NBME 28

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

178 Exam Section 4: Item 28 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 28. A 32-year-old woman is found to have panic disorder with agoraphobia. A drug is prescribed that activates benzodiazepine binding sites on the y-aminobutyric acid, (GABA) receptor. This drug is most likely which of the following? A) Alprazolam O B) Buspirone C) Flumazenil O D) Hydroxyzine E) Ramelteon

A. Alprazolam is a benzodiazepine medication that binds to benzodiazepine sites on the y-aminobutyric acidA (GABAA) receptor. Benzodiazepines increase the frequency of the opening of the ionotropic GABAA receptor, increasing the influx of chloride ions and decreasing the likelihood of neuronal action potentials by hyperpolarizing the neuron. Anxiety disorders are related to increased neuronal activity in the limbic system brain areas such as the amygdala, which mediates emotional responses such as fear. Benzodiazepine medications decrease fear responses by increasing GABA signaling in the amygdala. Panic disorder is characterized by recurrent panic attacks that are unexpected and associated with worry about future panic attacks or avoidance of panic attack triggers. Agoraphobia, or the fear of leaving one's home, frequently accompanies panic disorder. Treatment typically includes long-term selective serotonin reuptake inhibitors and short-term benzodiazepines. Long-term benzodiazepine use is associated with addiction. Incorrect Answers: B, C, D, and E. Buspirone (Choice B) is a serotonin modulator that does not interact with GABAA receptors. Buspirone is typically utilized to treat generalized anxiety disorder rather than panic disorder with agoraphobia. Flumazenil (Choice C) is a competitive antagonist at the benzodiazepine site on the GABAĄ receptor. Flumazenil is utilized to reverse accidental benzodiazepine overdose in patients who are not chronically dependent and benzodiazepine sedation after surgical procedures in select patients. Its use is limited as it can precipitate seizures. Hydroxyzine (Choice D) is an antihistamine medication with central nervous system activity. Hydroxyzine is used off-label for acute anxiety but does not bind the GABAA receptor. Ramelteon (Choice E) is a selective agonist of the melatonin receptors. Ramelteon is used for sleep-onset insomnia, not anxiety or panic attacks. Educational Objective: Benzodiazepines bind to and activate the GABAA receptor in the amygdala, decreasing neuronal action potentials and therefore attenuating the fear response. Benzodiazepines are utilized in conjunction with selective serotonin reuptake inhibitors to treat panic disorder. %3D Previous Next Score Report Lab Values Calculator Help Pause

119 Exam Section 3: Item 19 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 19. A 25-year-old woman with a 3-year history of celiac disease comes to the physician because of a markedly itchy rash on her scalp, arms, buttocks, and legs for 5 weeks. Examination of the skin shows many crusted papules surrounded by excoriations over the upper and lower extremities and a few vesicles on the buttocks. A skin biopsy specimen shows the vesicles to be subepidermal. Direct immunofluorescence of the vesicles shows granular deposits of IgA in the epidermal-dermal junction. Which of the following is the most likely diagnosis? A) Bullous pemphigoid B) Dermatitis herpetiformis C) Dyshidrotic eczema D) Epidermolysis bullosa acquisita E) Pemphigus

B. Dermatitis herpetiformis is a vesiculobullous disorder characterized by small, tense, vesicles located on extensor sites (elbows, knees, and buttocks), which are extremely pruritic. Often, no vesicles will be present as they will have been intensely scratched and replaced by erosions. Dermatitis herpetiformis is caused by the deposition of IgA along the epidermal basement membrane. On histopathologic examination, small collections of neutrophils focally separate the epidermis from the dermis at the dermal papillae, which correspond to a vesicle clinically. Direct immunofluorescence (DIF) demonstrates a granular pattern of IgA deposition along the basement membrane at the epidermal-dermal junction of the dermal papillae. Dermatitis herpetiformis has a strong association with celiac disease, or gluten-sensitive enteropathy, and the primary treatment is the avoidance of gluten. In severe presentations, dapsone can be used to provide near immediate relief. Incorrect Answers: A, C, D, and E. Bullous pemphigoid (Choice A) is caused by antibodies against the hemidesmosome and is characterized by tense bullae. It most often presents in elderly men with severe pruritus. The mucosa is usually spared. Early in the disease course, bullae may be absent and only urticarial plaques and severe itching will be present. On histopathology, the bullae will demonstrate eosinophils. Direct immunofluorescence demonstrates a linear pattern of IgG deposition along the basement membrane. Dyshidrotic eczema (Choice C) is characterized by small, firm vesicles on the palms and lateral fingers. It is commonly seen as a form of allergic contact dermatitis but can also be induced by dermatophyte infection. Epidermolysis bullosa acquisita (Choice D) is characterized by antibodies against the anchoring fibrils, which are another component of the hemidesmosome complex and necessary for anchoring the epidermis to the basement membrane and dermis. It is most commonly seen in elderly patients with tense bullae and can be a marker of underlying malignancy. Pemphigus (Choice E) is caused by antibodies against desmosomes, the protein complex that maintains cell to cell adhesion in the epidermis. Because the target of pemphigus vulgaris is more superficial, in the epidermis, the resultant blisters will be fragile and flaccid. Thus, the clinical manifestations of pemphigus include flaccid bullae and erosions. It often also involves mucosal surfaces. Educational Objective: Clinically, dermatitis herpetiformis is characterized by severely pruritic vesicles or erosions on the extensor surfaces of extremities. It is strongly associated with celiac disease. DIF demonstrates a coarse, granular pattern of IgA deposition at the epidermal-dermal junction of the dermal papillae, which confirms the diagnosis. %3D Previous Next Score Report Lab Values Calculator Help Pause

169 Exam Section 4: Item 19 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 19. Strips of tracheal muscle contract after administration of carbachol then relax after administration of epinephrine. The graph shows muscle tension as a function of the log dose of epinephrine under control conditions (solid curve) and in the presence of equal concentrations of antagonists X (dashed curve) and Y (dotted curve). X and Y are most likely to be which of the following drugs? Drug Y yohimbine Drug X A) Clonidine B) Metoprolol propranolol C) Phentolamine prazosin O D) Prazosin phentolamine Log (epinephrine] E) Propranolol metoprolol F) Yohimbine clonidine

B. Drugs X and Y are most likely metoprolol and propranolol. Metoprolol is a selective B1-adrenergic blocking agent while propranolol is a nonselective B-adrenergic blocking agent with activity at B, and B2 receptors. Tracheal and bronchial smooth muscle tone, and thus airway diameter, is primarily regulated by parasympathetic innervation and signaling through muscarinic receptors. However, the airway smooth muscle does express B2 receptors, and activation results in bronchodilation through G, protein-coupled receptor signal transduction. This mechanism is the rationale for the use of B-adrenergic agonists in the treatment of asthma exacerbations. Drug X on the graph is likely metoprolol, which is selective for B-receptors with minimal B2 activity, resulting in a slight increase in the concentration of epinephrine needed to outcompete for the binding site. Propranolol is active at both B, and B, receptors, resulting in a greatly increased concentration of epinephrine needed to stimulate smooth muscle contraction as represented by curve Y on the graph. Incorrect Answers: A, C, D, E, and F. Choices A and F are incorrect, as clonidine is an a-adrenergic agonist and yohimbine is an azadrenergic blocking agent. a-adrenergic receptors are primarily located in the central nervous system, and activation results in decreased sympathetic outflow. They would not be expected to affect the isolated tracheal smooth muscle response to epinephrine. Choices C and D are incorrect, as phentolamine is a nonselective a-adrenergic blocking agent and prazosin is a selective a-adrenergic blocking agent. Activation of a-adrenergic receptors on smooth muscle cells results in increased muscle tension through G. protein-coupled receptor signal transduction. These receptors are primarily located on smooth muscle cells of blood vessels and the genitourinary tract, not the trachea. Choice E is incorrect as metoprolol has decreased inhibitory activity at B, receptors compared to propranolol and thus has a decreased effect on tracheal smooth muscle tension. Educational Objective: Stimulation of Byadrenergic receptors results in airway smooth muscle relaxation and bronchodilation. Use of nonselective B-adrenergic blocking agents may worsen bronchoconstriction in the setting of asthma or chronic obstructive pulmonary disease. %3D Previous Next Score Report Lab Values Calculator Help Pause Muscle tension (g)

153 Exam Section 4: Item 3 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 3. A 35-year-old woman, gravida 1, para 1, comes to the physician for a follow-up examination 1 month after delivering a healthy male newborn. During her pregnancy, she was diagnosed with gestational diabetes. Her mother and sister also developed this condition during pregnancy. Today, laboratory studies show a serum glucose concentration of 100 mg/dL. This patient most likely has decreased activity of which of the following enzymes? A) D13 kinase B) Glucokinase C) Mevalonate kinase D) Phosphofructokinase E) Pyruvate kinase

B. Glucokinase catalyzes the phosphorylation of glucose to glucose-6-phosphate during glycolysis. While glucose can freely diffuse into and out of cells, glucose-6-phosphate is sequestered in the cell due to its charged phosphate group. Glucokinase is present in hepatocytes and the pancreatic B cells and is stimulated by high blood glucose concentration, causing uptake when circulating concentrations of glucose are high. Its action is also stimulated by insulin. In the B cells, the generation of adenosine triphosphate (ATP) through glycolysis leads to closure of ATP-dependent potassium channels, which leads to an increase in membrane potential and an influx of calcium. The calcium causes cellular secretory granules containing insulin to fuse with the cell membrane. Deficiencies of the enzyme lead to maturity-onset diabetes of the young (MODY), with an increased threshold for insulin release due to low cytoplasmic glucose. It is inherited in an autosomal dominant pattern and presents with mild hyperglycemia that can often be managed with diet alone. Incorrect Answers: A, C, D, and E. D13 kinase (Choice A), as part of the kinase family of proteins, acts to hydrolyze nucleotide triphosphate molecules in the phosphorylation of proteins in the cell. D13 kinase may relate to K-Ras D13, a downstream protein in the cell growth and differentiation pathway triggered by insulin signaling. The deficiency in gestational diabetes arises from intracellular glucose levels that are less than normal due to failure to phosphorylate glucose, not other proteins in the signal transduction cascade. Mevalonate kinase (Choice C) is an enzyme present in the cholesterol biosynthesis pathway. Deficiencies in this enzyme lead to mevalonic aciduria, an inborn error of metabolism marked by developmental delay, failure to thrive, hepatomegaly, and febrile crises. Hyperimmunoglobulin D syndrome also results from deficiency, which presents with increased levels of immunoglobulin D, recurrent fever, abdominal pain, and lymphadenopathy. It does not play a role in the development of diabetes mellitus. Phosphofructokinase (Choice D) converts fructose-6-phosphate to fructose-1,6-bisphosphate or fructose-2,6-bisphosphate. Deficiencies lead to glycogen storage disease VIII which presents with fatigue, myalgias, and rhabdomyolysis with exercise. It does not play a role in the development of diabetes mellitus. Pyruvate kinase (Choice E) converts phosphoenolpyruvate to pyruvate. Its deficiency presents with extravascular hemolysis. It typically presents in infancy, rather than in adulthood. Educational Objective: Glucokinase acts as a glucose sensor in hepatocytes and ß cells of the pancreas. Deficiencies of the enzyme lead to MODY, which presents with an autosomal dominant pattern of mild hyperglycemia. %3D Previous Next Score Report Lab Values Calculator Help Pause

163 Exam Section 4: Item 13 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 13. As part of an annual physical examination, a healthy 50-year-old man has laboratory studies done 8 hours after an overnight fast. His serum glucose concentration is 75 mg/dL. Activation of which of the following receptors in liver cells most likely maintained the serum glucose concentration in this patient? A) Cytokine B) G protein-coupled C) Nuclear D) Serine kinase E) Tyrosine kinase

B. The fasting state is dependent on hepatic and muscle glycogenolysis, hepatic gluconeogenesis, and lipolysis, which are all mediated by glucagon and epinephrine. Glucagon is produced by pancreatic islet a-cells and primarily works to increase serum glucose through glycogenolysis and gluconeogenesis. Glucagon acts on hepatocytes via a G protein-coupled receptor, resulting in activation of a cyclic adenosine monophosphate pathway that ends in the activation of glycogen phosphorylase. In turn, glycogen phosphorylase degrades glycogen polymers with the release of glucose monomers into serum following dephosphorylation by glucose-6-phosphatase. Glycogen phosphorylase is upregulated in states of starvation, and results in the degradation of glycogen stores. This metabolic pathway results in the release of glucose monomers for the maintenance of serum blood sugar during periods of fasting, as illustrated by this patient's normal blood glucose concentration following an overnight fast. Incorrect Answers: A, C, D, and E. Cytokine (Choice A) receptors are involved in the immune response. Cytokines such as interferons and interleukins act on cytokine receptors and mediate intracellular responses through a JAK-STAT pathway. Nuclear (Choice C) receptors are found intracellularly. Due to their intracellular location, they are acted upon by lipid-soluble ligands, such as vitamins A and D, steroid hormones (eg, estrogen, progesterone), and thyroid hormone. Nuclear receptors directly bind DNA and regulate expression of genes. Glucagon and epinephrine are hydrophobic peptide molecules that act at the cell surface, never crossing the plasma membrane. Serine kinase (Choice D) is an enzyme class that functions in protein phosphorylation and regulation of cell proliferation, cell differentiation, and apoptosis. Kinases are ubiquitous in cell signaling pathways, however the initial receptor targeted by glucagon or epinephrine is a G protein-coupled receptor at the cell surface. Insulin acts on a tyrosine kinase (Choice E) receptor. Insulin also functions in glucose homeostasis, but generally has the opposite effect of glucagon. Insulin acts to increase peripheral tissue uptake of glucose and promotes hepatic glucose storage and glycogenesis, not glycogenolysis. Educational Objective: Glucagon acts on hepatocytes via a G protein-coupled receptor, resulting in activation of a cyclic adenosine monophosphate pathway that ends in the activation of glycogen phosphorylase. This pathway results in the release of glucose monomers into the serum to stabilize serum blood sugar during periods of fasting. %3D Previous Next Score Report Lab Values Calculator Help Pause

196 Exam Section 4: Item 46 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 46. A 48-year-old man begins furosemide therapy for pedal edema associated with biventricular failure and hypertension. Five days later, the edema is not fully resolved, and his serum potassium concentration has decreased from 4.2 mEq/L to 3 mEq/L. A drug with which of the following actions should be added to this patient's medication regimen? A) Blocks basolateral K* channels in the collecting duct B) Decreases the luminal permeability to Na* in the collecting duct C) Increases the delivery of Na* and K+ to the collecting duct D) Increases the negative charge of the luminal tubule fluid E) Stimulates Na-K+ATPase in the collecting duct

B. The patient should be started on a potassium-sparing diuretic given his persistent edema and hypokalemia in the setting of furosemide therapy. Potassium-sparing diuretics work by decreasing the luminal permeability to Na+in the collecting duct. Triamterene and amiloride achieve this by direct inhibition of the epithelial sodium channel (ENAC) on the luminal surface. This reduces the amount of Na that can pass through the membrane and results in natriuresis and diuresis. Spironolactone and eplerenone are mineralocorticoid receptor antagonists (MRAS) that block the action of aldosterone. Aldosterone normally acts on the cells of the collecting duct by activation of transcription factors that increase the expression of ENaC on the luminal surface and the NaK+ATPase on the basolateral surface. The NaK+ATPase maintains an electrochemical gradient that promotes Na diffusion into the cell. Inhibition by MRAS results in decreased Na+ and K* exchange in the collecting duct, with the overall effect of eliminating more Na*in the urine and retaining K+. The major adverse effect of potassium-sparing diuretics is hyperkalemia. MRAS may also result in anti-androgen side effects such as gynecomastia. Incorrect Answers: A, C, D, and E. Blocks basolateral K+ channels in the collecting duct (Choice A) is incorrect as passive K* channels are located on the luminal surface of collecting duct cells. Modulation of K+ diffusion is achieved by decreasing Na+ entry into the cell, which results in a decreased electrochemical gradient for K* movement into the lumen. Loop diuretics increase the delivery of Na+ and K+to the collecting duct (Choice C) by blocking the Na+K+2CI cotransporter in the thick ascending loop of Henle. Furosemide is a loop diuretic. They are associated with hypokalemia, hypocalcemia, and hypomagnesemia. Increases the negative charge of the luminal tubule fluid (Choice D) is incorrect, as the decreased movement of Na+ out of the lumen decreases the negative charge of the luminal tubule fluid, resulting in a reduced electrochemical gradient for the diffusion of K* into the urine. Aldosterone stimulates Na-K+ATPase in the collecting duct (Choice E) resulting in increased Na+ diffusion through ENAC channels. MRAS block this activity. Educational Objective: Potassium-sparing diuretics may be added to a diuretic regimen in the setting of persistent edema with hypokalemia. They act by reducing the exchange of K+ for Na+ in the collecting ducts. %3D Previous Next Score Report Lab Values Calculator Help Pause

198 Exam Section 4: Item 48 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 48. A 22-year-old man comes to the physician because of a 2-month history of foul-smelling, watery diarrhea with significant flatulence; he also has had a 4.5-kg (10-lb) weight loss during this period. He recently returned from a trip to rural Indonesia, where he did not always have access to clean water. He appears thin. Physical examination shows a soft, mildly distended abdomen with active bowel sounds. A photomicrograph of a stool specimen is shown. The most appropriate pharmacotherapy has which of the following mechanisms of action? A) Enhancement of cell membrane permeability to chloride ions B) Formation of destructive free radicals C) Inhibition of DNA polymerase D) Inhibition of protein synthesis E) Prevention of microtubule assembly 1ομm

B. The photomicrograph displays a flagellated, pear-shaped trophozoite with two large central nuclei and a median axostyle, consistent with the appearance of Giardia lamblia. G. lamblia is a flagellated, parasitic organism that causes giardiasis. Giardiasis is usually acquired through the ingestion of cysts in contaminated water, which then mature into trophozoites in the small bowel. Parasitic overgrowth of the small bowel leads to intestinal malabsorption. Giardiasis presents with persistent diarrhea that is foul-smelling and fatty, as well as abdominal pain, bloating, cramping, and weight loss. Diagnosis is made with stool culture, microscopy, or stool antigen testing. Metronidazole is the first line treatment and asserts its antimicrobial effect through the formation of destructive free radicals. Incorrect Answers: A, C, D, and E. Enhancement of cell membrane permeability to chloride ions (Choice A) is the mechanism of action of the anthelminthic agent ivermectin. Ivermectin alters cellular membrane potentials, leading to paralysis and death of the target organism. Ivermectin is not useful for treatment of giardiasis. Inhibition of DNA polymerase (Choice C) is the mechanism of action of several important antivirals, including acyclovir, cidofovir, and foscarnet. While several acute viral syndromes can present with diarrhea, including cytomegalovirus and HIV, the photomicrograph identifies G. lamblia as the causative organism of this patient's diarrhea. Inhibition of protein synthesis (Choice D) is the mechanism of action of several important classes of antibiotics, including macrolides, aminoglycosides, and tetracyclines. None of these drugs are first line for the treatment of giardiasis. Prevention of microtubule assembly (Choice E) is the mechanism of action of the anthelminthic agents albendazole and mebendazole. While albendazole is sometimes used for the treatment of giardiasis, metronidazole is first line. Educational Objective: Metronidazole is the first line treatment of giardiasis and enacts its antimicrobial effect through the formation of destructive free radicals. I3D Previous Next Score Report Lab Values Calculator Help Pause

151 Exam Section 4: Item 1 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 1. A male newborn is delivered vaginally at 39 weeks' gestation to a 24-year-old primigravid woman. The pregnancy was complicated by polyhydramnios. Soon after birth, the newborn has frothing at the mouth, cyanosis, and respiratory distress. An unsuccessful attempt is made to pass a nasogastric tube. An x-ray shows an air-distended stomach. Which of the following is the most likely cause of these findings in this patient? A) Abnormality of aortic arch resulting in vascular ring B) Deviation of the tracheoesophageal septum C) Failure of re-formation of duodenal lumen D) Hypertrophy of pyloric muscle E) Incomplete formation of the tracheal cartilage rings

B. The typical presentation for a newborn with esophageal atresia is an inability to feed, with associated gagging and an inability to manage oral secretions. Esophageal atresia often presents with concomitant tracheoesophageal fistula (TEF), which occurs as a result of deviation of the tracheoesophageal septum during embryonic development. The presence of a TEF conveys increased risk of aspiration pneumonia and respiratory distress. Initial evaluation requires placement of a nasogastric tube and evaluation of its position with a chest x-ray. If the nasogastric tube remains coiled within the esophagus, this suggests the presence of esophageal atresia. Air within the gastrointestinal tract immediately after birth and prior to feeding suggests the existence of a TEF distal to the level of esophageal atresia. Definitive management requires surgical repair. As well, additional screenings for associated congenital malformations should occur. As the fetus is unable to swallow amniotic fluid, a pregnancy complicated by polyhydramnios suggests the diagnosis prior to birth. Incorrect Answers: A, C, D, and E. Abnormality of aortic arch resulting in vascular ring (Choice A) presents with stridor, wheezing, cough, dysphagia, and vomiting due to compression of the trachea and esophagus. Vascular rings can be distinguished from TEF by improvement of symptoms upon neck repositioning (eg, extension) and by absence of an air-distended stomach, which is more suggestive of TEF. Failure of re-formation of duodenal lumen (Choice C) is observed in duodenal atresia. Duodenal atresia presents with vomiting and abdominal distension, as well as distension of the stomach and duodenum on abdominal radiograph. Duodenal atresia is more common in children with Down syndrome. Hypertrophy of pyloric muscle (Choice D) is observed in pyloric stenosis. Pyloric stenosis presents in infancy with nonbilious vomiting around 4 to 8 weeks of life with a palpable abdominal mass. Incomplete formation of the tracheal cartilage rings (Choice E) may lead to compression of the trachea with associated respiratory distress, wheezing, and positional stridor, often worse when supine. The esophagus would not be affected and there would not be difficulty passing a nasogastric tube. Educational Objective: Esophageal atresia with tracheoesophageal fistula presents with an inability to feed, gagging, coughing, respiratory distress, and cyanosis in the neonatal period. Inability to pass a nasogastric tube with the presence of gastric air is a characteristic finding. Definitive management requires surgical repair. %3D Previous Next Score Report Lab Values Calculator Help Pause

180 Exam Section 4: Item 30 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 30. A premenopausal 49-year-old woman asks her physician about her risk for osteoporosis. Her mother had osteoporosis and disabling bone fractures after menopause. The patient works as a gardener for a landscape service and plays tennis 3 times a week. She does not smoke, drinks 1 glass of wine a week, and takes no medications. Her weight is 55 kg (121 Ib). Which of the following additional information is needed to evaluate her risk for osteoporosis? A) Age at first sexual intercourse B) Dietary history C) Exposure to ionizing radiation D) Exposure to organophosphate E) Family history of rheumatic diseases F) Socioeconomic history

B. There are numerous factors that contribute to the development of osteoporosis, which is defined as a decrease in bone mineral density 2.5 standard deviations below the normal mean. It typically occurs in the fifth through seventh decades of life and is approximately four times more common in females than in males. It is also more common in individuals of Asian or northern European descent. In addition to gender and genetic factors, there are many environmental factors that also contribute. Long-term therapy with corticosteroids, anticonvulsants, and proton pump inhibitors have been shown to increase the risk of osteoporosis. Sedentary lifestyle or immobilization has also been implicated in the development of osteoporosis. Additional environmental factors include malnutrition or malabsorption syndromes (specifically those that lead to low calcium and vitamin D levels), low body weight, chronic alcohol use, cigarette smoking, and decreased estrogen or testosterone production. The patient's dietary history should be investigated to understand any risk factors she may have for malnutrition or malabsorption, which would place her at a greater risk for deficiencies that predispose to osteoporosis. Incorrect Answers: A, C, D, E, and F. Age of first sexual intercourse (Choice A) does not contribute to osteoporosis. An overall longer duration of exposure to estrogen and testosterone can contribute to increased bone mineral density and decrease the risk of osteoporosis as these sex hormones lead to increased bone production. Exposure to ionizing radiation (Choice C) does not increase the risk of osteoporosis. Exposure to significant doses of ionizing radiation does increase the risk of cancer, particularly in the thyroid, bone marrow, and colon. Exposure to organophosphate (Choice D) leads to organophosphate poisoning, which is caused by increased levels of acetylcholine due to the inhibition of acetylcholinesterase. Symptoms of organophosphate poisoning include diarrhea, vomiting, urination, sweating, salivation, bronchospasm, bradycardia, fasciculations, lethargy, and possible coma. Family history of rheumatic diseases (Choice E) is a risk factor for autoimmune diseases, but not necessarily osteoporosis. The increased inflammatory state associated with rheumatic diseases may lead to decreased bone mineral density related to interleukin-1 activation of osteoclasts, however a family history of rheumatic disease does not directly increase the risk of osteoporosis. Socioeconomic history (Choice F) may indirectly lead to increased risk of osteoporosis. For example, persons of low socioeconomic status may have decreased access to appropriate nutrition that may lead to decreased levels of calcium and vitamin D. However, this is not a direct cause of osteoporosis. Educational Objective: Osteoporosis is a condition of significantly decreased bone mineral density, which increases the risk for fractures, commonly of the distal radius, hip, and vertebrae. There are numerous risk factors for this disease. Decreased intake or absorption of vitamin D and calcium significantly increases the risk for low bone mineral density. %3D Previous Next Score Report Lab Values Calculator Help Pause

146 Exam Section 3: Item 46 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 46. Humoral antibody produced against many viruses following natural infection mediates protection against reinfection with the same strain primarily by which of the following mechanisms? A) Activation of macrophages B) Blocking of viral attachment to cell receptors C) Inhibition of viral nucleic acid replication D) Killing of intracellular virus E) Prevention of release of virions from infected cells

B. Viral infection causes humoral antibody production through the process of T lymphocyte and B lymphocyte activation. This is part of the adaptive immune system response. When a dendritic cell encounters a viral antigen, or a cell is infected with a virus, antigens are presented on major histocompatibility complexes (MHC). These MHC-antigen complexes then bind to T-lymphocyte receptors, activating those T lymphocytes and causing them to proliferate, survive, and secrete cytokines. The active T lymphocytes then bind to B lymphocytes via the CD40 ligand and receptor pairing, which leads to B lymphocyte activation and antibody class switching, ultimately leading to humoral antibody production. The antibodies produced target a variety of viral particles, some of which include the proteins on the viral surface required to bind to and infect host cells. For example, influenza contains the protein hemagglutinin, which binds the sialic acid receptor on the outside of host cells and allows for viral entry. When antibodies are present to hemagglutinin, they bind to the protein and create a barricade between the ligand and receptor, preventing viral attachment to cell receptors and subsequent cell entry. Incorrect Answers: A, C, D, and E. Activation of macrophages (Choice A) is a component of the innate immune system, not the adaptive immune system. Activated macrophages may become histiocytes and may form granulomas in the case of some infections (eg, Mycobacterium tuberculosis). Inhibition of viral nucleic acid replication (Choice C) is the mechanism utilized by nucleoside reverse transcriptase inhibitors (NRTIS) and non-nucleoside reverse transcriptase inhibitors (NNRTIS) in the treatment of HIV. This is not the effect of circulating humoral antibodies. Killing of intracellular virus (Choice D) is a feature of the cytotoxic CD8+ T lymphocytes response to viral infection. When a cell has been infected with virus, it presents intracellular antigen on the cell surface through class I MHC. This is recognized by CD8+ cytotoxic T lymphocytes, which then cause cell death by apoptosis and subsequent death to the intracellular virus. Neuraminidase inhibitors, a type of medication utilized in treating influenza, prevent the release of virions from infected cells (Choice E). This is not an effect of antiviral antibodies. Educational Objective: Humoral antibody production by B lymphocytes occurs after viral infection leads to T-lymphocyte activation followed by associated B-lymphocyte activation. Once formed, antibodies surround viral particles, effectively blocking them from binding to cell receptors. Ultimately, this prevents the virus from entering the cell, the first step in viral infection of a cell. %3D Previous Next Score Report Lab Values Calculator Help Pause

123 Exam Section 3: Item 23 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 23. A 67-year-old woman with Crohn disease is admitted to the hospital for bowel rest and parenteral hyperalimentation. A central venous catheter is placed. Seven days after admission, she develops a temperature of 39.2°C (102.6°F) and shaking chills. Broad-spectrum antibiotics are administered pending blood cultures. She continues to have daily temperature spikes. Two days later, blood culture grows yeast forms. Which of the following is the most likely causal organism? A) Aspergillus fumigatus B) Blastomyces dermatitidis C) Candida albicans D) Cryptococcus neoformans E) Mucor ramosissimus

