked locus. Which of the following is the most likely cause of these findings? A) Insertion of a LINE sequence B) Random segregation C) Recombination D) Single nucleotide polymorphism E) Transduction

C. Recombination takes place in prophase I of meiosis and is the process by which two chromatids exchange a small portion of genetic material and separate, or unlink, two traits previously present on the same chromatid. In this example, all the individuals have one chromosome with alleles 1, 2, and 3 at three adjacent loci. Most of the affected individuals have the alleles 4, 5, and 6 on their second chromosome at these same loci, indicating that it is this chromosome which carries the genetic defect. Two affected individuals still have alleles 5 and 6 but have replaced 4 with 1 at the first locus. This suggests that recombination occurred between the first two gene loci causing allele 4, which had previously been linked to the locus with the mutation, to become unlinked. The chance that a recombination event will occur between two given loci is called the recombination fraction. Incorrect Answers: A, B, D, and E. Insertion of a long interspersed nuclear element (LINE) sequence (Choice A) is a method of bacterial genetic variation through transposable elements. This is seen in bacteria when a segment of DNA moves from a chromosome to a plasmid. When it occurs, regions of DNA on either side may also be transposed, conferring new properties to the recipient. Random segregation (Choice B) is a principle applying to meiosis, the separation of DNA into haploid gametes, which states that parental genes separate randomly and equally into gametes. Each gamete has an equal chance of getting each parental allele. It does not impact the linkage of alleles at two closely located gene loci. Single nucleotide polymorphism (Choice D) is the most common type of genetic variation among individuals and is characterized by a single nucleotide changing from one base to another, altering that site on the DNA. A single nucleotide polymorphism might cause an allele at a given gene locus to be altered but does not separate linked alleles from each other as occurs during recombination. Transduction (Choice E) is a method by which bacterial genes are taken from one bacterium and given to another. It occurs when a lytic phage infects a bacterium, cleaves bacterial DNA, and then packages some of those cleaved pieces into a new phage capsid. That phage then infects another bacterium, leading to transfer of the bacterial genes. This is not a mechanism of genetic variation in eukaryotic cells. Educational Objective: Recombination takes place in prophase I of meiosis and is the process by which two chromatids exchange a small portion of genetic material and separate, or unlink, two traits previously present on the same chromatid. I3D Previous Next Score Report Lab Values Calculator Help Pause 4. 3. -23 956 123 2. 2. -23 -23 123 456 -23 -23 123

87 Exam Section 2: Item 37 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 40 - 37. Which of the following is the mean number of episodes of urinary tract infections for children (N3D100) in the sample shown in the graph? 35 A) 1 30- 30 B) 1.55 25 C) 2.07 20 D) Cannot be determined from this graph 10 10 0 1 2 Number of urinary tract infections

B. This graph is a bar chart that demonstrates counts (number of children) on the Y-axis and number of urinary tract infections (UTIS) on the X-axis. The count (number of children) represented by each bar is also provided at the top of the bar. Thus, this graph communicates that within this study, 25 children had no UTI infections, 30 children had one UTI, 10 children had two UTIS, and 35 children had three UTIS. This information is sufficient for calculating the overall mean number of UTIS in the study sample. Calculation of the mean number of UTIS would be equal to the total number of UTIS divided by the total number of children. Formally, this is calculated as: (25x0) + (30x1) + (10x2) + (35x3)) / (25 + 30 + 10 + 35) = 1.55. Incorrect Answers: A, C, and D. 1 (Choice A) and 2.07 (Choice C) are incorrect numeric calculations. They are not the mean number of infections per child. 2.07 would be the mean if only non-zero data points were considered (if the average were computed considering children that had at least one UTI). Cannot be determined from this graph (Choice D) is incorrect as the graph and the additional numeric labels on the bars provide sufficient information for calculation of the mean number of infections. Educational Objective: Mean is a basic measure of central tendency and is calculated as a simple average. When calculating the mean number of outcomes per person, the total number of outcomes is divided by the total number of individuals. %3D Previous Next Score Report Lab Values Calculator Help Pause Number of children

23 Exam Section 1: Item 23 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 23. A 17-year-old boy is brought to the physician by his mother because she is concerned that his puberty is delayed. The mother states, "He is so short. His father is 6 feet 5 inches tall. I don't understand why he has not had his growth spurt." When the mother leaves the room, the patient states, "I'm fine. I don't know what's the matter with her. She wants me to be tall like my dad." The patient is 175 cm (5 ft 9 in) tall and weighs 70 kg (155 Ib); BMI is 23 kg/m2. Sexual development is Tanner stage 4. In addition to reassuring the mother that her son is fine, which of the following is the most appropriate initial statement by the physician to the mother? A) "Since your son is fine with his height, you should try to accept him as he is." B) "Tell me more about your concerns about your son's height." C) "We'll do some blood tests just to be sure that all your son's hormone levels are okay." D) "Your son is average for his height and weight." E) "Your son is not going to be any taller."

B. When patients or patients' family members express medical concerns, physicians should initially ask open-ended questions to explore the understanding and fears of the patient or family member. The physician can then tailor further discussion and reassurances to address these knowledge gaps and fears. Asking open-ended questions also invites the family member to elaborate on their concerns, as there may be medically or psychiatrically relevant details that the family member reveals on further discussion. Further, listening to the specific concerns of the patient or family members will improve therapeutic alliance. Incorrect Answers: A, C, D, and E. Giving parental advice (Choice A) would be unwarranted in this situation and outside of the physician's scope of practice. If a parent's behavior is clearly affecting the mental or physical health of the patient, the physician may tactfully bring the issue to the parent's attention. However, this patient's health is not clearly impacted by his mother's concern. Physicians should refrain from ordering tests that are medically unnecessary based on patient or family concern (Choice C). The physician should instead reassure and educate after listening to the patient or family's specific concerns. Saying that the patient is average for his height and weight (Choice D) would not address this mother's concern about the patient being shorter than his father. This statement would also prevent elaboration of the mother's specific concerns, understanding, and fears. Informing the patient's mother that her son is not going to be any taller (Choice E) would not be accurate or reassuring. This statement would also prevent elaboration of the mother's specific concerns, understanding, and fears. Educational Objective: When patients or families express medical concerns, physicians should ask open-ended questions to elucidate the specific nature of the concern and the patient's or family member's understanding. The physician can then tailor further discussion to address knowledge gaps and specific fears, and may learn additional, medically relevant details about the concern. %3D 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 16-year-old boy with moderate intellectual disability is brought to the physician for a routine examination. There is a family history of mild and moderate intellectual disability in his mother and brother, respectively. Physical examination shows a long face, prominent ears, and moderately enlarged testicles. Which of the following best describes the genetic mechanism of this patient's disorder? A) Mutation in a mitochondrial gene B) Presence of an extra sex chromosome C) Translocation of a portion of an autosome D) Trinucleotide repeat mutation on the X chromosome E) Trisomy of an autosome

D. This patient's constellation of findings, including intellectual disability, an elongated face, large ears, and macro-orchidism is suggestive of Fragile X syndrome. Patients may also present with hyperextensible joints and a high-arched palate. Fragile X syndrome is a common cause of inherited intellectual disability and is inherited 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 increased risk of 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, B, C, and E. Mutation in a mitochondrial gene (Choice A) 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 mitochondrial encephalopathy, lactic acidosis, and stroke-like episodes (MELAS) syndrome. Presence of an extra sex chromosome (Choice B) is observed in Klinefelter syndrome, in which patients have an XXY karyotype. Characteristic features include intellectual disability, eunuchoid body shape, tall stature, elongated extremities, and hypogonadism. Translocation of a portion of an autosome (Choice C) is observed in Robertsonian translocations and causes a small portion of cases of Down syndrome. Characteristic physical findings of Down syndrome include intellectual disability, broad, flat, facial features with prominent epicanthal folds, and a single palmar crease. Patients with Down syndrome are at increased risk of Alzheimer dementia, acute lymphoblastic and acute myeloid leukemia, cardiac septal defects, duodenal atresia, and Hirschsprung disease. Trisomy of an autosome (Choice E) is observed in Down syndrome (trisomy 21), Edwards syndrome (trisomy 18), and Patau syndrome (trisomy 13). Edwards syndrome presents with characteristic physical features including intellectual disability, a prominent occiput, low-set ears, micrognathia, clenched, overlapping fingers, and feet with a prominent calcaneus and convexly rounded soles. Patau syndrome presents with characteristic physical features including intellectual disability, cleft lip and palate, holoprosencephaly, microphthalmia, cutis aplasia, feet with a prominent calcaneus and convexly rounded soles, and polydactyly. Educational Objective: Fragile X syndrome is an X-linked dominant cause of inherited intellectual disability that presents due to 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 macro-orchidism. %3D Previous Next Score Report Lab Values Calculator Help Pause

76 Exam Section 2: Item 26 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 26. An 18-year-old woman is being evaluated for amenorrhea. She has never had a menstrual period. She is 183 cm (6 ft) tall. Breast development and external genitalia are normal. There is no axillary or pubic hair. Which of the following karyotypes is most likely? A) 45,X B) 46,XX C) 46,XY D) 46,X,i(Xq) E) 47,XXX

C. Androgen insensitivity syndrome is due to a defect in the androgen receptor complex resulting in a genetically 46,XY male developing a female phenotype. Testes are present and produce testosterone normally, but absence of a functioning androgen receptor prevents hormone binding and thereby prevents development of male sexual characteristics. Patients present with female external genitalia, scant pubic and axillary hair, absent uterus and fallopian tubes, and a rudimentary vagina. Patients demonstrate increased levels of testosterone, estrogen, and luteinizing hormone. Menses will not occur due to the lack of cycled progesterone and estrogen, and the lack of a functional uterus with endometrial lining. Incorrect Answers: A, B, D, and E. 45,X (Choice A), also known as Turner syndrome, 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. Turner syndrome commonly results from monosomy of the X chromosome, but may also result from partial deletion of the chromosome, as in patients with isochromosome Xq, 46,X,i(Xg) (Choice D). 46,XX (Choice B) represents a normal female phenotype. This patient's height, amenorrhea, and absence of axillary and pubic hair are more suggestive of androgen insensitivity. 47,XXX (Choice E), also known as trisomy X or triple X syndrome, presents with tall stature but also with epicanthal folds and intellectual disability. Absence of axillary and pubic hair is not a feature of this syndrome. Educational Objective: Androgen insensitivity syndrome occurs due to a defect in the androgen receptor, and results in persons with 46,XY chromosomes developing phenotypically female characteristics. %3D Previous Next Score Report Lab Values Calculator Help Pause

56 Exam Section 2: Item 6 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 6. A 33-year-old woman has had weakness of the right lower two thirds of the face for the past 2 months. Which of the following labeled regions in the normal brain shown is the most likely site of the lesion causing this symptom? Central sulcus EF A- OA) OB) C) D) E) F) G) OH) OJ)

C. Choice C identifies the left posterolateral frontal lobe, representing the primary motor cortex in the precentral gyrus. This brain region mediates motor function of the right-sided (contralateral) face. Lesions of this area result in weakness, facial droop, and an upper motor neuron pattern of dysfunction (eg, hyperreflexia). Importantly, the forehead would be spared, as the cranial nerve VII nucleus that controls forehead musculature is dually innervated by upper motor neurons from bilateral precentral gyri. Because of this, the patient would be expected to symmetrically raise her eyebrows. Alternatively, lesions involving the lower motor neurons of facial expression (eg, facial nerve inflammation in Bell palsy) affect the forehead and lower face together. A lesion of the precentral gyrus in this 33-year-old woman could be caused by autoimmune disease such as multiple sclerosis, along with malignancy, trauma, or stroke. Incorrect Answers: A, B, D, E, F, G, H, I, and J. Choice A identifies the left posterior, inferior frontal lobe. In the dominant hemisphere (typically the left hemisphere), this brain region represents the Broca area. Lesions can result in Broca aphasia. Choice B identifies the prefrontal cortex. This area is associated with functions including learning, reasoning, problem solving, emotion, behavioral control, memory, self-regulation, and personality. Lesions affecting this area may result in behavioral dysregulation and the development of psychiatric symptoms (eg, post-stroke depression). Choice D identifies the left posteromedial frontal lobe, representing the primary motor cortex in the precentral gyrus. This brain area controls the motor function of the right (contralateral) lower extremity. Lesions of this region would lead to an upper motor neuron pattern of weakness of the right leg. Choice E identifies the left anteromedial parietal lobe, representing the primary sensory cortex in the postcentral gyrus. This brain area controls sensation of the right (contralateral) leg. Lesions of this region would lead to sensory deficits of the right leg. Choice F identifies the left anterolateral parietal lobe, representing the primary sensory cortex in the postcentral gyrus. This brain region mediates sensation in the right (contralateral) distal upper extremity and face. Choice G identifies the left angular gyrus in the inferior parietal lobe. This brain area mediates multimodal sensory integration and assists in mental spatial rotation, paying attention, and solving problems. Patients with left angular gyrus damage may demonstrate agraphia, acalculia, finger agnosia, and left-right disorientation (known as Gerstmann syndrome). Choice H identifies the left posterior, superior temporal gyrus, an area that in the dominant hemisphere makes up one portion of Wernicke area, a brain region involved in the understanding of language. Lesions involving this area can result in Wernicke aphasia. Choice I identifies the middle temporal gyrus, which assists in semantic memory processing, visual perception, and sensory integration. Lesions in this brain area have been associated with deficits in complex visual perception and semantic memory processing. Choice J identifies the superior temporal gyrus, which represents the auditory association area. Lesions of this area may disrupt spoken word recognition. Educational Objective: The lateral aspect of the precentral gyrus mediates the motor function of the contralateral face. Lesions affecting this region can result in facial droop, weakness, and an upper motor neuron-pattern of dysfunction. II Previous Next Score Report Lab Values Calculator Help Pause www

74 Exam Section 2: Item 24 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 24. A 14-year-old boy is brought to the physician by his mother because of daily headaches for 2 months. The headaches are described as a bilateral aching in the temples. His mother states that he also "has not been himself" for the past few months. He seems more confused, often forgetting names, dates, and places, and he is clumsy with frequent falls. His school performance also has declined over the past quarter. Physical examination shows a broad-based, ataxic gait. He is alert and oriented to person, place, and time, but he is slow to answer questions. Chronic abuse of which of the following substances is the most likely cause of this patient's condition? A) Cocaine B) Ethanol C) Inhaled glue D) Methamphetamines E) PCP (phencyclidine)

C. Chronic inhalant abuse (eg, glue, spray paint, shoe polish, toluene, nitrous oxide) is common in children and adolescents and has effects on multiple organ systems: neuropsychiatric (headache, cognitive impairment, anosmia, cerebellar dysfunction, mood swings, irritability, hallucinations), dermatologic (perioral or perinasal dermatitis), otolaryngologic (nosebleeds, halitosis), ocular (conjunctival injection), cardiac (dysrhythmia, tachycardia), gastrointestinal (nausea, anorexia), and respiratory (wheezing, coughing, sneezing). This patient's recent confusion, broad-based ataxia consistent with cerebellar dysfunction, and poor school functioning are likely related to the neuropsychiatric effects of inhalants. Treatment is primarily supportive, though some inhalants have specific antidotes (eg, methylene blue for nitrites). Prevention is key; schools should monitor the use of solvent-based products and parents and children should be educated about the risks of inhalants. Incorrect Answers: A, B, D, and E. Chronic cocaine abuse (Choice A) may lead to otolaryngologic symptoms such as nasal septal perforation and nosebleeds as well as neuropsychiatric symptoms such as anosmia, cognitive impairment, and psychotic symptoms (delusions, hallucinations). Cerebellar dysfunction and headaches would be atypical of chronic cocaine use. Chronic ethanol abuse (Choice B) can lead to ventricular and sulcal enlargement and consequent cognitive impairment. Wernicke-Korsakoff syndrome may cause severe cognitive impairment as well as ophthalmoplegia and ataxia, and decades of ethanol use can lead to cerebellar degeneration. However, cerebellar dysfunction is unlikely to be seen after only two months of ethanol abuse, and headaches are more typical of inhalant abuse. Chronic methamphetamine abuse (Choice D) can lead to methamphetamine-induced psychotic disorder, which features chronic delusions, paranoia, and hallucinations. Chronic methamphetamine use may also lead to cognitive impairment. However, cerebellar dysfunction and headaches would be atypical. Chronic PCP (phencyclidine) abuse (Choice E) can lead to depression, psychotic symptoms, memory loss, and dysarthria. Headaches and cerebellar dysfunction would be atypical. Educational Objective: Chronic inhalant abuse can lead to several neuropsychiatric manifestations such as headache, cognitive impairment, anosmia, cerebellar dysfunction, mood swings, irritability, and hallucinations. Cerebellar dysfunction and headaches distinguish chronic inhalant abuse from the chronic abuse of many other substances, although ethanol also leads to cerebellar dysfunction after many years of abuse. %3D Previous Next Score Report Lab Values Calculator Help Pause

37 Exam Section 1: Item 37 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 37. A 5-year-old girl with premature sexual development is diagnosed with precocious puberty. Pelvic examination shows a mass consistent with an ovarian tumor. Laboratory studies show decreased serum concentrations of gonadotropins and a marked increase in circulating estrogens. The ovarian tumor is most likely derived from which of the following cell types? A) Endothelial cells B) Germinal epithelium C) Granulosa cells D) Stromal fibroblasts E) Thecal cells

C. Granulosa cell tumors are a type of malignant sex-cord stromal tumor. They generally occur in women in their sixth decade of life, though can occur at any age, including during childhood. Their presentation is marked by the effect of their functional production of estrogen, a typical product of granulosa cells. Thus, it may present with precocious puberty, as in this patient, or vaginal bleeding in younger or premenarcheal women. Precocious puberty is suspected when girls younger than 8 years old or boys younger than 9 years old develop secondary sexual characteristics. Sexual maturity rating stage 3 characteristics, including thickening of pubic and axillary hair and breast enlargement, are typically seen in girls ages 11 to 13. The presence of these findings at a significantly earlier age suggests precocious puberty and warrants evaluation. Granulosa cell tumors are typically indolent and may not be detected until large or advanced. On histology, granulosa cell tumors demonstrate Call-Exner bodies, which are granulosa cells arranged around eosinophilic fluid, resembling ovarian follicles. Treatment is through surgical excision, and if the tumor is early stage, prognosis is generally favorable. In the postmenopausal patient, granulosa cell tumors often present with postmenopausal vaginal bleeding, which should prompt investigation. Incorrect Answers: A, B, D, and E. Proliferation of endothelial cells (Choice A) leads to vascular neoplasms, such as angiosarcoma and Kaposi sarcoma. A mature or immature teratoma contains tissue derived from all three embryological tissue lines, which could include vascular tissue, but these are not primary malignancies of the endothelial cells. Germinal epithelium (Choice B) is the layer of cells that cover the surface of the ovary. Most ovarian tumors, including the common serous cystadenocarcinoma, arise from germinal epithelium. However, these do not typically produce hormones. The neoplastic proliferation of stromal fibroblasts (Choice D) results in the formation of a benign ovarian fibroma, which is the most common type of ovarian sex-cord stromal tumor. They are often asymptomatic and do not result in the production of active hormones. Meigs syndrome consists of an ovarian fibroma in association with ascites and a pleural effusion. A neoplastic proliferation of thecal cells (Choice E) is called a thecoma. These tumors may also produce estrogen or androgen and may present with postmenopausal bleeding but are less common than granulosa cell tumors. Educational Objective: Granulosa cell tumors are malignant sex-cord stromal tumors that typically occur in women and result in the unregulated production of estrogen. In children and young women, they often result in precocious puberty or premenstrual vaginal bleeding. %3D Previous Next Score Report Lab Values Calculator Help Pause

34 Exam Section 1: Item 34 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 34. A 22-year-old woman is brought to the emergency department in a semicomatose condition after collapsing near the end of running a marathon. Her prerace weight was 47 kg (103 Ib). She now weighs 50 kg (110 Ib). Her pulse is 115/min, respirations are 15/min, and blood pressure is 90/50 mm Hg. Physical examination shows cool, dry skin. She is responsive to painful stimuli. Laboratory studies show: Serum Na+ 116 mEq/L 4.8 mEq/L 89 mEq/L 22 mEq/L Urea nitrogen 22 mg/dL 101 mg/dL 1 mg/dL K+ CI- HCO,- Glucose Creatinine This patient's condition is most likely due to which of the following? A) Decreased ADH (vasopressin) B) Decreased aldosterone C) Excessive fluid intake D) Inadequate fluid intake E) Increased aldosterone

C. Hyponatremia is often asymptomatic if chronic and slowly developing, although when acute, it is associated with central nervous system symptoms such as headache, nausea, vomiting, confusion, delirium, weakness, seizures, and coma. Volume status and osmolality should be determined to evaluate the etiology of hyponatremia. In this patient who is hyponatremic and hypervolemic as evidenced by acute weight gain, etiologies can include syndromes of fluid overload (eg, cirrhosis, congestive heart failure, nephrotic syndrome) or excessive fluid intake. Excessive fluid intake and hypervolemia leads to dilutional hyponatremia. The abrupt decrease in serum osmolality leads to transcellular shifting of fluid by osmosis, with neuronal swelling (cerebral edema) in the central nervous system, which can lead to brain herniation. Emergency treatment involves administration of hypertonic saline to correct osmolar shifts and address severe neurologic symptoms and prevent seizures due to hyponatremia. Management for hypervolemic hyponatremia involves water restriction, diuretics, and dialysis in severe cases. Incorrect Answers: A, B, D, and E. ADH (vasopressin) acts at the level of the distal convoluted tubule and collecting duct to increase water reabsorption. Water reabsorption in absence of solute reabsorption, specifically sodium, will lead to euvolemic, hypoosmolar hyponatremia. Decreased ADH (Choice A) secretion would lead to decreased water reabsorption and hyperosmolar hypernatremia, not hyponatremia. Aldosterone promotes reabsorption of sodium in the distal convoluted tubule and collecting ducts, causing indirect reabsorption of water, and to a lesser extent than the water reabsorption that occurs with ADH. Due to its actions on a sodium-potassium pump, it also leads to the secretion of potassium in the urine. This generally leads to sodium reabsorption in excess of water reabsorption. Decreased aldosterone (Choice B) would lead to decreased sodium and water reabsorption and would also present with metabolic acidosis. It would be unlikely to cause acute, severe hyponatremia as seen in this patient due the proximate effect of increased fluid intake while running, though could cause chronic hyponatremia. Increased aldosterone (Choice E) would lead to the opposite effect (eg, mild hypernatremia, hypertension, and hypokalemia). Inadequate fluid intake (Choice D) would lead to weight loss and hypernatremia. During marathons, runners lose fluid through diaphoresis and respirations and require adequate hydration. These losses combined with inadequate hydration would lead to hypernatremia and hyperosmolar serum. Educational Objective: Acute hyponatremia is associated with central nervous system symptoms such as headache, nausea, vomiting, confusion, delirium, weakness, seizures, and coma. Excessive fluid intake can lead to hypervolemic hyponatremia and cerebral edema. %3D Previous Next Score Report Lab Values Calculator Help Pause

46 Exam Section 1: Item 46 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 46. A 22-year-old woman comes to the physician because of a 6-month history of difficulty swallowing. She says that she feels like she is choking on both solids and liquids. She has no pain with swallowing. She has had a 4.5-kg (10-lb) weight loss during this time. There is no history of fever or chills. She is not sexually active. She does not smoke cigarettes or use illicit drugs. She is 170 cm (5 ft 7 in) tall and weighs 59 kg (130 lb); BMI is 20 kg/m2 Her vital signs are within normal limits. Physical examination shows no abnormalities. An x-ray of the esophagus is shown. Which of the following is the most likely explanation for this patient's symptoms? H1:439 H2:256 E 15% A) Acid reflux into the lower esophagus B) Atrophy of the smooth muscle in the esophagus C) Inflammatory degeneration of esophageal wall neurons D) Longitudinal mucosal tear at the esophagogastric junction O E) Perforation of the esophageal wall Physt

C. Inflammatory degeneration of esophageal wall neurons has led to the development of achalasia in this patient with weight loss and dysphagia to both solids and liquids. Achalasia is an esophageal dysmotility disorder resulting from deficient peristalsis as a result of impaired neuromuscular transmission and impaired relaxation of the lower esophageal sphincter (LES). It manifests as dysphagia, odynophagia, weight loss, halitosis, and regurgitation of undigested food. It is diagnosed with a barium swallow and esophageal manometry. Destruction of nerves in the myenteric plexus impairs local nitric oxide production, prevents smooth muscle relaxation, and disproportionately affects inhibitory neurons, which function to relax the LES. An increase in LES pressure and dysfunctional peristalsis cause gradual dilation of the esophagus with retention of food, leading to dysphagia and regurgitation. On barium esophagography, achalasia classically appears as a dilated esophagus with a "bird beak" distal taper at the LES. Treatment includes pneumatic dilation or injection of botulinum toxin to relax the LES. Incorrect Answers: A, B, D, andE. Acid reflux into the lower esophagus (Choice A) describes gastroesophageal reflux disease (GERD). GERD commonly occurs as a result of increased intra-abdominal pressure related to obesity or diminished closing pressure of the LES permitting acidic contents of the stomach to reflux into the esophagus. Chronic GERD can lead to intestinal metaplasia of the lower esophagus, which predisposes to esophageal cancer. Esophageal cancer can present with dysphagia and odynophagia and can appear on x-rays as similar to achalasia but would be uncommon in a young patient. Atrophy of the smooth muscle in the esophagus (Choice B) describes the pathophysiologic mechanism by which scleroderma causes dysphagia. Gradual atrophy of smooth muscle leads to the loss of LES tone and dysfunctional peristalsis. Longitudinal mucosal tear at the esophagogastric junction (Choice D) is known as a Mallory-Weiss tear and results in hematemesis and chest or epigastric pain. It often occurs after prolonged retching or vomiting, and gradually heals without intervention. Perforation of the esophageal wall (Choice E), referred to as Boerhaave syndrome, develops commonly after forceful vomiting or retching. It presents with substernal chest pain and odynophagia; x-rays and CT scans typically reveal pneumomediastinum. Extraluminal contrast would be seen on a fluoroscopic esophagogram. Depending upon the location of the tear, gastric contents may invade the mediastinum or the pleura, leading to mediastinitis or pleural effusion. Surgical repair is required for definitive management although stenting via esophagogastroduodenoscopy can be used as a temporizing measure. Educational Objective: Achalasia results from the destruction of neurons in the myenteric plexus resulting in dysfunctional peristalsis and a chronically increased LES tone. Presenting symptoms include dysphagia and regurgitation; progressive dilation of the esophagus is seen on imaging. Treatment is with pneumatic dilation or botulinum toxin. Previous Next Score Report Lab Values Calculator Help Pause 00

68 Exam Section 2: Item 18 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 18. A 14-year-old boy has persistent leukocytosis and neutrophilia without evidence of a current infection. He has a history of recurrent infections of the skin, upper and lower airways, and perirectal area. Gram-negative and gram-positive rods have been isolated. The number and function of B and T lymphocytes are normal. Production of hypochlorous acid by neutrophils and the nitroblue tetrazolium reduction test are normal. Neutrophil chemotactic response to the formyl-MetLeuPhe (FMLP) peptide is diminished. Which of the following disorders of neutrophils is the most likely diagnosis? A) Chronic granulomatous disease B) Cyclic neutropenia C) Leukocyte adhesion deficiency D) Myeloperoxidase deficiency E) Neutrophil-specific granule deficiency

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. It is typically characterized by recurrent bacterial infections, impaired wound healing, a delayed detachment of the umbilical cord after birth, and a lack of leukocytes at sites of infections with an absence of pus. The actions of leukocyte phagocytosis and bacterial killing are otherwise unimpaired. Laboratory studies in patients with LAD will reveal increased leukocyte levels in the blood. Leukocyte migration studies may reveal decreased responsiveness to chemotactic agents. Incorrect Answers: A, B, D, and E. Chronic granulomatous disease (Choice A) results from a defect in the nicotinamide adenine dinucleotide phosphate (NADPH) oxidase complex. Diagnosis can be made by an abnormal nitroblue tetrazolium reduction test as functional NADPH oxidase is needed to reduce nitroblue. Cyclic neutropenia (Choice B) is a rare inherited immunodeficiency syndrome associated with dysfunctional neutrophil elastase. It is characterized by recurrent episodes of neutropenia and infection. Myeloperoxidase deficiency (Choice D) is an inherited immunodeficiency syndrome characterized by the inability to produce hypochlorous acid within phagolysosomes. Disease is typically mild and may present with recurrent Candida albicans infection. Neutrophil-specific granule deficiency (Choice E) results from the defective production of granules within neutrophils. The disorder is characterized by recurrent pyogenic infections that occur early in childhood as well as impaired production of defensins. Educational 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. %3D Previous Next Score Report Lab Values Calculator Help Pause

72 Exam Section 2: Item 22 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 22. A 14-year-old girl is brought to the physician because of a 1-month history of migraine-like headaches, vomiting, and multiple left-sided focal seizures. She has had hearing loss since the age of 11 years. Her mother and maternal grandmother have high- tone deafness. Physical examination shows loss of vision in one half of the visual field of the right eye and weakness of the right upper and lower extremities. Serum and cerebrospinal fluid concentrations of lactic acid are increased. This patient most likely has a mutation of which of the following? A) Endoplasmic reticulum glycosyltransferase B) Lysosomal a-glucosidase C) Mitochondrial tRNA Leu D) Nuclear proteasome activator E) Peroxisomal catalase

C. Mitochondrial encephalopathy, lactic acidosis, and stroke-like episodes (MELAS) syndrome is a family of mitochondrially inherited diseases that are caused by mutations in multiple mitochondrial genes, often those encoding mitochondrial tRNA. Patients present in childhood with muscular weakness and myalgia, stroke-like episodes characterized by hemiparesis and vision loss, seizures, headache, and lactic acidosis. A family history suggestive of mitochondrial inheritance is suggestive of the diagnosis. Like other mitochondrial myopathies, muscle biopsy may demonstrate ragged red fibers, although genetic testing confirms the diagnosis. Treatment is supportive but may involve supplementation of vitamins and carnitine. Other mitochondrial diseases include Leber hereditary optic neuropathy, maternally inherited diabetes and deafness, and myoclonic epilepsy with ragged red fibers. Incorrect Answers: A, B, D, and E. Endoplasmic reticulum glycosyltransferase (Choice A) is important for conjugation of sugar moieties with proteins prior to transport to the cellular surface. This process plays a role in the determination of the A, B, and O blood groups but is not a cause of MELAS syndrome. Lysosomal a-glucosidase (Choice B) is the enzyme whose deficiency underlies glycogen storage disease, type II (Pompe disease). Glycogen storage disease, type II (Pompe disease) leads to cardiomegaly, cardiomyopathy, hepatomegaly, and hypotonia. Nuclear proteasome activation (Choice D) occurs in the setting of the accumulation of misfolded protein requiring degradation. Proteasomes, when failing to activate, are implicated in many neurodegenerative diseases such as Alzheimer and Huntington disease. Peroxisomal catalase (Choice E) deficiency is characteristic of acatalasia, a peroxisomal disorder characterized by an accumulation of hydrogen peroxide within cells. Educational Objective: MELAS syndrome is caused by mitochondrial mutations in many genes, including those that encode mitochondrial tRNALeu Mitochondrial inheritance patterns suggest the diagnosis, which can be confirmed with genetic sequencing. Patients present in childhood with muscular weakness and myalgia, stroke-like episodes characterized by hemiparesis and vision loss, seizures, headache, and lactic acidosis. %3D Previous Next Score Report Lab Values Calculator Help Pause

40 Exam Section 1: Item 40 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 40. Lesch-Nyhan syndrome, an X-linked recessive disease, is seen in approximately 1/100,000 males. Which of the following is the expected prevalence of heterozygous females? A) 1/1000 B) 1/10,000 C) 1/50,000 D) 1/200,000 E) 1/10,000,000

C. The Hardy-Weinberg principle, known as Hardy-Weinberg equilibrium, proposes that allele frequencies will remain constant across generations in the absence of evolutionary change. The Hardy-Weinberg equilibrium equation is: p2 + 2pq + q? = 1, where p is the frequency of the dominant allele and q is the frequency of the recessive allele. 2pq is the probability of heterozygosity, p2 is the probability of dominant homozygote, and q? is the probability of being a recessive homozygote. Calculation of Hardy-Weinberg allele frequencies for X-linked recessive disorders requires special considerations, as affected males carry only a single allele. In this circumstance, the frequency of affected males represents q, rather than q? as in autosomal recessive disorders. The maternal carrier frequency, 2pq, can then be calculated in a straightforward manner as follows: q = 1/100,000 = 0.00001, p = 1-0.00001 = 0.99999, 2pq = 2 x 0.99999 x 0.00001 = 0.00002 = 1/50,000. %3D Incorrect Answers: A, B, D, and E. Assuming that the carrier frequency is either one hundred times greater (Choice A), ten times greater (Choice B), half as great (Choice D), or one hundred times lesser (Choice E) is incorrect and fails to employ the Hardy-Weinberg principle. Choices A, B, and E reflect mathematical errors involving the use of q (the recessive allele) without factoring for the presence of the dominant allele and involve errors on the order of magnitude. Choice D, 1/200,000, reflects a mathematical error wherein pq was divided by two instead of multiplied. pq/2 = 0.00001*.99999/2 = 0.000005 = 1/200,000. %3D Educational Objective: When X-linked inheritance occurs, the Hardy-Weinberg equilibrium must be factored for the imbalance in allele expression. Affected males only carry a single allele and therefore, q must be adjusted to calculate the maternal allele frequency. %3D 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. The diagram shows the major factors that determine blood pressure. Which of the following labeled factors is affected most by an a,-adrenergic antagonist? Central nervous system Peripheral resistance (arteriolar) Blood pressure = Cardiac output x Stroke volume Heart rate B Contractility Venous return Capacitance vessel tone (venular) Blood volume E) A) B) C) D) O E)