C. Candida albicans is the most likely pathogen responsible for this patient's fever and rigors, especially given the presence of yeast forms on blood culture, which is diagnostic of fungemia. C. albicans is a ubiquitous dimorphic fungus that causes a wide spectrum of clinical diseases in immunocompetent hosts including infections of moist skin folds (intertrigo), vaginal infections, and oral thrush. It also causes severe disease in patients who are immunocompromised such as those with HIVIAIDS infection, neutropenic patients on chemotherapy, patients with poorly controlled diabetes mellitus, and those with autoimmune disorders such as Crohn disease, such as this patient. Patients receiving parenteral nutrition are at substantially higher risk for invasive fungal disease from C. albicans, and any fungi isolated in blood cultures should be treated as evidence of disseminated disease. Initial treatment may include an echinocandin with broad activity against Candida species with narrowing of antifungals following speciation and sensitivity studies. Ophthalmologic examination in patients with candidemia is essential to rule out ocular involvement. Incorrect Answers: A, B, D, and E. Aspergillus fumigatus (Choice A) is a mold that causes aspergillosis, which can present with aspergilloma, allergic bronchopulmonary aspergillosis, chronic necrotizing aspergillosis, and airway invasive or angioinvasive aspergillosis. Aspergillomas (fungal masses) are the most common and arise within existing cavitary lesions of the lung such as those caused by previous tuberculosis, emphysematous bullae, or cavitary squamous cell lung carcinoma. Parenteral nutrition is not a risk factor for disease. Blastomyces dermatitidis (Choice B) is a yeast endemic to the Mississippi and Ohio river valleys. It can cause acute and chronic pneumonia in addition to skin lesions with verrucous borders, osteomyelitis, prostatitis, and meningoencephalitis in immunocompromised hosts. This patient has no known risk factors and fungemia with B. dermatitidis would be rare. Cryptococcus neoformans (Choice D) is an encapsulated yeast and opportunistic pathogen in immunocompromised patients, especially patients with HIVIAIDS infection and CD4+ counts less than 200 cells/mm3. Disseminated disease can cause cryptococcal meningitis and encephalitis. Histology reveals fungal cells with narrow-based budding and a bright red inner capsule on staining with mucicarmine. Parenteral nutrition is not a known risk factor for disease. Mucor ramosissimus (Choice E) is a ubiquitous fungus found commonly on decaying vegetation. Humans have frequent daily exposure to Mucor species without the development of disease, but in patients with immunocompromising conditions such as diabetes mellitus or stem cell transplant, Mucor species can cause devastating sinus, orbital, and central nervous system disease. Mortality is high. Parenteral nutrition is not known to predispose to infection. Educational Objective: Candida species can cause fungemia, and parenteral nutrition is a known risk factor. Yeast forms isolated on blood cultures should never be treated as a contaminant and treatment with a broad-spectrum antifungal should be started immediately. %3D Previous Next Score Report Lab Values Calculator Help Pause

190 Exam Section 4: Item 40 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 40. A 12-year-old girl is brought to the physician because of a rash on her left buttock for the past 2 days. The rash developed after the family returned from a 2-week-long early summer vacation in Maine. Her vital signs are within normal limits. A photograph of the lesion is shown. The likely cause of this patient's infection is taxonomically and morphologically most similar to the infectious agent of which of the following conditions? A) Bacillary angiomatosis B) Chancroid C) Leptospirosis D) Lymphogranuloma venereum E) Q fever

C. Erythema chronicum migrans, which can be identified by its classic targetoid patches, is often the first sign of Lyme disease. Lyme disease is caused by the spirochete Borrelia burgdorferi, whose vector is the Ixodes tick. The tick and organism are common in the northeastern United States but can be seen in other parts of the country as well. Early localized Lyme disease presents with erythema chronicum migrans and flu-like symptoms. Early disseminated infection presents with carditis, heart block, facial nerve palsy, and transient arthritis. Late disseminated Lyme presents with encephalopathy and arthritis. Treatment of this infection is with doxycycline. Ceftriaxone is an alternative in pregnant patients or children less than 8 years old and is generally the first line in central nervous system disease. Other spirochetes include Leptospira species and Treponema species. Infection with L. interrogans, found in water contaminated with animal urine, causes leptospirosis. Leptospirosis is characterized by flu-like symptoms, myalgias, jaundice, and photophobia with conjunctival injection. T. pallidum is the organism responsible for causing syphilis. Incorrect Answers: A, B, D, and E. Bartonella henselae is the cause of bacillary angiomatosis (Choice A). This infection is most commonly seen in immunocompromised individuals and presents with red, vascular papules. It may be mistaken for Kaposi sarcoma clinically but demonstrates neutrophils on histologic examination. Chancroid (Choice B) presents with an exudative, painful genital ulcer with inguinal lymphadenopathy. It is caused by infection with the organism Haemophilus ducreyi. T. pallidum, the spirochete that causes syphilis, results in a painless chancre instead. Lymphogranuloma venereum (Choice D) is caused by infection with Chlamydia trachomatis subtypes L1, L2, and L3. It presents as small painless genital ulcers with tender inguinal lymphadenopathy. C. trachomatis is not a spirochete. Q fever (Choice E) is another vector-borne illness caused by infection with Coxiella burnetti, which is not a spirochete. C. burnetti is acquired through the inhalation of spores from cattle or sheep amniotic fluid. Q fever commonly presents as pneumonia or endocarditis. Educational Objective: Borrelia, Leptospira, and Treponema are spirochete species that cause a diverse group of infections including Lyme disease, leptospirosis, and syphilis, respectively. Previous Next Score Report Lab Values Calculator Help Pause

113 Exam Section 3: Item 13 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 13. A newborn born at 36 weeks' gestation dies 3 days later. The photograph shown is of a section of brain as seen at autopsy. Which of the following is the most likely underlying disease? A) Biliary atresia B) Cyanotic congenital heart disease C) Hemolytic disease of the newborn D) Neonatal meningitis E) Respiratory distress syndrome 1 cm

C. Kernicterus occurs due to the deposition of unconjugated bilirubin in a newborn's brain, usually in the basal ganglia, pons, and cerebellum. It typically presents with poor feeding, lethargy, and seizures. Unconjugated or indirect bilirubin is circulating insoluble bilirubin that is largely bound to albumin. It has not yet undergone conjugation in the liver. Causes of increased unconjugated bilirubin include hemolysis, breast milk jaundice, Crigler-Najjar syndrome, Gilbert syndrome, and physiologic jaundice. In newborns, etiologies also include exposure of the mother to sulfonamide antibiotics, as they displace bilirubin from albumin, increasing the free plasma concentration of unconjugated bilirubin. Kernicterus typically occurs when the bilirubin level exceeds 25 to 30 mg/dL, which is commonly due to glucose-6-phosphate dehydrogenase deficiency or Rh hemolytic disease of the newborn. Risk factors include prematurity, cephalohematoma or significant bruising, poor breastfeeding, and East Asian ethnicity. Phototherapy or exchange transfusion can prevent severe unconjugated hyperbilirubinemia and subsequent kernicterus. However, mortality is high once kernicterus develops. Incorrect Answers: A, B, D, and E. Biliary atresia (Choice A) presents with worsening jaundice due to conjugated hyperbilirubinemia, pale stool, dark urine, poor weight gain, and, in untreated cases, cirrhosis, coagulopathy, and portal hypertension. It will not cause kernicterus, as conjugated bilirubin is water soluble and unable to deposit in the brain, and it is unlikely to cause death in the first 3 days of life. Cyanotic congenital heart diseases (Choice B) include truncus arteriosus, transposition of the great vessels, tricuspid atresia, tetralogy of Fallot, and total anomalous pulmonary venous return. They typically present with a cyanotic newborn and signs of global hypoxia on brain pathology but do not present with bilirubin deposits in the basal ganglia. Neonatal meningitis (Choice D) presents with fever or hypothermia, lethargy, irritability, a bulging fontanelle, hypotonia, and poor feeding. Leukocytosis, as well as evidence of bacterial infection on lumbar puncture, are typically present. Gross pathology may reveal purulent exudate in the leptomeninges. Respiratory distress syndrome (Choice E) presents with tachypnea, hypoxia, accessory muscle use, and cyanosis, often in premature infants. It is due to a surfactant deficiency that leads to alveolar collapse, and typically demonstrates ground glass opacities on chest x-ray. It does not typically cause pathologic changes within the brain unless it leads to severe hypoxemia. Educational Objective: Kernicterus is caused by high levels of unconjugated hyperbilirubinemia, commonly in the setting of hemolysis due to Rh alloimmunization or glucose-6-phosphate dehydrogenase deficiency. It is associated with poor feeding, lethargy, and seizures due to deposition of bilirubin in the basal ganglia, pons, and cerebellum. Previous Next Score Report Lab Values Calculator Help Pause

172 Exam Section 4: Item 22 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment * 22. A 60-year-old woman comes to the physician because of a 1-month history of bleeding from a lesion on her nose. She has no history of major medical illness and takes no medications. Physical examination shows a 1-cm lesion on the right naris. Microscopic examination of a biopsy specimen of the mass shows neoplastic cells that exhibit dense pigment granules. Which of the following is the most likely diagnosis? A) Actinic keratosis B) Basal cell carcinoma C) Melanoma D) Rhabdomyosarcoma E) Squamous cell carcinoma

C. 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. It is a neoplasm of melanocytes, which contain pigment in melanosomes. Histologically, it may appear as dendritic cells containing pigment granules with high nuclear to cytoplasmic ratios, mitotic figures, and increased size compared to normal melanocytes. 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. 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, D, and E. Actinic keratosis (Choice A) is a premalignant lesion that may progress to squamous cell carcinoma. They present as chronic rough, scaly patches of skin in areas of prolonged sun exposure (eg, face, ears, hands). Basal cell carcinoma (Choice B) is the most common form of skin cancer and is derived from the basal cells of the epidermis due to mutations in the sonic hedgehog pathway or p53 tumor suppressor. The primary risk factor for developing basal cell carcinoma is ultraviolet light exposure, though age, fair skin, and family history also contribute. Basal cell carcinomas typically present as pink, pearly papules or nodules with rolled borders and central ulceration most commonly on the sun exposed areas of the head and neck. Rhabdomyosarcoma (Choice D) is a neoplasm of mesenchymal tissue including smooth muscle. Sarcoma botryoides is a variant of embryonal rhabdomyosarcoma and presents in young girls with a grape-like, clear, polypoid mass protruding from the vagina. If rhabdomyosarcoma occurs within the skin, it appears as a subcutaneous nodule rather than a pigmented papule. Squamous cell carcinoma (Choice E) of the skin typically presents with a non-healing ulcerative lesion with a scale. It commonly occurs secondary to sun exposure, chronic draining sinus tracts, or immunosuppression. Educational Objective: Malignant melanoma is a neoplasm of melanocytes, which microscopically may appear as dendritic cells containing pigment granules with high nuclear to cytoplasmic ratios, mitotic figures, and increased size compared to normal melanocytes. %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

168 Exam Section 4: Item 18 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 18. A 28-year-old woman comes to the physician for advice on how to lose weight. She tells the physician that she binges on high-carbohydrate foods two to three times weekly, usually forcing herself to vomit after a binge. She is 168 cm (5 ft 6 in) tall and weighs 63 kg (140 lb); BMI is 23 kg/m2. Which of the following additional physical findings is most likely in this patient? O A) Bradycardia B) Decreased vibratory sensation at the ankles C) Parotid gland enlargement D) Sluggish deep tendon reflexes E) Sparse axillary and pubic hair

C. Patients with eating disorders such as anorexia nervosa and bulimia nervosa who purge by vomiting may demonstrate parotid gland enlargement on physical examination. This patient likely has bulimia nervosa (purging type). Bulimia nervosa (purging type) involves cycles of uncontrollable eating and compensatory behaviors such as vomiting, laxative, or diuretic overuse that occur at least once a week over three months or more. Únlike patients with anorexia nervosa, patients with bulimia nervosa typically have a normal BMI. Patients with the binging/purging type of anorexia or bulimia nervosa can demonstrate parotid gland enlargement, dental caries from gastric hydrochloric acid erosion of enamel, and scars on the knuckles secondary to abrasions from the incisors when inducing vomiting. Parotid gland enlargement typically manifests as painless bilateral enlargement. The loss of gastric hydrochloric acid leads to hypochloremia and metabolic alkalosis. In severe cases, signs of hypovolemia such as tachycardia and hypotension may be present. Treatment of bulimia nervosa is through a combined medical and psychiatric approach and involves correcting fluid and electrolyte derangements alongside behavioral and pharmacologic therapy. Incorrect Answers: A, B, D, and E. Bradycardia (Choice A) would be atypical for bulimia nervosa. Tachycardia may occur in severe cases of the purging type of anorexia or bulimia nervosa due to hypovolemia or torsades de pointes from metabolic alkalosis and hypokalemia. Severe anorexia nervosa can be associated with bradycardia as a physiological adaptation to decreased metabolism and increased vagal tone. Decreased vibratory sensation at the ankles (Choice B) and sluggish deep tendon reflexes (Choice D) may result from severe malnutrition in anorexia nervosa, as vitamin B12 deficiency can lead to subacute combined degeneration. Bulimia nervosa more commonly presents with physical exam signs of vomiting such as parotid gland enlargement than sequelae of severe malnutrition. Sparse axillary and pubic hair (Choice E) may be demonstrated in severely malnourished patients with anorexia nervosa due to decreased adrenocorticotropic hormone production. Bulimia nervosa more commonly presents with physical exam signs of vomiting such as parotid gland swelling than sequelae of severe malnutrition. Educational Objective: Bulimia nervosa (purging type) involves cycles of uncontrollable eating and compensatory behaviors such as vomiting, laxative, or diuretic overuse. Patients who purge by vomiting commonly demonstrate parotid gland swelling. %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

122 Exam Section 3: Item 22 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 22. A sexually active 20-year-old woman has had painful, erythematous, vesicular lesions on the vulva for 2 days. Examination by speculum also shows lesions in the vaginal vault and on the cervix. A photomicrograph of a scraping of the base of an unroofed vesicle is shown. Which of the following is the most likely diagnosis? A) Candidal vaginitis B) Condylomata acuminata C) Genital herpes D) Syphilis E) Trichomonas vaginalis vaginitis

C. This patient's vesicular lesions on the vulva are likely caused by herpes simplex virus (HSV) in the form of genital herpes, which also commonly causes fever, chills, and inguinal lymphadenopathy. Infection typically begins with the formation of vesicles, which lyse and progress to shallow, painful ulcers with an erythematous border. Genital herpes often causes vulvar pain with urination due to irritation of the vesicles and later, ulcers. The diagnosis may be confirmed by PCR testing but performing a Tzanck smear is another method of diagnosis. On Tzanck smear, herpes-infected multinucleated giant cells may be seen, as shown in this photomicrograph. The nuclei are close together and demonstrate nuclear molding. Genital herpes is most often caused by HSV-2 but can also be caused by HSV-1. HSV-1 and HSV-2 are both members of the herpesvirus family of double-stranded DNA enveloped viruses. Genital herpes is transmitted through sexual contact and perinatally. HSV may remain latent in the sacral ganglia until reactivation, when it causes painful vesicles and punched out erosions on the genitalia with associated inguinal lymphadenopathy. Systemic manifestations are possible, including viral meningitis and encephalitis. Treatment for herpetic infections involves drugs that inhibit viral DNA polymerase, classically by guanosine analogs such as acyclovir, valacyclovir, and famciclovir. Incorrect Answers: A, B, D, and E. Candidal vaginitis (Choice A) is caused by Candida albicans, a fungus that forms part of the normal vaginal flora. Overgrowth of Candida species may produce vulvovaginitis. Candidal vulvovaginitis results in pruritus and a thick, white vaginal discharge. Vaginal pH is normal, and hyphae may be visualized on wet mount. Vesicles and ulcers are not seen, and it is not painful. Condylomata acuminata (Choice B), or genital warts, are soft, pink, papillary lesions on the external genitalia associated with human papillomavirus that can appear on the penis, vagina, vulva, and/or anal canal. They are not painful. Diagnosis is made with examination and/or biopsy, which will show papillomatous lesions with virally infected cells called koilocytes in the epidermis, but not multinucleated giant cells. Syphilis (Choice D), caused by an infection with the spirochete Treponema pallidum, demonstrates multiple stages with varying symptoms, including primary with a painless chancre, secondary with fever, lymphadenopathy, and condylomata lata, and tertiary with tabes dorsalis, aortitis, and gummas. The chancre of primary syphilis typically presents as a painless ulcerative genital lesion, unlike the painful lesions seen in this patient. Trichomonas vaginalis vaginitis (Choice E) is a sexually transmitted infection that causes vaginitis and cervicitis in women and urethritis in men. Trichomoniasis can lead to overactive bladder (frequency, urgency, urge incontinence) and dysuria. Trichomoniasis presents with green, malodorous vaginal discharge and motile trichomonads on wet mount, not painful genital vesicles or ulcers. Educational Objective: Genital herpes infections present with painful vesicles and erosions. HSV-2 followed by HSV-1 are the most common viral causes and the diagnosis can be confirmed with a Tzanck smear, which demonstrates multinucleated giant cells with nuclear molding. II 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

185 Exam Section 4: Item 35 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 35. A matched case-control study is conducted to assess the association between the frequent use of tanning beds during teenage years and the development of melanoma. A group of women with melanoma are matched with a group of cancer-free women of the same age and ethnicity. An odds ratio (OR) of 1.4 is obtained. Which of the following 95% confidence intervals for the OR is the most precise statistically significant interval? A) -0.9 to 0.9 B) 0.9 to 1.7 C) 0.98 to 1.01 D) 1.01 to 1.73 E) 1.2 to 1.7

E. A confidence interval proposes a range of values within which a true value is likely to fall. Formally, a 95% confidence interval is the range of values within which the calculated statistic would fall between in 95% of cases if the sample and calculation were repeated, reflecting a reasonable margin of error (5%). In a normal distribution, the 95% confidence interval extends from two standard deviations below the mean to two standard deviations above the mean. Thus, a narrower confidence interval indicates a smaller standard deviation, and increased precision of the data. In addition, when evaluating odds ratios or relative risk, a value of 1.0 indicates that the two groups are similar with regard to the outcome of interest. Thus, a confidence interval that does not include 1.0 (above or below), would indicate a statistically significant difference between the two groups. Of the provided confidence intervals, 1.2 to 1.7 is the narrowest, or most precise, interval that does not include 1.0, indicating statistical significance. Incorrect Answers: A, B, C, and D. -0.9 to 0.9 (Choice A) is not a valid; the confidence interval for an odds ratio cannot be a negative number. 0.9 to 1.7 (Choice B) is a confidence interval that includes 1.0. This would indicate the difference between the two groups is not statistically significant. 0.98 to 1.01 (Choice C) describes a narrow confidence interval, however, the confidence interval includes 1.0 and therefore indicates the difference between the groups is not statistically significant. 1.01 to 1.73 (Choice D) describes a relatively wide confidence interval, indicating decreased precision. It does indicate a statistically significant difference between the groups, as it does not contain 1.0. Educational Objective: Confidence intervals are a measure of the precision of a statistical estimate. Smaller confidence intervals indicate greater precision. When interpreting an odds ratio or a relative risk, a confidence interval that does not include 1.0 indicates a statistically significant difference between the groups. %3D Previous Next Score Report Lab Values Calculator Help Pause

181 Exam Section 4: Item 31 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 31. A5-year-old boy is brought to the physician by his parents because of lethargy for 1 week. He has a history of occasional colds and ear infections. Physical examination shows mild pitting edema of the lower extremities. Auscultation of the chest and palpation of the abdomen show no abnormalities. Laboratory studies show: 15 g/dL 45% 7300/mm3 380 mg/dL Hemoglobin Hematocrit Leukocyte count Serum cholesterol Urine Blood Protein absent 3+ If a kidney biopsy specimen were obtained, light microscopic examination would most likely show which of the following? A) Crescents within the glomerular space B) Focal segmental glomerulosclerosis C) Neutrophilic infiltration of the glomerular tuft D) Thickened glomerular basement membranes E) No pathologic abnormality

E. Nephrotic syndrome is characterized by edema secondary to excessive proteinuria resulting in hypoalbuminemia and diminished intravascular oncotic pressure. This occurs because of loss of normal size and charge filtration of the glomerular capillary and Bowman capsule interface. Etiologies of nephrotic syndrome include minimal change disease, membranous nephropathy, and focal segmental glomerulosclerosis. Minimal change disease is the most common cause of nephrotic syndrome in children and can be idiopathic or triggered by a recent infection or immune stimulus. Patients classically present with dependent edema in the buttocks, lower back, and legs, foamy or dark-colored urine, hypoalbuminemia, and proteinuria. Minimal change disease appears normal on light microscopy. Incorrect Answers: A, B, C, and D. Crescents within the glomerular space (Choice A) is typical of crescentic glomerulonephritis, which is characterized by a crescent of eosinophilic fibrin and plasma proteins on biopsy of the glomerulus, adjacent to glomerular parietal cells, monocytes, and macrophages. It is generally seen in nephritic syndrome, which is characterized by hematuria and red blood cell casts in urine, oliguria, and hypertension. It can be associated with proteinuria, but to a lesser extent than nephrotic syndrome. Focal segmental glomerulosclerosis (FSGS) (Choice B) is a cause of nephrotic syndrome that can result in proteinuria and pitting edema. FSGS is most commonly associated with sickle cell disease, opioid abuse, and HIV infection, and is characterized by segmental sclerosis of the glomeruli, visible on light microscopy. Neutrophilic infiltration of the glomerular tuft (Choice C) is seen in poststreptococcal glomerulonephritis, which is a nephritic syndrome characterized by dark urine, hypertension, and mild proteinuria following an infection with streptococci, group A. Diabetic glomerulosclerosis can cause thickened glomerular basement membranes (Choice D). It occurs following nonenzymatic glycosylation of the glomerular basement membrane and efferent arterioles, characteristically presenting as Kimmelstiel-Wilson lesions on light microscopy. Diabetic nephropathy progresses over time, initially beginning as microalbuminuria, which leads to macroalbuminuria and end-stage kidney disease. Educational Objective: Minimal change disease is the most common cause of nephrotic syndrome in children and can be idiopathic or triggered by a recent infection or immune stimulus. Patients classically present with dependent edema in the buttocks, lower back, and legs, foamy or dark-colored urine, hypoalbuminemia, and proteinuria. Minimal change disease appears normal on light microscopy. %3D Previous Next Score Report Lab Values Calculator Help Pause

148 Exam Section 3: Item 48 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 48. A 45-year-old man is brought to the physician because of a 4-week history of mild confusion, memory difficulties, generalized muscle aching, and intermittent tingling of his lips and fingers. He says he has been sleeping more than usual, and that he feels "down in the dumps." Vital signs are normal. Physical examination shows no abnormalities except for dry, scaly skin and dry hair. Tapping with a reflex hammer over the left side of the face just anterior to the ear causes left facial muscle contractions. Which of the following is the most likely diagnosis? A) Addison disease B) Cushing disease C) Hyperparathyroidism O D) Hyperthyroidism E) Hypoparathyroidism O F) Hypothyroidism

E. The patient's presentation of confusion, memory difficulties, myalgias, and perioral and/or peripheral paresthesia raises suspicion for hypocalcemia. Facial muscle contractions elicited by tapping the facial nerve anterior to the ear on examination is a classic sign of hypocalcemia known as Chvostek sign. Patients may also have carpal spasm during blood pressure cuff measurement and arterial occlusion. Common causes of hypocalcemia include vitamin D deficiency, hypomagnesemia, acute pancreatitis, citrate chelation from blood product administration, chronic kidney failure, malnutrition, hypomagnesemia, and post-surgical or idiopathic hypoparathyroidism. Parathyroid hormone is a peptide hormone that increases osteoclast activity, vitamin D synthesis, and renal calcium reabsorption in order to maintain calcium homeostasis. Deficiency of parathyroid hormone results in hypocalcemia. Incorrect Answers: A, B, C, D, and F. Addison disease (Choice A) or primary adrenal insufficiency may be due to autoimmune destruction or tuberculosis. It presents with hypotension, hyperkalemia, non-anion gap metabolic acidosis, and skin hyperpigmentation secondary to a deficiency of aldosterone and cortisol. Cushing disease (Choice B) occurs due to increased cortisol levels and presents with truncal weight gain, striae, muscle weakness, and fatty deposits around the face and neck. Vital signs generally demonstrate hypertension, and laboratory studies may show hypokalemia and hyperglycemia. Hyperparathyroidism (Choice C) and excessive production of parathyroid hormone results in increased serum calcium from increased osteoclast activity, intestinal absorption, and renal tubular absorption. Patients with hypercalcemia would present with nephrolithiasis, chronic bone pain, abdominal discomfort, and psychiatric disturbances. Hyperthyroidism (Choice D) classically presents with heat intolerance, weight loss, warm, flushed skin, hair loss, loose stools. Female patients may also present with oligomenorrhea or amenorrhea. Hypothyroidism (Choice F) classically presents with cold intolerance, weight gain, dry skin, constipation, and menorrhagia in female patients. Educational Objective: Hypoparathyroidism can result in hypocalcemia, which can present with confusion, memory difficulties, myalgias, perioral and/or peripheral paresthesias, and facial muscle spasm upon tapping the facial nerve on examination. %3D Previous Next Score Report Lab Values Calculator Help Pause

141 Exam Section 3: Item 41 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment V 41. An 8-year-old girl is brought to the physician for a well-child examination. Her mother says that she has been well except for an occasional cold. Her immunizations are up to date. She is at the 50th percentile for height and 60th percentile for weight. Physical examination shows breast bud development and a few pubic hairs. The mother asks whether her daughter's development is normal. Which of the following is the most appropriate initial response by the physician? A) "I am concerned that your daughter may have precocious puberty." B) "I think that your daughter is likely to begin menstruating in the next few months." C) "I would like to order laboratory studies to ensure that your daughter's development is normal." 17 D) "Your daughter's breast development is somewhat less than might be expected for her age." E) "Your daughter's development is normal."

E. This 8-year-old female patient appears to be undergoing normal sexual maturation. Two physiological events explain the onset of puberty: adrenarche, or the production of androgens by the adrenal gland, and gonadarche, the stimulation of the gonads to produce estradiol by follicle-stimulating hormone and luteinizing hormone. Thelarche, or the development of breast buds, is mediated by gonadal estradiol and is typically the first physical change in puberty, starting as early as 8 years old. Pubarche, or the development of pubic hair as a result of adrenal androgens, is typically the second physical change. Both thelarche and pubarche usually occur in sexual maturity rating (SMR) stage 2 of sexual maturity. Alternatively, menarche, or the first menstruation, which results from the effects of gonadal estradiol on the endometrial lining, typically happens after thelarche in late SMR 3. The growth spurt typically occurs a few months before menarche. This patient's average height and weight reflect that she has not yet started her growth spurt. Incorrect Answers: A, B, C, and D. This patient's sexual maturity is not precocious or delayed (Choices A and D), so no laboratory testing is indicated (Choice C). This patient has entered SMR stage 2, which may develop as early as 8 years old (or as late as 15 years old). Physicians should consider laboratory evaluation of precocious puberty for children presenting with secondary sexual development earlier than 7 or 8 years old. Menarche typically follows thelarche after a few years, not a few months (Choice B). The first menstrual period is related to estradiol stimulation of the endometrial lining rather than ovulation, whereas subsequent, regularly occurring menstrual periods typically result from ovulation. Educational Objective: Early sexual maturation, which begins as early as 8 years old, typically includes thelarche and pubarche. Menarche usually occurs a few years after thelarche. %3D Previous Next Score Report Lab Values Calculator Help Pause

166 Exam Section 4: Item 16 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 16. A 25-year-old woman comes to the physician because of general malaise and a facial rash for 1 week. She has a 10-year history of episodes of pleurisy and arthritic pain in peripheral joints. Physical examination shows an erythematous malar rash that does not involve the nasolabial folds. Results of cardiolipin antibody, anti-dsDNA, and anti-Sm antibody assays are positive. Which of the following hematologic abnormalities is most likely in this patient? A) Hemolytic uremic syndrome B) Macrocytic anemia C) Multiple nucleated erythrocytes D) Rouleaux formations E) Thrombocytopenia

E. Thrombocytopenia is a common finding in patients with systemic lupus erythematosus (SLE) and is often due to immune thrombocytopenic purpura (ITP), a condition defined by the autoantibody-mediated destruction of circulating platelets. Bone marrow biopsy, if obtained, will classically show an increased number of megakaryocytes indicating adequate platelet production but increased peripheral destruction of platelets. It usually presents with petechiae and purpura, and laboratory studies will show prolonged bleeding time. Severe ITP may lead to uncontrolled hemorrhage. Given this patient's positive anticardiolipin antibody, thrombocytopenia may also be secondary to antiphospholipid antibody syndrome, but this is less common than ITP. Other causes of thrombocytopenia in patients with SLE include splenomegaly and drug-induced thrombocytopenia. There are criteria to diagnose SLE, with criterion divided into clinical and immunologic categories. This patient meets criteria given the presence of arthritis, pleuritis, malar rash, anticardiolipin antibodies, anti-dsDNA antibodies, and anti-Sm antibodies. Other potential findings in patients with SLE include photosensitivity, discoid rash, oral ulcers, neurologic phenomena, and leukopenia. Given the clinical diversity of patients with SLE it is common for patients to go many years without meeting enough criteria to receive the diagnosis, and ITP can often precede a formal diagnosis of SLE by several years. Incorrect Answers: A, B, C, and D. Hemolytic uremic syndrome (Choice A) is classically associated with infection from Escherichia coli 0157:H7, a Gram-negative rod. It often presents in children with bloody diarrhea, microangiopathic hemolytic anemia, thrombocytopenia, and renal failure. Macrocytic anemia (Choice B) describes the presence of large erythrocytes and is most commonly a result of folate or vitamin B12 deficiency. It is often seen in patients with malnutrition, alcoholism, pernicious anemia, or Crohn disease. While patients with SLE may develop macrocytic anemia from nutritional deficiency, it is not a classically associated condition. Multiple nucleated erythrocytes (Choice C) on the peripheral blood smear is seen in conditions such as myelofibrosis, where erythroid precursors still possessing a nucleus are prematurely released from the bone marrow. It can also be seen in severe hemolytic anemia as a result of increased release of immature erythrocytes from the bone marrow in response to anemia. It is not classically seen in SLE. Rouleaux formation (Choice D) is a phenomenon seen in conditions such as multiple myeloma and Waldenstrom macroglobulinemia, where there is an increase in serum immunoglobulins. It is occasionally seen in inflammatory states where there are increased levels of acute phase reactants such as fibrinogen. These molecules can interact with sialic acid on the surface of erythrocytes and cause the cells to aggregate in rouleaux formation. It is not classically associated with SLE. Educational Objective: Thrombocytopenia is a common presenting feature of SLE and may precede the diagnosis by several years. It is most often caused by ITP, a condition defined by the autoantibody-mediated destruction of circulating platelets. %3D Previous Next Score Report Lab Values Calculator Help Pause

114 Exam Section 3: Item 14 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 14. A 50-year-old woman is brought to the emergency department 20 minutes after the sudden onset of pain in her epigastrium that radiates to her back and is associated with nausea. She has a history of cholelithiasis. She appears restless. Her temperature is 37.6°C (99.7°F), pulse is 120/min, and blood pressure is 115/60 mm Hg. Abdominal examination shows distention with rebound tenderness and guarding. The pain is partially relieved when the patient bends forward. A CT scan of the abdomen shows fluid surrounding the pancreas. Obstruction of which of the following is the most likely cause of the findings in this patient? A) Ampulla of Vater B) Cisterna chyli C) Common hepatic duct D) Cystic duct E) Superior mesenteric artery

A. Ampulla of Vater obstruction secondary to gallstones is the most likely cause of this patient's acute pancreatitis, which is diagnosed when at least two of three criteria are met: epigastric pain that radiates to the back, lipase level at least three times the upper limit of normal, and/or evidence of acute pancreatitis on abdominal imaging. This patient meets two of these criteria, which is sufficient for a diagnosis of acute pancreatitis. The most common causes of acute pancreatitis are alcohol use disorder and gallstones. In this instance, the patient has known gallstone disease, and it is likely that one or more gallstones has traveled out of the gallbladder via the cystic duct, down the common bile duct, and has become lodged at the ampulla of Vater, where the main pancreatic duct and the common bile duct meet. Obstruction at this point blocks both the common bile duct, predisposing to ascending cholangitis, and the main pancreatic duct, predisposing to acute pancreatitis. Treatment is with endoscopic retrograde cholangiopancreatography to remove the obstructing stone. Incorrect Answers: B, C, D, and E. Cisterna chyli (Choice B) is a dilated area within the thoracic duct at the level of L1-L2 that collects lymph from the intestinal and lumbar lymphatic trunks. It is not involved in the pathogenesis of acute pancreatitis. Obstruction would interfere with normal lymphatic flow. Common hepatic duct (Choice C) is a conduit for bile released from the liver. It joins with the cystic duct from the gallbladder to form the common bile duct. As the hepatic duct is superior to the pancreatic duct and ampulla of Vater, obstruction by a gallstone would not cause pancreatitis. Cystic duct (Choice D) is the conduit for bile to and from the gallbladder. The cystic duct meets the common hepatic duct to form the common bile duct. Obstruction of the cystic duct may cause cholecystitis, but it does not cause pancreatitis. Superior mesenteric artery (Choice E) is a branch of the abdominal aorta that supplies the gastrointestinal tract from the duodenum through the proximal two-thirds of the transverse colon. The inferior pancreaticoduodenal branch supplies blood to the head of the pancreas. Obstruction causes acute or chronic mesenteric ischemia. Educational Objective: Acute gallstone pancreatitis occurs when a gallstone becomes lodged at a point in the common bile duct distal to the main pancreatic duct insertion, which leads to obstruction and pancreatic enzyme autoactivation. A common location for obstruction is the ampulla of Vater, which is the site of confluence of the common bile duct and the main pancreatic duct. %3D Previous Next Score Report Lab Values Calculator Help Pause

173 Exam Section 4: Item 23 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 23. A 13-year-old girl is brought to the physician because of a mass in the thoracic region that has been enlarging during the past month. She was placed in a foster home 1 year ago. She is at the 70th percentile for height and weight. Breast and pubic hair development is Tanner stage 2. Physical examination shows a 3-mm, mildly pigmented mass located in the midclavicular line at the 10th rib. The mass has a smooth surface, uniform outer border, and central papule. There are no other pigmented lesions. Which of the following is the most likely diagnosis? A) Accessory nipple B) Benign pigmented nevus C) Melanoma D) Neurofibroma E) Seborrheic keratosis

A. An accessory nipple is an uncommon finding due to an aberrance in the embryologic migration of mammary tissue. Clinically, it will be characterized by a small, slightly hyperpigmented papule surrounded by an areola-like, round patch located at the midclavicular line, as this line marks the normal migration of mammary tissue. The papule may grow in size during puberty and thus come to clinical attention. On biopsy, histopathologic evaluation will show typical findings of mammary tissue including a papillomatous epidermis, smooth muscle bundles in the dermis, mammary glands in the dermis, and a central duct leading to an opening in the epidermis. An accessory nipple is typically an isolated finding and is not commonly seen in conjunction with other embryological abnormalities. Treatment is simple excision of the lesion. Incorrect Answers: B, C, D, and E. Benign pigmented nevus (Choice B), also known as a compound nevus, is a benign proliferation of melanocytes located in both the epidermis and dermis. They are very common. However, they will not typically present as a hyperpigmented mass with a central papule. Compound nevi also should not display asymmetry, border irregularity, or multiple colors, as this would be more suggestive of melanoma. Melanoma (Choice C) 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. While this lesion has been enlarging, this is due to its response to the hormones of puberty, not malignant invasion. Neurofibroma (Choice D) is characterized by a fleshy, skin colored papule or nodule that is able to be pushed back into the dermis on palpation. It is a finding of neurofibromatosis type I and type II. It is not usually pigmented. 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. They also tend to occur later in life and would be uncommon in an adolescent. Educational Objective: An accessory nipple is characterized by a round hyperpigmented patch with a central papule, which tends to grow around the time of puberty secondary to hormonal influences. It is usually located along the mammary migration line (midclavicular line) as it is a remnant of a mammary gland that failed to fully migrate. %3D Previous Next Score Report Lab Values Calculator Help Pause

124 Exam Section 3: Item 24 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 24. A 25-year-old woman comes to the physician because of a 6-month history of a white discharge from her breasts. Her last menstrual period was 6 months ago. She has had no headaches or visual field changes. She has never been sexually active. Gentle palpation of both breasts produces expression of a milky discharge. Serum studies show: Thyroid-stimulating hormone Follicle-stimulating hormone Luteinizing hormone Prolactin 1.5 μυImL 6 mlU/mL 6 mlU/mL 200 ng/mL A 5-day course of medroxyprogesterone fails to induce menses. An MRI of the brain shows an 8-mm lesion in the pituitary gland. The most appropriate treatment for this patient is a drug from which of the following classes? A) Dopamine agonist B) Dopamine antagonist C) Oxytocin agonist D) Oxytocin antagonist E) Serotonin (5-HT,) agonist F) Serotonin (5-HT,) antagonist

A. An inappropriate increase in serum prolactin is defined as hyperprolactinemia and causes the production of breast milk (galactorrhea), growth of glandular breast tissue, and amenorrhea. Amenorrhea is caused by the suppression of gonadotropin-releasing hormone. Increases in prolactin often arise from a hyperfunctioning anterior pituitary microadenoma. Since dopamine inhibits prolactin secretion, dopamine antagonists can cause galactorrhea as an adverse effect, which may be seen with the use of typical antipsychotic medications such as haloperidol. In contrast, prolactin is positively regulated by thyrotropin-releasing hormone (TRH) and, in states of hypothyroidism with upregulated TRH, galactorrhea can occur. With the exclusion of thyroid pathology and any contributions from dopamine antagonists, further diagnostic evaluation should include an MRI of the brain to assess for the presence of a pituitary adenoma. If a pituitary microadenoma is present, as is noted in this clinical scenario, treatment is with dopamine agonists such as bromocriptine or cabergoline. If such medications fail, or if a macroadenoma is found, surgical resection of the adenoma is the next step. Incorrect Answers: B, C, D, E, and F. Dopamine antagonists (Choice B) decrease the level of dopamine in the brain, which reduces the amount of negative feedback on prolactin. They would lead to further increases in this patient's symptoms. Oxytocin agonists (Choice C) would lead to uterine contractions and would also promote milk letdown. They would not be useful in treatment of a prolactinoma. Oxytocin antagonists (Choice D) would cause uterine atony and decrease the milk letdown reflex. They would not be useful in the treatment of a prolactinoma. Serotonin (5-HT) agonists and antagonists (Choices E and F) include antidepressants and atypical antipsychotics. These medications are not effective in the treatment of prolactinoma. Educational Objective: Prolactinomas present with galactorrhea and amenorrhea. If a pituitary microadenoma is present, treatment is with dopamine agonists such as bromocriptine or cabergoline. If such medications fail, or if a macroadenoma is found, surgical resection of the adenoma is the next step. %3D Previous Next Score Report Lab Values Calculator Help Pause

188 Exam Section 4: Item 38 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 38. When control lymphocytes are treated with corticosteroids, a majority of cells shrink in size and develop peripheral chromatin condensation; cytoplasmic organelles are intact. DNA isolated from the control lymphocytes, electrophoresed on an agarose gel and stained with ethidium bromide, shows a ladder of regularly spaced bands. When lymphocytes from the same culture are transfected so that they overexpress Gene X, which encodes a normal protein product, the cells continue to proliferate and do not undergo any of the morphologic changes seen in the control cells. Gene X is most likely to encode which of the following? A) BCL2 protein B) ERBB2 protein C) p53 protein D) Platelet-derived growth factor E) Rb protein

A. BCL2 is the most likely protein encoded by Gene X. BCL2 is a normal cellular protein that plays a role in the regulation of apoptosis. It is found in the outer mitochondrial membrane, nuclear envelope, and endoplasmic reticulum. BCL2 functions by preventing the oligomerization of BAX/BAK in the outer mitochondrial membrane, which would normally cause release of cytochrome c and activation of caspases in the cytoplasm via interaction with apoptotic protease activating factors. There are two primary morphologic stages of apoptosis that can be recognized at the cellular level. In the first stage, nuclei appear wrinkled and develop peripheral chromatin condensation as a result of fragmentation of high-molecular-weight DNA. This is followed by increased chromatin condensation and the development of nuclear bodies, classified as stage II. Inhibition of caspases, as occurs with upregulation of BCL2, inhibits apoptosis, so cells with increased BCL2 expression will not demonstrate these characteristic morphologic changes. Upregulation or constitutive expression of BCL2 leads to prolonged cellular survival. BCL2 overexpression is seen in several types of B cell non-Hodgkin lymphoma. Incorrect Answers: B, C, D, and E. ERBB2 protein (Choice B) is a growth factor receptor with intracellular tyrosine kinase activity that signals through the PI3K/Akt/MTOR pathway to upregulate genes involved in cellular proliferation. Constitutive expression does increase cellular proliferation but does not directly affect apoptosis. p53 protein (Choice C) is a potent tumor suppressor gene that is commonly inactivated in a variety of malignancies. Increased or constitutive activation of p53 would be more likely to induce apoptosis, not inhibit it. Platelet-derived growth factor (PDGF) (Choice D) is a substance secreted by platelets that stimulates cellular proliferation and angiogenesis. The PDGF receptors, not the growth factor itself, are often mutated in various malignancies including colon and breast cancer. Rb protein (Choice E) is a tumor suppressor protein that blocks cellular entry into S-phase and thereby preventing cellular division. Increased expression would lead to reduced cellular growth. Educational Objective: BCL2 is an antiapoptotic protein that blocks the oligomerization of BAX/BAK in the outer mitochondrial membrane and prevents release of cytochrome c, which normally activates caspases and promotes apoptosis. Overexpression of BCL2 prevents apoptosis. This is seen visually in lymphocytes as an absence of peripheral chromatin condensation, an indicator of apoptosis, when exposed to corticosteroids. %3D Previous Next Score Report Lab Values Calculator Help Pause

162 Exam Section 4: Item 12 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 12. A 68-year-old man with type 2 diabetes mellitus, hypertension, and hyperlipidemia comes to the physician because of pain with urination, blood in his urine, and fatigue during the past week. He has a 6-month history of increased urinary frequency with small volume and nighttime urination. He has had two episodes of acute cystitis during the past 8 months. His temperature is 38.3°C (100.9°F), pulse is 92/min, respirations are 20/min, and blood pressure is 108/72 mm Hg. Physical examination shows suprapubic tenderness and an enlarged, soft prostate. Ultrasonography shows a bladder diverticulum and enlarged kidneys with dilation of renal pelves and calyces. A urine culture grows Escherichia coli. Which of the following is the most likely underlying cause of this patient's condition? A) Benign prostatic hyperplasia B) Hypertension C) Nephrolithiasis D) Type 2 diabetes mellitus E) Xanthogranulomatous cystitis

A. Benign prostatic hyperplasia (BPH) affects middle-aged and elderly male patients and occurs due to excess growth of tissue in the central zone of the prostate causing lower urinary tract obstruction. BPH presents with urinary dribbling and incontinence, difficulty starting or maintaining a urinary stream, nocturia, the need to frequently urinate, and incomplete voiding. These symptoms are typically insidious in onset, chronic, and progressive. Chronic outlet obstruction leads to urinary retention and may cause the formation of bladder diverticula or lead to hydronephrosis. Patients with urinary retention may develop urinary tract infections due to the loss of hydrodynamic flushing of bacteria out of the urethra. Diagnosis is made by history, physical, and urodynamic testing, along with ruling out alternate diagnoses (eg, prostatic malignancy). Treatment is with supportive care, a-blockers (eg, tamsulosin), 5-a-reductase inhibitors (eg, finasteride), or surgery. Incorrect Answers: B, C, D, and E. Hypertension (Choice B) is a common cause of chronic kidney disease and end-stage kidney disease, which may present with oliguria. However, this would not produce lower urinary tract obstruction with urinary retention and hydronephrosis. Nephrolithiasis (Choice C) is more likely to cause urinary tract obstruction proximal to the bladder, leading to unilateral hydronephrosis. Bladder outlet obstruction is more frequently due to prostatic hyperplasia, urethral stricture, or bladder outlet malignancy. Type 2 diabetes mellitus (Choice D) can cause neurogenic bladder that presents with overflow incontinence. This patient's enlarged prostate suggests prostatic hyperplasia as a more likely cause of his symptoms. Xanthogranulomatous cystitis (Choice E) is an exceedingly rare entity that may present with symptoms of cystitis and is characterized by the presence of lipid-laden macrophages and Touton giant cells. This rare diagnosis is much less likely than benign prostatic hyperplasia. Educational Objective: Benign prostatic hyperplasia presents with urinary dribbling and incontinence, difficulty starting or maintaining a urinary stream, nocturia, frequent urination, and incomplete voiding. Chronic outlet obstructic cause the formation of bladder diverticula or lead to hydronephrosis. Patients with urinary retention may develop frequent urinary tract infections due to the limited clearance of urethral bacteria. eads to urinary retention and may %3D Previous Next Score Report Lab Values Calculator Help Pause

184 Exam Section 4: Item 34 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 34. A 55-year-old man comes to the physician because of a 3-week history of night sweats, weakness, fatigue, loss of appetite, and lumps in his neck. He owns a printing business and works with paints and solvents. Physical examination shows edematous, tender axillary and inguinal lymph nodes. His leukocyte count is 45,570/mm3 (80% large, pleomorphic blasts with Auer rods). Occupational exposure to which of the following most likely contributed to the development of this patient's disease? A) Benzene B) Cadmium C) Chlordane D) Naphthylamine E) Vinyl chloride

A. Benzene is known to increase the risk for developing acute myelogenous leukemia (AML), which is the most likely diagnosis in this patient. The risk is thought to be dose-dependent and individuals in industries such as plastic manufacturing and printing are at a higher risk of exposure. Benzene is metabolized primarily in the lung and liver via the action of cytochrome P450 (CYP) enzymes to form benzene oxide, which spontaneously converts to phenol and is followed by conversion to hydroquinone by CYP2Ē1. Hydroquinone is converted in the bone marrow by myeloperoxidase to reactive benzoquinones, which are toxic to hematopoietic cells. Common mutations that arise from benzene exposure include 5q deletions and t(8;21) translocations. AML is considered an oncologic emergency. Treatment with cytarabine and idarubicin is started immediately following diagnosis unless the patient is found to have acute promyelocytic leukemia (APML) in which case treatment is with all-trans-retinoic acid to induce differentiation of malignant blasts. Auer rods suggest a diagnosis of APML but can be seen in other forms of AML. Cytogenetic studies determine the likelihood that the patient will require subsequent stem cell transplant, but therapy is rarely delayed while awaiting these studies. Incorrect Answers: B, C, D, and E. Cadmium exposure (Choice B) is associated with lung, renal, and prostate cancer. It does not predispose to AML. Occupational exposure is common in patients who work making batteries, plastics, and pigments. Chlordane exposure (Choice C) is associated with central nervous system disease including seizures, headaches, and vision changes. When ingested, nausea and vomiting are common. It was previously used as a pesticide and is not found naturally in the environment. It is not associated with any malignancy. Naphthylamine exposure (Choice D) can occur in the dye industry. Occupational exposure is associated with bladder cancer but not AML. Vinyl chloride exposure (Choice E) predisposes to hepatic angiosarcoma and is often seen in patients who work with polyvinylchloride pipes. Educational Objective: AML is a malignancy of the myeloid stem cell and is associated with occupational exposure to benzene. Industrial exposures often occur in the dye, printing, or plastic industry. %3D Previous Next Score Report Lab Values Calculator Help Pause

112 Exam Section 3: Item 12 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 12. A7-year-old girl is brought to the physician because of a 2-week history of painful swelling under her right arm, which has become increasingly severe during the past 3 days. Physical examination shows a 2 x 1-cm, tender, right axillary lymph node and a small papule on the dorsum of the right hand. A silver-stained biopsy specimen of the papule shows pleomorphic bacilli. Which of the following is the most likely causal organism? A) Bartonella henselae B) Brucella melitensis C) Burkholderia mallei D) Francisella tularensis E) Streptobacillus moniliformis

A. Cat scratch disease is caused by infection with the Gram-negative pleomorphic coccobacillus Bartonella henselae. Infection is acquired via scratches or bites from domestic or feral cats, may be localized or disseminated, but most commonly causes local lymphadenitis in the lymphatic drainage pattern of the scratch location. The most commonly involved lymph nodes are the axillary and cervical lymph nodes. Histologic examination shows necrotizing granulomas with stellate (star-shaped) abscesses. Multinucleated giant cells may or may not be present. Other causes of lymphadenitis in children include viruses (eg, adenovirus, respiratory syncytial virus, rhinovirus, Epstein-Barr virus), Mycobacteria species, Streptococcus species, Staphylococcus species, malignancy, fungal infections, and Kawasaki disease. The small papule along the dorsum of the hand suggests a site of inoculation, whereas viral infection would likely demonstrate no such lesion. Incorrect Answers: B, C, D, and E. Brucella melitensis (Choice B) is one of the species of Gram-negative coccobacilli that cause brucellosis, also known as undulant fever. The other species that infect humans are B. abortus, B. canis, and B. suis. Brucellosis is commonly contracted by the consumption of unpasteurized milk and would not be expected to present with a draining punctum. Burkholderia mallei (Choice C) is a Gram-negative rod that causes the zoonotic disease glanders, characterized by nodular lesions and ulcerations in mucous membranes, upper respiratory tract, and lungs. It rarely affects humans, and typically is found in horses, mules, donkeys, and other animals. Francisella tularensis (Choice D) is a Gram-negative coccobacillus that causes tularemia. Signs and symptoms of tularemia include fever, skin ulcers, and lymphadenopathy. It can also cause pharyngitis and severe, fatal pneumonia. It is characteristically associated with rabbits. Streptobacillus moniliformis (Choice E) is a Gram-negative rod that causes rat-bite fever, a zoonotic illness transmitted by the bite or scratch of an infected rat, or the consumption of food or water contaminated by rodent urine or feces carrying the bacteria. Symptoms can include fever, vomiting, rash, arthralgias, myalgias, and headache. It can be complicated by disseminated spread, and can cause abscesses, pneumonia, meningitis, endocarditis, or hepatitis. Educational Objective: Cat scratch disease is caused by infection with the Gram-negative pleomorphic coccobacillus B. henselae. It is a common cause of pediatric lymphadenitis. Histologic examination shows necrotizing granulomas with stellate (star-shaped) abscesses. Multinucleated giant cells may or may not be present. %3D Previous Next Score Report Lab Values Calculator Help Pause

110 Exam Section 3: Item 10 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 10. A 43-year-old woman has had increasing fatigue, intolerance to cold, and dry thickened skin for 7 weeks. The thyroid gland is firm, nodular, and diffusely enlarged. Examination of tissue obtained on biopsy of the thyroid gland shows a diffuse infiltration of lymphocytes with occasional lymphoid follicles. Which of the following is the most likely diagnosis? A) Chronic autoimmune (Hashimoto) thyroiditis B) Diffuse toxic goiter (Graves disease) C) Granulomatous thyroiditis D) Papillary carcinoma E) Toxic multinodular goiter F) Viral thyroiditis

A. Chronic autoimmune (Hashimoto) thyroiditis is the most common form of thyroiditis and is characterized by the presence of antithyroid peroxidase and antithyroglobulin antibodies. Women are disproportionately affected compared to men. The early stages of chronic lymphocytic (Hashimoto) thyroiditis can present with neck swelling and a diffusely enlarged, nontender thyroid gland. The patient may display evidence of hyperthyroidism from destruction of thyroid follicular cells and the unregulated release of thyroid hormone. During the chronic stages of chronic lymphocytic (Hashimoto) thyroiditis, the patient typically develops primary hypothyroidism from the underlying inflammatory destruction of normal thyroid parenchyma and resultant insufficient production of thyroid hormone. Signs and symptoms during this stage include fatigue, cold intolerance, weight gain, hyporeflexia, myxedema, and dry, cool skin. Fine needle biopsy demonstrates a lymphoid infiltrate with the presence of follicles and Hürthle cells. Treatment is with the oral replacement of thyroid hormone. Incorrect Answers: B, C, D, E, and F. Diffuse toxic goiter (Graves disease) (Choice B) is the most common cause of hyperthyroidism and is caused by an autoantibody that activates thyroid stimulating hormone receptors on the thyroid. It presents with symptoms of hyperthyroidism, pretibial myxedema, and thyroid ophthalmopathy, which can cause diplopia, proptosis, and restrictive strabismus. Granulomatous thyroiditis (Choice C) and viral thyroiditis (Choice F), both refer to subacute granulomatous thyroiditis, also known as subacute (de Quervain) thyroiditis, which is a self-limited thyroiditis that often follows an acute viral illness and presents with symptoms of hyperthyroidism. Histology demonstrates granulomatous inflammation. A painful, tender thyroid is a characteristic finding. Papillary carcinoma (Choice D) is the most common thyroid cancer. It typically presents with a solitary thyroid nodule and is characterized by the presence of nuclear grooves with empty-appearing nuclei and psammoma bodies on histological evaluation. It typically carries a good prognosis. Toxic multinodular goiter (Choice E) presents with symptoms of hyperthyroidism and multiple palpable nodules. Histology demonstrates isolated clusters of enlarged, hyperfunctional follicular cells. Educational Objective: Chronic lymphocytic (Hashimoto) disease is the most common form of thyroiditis and is characterized by the presence of antithyroid peroxidase and antithyroglobulin antibodies. Patients with chronic disease present with signs and symptoms of hypothyroidism, including fatigue, cold intolerance, weight gain, hyporeflexia, myxedema, and dry, cool skin. Fine needle biopsy demonstrates a lymphoid infiltrate with the presence of follicles and Hürthle cells. %3D Previous Next Score Report Lab Values Calculator Help Pause

175 Exam Section 4: Item 25 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 25. A 5-month-old boy is brought to the emergency department because of a 3-day history of fever and severe wet cough. His temperature is 40.5°C (104.9°F), and respirations are 65/min. Crackles are heard over all lung fields. A photomicrograph of a silver-stained specimen obtained via bronchoalveolar lavage is shown. In addition to a lack of expression of human leukocyte antigen-DR molecules by lymphocytes, flow cytometry of a peripheral blood specimen will most likely show markedly decreased populations of which of the following cell types? A) CD4+ T lymphocytes B) Dendritic cells C) Monocytes D) Natural killer cells E) Segmented neutrophils

A. Pneumocystis jirovecii is an opportunistic, yeast-like fungal organism that can cause pneumonia in immunocompromised patients. Chest imaging typically reveals diffuse, bilateral infiltrates often prominently about the hila. Methenamine silver, shown here, or toluidine blue stains, selectively stain the cyst walls and are used to confirm the diagnosis. The presence of a pneumonia caused by P. jirovecii indicates that this patient is immunodeficient. Immunodeficiencies can be divided based on the cell type affected: T lymphocytes, B lymphocytes, both T and B lymphocytes, and phagocytes. In addition, the organisms causing the infection may also provide a clue as to what type of cell is deficient. For example, fungal infections due to Candida species, P. jirovecii, and Cryptococcus species are characteristic of T-lymphocyte deficiency. In this case, the absence of the human leukocyte antigen-DR (HLA-DR) molecules on the surface of lymphocytes suggests bare lymphocyte syndrome, a form of severe combined immunodeficiency (SCID). SCID is a combined T-lymphocyte and B-lymphocyte immunodeficiency. Findings include low T lymphocytes on flow cytometry, lack of germinal centers on lymph node biopsy, and the absence of a thymic shadow on chest radiograph. HIV-positive patients with CD4+ counts below 200 cells/mm3 are also at risk for developing pneumonia from P. jirovecii. However, it has become less common following the widespread use of prophylaxis with trimethoprim-sulfamethoxazole in these vulnerable patients. Incorrect Answers: B, C, D, and E. Decreased populations of dendritic cells (Choice B) would lead to decreased antigen presenting to T lymphocytes and secondarily cause a lack of T lymphocyte activation. This is a less likely explanation of this child's P. jirovecii pneumonia. Monocytes (Choice C), or macrophages that have not yet left the bloodstream, are deficient or dysfunctional in Chediak-Higashi syndrome, chronic granulomatous disease, and leukocyte adhesion deficiency. These immunodeficiencies are more likely to present with recurrent bacterial infections than infection with P. jirovecii. Natural killer cells (Choice D) are deficient in the rare condition, natural killer deficiency (NKD). It is commonly seen in children with recurrent viral infections or cancer. P. jirovecii pneumonia is not a common complication. Segmented neutrophils (Choice E) are deficient in extravascular tissues in leukocyte adhesion deficiency due to an inability to extravasate from the peripheral bloodstream. LFA-1integrin is the characteristic defective protein, which impairs intravascular neutrophil adhesion. Recurrent skin bacterial abscesses without pus and delayed separation of the umbilical cord at birth are typical findings. Educational Objective: P. jirovecii is an opportunistic, yeast-like fungal organism that can cause pneumonia in immunocompromised patients, particularly those with T-lymphocyte dysfunction or deficiency. %3D Previous Next Score Report Lab Values Calculator Help Pause

195 Exam Section 4: Item 45 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 45. A 65-year-old woman with severe sensorineural hearing loss undergoes surgical placement of a cochlear implant. This neural prosthesis converts sound energy to electrical signals, which results in stimulation of which of the following structures? A) Auditory nerve endings in the cochlea B) Cochlear nerve as it enters the pons C) Inner hair cells of the cochlea D) Olivocochlear efferent axons that innervate outer hair cells E) Oval window of the cochlea

A. Cochlear implants directly stimulate the auditory nerve endings in the cochlea. Sensorineural hearing loss typically results from damage of the hair cells in the cochlea. Cochlear implants bypass the damaged cochlea. The external microphone of the implant converts mechanical sound waves to an electrical signal that is transmitted to an electrode within the cochlea located adjacent to the auditory nerve endings. Normally, sound transmission in the cochlea begins when the stapes footplate of the middle ear vibrates against the oval window of the cochlea (inner ear), which creates vibrations in the perilymph of the cochlea. These vibrations are ultimately transmitted to the organ of Corti, which contains hair cells that generate electrical impulses and send the impulses to the spiral ganglion of the cochlea (auditory nerve endings). Axons of the spiral ganglion unite to form the auditory nerve, and ultimately the signal is transmitted to the auditory cortex. In sensorineural hearing loss, the damaged hair cells can no longer convert mechanical vibrations to electrical impulses. Incorrect Answers: B, C, D, and E. Stimulating the cochlear nerve as it enters the pons (Choice B) is not the location of a cochlear implant, as distal placement is more invasive and less effective. Stimulating the inner hair cells of the cochlea (Choice C) or the olivocochlear efferent axons that innervate outer hair cells (Choice D) would not be effective, as the hair cells are defective in sensorineural hearing loss and would be unable to transmit electrical signals. The olivocochlear efferent axons provide constant stimulation to the hair cells and are thought to protect the cochlea from loud sounds. Stimulating the oval window of the cochlea (Choice E) would not be effective. The oval window normally transmits mechanical vibrations to the perilymph and ultimately to the hair cells. If the hair cells are defective, oval window stimulation would not solve the problem. Educational Objective: In sensorineural hearing loss, the hair cells within the cochlea are damaged. Cochlear implants utilize an external microphone that transmits electrical impulses to an electrode that directly stimulates the auditory nerve endings. In this way, cochlear implants bypass the damaged cochlea. %3D Previous Next Score Report Lab Values Calculator Help Pause

164 Exam Section 4: Item 14 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 14. A 73-year-old man with stage IV colon cancer begins chemotherapy with 5-fluorouracil, irinotecan, and leucovorin. This patient is at greatest risk for developing signs of drug toxicity in which of the following tissues? A) Bone marrow, gut mucosa, and hair follicles B) Liver, kidney, and endocrine organs C) Nerves, bone, and heart muscle D) Skeletal muscle, heart, and bone E) Trachea, bronchial epithelium, and liver

A. Cytotoxic chemotherapy is commonly employed in the treatment of malignancy. These medications are effective by targeting actively dividing cells. These include malignant neoplasms as well as normal host cells that demonstrate high rates of replication such as the bone marrow, gut mucosa, and hair follicles. These tissues have a high rate of cell turnover, referred to as a high growth fraction. 5-fluorouracil is an antimetabolite that impairs DNA synthesis in the S phase of the cell cycle by decreasing thymidine synthesis. Irinotecan is an inhibitor of topoisomerase I. Leucovorin (folinic acid) enhances the activity and binding of 5-fluorouracil. Common side effects of cytotoxic chemotherapy include fatigue, alopecia, gastrointestinal distress, nausea, vomiting, diarrhea, rashes, and pancytopenia. Neoplasms may develop resistance to chemotherapeutic agents. Mechanisms of resistance include decreased binding affinity through binding site mutations, expression of efflux pumps, enzymatic degradation of the drug, and inhibitor binding to the drug. Incorrect Answers: B, C, D, and E. The other answer choices (Choices B, C, D, and E) refer to tissue types that have a low growth fraction and are not affected to the same degree by general cytotoxic chemotherapy. Certain chemotherapeutic agents are associated with specific organ system toxicities. Doxorubicin is a DNA-intercalating agent associated with cardiotoxicity. Bleomycin causes oxidative damage to DNA and is associated with pulmonary fibrosis. Cisplatin and carboplatin result in cross-linking of DNA, which disrupts DNA repair and may cause ototoxicity and nephrotoxicity. Vincristine inhibits microtubule formation and is associated with peripheral neuropathy. Methotrexate, 5-fluorouracil, and mercaptopurine inhibit nucleotide synthesis and result in bone marrow suppression. Cyclophosphamide is an alkylating agent associated with hemorrhagic cystitis. Educational Objective: Cytotoxic chemotherapy targets cells with a high growth fraction, which include rapidly dividing malignant neoplasms as well as normal cells of the bone marrow, gastrointestinal tract, and hair follicles. %3D Previous Next Score Report Lab Values Calculator Help Pause