C. The autonomic nervous system plays a primary role in the maintenance of blood pressure via its effects on peripheral arteriolar resistance, heart rate, myocardial contractility, venous capacitance, and to an indirect degree, blood volume. The sympathetic nervous system primarily acts through a- and B-adrenergic receptors, which are stimulated by dopamine, epinephrine, and norepinephrine. In relation to blood pressure regulation, the a, receptor leads to smooth muscle contraction, especially of the vasculature, which increases blood pressure by increasing systemic vascular resistance. The B, receptor, in comparison, causes increases in heart rate, myocardial contractility, and renin release, which all lead to increases in blood pressure. An antagonist at the a, receptor would decrease vascular smooth muscle contraction, to a greater extent around arteries, leading to vascular dilation and decreased blood pressure as mean arterial pressure is directly related to cardiac output and systemic vascular resistance. Incorrect Answers: A, B, D, and E. Central nervous system inputs to cardiac output (Choice A) can occur through the limbic system in response to strong emotions or anticipation of physical activity or through central components of the autonomic nervous system (eg, brainstem, spinal autonomic preganglionic neurons). These autonomic stimuli to the heart would be less affected by an a1 receptor antagonist than peripheral resistance, as the stimuli would also include effects on the B, receptor. By contrast, central a2-adrenoreceptors are the target of the agonist clonidine, which results in diminished sympathetic tone, forming the basis for the use of clonidine in hypertensive emergency. Heart rate (Choice B) is primarily affected by the B1 receptor, with agonism leading to increases in heart rate. As part of the parasympathetic nervous system, heart rate is decreased by M2 receptors. a, receptor antagonists do not affect heart rate directly, though can cause reflex tachycardia. Blood volume (Choice D) is affected by changes in red blood cells and plasma volume, such that anemia or hemorrhage would decrease blood volume. However, changes in plasma volume can also be mediated by diuretics (decreasing plasma volume) or renin release (increasing plasma volume). a, receptors do not moderate blood volume. Capacitance vessel tone (Choice E) describes the ability of the vessel to hold a volume of blood at a specific blood pressure. The capacity of the venular compartment increases with decreased somatic muscle movement, valvular dysfunction, and nitroglycerin administration. Venule tone decreases with a, receptor blockade, but to a lesser extent than peripheral arteriolar resistance as there is minimal vascular smooth muscle in venules compared to arterioles. Educational Objective: The autonomic nervous system plays a primary role in the maintenance of blood pressure via its effects on peripheral arteriolar resistance, heart rate, myocardial contractility, venous capacitance, and to an indirect degree, blood volume. a-adrenergic receptors primarily modulate mean arterial pressure by increasing peripheral arteriolar resistance. Previous Next Score Report Lab Values Calculator Help Pause

84 Exam Section 2: Item 34 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 34. A 32-year-old woman comes to the physician for a follow-up examination after atypical cells were noted on a recent Pap smear. Physical examination shows a 1 x 1-cm area of leukoplakia on the cervix. A biopsy specimen of the lesion shows invasive squamous cell carcinoma. Malignant cells from this site will most likely drain first to which of the following lymph nodes in this patient? A) Femoral B) Inferior mesenteric C) Internal iliac D) Lumbar OE) Superficial inguinal

C. The internal iliac 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. Internal iliac lymph nodes drain the lower rectum to the anal canal above the pectinate line, bladder, middle third of the vagina, cervix, and prostate. After reaching the internal iliac 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. In 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. Knowledge of the first draining lymph node of an anatomical site, or sentinel node, is important in patients with invasive neoplasms as the sentinel node is often biopsied to generate prognostic data and guide treatment. Incorrect Answers: A, B, D, and E. The lymph nodes in the femoral (Choice A) triangle include the superficial and deep inguinal lymph nodes. The superficial inguinal (Choice E) lymph nodes drain lymphatic fluid from the labia majora, vulva, scrotum, anal canal below the pectinate line, and the skin below the umbilicus except the popliteal fossa. The deep inguinal lymph nodes drain the glans penis. The inferior mesenteric (Choice B) lymph nodes drain lymph from the large bowel, extending from the region of the splenic flexure to the upper rectum. The portion of the large bowel proximal to the splenic flexure and small bowel are drained by the superior mesenteric lymph nodes. The lumbar (Choice D) lymph nodes, or periaortic lymph nodes, are comprised of the preaortic, paraaortic, and retroaortic groups. The internal iliac nodes drain to the paraaortic, as do the ovaries, testes, uterus, and kidneys. However, it is not the first drainage site of the cervix. Educational Objective: Lymph from the lower rectum to the anal canal above the pectinate line, bladder, middle third of the vagina, cervix, and prostate drains to the internal iliac lymph nodes. %3D Previous Next Score Report Lab Values Calculator Help Pause

33 Exam Section 1: Item 33 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 33. The synthesis of the enzymes necessary for the replication of the genome occurs during which of the following phases of the cell cycle? E A G, M 2. CG, G B 1 Ds O A) B) C) D) E)

C. The normal cell cycle in somatic cells involves several stages: G, S, G, and M phase. G0 phase denotes cell cycle arrest. During the G1 phase, the cellular contents, except the chromosomes, are duplicated. In S phase, chromosomal duplication occurs; in G2 phase, the fidelity of replication is checked. M phase involves the attachment of chromosomes to spindles followed by their separation and division into two identical cells, with the stages of M phase denoted as prophase, prometaphase, metaphase, anaphase, and telophase. The synthesis of enzymes necessary for replication of the genome occurs during the G1 phase in preparation for the S phase of genome duplication. Such enzymes include helicase (unwinds DNA at the replication fork), primase (creates an RNA primer for replication to start with), DNA polymerase (elongates the DNA strand being created and removes the RNA primer), DNA topoisomerases (remove supercoils in the DNA), ligase (connects small fragments of DNA or Okazaki fragments), and telomerase (adds DNA at the 3' end with each duplication to protect the genetic material). The phases of the cell cycle are regulated by checkpoints and regulatory proteins called cyclins and cyclin-dependent kinases to prevent indefinite cell proliferation. Incorrect Answers: A, B, D, and E. M, or mitosis, (Choice A) is the process by which the duplicated genome is divided into two cells. The phases of mitosis are prophase, prometaphase, metaphase, anaphase, and telophase. Protein synthesis does not occur during this phase. Mitosis is followed by cytokinesis during which the cell, its contents, and its cytoplasm are divided in two. Go (Choice B) is the arrest phase of the cell. Cells that are not actively dividing reside in G, and thus do not have a need to synthesize the enzymes necessary for DNA replication. S (Choice D) is the phase of the cell in which DNA replication occurs. The enzymes necessary for replication must be produced before this phase begins. G2 (Choice E) is the phase in which the integrity of the duplicated genome is checked and corrected. It is also a phase of growth and protein synthesis in preparation for the phases of mitosis and cell division. Educational Objective: The normal cell cycle in somatic cells involves several stages: Go, G, S, G, and M phase. DNA duplication occurs during the S phase and thus the enzymes necessary for duplication must be synthesized prior to the start of this phase during G1. Previous Next Score Report Lab Values Calculator Help Pause

100 Exam Section 2: Item 50 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 50. A 35-year-old woman is brought to the emergency department after she sustains a fracture of the neck of the fibula of her right leg. She was struck by a car while crossing the street. Which of the following findings is most likely on examination of the affected leg? Pain Over Pain Over Proximal Fibula Distal Fibula Dorsiflexion Plantar Flexion Achilles Reflex A) Absent absent 4/5 1/5 1+ B) Absent present 0/5 4/5 1+ C) Present absent 1/5 4/5 2+ D) Present absent 5/5 0/5 absent E) Present present 4/5 4/5 2+

C. There are number of key structures along the lateral aspect of the lower leg that can be damaged during trauma. The fibula provides lateral support and structure to the lower leg. The common peroneal nerve is located superficially, just distal to the head of the fibula below the knee joint. This nerve branches into the superficial peroneal nerve and deep peroneal nerve as it travels in the lateral compartment of the lower leg. The superficial peroneal nerve provides motor innervation to peroneus brevis and longus, as well as sensory innervation to the dorsolateral aspect of the foot. The deep peroneal nerve provides motor innervation to the ankle dorsiflexors as well as sensory innervation to the first dorsal webspace of the foot. Other important structures of the lateral lower leg include the lateral or fibular collateral ligament, which provides stability to the knee as well as the syndesmosis ligaments, which support the distal tibiofibular joint, providing stability to the ankle. In this case, the patient has a fracture of the fibular neck, which is located at the proximal aspect of the fibula, just distal to the fibular head and adjacent to the common peroneal nerve as it wraps around the fibula. Injury in this location typically presents with pain at the proximal fibula and weakness and numbness in the distribution of the peroneal nerve. With this injury, pain over the distal fibula would not be present. Function of the gastrocnemius and soleus muscles would be unaffected as these muscles are innervated by the tibial nerve, which is in the posterior aspect of the lower leg (popliteal fossa) leading to an intact Achilles reflex. Incorrect Answers: A, B, D, and E. Plantar flexion weakness without bony pain in the lower leg (Choice A) may represent a compressive lesion of the tibial nerve or compression of the S1 sacral nerve root. These injuries are not consistent with the injury mechanism of this patient. Pain over the distal fibula and weakness in dorsiflexion (Choice B) can be seen in a penetrating traumatic injury to the fibula that also severed the extensor tendons of the toes and dorsiflexors of the ankle. Such an injury is unlikely to lead to severing of the motor nerves of the anterior compartment as the muscle bellies of these muscles are proximal to the level of the injury. Pain over the proximal fibula with weakness in plantar flexion (Choice D) is an uncommon injury pattern as the proximal fibula is not located near the tibial nerve. Although, it should be noted that in high energy injuries such as highway-speed motorcycle accidents, massive damage to bone, soft tissue, and nerves may occur leading to injuries to structures that are spatially separated. A traumatic injury causing pain over the proximal fibula and distal fibula (Choice E) would be concerning for a Maisonneuve injury, which consists of an injury of the distal tibiofibular syndesmosis, typically associated with a fracture of the medial or lateral malleolus, combined with an injury of the proximal fibular shaft or neck. Educational Objective: An injury to the lateral lower extremity can result in a proximal fibular fracture. The common peroneal nerve wraps around the fibular neck as it enters the lateral compartment of the lower leg. Injury to this area can lead to peroneal nerve injury and a consequent foot drop. %3D Previous Next Score Report Lab Values Calculator Help Pause

83 Exam Section 2: Item 33 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 33. A 62-year-old man is brought to the emergency department 2 hours after the sudden onset of pain and coolness of his right leg. He is otherwise healthy except for mild hyperthyroidism treated with propylthiouracil. Examination of the lower extremities shows normal skin, nails, and hair growth patterns. Pulses are absent in the right lower extremity and normal on the left. Which of the following is the most likely diagnosis? A) Cellulitis B) Deep venous thrombosis C) Embolic arterial occlusion D) Lumbar disc herniation O E) Rhabdomyolysis

C. Thromboembolic events can cause critical limb ischemia in persons with otherwise relatively normal peripheral vasculature. The most common cause of emboli to the lower extremity is from the heart, related to a variety of underlying pathology. Thrombotic emboli from the left atrium to the lower extremity are associated with atrial fibrillation, which causes uncoordinated, inadequate contractions of the left atrium and consequent impaired blood flow, stasis of which serves as a trigger for thrombus formation. Thrombotic emboli from the left ventricle have been associated with abnormal flow patterns in the left ventricle secondary to myocardial infarction and ventricular aneurysm formation, which also results in impaired contractility and pooling of blood. Less commonly, cardiac tumors and thrombotic debris from the valves (endocarditis) can cause acute arterial occlusion. Atherosclerotic plaque from a diseased aorta may also embolize causing a distal ischemic event. Symptoms of a thromboembolic event of the extremity includes acute pain, swelling, pallor, and pulselessness. Late manifestations include muscle and nerve death with extreme pain and possible compartment syndrome. Incorrect Answers: A, B, D, and E. Cellulitis (Choice A) is a superficial infection of the skin and subcutaneous tissue. It typically presents as a warm, erythematous, painful region of skin that may enlarge in size and may originate at an area of trauma or skin breakdown. It does not result in diminished pulses. Deep venous thrombosis (Choice B) commonly occurs in the lower extremities after surgery, during periods of immobilization, and in patients with increased inflammatory states such as cancer. Although this would present with acute pain in the extremity with associated swelling, a venous thrombosis would not cause pulselessness unless advanced and large (eg, phlegmasia cerulea dolens). Lumbar disc herniation (Choice D) occurs most often following chronic disc degeneration combined with a traumatic event such as bending and lifting a heavy object causing an acute increase in the intradiscal pressure. Acute disc herniation presents as acute back pain along with acute extremity pain and/or weakness in the distribution of the impinged spinal nerve. Rhabdomyolysis (Choice E) is an acute breakdown of muscle tissue leading to increased serum levels of myoglobin and electrolyte derangements (eg, hyperkalemia). This may occur in healthy adults after periods of intense physical activity. It may also occur in patients with seizure disorders and in elderly patients who have fallen and sustained prolonged compression of muscle. Educational Objective: Acute embolic arterial occlusion can occur in the extremities and in the internal organs such as the kidney and intestine. Embolization of thrombus, cholesterol plaque, or tumor tissue, most commonly from the heart, leads to arterial occlusion that can be detected on physical examination as pulselessness, pallor, and pain. %3D 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 34-year-old woman is admitted to the hospital for treatment of pulmonary tuberculosis. Infliximab therapy was initiated 6 months ago for severe Crohn disease. This pharmacotherapy most likely inhibited which of the following immunologic functions in this patient? A) Activation of nuclear factor KB to induce expression of interleukin-10 (IL-10) B) Direct toxicity to the causal organism C) Maintenance of granulomas D) Recruitment of segmented neutrophils to ingest and kill the bacteria E) Stimulation of B lymphocytes to produce neutralizing antibodies against the causal organism

C. Tumor necrosis factor-a (TNF-a) is a cytokine secreted by macrophages, which supports granuloma formation. Granulomas are collections of histiocytes, or macrophages with abundant pink cytoplasm that often contain multi-nucleated giant cells and are surrounded by lymphocytes. One purpose of granuloma formation is to sequester an infection or foreign body. Monoclonal antibody therapy targeted against TNF-a increases a patient's risk for active mycobacterial infection because it stops the production of TNF-a, leading to breakdown of the granuloma and release of any contained organism. Monoclonal antibodies against TNF-a that are specifically used to treat Crohn disease include infliximab, adalimumab, and certolizumab. These medications are known to increase the risk of reactivating latent Mycobacterium tuberculosis because of their deleterious effect on granuloma formation and maintenance. All patients who are considered for treatment with these agents must undergo screening for latent M. tuberculosis infection. If positive, treatment for latent M. tuberculosis with nine months of isoniazid is indicated. Incorrect Answers: A, B, D, and E. Activation of nuclear factor KB to induce expression of interleukin-10 (IL-10) (Choice A) occurs within the Th2 cell type. IL-10 attenuates the immune response by inhibiting activated macrophages and decreasing expression of Th, cytokines. Inhibition of this IL-10 would therefore stimulate the immune process against infectious organisms, not decrease it. TNF-a is a cytokine that maintains granulomas and assists in walling off infections such as tuberculosis. It does not have any direct toxicity to the causal organism (Choice B), and thus this is not the mechanism by which anti-TNF-a monoclonal antibodies increase the risk of latent tuberculosis reactivation. IL-8, not TNF-a, is responsible for the recruitment of segmented neutrophils to ingest and kill the bacteria (Choice D). This process is not affected by the inhibition of TNF-a. Rituximab, an anti-CD20 monoclonal antibody, diminishes B lymphocytes and is used in chemotherapy regimens for lymphoma. It is also used in the treatment of autoimmune conditions including vasculitis, rheumatoid arthritis, and pemphigus vulgaris. This monoclonal antibody, not a TNF-a inhibitor, prevents stimulation of B lymphocytes to produce neutralizing antibodies against the causal organism (Choice E). Educational Objective: The use of monoclonal antibody therapy against TNF-a is associated with an increased risk of reactivated latent M. tuberculosis secondary to inadequate granuloma maintenance. Because of this, all prospective candidates for this therapy should be screened for latent M. tuberculosis infection and accordingly treated if positive. %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. A 5-year-old boy is brought to the emergency department after ingesting 10 oz of a household cleaning solvent. He is treated for acute hepatic and renal failure for 1 week and then discharged. During the next month, regeneration of this boy's mature hepatocytes and renal tubular epithelial cells will be accomplished mostly by which of the following mechanisms? A) Activation of stem cells to enter G, phase of the cell cycle B) Decreased apoptosis at G,M transition of the cell cycle C) Recruitment of cells from G, into the cell cycle D) Shortened time for progression of cells through the cell cycle E) Terminal differentiation by cells exiting from the cell cycle