157 Exam Section 4: Item 7 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 7. A 28-year-old woman comes to the physician because of a 1-week history of fatigue, muscle pain, and generalized weakness. She is unable to climb stairs because of the symptoms. She also has a 1-month history of a rash on her face, elbows, and knuckles. Her temperature is 38.3°C (100.9°F), pulse is 100/min, respirations are 20/min, and blood pressure is 115/65 mm Hg. Physical examination shows the findings in the photographs. There is also proximal muscle weakness of the lower extremities bilaterally. Which of the following is the most likely diagnosis? OA) Dermatomyositis B) Guillain-Barré syndrome C) Nephrotic syndrome D) Psoriasis OE) Systemic lupus erythematosus

A. Dermatomyositis is an autoimmune condition affecting the skin and the proximal muscles. Cutaneous manifestations include pink papules over the dorsal fingers (Gottron papules), pink to light purple patches over the upper eyelids (heliotrope sign), light pink to lilac colored patches with telangiectasias on the upper back and sun-exposed chest, and dilated capillaries with drop out on the proximal nail folds. Amyopathic dermatomyositis is a rare form in which only cutaneous manifestations are present, but a concomitant inflammatory myopathy with proximal muscle weakness is more typical. An inflammatory infiltrate of the muscle occurs causing damage to the myocytes, leading to release of creatine kinase, an intracellular molecule. This can be used to monitor disease progress in dermatomyositis. Dermatomyositis is also characterized by several autoantibodies. It demonstrates a positive antinuclear antibody (ANA) and anti-Jo-1, anti-SRP, and anti-Mi-2 antibodies are commonly detected. Dermatomyositis is also associated with occult malignancy and patients with a new diagnosis of dermatomyositis should undergo age-appropriate screening for malignancy. Treatment of dermatomyositis involves systemic steroids followed by long-term immunosuppressive therapy with medications such as methotrexate or azathioprine. Incorrect Answers: B, C, D, and E. Guillain-Barré syndrome (Choice B) is typically a self-limited demyelinating polyneuropathy that starts with the distal extremities and progresses proximally. It may affect the lungs through diaphragmatic paralysis, which is the most serious complication of the disease. While proximal muscle weakness can ensue, this would only occur after involvement of the distal extremities in the classic form. Nephrotic syndrome (Choice C) is characterized by proteinuria, edema, hypoalbuminemia, and hypercholesterolemia. Due to damage to the glomerular filtration barrier, large proteins such as albumin are filtered into the urine causing hypoalbuminemia. Membranous nephropathy is the most common nephrotic syndrome in white adults and can be primary (idiopathic) or secondary to conditions such as systemic lupus erythematosus, autoimmune disease against phospholipase A2-receptors, chronic hepatitis infection (especially hepatitis B or C virus), or pharmaceuticals such as nonsteroidal anti-inflammatory drugs or penicillamine. It is not a feature of dermatomyositis. Psoriasis (Choice D) is characterized by thick, salmon-colored plaques with silvery white scale classically present on the extensor extremities. It can be accompanied by psoriatic arthritis, an inflammatory joint disease, which presents with morning stiffness and swelling of the small joints of the fingers and toes. Psoriasis does not typically cause muscle pain, generalized weakness, or fever. Systemic lupus erythematosus (Choice E) is an autoimmune disorder with a wide range of presentation. Cutaneous manifestations include an erythematous rash over the malar cheeks that spares the nasolabial folds after sun exposure or discoid lesions with hyperpigmented edges and pink, atrophic centers. Painless oral ulcers and hair loss may occur. Fatigue, joint pain, and fever are also common presenting signs. However, proximal muscle weakness is more specific for an inflammatory myositis. Systemic lupus erythematosus does not typically present with Gottron papules on the dorsal hands or the heliotrope rash on the eyelids. Educational Objective: Dermatomyositis is an autoimmune disease affecting the skin and muscles, which demonstrates a positive ANA, anti-Jo-1, anti-SRP, and anti-Mi-2 antibodies. Classic cutaneous findings include Gottron papules, heliotrope rash on the eyelids, and a light pink rash with telangiectasias on the upper back and chest. These findings are usually accompanied by proximal muscle weakness and increased levels of creatine kinase. Previous Next Score Report Lab Values Calculator Help Pause

135 Exam Section 3: Item 35 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 35. A 2-month-old girl is brought to the emergency department by her mother because of difficulty breathing for 8 hours. Physical examination shows cyanosis. A loud systolic murmur is heard. Laboratory studies show hypocalcemia and lymphopenia. Echocardiography shows tetralogy of Fallot. The diagnosis of DiGeorge syndrome is made, and a karyotype is ordered. Chromosomal analysis by high-resolution G-banding is reported as 46,XX. Which of the following genetic mechanisms is the most likely cause of this patient's anomalies? A) Deletion B) Nondisjunction C) Paracentric inversion D) Pericentric inversion E) Reciprocal translocation F) Robertsonian translocation G) Trinucleotide repeat H) Uniparental disomy

A. DiGeorge syndrome, a 22q11 deletion syndrome, presents as a result of failure of the third and fourth branchial pouches to develop. The third branchial pouch gives rise to the thymus and inferior parathyroid glands, and the fourth branchial pouch gives rise to the superior parathyroid glands. Classically, DiGeorge syndrome presents with immunodeficiency due to failure of thymus development (thymic aplasia), hypocalcemia from failure of parathyroid gland development, structural cardiac defects such as tetralogy of Fallot, and craniofacial malformations. Common physical examination findings include a cleft palate and low-set ears; however, the expressivity varies and combinations of abnormal palate, facial, and cardiac malformations are observed. An increased susceptibility to predominantly viral and fungal infections is the result of immunodeficiency from aplasia of the thymus (the site of maturation of T lymphocytes). The definitive diagnosis requires fluorescence in-situ hybridization to detect the deleted DNA segment with a fluorescent probe. Treatment is supportive, as there is no cure for the condition. Incorrect Answers: B, C, D, E, F, G, and H. Nondisjunction (Choice B) is the mechanism by which errors in chromosome numbers occur due to failure of paired chromosomes to separate appropriately during cell division. Examples of disorders include Down syndrome (trisomy 21), Edwards syndrome (trisomy 18), and Patau syndrome (trisomy 13). Paracentric inversions (Choice C) and pericentric inversions (Choice D) occur when two breaks in a chromosome result in an excised fragment that then rejoins the same chromosome with the fragment reversed from its original orientation. A paracentric inversion does not involve the centromere (the fragment is a section of one of the chromosome arms) while a pericentric version does involve the centromere. Reciprocal translocation (Choice E) occurs when fragments from two chromosomes exchange places, which results in a unique derivative pair of chromosomes. Robertsonian translocation (Choice F) occurs when two different chromosomes break at the centromere and recombine with the break from the other. This most often happens with acrocentric chromosomes, which have a long arm containing most of the genetic material and a short arm. A derivative pair forms from the long arms joining together and short arms joining together. Trinucleotide repeat (Choice G) describes three nucleotides in repetitive series that are susceptible to replication errors. Genetic conditions that arise from these errors tend to display genetic anticipation. Examples include Huntington disease, myotonic dystrophy, and fragile X syndrome. Uniparental disomy (Choice H) is the inheritance of a pair of nonidentical chromosomes from a single parent. Genomic imprinting results in methylation and suppression of genes dependent on the parent-of-origin. Uniparental disomy occurs in Angelman syndrome and Prader-Willi syndrome. Educational Objective: Deletion of genetic material from a chromosome results in missing genes that may cause congenital disorders. DiGeorge syndrome is an example that results from a missing segment of chromosome 22. II Previous Next Score Report Lab Values Calculator Help Pause

161 Exam Section 4: Item 11 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 11. A 27-year-old man comes to the physician because of fever and cough for 1 month. An x-ray of the chest shows diffuse infiltrates. Tiny intracellular yeast forms are seen on direct smears, and tuberculate macroconidia are seen after culture at room temperature. Which of the following exposures is most consistent with this patient's illness? A) Bird droppings B) Desert sandstorm C) Insect bite D) Moldy hay E) Rose thorn puncture

A. Histoplasmosis is an endemic mycosis secondary to Histoplasma capsulatum. Pulmonary histoplasmosis can present in acute, subacute, or chronic phases, generally characterized by fever, malaise, myalgia, cough, and hemoptysis with the development of calcified nodules and mediastinal or hilar lymphadenopathy. Transmission occurs through inhalation of fungal spores. It is endemic to the Mississippi and Ohio River valleys in the United States, as well as parts of Central and South America, sub-Saharan Africa, and Southeast Asia. The virulence of H. capsulatum is dependent on its growth inside nonactivated macrophages, which protect the fungal cells from detection and elimination. Risk factors for histoplasmosis include exposure to bird or bat droppings. Diagnosis is typically made via the detection of histoplasma antigen in the serum and/or urine. Histology will reveal oval yeast cells within macrophages. Histoplasma species on culture grows with septate mycelia and characteristic tuberculate macroconidia. Incorrect Answers: B, C, D, and E. Desert sandstorms (Choice B) are a vector by which microorganisms may be disseminated to distant environments. Coccidioidomycosis is an endemic mycosis prevalent in the Southwestern United States associated with dust exposure that may be transmitted by sandstorms. Histology reveals spherules filled with Coccidioides species endospores. Insect bites (Choice C) are a common vector for infectious disease transmission. H. capsulatum is transmitted via inhaled spores from the environment, not insect bites. Moldy hay (Choice D) is associated with hypersensitivity pneumonitis secondary to the inhalation of dust and mold spores. H. capsulatum is transmitted via inhaled spores from sources such as bird or bat droppings, not from hay. Rose thorn puncture (Choice E) is associated with sporotrichosis, which is caused by the dimorphic fungus Sporothrix schenckii. Transmission is via traumatic inoculation into the skin. Educational Objective: Histoplasmosis is an endemic mycosis caused by H. capsulatum that causes pulmonary illness when spores are inhaled from the environment. Histoplasmosis is most associated with exposure to bird or bat droppings. %3D Previous Next Score Report Lab Values Calculator Help Pause

127 Exam Section 3: Item 27 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 27. The tracing shows contractile responses of a skeletal muscle preparation to electrical stimulation at five different frequencies. The amount of calcium sequestration in sarcoplasmic reticulum is highest during which of the following labeled responses? A B Авс D C Stimulus 1-Hz 3-Hz 6-Hz 9-Hz 12-Hz frequency Time A) B) C) D) E)

A. Muscle fibers utilize electromechanical coupling to allow for the locomotion of the human skeleton. Normally this process occurs via acetylcholine released from the postsynaptic neuron, which results in the occurrence of endplate potentials at the neuromuscular junction. Once a threshold is reached, muscle cell depolarization and action potential propagation occur. The T tubules, infoldings of the sarcolemma, transmit the action potential deep into the muscle cell. Dihydropyridine receptors are voltage-gated calcium channels that are mechanically coupled to ryanodine receptors. Depolarization results in conformational changes of these receptors, which results in the subsequent release of calcium from the sarcoplasmic reticulum. Calcium in the cytoplasm of the muscle cell binds to troponin C, which induces a conformational change that moves tropomyosin, allowing for myosin-actin binding and mechanical contraction. In this experiment, electrical stimuli are applied at numerous frequencies to a skeletal muscle and the tension is measured. At 1 Hz, muscle contraction can be demonstrated after the stimulus followed by relaxation of the muscle as evidenced by the descending slope of the tension measurement. During muscle relaxation, calcium is sequestered back into the sarcoplasmic reticulum. This cycle then occurs again as another stimulus is applied and tension once again rises and falls. It can be noted that the tension does not fully return to baseline, as not enough time occurs between each stimulus to allow for complete relaxation of the muscle and full calcium sequestration into the sarcoplasmic reticulum. As the frequency increases, the amount of time allowed for relaxation and calcium sequestration decreases, leading to the inability of the muscle to approach its baseline tension. Therefore, stimulation of the muscle at 1 Hz allows for the highest level of calcium sequestration into the sarcoplasmic reticulum between each stimulus as this permits the greatest relaxation time. By comparison, at 12 Hz, the stimulus frequency does not allow for any relaxation of the muscle and tension continues to build. This is also known as tetany. In this scenario, there is minimal sequestration of calcium and no relaxation of the muscle at 12 Hz. Incorrect Answers: B, C, D, and E. Increasing frequency of stimuli leads to decreased time for calcium sequestration into the sarcoplasmic reticulum. Cytoplasmic calcium concentrations remain increased, which prevents mechanical relaxation of the muscle fibers. Stimulation at 3 Hz (Choice B), 6 Hz (Choice C), 9 Hz (Choice D), and 12 Hz (Choice E) sequentially decreases the time for calcium sequestration into the sarcoplasmic reticulum, resulting in eventual tetany. Educational Objective: Skeletal muscle is comprised of contractile cells called myocytes. Myocytes respond to action potentials by releasing calcium into the cytoplasm that results in tropomyosin conformation changes and actin-myosin binding. The generation of mechanical tensile force results. Between each stimulus, calcium is sequestered into the sarcoplasmic reticulum, allowing for muscle relaxation. Increasing stimulus frequency leads to inadequate time for calcium sequestration resulting in tetany, which is a state of high intracellular calcium concentration and constant force. Previous Next Score Report Lab Values Calculator Help Pause Tension

152 Exam Section 4: Item 2 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 2. A 22-year-old man is brought to the emergency department 30 minutes after a friend found him unconscious on the floor. A drug overdose is suspected. His pulse is 120/min, respirations are 4/min with apparent low tidal volumes, and blood pressure is 120/100 mm Hg. Physical examination shows mild cyanosis. He is intubated and mechanically ventilated with positive pressure ventilation (rate = 12/min; V-= 7 mL/kg) using 10 cm H,0 positive end-expiratory pressure (PEEP). Which of the following sets of findings best describes the effects of the mechanical ventilation with PEEP on this patient's alveolar (PA) and intrapleural (P) pressures? ITTTT Peak Inspiratory PA End-TidalPA Peak Inspiratory Pip End-Tidal Pip A) Positive positive positive positive B) Positive zero positive negative C) Zero zero zero negative D) Zero negative zero zero E) Negative positive positive positive F) Negative zero negative negative

A. Spontaneous breathing is accomplished by expansion of the chest wall, which pulls the parietal pleura outward and generates a negative intrapleural pressure relative to the atmosphere. The pressure gradient that results drives air into the lungs. Normal end- tidal intrapleural pressure is around -5 cm H ,0 relative to atmospheric pressure. Mechanical ventilation is indicated for patients who are unable to breathe on their own, which may be due to central nervous system depression, neuromuscular disease, or decreased lung compliance (requiring greater force to expand lung against increased lung recoil). Modern mechanical ventilation utilizes a positive pressure system. The ventilator circuit is connected to the airway via an endotracheal tube and sealed with a cuff. Positive pressure is generated by the ventilator, which drives air into the lungs. Positive end-expiratory pressure (PEEP) may be utilized to prevent the pressure in the alveoli from dropping to zero or becoming negative during the respiratory cycle due to chest wall recoil. This helps keep alveoli stented open and able to participate in gas exchange. The positive pressure is transmitted to the pleural space. The peak inspiratory and end-tidal alveolar pressures will be positive relative to atmosphere, as well as the peak- inspiratory and end-tidal intrapleural pressures. Incorrect Answers: B, C, D, E, and F. Choices C, D, E, and F are incorrect as the peak inspiratory pressure will be positive to drive air into the lungs in the absence of spontaneous chest wall movement. The peak inspiratory pressure represents the pressure needed to overcome airway resistance and deliver the full tidal volume to the lungs. Choices B, C, D and F are incorrect as the end-tidal alveolar pressure will be positive relative to atmosphere in a positive-pressure ventilatory circuit utilizing PEEP. This may help prevent atelectasis and can improve oxygenation in disorders with reduced lung compliance such as acute respiratory distress syndrome. Choices C, D, and F are incorrect in that the peak inspiratory intrapleural pressure will be positive in this system. A negative peak inspiratory intrapleural pressure occurs in spontaneous breathing. Similarly, choices B, C, D, and F are incorrect in that the end-tidal intrapleural pressure is positive in this system. In spontaneous breathing, the end-tidal intrapleural pressure is normally around -5 cm H ,0 due to the outward pull from the chest wall and inward pull from lung elastic recoil. Educational Objective: Modern mechanical ventilation utilizes a positive pressure system to drive air into the lungs. This is in contrast to normal, spontaneous breathing, in which outward expansion of the chest wall generates negative intrapleural and alveolar pressures relative to atmosphere, creating a pressure gradient for the movement of air into the lung. %3D Previous Next Score Report Lab Values Calculator Help Pause

131 Exam Section 3: Item 31 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 31. A 28-year-old woman who works as a glassblower has recently sustained burns on her hands but feels little pain. Neurologic examination shows decreased pain sensation in both upper extremities and weakness and atrophy of the intrinsic muscles of both hands. Which of the following best explains these findings? A) Cystic lesion involving midline structures and both ventral horns in the cervical enlargement B) Demyelination of the left corticospinal tract at all levels of the cord C) Lesion of the left anterolateral quadrant at C-2 D) Loss of myelin in the dorsal column white matter E) Loss of myelin in the lateral column white matter and loss of ventral horn motoneurons

A. This patient presenting with hypoalgesia, weakness, and atrophy of the upper extremities most likely has syringomyelia, which features a cystic lesion involving midline spinal cord structures and both ventral horns in the cervical enlargement. Syringomyelia involves a progressively expanding cavity in the cervical central spinal canal and may be associated with congenital malformations (eg, Chiari malformation type I) spinal cord infections, autoimmune diseases (eg, multiple sclerosis), neoplasms, or spinal cord trauma. Syringomyelia typically damages spinothalamic fibers decussating in the anterior white commissure. Consequently, patients typically present with deficits in pain and temperature sensation in the bilateral upper extremities. Since the ascending spinothalamic tracts in the lateral spinal cord are unaffected, sensation in the lower body is intact. The cystic lesion can expand to the bilateral anterior horns (the location of the cell bodies of lower motor neurons). Therefore, patients may also present with a lower motor neuron-pattern of dysfunction in the bilateral upper extremities (eg, decreased muscle tone, hyporeflexia, early muscle atrophy). Incorrect Answers: B, C, D, and E. Demyelination of the left corticospinal tract at all levels of the cord (Choice B) would result in hemiparesis that spares the face. Upper motor neuron signs (eg, increased muscle tone, hyperreflexia, delayed atrophy) would predominate, and sensory deficits would not be demonstrated. A lesion of the left anterolateral quadrant at C-2 (Choice C) would affect the left spinothalamic tract and left anterior horn. Patients would present with hemisensory deficits in pain and temperature inferior to the neck and weakness of the trapezius and rectus capitis muscles. The loss of myelin in the dorsal column white matter (Choice D) would affect the dorsal column-medial lemniscus pathway and lead to the loss of sensation of fine touch, pressure, proprioception, and vibration. This patient instead demonstrated localized motor dysfunction and loss of pain and temperature sensation. The loss of myelin in the lateral column white matter and loss of ventral horn motoneurons (Choice E) would affect the spinothalamic tracts and lower motor neuron cell bodies. Patients would present with a lower motor neuron-pattern of dysfunction (eg, decreased muscle tone, hyporeflexia, early muscle atrophy) and deficits in pain and temperature sensation. Educational Objective: Syringomyelia features a cystic cavity in the cervical central spinal canal, which affects midline spinal cord structures such as the anterior white commissure (decussation of the spinothalamic tracts) and the anterior horns (cell bodies of lower motor neurons). Patients with syringomyelia may present with deficits in pain and temperature sensation and a lower motor neuron-pattern of dysfunction in the bilateral upper extremities. %3D Previous Next Score Report Lab Values Calculator Help Pause

107 Exam Section 3: Item 7 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 7. A 54-year-old man with a 1-year history of multiple sclerosis has increasingly painful muscle spasms. An agonist that acts at which of the following receptors is most likely to alleviate the increased extensor tone and clonus? A) y-Aminobutyric acid; O B) Histamine-1 (H,) C) Histamine-2 (H2) OD) Muscarinic-1 (M,) E) Muscarinic-2 (M2) F) Nicotinic

A. V-aminobutyric acid (GABA)B receptors are metabotropic G-protein coupled receptors that increase efflux of potassium and thus hyperpolarize skeletal muscle cells and decrease action potential frequency. Baclofen is a GABA analog that acts as an agonist at GABA-B receptors and increases muscle relaxation. Therefore, baclofen is a first-line treatment for muscle spasticity in multiple sclerosis and following neurologic injury. GABA-g receptors are also found in the central nervous system (CNS); therefore, baclofen can cause sedation, especially when used in conjunction with other CNS depressants. Incorrect Answers: B, C, D, E, and F. Histamine-1 (H1) (Choice B) receptors are involved in acute inflammatory reactions, sedation, and allergic responses and are distributed widely in the brain, vascular endothelium, and smooth muscle in the heart and lungs. H, agonism results in increased central pain, vascular permeability, and nasal and bronchial mucus production along with pruritus. Histamine-2 (H2) (Choice C) receptor agonism results in increased gastric acid secretion. H, and H2 receptors are not highly expressed on skeletal muscle. Muscarinic-1 (M1) (Choice D) receptors are primarily found in the CNS and enteric nervous system. M1 receptor agonists are under investigation in Alzheimer dementia. M, receptors are not expressed on skeletal muscle. Muscarinic-2 (M2) (Choice E) receptors are primarily demonstrated by sinoatrial and atrioventricular nodal cells. M2 receptor agonists increase potassium efflux and decrease heart rate. M2 receptors are not highly expressed in skeletal muscle cells. Nicotinic (Choice F) acetylcholine receptors are highly expressed on skeletal muscle and respond to acetylcholine release by lower motor neurons. Varenicline, which targets tobacco dependence, is a nicotinic acetylcholine receptor agonist. As nicotinic acetylcholine receptor stimulation leads to muscle contraction, nicotinic acetylcholine receptor agonists would not be helpful for chronic muscle spasticity. Educational Objective: Baclofen is a GABA analog that acts as an agonist at GABA-B receptors, which hyperpolarizes muscle cells and relieves muscle spasticity. %3D Previous Next Score Report Lab Values Calculator Help Pause

174 Exam Section 4: Item 24 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 24. A 52-year-old woman comes to the emergency department because of a 2-week history of progressive shortness of breath and fatigue. Her pulse is 102/min, respirations are 22/min, and blood pressure is 100/80 mm Hg. Physical examination shows muffled breath sounds. Echocardiography shows a large pericardial effusion. Pericardiocentesis yields cloudy, serosanguineous fluid. Analysis of the fluid shows an increased protein concentration, numerous RBCS, and a small number of WBCS, indicative of malignancy. Metastasis from which of the following sites is the most likely cause of the findings in this patient? A) Bladder B) Breast C) Colon D) Ovaries E) Stomach

B. Breast cancer encompasses many types of breast carcinoma, including invasive ductal and lobular carcinoma, medullary carcinoma, and inflammatory breast carcinoma. The most common type is invasive ductal carcinoma, which typically presents as a palpable, immobile breast mass, which may demonstrate features such as spiculated margins and microcalcifications on mammography. Metastases occur when the breast carcinoma invades beyond the ductal structures of the breast, into the lymphatics, and to distant anatomical sites. Metastatic breast cancer most commonly spreads to the bone, lung, and brain. It may result in hypercalcemia secondary to metastatic osteolytic lesions causing destruction of bone. In this case, breast cancer involvement of the mediastinum and pericardial space, whether by direct invasion or metastasis, has caused a pericardial effusion. While lung cancer is the most common cause of malignancy-induced pericardial effusion, breast cancer is also frequently implicated, as in this female patient. The treatments of malignancy (eg, chemotherapy, radiation) may also potentially cause pericardial effusions, though not in the case of this patient whose pericardial effusion was the initial presentation of malignancy. Incorrect Answers: A, C, D, and E. Bladder cancer (Choice A) commonly metastasizes to the lung, liver, and bone. Pericardial involvement is uncommon. Colon (Choice C) and stomach (Choice E) cancer commonly metastasize to the liver, lung, lymph nodes, and peritoneum. Involvement of these sites would not be likely to cause a pericardial effusion. Ovarian cancer (Choice D) most commonly metastasizes to the lung, brain, liver, and distant lymph nodes. It also has the possibility of involving the peritoneum, causing peritoneal carcinomatosis. Metastases to the lungs may cause pleural effusions, but pericardial involvement is uncommon. Educational Objective: Pericardial effusions secondary to malignancy may occur from direct invasion of the tumor or metastatic spread. %3D Previous Next Score Report Lab Values Calculator Help Pause

149 Exam Section 3: Item 49 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 49. A 38-year-old man comes to the physician because of a 1-week history of a discharge from his penis and testicular pain. He is sexually active and uses condoms inconsistently. Physical examination shows a thick green urethral discharge and exquisite tenderness of the right testis. The pain is relieved by elevation of the scrotum. A photomicrograph of a Gram stain of the discharge is shown. Which of the following is the most likely causal organism? A) Chlamydia trachomatis B) Mycoplasma genitalium C) Neisseria gonorrhoeae D) Trichomonas vaginalis OE) Ureaplasma urealyticum

C. This patient presents with signs and symptoms suggestive of epididymitis, which frequently occurs due to infection with Neisseria gonorrhoeae or Chlamydia trachomatis. This patient's symptoms of thick, green urethral discharge and photomicrograph demonstrating accumulation of Gram-negative diplococci within neutrophils is most consistent with N. gonorrhoeae. Epididymitis typically presents with fever, purulent urethral discharge, and pain localized to one testis, which may be alleviated with elevation of the hemiscrotum (distinguishing the pain of epididymitis from that of testicular torsion, which is not alleviated by this maneuver). N. gonorrhoeae usually responds to treatment with ceftriaxone, however, resistance to multiple antibiotics is becoming increasing prevalent amongst N. gonorrhoeae. Coinfection with C. trachomatis is frequent, and patients should be treated empirically for both infections. Sexual partners require testing and treatment due to the high rate of asymptomatic infection. Incorrect Answers: A, B, D, and E. Chlamydia trachomatis (Choice A) is a Gram-negative bacterium that may cause epididymitis. However, it is less likely to produce green discharge and appears microscopically as an obligate intracellular organism with elementary and reticulate bodies, rather than as a diplococcus. Mycoplasma genitalium (Choice B) is a rare cause of urethritis in men and does not commonly cause epididymitis. M. genitalium is also notable for its small genome, leading it to be chosen as the first organism to have its entire genome reproduced synthetically. Trichomonas vaginalis (Choice D) is an anaerobic, flagellated protozoan that causes vaginal trichomoniasis. Trichomoniasis presents with pruritus, dysuria, dyspareunia, cervical erythema, and abnormal discharge that is described as green, frothy, and malodorous. Men affected by trichomoniasis are often asymptomatic, and T. vaginalis does not commonly cause epididymitis. Ureaplasma urealyticum (Choice E) may cause urethritis in men. It is characterized by its lack of a cell wall on microscopy and by its production of urease. Educational Objective: N. gonorrhoeae is a Gram-negative diplococcus that is a common cause of epididymitis. Epididymitis presents with fever, purulent urethral discharge, and pain localized to a testis that is alleviated with elevation of the hemiscrotum. %3D Previous Next Score Report Lab Values Calculator Help Pause

194 Exam Section 4: Item 44 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 44. A 52-year-old woman is admitted to the hospital because of chest pain and shortness of breath for 2 hours. Her pulse is 102/min, respirations are 35/min, and blood pressure is 110/65 mm Hg. Cardiac examination shows a laterally displaced apex, a systolic thrill, and a grade 4/6, crescendo-decrescendo systolic ejection murmur. Cardiac catheterization shows a pulmonary capillary wedge pressure of 40 mm Hg (N=5-16). Which of the following sets of changes in this patient's left ventricle is most likely? Myocardial Oxygen Consumption Myocardial Oxygen Tension Stroke Adenosine Concentration Work A) Increased increased increased decreased B) Increased increased decreased increased C) Increased decreased increased increased D) Increased decreased decreased increased E) Decreased increased increased increased F) Decreased increased increased decreased G) Decreased decreased increased increased H) Decreased decreased decreased decreased

B. Aortic stenosis (AS) is a common disorder of the aortic valve that results from calcification. Calcification of the aortic valve is a pathologic consequence of mechanical stresses on heart valves, and results from repetitive microtrauma from the opening and closing of valve leaflets with associated chronic inflammation. Many people will develop some degree of AS over time, but structural abnormalities of the valve, such as a bicuspid aortic valve, alter the biomechanics of valve opening and closing and increase the likelihood of earlier calcification and resultant stenosis. While many patients may be asymptomatic, those with severe AS may complain of fatigue, shortness of breath, cough, diminished exercise tolerance, angina, or syncope with exertion. Exam findings include a crescendo-decrescendo systolic murmur best heard at the upper right sternal border, and pulsus parvus et tardus (weak and delayed) may be noted on examination of peripheral pulses. Due to the chronic increased afterload from a fixed obstruction by the valve, left ventricular hypertrophy and resultant diastolic dysfunction can occur with evidence of increased pulmonary capillary wedge pressure on cardiac catheterization. Contracting against a stenosed valve requires increased stroke work, resulting in increased myocardial oxygen consumption, decreased myocardial oxygen tension, and increased adenosine concentration. Increased adenosine concentration, which occurs as adenosine triphosphate (ATP) is dephosphorylated for contraction, and decreased oxygen tension promote coronary artery vasodilation to increase perfusion to regions of high cardiomyocyte demand. Incorrect Answers: A, C, D, E, F, G, and H. Choices A, F, and H are incorrect as the adenosine concentration increases due to the breakdown of adenosine-phosphate complexes for energy utilization. Choices C, D, G, and H are incorrect because myocardial oxygen consumption is increased in order to generate the required contractile force. Myocardial oxygen consumption falls as myocardial workload falls. Similarly, choices A, C, E, F, and G are incorrect as the myocardial oxygen tension decreases due to increased metabolic demand for energy in the form of ATP and increased oxygen utilization by mitochondria. Choices E, F, G, and H are incorrect as the stroke work of the left ventricle increases to generate a pressure gradient across a stenosed aortic valve to drive blood flow forward. Surgery to replace the stenotic valve would reduce the myocardial stroke work necessary to propel blood forward by decreasing afterload on the ventricle. Educational Objective: AS may lead to left ventricular hypertrophy due to chronically increased afterload. Increased afterload results in increased ventricular stroke work and myocardial oxygen consumption. II Previous Next Score Report Lab Values Calculator Help Pause

125 Exam Section 3: Item 25 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 25. A 23-year-old woman is admitted to the hospital after taking an overdose of 16 acetaminophen tablets. She ingested the acetaminophen after she broke up with her boyfriend of 2 months. She has been admitted to the hospital several times during the past 5 years for treatment of suicide attempts and other self-harm attempts. Her history includes childhood sexual abuse, and she has difficulty holding a job. She tells the physician that she has received substandard care from this hospital in the past. She says, "But now that you're my physician I know I'm going to get better. I can tell because we have a connection." Her affect is bright and cheerful. This patient most likely has which of the following types of personality disorders? A) Antisocial B) Borderline C) Dependent D) Narcissistic E) Schizoid

B. Borderline personality disorder (BPD) is a cluster B personality disorder, the emotional or dramatic cluster, that features an unstable sense of self and tumultuous relationships. Likely due to a combination of genetic polymorphisms in serotonin and dopamine receptors and emotional invalidation or abuse during childhood, patients with BPD unconsciously learn to make impulsive and dramatic gestures, including self-harm, to obtain attention and emotional fulfilment from others. Chronic invalidation also leads to poor self-esteem and the consequent reliance on others for self-esteem needs, resulting in an intense fear of abandonment and severe distress when abandonment happens. Due to inadequate emotional attunement from caregivers, patients with BPD do not learn to label, understand, or regulate their emotions. Their negative emotional experiences are so intensely painful that people with BPD must dissociate these experiences from their consciousness. This splitting leads people with BPD to black-and-white thinking that includes seeing people as all good and others as all bad, as in this patient. Incorrect Answers: A, C, D, and E. Antisocial personality disorder (Choice A) is a cluster B personality disorder that arises from a deficit in empathy, resulting in pervasive violations of others' rights, aggression, and a hostile and manipulative attitude toward others. These patients may also be chronically bored and hence sensation seeking, leading to a high rate of substance abuse and gambling. Patients with BPD may adopt a hostile attitude toward others at times but are more likely to self-harm than harm others. Dependent personality disorder (Choice C) is a cluster C personality disorder, which is the anxious cluster. Dependent personality disorder presents with an excessive need to be cared for by others that manifests as severe separation anxiety and clinging behavior. Though this patient does illustrate abandonment-related distress, she also illustrates difficulty with emotional regulation, black-and-white thinking, and suicidal gestures that are more typical of BPD. Narcissistic personality disorder (Choice D) is a cluster B personality disorder that is characterized by fragile self-esteem and compensatory arrogant, self-aggrandizing behavior to gain admiration, sometimes at others' expense. Patients with BPD do not seek admiration from others, instead alternating between seeking emotional closeness and emotional distance. Patients with narcissistic personality do not typically experience the intense emotions or fear of abandonment experienced by patients with BPD. Schizoid personality disorder (Choice E), a cluster A personality disorder, manifests as extreme social detachment and a cold, restricted affect. Patients with BPD are instead dependent on others and present with labile affect. Educational Objective: BPD is a cluster B personality disorder that features an unstable sense of self, emotional dysregulation, and tumultuous relationships. To unconsciously fulfill their emotional needs, patients with BPD make suicidal gestures, desperately avoid abandonment, and either idealize or devalue other people. %3D Previous Next Score Report Lab Values Calculator Help Pause

136 Exam Section 3: Item 36 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 36. A 68-year-old woman has had progressive shortness of breath with exertion since she started treatment with eyedrops for glaucoma 2 weeks ago. Pulmonary function tests showa decreased forced expiratory volume in 1 sec (FEV,). The lungs are clear to auscultation. The eyedrops are most likely from which of the following classes of drugs? A) a-Adrenergic agonist B) B-Adrenergic agonist C) B-Adrenergic blocker D) Carbonic anhydrase inhibitor E) Cholinesterase inhibitor F) Prostaglandin analog

C. Timolol is a non-selective B-adrenergic receptor blocker and is useful for the treatment of primary open angle glaucoma. Blockade of B-adrenergic receptors in the ciliary body leads to the reduced formation of aqueous humor and a decreased intraocular pressure, which is useful for slowing the progression of optic nerve damage. Systemic side effects of topical timolol eye drops include the exacerbation of asthma, as in this patient, as well as bradycardia, decreased exercise tolerance, erectile dysfunction, and masking of hypoglycemia in patients with diabetes mellitus. Incorrect Answers: A, B, D, E, and F. a-Adrenergic agonists (Choice A) include the medications brimonidine and apraclonidine (which target az receptors), which are used in the treatment of primary open angle glaucoma. a2-Adrenergic agonists mediate their effects by decreasing aqueous humor production. Systemic side effects of these medications include dry mouth, bradycardia, and hypotension. a2-Adrenergic agonists can cause the blood brain barrier and can cause hypotonia and apnea in children. By contrast, a,-adrenergic agonists when used in the eye promote mydriasis. B-Adrenergic agonists (Choice B) include drugs such as albuterol, which are useful for the treatment of asthma. B-Adrenergic agonists are not commonly used in topical formulations and would not be useful for the treatment of glaucoma. Carbonic anhydrase inhibitors (Choice D) include topical glaucoma medications such as dorzolamide and brinzolamide, as well as oral acetazolamide. These drugs reduce the formation of aqueous humor. Systemic side effects include paresthesia, hypokalemia, acidosis, and increased risk of nephrolithiasis. Cholinesterase inhibitors (Choice E) that are used for glaucoma include echothiophate, which lowers intraocular pressure by increasing outflow of aqueous humor through the trabecular meshwork. This drug class is rarely used for the treatment of glaucoma. Prostaglandin analogs (Choice F) include latanoprost, bimatoprost, and travaprost, which lower intraocular pressure by increasing outflow of aqueous humor through the uveoscleral pathway. Topical prostaglandin therapy has few systemic side effects. Local ocular side effects include lengthening of eyelashes, change in iris color, and atrophy and hyperpigmentation of periorbital soft tissues. Educational Objective: B-adrenergic receptor blockers, such as timolol, are useful for the treatment of glaucoma by lowering intraocular pressure. Systemic side effects include the exacerbation of asthma, bradycardia, decreased exercise tolerance, erectile dysfunction, and masking of hypoglycemia. Non-selective B-adrenergic receptor blockers are relatively contraindicated in patients with a history of asthma. %3D Previous Next Score Report Lab Values Calculator Help Pause

154 Exam Section 4: Item 4 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 4. An 18-month-old boy is brought to the physician by his parents because of eye pain for 3 days. He has a history of multiple infections since the age of 3 weeks, including upper respiratory tract infections, staphylococcal osteomyelitis, and two episodes of pneumonia (one due to Staphylococcus aureus and the other due to Aspergillus fumigatus). Ophthalmologic examination shows erythema of the conjunctiva of the left eye with a small amount of purulent discharge. Physical examination shows multiple ulcerated lesions with erythematous borders over the face, neck, and back; cervical and axillary lymphadenopathy; and hepatosplenomegaly. Serum IgE, IgG, and IgM concentrations are within the reference ranges. Culture of the conjunctival secretions grows Staphylococcus epidermidis, and culture of several skin lesions grows Staphylococcus aureus. This patient most likely has which of the following immunodeficiency diseases? A) Chédiak-Higashi syndrome B) Chronic granulomatous disease C) DiGeorge syndrome D) Perinatal HIV infection E) Wiskott-Aldrich syndrome

B. Chronic granulomatous disease (CGD), an immunodeficiency syndrome caused by a defect in the nicotinamide adenine dinucleotide phosphate (NADPH) oxidase complex required for phagocytic killing of catalase-positive organisms, is characterized by recurrent pyogenic infections caused by Gram-positive and Gram-negative bacteria. Recurrent pneumonia is the most common presenting infection in patients with CGD, but skin and soft tissue infections are also frequently encountered. Common infecting bacteria include Staphylococcus species, Burkholderia cepacia, and Nocardia species. Patients with CGD are at risk for fungal infections, especially Aspergillus species. NADPH oxidase uses oxygen as a substrate for the generation of free radicals (superoxide anions). Free radical oxygen species are subsequently used for the creation of hydrogen peroxide and hypochlorous acid. Activation of this pathway leads to the respiratory burst, which results in bacterial death. Deficiency of NADPH oxidase renders phagocytes unable to neutralize catalase-positive bacteria, which can neutralize their own hydrogen peroxide, thus leaving the host cells without a substrate necessary to complete the respiratory burst. Diagnosis is made by an abnormal dihydrorhodamine test or a nitroblue tetrazolium reduction test. In this latter test, normal phagocytes use the action of NADPH to reduce nitroblue, which leads to a color change from yellow to blue. Patients with CGD will not demonstrate color change. Incorrect Answers: A, C, D, and E. Chédiak-Higashi syndrome (Choice A) is a rare, autosomal recessive disorder of the immune system caused by mutations in the lysosomal trafficking regulator gene (LYST) that encodes a protein essential for normal formation and transportation of lysosomes within the cell. The clinical manifestations include frequent bacterial infections, oculocutaneous albinism, peripheral neuropathy, and progressive neurologic dysfunction. DiGeorge syndrome (Choice C) is caused by a 22q11 deletion that leads to failure of the third and fourth branchial pouches to develop. These structures give rise to the thymus and parathyroid glands, and their absence leads to thymic aplasia and hypocalcemia in the setting of cardiac and craniofacial defects. Patients are susceptible to viral and fungal infections. The definitive diagnosis requires fluorescence in-situ hybridization to detect the missing DNA segment. Treatment is supportive. Perinatal HIV infection (Choice D) is likely to occur in situations in which the mother did not receive antiretroviral therapy prior to or during pregnancy, or in instances when the HIV viral load is detectable. Most patients do not develop clinical manifestations of HIV infection in resourced settings as diagnosis is made quickly and patients are started on antiretroviral therapy to prevent the development of AIDS. Wiskott-Aldrich syndrome (Choice E) is caused by a mutation in the WAS gene on the X-chromosome that encodes a protein essential for actin cytoskeleton rearrangement that occurs during interactions between T lymphocytes, antigen-presenting cells, and B lymphocytes, leading to an impaired innate and adaptive immune system. The phenotype is variable but classically includes eczema, thrombocytopenia, and infections with encapsulated bacteria and opportunistic pathogens. Educational Objective: CGD is caused by NADPH oxidase deficiency and results in impaired intracellular killing of pathogens by phagocytes. This presents as recurrent pyogenic infections with catalase-positive organisms such as Staphylococcus aureus. %3D Previous Next Score Report Lab Values Calculator Help Pause

134 Exam Section 3: Item 34 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 34. A healthy 24-year-old woman participates in a study to determine the role of ghrelin in appetite. She is given free access to food during the study. The graph shows plasma concentrations of ghrelin in the woman during the study period. Which of the following labeled points on the curve most likely represents the consumption of a meal? IN A E Time A) B) C) D) E)

B. Ghrelin is an orexigenic peptide hormone produced by gastrointestinal enteroendocrine cells. Ghrelin activity stimulates hunger and activates hypothalamic and pituitary neural reward circuits by binding to growth hormone secretagogue receptor 1A. Ghrelin concentration increases during the preprandial period and begins to decline immediately after the ingestion of food during the postprandial period. Ghrelin concentration may also be influenced by circadian rhythm. Ghrelin plays a role in obesity in patients with Prader-Willi syndrome, in whom ghrelin concentration is increased. Ghrelin is antagonized by leptin, which is produced by adipocytes and enterocytes in the fed state in order to reduce hunger and increase satiety. Incorrect Answers: A, C, D, and E. Choice A demonstrates an increasing but not yet maximal ghrelin plasma concentration, which is observed prior to the ingestion of a meal in order to stimulate hunger. Choice C demonstrates a decreasing ghrelin plasma concentration, which is observed immediately after ingestion of a meal in order to promote satiety. Choices D and E represent non-maximal fluctuations in ghrelin serum concentration; ghrelin concentration is the highest in the preprandial period. Educational Objective: Ghrelin is an orexigenic peptide hormone that stimulates hunger and activates hypothalamic and pituitary neural reward circuits. Ghrelin concentration increases during the preprandial period prior to a meal and begin to decline immediately after the ingestion of food during the postprandial period. II Previous Next Score Report Lab Values Calculator Help Pause Plasma ghrelin concentration

133 Exam Section 3: Item 33 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 33. A78-year-old man is brought to the emergency department after falling from a ladder. On arrival, he is lucid. Physical examination shows a severe scalp contusion and laceration. Neurologic examination shows no abnormalities. On the second day, he is stuporous and has a headache. Examination shows slight weakness, hyperreflexia, and an extensor plantar reflex, all on the right. ACT scan of the head is shown. The most likely cause of these findings is trauma to which of the following vascular structures? A) Artery in the subarachnoid space B) Dural bridging veins C) Middle cerebral artery D) Middle meningeal artery E) Superior sagittal sinus

B. Head trauma, the use of anticoagulants, physical abuse, alcohol use disorder, cerebral atrophy, and malignancy are associated with the risk of developing a subdural hematoma (SDH). On CT scan, SDH appears as a crescent-shaped fluid collection abutting the internal surface of the skull, not bound by suture lines. In some cases, the SDH may be large enough to compress adjacent brain parenchyma; the displacement of the cerebral hemisphere can result in neurologic deficits, altered mental status, seizures, herniation syndromes, and signs of increased intracranial pressure. Patients may manifest upper motor neuron exam findings such as spasticity, hyperreflexia, and extensor plantar reflex (Babinski sign). Elderly patients are at risk for cerebral atrophy, which places traction on dural bridging cortical veins. In this setting, minor trauma can result in shearing of the veins leading to hemorrhage. Untreated, the subdural hematoma presents the risk of continuing to grow. The intracranial compartment is a fixed space within a rigid skull; pathologic lesions such as SDH can result in midline shift and herniation of central nervous system structures if large enough. When such mass effect is present or at risk of developing, surgical interventions such as a decompressive burr hole or craniotomy may be indicated. Incorrect Answers: A, C, D, and E. Subarachnoid hemorrhages can be due to aneurysmal rupture of an artery in the subarachnoid space (Choice A) or from traumatic injury. Patients may present with an acute, severe headache, stupor, or coma in severe cases. CT scan of the head demonstrates hyperdense blood products in the cerebral sulci and basal cisterns, not the crescent-shaped collection seen in SDH. The middle cerebral artery (Choice C) is a branch of the internal carotid artery that supplies blood to the temporal lobe, frontal lobe, and parietal lobe. It is unlikely to be injured in trauma. In the case of dissection or thrombosis of the middle cerebral artery, it classically presents with contralateral motor and sensory deficits of the face and upper extremity, and aphasia (eg, Wernicke, Broca) if involving the dominant hemisphere. Epidural hematoma is classically associated with trauma to the middle meningeal artery (Choice D). Epidural hematomas present on CT scan as lens-shaped, bi-convex hyperdense collections abutting the skull. They often compress adjacent brain and are bound by dural sutures. The superior sagittal sinus (Choice E) functions in venous drainage of the brain and lies along the superior midline between the two cerebral hemispheres. Superior sagittal sinus injury is rare and can be due to skull fracture. It can have a similar appearance to an epidural hematoma. Educational Objective: Elderly patients are at risk for cerebral atrophy, which causes traction on dural bridging cortical veins such that minor trauma can result in venous injury leading to a subdural hematoma. Subdural hematomas appear as crescent-shaped fluid collections abutting the internal surface of the skull, not bound by suture lines. II Previous Next Score Report Lab Values Calculator Help Pause

138 Exam Section 3: Item 38 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 38. An 18-year-old man comes to the physician because of nonpitting facial edema and difficulty breathing several hours after a dental procedure. He has a history of unexplained episodes of abdominal pain. His serum C2 concentration is found to be markedly decreased. Which of the following best explains this patient's illness? A) Absent pyrin B) Decreased C1 esterase inhibitors C) Increased urine porphyrins D) Penicillin allergy E) Superantigen production

B. Hereditary angioedema is a disorder of the complement system that results from decreased C1 esterase inhibitors and is characterized by overactivation of the kallikrein-kinin pathway. Kallikrein activates bradykinin, which causes vasodilation and increases vascular permeability. C1 esterase is responsible for kallikrein inhibition. Unregulated kallikrein activity results in an excess of bradykinin and angioedema. The disorder typically presents early in life with recurrent episodes of angioedema, which may include the face, oropharynx, extremities, or gastrointestinal tract. Symptoms depend upon the affected area and include dyspnea secondary to upper airway obstruction, abdominal pain, and gastrointestinal distress. Physical examination displays edema and the absence of urticaria, which can help distinguish hereditary angioedema from an anaphylactic reaction. Laboratory analysis classically reveals decreased levels of C2 and C4. Diagnosis can be confirmed by direct measurement of C1 esterase inhibitor concentration and functional assays. Airway compromise can be life-threatening and immediate treatment is indicated to secure the airway. Epinephrine and antihistamines are typically administered until anaphylaxis has been ruled out. Patients should be treated with plasma-derived or recombinant C1 inhibitor as soon as possible. Incorrect Answers: A, C, D, and E. Absent pyrin (Choice A) is the cause of familial Mediterranean fever. Patients typically present in childhood with recurrent fever, abdominal pain, and arthralgias. It is not associated with facial edema. Increased urine porphyrins (Choice C) suggest the presence of porphyria, which is a group of disorders of heme synthesis. These disorders are characterized by accumulation of heme precursors and may present with dark colored urine, gastrointestinal distress, and blistering of the skin. C2 levels would not be affected. Penicillin allergy (Choice D) may present with anaphylaxis, which can be difficult to distinguish from hereditary angioedema in the acute setting. The presence of decreased C2 levels is more suggestive of hereditary angioedema. Superantigen production (Choice E) occurs in severe complications of staphylococcal and streptococcal infections such as toxic shock syndrome, scalded skin syndrome, and scarlet fever. Superantigens stimulate a massive immune response resulting in widespread inflammation. Educational Objective: Hereditary angioedema results from a deficiency of C1 esterase inhibitors and may present with angioedema of the face, extremities, oropharynx, and gastrointestinal tract due to increased levels of bradykinin. %3D Previous Next Score Report Lab Values Calculator Help Pause

117 Exam Section 3: Item 17 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 17. A 23-year-old man cuts his lip inadvertently while shaving. Seconds after the injury, the bleeding nearly stops. Which of the following mechanisms is the most likely cause of the early rapid control of blood loss in this man? A) Activation of antithrombin III B) Localized secretion of endothelin C) Oxygen-stimulated cleavage of thromboplastin D) Polymerization of fibrin

B. Localized secretion of endothelin from endothelial cells after tissue injury is stimulated by thrombin, which accumulates at the site of bleeding as a normal part of the coagulation cascade. Thrombin is activated by factor Xa at the convergence of the intrinsic and extrinsic clotting cascades after which it converts fibrinogen to fibrin, allowing for fibrin polymerization and clot formation. Thrombin also stimulates the release of endothelin from endothelial cells. Endothelin is a short peptide made by the vascular endothelium via the action of endothelin converting enzyme and binds to its G protein-coupled receptors (ÉTA and ETB) on vascular smooth muscle. This induces vasoconstriction via the intracellular release of calcium. Vasoconstriction, in addition to clot formation, is an important step in preventing further blood loss by increasing vascular resistance and thereby diminishing flow. Incorrect Answers: A, C, and D. Activation of antithrombin III (AT-III) (Choice A) allows AT-III to bind to factors Xa and lla (thrombin), as well as to factors IXa, Xla, XIla, VIlla, kallikrein, and plasmin, thereby exerting a broad inhibitory effect on the clotting cascade. Heparin, a medication used commonly to treat blood clots, augments the activity of AT-III. Therefore, activation would result in prolonged bleeding. Oxygen-stimulated cleavage of thromboplastin (Choice C) would result in inactivation of thromboplastin and inhibition of the clotting cascade. Thromboplastin is a complex of lipids and tissue factor that has binding sites for factor VII. Endothelial or tissue damage exposes tissue factor and subendothelial collagen, which bind factor VII leading to its activation to factor Vlla. Activation is stimulated by hypoxic conditions, not by the presence of oxygen. Polymerization of fibrin (Choice D) is the final step in the coagulation cascade and serves to build the fibrin superstructure to which platelets and erythrocytes can bind to form a mature clot. This is important in stopping bleeding but occurs later than the immediate vasoconstriction that results from the release of endothelin. Educational Objective: Endothelin release from endothelial cells in response to tissue injury causes vasoconstriction that aids in reducing blood flow. It is an early response to injury and occurs before other components of the clotting cascade, such as the polymerization of fibrin. %3D Previous Next Score Report Lab Values Calculator Help Pause

144 Exam Section 3: Item 44 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 44. A 65-year-old man dies of renal failure. Urinalysis shows: Blood Protein negative negative negative positive Glucose Bence Jones protein At autopsy, the kidneys are found to be enlarged. A histologic section of renal tissue is shown in the photomicrograph. Which of the following is the most likely diagnosis? A) Acute tubular necrosis B) Amyloidosis C) Lipoid nephrosis D) Membranous glomerulonephritis E) Pyelonephritis

B. Multiple myeloma is a malignancy caused by the neoplastic proliferation of plasma cells. Neoplastic plasma cells overproduce monoclonal immunoglobulin and light or heavy chains, which may result in acute kidney injury. Patients also commonly present with symptoms of fatigue and weight loss, symptoms of hypercalcemia (eg, abdominal cramping, kidney stones, psychiatric disturbance), symptoms of anemia (eg, pallor, lightheadedness, dyspnea on exertion), or with opportunistic infections secondary to immune dysfunction. Multiple myeloma results in primary amyloidosis due to overproduction and deposition of immunoglobulin chains in the kidneys, heart, brain, and other organs. Renal amyloid deposition on biopsy reveals deposits in the mesangium. Evaluation under polarized light with Congo red stain reveals green birefringent crystals. Evaluation for multiple myeloma may also include urinalysis revealing Bence Jones protein, or immunoglobulin light chains, peripheral blood smear showing rouleaux formation of erythrocytes, and an M spike on serum protein electrophoresis. Incorrect Answers: A, C, D, and E. Acute tubular necrosis (Choice A) typically occurs following an ischemic or nephrotoxic insult to the kidneys, which results in necrosis of the tubular epithelium. Granular, muddy brown casts are typical on urinalysis. Lipoid nephrosis (Choice C) or minimal change disease and membranous glomerulonephritis (Choice D) or membranous nephropathy are nephrotic syndromes, which are characterized by proteinuria (>3g/day), edema, hypoalbuminemia, and hypercholesterolemia. Minimal change disease appears normal on light microscopy and is the most common cause of nephrotic syndrome in children. It can be idiopathic or triggered by a recent infection or immune stimulus. A biopsy of membranous nephropathy usually shows diffuse thickening of the glomerular capillary wall on light microscopy. On immunofluorescence, IgG deposition will occur along the glomerular basement membrane. Pyelonephritis (Choice E) typically presents with fever, nausea, vomiting, dysuria, and flank pain with associated costovertebral angle tenderness on physical examination. This most commonly occurs from ascending urinary tract infections. Educational Objective: Multiple myeloma is a malignancy caused by the neoplastic proliferation of plasma cells, which results in the abnormal production of immunoglobulin light and heavy chains. Deposition of immunoglobulin leads to renal amyloidosis and Bence Jones protein on urinalysis. %3D Previous Next Score Report Lab Values Calculator Help Pause

143 Exam Section 3: Item 43 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 43. A 52-year-old man with lung cancer comes to the physician for a follow-up examination. Smoking cessation has been recommended, but the patient says that he is unable to quit because he experiences a pleasurable sensation as he inhales the smoke. He has smoked 2 packs of cigarettes daily for 30 years. Which of the following mechanisms best explains how nicotine produces this sensation leading to addiction in this patient? A) Activation of opioid receptors in the midbrain B) Increased release of dopamine in the nucleus accumbens C) Inhibition of y-aminobutyric acid (GABA) release in the hypothalamus D) Inhibition of glutamate receptors in the amygdala E) Potentiation of GABA receptors in the cerebral cortex

B. The release of dopamine in the nucleus accumbens mediates anticipatory reward. This patient's pleasure immediately after inhaling is likely related to this anticipatory dopamine signaling rather than the effect of the nicotine itself. The ventral tegmental area in the midbrain projects dopaminergic fibers to the nucleus accumbens, which is part of the limbic system, called the mesolimbic pathway. All drugs of abuse increase extracellular dopamine in the nucleus accumbens. The mesolimbic dopamine pathway is therefore thought to underlie all addictions, as anticipatory reward can manifest as craving and lead to drug-seeking behavior at the expense of a patient's health or social functioning. Incorrect Answers: A, C, D, and E. Activation of opioid receptors in the midbrain (Choice A) may indirectly mediate reward by increasing midbrain dopamine signaling. However, dopamine itself is the direct mediator of reward and addiction. Inhibition of y-aminobutyric acid (GABA) release in the hypothalamus (Choice C) would result in the modulation of appetite, sleep, and other homeostatic activities. The relationship between environmental stress and relapse on drugs of abuse may be mediated by the hypothalamus. However, the mesolimbic pathway plays a larger role in maintaining addictive behaviors. The inhibition of glutamate receptors in the amygdala (Choice D) would lead to decreased fear learning. Repeated alcohol use leads to upregulation of glutamate receptors in the amygdala, which may increase anxiety during withdrawal and lead to relapse. Thus, inhibiting glutamate signaling may decrease substance use. Potentiation of GABA receptors in the cerebral cortex (Choice E) mediates the excitatory-inhibitory balance in the cortex (important for preventing epilepsy). Addiction is postulated to result from insufficient GABAergic signaling in several brain regions. Educational Objective: The mesolimbic pathway refers to the projection of dopaminergic neurons from the ventral tegmental area to the nucleus accumbens. This dopaminergic pathway mediates the anticipatory reward of drugs of abuse and addictive behaviors. %3D Previous Next Score Report Lab Values Calculator Help Pause

200 Exam Section 4: Item 50 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 50. A 34-year-old woman comes to the physician because she has been feeling weepy and overwhelmed since delivering a healthy male newborn 6 weeks ago. She has had fatigue and irritability during this period, and she has had no interest in engaging in activities she used to enjoy prior to the birth of the baby. She also has had difficulty sleeping because she is always listening for sounds indicating that her baby is awake. She says, "My husband doesn't help me at all with the baby. We argue all the time now." She then says, "I feel guilty because l'm not enjoying my baby more. I don't know if I will be a good mother or not." Physical examination shows no abnormalities. Which of the following statements by the physician is most appropriate? A) "Has it come as a surprise to you how hard parenting is? Many people feel that way." B) "I'm concerned about how bad you've been feeling lately. Have you had any thoughts about death or wanting to be dead?" C) "I'm sure that you'll feel better soon. After all, look at your beautiful baby." D) "Most new mothers feel this way, and sometimes it helps to see a therapist. Would you like me to give you a referral?" E) "Your relationship with your husband sounds strained. I don't blame you for being angry with him."