C. When hepatocytes or renal tubular epithelial cells are destroyed, the remaining cells are recruited from quiescence (the Go phase) to re-enter the cell cycle. In healthy patients, the majority of hepatocytes and renal tubular epithelial cells are in the Go phase. When cells are destroyed such as in acute liver or renal failure, genes are induced that prime remaining cells to re-enter the cell cycle from quiescence. These cells transition from Go phase to G, phase, where growth factors engender cell growth. Cells that grow sufficiently surpass the G, restriction point, at which point they are committed to DNA replication and cell division via mitosis. After the G, restriction point, the cells transition to the S phase, when DNA replicates. The cells enter another growth phase, the G2 phase, and finally the M phase, when mitosis occurs and hepatocytes and renal tubular epithelial cells regenerate. Incorrect Answers: A, B, D, and E. Activation of stem cells to enter the G, phase of the cell cycle (Choice A) does not play a major role in the regeneration of hepatocytes or renal tubular epithelial cells. Stem cells from the bone marrow or within the liver/kidney itself may minorly contribute to regeneration, but the recruitment of the large population of quiescent cells into the cell cycle is more crucial. Decreased apoptosis at G1-M transition of the cell cycle (Choice B) and shortened time for progression of cells through the cell cycle (Choice D) would not lead to regeneration of hepatocytes or renal tubular epithelial cells. Most cells are quiescent (not in the cell cycle) at baseline so decreasing apoptosis or shortening the cell cycle time would not lead to an appreciable increase in cells. Terminal differentiation by cells exiting from the cell cycle (Choice E) does not occur after hepatic or renal damage and would prevent cells' future ability to regenerate. Many cell types (eg, skeletal muscle cells) terminally differentiate and lose their ability to regenerate. Educational Objective: The vast majority of hepatocytes and renal tubular epithelial cells are in the Go phase (quiescence) at baseline. When cells are destroyed, the remaining cells re-enter the cell cycle at the G1 phase to grow and divide, leading to regeneration. %3D Previous Next Score Report Lab Values Calculator Help Pause

62 Exam Section 2: Item 12 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 12. Drug X is used to treat pain associated with rheumatoid arthritis. The drug is a weak acid, with a pKa of 4.4; it is absorbed principally through the stomach. It is determined that Drug X is absorbed efficiently by the body because of the ionization conditions under which it exists at gastric and blood pH. Which of the following sets of physical chemical states of the drug is most likely? At Gastric pH At Blood pH A) lonized ionized B) lonized nonionized C) Nonionized ionized D) Nonionized nonionized

C. pKa, the negative log of the acid dissociation constant, describes the tendency of a given acid to exist in a protonated or deprotonated form in a solution of given pH. pKa indicates the pH at which 50% of the molecules will exist in each protonated and deprotonated forms. pKa is inversely related to the strength of an acid, such that a strong acid will have a low pKa value and a weak acid will have a high pKa value. The relationship between pKa and the pH of a buffering solution is described with the Henderson- Hasselbalch equation. In general, if an acid exists in a solution with a pH that is less than its pKa the acid will exist in a nonionized, protonated form. In contrast, if an acid exists in a solution with a pH that exceeds its pKa then the acid will exist predominantly in its ionized, conjugate base form. Drug X is a weak acid with a pka of 4.4. The pH of gastric acid is less than 2, therefore, due to the excess of protons in solution, Drug X will bind its proton in the stomach and will be predominantly nonionized, which permits its absorption across the hydrophobic intestinal membrane. The pH of the bloodstream is approximately 7.35 - 7.45, and under these conditions Drug X will be predominantly ionized. Incorrect Answers: A, B, and D. Choice A describes a strong acid with essentially complete dissociation (pKa less than 1). A strong acid would be capable of dissociation even under the acidic gastric pH. Examples include sulfuric acid, hydrobromic acid, and hydroiodic acid. Choice B does not describe the behavior of an acid as it is ionized at a low gastric pH but nonionized at a higher blood pH. Choice D describes a very weak acid with a pKa greater than the pH of blood and instead likely describes a base, a molecule accepting a proton at any acidic or neutral pH. Educational Objective: pKa describes the tendency of a given acid to exist in a protonated or deprotonated form in a solution of given pH. Acidic molecules dissociate into protons and conjugate base when they are found in solutions with a pH greater than their pka. %3D Previous Next Score Report Lab Values Calculator Help Pause

36 Exam Section 1: Item 36 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 36. A 23-year-old man in the emergency department has apnea and pinpoint pupils. Needle tracks are present on his arms. Activation of which of the following opioid receptors in the central nervous system is most likely to be responsible for the apnea? A) õ B) к C) H D) o

C. µ opioid receptor agonism likely explains this patient's apnea. Opioid receptors are G-protein-coupled receptors located throughout the central nervous system (CNS) and peripheral nervous system (PNS). Opioid intoxication causes euphoria, altered mental status, sedation, bradycardia and hypotension, depressed respiratory drive (or apnea), and constricted pupils. µ opioid receptor activity mediates analgesia, reward, and adverse CNS and PNS effects. Specifically, the µ2 opioid receptor, located in the CNS and on peripheral chemoreceptors and baroreceptors, controls respiratory depression. Miosis is a distinctive finding that is less common in intoxication with other CNS depressants and is caused by direct µ2 opioid receptor activity in brain areas responsible for pupillary control. Opioids also act on 42 receptors within the enteric nervous system, reducing gut motility and causing constipation. The treatment of opioid toxicity includes supportive care (eg, respiratory support) and naloxone, a short-acting opioid receptor antagonist. Incorrect Answers: A, B, and D. ō opioid receptors (Choice A) mediate analgesia and reward alongside u opioid receptors and may be responsible for neuronal adaptations that lead to addiction. Ō opioid receptors have been implicated in other diverse CNS roles such as modulation of motor function, epileptogenesis in absence seizures, and post-ischemic neuroprotection. Ō opioid receptors are not known to mediate respiratory depression. K opioid receptors (Choice B) synergistically alleviate pain with u opioid receptors but oppose the reward signaling of u opioid receptors, leading to dysphoria under stressful conditions. Antagonism of K opioid receptors is being investigated as a therapeutic target in mood disorders. K opioid receptors are not known to mediate respiratory depression. o receptors (Choice D) were originally thought to be a subtype of opioid receptor but are now theorized to represent their own receptor class. o receptors are thought to regulate higher-order functions such as memory and drug dependence. For example, upregulation of o receptors has promoted stimulant-seeking behavior. Educational Objective: In opioid intoxication, µ opioid receptors mediate analgesia, reward, and adverse CNS and PNS effects. Specifically, the u2 opioid receptor, located in the CNS and on peripheral chemoreceptors and baroreceptors, controls respiratory depression. %3D 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 25-year-old man is brought to the emergency department because of severe abdominal pain, nausea, and vomiting for 1 hour. The pain originates in the left flank and radiates to his groin. His pulse is 100/min, respirations are 18/min, and blood pressure is 150/100 mm Hg. Physical examination shows tenderness of the left flank and the left lower quadrant of the abdomen. Bowel sounds are mildly hypoactive. Test of the stool for occult blood is negative. Which of the following best explains these findings? A) Colon neoplasm B) Diverticulitis C) Epididymitis D) Renal infarction E) Torsion of the testis F) Ureteral calculus

F. Ureteral calculus typically presents with colicky, unilateral flank pain radiating to the groin, and with gross or microscopic hematuria. Pain may be significant enough to trigger nausea, as in this case. The common types of urinary tract calculi are calcium oxalate or phosphate, ammonium magnesium phosphate, uric acid, and cystine. On urinalysis, red blood cells without casts are common. Fever, dysuria, and pyuria would not be expected unless there was a concomitant infection. Treatment for ureteral calculus is symptomatic with pain control and nausea relief. Most ureteral calculi pass spontaneously after a period of observation for patients with well-controlled pain and no signs of sepsis or infection. Stone removal by shock wave lithotripsy or endoscopic removal is an option for patients requiring emergency therapy. It is also an option for patients with persistent obstruction, uncontrolled symptoms, or failure of stone progression. In general, stones smaller than 5 mm will pass without operative assistance. Obstructing stones may require temporary placement of a ureteral stent to prevent hydronephrosis and renal parenchymal injury. Incorrect Answers: A, B, C, D, and E. Colon neoplasm (Choice A) would be unlikely in an otherwise healthy young patient with no family history of polyposis syndromes and acute, severe, flank pain. It would typically present with insidious weight loss, anemia, constipation, or blood per rectum. In addition, test for stool for occult blood is negative, making this diagnosis unlikely. Diverticulitis (Choice B) can present with left lower quadrant abdominal pain and tenderness but would be less abrupt in presentation and typically present with fever, diarrhea, and hyperactive bowel sounds. It would be unlikely to cause flank pain. Epididymitis (Choice C) is a common cause of painful scrotal swelling and refers to acute infection and inflammation of the epididymis. In younger males, this is commonly secondary to sexually transmitted infections such as Chlamydia trachomatis or Neisseria gonorrhoeae. In older males, Escherichia coli is more common. Renal infarction (Choice D) can cause flank pain, nausea, and vomiting, and can be due to thromboembolic disease, renal artery dissection, or a hypercoagulable state. However, it is rare and ureteral calculus is more common and likely in this patient. Torsion of the testis (Choice E) occurs when the testicle twists on the spermatic cord resulting in subsequent loss of testicular blood supply. Patients typically present with acute, severe testicular pain, swelling, and erythema. On physical examination, the testicle typically demonstrates an abnormal lie (eg, transverse), extreme tenderness to palpation, absent cremasteric reflex, and pain that does not improve with elevation of the scrotum (as it does in epididymitis). Educational Objective: Ureteral calculus typically presents with colicky, unilateral flank pain radiating to the groin, along with gross or microscopic hematuria. %3D Next Score Report Lab Values Calculator Help Pause

67 Exam Section 2: Item 17 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 17. An 88-year-old man who lives alone is brought to the physician by his daughter because she is concerned that he has not been eating a well-balanced diet for 9 months. He is 170 cm (5 ft 7 in) tall and weighs 50 kg (110 Ib); BMI is 17 kg/m2. Physical examination shows multiple ecchymoses on the upper and lower extremities. Laboratory studies show: 160,000/mm3 12 sec (INR=1) Platelet count Prothrombin time Serum Vitamin B, (pyridoxine) Vitamin C (ascorbic acid) Folic acid 9 ng/dL (N=5-30) 0.1 mg/dL (N=0.4-2) 5 ng/dL (N=2-20) The ecchymoses in this patient are most likely due to a disorder of which of the following? A) Arachidonic acid production B) Binding of carboxyglutamic acid to phospholipid OC) Carboxylation of factor II (prothrombin) D) Proline hydroxylation E) Transfer of methyl groups to organic acids

D. Vitamin C 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 scurvy, which presents with signs and symptoms of impaired collagen synthesis including swollen, bleeding gums, easy bruising and bleeding (eg, hemarthrosis), petechiae, impaired wound healing, and short, fragile, curly hair. It should not be confused with hemophilia, as it does not result in a deficiency of factors. Instead, the collagen and connective tissue deficiency weakens blood vessel walls 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 bruising and diet limited in vitamin C plus normal prothrombin time suggests a diagnosis of scurvy due to impaired collagen synthesis. Incorrect Answers: A, B, C, and E. Arachidonic acid is produced in inflammatory states from membrane phospholipids and is converted to thromboxane-A, by cyclooxygenase. Thromboxane-A, is involved in platelet activation. Decreased arachidonic acid production (Choice A) leading to decreased synthesis of thromboxane-A, would result in diminished platelet activation. Anti-inflammatory agents that decrease arachidonic acid production include corticosteroids such as dexamethasone, hydrocortisone, and prednisone. Carboxyglutamic acid is a modified amino acid frequently found in coagulation factors. Vitamin K is required for its synthesis. Carboxylation of glutamic acid residues occurs on procoagulant factors II, VII, IX, and X. A deficiency in vitamin K leading to decreased carboxylation results in impaired binding of carboxyglutamic acid to phospholipid (Choice B) and decreased carboxylation of factor II (prothrombin) (Choice C). The net effect is coagulopathy, manifest as easy bruising with increased prothrombin time and activated partial thromboplastin time. Vitamin Bg is converted to tetrahydrofolic acid which serves as a coenzyme for methylation reactions. Folate deficiency resulting in impaired transfer of methyl groups to organic acids (Choice E) leads to impaired DNA and RNA synthesis and a macrocytic, megaloblastic anemia. Educational Objective: Vitamin C 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) and disordered hair growth. II Previous Next Score Report Lab Values Calculator Help Pause

54 Exam Section 2: Item 4 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 4. A 32-year-old primigravid woman delivers a healthy 3402-g (7-lb 8-oz) male newborn after an uncomplicated cesarean delivery because of a nonreassuring fetal stress test. Two days prior to discharge from the hospital, she has persistent numbness of the area surrounding the abdominal incision. The physician assures the patient that sensation will gradually return as the nerves regenerate. Which of the following best describes the rate-limiting step in this patient's return to normal sensation? A) Dorsal root ganglion cell proliferation B) Fast anterograde axonal transport C) Fibroblast proliferation D) Retrograde axonal transport E) Slow anterograde axonal transport

E. After peripheral nerve injury, nerves regenerate with the help of slow anterograde axonal transport. When an axon is injured or severed, the axon and associated myelin distal to the site of injury are degraded in a process called Wallerian degeneration. Despite this degradation, the Schwann cells remain in the same orientation and position to guide nerve regeneration. Protein synthesis in the dorsal root ganglia cell bodies and axons and subsequent anterograde axonal protein transport to the axon terminal mediate peripheral nerve regeneration. Slow anterograde axonal transport occurs when microtubule motor proteins transport cytoskeletal proteins such as neurofilaments and cytoplasmic proteins such as actin and glycolytic enzymes to the axon terminal. Slow anterograde axonal transport occurs at approximately 1 mm per day and is slower than fast anterograde axonal transport or retrograde axonal transport. Therefore, slow anterograde axonal transport is the rate-limiting step in peripheral nerve regeneration. Incorrect Answers: A, B, C, and D. Dorsal root ganglion cell proliferation (Choice A) is not a known mechanism of peripheral nerve regeneration. Alterations in regeneration-associated gene expression in dorsal root ganglia assist in axonal regeneration. Fast anterograde axonal transport (Choice B) in peripheral nerve regeneration involves transport of proteins along microtubules to the axon terminal, similar to slow anterograde axonal transport. Fast anterograde axonal transport moves vesicles and membranous organelles. It involves a faster transport rate than slow anterograde axonal transport. Fibroblast proliferation (Choice C) would result in the formation of a scar, which is not innervated and would therefore not explain the return of normal sensation. Peripheral nerve regeneration should ideally overcome fibroblast infiltration. Retrograde axonal transport (Choice D) refers to the microtubule-dependent transport of proteins that serve as injury signals to the dorsal root ganglion cell body. These injury signals lead the cell body to upregulate transcription factors that increase the transcription of regeneration-associated genes. Educational Objective: After peripheral nerve injury, microtubule motor proteins transport newly synthesized cytoskeletal and cytoplasmic proteins to the axon terminal in a process called slow anterograde axonal transport. Slow anterograde axonal transport is the rate-limiting step of peripheral nerve regeneration. %3D Previous Next Score Report Lab Values Calculator Help Pause

42 Exam Section 1: Item 42 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 42. A 45-year-old man who is able to bicycle 4 5 minutes a day has switched to a rowing machine. After 5 minutes on the machine, he experiences vertigo, lightheadedness, and fatigue of the left upper extremity. Within a few minutes of stopping the rowing exercise, all symptoms resolve. Which of the following findings is most likely on physical examination? A) Diastolic murmur at the cardiac apex B) Increased jugular venous pressure C) Pansystolic murmur at the cardiac apex D) Right carotid bruit E) Supraclavicular bruit

E. This patient with exercise-induced vertigo, lightheadedness, and upper extremity fatigue is most likely experiencing subclavian steal syndrome, which manifests with a supraclavicular bruit on physical exam. Atherosclerotic disease of the subclavian artery proximal to the origin of the vertebral artery leads to decreased pressure in the distal subclavian artery. During exercise of the ipsilateral upper extremity, atherosclerotic disease of the ipsilateral subclavian artery may initially prevent sufficient blood supply to the arm. The arm may be perfused through retrograde flow in the ipsilateral vertebral artery, diminishing posterior cerebral circulation and resulting in symptoms of vertebrobasilar insufficiency. Blood from the contralateral subclavian and vertebral artery may flow in the retrograde direction at the vertebrobasilar confluence instead of continuing to ascend the basilar artery. Symptoms of vertebrobasilar insufficiency can include vertigo, lightheadedness, disequilibrium, ataxia, and nystagmus (due to transiently decreased blood supply to the vestibular nuclei). Subclavian steal syndrome is a compensatory mechanism that is typically not in itself dangerous. Instead, it indicates significant atherosclerotic disease that should be managed with surgical bypass or angioplasty with or without stenting along with lifestyle changes to prevent further atherosclerosis. Incorrect Answers: A, B, C, and D. A diastolic murmur at the cardiac apex (Choice A) is typical of mitral stenosis while a pansystolic murmur at the cardiac apex indicates mitral regurgitation (Choice C). Mitral stenosis or mitral regurgitation can lead to heart failure, atrial fibrillation, thromboembolic events, and secondary pulmonary hypertension. They do not lead to vertebrobasilar insufficiency or exercise-induced upper extremity fatigue. Increased jugular venous pressure (Choice B) is a nonspecific physical examination finding that may indicate states of fluid overload, right heart failure, or restriction of right heart filling as in pulmonary hypertension or constrictive pericarditis. In acute cardiogenic shock, patients may present with focal neurological deficits from ischemia of watershed areas of the brain and brain regions vulnerable to anoxic brain injury (eg, hippocampus). Symptoms include cortical blindness, stupor, weakness of the bilateral proximal upper and lower extremities, and/or an amnesic syndrome. A right carotid bruit (Choice D) reflects atherosclerotic disease of the right carotid artery, which could produce transient ischemic attacks that could manifest as transient loss of monocular vision or hemispheric signs of cerebral infarction (eg, contralateral hemiparesis, hemisensory loss, or homonymous hemianopsia) rather than signs of vertebrobasilar insufficiency. Patients with atherosclerotic disease of the subclavian artery may additionally demonstrate atherosclerotic disease of the carotid artery, but subclavian atherosclerosis is the most direct explanation of this patient's symptoms. Educational Objective: Subclavian steal syndrome occurs when atherosclerotic disease of one subclavian artery leads to retrograde, collateral blood flow from the vertebrobasilar system, thus decreasing cerebral blood flow from the basilar artery. Symptoms of transient vertebrobasilar insufficiency such as vertigo and lightheadedness may result. %3D Previous Next Score Report Lab Values Calculator Help Pause