B. This patient meets criteria for major depressive disorder, postpartum onset and should be immediately evaluated for suicidal ideation. Major depressive disorder, postpartum onset presents with the typical symptoms of major depressive disorder (depressed mood, decreased energy, anhedonia, sleep disturbances, appetite disturbances, guilt, poor concentration) and may be delayed by as much as 12 months after delivery. Risk factors for postpartum depression include sleep disturbance, childcare stress, decreased estrogen levels, breastfeeding difficulty, body image dissatisfaction, and poor postpartum social support. Consequences of postpartum depression can be severe and include impaired bonding with the child, child neglect, suicide, and infanticide. Patients with postpartum depression should be proactively and expeditiously monitored for these sequelae. Treatment of moderate to severe postpartum depression typically includes antidepressant medication, which poses minimal risk to breastfeeding infants, and inpatient admission in high-risk situations. Incorrect Answers: A, C, D, and E. Normalizing the patient's experience (Choices A and D) may help the patient feel emotionally validated in the moment, and referral to a therapist (Choice D) may assist in long-term improvement. However, the patient and infant's safety are the most acute issues and should be immediately assessed. Telling the patient that she will feel better soon (Choice C) is a less effective way of validating a patient's emotions than normalizations or reflective statements. Additionally, the patient and infant's safety are the most acute issues and should be immediately assessed. Exploring the patient's relationship with her husband (Choice E) would not address the safety of this patient and her infant in an expeditious manner. Further, interpersonal dynamics appear to play only a small role in this patient's depression. Educational Objective: Major depressive disorder, postpartum onset presents with symptoms of major depressive disorder in the 12 months following delivery and can be associated with serious consequences such as suicide and infanticide. All patients with major depressive disorder, postpartum onset should be assessed expeditiously for suicidal and homicidal ideation. %3D Previous Next Score Report Lab Values Calculator Help Pause

179 Exam Section 4: Item 29 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 29. A 1-year-old African American girl is brought to the physician because of a slow growth rate. At birth she was at the 25th percentile for length and 30th percentile for weight. She has been exclusively breast-fed since birth. Her length and weight are now at the 5th percentile. Physical examination shows bowed legs, swelling of the wrists, and a bulging of the costochondral junctions. Her serum calcium concentration is within the reference range, and her serum phosphorus concentration is decreased. Which of the following sets of serum findings is most likely in this patient? Parathyroid Hormone 25-Hydroxyvitamin D A) ↑ B) C) Normal OD) normal OE) ↑ OF)

B. Vitamin D plays a role in serum calcium and phosphate homeostasis by promoting intestinal absorption of calcium and phosphate, increasing bone mineralization at low levels, and bone resorption at higher levels. Parathyroid hormone (PTH) also has a role in calcium and phosphate regulation by stimulating osteoclastic bone reabsorption and distal convoluted tubular calcium reabsorption and phosphate excretion in the kidney. Vitamin D deficiency can be caused by malabsorption in the intestinal tract, malnutrition or insufficient dietary intake, and decreased sun exposure. Decreased levels of vitamin D result in decreased intestinal calcium absorption and hypocalcemia, sensed by the parathyroid gland via calcium-sensing receptors, leading to an increase in secretion of PTH to normalize serum calcium levels. Increased PTH results in increased serum calcium from increased osteoclast activity, intestinal absorption, and renal tubular absorption. Serum phosphorus concentration is decreased secondary to increased excretion by the renal tubules. This leads to rickets in children, characterized by bone pain and deformities such as bowing of the legs, bulging of costochondral junctions, and growth delay. Because of the risk for vitamin D deficiency, breast-fed infants are recommended to have oral vitamin D supplementation. Incorrect Answers: A, C, D, E, and F. Vitamin D deficiency and rickets are characterized by low serum concentration of vitamin D, not normal or increased (Choices A, D, and E). Decreased levels of vitamin D result in decreased intestinal calcium absorption and hypocalcemia, sensed by the parathyroid gland via calcium-sensing receptors, leading to an increase in the secretion of PTH to normalize serum calcium concentration, not a decrease in PTH or normal PTH concentration (Choices C, D, E, and F). Educational Objective: Vitamin D deficiency in breast-fed infants can lead to rickets, characterized by bone pain and deformities such as bowing of the legs, bulging of costochondral junctions, and growth delay. Decreased concentration of vitamin D result in decreased intestinal calcium absorption and hypocalcemia, sensed by the parathyroid gland via calcium-sensing receptors, leading to an increase in the secretion of PTH to normalize serum calcium concentration. %3D Previous Next Score Report Lab Values Calculator Help Pause

104 Exam Section 3: Item 4 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 4. An 18-year-old woman with mild intellectual disability is brought to the physician because of a 3-day history of decreased ability to see in reduced light. She has a lifelong history of chronic diarrhea. Two years ago, she developed a lack of muscle control of her arms and legs, and generalized weakness. Her 16-year-old brother has had similar symptoms. Ophthalmologic examination shows bilateral retinitis pigmentosa. There is ataxia and loss of deep tendon reflexes. Laboratory studies show erythrocytes with spiny projections and a serum total cholesterol concentration of 40 mg/dL. Which of the following apolipoproteins is most likely deficient in this patient? A) Apo A-I B) Apo A-II C) Apo B D) Apo C E) Apo E

C. Apo B deficiency resulting in abetalipoproteinemia, a rare autosomal recessive disorder that is caused by a mutation in the microsomal triglyceride transfer protein (MTP) gene, is the most likely explanation for this patient's symptoms. Most mutations result in reduced function or complete absence. As MTP is required for the assembly and secretion of Apo B in the liver and gastrointestinal tract, an absence of this protein results in an absence of Apo B. Apo B is a necessary component of intermediate-density lipoprotein (IDL), VLDL, and LDL, so deficiency results in an inability to assemble these critical lipoprotein particles. This causes severe malabsorption of dietary fats and fat-soluble vitamins such as vitamin A, D, E, and K. Symptoms are present in infancy as failure to thrive and mental retardation. Subsequent symptoms can include peripheral neuropathies, retinitis pigmentosa leading to complete blindness, and muscular weakness. Additionally, chronic malabsorptive diarrhea is characteristic and is a component of the pathophysiology of this disease. Incorrect Answers: A, B, D, and E. Apo A-I (Choice A) is required for assembly of HDL and acts as a ligand for ABCA1 binding. Deficiencies result in absent HDL, corneal opacities, xanthomas, and early atherosclerosis. Apo A-IlI (Choice B) is also a component of HDL. Deficiency would result in a reduced amount of circulating HDL. Neither deficiency results in the symptoms of this patient. Apo C (Choice D) consists of three separate proteins: C-I, C-II, and C-III. Deficiency of C-Il results in hyperlipoproteinemia type IB. Clinical manifestations closely resemble lipoprotein lipase deficiency with profound hypertriglyceridemia. Apo E (Choice E) deficiency, such as limited Apo E2, can lead to dysbetalipoproteinemia as a result of reduced clearance of VLDL and chylomicrons whereas Apo E4 deficiency results in hypercholesterolemia and premature atherosclerotic heart disease. Educational Objective: Apo B is an essential component of LDL, IDL, and VLDL. It is required for the normal absorption and trafficking of lipids to and from the intestines and the liver. Mutations in the MTP gene lead to absent or reduced levels of Apo B, with consequent chronic malabsorptive diarrhea, fat-soluble vitamin deficiencies, and neurologic manifestations such as peripheral neuropathy, retinitis, and mental retardation. %3D Previous Next Score Report Lab Values Calculator Help Pause

171 Exam Section 4: Item 21 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 21. A21-year-old woman with primary pulmonary hypertension begins treatment with bosentan. As a result, blockade of which of the following is most likely to occur? A) Angiotensin |l receptors B) Calcium channels C) Endothelin receptors D) Production of phosphodiesterase 5 E) Voltage-gated sodium channels

C. Bosentan is a competitive endothelin-1 receptor antagonist, which is used for the treatment of primary pulmonary arterial hypertension (PAH). Two clinically relevant subtypes of endothelin receptors are the endothelin A (ETA) and the endothelin B (ETB) receptors. Binding of endothelin-1 to ETA results in vasoconstriction, whereas binding of endothelin-1 to ÉTB results in production of nitric oxide and vasodilation. In the context of PAH, the aggregate effect of endothelin-1 results in vasoconstriction and an increase in pulmonary vascular resistance. Bosentan competitively antagonizes the binding of endothelin-1 to both ETA and ETB receptors. The net effect of blockade is vasodilation with reduced pulmonary vascular resistance, limiting pulmonary hypertension and relieving strain on the right side of the heart. Bosentan is hepatotoxic, and patients treated with bosentan should be monitored for elevations of serum transaminases. Incorrect Answers: A, B, D, and E. Blockade of angiotensin II receptors (Choice A) is the mechanism of action of drugs such as losartan, valsartan, and candesartan. Angiotensin receptor blockers are not useful for the treatment of PAH. Blockade of calcium channels (Choice B) is the mechanism of action of drugs such as amlodipine, nifedipine, verapamil, and diltiazem. While these drugs are useful for the treatment of systemic hypertension and of supraventricular dysrhythmias (eg, atrial fibrillation), they are not useful for treatment of PAH. Blocking production of phosphodiesterase 5 (Choice D) is the mechanism of action of sildenafil and vardenafil. While sildenafil is used for the treatment of PAH, blockade of phosphodiesterase 5 is not the mechanism of action of bosentan. Blockade of voltage-gated sodium channels (Choice E) is the mechanism of action of antiarrhythmic agents such as lidocaine. Sodium channel blockers are also useful for the treatment of epilepsy and for local anesthesia. Educational Objective: Bosentan is a competitive endothelin-1 receptor antagonist and causes reduced pulmonary vascular resistance. It is effective in the treatment of pulmonary arterial hypertension. %3D Previous Next Score Report Lab Values Calculator Help Pause

147 Exam Section 3: Item 47 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 47. An investigator is studying a T-lymphocyte clone that recognizes a peptide from the hemagglutinin glycoprotein of influenza virus bound by class II human leukocyte antigen. This clone most likely recognizes a peptide that was combined with a human leukocyte antigen in which of the following cellular compartments? A) Cytosol B) Endoplasmic reticulum C) Endosomes D) Mitochondria E) Nucleus

C. Endosomes are cytoplasmic organelles composed mainly of microtubules and vesicles that traffic and sort endocytosed material, including vaccines and live viruses. Endosomes exist in several forms (early, late, recycling) and can traffic endocytosed material to the Golgi complex, to lysosomes, or to the extracellular membrane. The class II human major histocompatibility complex (MHC) proteins, also known as class II human leukocyte antigen (HLA) proteins, are synthesized in the rough endoplasmic reticulum and are trafficked to the Golgi complex and then to late endosomes, where they form high-affinity combinations with degraded, endocytosed proteins. Bound protein is then recycled to the cell surface membrane where the peptide fragments are presented on class II HLA/MHC proteins to T lymphocytes. The T-lymphocyte receptor identifies the peptide bound to the human leukocyte antigen and initiates or sustains an immune response. Incorrect Answers: A, B, D, and E. The cytosol (Choice A) is the site of degradation of cytoplasmic peptides via the proteasome, which are then combined with class I MHC proteins in the endoplasmic reticulum and shuttled to the plasma membrane for presentation. An absence of class I MHC proteins on the cell surface is an initiating trigger for natural killer cell-induced cell destruction. Endoplasmic reticulum (Choice B) is the site of combination of target peptides with class I MHC proteins. Mitochondria (Choice D) do not play a significant role in the processing of endocytosed material. The nucleus (Choice E) is the site of MRNA synthesis but is not a major site of trafficking of endocytosed material. Educational Objective: Endosomes are cytoplasmic organelles composed mainly of microtubules and vesicles and serve to traffic and sort endocytosed material. Endocytosed viral protein activates the immune system through combination with HLA proteins and subsequent presentation to T lymphocytes. %3D Previous Next Score Report Lab Values Calculator Help Pause

140 Exam Section 3: Item 40 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 40. A 54-year-old woman with chronic kidney disease comes to the physician because of a 2-week history of itchy skin, increasingly severe nausea, and decreased urination. Current medications are hydrochlorothiazide and lisinopril. Her pulse is 80/min, respirations are 14/min, and blood pressure is 160/90 mm Hg. Physical examination shows no abnormalities except for excoriations from itching. Laboratory studies show: 9.8 g/dL 30.1% 89 um3 5600/mm3 Hemoglobin Hematocrit Mean corpuscular volume Leukocyte count Platelet count 287,000/mm3 Which of the following is the most likely cause of the findings in this patient? A) Hypoalbuminemia OB) Hypothyroidism C) Uremia D) Vitamin C deficiency E) Vitamin K deficiency

C. Gradual decline in kidney function in patients with chronic kidney disease can be asymptomatic until patients may present with volume overload, electrolyte abnormalities, anemia, hypertension, and uremia. The patient takes hydrochlorothiazide, a diuretic that inhibits sodium chloride reabsorption in the distal collecting tubule of the nephron. An adverse effect of hydrochlorothiazide is increased levels of uric acid. Uremia can present with nonspecific symptoms including nausea, vomiting, and pruritus. On physical examination, patients may have asterixis, encephalopathy, and pericardial rub in pericarditis. Platelet count may be normal, but uremia leads to platelet dysfunction and prolonged bleeding time which can manifest with easy bruising and mucocutaneous bleeding. Incorrect Answers: A, B, D, and E. Hypoalbuminemia (Choice A) can be associated with nephrotic syndrome. Due to the excessive proteinuria in nephrotic syndrome, serum oncotic pressure is decreased, resulting in dependent edema in the buttocks, lower back, and legs. Hypothyroidism (Choice B) presents with weight gain, fatigue, constipation, cold intolerance, and menstrual irregularities. Physical examination may reveal bradycardia, dry, edematous skin, and delayed relaxation of deep tendon reflexes. This patient does not demonstrate any signs or symptoms to suggest hypothyroidism. Vitamin C deficiency (Choice D) causes the constellation of symptoms known as scurvy, which is characterized by petechiae, perifollicular hemorrhage, bruising, poor wound healing, and short, curly, fragile hairs. Vitamin C deficiency results from dietary insufficiency, which is rare in resource-rich settings. Vitamin K plays a critical role in the synthesis of hepatic coagulation proteins; it is oxidized in the liver during carboxylation of glutamic acid residues on coagulation factors II, VII, IX, X, and proteins C and S. Vitamin K deficiency (Choice E) leads to coagulopathy with increased prothrombin time and activated partial thromboplastin time and is characterized by easy bruising. Educational Objective: Chronic kidney disease and a gradual decline in kidney function can lead to uremia. Uremia can present with nonspecific symptoms including nausea, vomiting, and pruritus. %3D Previous Next Score Report Lab Values Calculator Help Pause

115 Exam Section 3: Item 15 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 15. A74-year-old man comes to the office because of a 3-week history of progressive shortness of breath with exertion. He has had difficulty climbing one flight of stairs. He also has a 20-year history of poorly controlled hypertension, for which he rarely takes his prescribed medication, and osteoarthritis treated with naproxen daily. His temperature is 37.1°C (98.8°F), pulse is 76/min, respirations are 20/min, and blood pressure is 180/98 mm Hg. Pulse oximetry on room air shows an oxygen saturation of 92%. Physical examination shows 10-cm jugular venous distention above the sternal angle. Crackles are heard over both lung bases. An S, is heard. Echocardiography shows a normal ejection fraction. Which of the following best explains this patient's symptom? A) Decreased myocardial oxygen supply B) External compression of the right ventricle C) Impaired left ventricular relaxation D) Increased turbulent flow across the mitral valve E) Obstruction of the left ventricular outflow tract

C. Hypertension is a common cause of chronic systolic and diastolic left-sided heart failure and is considered a modifiable risk factor. Chronic hypertension may result in left ventricular hypertrophy due to constant work against an increased afterload. The thickened myocardium results in impaired left ventricular relaxation and increased diastolic filling pressures that lead to left atrial enlargement and right-sided heart failure over time. This patient's history, signs, symptoms, and exam findings are consistent with chronic left- sided and right-sided heart failure. On physical exam, there is an S4 gallop suggestive of a stiff and noncompliant left ventricle. Jugular venous distension and pulmonary crackles are suggestive of impaired forward flow of blood. An ECG may show evidence of left ventricular hypertrophy and left atrial enlargement. A normal ejection fraction on echocardiography indicates that this is diastolic heart failure (also called heart failure with preserved ejection fraction). Adequate control of blood pressure decreases afterload and reduces the degree of cardiac myocyte hypertrophy, which decreases the risk of developing heart failure. Incorrect Answers: A, B, D, and E. Decreased myocardial oxygen supply (Choice A) results in angina and may be due to a non-obstructing coronary plaque (with luminal narrowing), acute coronary syndrome, or coronary artery dissection, aneurysm, or vasospasm. Echocardiography may show hypokinetic or akinetic regions in the territory experiencing the decreased supply. External compression of the right ventricle (Choice B) may occur in the setting of a mediastinal mass, pectus excavatum, constrictive pericarditis, or cardiac tamponade and would result in right-sided heart failure due to the inability of the right ventricle to expand and contract fully. In an otherwise unrestricted pericardium, left ventricular hypertrophy does not impede on right ventricular expansion. Increased turbulent flow across the mitral valve (Choice D) may occur in mitral stenosis or regurgitation, generating the characteristic murmurs. Mitral regurgitation is commonly associated with mitral valve prolapse or prior myocardial infarction and presents with a holosystolic murmur best heard in the left fourth or fifth intercostal space along the midclavicular line with radiation to the left axilla. Obstruction of the left ventricular outflow tract (Choice E) may occur in hypertrophic cardiomyopathy. Diastolic dysfunction is common due to a hypertrophic and noncompliant left ventricle. Echocardiography will reveal a thickened interventricular septum and ventricular walls. Left ventricular hypertrophy due to hypertension does not typically obstruct the outflow tract. Educational Objective: Left ventricular hypertrophy is a risk factor for diastolic heart failure due to a stiff and noncompliant left ventricle. %3D Previous Next Score Report Lab Values Calculator Help Pause

177 Exam Section 4: Item 27 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 27. A study is conducted to determine if concomitant oral administration of ciprofloxacin and sucralfate results in a decreased plasma ciprofloxacin concentration. Twelve subjects are enrolled in the study; six receive ciprofloxacin only, and six receive both treatments. The half-life is determined, and a 1-week washout period follows. Each subject then receives the other treatment, and half-life is measured again. Which of the following best describes the design of this study? A) Case-control B) Case series C) Crossover D) Historical cohort E) Prospective cohort

C. In a crossover study, patients will be switched between treatment cohorts during the study duration. In this method, each patient can serve as their own control. As described in the study here, a washout period is used to allow for residual effects of the prior treatment to be adequately removed from the body. Because each patient acts as their own independent control, this study design is efficient, as it requires a smaller number of patients than would a randomized controlled trial to achieve the same power. 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 conduction of larger cohort or randomized trials with sufficient power. A historical cohort study (Choice D) (retrospective cohort study) identifies a group of patients and looks back over time to identify whether an exposure is associated with an outcome of interest. In this retrospective design, the hypothesis or question is designed after the study time period has passed. For example, an orthopedic surgeon seeks to identify the risk factors associated with prosthetic joint infection. In order to accomplish this, the investigators review the records of total joint arthroplasty patients from the previous 10 years and identify diabetes mellitus and obesity as risk factors. A prospective cohort study (Choice E) identifies a group of patients and follows them over time to identify whether an exposure is associated with an outcome of interest. In a prospective design, the hypothesis and analysis protocols are established prior to the start of the study period. An example of a prospective cohort study would be following a group of 1,000 smokers for a time period of 10 years and identifying the proportion of these patients who develop pancreatic cancer to identify the risk of pancreatic cancer in smokers as compared to a control group of nonsmokers. Educational Objective: A crossover study is an efficient methodology as each patient undergoes multiple treatment strategies and serves as their own control. This study design often uses a washout period between treatment arms in order to avoid an overlap of treatment effects. %3D Previous Next Score Report Lab Values Calculator Help Pause

105 Exam Section 3: Item 5 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 5. An investigator is studying obesity in an adolescent population. Five thousand normal-weight patients are followed from the age of 10 years to the age of 15 years. At the conclusion of the study, 1100 patients meet the criteria for obesity. Which of the following best represents the incidence per 1000 patient-years of obesity during the course of this study? A) 10 B) 25 C) 44 D) 50 E) 64

C. Incidence rate is defined as the number of new cases of disease divided by the total population at risk per unit of time. It is an important epidemiologic measure that provides an understanding of how quickly the number of new cases of a disease are occurring. In this case, the incidence rate of obesity per 1000 patient years would be calculated as follows. 1100 cases of obesity divided by 5000 patients at risk, divided by the 5-year duration of the study period (1100 cases/5000 patients at risk/5 years = 0.044 cases/patient/year). This number can then be multiplied by 1000 in order to convert the units from cases per patient per year to cases per 1000 patient-years (0.044 x 1000 = 44 cases per 1000 patient-years). Incorrect Answers: A, B, D, and E. Each of these numbers (Choices A, B, D, and E) represent erroneous calculations or the use of random, incorrect numbers that do not represent the appropriate calculation of the incidence rate. Educational Objective: Incidence per 1000 patient-years is a common method of presenting incidence data. It can be calculated by dividing the number of new cases by the total number of persons in the population at risk divided by the study duration in years, and then multiplying by 1000. %3D Previous Next Score Report Lab Values Calculator Help Pause

130 Exam Section 3: Item 30 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 30. A 5-month-old girl is brought to the physician because of a 3-day history of "floppiness" when she sits. She also has had difficulty feeding and irritability since the age of 1 month. She is at the 3rd percentile for length, 50th percentile for weight, and 3rd percentile for head circumference. Her temperature is 36°C (96.8°F), pulse is 100/min, respirations are 29/min, and blood pressure is 80/50 mm Hg. Neurologic examination shows bilateral horizontal nystagmus, decreased extensor tone, hyperreflexia, and ankle clonus. Babinski sign is present bilaterally. Respiratory chain enzyme studies on cultured fibroblasts show decreased complex II activity. The most likely cause of these findings is a defect in which of the following processes? A) Coupling of electron transport to proton gradient formation B) Coupling of proton gradient to ATP synthesis C) Oxidation of succinate to fumarate D) Reduction of pyruvate to lactate E) Transfer of electrons from NADH to coenzyme Q F) Transfer of electrons from ubiquinol to cytochrome c

C. Leigh syndrome, or subacute necrotizing encephalopathy, is caused by a variety of genetic alterations, including mutations of complexes I, II, IV, or V, coenzyme Q, or the pyruvate dehydrogenase complex. It commonly presents early in life with severe neurological dysfunction and is associated with seizures, developmental delay, psychomotor regression, nystagmus, ataxia, dystonia, and weakness. Laboratory studies may show lactic acidosis, and CT scan may show necrosis in the spinal cord, brain stem, thalamus, and basal ganglia. Mortality is high, with life expectancy only being several months after diagnosis. In this patient's case, her diagnosis is due to a genetic mutation in complex II, or succinate dehydrogenase, which catalyzes the oxidation of succinate to fumarate with the assistance of FAD*. Typically, the FADH, formed from the conversion transfers electrons to an iron-sulfur protein, where they are then transferred to cytochrome b and subsequently ubiquinone. Ubiquinone (coenzyme Q) carries the electrons to complex IIII of the electron transport chain (ETC). The electrons are then used to generate ATP via ATP synthase. As this addition of electrons occurs later in the electron transport cycle than the electrons of NADH (which join at complex I), one molecule of FADH2 creates 1.5 molecules of ATP, whereas one molecule of NADH creates 2.5 molecules of ATP. Incorrect Answers: A, B, D, E, and F. Coupling of electron transport to proton gradient formation (Choice A) occurs at complex I and complex II, which are responsible for building the proton gradient that allows ATP synthase to generate ATP. It does not occur at complex II, which accounts for the decreased energy production from FADH2 introduction to the ETC. Coupling of proton gradient to ATP synthesis (Choice B) occurs at complex V, or ATP synthase. The flow of protons down their electrochemical gradient into the mitochondrial matrix provides the energy required for production of ATP via oxidative phosphorylation. Reduction of pyruvate to lactate (Choice D) is catalyzed by lactate dehydrogenase, utilizing NADH as a proton donor. Pyruvate is reduced to lactate in anaerobic metabolism, whereas it enters the tricarboxylic acid cycle in aerobic metabolism to generate NADH, FADH2, carbon dioxide, and guanosine triphosphate. Transfer of electrons from NADH to coenzyme Q (Choice E) occurs in complex I, with the transfer generating a proton gradient. NADH does not enter the ETC via complex II. Transfer of electrons from ubiquinol to cytochrome c (Choice F) occurs at complex Il, with the transfer generating a proton gradient. This patient has reduced activity of complex II, rather than complex III. Educational Objective: Complex II of the ETC, or succinate dehydrogenase, generates FADH2 via the oxidation of succinate to fumarate. The electrons from FADH2 then enter the ETC to generate ATP via oxidative phosphorylation. Deficiencies in complex II can cause severe neurological disease, such as Leigh syndrome. %3D Previous Next Score Report Lab Values Calculator Help Pause

186 Exam Section 4: Item 36 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 36. A 6-week-old male infant has a persistent discharge from a swollen umbilical stump. If the discharge contains intestinal fluid, to which of the following portions of the gastrointestinal tract is the fistula most likely connected? A) Cecum B) Duodenum C) lleum D) Jejunum E) Sigmoid colon

C. Meckel diverticulum results from persistence of the vitelline duct. In some cases, Meckel diverticulum is associated with umbilical anomalies, including formation of a fistulous tract and discharge of intestinal fluid in cases when the duct did not completely obliterate during development. Meckel diverticulum is commonly located in the ileum, two feet proximal to the ileocecal valve. The diverticulum is approximately two inches in length. In children, it may serve as a potential lead-point for intussusception. A technetium 99m pertechnetate scan demonstrates uptake in the right lower quadrant, corresponding with ectopic gastric mucosa within the Meckel diverticulum. Treatment of symptomatic cases is through surgical resection. Incorrect Answers: A, B, D, and E. The cecum (Choice A) is the site of the ileocecal valve and the appendix. It is located in the right lower quadrant of the abdomen. The ileocecal valve is a useful landmark for identifying Meckel diverticulum, which occurs in the adjacent ileum. The duodenum (Choice B) is the first segment of the small intestine and is the site of the pyloric valve and the sphincter of Oddi. Meckel diverticulum occurs in the ileum, rather than the duodenum. The jejunum (Choice D) may be affected in cases of pediatric midgut malrotation or volvulus. It can be distinguished from other segments of the small intestine by its prominent plicae circulares, larger caliber, thicker muscular walls, and long vasa recta. Meckel diverticulum occurs in the ileum, rather than the jejunum. The sigmoid colon (Choice E) is a distal segment of the large intestine and is found proximal to the rectum. It is a common site of polyps, colon carcinoma, and non-congenital diverticulosis, but is not the site of Meckel diverticulum. Educational Objective: Meckel diverticulum results from persistence of the vitelline duct. In some cases, Meckel diverticulum is associated with umbilical anomalies, including formation of a fistulous tract and discharge of intestinal fluid when the duct persists. %3D Previous Next Score Report Lab Values Calculator Help Pause

183 Exam Section 4: Item 33 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 33. A72-year-old woman with coronary artery disease comes to the physician because of a 2-month history of progressive angina symptoms with exertion. Five months ago, the patient underwent stent placement for significant stenoses of the proximal anterior interventricular (left anterior descending) and right coronary arteries. She remained symptom-free for 3 months. Her pulse is 76/min and regular, and blood pressure is 135/85 mm Hg. An ECG at rest shows no abnormalities; an exercise stress test shows ST-segment changes in leads II, III, and aVE. Which of the following is the most likely cause of this patient's recurrent angina symptoms? O A) Aneurysm of the right coronary artery B) Dissection of the right coronary artery C) Neointima formation in the right coronary stent D) Thrombosis of the right coronary stent E) Vasospasm in the stented right coronary artery

C. Neointima formation in the right coronary stent is the most likely cause of the patient's recurrent symptoms. The location of ST-segment changes in the exercise stress test suggests impairment of blood flow to the right coronary artery territory. Neointima formation is a common cause of vascular restenosis after percutaneous coronary intervention (PCI). It is primarily composed of vascular smooth muscle cells in a rich extracellular matrix that have migrated into the stent to form a layer that shares some features with the tunica intima. Migration is promoted by macrophages and inflammatory cells in the setting of a foreign body reaction. The resulting neointima formation decreases the lumen diameter may cause inadequate myocardial perfusion during exertion with recurrence of angina. Drug-eluting stents (DES) attempt to reduce the risk of neointima formation by the continuous, slow release of drugs that inhibit cell proliferation. Incorrect Answers: A, B, D, and E. Aneurysm of the right coronary artery (Choice A) is less likely than neointimal hyperplasia. An aneurysm is a dilation of a blood vessel involving all three lays of the vessel wall. Coronary artery aneurysms are associated with atherosclerosis and certain forms of vasculitis, such as mucocutaneous lymph node syndrome (Kawasaki disease). They are usually asymptomatic, but may be susceptible to rupture, dissection, or thrombus formation. Dissection of the right coronary artery (Choice B) is an uncommon occurrence, but typically presents with symptoms of acute coronary syndrome, dysrhythmias, or sudden cardiac death. Coronary artery dissection is a potential complication of PCI but would present in the immediate periprocedural setting rather than five months later. Thrombosis of the right coronary stent (Choice D) is a potential complication of PCI, and the indication for use of antiplatelet agents following stent placement. In-stent thrombosis is less common than neointima formation. Vasospasm in the stented right coronary artery (Choice E) is another potential complication of stent placement. Vasospasm symptoms are not typically reserved to exertion and may occur while the patient is at rest or may wake the patient from sleep. Educational Objective: Neointima formation is a common complication of coronary stenting and the most common cause of recurrent angina. DES attempt to reduce the risk by releasing antiproliferative agents. %3D Previous Next Score Report Lab Values Calculator Help Pause

156 Exam Section 4: Item 6 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 6. A 5-year-old boy has severe cholestatic liver disease and poor bile secretion. This patient is also likely to have a deficiency of vitamin E because its absorption requires which of the following? A) Binding to cholesterol B) Conversion to an active form by the liver C) Formation of micelles D) Intrinsic factor E) Trypsin

C. The formation of micelles is a necessary step in the absorption of fat-soluble vitamins such as vitamin E. A micelle is a sphere containing bile acids, which are amphipathic. The hydrophilic heads orient at the periphery of the sphere, while the hydrophobic contents remain on the inside. When ingested, vitamin E is incorporated into micelles via affinity with bile acids secreted from the liver. Their amphipathic nature allows them to remain soluble in aqueous solution, while transporting lipids and fat-soluble vitamins in their hydrophobic cores. Bile acids and micelles thus are necessary for transporting hydrophobic molecules into enterocytes. Once within an enterocyte, vitamin E is incorporated into chylomicrons and transported to the liver for storage where it can be secreted in VLDL particles. A deficiency in bile production or secretion such as in this patient with cholestatic liver disease would greatly alter the absorption of vitamin E as a result of failure to form micelles. Deficiency of vitamin E can cause myopathy, anemia, and neurological dysfunction. Incorrect Answers: A, B, D, and E. Binding to cholesterol (Choice A) is not the mechanism by which vitamin E is absorbed. Cholesterol is a necessary component of bile acids, which are needed to solubilize vitamin E in micelles, but cholesterol doesn't directly bind vitamin E. Conversion to an active form by the liver (Choice B) is not impaired in patients who cannot adequately secrete bile. Vitamin E is primarily stored in the liver and secreted in VLDL. It does not require modification by the liver to become active. Intrinsic factor (Choice D) is a protein found in the stomach that is required for the absorption of vitamin B12, not vitamin E. Antibodies directed against this protein result in pernicious anemia. Trypsin (Choice E) is a digestive enzyme that degrades proteins. Trypsinogen is created in the pancreas and secreted into the duodenum after a meal where it is activated to trypsin. It does not play a role in the absorption of vitamin E. Educational Objective: Vitamin E is a fat-soluble vitamin that must be incorporated into micelles in the presence of bile acids before it can be absorbed by enterocytes and transported to the liver. An absence of bile acids, as might occur in conditions where bile secretion is impaired, would result in vitamin E malabsorption due to the failure to form micelles. %3D Previous Next Score Report Lab Values Calculator Help Pause