55 Exam Section 2: Item 5 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 5. An 18-year-old woman comes to the physician because of progressive fever, general malaise, and blood in her urine since she began oral antibiotic therapy for a urinary tract infection 5 days ago. She also has a 3-day history of a rash. Her temperature is 38°C (100.4°F), pulse is 75/min, respirations are 12/min, and blood pressure is 125/80 mm Hg. Physical examination shows a petechial rash over the chest, back, and upper and lower extremities. Urinalysis shows: Blood Protein 3+ 1+ Leukocytes Eosinophils 150/hpf 30% Which of the following is the most likely diagnosis? A) Acute tubular necrosis O B) Glomerulonephritis C) IgA nephropathy D) Interstitial nephritis E) Papillary necrosis

D. Acute interstitial nephritis (AIN) is caused by a hypersensitivity reaction to medications (eg, nonsteroidal anti-inflammatory drugs, diuretics, sulfonamides, rifampin, proton pump inhibitors, antibiotics), infections, or autoimmune disorders such as sarcoidosis and systemic lupus erythematosus. Patients may be asymptomatic, but common signs, symptoms, and laboratory findings include rash, azotemia, sterile pyuria, hematuria, and eosinophilia. Patients may have proteinuria, but typically not to the extent of nephrotic syndrome. AIN can be complicated by acute kidney injury and a decline in kidney function. Treatment of AIN includes supportive care and discontinuing the offending drug when there is one. Eosinophils are often detected on urinalysis, as demonstrated in this case. Incorrect Answers: A, B, C, and E. Acute tubular necrosis (Choice A) occurs following an ischemic or nephrotoxic insult to the kidneys, which results in necrosis of the tubular epithelium. Granular, muddy brown casts are typical on urinalysis. It would be unlikely to cause systemic symptoms, fever, and hematuria as seen in this patient. Glomerulonephritis (Choice B) refers to a variety of glomerular diseases, including nephritic and nephrotic syndromes. Nephritic syndromes typically present with acute renal failure associated with hematuria, red blood cell urine casts, and hypertension. Nephrotic syndrome typically presents with excessive proteinuria (>3g/day) hyperlipidemia, hypoalbuminemia, and edema. Antibiotics, rash, and urine eosinophils are more consistent with AIN. IgA immune complex deposition in small vessels can lead to IgA nephropathy (Choice C). When IgA deposition occurs in the renal mesangium, glomerulonephritis may ensue, causing microscopic hematuria, red cell casts, and proteinuria. Renal papillary necrosis (RPN) (Choice E) occurs following ischemic, inflammatory, infectious, or toxin-mediated damage to the renal papilla and describes the sloughing and loss of the papillae including substructures such as the distal collecting tubule. RPN can be triggered by infections (eg, acute pyelonephritis), diabetes mellitus, sickle cell disease, or nonsteroidal anti-inflammatory drugs. It typically presents with hematuria and acute flank pain. Educational Objective: AIN is caused by a hypersensitivity reaction to medications (eg, nonsteroidal anti-inflammatory drugs, diuretics, sulfonamides, rifampin, proton pump inhibitors, antibiotics), infections, or autoimmune disorders. Patients may be asymptomatic, but common signs, symptoms, and laboratory findings include rash, azotemia, sterile pyuria, hematuria, eosinophilia, and urine eosinophils. %3D Previous Next Score Report Lab Values Calculator Help Pause

96 Exam Section 2: Item 46 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 46. A 25-year-old woman with a history of rheumatic fever and mitral valve dysfunction comes to the physician because of a 2-week history of fever and fatigue. She underwent a root canal procedure 1 month ago, before which she had taken a single dose of amoxicillin. Her temperature is 38.4°C (101.2°F). A grade 4/6 blowing murmur is heard on auscultation under the left axilla. A photomicrograph of a Gram stain of the organism recovered from a blood culture specimen is shown. On blood agar plates, the organism shows alpha hemolysis. Which of the following is the most likely causal organism? A) Enterococcus faecalis B) Group A beta-hemolytic streptococci C) Staphylococcus aureus D) Streptococcus mitis E) Streptococcus pneumoniae and

D. Bacterial endocarditis refers to acute or subacute bacterial infection of the cardiac valvular endocardium, most commonly from an acute infection with Staphylococcus aureus and/or a subacute infection with viridans streptococci. It can present with an array of symptoms, most commonly fever, new cardiac murmur, anemia, glomerulonephritis, Roth spots, Osler nodes, and splinter hemorrhages in the nailbeds and conjunctiva. Risk factors for infection include prior valvular damage (eg, rheumatic heart disease), intravenous drug use, immunosuppression, prosthetic cardiac valves, and congenital heart disease. The mitral valve is most commonly affected, and mitral regurgitation may present with a holosystolic murmur best heard in the left fourth or fifth intercostal space along the midclavicular line with radiation to the left axilla. Viridans streptococci, such as Streptococcus mitis, are commensal organisms of the human oropharynx. Dental procedures may cause transient bacteremia, where they reach the heart valves via hematogenous spread. Viridans streptococci are a-hemolytic (partial or incomplete lysis of red cells) cocci that grow in chains. They may be distinguished from Str. pneumoniae by demonstration of optochin-resistance on antibiotic sensitivity analysis. Incorrect Answers: A, B, C, and E. Enterococcus faecalis (Choice A) are Gram-positive cocci that normally reside in the colon and may cause disease in the setting of genitourinary or gastrointestinal disruption. They demonstrate a variable hemolytic pattern (a or y) on blood agar. They are classically associated with urinary tract infections, biliary tree infections, and endocarditis. Group A beta-hemolytic streptococci (Choice B) cause a variety of human disease, most notably skin and soft tissue infections, streptococcal pharyngitis, and rheumatic fever. They demonstrate a B-hemolytic pattern (complete lysis of red cells) on blood agar. Staphylococcus aureus (Choice C) commensally occupy human skin and the nasopharynx but have virulent potential. They are Gram-positive cocci that tend to grow in clusters. S. aureus is a common cause of acute, severe infective endocarditis, bacteremia, and post-viral pneumonia. Streptococcus pneumoniae (Choice E) are a-hemolytic, Gram-positive cocci that grow in chains. They are a less common cause of infective endocarditis than viridans streptococci, especially in the setting of a recent dental procedure. Educational Objective: Bacterial endocarditis with viridans streptococci is a risk following dental procedures due to transient bacteremia. The presence of previously damaged or prosthetic valves increases the risk of infection. II Previous Next Score Report Lab Values Calculator Help Pause

59 Exam Section 2: Item 9 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 9. A 19-year-old man who is a college student is brought to the emergency department because of the sudden onset of right-sided chest pain and difficulty breathing after an accident in which he was thrown from his bicycle. He has difficulty walking and cannot climb stairs because of pain and shortness of breath. He is slightly cyanotic, afebrile, and tachypneic. Which of the following is most suggestive that fractured ribs caused the respiratory problem? A) Bronchophony B) Expiratory stridor C) Inspiratory stridor D) Subcutaneous crepitus E) Succussion splash

D. Chest pain, shortness of breath, and tachypnea after a blunt traumatic injury raises concern for rib fracture and pneumothorax. Rib fractures can be complicated by tearing of the adjacent visceral or parietal pleura, allowing air in the lung to escape into the intrapleural space and to subcutaneous tissues. Air in subcutaneous tissues is demonstrated by subcutaneous crepitus upon palpation. Pain from rib fractures, along with reduced pulmonary expansion and ventilation in the setting of pneumothorax, presents with shortness of breath, tachypnea, and respiratory distress. Rib fractures are typically treated supportively, with oxygen supplementation and pain control to reduce complications of atelectasis and pneumonia. In the presence of an associated pneumothorax or hemothorax, tube thoracostomy may be necessary to promote lung expansion, exclude significant hemorrhage, and prevent residual pneumothorax, empyema, or fistula. Incorrect Answers: A, B, C, and E. Bronchophony (Choice A) refers to an abnormal or increased sound of voice auscultated with a stethoscope over an area of lung consolidation. It is expected in pneumonia. In the absence of fever or productive cough suggestive of pneumonia, it would be unlikely in this patient. Expiratory stridor (Choice B) can be seen in tracheomalacia, which results in excessive tracheal end-expiratory collapse due to atrophy and/or reduction of the tracheal elastic fibers and decreased integrity of tracheal cartilage. In adults, this most commonly occurs after prolonged endotracheal intubation, which damages the tracheal cartilage directly. Inspiratory stridor (Choice C) can be seen in illnesses obstructing the airway such as croup and epiglottitis. Croup is characterized by acute viral inflammation of the larynx causing upper respiratory tract symptoms and a barking cough. Epiglottitis presents with fever and acute, severe pharyngitis, drooling, hoarseness, and dysphagia. Succussion splash (Choice E) describes the auscultation finding upon back-and-forth movement of fluid in the stomach or intestines, such as seen in gastroparesis, gastric outlet obstruction, or bowel obstruction. Educational Objective: Rib fractures after blunt traumatic injury to the chest can be complicated by tearing of the visceral or parietal pleura, allowing air in the lung or pleural space to escape to the subcutaneous tissues, which results in subcutaneous crepitus upon examination. %3D Previous Next Score Report Lab Values Calculator Help Pause

50 Exam Section 1: Item 50 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 50. A cohort study assessing risk factors for acquisition of infection with a newly identified agent is performed. Only newly diagnosed subjects are eligible, and controls are selected on the basis of age. The results of this study are shown: Infection Present Infection Absent Turtle exposure No turtle exposure 60 40 20 80 Which of the following is the relative risk for the exposure variable? OA) 0.7 OB) 1.0 OC) 1.7 D) 2.2 OE) 3.1

D. Cohort studies are often used to define the relationship between an exposure and an outcome of interest. This is often calculated as relative risk. The definition of relative risk is the incidence rate in the exposed group divided by the incidence rate of the unexposed group. In this question, the incidence of infection in the turtle exposure group would be calculated as 60 infections divided by 80 total persons in the turtle exposure group (60/80 0.75). The incidence of infection in the non-turtle exposure group would be 40 infections divided by 120 total persons without turtle exposure (40/120 = 0.33). Relative risk would be calculated as the incidence of infection in the turtle exposed group divided by the incidence of infection in the non-turtle exposed group ((60/80)/(40/120) = 2.25). This signifies that the risk of infection in the group exposed to turtles is 2.25 times the risk of infection in the group not exposed to turtles. Incorrect Answers: A, B, C, and E. These answer choices (Choices A, B, C, and E) represent erroneous calculations or the use of random, incorrect numbers that do not represent the appropriate calculation of the relative risk or other epidemiologic measures. Educational Objective: Relative risk is a measure of the likelihood of an outcome in a group with a particular exposure versus the likelihood of the same outcome in a group without that exposure. It is calculated as the incidence of the outcome in the exposed group divided by the incidence of the outcome in the non-exposed group. %3D Previous Next Score Report Lab Values Calculator Help Pause

71 Exam Section 2: Item 21 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 21. A 25-year-old woman comes to the physician for a routine health maintenance examination. She is currently preparing for an 8-km (5-mile) race, and she sprint trains twice weekly. During this training, she runs 200-meter sprints in two groups of 10 sprints separated by a 30-second rest between each sprint in a group, and a 2-minute rest between each group of 10. After the workout her legs feel weak, and her muscles burn and sometimes cramp. Which of the following best explains her symptoms? A) Decreased activity of the sodium-proton antiporter, resulting in an acidic sarcoplasm B) Increased activity of the sodium-proton antiporter, resulting in an acidic sarcoplasm C) Increased oxygen delivery to the muscle, leading to increased metabolism and acid production D) Regeneration of NAD+ from NADH, which produces acid

D. During vigorous exercise activity, adenosine triphosphate (ATP) stores in skeletal muscle are depleted during the cyclic mechanical excitation-contraction coupling of myosin and actin. There are a number of mechanisms by which skeletal muscle regenerates ATP. When oxygen concentrations are high, aerobic respiration allows for the conversion of glucose to pyruvate, which enters the citric acid cycle producing ATP and NADH. NADH shuttles protons and electrons to the electron transport chain, where oxidative phosphorylation produces ATP. When oxygen concentrations are low, cells undergo anaerobic respiration, which regenerates NAD* from NADH through the production of lactic acid from pyruvate. This is a quicker, although less efficient process than aerobic respiration. As well, the runner is utilizing primarily type 2 muscle fibers during her sprints (fast twitch), which exhibit a lower concentration of mitochondria, making them dependent on anaerobic glycolysis. The subsequent buildup of lactic acid results in increased acidity of the tissue environment, which leads to tissue vasodilation and increased oxygen delivery to the muscle. Clinically, this manifests as muscle pain and a burning sensation. Incorrect Answers: A, B, and C. Activity of the sodium-proton antiporter (Choice A) does not decrease during intense exercise. The sodium-proton antiporter activity increases during rigorous exercise in order to remove H* from the sarcoplasm and maintain homeostasis. Increased activity of the sodium-proton antiporter (Choice B) does occur in exercise states, however this function serves to reduce the acidity of the sarcoplasm, not increase sarcoplasmic acidity. Increased oxygen delivery to the muscle would facilitate increased aerobic respiration, which would decrease lactic acid production (Choice C) as the muscle cells would optimize energy efficiency by using aerobic cellular respiration. Educational Objective: During vigorous exercise and times of increased metabolic demand of myocytes, there may be insufficient oxygen for aerobic respiration. Myocytes consequently depend on anaerobic respiration, which results in the regeneration of NAD+ from NADH through the production of lactic acid. This increased sarcoplasmic acidity results in muscle pain and burning. %3D Previous Next Score Report Lab Values Calculator Help Pause

91 Exam Section 2: Item 41 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 41. A 10-year-old boy who has begun chemotherapy for acute myelogenous leukemia awakens at night with fever and severe pain in the ankles. Treatment with over-the-counter analgesics does not resolve the pain. The next morning, he has pain with urination and blood in the urine. An increased serum concentration of which of the following compounds is the most likely cause of this patient's symptoms? A) Cystine B) Glycine C) Magnesium D) Urea E) Uric acid

E. Tumor lysis syndrome most commonly occurs following chemotherapy initiation for leukemia or lymphoma. The abrupt destruction of a large number of tumor cells results in interstitial and serum release of their contents, manifesting as hyperphosphatemia, hypocalcemia, hyperkalemia, and hyperuricemia. Patients with tumor lysis syndrome can present with fever, nausea, vomiting, diarrhea, lethargy, cardiac dysrhythmias, seizures, tetany, muscle cramps, and/or syncope. They are at risk for sudden death secondary to the electrolyte and metabolic abnormalities. Purine nucleic acids are metabolized to hypoxanthine and xanthine, which are converted to uric acid by xanthine oxidase, resulting in hyperuricemia. Uric acid can crystallize in the renal tubules, leading to nephropathy and acute kidney injury. It can also be deposited in the joints, leading to arthralgias (eg, gout). Treatment of tumor lysis syndrome involves supportive care and correction of electrolyte and metabolic abnormalities. Rasburicase is used to promote the degradation of uric acid and is used in both prevention and treatment of tumor lysis syndrome. In patients with severe kidney injury and refractory electrolyte derangements, renal replacement therapy with dialysis may be indicated. Incorrect Answers: A, B, C, and D. Cystine (Choice A) is an amino acid that is metabolized to urea (Choice D) and does not result in hyperuricemia. Urea, in contrast to uric acid, is soluble in water and is normally reabsorbed in the nephron, contributing to the hyperosmolar gradient surrounding the loop of Henle. Urea does not cause acute kidney injury or contribute to hyperuricemia. Glycine (Choice B) is an amino acid that is found in many polypeptides, especially collagen, and is also an inhibitory neurotransmitter in the central nervous system. It is not appreciably increased in tumor lysis syndrome. Tumor lysis syndrome can result in high levels of magnesium (Choice C), but hypermagnesemia does not result in crystal nephropathy or arthropathy. Educational Objective: Tumor lysis syndrome most commonly occurs following chemotherapy initiation for leukemia or lymphoma. The abrupt destruction of a large number of tumor cells results in interstitial and serum release of their contents, manifesting as hyperphosphatemia, hypocalcemia, hyperkalemia, and hyperuricemia, the latter of which results in acute kidney injury and gout. %3D 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. Failure of normal differentiation of the endoderm in the embryonic lung bud is most likely to affect the development of which of the following? A) Capillary patterns B) Cartilage in bronchi C) Smooth muscle on the bronchi D) Surfactant secretion E) Tracheal rings

D. Endoderm is one of the three primary embryonic germ layers and composes the innermost layer of the early developing organism. Endoderm derivatives include the epithelial linings of the respiratory tract, gastrointestinal tract, biliary system, genitourinary tract, vagina, and middle ear. Organs that arise from the endoderm include the liver, parathyroid glands, thymus, pancreas, and the follicular and parafollicular cells of the thyroid. Surfactant secretion in the mature lung is achieved by type Il pneumocytes, a component of the respiratory epithelium. Defective differentiation of the endoderm in the embryonic lung bud would most likely result in impaired development of type Il pneumocytes and reduced secretion of surfactant. Incorrect Answers: A, B, C, and E. Capillary patterns (Choice A), cartilage in bronchi (Choice B), smooth muscle on the bronchi (Choice C), and tracheal rings (Choice E) are all derivatives of the mesoderm. Mesoderm is the middle embryonic germ layer and primarily responsible for development of connective tissue structures, including muscle, dermis, bone, cartilage, dura mater, the cardiovascular system, lymphatic system, blood components, kidneys, adrenal cortex, and reproductive organs. Educational Objective: Cells of the respiratory epithelium arise from the embryonic endoderm germ layer. This includes type I pneumocytes, which form the simple squamous epithelium of the alveoli, and type Il pneumocytes, which secrete surfactant. %3D 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 75-year-old woman comes to the physician because of a 3-month history of an enlarging lesion on her face. Physical examination shows a 1.5-cm, brown-black, mottled, scaly lesion with irregular borders. Microscopic examination of a biopsy specimen of the lesion shows atypical melanocytes spread along the basilar layer of the epidermis. Which of the following is the most likely cause of these findings? A) Acanthosis nigricans B) Actinic keratosis C) Compound nevus D) Lentigo maligna E) Seborrheic keratosis

D. Malignant melanoma is likely to be present when a lesion demonstrates asymmetry, irregular appearing borders, variable coloration, a diameter greater than 6 mm, and rapid evolution in characteristics. Malignant melanoma could rapidly invade and metastasize, which carries a poor prognosis when diagnosed late. Subtypes include superficial spreading, nodular, lentigo maligna, and acral lentiginous. The lentigo maligna type is classically seen in elderly individuals in areas of extensive sun damage, such as the face. Lentigo maligna typically grows slowly and superficial along the dermal-epidermal junction. Any lesion with features suggestive of malignant melanoma should be surgically excised with negative margins and pathologically examined for the depth of dermal invasion. Incorrect Answers: A, B, C, and E. Acanthosis nigricans (Choice A) is characterized by hyperpigmented, velvety patches seen on the neck, upper back, breasts, and axillae which is a marker of metabolic syndrome and diabetes. It is neither pre-malignant nor malignant and there are no atypical cells on histopathologic examination. Actinic keratosis (Choice B) 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. Compound nevus (Choice C) is a benign proliferation of melanocytes located in both the epidermis and dermis. They are very common. Compound nevi should not display asymmetry, border irregularity, or multiple colors. Development of compound and other benign nevi should cease in the fourth to fifth decade. New or changing nevi after this time are concerning for melanoma. Seborrheic keratosis (Choice E) is a benign proliferation of the epidermis; lesions exhibit a greasy, adherent appearance. While seborrheic keratoses are often brown, this is due to the keratin produced by the epidermis rather than melanin. Educational Objective: Lentigo maligna is a subtype of malignant melanoma that most commonly manifests on the sun-exposed skin of elderly patients. It is often slow growing and typically confined to the dermal-epidermal junction. %3D Previous Next Score Report Lab Values Calculator Help Pause

94 Exam Section 2: Item 44 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 44. A previously healthy 40-year-old man is brought to the emergency department 1 hour after the sudden onset of severe pain in his left leg while playing tennis. He is found to have ruptured the left Achilles tendon and undergoes operative repair and long leg cast immobilization. Six months later, the left calf shows a 2-cm decrease in circumference compared with the right calf. Which of the following is the most likely cause of this decrease? A) Decreased glycogen synthesis B) Decreased myosin light chain phosphatase activity C) Increased phosphatidyl degradation D) Increased protein degradation E) Mitochondria damage F) Necrosis of muscle fibers

D. Muscles are dynamic tissues that respond to use over time with hypertrophy and disuse over time with atrophy. Muscle disuse atrophy occurs when an extremity is non-weight-bearing for an extended length of time, in cases of limited mobility due to frailty, or during periods of gross immobilization such as long periods of time in the intensive care unit. In the chronic disuse, there will be decreased input from the motor neuron and consequent decreased muscle contractions. This diminished activity leads to a decrease in muscle protein synthesis. Typically, there is a balance between protein synthesis and protein degradation in order to maintain appropriate sarcomere architecture and functionality, however, during disuse atrophy there is preferential muscle protein degradation. It has been shown that the Akt-mTOR pathway is involved in attenuated protein synthesis in muscle disuse. Similarly, the ubiquitin-proteasome pathway is hypothesized to be involved in the increased protein degradation in disuse atrophy. Incorrect Answers: A, B, C, E, and F. Decreased glycogen synthesis (Choice A) occurs in times of relative energy deficiency and strenuous exercise in skeletal muscle. Decreased insulin levels and increased levels of glucagon and epinephrine during times of stress signal the breakdown of glycogen in skeletal muscle to provide glucose for glycolytic metabolism. Myosin light chain phosphatase (Choice B) dephosphorylates the light chain of myosin in smooth muscle leading to relaxation. This enzyme does not play a role in the pathophysiology of disuse atrophy. Phospholipase C is an enzyme that catalyzes the hydrolysis of phosphatidylinositol 4,5-bisphosphate leading to the formation of diacylglycerol. This process is regulated by the Gg protein-coupled receptor and plays a role in smooth muscle contraction, not skeletal muscle. Increased phosphatidyl degradation (Choice C) does not play a role in the pathophysiology of disuse atrophy. During normal cell respiration, the electron transport chain produces superoxide radicals and reactive oxygen species such as hydrogen peroxide. Under conditions of inadequate antioxidant concentration, these reactive oxygen species may lead to damage of mitochondrial DNA as well as membrane lipids and proteins (Choice E). This may eventually lead to cell death but is not the main mechanism of disuse atrophy. Necrosis of muscle fibers (Choice F) is seen in conditions such as muscular dystrophy, autoimmune myopathies, and rhabdomyolysis. It is not a causal factor in disuse atrophy. Educational Objective: Muscle disuse atrophy occurs when an extremity is non-weight-bearing for an extended length of time, in cases of limited mobility due to frailty, or during periods of gross immobilization such as long periods of time in the intensive care unit. The pathophysiology of disuse atrophy is related to an imbalance between protein production and protein degradation. %3D Previous Next Score Report Lab Values Calculator Help Pause

93 Exam Section 2: Item 43 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 43. A viral gene product is found to decrease expression of class I MHC molecules on the surfaces of infected cells. A mutant strain of virus is isolated that has a nonfunctional form of this gene. Which of the following types of cells are likely to contribute more effectively to control of the parental strain of the virus than to control of the mutant virus? A) CD4+ T lymphocytes B) CD8+ T lymphocytes C) Follicular dendritic cells D) Natural killer cells E) Plasma cells

D. Natural killer (NK) cells are more likely to eradicate the parental strain of the virus as this strain decreases class I MHC expression on the surface on infected cells. NK cells target virally infected cells in response to an absence of surface class I MHC molecules. When activated, NK cells synthesize perforin and granzyme, which are both proapoptotic. The mutant strain of the virus does not change the expression of class I MHC on virally infected cells and is therefore unlikely to result in activation of NK cells, effectively leading to evasion of this part of the innate immune system. In addition to activation as a result of absent class I MHC molecules, NK cells are also activated by an antibody-dependent mechanism in which pathogens opsonized by immunoglobulin (Ig) bind to ČD16 on the surface of NK cells. Incorrect Answers: A, B, C, and E. CD4+ T lymphocytes (Choice A) recognize class II MHC molecules, not class I MHC molecules. They recruit macrophages, cytotoxic T cells, plasma cells, and eosinophils. CD8+ T lymphocytes (Choice B) induce apoptosis in cells that express class I MHC. They would be more likely to play a role in control of cells affected by the mutant virus, which does not attenuate the expression of class I MHC molecules on the surface of infected cells. Follicular dendritic cells (FDCS) (Choice C) have receptors that bind and endocytose antigens via specific receptors (eg, CR1 and CR2) to allow presentation of these antigens to B lymphocytes. B lymphocytes that recognize particular antigens are induced to proliferate and produce antibodies. FDCS do not directly interact with T lymphocytes via MHC molecules. Plasma cells (Choice E) are terminally differentiated B lymphocytes that secrete a specific immunoglobulin against a particular antigen. Interaction with class II MHC molecules is a component of B lymphocyte activation, proliferation, and differentiation, but B lymphocytes do not respond to an absence of class I MHC molecules. Educational Objective: NK cells are activated to induce apoptosis in cells that do not express class I MHC molecules on their cell surface. The parental strain of the virus described induces infected cells to downregulate expression of class I MHC molecules, which would result in NK cell-induced apoptosis. %3D Previous Next Score Report Lab Values Calculator Help Pause

92 Exam Section 2: Item 42 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 42. A 27-year-old woman is brought to the physician by her family because she has been progressively lethargic and unwilling to leave her apartment over the past week. She has been receiving treatment in a mental health center for 10 years but missed her last appointment 18 days ago because of a snowstorm. She is now reluctant to return because she believes the staff is involved in an extraterrestrial plot. Three years ago, she had similar symptoms treated with electroconvulsive therapy. She appears disheveled. She is having auditory hallucinations of several people talking about her. Physical examination shows normal findings. She has poor eye contact, a flat affect, and slow speech. She describes an elaborate delusional system about the plot at the mental health center. Thought content is otherwise impoverished. Which of the following is the most likely diagnosis? A) Bipolar disorder, depressed B) Borderline personality disorder C) Delusional disorder D) Schizophrenia E) Schizotypal personality disorder

D. Patients with chronic psychotic disorders such as schizophrenia or schizoaffective disorder commonly demonstrate delusional thinking, auditory hallucinations, disheveled appearance, a flat affect (severely decreased range of affect), an impoverished thought content (decreased incidence of thoughts), and avolition (decreased ability to perform tasks). According to the DSM-5, patients with schizophrenia have two clusters of symptoms: positive symptoms (addition of mental dysfunction including delusions, hallucinations, and disorganized thinking and behavior) and negative symptoms (absence of certain mental functions such as affect, volition, thought, and speech). These symptoms lead to functional impairment. This patient illustrates several positive and negative symptoms and functional impairment (eg, inability to leave the home). Schizophrenia symptoms are typically chronic but can feature episodic exacerbations of both positive and negative symptoms in the setting of stress or medication nonadherence (as in this patient who missed her appointment). Treatment centers around antipsychotic medications, but in severe or treatment-refractory cases, electroconvulsive therapy may be considered. Incorrect Answers: A, B, C, and E. Depressive episodes in bipolar disorder (Choice A) would present with depressed mood, anhedonia, psychomotor retardation, and neurovegetative symptoms such as sleep, appetite, and energy disturbances. Severe depressive episodes may be associated with mood-congruent psychotic symptoms such as delusions of guilt as opposed to this patient's complex paranoid delusions and auditory hallucinations. Additionally, mental status examination would likely demonstrate a dysphoric affect rather than a flat affect. A disheveled appearance is also more typical of a chronic psychotic disorder such as schizophrenia than a depressive episode of bipolar disorder. Borderline personality disorder (Choice B) is a cluster B personality disorder, the emotional or dramatic cluster, that features an unstable sense of self and tumultuous relationships. Though patients with borderline personality disorder may experience transient psychosis when emotionally dysregulated, several days of a delusional belief system and auditory hallucinations would be atypical. A flat affect is also atypical of borderline personality disorder. Delusional disorder (Choice C) features the presence of one or more delusions for a month or longer without other psychotic symptoms. Prominent hallucinations, negative symptoms, and functional impairment, as seen in this patient, would be atypical of delusional disorder. Schizotypal personality disorder (Choice E) is one of the cluster A personality disorders, the odd or eccentric cluster. The disorder is characterized by odd or magical behavior and thinking, as well as a constricted affect. Complex, rigid paranoid delusions as in this patient would be atypical for patients with schizotypal personality disorder. Additionally, this patient's negative symptoms and functional impairment are more consistent with schizophrenia. Educational Objective: Patients with schizophrenia have two clusters of symptoms: positive symptoms (addition of mental dysfunction including delusions, hallucinations, and disorganized thinking and behavior) and negative symptoms (absence of certain mental functions such as affect, volition, thought, and speech). Schizophrenia symptoms are typically chronic but can feature episodic exacerbations of both positive and negative symptoms in the setting of medication nonadherence. %3D Previous Next Score Report Lab Values Calculator Help Pause

35 Exam Section 1: Item 35 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 35. A 36-year-old woman with type 2 diabetes mellitus comes to the physician for a follow-up examination. Current medications include a sulfonylurea. She is 173 cm (5 ft 8 in) tall and weighs 95 kg (210 Ib); BMI is 32 kg/m2. Physical examination shows acanthosis nigricans. Treatment with metformin is most likely to produce which of the following effects in this patient? A) Decrease intestinal carbohydrate digestion B) Increase beta-cell insulin secretion C) Increase deposition of adipocyte fat D) Increase hepatic triglyceride synthesis E) Inhibit hepatic gluconeogenesis

E. Type 2 diabetes mellitus is a common disorder characterized by peripheral insulin resistance and hyperglycemia. Uncontrolled diabetes increases the risk for macrovascular complications (eg, coronary artery disease, peripheral arterial disease, and stroke) and microvascular complications (eg, nephropathy, retinopathy, neuropathy). Treatment with metformin is a first-line therapy along with diet modification, increased activity, and weight loss. Metformin is an oral biguanide antihyperglycemic that has multiple glucose- lowering effects. It inhibits hepatic gluconeogenesis and increases peripheral sensitivity to insulin resulting in increased glucose uptake from the circulation. Metformin is generally well-tolerated and not associated with hypoglycemic episodes or weight gain. Adverse effects include gastrointestinal disturbance and less frequently, lactic acidosis, which may be exacerbated in renal failure. Incorrect Answers: A, B, C, and D. Decreased intestinal carbohydrate digestion (Choice A) is achieved through the administration of a-glucosidase inhibitors such as acarbose and miglitol. These medications function to decrease carbohydrate hydrolysis at the intestinal brush border, resulting in decreased glucose absorption. Sulfonylureas, glucagon-like peptide-1 (GLP-1) agonists, and dipeptidyl peptidase-4 (DPP-4) inhibitors increase beta-cell insulin secretion (Choice B) through different mechanisms. Sulfonylureas inhibit pancreatic beta-cell potassium channels resulting in membrane depolarization and insulin release. GLP-1 agonists mimic the activity of GLP-1, which is an incretin produced by the gastrointestinal tract that stimulates insulin release. DPP-4 inhibitors decrease the enzymatic clearance of endogenous GLP-1. Metformin does not increase the deposition of adipocyte fat (Choice C) or increase hepatic triglyceride synthesis (Choice D). Insulin stimulates fatty acid synthesis in adipocytes and triglyceride synthesis in the liver. Treatment with insulin or antihyperglycemic agents that promote insulin release are associated with weight gain. Educational Objective: Metformin is a first-line therapy for the treatment of type 2 diabetes mellitus in addition to diet and activity modification. Metformin inhibits hepatic gluconeogenesis and increases glucose uptake by peripheral tissues. %3D Previous Next Score Report Lab Values Calculator Help Pause

90 Exam Section 2: Item 40 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 40. A 65-year-old man develops urinary incontinence immediately after an operation for prostate cancer. The most likely cause of his condition is damage to the prostatic nerve plexus that resulted in denervation of the internal urethral sphincter. The function of which of the following types of tissue is most likely impaired as a result of this damage? A) Dense irregular connective tissue B) Dense regular connective tissue C) Skeletal muscle D) Smooth muscle E) Transitional epithelium