159 Exam Section 4: Item 9 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 9. An investigator is planning to create gene therapy for Leigh syndrome, which is caused by an A G mutation in the mitochondrial tRNALeu gene. Which of the following is the most likely reason why mitochondria encode their own tRNA? A) Mitochondria cannot import proteins or RNA O B) Mitochondria produce large amounts of reactive oxygen species C) Mitochondria use a non-standard genetic code D) The unusually high mitochondrial pH denatures nuclear-encoded tRNA E) The unusually low mitochondrial pH hydrolyzes nuclear-encoded tRNA

C. The mitochondrial genome differs from the nuclear genome in that it is small, circular, and present in numerous copies within each mitochondrion. Additionally, the mitochondrial genome relies on different coding sequences for stop codons and for some amino acids, such as tryptophan. The use of different coding sequences necessitates the use of a separate system of tRNA. The mitochondrial genome encodes several proteins for the oxidative phosphorylation electron transport chain proteins (eg, cytochromes), but mitochondrial function remains heavily dependent on importing proteins coded by the cell nucleus. Incorrect Answers: A, B, D, and E. The mitochondria do rely on importing proteins and RNA (Choice A) from the cell cytoplasm. The mitochondrial genome is not capable of producing all of the required proteins for mitochondrial function. Mitochondrial coding of tRNA does not impact the mitochondrial ability to import proteins. Mitochondria produce large amounts of reactive oxygen species (Choice B) but this does not contribute to the need for a separate system of TRNA. Many important mitochondrial antioxidants, such as glutathione, are synthesized in the endoplasmic reticulum using standard cellular tRNA. The mitochondrial matrix pH is not unusually low or high, nor is the inner mitochondrial membrane space, despite the presence of a proton gradient (Choices D and E). The inner membrane is slightly acidic, and the matrix is slightly alkaline compared to the cytoplasmic pH. Nuclear-encoded tRNA is relatively stable and is neither hydrolyzed nor denatured under slightly acidic or alkaline conditions. Educational Objective: The mitochondrial genome relies on different coding sequences for stop codons and for some amino acids, such as tryptophan. The use of different coding sequences necessitates the use of a separate system of tRNA. %3D Previous Next Score Report Lab Values Calculator Help Pause

103 Exam Section 3: Item 3 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 3. A 24-year-old woman comes to the physician for an examination prior to employment. Physical examination shows no abnormalities. An ECG shows a sinus rhythm of 70/min, PR interval of 152 msec, and QRS complex of 84 msec; there are no ectopic beats. Which of the following best represents the longest conduction time in this patient? A) Anterior left bundle branch B) Intra-atrial C) Low right atrium to bundle of His D) Proximal bundle of His to ventricular myocardium E) Right bundle branch

C. The normal electrical conduction pathway of heart starts in the sinoatrial (SA) node with spontaneous slow depolarization of the SA nodal pacemaker cells located in the superior right atrium. The atria then depolarize which results in the P-wave on the ECG. The action potential spreads to the atrioventricular (AV) node, where the signal is slowed and conducted to the bundle of His which runs along the interventricular septum. Slowing of conduction through the AV node allows the atria to fully contract and complete filling of the ventricles prior to systole. The action potential is conducted along fascicular branches and Purkinje fibers to stimulate the coordinated contraction of ventricular cardiomyocytes. Depolarization of the ventricles creates the QRS complex on the ECG. The PR interval represents the time from initial depolarization of the atria to the start of ventricular depolarization. The normal range is 120-200 msec. Conduction time from the low right atrium to the bundle of His therefore takes the longest due to the physiologic delay in conduction through the AV node. Incorrect Answers: A, B, D, and E. Conduction through the anterior left bundle branch (Choice A), proximal bundle of His to ventricular myocardium (Choice D), and right bundle branch (Choice E) is rapid and occurs within the initial period of the QRS complex in order for coordinated contraction of the ventricles to occur. Delays in these conduction pathways may be due to structural abnormalities, fibrosis from ischemic or inflammatory injury, toxic-metabolic syndromes, electrolyte derangements, or normal anatomic variants, and appear as a widened QRS complex (>120 msec) on the ECG. Intra-atrial (Choice B) conduction is represented by the duration of the P-wave. Conduction is rapid in order for coordinated atrial contraction to occur. Educational Objective: The AV node delays the electrical impulse conducted from the atria in order for atrial contraction to complete and the ventricles to fill entirely prior to ventricular depolarization. %3D Previous Next Score Report Lab Values Calculator Help Pause

102 Exam Section 3: Item 2 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 2. A 43-year-old woman comes to the physician because of a 4-week history of intermittent episodes of severe abdominal pain, especially after eating fatty foods. Physical examination shows no abnormalities. Abdominal ultrasonography shows several 3- to 4-mm gallstones contained within the gallbladder. The patient declines surgery. A drug with which of the following mechanisms of action is most appropriate for this patient? A) Decreased gallbladder muscle contractility B) Increased gastrin production C) Increased metabolism of bile salts D) Inhibition of biliary cholesterol secretion E) Inhibition of cytochrome P450 activity

D. Inhibition of biliary cholesterol secretion is achieved with ursodeoxycholic acid (ursodiol), a medication used to treat biliary colic, primary biliary cirrhosis, and intrahepatic cholestasis of pregnancy. In patients who have atypical symptoms that are not definitively attributable to gallstones, improvement with ursodiol suggests that cholecystectomy may be beneficial, but in patients with classic biliary colic as in this patient, ursodiol is considered a second line to surgery. It works by reducing cholesterol secretion from the liver by 40-60%, and thereby reducing the cholesterol content of the bile. It also reduces the amount of cholesterol absorbed in the gut. As most gallstones are made of cholesterol, with a minority formed from heme as a result of chronic hemolysis, gradually reducing the amount of cholesterol in the bile prevents formation of new cholesterol gallstones. Incorrect Answers: A, B, C, and E. Decreased gallbladder muscle contractility (Choice A) is a feature of many medications including atropine (cholinergic effect), somatostatin, calcium channel blockers, ondansetron, and opiates. These medications are not used to treat cholelithiasis. Increased gastrin production (Choice B) is a characteristic of medications that increase the pH of the stomach, including proton-pump inhibitors and H2-blocking medications. Gastrin secretion is under inhibitory control by gastric pH and secretion is low when the pH is low. Excess gastrin secretion is also a defining feature of Zollinger-Ellison syndrome. Increased metabolism of bile salts (Choice C) is achieved with the use of bile acid sequestrants such as cholestyramine. This class of medications forms a nonabsorbable complex with bile acids in the gastrointestinal tract and facilitates their excretion in the stool. Inhibition of cytochrome P450 activity (Choice E) is the result of numerous medications and can sometimes account for severe medication interactions as inhibition of this enzyme affects the metabolism of certain drugs. Cytochrome P450 refers to numerous subtypes of enzymes, including CYP3A4 and CYP2D6. Common medications known to interact with this class of enzymes include fluconazole, ciprofloxacin, and fluoxetine. Educational Objective: Ursodiol decreases the cholesterol content of bile and thereby prevents the formation of cholesterol gallstones. It is considered second line therapy when compared to cholecystectomy in patients with cholelithiasis and biliary colic. %3D Previous Next Score Report Lab Values Calculator Help Pause

150 Exam Section 3: Item 50 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 50. One thousand office workers are surveyed to determine whether or not they use computers and whether or not they have symptoms of carpal tunnel syndrome. Which of the following best describes this study? A) Case-control study B) Case series C) Cohort study D) Cross-sectional survey E) Randomized clinical trial

D. A cross-sectional survey seeks to identify the prevalence of a condition at a single point in time. 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. Causality between the risk factor and outcome cannot be concluded. It can be used to assess associations with prior exposures. Associations with prior exposures may be subject to recall bias as survey participants may not accurately represent or recall prior exposures. In the study described, the outcome data (carpal tunnel syndrome) and the risk factor data (computer use) are collected at the same time and the patients are not followed over time, making this a cross-sectional survey study design. Incorrect Answers: A, B, C, and E. A case-control study (Choice A) investigates an association between an exposure and an outcome. In this study design, a group of patients with the disease (cases) are identified. A group of patients without the disease (controls) are matched on baseline characteristics to the cases. Exp ure 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 conduction of larger cohort or randomized trials with sufficient power. A cohort study (Choice C) identifies a group of patients and follows them over time to identify whether an exposure is associated with an outcome of interest. Cohort studies may be retrospective or prospective in design. In a prospective design, the hypothesis and analysis protocols are established prior to the start of the study period. In a retrospective design the hypothesis or question is designed after the study time period has passed. An example of a prospective cohort study would be following a group of 1,000 smokers for a time period of 10 years and identifying the proportion of the patients who develop pancreatic cancer to identify the risk of pancreatic cancer in smokers as compared to a control group of nonsmokers. 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 the 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 cross-sectional survey study design collects data about both the exposure and outcome at the same time point and does not follow patients over time. It can be used to assess the prevalence of a specific disease and also assess associations with prior exposures. Associations with prior exposures may be subject to recal bias as survey participants may not accurately represent or recall prior exposures. %3D Previous Next Score Report Lab Values Calculator Help Pause

145 Exam Section 3: Item 45 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 45. A 56-year-old man comes to the physician because of a 3-month history of a mild, nonproductive cough. He has a 20-year history of alcoholism. Physical examination shows no abnormalities. A PPD skin test result is positive. A chest x-ray and a Gram stain of induced sputum show no abnormalities. Treatment with isoniazid and a daily multivitamin is begun. The patient is compliant with the isoniazid therapy but not with the multivitamin supplementation. Two months later, he develops tingling and numbness of both feet. Deficiency of which of the following is the most likely cause of the foot symptoms in this patient? A) Folic acid B) Niacin C) Vitamin B2 (riboflavin) D) Vitamin B, (pyridoxine) E) Vitamin B12 (cobalamin)

D. A positive PPD skin test with no evidence of active tuberculosis (normal chest x-ray and sputum gram stain) is diagnostic of latent tuberculosis infection. Treatment for latent tuberculosis infection involves nine months of isoniazid. Isoniazid binds vitamin B6 (pyridoxine) which may result in early excretion and inactivation of pyridoxine. Isoniazid also inhibits pyridoxine phosphokinase, which is necessary to activate pyridoxine, thereby causing pyridoxine deficiency. Therefore, all patients taking isoniazid also concurrently take pyridoxine supplementation. Pyridoxine deficiency limits synthesis of histamine, hemoglobin, and neurotransmitters including epinephrine, norepinephrine, dopamine, serotonin, and y-aminobutyric acid. Deficiency commonly presents with peripheral neuropathy, dermatitis, sideroblastic anemia, glossitis, and seizures (especially with isoniazid use). Incorrect Answers: A, B, C, and E. Folic acid (Choice A) is converted to tetrahydrofolic acid and used as a coenzyme in the synthesis of nucleotides and nucleosides. Folate is contained in leafy vegetables and absorbed in the jejunum. Folate deficiency is often seen in patients with malnutrition, alcohol use disorder, and patients taking anti-folate medications (eg, phenytoin, methotrexate). Megaloblastic anemia occurs in the setting of impaired DNA synthesis. Niacin (Choice B) deficiency can result in glossitis, diarrhea, neuropsychological disturbances such as dementia and hallucinations, and dermatitis. 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 isoniazid use. Vitamin B12 (cobalamin) (Choice E) deficiency, similar to folic acid deficiency, is classically characterized by erythrocyte macrocytosis and hypersegmented neutrophils. In addition, it also results in posterior column and spinocerebellar tract demyelination. It is most often seen in patients with malnutrition, alcohol use disorder, pernicious anemia, or Crohn disease. Educational Objective: Isoniazid therapy for latent tuberculosis without sufficient pyridoxine supplementation can lead to pyridoxine deficiency, which is characterized by peripheral neuropathy, dermatitis, sideroblastic anemia, glossitis, and seizures. %3D Previous Next Score Report Lab Values Calculator Help Pause

191 Exam Section 4: Item 41 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 41. A single, oral, 6-mg dose of a short-acting B zagonist is studied to determine its pharmacokinetics and pharmacodynamics in young subjects with stable mild asthma. In this group, the Cma/AUC following oral dosing varies by less than 8%. However, in 30% of subjects, the maximal FEV, increase is 25% lower than that of the other subjects. Which of the following mechanisms best explains these observations in this subgroup? A) Decreased catechol O-methyltransferase activity in bronchial smooth muscle B) Decreased phosphodiesterase IV in bronchial smooth muscle C) Increased gastrointestinal P-glycoprotein D) Receptor polymorphism with reduced signal transduction O E) Receptor upregulation

D. Airway smooth muscle cells express B2 receptors, and activation results in bronchodilation through Gs protein-coupled receptor signal transduction. This mechanism is the rationale for the use of B-adrenergic agonists in the treatment of asthma. The difference in response between the subgroups is best explained by receptor polymorphism with reduced signal transduction in the lower maximal FEV1 group. The minimal variation in Cmax/AUC following oral dosing indicates that similar absorption occurred between the groups, and it can be extrapolated that the airway smooth muscle cells were exposed to similar concentrations of the drug. Receptor polymorphism refers to minor variations in the genes encoding receptors (most often single nucleotide polymorphisms) that may result in functional proteins with differing amplitudes of downstream response upon activation. Incorrect Answers: A, B, C, and E. Decreased catechol O-methyltransferase (COMT) activity in bronchial smooth muscle (Choice A) does not adequately explain the findings in this experiment. COMT is responsible for the enzymatic degradation of catecholamines, and decreased activity may result in a reduced susceptibility to bronchoconstriction but would not be expected to alter the response to a B2-adrenergic agonist. Decreased phosphodiesterase IV in bronchial smooth muscle (Choice B) results in reduced degradation of CAMP and increased bronchodilation. A polymorphism in phosphodiesterase IV that results in reduced levels would be expected to increase the maximal FEV, and would not cause a variable response to the B2-adrengeric agonist. Increased gastrointestinal P-glycoprotein (Choice C) decreases serum concentrations of orally administered drugs through increased efflux out of enterocytes. The similar Cmax/AUC levels in the subgroups suggests that drug absorption in the gastrointestinal tract was not the cause of the different results, and it can be extrapolated that the airway smooth muscle cells were exposed to similar concentrations of the drug. Receptor upregulation (Choice E) is less likely than receptor polymorphism in the experiment subgroups assuming similar exposures and controls for confounders. Educational Objective: Minor variations in nucleotide sequences can result in functional proteins that respond to activation with different amplitudes of signaling effects. This is called receptor polymorphism. %3D Previous Next Score Report Lab Values Calculator Help Pause

118 Exam Section 3: Item 18 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 18. A 21-year-old woman comes to the physician for a routine health maintenance examination. Her 3-year-old sister was recently diagnosed with cystic fibrosis. Prior to any genetic testing, which of the following best approximates this patient's risk of being a heterozygote for cystic fibrosis? A) 0 B) 1/2 C) 1/4 D) 2/3 E) 3/4

D. Inheritance of disease is based on autosomal, mitochondrial, and X-linked genetics of the parents, genes distributed into the gametes, de novo mutations acquired during embryonic cell division and differentiation, penetrance and expressivity of the condition, mosaicism, and epigenetic modifications of expression. In this case, cystic fibrosis is present in one sibling but not another. Cystic fibrosis is inherited in an autosomal recessive pattern and follows the principles of Mendelian genetics. The wild type allele can be represented by A and the mutated allele by a. If the offspring of two phenotypically unaffected individuals has cystic fibrosis, then both parents must be heterozygous (Aa). An offspring of two heterozygous parents has four potential outcomes: they could receive both mutated alleles and be homozygous for the recessive mutated gene (aa) in which case they would demonstrate the disease assuming complete penetrance, they could receive one wild type allele from the father and one mutated allele from the mother (Aa) or vice versa, or they could receive both wild type alleles (AA). In an individual who is known not to have the disease (phenotypically normal as in this patient), the outcome of aa is ruled out, leaving three other possible genotypes. There are two ways to be heterozygous (Aa) and one way to be homozygous for the dominant allele (AA). Thus, for this individual, the likelihood of heterozygosity for the allele causing cystic fibrosis is 2/3. Incorrect Answers: A, B, C, and E. The individual in question has a sister with cystic fibrosis, meaning her parents must both carry recessive alleles. This means there is a non-zero likelihood that the individual in question is heterozygous at the gene locus. The likelihood of her being heterozygous cannot be 0 (Choice A). 1/2 (Choice B) is the likelihood that an offspring with an unknown phenotype will be heterozygous. However, the individual in this case is 21 years old and has not demonstrated any manifestations of cystic fibrosis, so the possibility of her being homozygous recessive (aa) can be ruled out. 1/4 (Choice C) is the likelihood that an offspring with an unknown phenotype would be either homozygous recessive or homozygous dominant. As this individual's phenotype is known, there are only three possible genotypes, not four. 3/4 (Choice E) is the likelihood that an individual with an unknown phenotype is not homozygous recessive. For example, if the parents were wanting to conceive again and asked the likelihood of their child being healthy, that probability would be 3/4. Educational Objective: Inheritance of cystic fibrosis follows a Mendelian pattern. It is inherited in an autosomal recessive pattern. Homozygous recessive individuals demonstrate the disease while heterozygous individuals do not. %3D Previous Next Score Report Lab Values Calculator Help Pause

192 Exam Section 4: Item 42 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 42. A 19-year-old man comes to the emergency department because of increasingly severe shoulder and abdominal pain for 3 days. His temperature is 39°C (102.2°F). Physical examination shows signs of acute peritonitis. An abdominal x-ray is shown. Which of the following is the most likely cause of this patient's current condition? A) lleus B) Intraperitoneal abscess C) Nephrolithiasis D) Perforated viscus E) Retained foreign body 86代

D. Fever, abdominal pain, peritonitis, and free air seen under the diaphragm on upright chest x-ray raises suspicion for a perforated viscus. Patients with perforated viscus can demonstrate intraabdominal free air and leakage of enteric contents, resulting in inflammation, irritation, and infection of the peritoneum. Inflammation involving the diaphragm can lead to referred pain to the scapula or shoulder due to shared innervation by cervical spinal nerves C3-5. Perforated viscus can occur secondary to peptic ulcer disease, diverticulitis, appendicitis, or blunt or penetrating abdominal trauma among other etiologies. Resulting peritonitis is a surgical emergency, and patients are treated with broad-spectrum antibiotics and surgery to repair the perforated viscus and lavage the peritoneal cavity. Incorrect Answers: A, B, C, and E. lleus (Choice A) refers to intestinal hypomotility without a mechanical obstruction, most commonly occurring following surgery or in the setting of electrolyte abnormalities, and typically presents with diffuse small and large bowel dilatation. Intraperitoneal abscess (Choice B) can also cause abdominal pain and fever and can result from areas of inflammation and infection such as appendicitis or diverticulitis. Abscesses are contained structures and do not result in intraabdominal free air. Nephrolithiasis (Choice C) typically presents with unilateral flank pain that is colicky and sometimes radiates to the groin with associated hematuria. The common types of nephroliths are calcium oxalate or phosphate, ammonium magnesium phosphate, uric acid, and cystine. It would not cause fever unless complicated by a urinary tract infection. Retained foreign body (Choice E) can cause abdominal pain depending on the foreign body, but typically does not cause fever or free air under the diaphragm unless it has perforated the intestine causing leakage of enteric contents. Educational Objective: A perforated viscus results in the leakage of enteric contents, fever, abdominal pain, peritonitis, and pneumoperitoneum. Diaphragmatic irritation can cause referred shoulder or scapular pain. %3D Previous Next Score Report Lab Values Calculator Help Pause ORTABLE

121 Exam Section 3: Item 21 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 21. A 68-year-old man with recurrent urinary tract infections comes to the physician because of a 2-day history of fever and pain with urination. Urinalysis shows pus. He is treated with levofloxacin. His symptoms worsen, and a blood culture grows Pseudomonas aeruginosa that is resistant to levofloxacin. Alteration of which of the following best explains the resistance of P. aeruginosa to levofloxacin in this patient? A) Dihydrofolate reductase B) B-Lactamase C) 30S ribosomal subunit D) Topoisomerase E) Transpeptidase

D. Fluoroquinolone antibiotics inhibit bacterial topoisomerase II (DNA gyrase) and topoisomerase IV. These enzymes function to unwind supercoils in DNA through purposeful single- or double-stranded breaks, which prevent damage to the DNA and allow it to continue its normal replicative processes and functionality. Examples of fluoroquinolone antibiotics include ciprofloxacin, ofloxacin, levofloxacin, and moxifloxacin. They are most effective against Gram-negative infections of the respiratory tract, urinary tract, and gastrointestinal tract, and they are also are used for certain eye and ear infections. They exhibit anti-bacterial activity against Pseudomonas aeruginosa. Bacteria develop resistance to fluoroquinolones through bacterial chromosome-encoded mutations in topoisomerase, which disrupts the typical binding of fluoroquinolones to the enzyme. A potential secondary mechanism of resistance is the overexpression of cellular efflux pumps, which actively remove fluoroquinolone molecules from the bacteria cells, reducing their intracellular concentration and effectiveness. Incorrect Answers: A, B, C, and E. Dihydrofolate reductase (Choice A) functions to convert dihydrofolate to tetrahydrofolate in order to shuttle methyl groups for the synthesis of purine nucleotides. Trimethoprim has an extremely high affinity for bacterial dihydrofolate reductase and a much lower affinity for human dihydrofolate reductase, thus, it is an effective antibacterial agent. The main modality of resistance to B-lactam antibiotics in Gram-negative organisms is the production of B-lactamases (Choice B), also known as penicillinases. These B-lactamases cleave the B-lactam ring in these antibiotics, rendering them inactive. The 30S ribosomal subunit (Choice C) is the binding site for tetracycline and aminoglycoside antibiotics. The main mode of resistance to tetracycline antibiotics is the production of bacterial efflux pumps. Aminoglycoside resistance is typically derived from bacterial transferase enzymes that inactive the antibiotic through acetylation, adenylation, or phosphorylation. The main modality of resistance to B-lactam antibiotics in Gram-positive organisms is an altered structure of transpeptidase penicillin-binding proteins (Choice E), rendering B-lactam antibiotics unable to bind and inactivate them. Educational Objective: Fluoroquinolone antibiotics act by binding DNA gyrase and topoisomerase IV, both of which are topoisomerases. Genetic alterations in these enzymes, which reduce the ability of the fluoroquinolone to bind, facilitate resistance to this class of antibiotics. %3D Previous Next Score Report Lab Values Calculator Help Pause

116 Exam Section 3: Item 16 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 25 16. A study is conducted to assess the effect of a new selective estrogen receptor modulator (SERM) on bone fractures. Two thousand women with a history of vertebral fracture and 5000 women with no fracture history are randomly assigned to a placebo or to SERM at a dose of 60 mg or 120 mg. The graph shows the percent of women who had a new fracture during 5 years of follow-up (± standard error of the mean). Which of the following is the most accurate conclusion about preventing new fractures based on these data? T 20 A) High-dose SERM treatment is more effective than low-dose SERM treatment 15 T T B) Low-dose SERM treatment is more effective than high-dose SERM treatment C) Placebo treatment is effective only in women with a history of vertebral fracture D) SERM treatment is effective only in women with a history of vertebral fracture 10- E) SERM treatment is effective only in women with no history of fracture T T T No fracture history Fracture history Placebo SERM 60 mg SERM 120 mg

D. In the graph, the left group of bars demonstrate that in women with no vertebral fracture history, the administration of placebo, 60 mg selective estrogen receptor modulator (SERM), and 120 mg SERM demonstrated equivalent fracture risk over the 5-year period of approximately 4.5%. This data suggests that treatment with SERM of either dose does not decrease the 5-year risk of fracture in patients without vertebral fracture history. When interpreting the right portion of the graph, treatment with placebo in patients with a history of vertebral fracture demonstrates a 5-year risk of new fracture of approximately 22%. Treatment with 60 mg SERM or 120 mg SERM in patients with a history of vertebral fracture demonstrated a risk of new fracture of 14.5% at 5 years. Thus, in a randomized, placebo-controlled trial, SERM decreased fracture risk by approximately 7.5% in patients with prior vertebral fractures. Therefore, it is accurate to state that SERM treatment is effective only in women with a history of vertebral fracture. Incorrect Answers: A, B, C, and E. High-dose SERM treatment is more effective than low-dose SERM treatment (Choice A) and low-dose SERM treatment is more effective than high-dose SERM treatment (Choice B) are incorrect as high and low-dose SERM treatment demonstrated equivalent risk of new fracture in patients both with and without vertebral fracture history. Placebo treatment is effective only in women with a history of vertebral fracture (Choice C) is not correct as placebo is generally understood to be a control, not an intervention. Moreover, the SERM was shown to be more effective than the placebo in the patient group with prior vertebral fractures. SERM treatment is effective only in women with no history of fracture (Choice E) is incorrect as fracture risk in this group was not shown to be different from the risk in patients treated with placebo. Educational Objective: Placebos are often used as a control for interventional studies. Demonstrating a statistically significant reduction in the risk of disease relative to a placebo is the gold standard for the evaluation of a drug or intervention. The stratification of patients into groups within a clinical trial is useful for identifying the effect of an intervention in patients at high and low risk of the disease under investigation. I3D Previous Next Score Report Lab Values Calculator Help Pause % With new fracture

176 Exam Section 4: Item 26 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 26. A 29-year-old woman comes to the physician because of irregular menstrual periods since menarche at the age of 12 years. She is 160 cm (5 ft 3 in) tall and weighs 86 kg (190 lb); BMI is 34 kg/m2 She is evaluated, and the diagnosis of polycystic ovarian syndrome is made. After explaining the diagnosis, the physician discusses behavioral changes, including dietary modification and exercise as part of her treatment plan. Which of the following is most likely to result in patient adherence to this plan? A) Ascertain the patient's educational level and provide appropriate publications B) Ask the patient to bring a family member or friend to the next appointment C) Inform the patient of the health consequences of not treating her condition D) Provide follow-up appointments to assess the patient's progress in attaining her goals E) Refer the patient to a support group

D. Making goals that are specific, measurable, attainable, relevant, and time-bound can increase patients' motivation to make behavioral changes such as dietary modifications and exercise. Providing follow-up appointments to track patients' incremental progress is the best way to increase patients' intrinsic motivation to attain these goals. Further, the formation of a trusting partnership between the physician and patient through regular follow-up appointments has been shown to improve adherence. During these appointments, physicians can also improve patients' readiness to change health behavior with an interview technique called motivational interviewing. Motivational interviewing utilizes open-ended, non-judgmental questions to help the patient explore their reasons for wanting to change or maintain the habit. Incorrect Answers: A, B, C, and E. Educating the patient (Choices A and C) may modestly improve the patient's extrinsic motivation to change her behavior. However, improving the therapeutic alliance and tracking progress toward goals can increase a patient's intrinsic motivation, which is more likely to result in behavior change. Improving a patient's support system (Choices B and E) may increase the patient's adherence to the plan. However, fostering the physician-patient relationship and tracking progress toward a goal are more likely to improve adherence. Educational Objective: Making regular follow-up appointments to track patients' incremental progress toward their goals can improve patients' intrinsic motivation to adhere to treatment recommendations. These follow-up appointments can foster trust and collaboration within the physician-patient relationship, which also improves treatment adherence. %3D Previous Next Score Report Lab Values Calculator Help Pause

111 Exam Section 3: Item 11 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 11. A 3-year-old girl has a history of recurrent infections. In vitro, neutrophils isolated from this patient are capable of phagocytosis and can kill Lactobacillus species but not Staphylococcus aureus. This patient most likely has a defect involving which of the following enzymes? A) Catalase B) Elastase OC) Myeloperoxidase D) NADPH oxidase E) Superoxide dismutase

D. NADPH oxidase deficiency resulting in chronic granulomatous disease (CGD) most likely explains this patient's recurrent infections and normal neutrophil phagocytosis but with impaired killing of Staphylococcus aureus. NADPH oxidase within neutrophils uses oxygen as a substrate for the generation of free radicals (superoxide anions). Free radical oxygen species are subsequently used for the creation of hydrogen peroxide and hypochlorous acid. Activation of this pathway leads to the respiratory burst which results in bacterial death. Deficiency of NADPH oxidase renders phagocytes incapable of neutralizing catalase-positive bacteria (eg, S. aureus) which can neutralize their own hydrogen peroxide, thus leaving the host cells without the substrate necessary to complete the respiratory burst. Lactobacillus species are catalase-negative. Diagnosis is made by an abnormal dihydrorhodamine test or a nitroblue tetrazolium reduction test. In the latter test, normal phagocytes use the action of NADPH to reduce nitroblue, which leads to a color change from yellow to blue. Patients with CGD will not demonstrate color change. Recurrent pneumonia is the most common presenting infection in patients with CGD, and the most common infecting bacteria include Staphylococcus species, Aspergillus species, Burkholderia cepacia, and Nocardia species. Patients with CGD are also at risk for fungal infections, especially Aspergillus species. Incorrect Answers: A, B, C, and E. Catalase (Choice A) is a ubiquitous enzyme present in both eukaryotic and prokaryotic organisms that catalyzes the conversion of hydrogen peroxide to water and oxygen. Catalase-positive bacteria are more likely to cause disease in patients with CGD, but human catalase deficiency is often benign. Elastase (Choice B) is a serine protease secreted by the pancreas and functions to digest proteins in the gastrointestinal tract. Patients with chronic pancreatitis and pancreatic insufficiency will often have an absence of stool elastase, which indicates inadequate pancreatic enzyme function. Myeloperoxidase (Choice C) is an autosomal recessive immune disorder caused by mutations in the MPO gene on chromosome 17. Dysfunctional myeloperoxidase results in an inability to produce hydroxy-halide radicals, namely hypochlorite from hydrogen peroxide and chloride. Myeloperoxidase deficiency results in impaired, but not absent bacterial killing as the enzymatic products of NADPH oxidase and superoxide dismutase are also bactericidal. Patients typically present with recurrent fungal infections. Superoxide dismutase (Choice E) is an enzyme that converts reactive oxygen species to oxygen and hydrogen peroxide, thus preventing oxidative damage to cells. Deficiency has been linked to familial amyotrophic lateral sclerosis. Educational Objective: NADPH oxidase is necessary for the formation of reactive oxygen species in phagocytes, which allows for bactericidal activity. Mutations in the proteins that make up the NADPH oxidase complex result in CGD, typically characterized by recurrent pyogenic bacterial infections. %3D Previous Next Score Report Lab Values Calculator Help Pause