D. Pelvic parasympathetic nerves in the pelvic plexus function to excite and contract the detrusor muscle, a smooth muscle of the bladder, via muscarinic acetylcholine receptors, while sympathetic nerves mediate relaxation of the internal urethral sphincter, also smooth muscle, via a1-adrenergic receptors, leading to normal urination. Damage to the pelvic plexus can affect both bladder contraction and urethral sphincter relaxation, leading to overflow incontinence. Overflow incontinence is characterized by chronic urinary retention and a chronically distended bladder. When intravesical pressure exceeds outlet closing pressure, incontinence results as urine flows. Overflow incontinence is treated with managing inciting conditions, timed voiding, or placement of a urethral catheter. Incorrect Answers: A, B, C, and E. Dense irregular connective tissue (Choice A) is found in the dermis and consists of a large proportion of collagen fibers. It consists of fibers that are arranged in a matrix, whereas dense regular connective tissue (Choice B) is arranged in parallel fibers. Dense regular connective tissue is found in ligaments and tendons, areas of high tensile strength. Neither are involved in control of urination or urinary incontinence. The pudendal nerve supplies sensory neurons to the external genitalia and somatic motor neurons to the pelvic muscles such as the external urethral sphincter. The external urethral sphincter is a skeletal muscle (Choice C) that provides voluntary control of urination. Damage to the pudendal nerve would result in loss of voluntary control over voiding. The transitional epithelium (Choice E) of the bladder is specialized stratified epithelium that is able to expand and contract depending on the bladder volume. This allows for bladder distension. It does not have a role in control of urination or incontinence. Educational Objective: The internal urethral sphincter is a smooth muscle innervated by the pelvic plexus. Damage to the pelvic plexus can affect both bladder contraction and internal urethral sphincter relaxation, leading to overflow incontinence. %3D Previous Next Score Report Lab Values Calculator Help Pause

51 Exam Section 2: Item 1 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 1. A 36-year-old man with HIV infection has been treated with a combination of antiretroviral drugs, including zidovudine (AZT), for 3 years. Laboratory studies show a marked increase in his plasma HIV viral load during the past 3 months. Viral resistance to zidovudine is suspected. A mutation in which of the following is most likely to explain the resistance to zidovudine in this patient? A) Integrase B) Neuraminidase C) Protease D) RNA-dependent DNA polymerase E) Thymidine kinase

D. RNA-dependent DNA polymerase (HIV reverse transcriptase) mutations are most likely to account for this patient's resistance to zidovudine (AZT). Drugs that block HIV reverse transcriptase are either nucleoside reverse transcription inhibitors (NRTIS) or non- nucleoside reverse transcription inhibitors (NNRTI). AZT is an NRTI while drugs such as efavirenz are NNRTIS. The process of HIV reverse transcription is error prone and can result in mutations of the HIV DNA as a result of inaccurate transcription, but selective pressure garnered by use of NRTIS can lead to mutations of the HIV reverse transcriptase enzyme that allow it to evade drugs such as AZT. AZT is a thymidine analog and the mutations that result in resistance to AZT and similar drugs are referred to as thymidine analog mutations (TAMS). Depending upon the specific mutations in HIV reverse transcriptase, the mutation can confer resistance to all drugs within the NRTI class, but usually at least three TAMS are necessary for HIV resistance to emerge. Continued use of the same drug will lead to accumulation of additional mutations due to selective pressure. Resistance is most often prevented by using a combination of medications from different classes, but patients who were treated in the early days of the HIV pandemic before the wide availability of combination regimens are more likely to have mutations that developed as a result of monotherapy with drugs such as AZT. Incorrect Answers: A, B, C, and E. Integrase (Choice A) inhibitors include medications such as dolutegravir and raltegravir. Integration into the host genome occurs via the action of an HIV-encoded enzyme, so mutations in this enzyme can lead to resistance to integrase inhibitors. Mutations often confer resistance to all medications within the integrase inhibitor class. AZT is not an integrase inhibitor and would be unlikely to select for such a mutation. Neuraminidase (Choice B) is an enzyme that breaks glycosidic linkages of neuraminic acids and is most commonly associated with the influenza virus. Protease (Choice C) inhibitors such as darunavir and lopinavir will select for mutations in HIV protease, which is necessary in the cleavage of immature viral proteins into mature viral proteins prior to exit from the cell. AZT would not select for mutations of HIV protease. Thymidine kinase (Choice E) is an enzyme that phosphorylates guanosine analog medications such as acyclovir and valacyclovir that are used to treat herpetic infections. These drugs inhibit herpesvirus DNA polymerase but are not activated until phosphorylated. These drugs do not play a role in the treatment of HIV unless there is concomitant infection with herpesvirus and would not select for mutations in HIV reverse transcriptase. Educational Objective: HIV medications that inhibit the RNA-dependent DNA polymerase (HIV reverse transcriptase) are either NRTIS or NNRTIS. When used as monotherapy, they can potentially induce resistance via mutation of the HIV reverse transcriptase. This issue has been largely attenuated by using combination regimens to treat HIV. %3D Previous Next Score Report Lab Values Calculator Help Pause

86 Exam Section 2: Item 36 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 36. A61-year-old man has erectile dysfunction due to spinal cord injury at L-2. Sildenafil is likely to markedly correct the dysfunction by acting at which of the following labeled structures in the transverse section of the penis? A A) B) C) D)

D. Sildenafil is a selective inhibitor of cyclic guanosine monophosphate (CGMP)-specific phosphodiesterase-5 (PDE5). Its site of effect in treating erectile dysfunction is primarily at the corpora cavernosa, a pair of richly vascularized erectile bodies within the shaft of the penis. By inhibiting PDE5, sildenafil increases the concentration of nitric oxide within the corpus cavernosum (indicated by letter D), which in turn promotes corpus cavernosum vascular dilation and tumescence. Adverse effects of sildenafil include headache, flushing, cyanopsia (a transient blue tint to vision), and priapism. Sildenafil is contraindicated in patients who are using nitrate medications, as these two drug classes may act synergistically and produce life-threatening hypotension. Incorrect Answers: A, B, and C. Choice A indicates the superficial dorsal vein of the penis. This vein drains many of the superficial structures of the penis but plays no role in the initiation of erection. Choice B indicates the connective tissues surrounding the corpora cavernosa, which include the tunica albuginea and the Buck fascia. Connective tissues limit the egress of venous blood from the erect penis but are not the site of action of sildenafil. Choice C indicates the urethra, which plays no role in erections. Educational Objective: Sildenafil selectively inhibits CGMP-specific phosphodiesterase-5 (PDE5), leading to an increase in nitric oxide within the corpora cavernosa, which results in vascular dilation and tumescence of the erectile bodies. Sildenafil should not be prescribed to patients using nitrates due to the potential for life-threatening hypotension. %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. Which of the following types of sensory information is compromised by lesions of the structure at site X in the photograph shown? A) Conscious proprioception B) Pain sensation C) Two-point discrimination D) Unconscious proprioception E) Vibration sense

D. The anterior lobe of the cerebellum (labeled X, pictured in cross-section as an arborized brain area posterior to the brainstem and anterior to the primary fissure of the cerebellum) mediates unconscious proprioception. The anterior lobe of the cerebellum receives information from the spinocerebellar tract about proprioception, or body position, that is gathered from muscle stretch and tension receptors on the ipsilateral side of the body. This proprioceptive information is transmitted outside of conscious awareness. The deep cerebellar nuclei use this proprioceptive information to control motor learning, movement course changes, and balance. Damage to the anterior lobe of the cerebellum, which commonly occurs in chronic alcoholism, may lead to broad-based gait ataxia. Incorrect Answers: A, B, C, and E. Conscious proprioception (Choice A), two-point discrimination (Choice C), and vibration sense (Choice E) are mediated by the dorsal column-medial lemniscus pathway, which relays this sensory information up the spinal cord to the thalamus and terminates in the primary sensory cortex in the parietal lobe. The cortex is a high-order brain area involved in several conscious brain functions, which reflects this pathway's mediation of the conscious (rather than unconscious) awareness of proprioception. Pain sensation (Choice B) is mediated by the spinothalamic pathway. The spinothalamic pathway transmits information about pain, temperature, and crude touch up the spinal cord to the thalamus, terminating in the primary sensory cortex. Educational Objective: The anterior lobe of the cerebellum mediates unconscious proprioception, whereas conscious proprioception is controlled by the dorsal column-medial lemniscus pathway. Lesions of the anterior lobe of the cerebellum can result in broad- based gait ataxia. %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. A76-year-old man undergoes laparotomy for resection of an abdominal aortic aneurysm. During the procedure, an incidental finding of acquired colonic diverticula is made. The diverticula in this patient are most likely present in which of the following? A) Ascending colon B) Cecum C) Descending colon D) Sigmoid colon E) Transverse colon

D. The sigmoid colon is the most common location for diverticula to form. Diverticula are outpouchings of the mucosal and submucosal layers into the muscular layer that occur at weak points in the gut wall where the small arterioles of the vasa recta penetrate. It is hypothesized that abnormal motility in the colon causes increased intraluminal pressure with subsequent herniation of the mucosa through weak points in the colonic wall. This may happen in the sigmoid colon because the diameter of the sigmoid is smaller than other parts of the colon, so abnormal peristalsis in this area causes higher intraluminal pressure compared to other segments. Risk factors for development of diverticula include obesity and a diet high in red meat and low in fiber. Diverticulosis predisposes to lower gastrointestinal bleeding (GI). It also predisposes to diverticulitis, a bacterial infection of a diverticulum that leads to a local inflammatory response. Diverticulitis presents classically with fever, left lower quadrant abdominal pain, and occasionally with bloody diarrhea. Incorrect Answers: A, B, C, and E. The ascending colon (Choice A) can be a site of colorectal carcinoma (CRC). Because stool is generally liquid in the ascending colon and tumors in this location tend not to be exophytic, CRC can present late as symptoms are less common. While diverticula may occur in the ascending colon, it is a less common location than the sigmoid. Similarly, the transverse colon (Choice E) and descending colon (Choice C) may develop diverticula, but these sites are less common than the sigmoid colon. Cecum (Choice B) is the junction between the ileum and the ascending colon. It is not a common site of diverticulosis, but since the appendix lies in close proximity to it, the cecum can occasionally become inflamed in severe, acute appendicitis. Educational Objective: Diverticulosis refers to a condition in which the colonic mucosa and submucosa herniate into the muscular layer at weak points where the vasa recta penetrate the colonic wall. Risk factors include chronic constipation and a low fiber diet. The sigmoid colon is the most frequently involved part of the colon as its smaller diameter predisposes to greater intraluminal pressure during peristalsis, thereby increasing the risk for diverticula formation. %3D Previous Next Score Report Lab Values Calculator Help Pause

53 Exam Section 2: Item 3 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 3. A 16-year-old boy is brought to the emergency department because of a 2-day history of increasingly severe abdominal pain. His temperature is 39°C (102.2°F), pulse is 86/min, respirations are 18/min, and blood pressure is 120/60 mm Hg. Abdominal examination shows exquisite tenderness of the right lower quadrant. His leukocyte count is 16,000/mm3. An appendectomy is done; the appendix is swollen with a tan exudate on the serosal surface. Which of the following best characterizes the leukocytosis in this patient? A) Basophilia B) Eosinophilia C) Lymphocytosis D) Monocytosis E) Neutrophilia

E. Appendicitis is acute inflammation of the appendix and usually results from appendiceal obstruction by a fecalith or by lymphoid hyperplasia. Appendiceal obstruction leads to bacterial proliferation within the lumen and wall of the appendix, eventually leading to appendiceal inflammation with potential necrosis and perforation if untreated. Bacterial proliferation provokes a leukocytosis, which is predominantly neutrophilic. Appendicitis presents with right lower quadrant abdominal pain, fever, anorexia, nausea, vomiting, and leukocytosis. Incorrect Answers: A, B, C, and D. Basophilia (Choice A) is a non-specific hematologic feature of various myeloproliferative disorders, such as acute myeloid leukemia, myelofibrosis, or polycythemia vera. It would not be observed as a predominant feature in the setting of an acute inflammatory response. Eosinophilia (Choice B) is typically observed in the setting of allergic or parasitic responses, neither of which are likely to be present because of appendicitis. Lymphocytosis (Choice C) is observed in response to viral infections, some of which feature gastrointestinal symptoms, such as cytomegalovirus or hepatitis. While viral infections may cause lymphoid tissue proliferation that occludes the appendix, acute appendicitis ensues due to bacterial proliferation within the appendix and is characterized by a neutrophilic response. Monocytosis (Choice D) is characteristic of response to chronic infections, including tuberculosis, ehrlichiosis, and leishmaniasis. It is not a prominent feature of acute appendicitis. Educational Objective: Appendicitis results from obstruction of the appendix and resultant bacterial proliferation within the lumen and wall of the appendix. Bacterial proliferation provokes a leukocytosis, which is predominantly neutrophilic. If untreated, appendiceal necrosis and perforation may result. %3D Previous Next Score Report Lab Values Calculator Help Pause

99 Exam Section 2: Item 49 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 49. A 21-year-old woman of Japanese descent comes to the emergency department because of a 3-hour history of facial flushing. Her symptoms began shortly after she drank a glass of champagne for the first time at a wedding reception. Physical examination shows profound erythema over the face. The most likely cause of these findings is a mutation in the gene for which of the following enzymes? A) Acetyl-CoA reductase B) Alcohol catalase C) Alcohol dehydrogenase D) Alcohol reductase E) Aldehyde dehydrogenase

E. Facial flushing due to alcohol consumption is caused by deficient activity of the enzyme aldehyde dehydrogenase, which converts acetaldehyde to acetic acid. Deficiency of aldehyde dehydrogenase leads to an accumulation of acetaldehyde, which stimulates to the release of histamine from mast cells and causes symptoms of facial flushing. In severe cases, histamine release may cause symptoms that mimic asthma, including wheezing and cough. Aldehyde dehydrogenase is also the target of the medication disulfiram, which utilizes the unpleasant physical effects of acetaldehyde accumulation to treat alcohol use disorder. Additional symptoms of aldehyde toxicity include headache, fatigue, nausea, vomiting, tachycardia, and hypotension. Antagonism of aldehyde dehydrogenase is also the mechanism underlying the disulfiram-like side effects of drugs such as metronidazole, ketoconazole, and nitrofurantoin. Incorrect Answers: A, B, C, and D. Acetyl-CoA reductase (Choice A) and similar enzymes in the family of fatty acyl CoA reductases act in fatty acid metabolism. They do not play a role in the metabolism of alcohol. Alcohol catalase (Choice B) converts ethanol to acetaldehyde using hydrogen peroxide as an oxygen donor. This is an infrequent mechanism of alcohol metabolism as compared to alcohol dehydrogenase and occurs in peroxisomes. This step precedes the conversion of acetaldehyde to acetic acid by aldehyde dehydrogenase. Alcohol dehydrogenase (Choice C) plays the primary role in the metabolism of alcohols such as ethanol, methanol, and ethylene glycol. Alcohol dehydrogenase is the target of the drug fomepizole, which is used to treat methanol or ethylene glycol poisoning. Alcohol reductase (Choice D) may refer to aromatic alcohol reductase, a gene that reduces alcohol groups associated with aromatic hydrocarbons. It is not significant in the pathophysiology of facial flushing in aldehyde metabolism. Educational Objective: Alcohol-associated facial flushing is caused by a deficiency of aldehyde dehydrogenase, which converts acetaldehyde to acetic acid. Deficiency of aldehyde dehydrogenase leads to accumulation of acetaldehyde, which stimulates to the release of histamine from mast cells. %3D Previous Next Score Report Lab Values Calculator Help Pause

25 Exam Section 1: Item 25 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 25. An 8-year-old boy continues to bleed excessively after tooth extraction. Prothrombin time, bleeding time, and platelet count are within the reference range. Partial thromboplastin time is prolonged but corrects after addition to the assay chamber of plasma from a patient with hemophilia A. Which of the following is the most likely diagnosis? A) Acute disseminated intravascular coagulation B) Factor V (proaccelerin) deficiency C) Factor VII (proconvertin) deficiency D) Hemophilia A E) Hemophilia B F) Immune thrombocytopenic purpura G) von Willebrand disease

E. Hemophilia B is an X-linked bleeding disorder caused by absent, decreased, or dysfunctional factor IX and most likely explains this patient's prolonged bleeding and increased partial thromboplastin time (PTT). As patients with hemophilia A lack factor VIII but have normal levels of factor IX, addition of plasma from a patient with hemophilia A would correct the PTT as demonstrated in this patient with hemophilia B. Factor IX is a component of the intrinsic clotting cascade and serves to activate factor X to Xa, which subsequently converts prothrombin to thrombin and facilitates the formation of a fibrin clot. The activity of the coagulation factors in the intrinsic coagulation cascade is measured by the PTT while the activity of the extrinsic pathway is measured by the prothrombin time (PT), which is normal in this patient. The clinical severity of hemophilia B is variable. Patients with severe disease present early in life with easy bruising, bleeding following a minor procedure, or hemarthrosis. Patients with less severe disease may not present until they experience an event such as trauma or surgery. Treatment includes replacement of the deficient factor. Incorrect Answers: A, B, C, D, F, and G. Acute disseminated intravascular coagulation (Choice A) is a syndrome characterized by overwhelming activation of the clotting cascade often precipitated by malignancy, sepsis, or obstetrical emergencies. Endothelial dysfunction leads to the formation of microthrombi and depletion of coagulation factors. Microthrombi cause shearing stress on erythrocytes leading to microangiopathic hemolytic anemia, while the depletion of coagulation factors manifests as a prolonged PT and PTT and increases the risk of major bleeding. Factor V (proaccelerin) deficiency (Choice B) is a rare inherited bleeding disorder that may present with mucocutaneous bleeding or major bleeding following trauma or surgery. Factor V is required as a cofactor for the formation of thrombin in the common pathway of the coagulation cascade. Deficiency is treated with fresh frozen plasma (FFP), which contains factor V. Factor VII (proconvertin) deficiency (Choice C) is a rare bleeding disorder with a spectrum of clinical severity. Patients who are most affected present with heavy menstrual bleeding or bleeding following invasive procedures. Treatment is with factor replacement, prothrombin complex concentrate, or FFP. Hemophilia A (Choice D) is an X-linked bleeding disorder that presents similarly to hemophilia B and is caused by an absent or reduced level of factor VII. Addition of plasma from another patient with hemophilia A would not correct the PTT. Immune thrombocytopenic purpura (Choice F) is caused by circulating antibodies against platelets that leads to thrombocytopenia. The PT and PTT are normal. Bleeding, if it occurs, tends to be mucocutaneous. von Willebrand disease (Choice G) is one of the most common hereditary bleeding disorders and is due to quantitative or qualitative abnormality of von Willebrand factor, which binds platelets and subendothelial collagen in primary hemostasis. Impaired platelet adherence leads to a prolonged bleeding time. It can present with epistaxis, gingival bleeding, petechiae, easy bruising, and menorrhagia. Educational Objective: Hemophilia B is an X-linked bleeding disorder that is caused by a deficiency in factor IX leading to an increased PTT. Symptoms depend on severity but include prolonged bleeding following invasive procedures or trauma, easy bruising, and hemarthrosis. Treatment is with recombinant factor IX. %3D Previous Next Score Report Lab Values Calculator Help Pause

47 Exam Section 1: Item 47 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 47. A 48-year-old woman comes to the office because of a 4-month history of headaches, itchy skin, difficulty swallowing, heartburn, chest tightness, pain in her arms and legs, and a burning sensation with urination. She has a history of similar symptoms since the age of 14 years, but previous examinations showed no abnormalities. Her vital signs are within normal limits. Physical examination and laboratory studies today show no abnormalities. Which of the following is the most likely diagnosis? A) Conversion disorder B) Factitious disorder C) Ilness anxiety disorder (hypochondriasis) D) Malingering E) Somatic symptom disorder

E. In somatic symptom disorder, the patient is preoccupied with one or more somatic symptoms such that these symptoms disrupt the patient's daily life. These symptoms may or may not originate from an underlying disease though are typically unconsciously produced or exacerbated by psychological factors such as stress. Patients with somatic symptom disorder persistently devote excessive time and energy to these symptoms or related health concerns (eg, repeatedly going to the doctor). Somatic symptom disorder typically demonstrates a chronic and fluctuating course. Management of somatic symptom disorder centers around regular primary care follow-up that targets coping skills, provides reassurance, and avoids unnecessary tests. Incorrect Answers: A, B, C, and D. Conversion disorder (Choice A), more commonly called functional neurologic disorder, involves neurologic symptoms such as sensory or motor dysfunction that are not fully explained by objective findings on physical examination or imaging. Though the symptoms disrupt daily life, patients with functional neurologic disorder may or may not be preoccupied with their symptoms. This patient's symptoms are not solely neurologic. In factitious disorder (Choice B), patients consciously produce symptoms (eg, purposely injuring themselves) for primary gain. Primary gain is the motivation to be cared for, which constitutes an unconscious motivator for the patient's conscious production of symptoms. This patient's symptoms are instead unconsciously produced. Patients with illness anxiety disorder (hypochondriasis) (Choice C) demonstrate persistently excessive anxiety and preoccupation about the possibility that they may have or acquire a serious illness such that they perform frequent health-related behaviors (eg, repeatedly going to the doctor) or exhibit maladaptive avoidance (eg, completely avoiding going to the doctor). Patients with somatic symptom disorder are anxious about their existing symptoms, while patients with illness anxiety disorder typically have milder somatic symptoms relative to their concern about developing a serious illness. In malingering (Choice D), patients consciously produce symptoms due to the conscious motivation of secondary gain. Secondary gain is an extrinsic motivator such as procuring disability payments. Educational Objective: In somatic symptom disorder, the patient is excessively preoccupied with one or more somatic symptoms such that these symptoms disrupt the patient's daily life. These symptoms may or may not originate from an underlying disease though are typically unconsciously produced or exacerbated by psychological factors such as stress. %3D Previous Next Score Report Lab Values Calculator Help Pause

57 Exam Section 2: Item 7 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment V 7. A 46-year-old woman comes to the physician because of temperatures of 38 to 39°C (100.4 to 102.2°F) and malaise for the past 2 days. She has chronic alcohol dependence. There are several spider angiomas on the face, chest, and back; she is not jaundiced. The abdomen is protuberant and nontender. There is shifting dullness to percussion. A firm liver edge is ballotable 3 cm below the right costal margin. Routine laboratory studies show mild anemia, mildly increased hepatic transaminase activities, and a decreased serum albumin concentration. Peritoneal aspiration yields serous fluid with 1100 leukocytes/mm3 (80% neutrophils) and 100 erythrocytes/mm3. Which of the following processes best accounts for the patient's febrile illness? O A) Acute cholecystitis B) Chronic pancreatitis C) Exacerbation of autoimmune hepatitis D) Pelvic inflammatory disease E) Spontaneous bacterial peritonitis

E. Spontaneous bacterial peritonitis (SBP) is the most likely etiology of this patient's fever. SBP refers to the development of a peritoneal infection in patients with ascites that is not due to instrumentation or introduction of bacteria into the peritoneum from an alternative source. This patient most likely has cirrhosis from alcohol use disorder as evidenced by a firm liver, spider angiomas, and ascites. SBP is a common complication in patients with cirrhosis and ascites regardless of the etiology. Portal hypertension (PH) leads to reduced intestinal motility with bacterial stasis and overgrowth in addition to impaired gastrointestinal immunity. PH also causes gut wall edema and this confluence of factors can predispose to translocation of bacteria from the gut lumen into the blood stream or across the wall into the peritoneal cavity. These pathogens can seed the peritoneal cavity which, in patients with ascites, provides a perfect growth medium for bacteria. Symptoms can include fever, hepatic encephalopathy, abdominal pain, and signs and symptoms of sepsis. Risk is higher in patients with lower serum albumin concentrations and in those who have experienced prior episodes of SBP. Diagnosis is made by diagnostic paracentesis revealing an absolute neutrophil count of at least 250 cells/mm3. Gram stain is not sensitive for the diagnosis, but common pathogens include enteric flora, Escherichia coli, Streptococcus species, and Klebsiella pneumoniae. Treatment is with antibiotics, followed by prophylaxis with ciprofloxacin or trimethoprim-sulfamethoxazole. Incorrect Answers: A, B, C, and D. Acute cholecystitis (Choice A) presents with fever and right upper quadrant pain. If gallstones migrate into the common bile duct, obstruction can lead to increased transaminase levels and a cholestatic pattern of liver injury with increased alkaline phosphatase. Right upper quadrant ultrasound reveals gallstones, a thickened gallbladder wall, and pericholecystic fluid, although the latter two findings can be seen in patients with cirrhosis in the absence of cholecystitis. A peritoneal fluid sample showing 880 neutrophils is diagnostic of SBP in this case. Chronic pancreatitis (Choice B) is often a consequence of recurrent acute pancreatitis, which may be secondary to alcohol use, hypertriglyceridemia, or autoimmune pancreatitis. It presents with malabsorption, steatorrhea, deficiencies of fat-soluble vitamins, and chronic abdominal pain. Exacerbation of autoimmune hepatitis (AIH) (Choice C) would require a prior diagnosis of AIH and would usually present with jaundice and liver failure as evidenced by an increased prothrombin time and encephalopathy. Pelvic inflammatory disease (Choice D) is frequently caused by Chlamydia trachomatis and Neisseria gonorrhoeae, although is sometimes caused by E. coli, Bacteroides fragilis, Staphylococcus or Streptococcus species. Symptoms include pelvic pain and cervical discharge; physical examination reveals cervical motion or adnexal tenderness. This patient's presentation is more consistent with SBP. Educational Objective: SBP is diagnosed in patients with ascites who have an ascitic fluid absolute neutrophil count of at least 250 cells/mm³ in the absence of other causes of peritonitis such as bowel perforation. It commonly causes fever, encephalopathy, and abdominal pain. Treatment is with antibiotics directed against the most likely pathogen and most patients are subsequently placed on antibiotic prophylaxis. %3D 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 65-year-old woman comes to the physician because of a 3-month history of headache, weakness of her arms, and left flank pain; she also has had a 14-kg (31-lb) weight loss during this period. Physical examination shows weakness of the proximal upper and lower extremity muscles. There is augmentation of strength with repetitive testing of the deltoid muscles. An MRI of the brain shows a single well-demarcated mass surrounded by edema in the right frontal lobe. A stereotactic biopsy specimen of the lesion shows a malignant, small blue cell neoplasm that expresses cytokeratin, chromogranin, and synaptophysin. Which of the following is the most likely diagnosis? A) Anaplastic ependymoma B) Extranodal primary central nervous system lymphoma C) Glioblastoma multiforme D) Primary cerebral neuroblastoma E) Pulmonary small cell carcinoma metastatic to the brain

E. The patient's presentation is most consistent with pulmonary small cell carcinoma metastatic to the brain. Pulmonary small cell tumors are typically centrally located in the lungs and associated with tobacco use. They are neoplasms of neuroendocrine cells and may be associated with numerous paraneoplastic syndromes, including Cushing syndrome due to adrenocorticotropic hormone production, syndrome of inappropriate antidiuretic hormone, Lambert-Eaton myasthenic syndrome due to presynaptic calcium channel antibody production, and paraneoplastic myelitis, encephalitis, and subacute cerebellar degeneration. Proximal extremity weakness and augmentation of strength with repetitive testing of the deltoid muscles are suggestive of Lambert-Eaton myasthenic syndrome. Histologic features of pulmonary small cell carcinoma include small dark blue tumor cells lacking nucleoli with a high nuclear to cytoplasm ratio. The brain is a common site for metastatic disease. Incorrect Answers: A, B, C, and D. Anaplastic ependymoma (Choice A) is a central nervous system neoplasm formed from ependymal cells. Location typically involves the fourth ventricle. Histologic characteristics include perivascular pseudorosettes formed by malignant cells arranged around a blood vessel. Extranodal primary central nervous system lymphoma (Choice B) is a rare type of malignant non-Hodgkin lymphoma, often associated with an underlying immunodeficiency syndrome (eg, AIDS). Histologic examination may reveal large, atypical lymphocytes. Glioblastoma multiforme (Choice C) is a malignant primary brain tumor characterized by central necrosis on histology. Imaging features include an expansile mass crossing the corpus callosum with surrounding vasogenic edema. Seizures are a common presenting symptom, as are headaches and focal neurologic deficits. Primary cerebral neuroblastoma (Choice D) is a malignancy of neuroendocrine cells associated with sympathetic nervous tissue. Neuroblastomas may arise from the adrenal glands, the sympathetic chain, or the central nervous system, and may secrete catecholamines. Metastatic pulmonary small cell carcinoma is more common in adults than a primary cerebral neuroblastoma. Educational Objective: Solitary brain lesions may be secondary to a primary central nervous system malignancy, metastatic disease, infection, or abscess. Histologic analysis and clinical features can help narrow the diagnosis. Pulmonary small cell carcinomas are neuroendocrine tumors associated with paraneoplastic syndromes and brain metastases. %3D Previous Next Score Report Lab Values Calculator Help Pause

31 Exam Section 1: Item 31 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 31. A female newborn delivered at 38 weeks' gestation to a 28-year-old woman, gravida 1, para 1, develops respiratory distress. Pregnancy and delivery were uncomplicated; amniotic fluid was clear, and the placenta was normal. Fetal ultrasonography and MRI at 34 weeks' gestation showed a congenital diaphragmatic hernia, with evidence of small bowel and stomach herniation into left hemithorax. She is 52 cm (20.4 in) long and weighs 3500 g (7 lb 11 oz). Her temperature is 37.5°C (99.5°F), pulse is 138/min, respirations are 50/min, and blood pressure is 70/55 mm Hg. Physical examination shows peripheral cyanosis that improves after administration of oxygen via endotracheal intubation. Breath sounds are decreased on the left. Cardiac examination shows normal heart sounds without murmurs. Which of the following complications of the described pathology is most likely to be life threatening to this newborn? A) Active pneumonia B) Alveolar edema C) Amniotic embolism D) Inadequate surfactant synthesis E) Pulmonary hypoplasia

E. This newborn presents with a congenital diaphragmatic hernia, which places the newborn at risk for pulmonary hypoplasia. Herniation of abdominal organs into the thorax compresses the lungs while they are developing in utero. Pulmonary hypoplasia results through a complex mechanism that involves direct compression of abdominal organs upon the developing lungs, which affects bronchial and vascular architecture with reduced branching, blood supply, and increased thickness of alveolar septa. Newborns with pulmonary hypoplasia commonly develop persistent pulmonary hypertension due to increased pulmonary vascular resistance, resulting in potentially severe heart failure. This collection of respiratory and circulatory difficulties may be rapidly life-threatening to the newborn. Treatment involves surgical repair of the hernia, although the management of congenital diaphragmatic hernia must also consider the multitude of secondary pulmonary and cardiovascular problems that result from the herniation of abdominal viscera into the developing thorax. Incorrect Answers: A, B, C, and D. Active pneumonia (Choice A) may develop in newborns who survive the immediate postpartum period. This late complication, while possible, is less important than the immediately life-threatening consequences of pulmonary hypoplasia. Alveolar edema (Choice B) occurs in neonatal respiratory distress syndrome and in congenital heart disease. Alveolar edema does not necessarily result from congenital diaphragmatic hernia. Amniotic embolism (Choice C) is a potentially dangerous obstetric complication wherein amniotic fluid enters the maternal circulation. Amniotic embolism may cause maternal disseminated intravascular coagulation. Inadequate surfactant synthesis (Choice D) occurs in premature newborns and underlies the pathogenesis of neonatal respiratory distress syndrome. Inadequate surfactant produced by type Il pneumocytes leads to alveolar collapse. This is a separate disease process from the pulmonary changes that occur in pulmonary hypoplasia. Educational Objective: Pulmonary hypoplasia occurs as a sequela of congenital diaphragmatic hernia and results in serious and potentially life-threatening consequences to pulmonary and cardiac function. Pulmonary hypoplasia is caused by compression of the lungs by herniated abdominal organs with secondary changes in bronchial and vascular development. %3D Previous Next Score Report Lab Values Calculator Help Pause

38 Exam Section 1: Item 38 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 38. A healthy 22-year-old woman undergoes testing to determine whether she is a suitable kidney donor for her 25-year-old brother with end-stage renal disease caused by type 1 diabetes mellitus. Immunologic studies show that she is human leukocyte antigen (HLA)-DR3/DR6 positive, and her brother is HLA-DR3/DR4 positive. To determine the extent of alloreactivity, a mixed lymphocyte reaction is done using irradiated stimulator cells isolated from the donor and responder cells isolated from the recipient. The T lymphocytes that proliferate in these cultures will most likely react with which of the following HLA types? A) DR3 only B) DR3 and DR4 C) DR3 and DR6 D) DR4 only E) DR4 and DR6 F) DR6 only

F. With allogeneic donation, ideally, donors and recipients should match at all human leukocyte antigen (HLA) loci (six out of six). The patient receiving the transplant and the sibling in the question have a single mismatch. They are both HLA-DR3 positive, but the donor sibling is HLA-DR6 positive, whereas the patient is not. Due to this difference, immune cells from the recipient will likely react to HLA-DR6. Tlymphocytes develop in the thymus, where they are selected against if they react to self-antigens (negative selection). They are permitted to persist if they react to non-self-antigens, which promotes immunity against non-native proteins. Because the recipient is already HLA-DR3/DR4 positive, his T lymphocytes would not react to these HLA types assuming that no autoimmune condition is present. Mismatches of any kind predispose to T-lymphocyte mediated transplant rejection, although HLA-A, B, and DR appear to be the most important. In this case, an immune response would be expected against the donor HLA-DR6 by the recipient's T-lymphocytes that bind the antigen. Incorrect Answers: A, B, C, D, and E. Because the recipient is already HLA-DR3/DR4 positive, his T lymphocytes would not react to HLA types DR3 or DR4 (Choices A, B, C, D, and E). During immune development, cells containing these antigens would have been presented to his T lymphocytes in the thymus. Any T lymphocytes containing receptors that bind and promote reaction against these native antigens would have been destroyed, preventing autoimmunity (negative selection). Educational Objective: Donors and recipients of transplants should ideally match at all HLA loci. In a mixed lymphocyte reaction, recipient T lymphocytes will react to donor HLA types that are absent in the recipient. %3D Previous Next Score Report Lab Values Calculator Help Pause