120 Exam Section 3: Item 20 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 20. A 1-year-old boy is brought to the physician for a well-child examination. The mother is concerned because her son's fine, pale hair has not changed color since birth. His eyes are blue. During ophthalmologic examination, the patient turns away from the flashlight and starts crying. Which of the following is the most likely cause of the pale skin color in this patient? A) Aberrant migration of neural crest cells B) Decreased number of epidermal melanocytes C) Immune destruction of melanocytes D) Inability to produce melanin E) Melanin dropout to the dermis

D. Oculocutaneous albinism is the partial or complete absence of melanin pigment from the melanocytes in the skin, hair, and eyes. The findings of albinism all relate to this lack of melanin pigment: white hair, milky skin, blue-gray eyes at birth, extreme sensitivity to ultraviolet (UV) radiation, and an increased risk of skin cancers throughout life. The findings are found at birth and present diffusely throughout the body, unless the mutation is present in a mosaic distribution. This is in contrast to other pigmentary aberrancies, which may affect melanocyte migration to some areas of the body on a background of normal skin. Oculocutaneous albinism has three subtypes, each of which is caused by a different mutation in the process of melanin production. In the most common form, oculocutaneous albinism type I, decreased or absent tyrosinase is the primary abnormality. Incorrect Answers: A, B, C, and E. Melanocytes are derived from the neural crest cells, which begin at the dorsal neural tube. During embryologic development they migrate from dorsal to ventral, and then to the epidermis. Aberrant migration of neural crest cells (Choice A) presents as piebaldism. In this condition, a white forelock, or a white patch of hair usually on the frontal hairline, occurs when the melanocytes do not fully migrate cephalad. The remaining skin will demonstrate its usual coloring. Decreased melanin production, rather than a decreased number of epidermal melanocytes (Choice B), is the primary abnormality of oculocutaneous albinism. A decreased number of epidermal melanocytes is demonstrated in idiopathic guttate hypomelanosis. In this condition, patients with chronic sun damage develop numerous small, benign, depigmented macules in sun exposed areas. Immune destruction of melanocytes (Choice C) characterizes vitiligo. In vitiligo, depigmented patches have a predilection for developing around the mouth, nose, eyes, hands, or genitalia, but any other location of the body can be involved. Autoimmune destruction of the melanocytes induces these patches to form. It is also associated with other autoimmune conditions, such as chronic lymphocytic (Hashimoto) thyroiditis. Melanin dropout to the dermis (Choice E) is the mechanism of post-inflammatory hyperpigmentation. When inflammation occurs in the epidermis at the basal junction, the melanocytes are also affected and die. As melanocytes and keratinocytes die, melanin is released and settles in the dermis where it can be challenging to remove, resulting in chronic hyperpigmentation. Educational Objective: Oculocutaneous albinism type I is caused by a decreased production of melanin by the melanocytes in the skin. Without melanin to counter the effects of UV radiation, patients are at high risk for both melanoma and non-melanoma skin cancers. %3D Previous Next Score Report Lab Values Calculator Help Pause

142 Exam Section 3: Item 42 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment V 42. An 11-year-old boy has had persistent pain in his right knee since he twisted it in a soccer game 3 weeks ago. Physical examination shows a tender area in the distal femur but no perceptible mass. An x-ray of the femur shows an osteolytic mass that has eroded through the cortex and elevated the periosteum; the mass is surrounded by reactive bone in some areas. Biopsy of the mass shows atypical cells with hyperchromatic pleomorphic nuclei. The cells are surrounded by an eosinophilic matrix, some of which is calcified. This tumor is most likely to spread first to which of the following areas? O A) Brain B) Cervical lymph nodes C) Liver D) Lung E) Vertebrae

D. Osteosarcoma is the most common osseous malignancy in children. It is characterized by cells with enlarged nuclei and a disorganized tissue structure that demonstrates areas of osteoid production (mineralized collagen that provides the structure of bone). This malignancy is locally destructive to bone and can erode through the cortex resulting in a periosteal reaction with elevation of the periosteum, which can be seen on x-rays (eg, Codman triangle). Treatment is typically surgical resection combined with chemotherapy. Osteosarcoma has a high mortality rate and high rate of metastasis to the lungs, even with adequate surgical resection. Even when initial staging does not detect metastatic disease, individual or small groups of cells may have already metastasized yet remain below the threshold of detection in staging. This may lead to high rates of lung metastases even after treatment. Incorrect Answers: A, B, C, and E. Malignancies that commonly metastasize to the brain (Choice A) include lung, breast, kidney, and colon cancers. The brain is not a common location for osteosarcoma metastases. The cervical lymph nodes (Choice B) are a common location for metastases from head and neck cancers such as squamous cell carcinoma of the face and oropharynx. This is not a common location for osteosarcoma metastases. The liver (Choice C) is a common location for metastases of colon cancer. The vertebrae (Choice E) are common sites of metastases of breast cancer and prostate cancer. The valveless venous system, known as the Batson system, which connects the pelvic veins, the thoracic veins, and the venous vertebral plexus may allow for the deposition of malignant cells in the vertebral bodies. Educational Objective: Osteosarcoma is the most common malignancy of bone. It is an aggressive malignancy that occurs in a bimodal distribution in children and in older adults. It commonly metastasizes to the lungs. %3D Previous Next Score Report Lab Values Calculator Help Pause

165 Exam Section 4: Item 15 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 15. An investigator is conducting a study of the lipopolysaccharide synthesis pathway. During the study, a polymerase chain reaction test is developed that detects a gene involved in lipid A biosynthesis. Which of the following organisms is most likely to yield a positive result using this polymerase chain reaction test? A) Candida albicans B) Clostridium difficile C) Mycoplasma pneumoniae D) Pasteurella multocida E) Staphylococcus aureus

D. Pasteurella multocida is the most likely organism to demonstrate presence of the gene coding for lipid A biosynthesis, a part of the outer membrane in Gram-negative bacteria. In comparison to Gram-positive organisms, which possess a cytoplasmic membrane and a large outer peptidoglycan wall, Gram-negative bacteria have an inner cytoplasmic membrane, a small peptidoglycan wall, and an outer membrane. Lipid A constitutes part of the lipopolysaccharide (LPS) moiety that lies within the outer membrane of Gram- negative bacteria. LPS is thought to account for many of the manifestations of Gram-negative sepsis, including severe rigors and high fevers, as it is highly immunogenic. Lipid A is made of two glucosamine units linked to fatty acids. Immune activation is thought to occur via signaling through toll-like receptors. P. multocida, a Gram-negative organism, is likely to exhibit the presence of the lipid A gene. Incorrect Answers: A, B, C, and E. Candida albicans (Choice A) is a fungus that causes a variety of infections in both immunocompetent and immunocompromised patients. C. albicans often causes skin infections in moist intertriginous areas such as the groin and inframammary folds but can also cause invasive infections with high morbidity and mortality, including blood stream and urinary tract infections. As it is a fungus, it does not possess lipid A. Clostridium difficile (Choice B) is a Gram-positive anaerobe that causes colitis. It does not possess lipid A, but does secrete toxin A and B which are largely responsible for its pathogenicity. Mycoplasma pneumoniae (Choice C) is a rod-shaped, intracellular, Gram-indeterminate bacteria that most commonly causes pneumonia. It does not possess a cell wall or lipid A. Staphylococcus aureus (Choice E) is a Gram-positive bacterium that causes multiple types of infections including skin and soft tissue infections, pneumonia, endocarditis, and bacteremia. It is a Gram-positive organism that does not possess lipid A in its cell wall. Educational Objective: Gram-negative bacteria possess lipid A as a part of LPS, a highly immunogenic component of the Gram-negative outer membrane that does not exist in Gram-positive bacteria. %3D Previous Next Score Report Lab Values Calculator Help Pause

160 Exam Section 4: Item 10 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 10. A 47-year-old man has microscopic blood on urinalysis. A CT scan of the abdomen shows a 5-cm mass in the left kidney. A photomicrograph of tissue obtained on renal biopsy is shown. Which of the following is the most likely diagnosis? A) Angiomyolipoma B) Nephroblastoma C) Oncocytoma D) Renal cell carcinoma E) Transitional cell carcinoma

D. Renal cell carcinoma (RCC) is an adenocarcinoma of tubular epithelial cells. RCC is a common primary malignancy of the kidney and generally occurs in older male smokers. It can present with gross or microscopic hematuria, flank pain, weight loss, or fever. Laboratory analysis may reveal polycythemia or hypercalcemia due to associated paraneoplastic syndrome production of erythropoietin or parathyroid hormone-related peptide. Hypercalcemia can result from bony metastasis. Diagnosis of RCC typically occurs with contrast-enhanced CT scan or MRI and is confirmed by biopsy at the time of nephrectomy. The histology of RCC is uniquely characterized by the presence of polygonal clear cells (seen in the photomicrograph) due to the accumulation of lipid and carbohydrate within the cells. It spreads hematogenously, and commonly presents as a metastatic neoplasm. The brain is a frequent site of metastasis. Incorrect Answers: A, B, C, and E. Angiomyolipomas (Choice A) are tumors derived from perivascular epithelioid cells. While typically benign, they can be associated with tuberous sclerosis. They present as a mass, which can be composed of smooth muscle, adipocyte, and epithelioid cells. Nephroblastoma (Choice B) is the most common renal malignancy in childhood due to mutations in tumor suppressor genes WT1 or WT2. It is characterized by a large, often palpable, unilateral flank mass and hematuria. It is not associated with clear cells on biopsy. Oncocytomas (Choice C) are uncommon renal malignancies and on histology, appear as large neoplastic cells with an eosinophilic granular cytoplasm due to presence of numerous mitochondria. The transitional epithelium of the collecting system and bladder is specialized stratified epithelium that is able to expand and contract depending on the bladder volume. Transitional cell carcinoma (Choice E) occurs within the collecting system and/or bladder, and the malignant cells do not appear clear. Educational Objective: RCC often presents in older male smokers with gross or microscopic hematuria, flank pain, weight loss, or fever. The histology of RCC is uniquely characterized by polygonal clear cells due to the accumulation of lipid and carbohydrate content within the cells. %3D Previous Next Score Report Lab Values Calculator Help Pause

197 Exam Section 4: Item 47 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 47. A70-year-old man comes to the physician because of a 4-month history of fatigue and cough productive of blood-tinged sputum. He has a history of hyperlipidemia. He had smoked 1 pack of cigarettes daily for 30 years until he quit 5 years ago. His temperature is 37.2°C (99°F), pulse is 120/min, respirations are 20/min, and blood pressure is 130/85 mm Hg. Diffuse crackles and wheezes are heard on inspiration. A chest x-ray is shown. Serum studies show: TEP 138 mEq/L 5.0 mEq/L 102 mEq/L 24 mEq/L 15 mg/dL Na+ K+ CI- HCO3- Ca2+ In addition to intravenous hydration and induction of diuresis to treat his hypercalcemia, it is most appropriate for this patient to begin therapy with a drug that inhibits the activity of which of the following cells? A) Chondroblasts B) Chondrocytes C) Osteoblasts D) Osteoclasts E) Parathyroid cells

D. Squamous cell carcinoma of the lung commonly occurs in smokers and presents with symptoms of weight loss, malaise, cough, blood-tinged sputum, and possible recurrent pneumonia due to blockage of bronchi by the malignant growth. Tumors that occur in the lung apices and impinge on the thoracic outlet are referred to as Pancoast tumors. A common paraneoplastic syndrome that occurs in squamous cell carcinoma of the lung is hypercalcemia due to secretion of parathyroid hormone- related peptide (PTHPP). This protein acts similarly to parathyroid hormone (PTH), causing an increase in osteoclast activity with resultant osteolysis, liberating free calcium and increasing the serum calcium concentration. PTHIP also increases the resorption of calcium and excretion of phosphate by the kidney. Laboratory values typically demonstrate a hypercalcemia and hypophosphatemia with a low PTH level. Treatment of hypercalcemia related to PTHTP requires the administration of bisphosphonates, which inhibit the activity of osteoclasts. Bisphosphonates are similar in structure to pyrophosphate and thus incorporate into the bone mineral matrix. When this bone mineral matrix is broken down by osteoclasts, the bisphosphonates enter the osteoclasts and lead to the initiation of osteoclast apoptosis. Definitive management requires treatment of the primary squamous cell carcinoma of the lung with a combination of either surgery, chemotherapy, and/or radiation depending on the stage. Incorrect Answers: A, B, C, and E. Inhibition of chondroblasts (Choice A) would not decrease serum calcium as chondroblasts do not regulate calcium homeostasis. Inhibition of chondroblasts would lead to the decreased ability of cartilage to remodel and heal. Inhibition of chondrocytes (Choice B) would lead to the decreased production of cartilage matrix such as type Il collagen and aggrecan, leading to a decreased volume of cartilage. Inhibition of osteoblasts (Choice C) would lead to decreased bone formation during the natural process of bone turnover. Decreased osteoblast activity naturally occurs with aging and leads to an imbalance of relatively increased osteoclast activity resulting in decreased bone mineral density with age. Inhibition of parathyroid cells (Choice E) would lead to a decreased production of parathyroid hormone. Although this would lead to decreased PTH-dependent osteoclast activity and renal tubular calcium absorption, this would not inhibit the production and activity of PTHRP by the malignant squamous cells. In this patient, the release of PTH by parathyroid cells would likely already be downregulated due to the high calcium levels driven by PTHPP. Direct inhibition of osteoclasts is therefore required. Educational Objective: Paraneoplastic production of PTHPP is commonly caused by squamous cell carcinoma of the lung. Osteolytic metastases of multiple myeloma and breast cancer can also lead to hypercalcemia. Hypercalcemia associated with malignancy is often treated with bisphosphonates to induce osteoclast inhibition. %3D Previous Next Score Report Lab Values Calculator Help Pause

199 Exam Section 4: Item 49 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 49. A 70-year-old woman is brought to the emergency department 30 minutes after she was found unresponsive at home. She appears stuporous. Physical examination shows 2-mm-diameter pupils that are reactive to light. Eye movements are full with ice water caloric stimulation. She withdraws the extremities symmetrically to painful stimuli. An MRI of the brain is shown; the arrowheads indicate abnormalities. Which of the following is most likely to be present in this patient 2 months later? A) Difficulty recognizing familiar faces B) Hesitant, telegraphic language output C) Impaired abstraction and problem solving D) Inability to learn and recall new facts E) Inappropriate, disinhibited behavior

D. The hippocampus, located in the mesial temporal lobe, mediates learning and memory. The CA1 region of the hippocampal cortex is frequently damaged in anoxic brain injury due to the high metabolic demand of this region. The hippocampus is very vulnerable to ischemic injury. The patient's brainstem and spinal cord are functioning on physical examination, but she still demonstrates hippocampal damage on MRI of the brain. The hippocampus forms anterograde, declarative memories (eg, memories about facts) and consolidates these memories into long-term memories. The hippocampus also assists with spatial navigation. Additionally, the hippocampus is part of the limbic system and forms strong connections with the brain regions that generate emotions such as the amygdala. As such, the hippocampus mediates the recall of emotional memories. Bilateral lesions of the hippocampus classically lead to an amnesic syndrome in which new facts cannot be learned or recalled. Incorrect Answers: A, B, C, and E. Difficulty recognizing familiar faces (Choice A), or prosopagnosia, results from damage to the non-dominant fusiform gyrus, which mediates facial recognition. The fusiform gyrus is located on the inferior aspect of the brain between the temporal and occipital lobes, lateral to the parahippocampal gyrus and hippocampus. The fusiform gyrus is less vulnerable than the hippocampus to ischemic damage due its lower metabolic needs. Hesitant, telegraphic language output (Choice B) would result from a lesion of Broca area. Broca area is located within the left posterior, inferior frontal lobe of the dominant hemisphere. This patient has bilateral lesions of the mesial temporal lobe. Impaired abstraction and problem solving (Choice C) would result from a lesion of the prefrontal cortex or another higher-order association area such as the angular gyrus. These brain areas are located outside of the mesial temporal lobe. Inappropriate, disinhibited behavior (Choice E) would result from bilateral lesions of the amygdala, which causes Kluver-Bucy syndrome. Kluver-Bucy syndrome features hyperorality, hyperphagia, and hypersexuality. The amygdala is less vulnerable than the hippocampus to ischemic damage during anoxic brain injury and would appear smaller in coronal cross-section than the hippocampus. Educational Objective: The hippocampal cortex is vulnerable to ischemic damage in anoxic brain injury. Bilateral hippocampal damage leads to an amnesic syndrome in which patients cannot form or consolidate anterograde, declarative memories. Previous Next Score Report Lab Values Calculator Help Pause

128 Exam Section 3: Item 28 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 28. A 33-year-old man comes to the physician because of intermittent numbness and pain of his right arm during the past 6 months. His blood pressure is 120/80 mm Hg in both upper extremities; the pulse examination shows no abnormalities. A chest x-ray shows a right complete cervical rib. A CT scan of the neck and chest shows compression of the brachial plexus. Which of the following vessels is at greatest risk for injury during operative removal of the right cervical rib? O A) Brachiocephalic artery B) Right common carotid artery C) Right internal carotid artery D) Right subclavian artery

D. The right subclavian artery originates from the brachiocephalic artery in the mediastinum and courses laterally deep to the clavicle. As this artery courses past the first rib, it becomes the axillary artery, surrounded by the brachial plexus, then courses distally to supply the arm. A cervical rib is an extra rib that articulates with the C7 vertebra. The cervical rib can extend into the area between the rib of the first thoracic vertebra and the clavicle, which may result in abnormal pressure placed on the structures coursing through this region. These structures include the divisions of the brachial plexus, the subclavian artery, and the subclavian vein. This can result in associated extremity numbness, tingling, pain, and potential muscle wasting if chronic. This constellation of findings is termed thoracic outlet syndrome. Treatment requires removal of the cervical rib to alleviate compression of adjacent structures and the associated symptoms. Care must be taken not to injure the underlying subclavian artery, which typically lies adjacent to the cervical rib. Incorrect Answers: A, B, and C. The brachiocephalic artery (Choice A) originates from the aorta and courses only a few centimeters prior to bifurcating into the right subclavian artery and the right common carotid artery. This structure is intrathoracic and is at low or risk of injury during cervical rib surgery. The right common carotid artery (Choice B) originates from the brachiocephalic artery and courses cranially to supply structures of the head. It terminates by bifurcating into the internal and external carotid arteries. This structure is medial, placing it at lower risk of injury during cervical rib removal. The right internal carotid artery (Choice C) originates from the right common carotid artery just below the angle of the jaw and proceeds cranially to perfuse the brain. This structure is too superior to be at risk during resection of a rib at the level of the seventh cervical vertebra. Educational Objective: The right subclavian artery courses from the brachiocephalic trunk laterally to the border of the first rib, deep to the clavicle. This structure can be compressed during extreme ranges of motion of the upper extremity and when an intervening cervical rib is present. The subclavian artery is at risk of injury during removal of a cervical rib due to its position. %3D Previous Next Score Report Lab Values Calculator Help Pause

108 Exam Section 3: Item 8 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment MALE 8. A 27-year-old primigravid woman at 36 weeks' gestation comes to the physician for a prenatal visit. Physical examination shows a uterus consistent in size with a 33-week gestation, so the patient is referred for ultrasonography, which shows normal fetal measurements. The ultrasonography shown is an image of the male fetal scrotum; the testes are indicated by the arrows. Which of the following is the most likely underlying cause of this finding? A) High attachment of the tunica albuginea B) Lack of scrotal attachment of the gubernaculum C) Obstruction of the posterior urethra D) Patent processus vaginalis E) Twisting of the spermatic cord

D. The ultrasonographic findings demonstrate testes that are surrounded by hypoechoic fluid, consistent with a diagnosis of hydrocele in the fetus. The testes descend from the abdomen into the scrotum by passing through the anterior abdominal wall and processus vaginalis, guided by the gubernaculum. Hydrocele results from a patent processus vaginalis that does not obliterate following descent of the testis, allowing peritoneal fluid to collect in the scrotal space. While typically unilateral, cases may be bilateral. Most cases resolve spontaneously prior to birth. Persistent cases may require surgical resection. Incorrect Answers: A, B, C, and E. High attachment of the tunica albuginea (Choice A) when referencing the fibrous tissue that invests the testis (the penis also contains a tunica albuginea), results in increased motility of the testis and spermatic cord within the tunica vaginalis, which may lead to testicular torsion. Lack of scrotal attachment of the gubernaculum (Choice B) may lead to an undescended testis due to inability of the gubernaculum to guide the testis through the inguinal canal to the scrotum. Ultrasonography would not demonstrate the testes inside the scrotum, as is found in this fetus. Obstruction of the posterior urethra (Choice C) would result in oligohydramnios and Potter sequence. Fluid would not accumulate in the scrotal space. Twisting of the spermatic cord (Choice E) would result in in utero testicular torsion and infarction of the testes. Bilateral in utero testicular torsion is rare; a patent processus vaginalis leading to hydrocele is more likely. However, it would present with a large accumulation of fluid within the scrotum. Educational Objective: Fetal hydrocele results from patency of the processus vaginalis, which allows peritoneal fluid to collect in the scrotum. Ultrasonography demonstrates scrotal accumulation of hypoechoic fluid, which may be bilateral. Most cases resolve spontaneously. II Previous Next Score Report Lab Values Calculator Help Pause

182 Exam Section 4: Item 32 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 32. Feedback inhibition of the activity of the enzyme that catalyzes this reaction is critical in the control of the de novo synthesis of which of the following? -0-CH2 A) Complex carbohydrates HH H. B) Folates OH OH C) Pentose phosphates Glutamine D) Purine nucleotides Glutamate PPi O E) Pyrimidine nucleotides P-0-CH2 NH2 HH H. ОН OH

D. This chemical reaction demonstrates de novo synthesis of purine nucleotides. Purine and pyrimidine nucleotides are the building blocks of DNA. Purine and pyrimidine nucleotides can be synthesized de novo or salvaged from intermediate byproducts of DNA or RNA degradation. In the de novo synthesis of both purines and pyrimidines, ribose 5-phosphate (provided by the pentose phosphate pathway) is initially converted to 5-phospho-a-D-ribosyl 1-pyrophosphate (PRPP) by PRPP synthetase. In the first step that commits the nucleotide precursor to the purine synthesis pathway, PRPP is then converted to 5'-phosphoribosylamine (PRA) by amidophosphoribosyltransferase (also called PRPP amidotransferase). In this reaction, glutamine serves as a nitrogen donor and is converted to glutamate. PRA is ultimately converted to inosine monophosphate (IMP), which can then be converted to adenine or guanine monophosphate (AMP or GMP). AMP and GMP can then form phosphodiester bonds with other nucleotides to form DNA. IMP, AMP, and GMP inhibit PRPP amidotransferase activity. Immune modulating and cancer medications inhibit enzymes in the purine and pyrimidine synthesis pathways. Incorrect Answers: A, B, C, and E. Complex carbohydrates (Choice A) are oligosaccharides in which monosaccharides are linked by glycosidic bonds rather than the phosphodiester bonds of nucleotide chains. Complex carbohydrate synthesis is not regulated by feedback inhibition. Though folates (Choice B) are involved in the enzymatic reactions that synthesize pyrimidine and purine bases, humans cannot synthesize folates. Humans instead rely on dietary sources of folates. The pentose phosphate pathway (Choice C) ultimately converts glucose-6-phosphate to pentoses, NADPH, and ribose-5-phosphate. Ribose-5-phosphate is a nucleotide precursor that is used to synthesize purines and pyrimidines. However, glutamine is not involved in the synthesis of pentoses. Additionally, pentoses do not contain nitrogen as the purine precursor in this reaction does. In the synthesis of pyrimidine nucleotides (Choice E), PRPP (the reactant) is used only to donate a ribose group to a preformed pyrimidine ring to form a nucleotide. In the purine synthesis reaction above, atoms and molecules are incrementally added to PRPP (eg, nitrogen from glutamine), which ultimately results in nucleotide formation. Educational Objective: The first committed step of the de novo synthesis of purine nucleotides involves the enzyme PRPP amidotransferase. PRPP amidotransferase converts PRPP to a molecule called PRA using glutamine as a nitrogen donor, and the nucleotides IMP, AMP, and GMP are ultimately produced. These nucleotides then inhibit further PRPP amidotransferase activity. %3D Previous Next Score Report Lab Values Calculator Help Pause

158 Exam Section 4: Item 8 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 8. A 58-year-old woman is brought to the emergency department 6 hours after the sudden onset of left-sided facial, arm, and leg weakness. She has hypertension treated with a diuretic. She reports the sensation of a fluttering heart for 1 week. Her pulse is 125/min and irregular, and blood pressure is 135/80 mm Hg. Physical examination shows a left-sided facial droop. There is flaccid weakness of the left upper and lower extremities. An ECG shows a rapid, irregular ventricular rate with no discernable P waves. Which of the following therapies would best improve function of the weakened muscles over the next 3 hours? A) Acetylcholine agonist B) Acetylcholine antagonist C) Antifibrinolytic D) Fibrinolytic E) Procoagulant

D. This patient with a likely thromboembolic cerebral infarction (stroke) from atrial fibrillation would benefit from fibrinolytic therapy. Strokes occur due to ischemic or hemorrhagic loss of blood supply to the brain and manifest as focal neurological deficits related to the dysfunction of the affected brain region. Approximately 80 to 85% of strokes are ischemic, commonly arising from thromboembolic disease, for which atrial fibrillation is a major risk factor. Atrial fibrillation typically presents with palpitations, fatigue, lightheadedness, and mild dyspnea if symptomatic. On ECG, atrial fibrillation demonstrates irregularly irregular RR intervals without discernible P waves. Prolonged atrial fibrillation leads to left atrial hemostasis and increases the risk of thrombosis. A thrombus from this patient's left atrium likely embolized to the right internal carotid artery supplying the right precentral gyrus, resulting in left-sided hemiparesis. Alteplase (tissue plasminogen activator) is a fibrinolytic medication utilized in ischemic strokes that binds to fibrin clots and converts plasminogen to plasmin, which lyses clots. If given early (generally within 3 to 4.5 hours after onset of the event), alteplase may promote neurological recovery. Incorrect Answers: A, B, C, and E. Acetylcholine agonists (Choice A) and antagonists (Choice B) are unlikely to be helpful in lysing this patient's thromboembolic clot, as the interaction of acetylcholine with the coagulation cascade is poorly defined. Acetylcholine agonists (eg, bethanechol) may improve cognition in patients with dementia, decrease heart rate, improve gut peristalsis, increase bladder contraction, and increase exocrine gland secretions. Acetylcholine antagonists (eg, benztropine) typically increase heart rate, decrease gut and bladder activity, and worsen cognitive function. They also act on muscle at the motor end-plate; however, this patient's weakness results from central nervous system dysfunction without pathology at the muscle fiber itself. Antifibrinolytic (Choice C) therapy (eg, tranexamic acid) displaces plasminogen from fibrin clots to promote hemostasis during intraoperative bleeding, heavy menstrual bleeding, and traumatic hemorrhage. Procoagulant (Choice E) therapy (eg, protamine, coagulation factors) increases activation of the coagulation cascade. Both antifibrinolytic and procoagulant therapy would not lyse this patient's clot and may lead to further thrombosis. Educational Objective: Patients with atrial fibrillation are at risk for thromboembolism due to left atrial hemostasis, which may result in ischemic strokes. Strokes manifest as focal neurological deficits related to the dysfunction of the affected brain region. Ischemic strokes are treated with fibrinolytic therapy, which promotes neurological recovery. %3D Previous Next Score Report Lab Values Calculator Help Pause

193 Exam Section 4: Item 43 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 43. A 28-year-old man is participating in a study of the effect of altitude and conditioning on combat readiness. He is a Special Forces sergeant in the army. While he exercises in an environmental chamber, the ambient Po, is decreased from 160 mm Hg to 60 mm Hg. The decrease in ambient Po, will most likely cause an increase in which of the following in this patient? A) Arterial diastolic pressure B) Arterial pulse pressure C) Arterial systolic pressure D) Mean arterial pressure E) Mean pulmonary artery pressure

E. A decrease in the ambient partial pressure of oxygen (Po2) results in hypoxic pulmonary vasoconstriction with a resultant increase in the mean pulmonary artery pressure. Acute hypoxia stimulates contraction of smooth muscle cells in the pulmonary vasculature. This mechanism diverts blood flow away from regions of the lung that have low alveolar oxygen content in what is called ventilation-perfusion matching. Ventilation-perfusion matching serves to maximize the functional surface area of the alveolar-capillary interface that is participating in gas exchange. Chronic hypoxic pulmonary vasoconstriction may occur from living at altitude or in the setting of chronic lung disease (eg, chronic obstructive pulmonary disease). Complications include right ventricular hypertrophy and tricuspid regurgitation that may progress to right-sided heart failure. Incorrect Answers: A, B, C, and D. An acute change in ambient oxygen content is not expected to cause a significant change in the arterial diastolic pressure (Choice A), arterial pulse pressure (Choice B), arterial systolic pressure (Choice C), or the mean arterial pressure (Choice D). A healthy patient is able to compensate for the reduced oxygen content in the alveoli by hyperventilating and increasing cardiac output through increased heart rate to maintain adequate oxygen delivery to the organs. Localized hypoxemia induces vasodilation in systemic arterial blood vessels, which promotes increased blood flow and oxygen delivery to the area. Educational Objective: Low oxygen content in the blood induces vasoconstriction in the pulmonary vasculature in order to direct blood flow to regions with a higher oxygen content. The opposite effect occurs in the systemic vasculature, with vasodilation in response to hypoxemia. This promotes increased delivery of oxygen to hypoxic tissue. %3D Previous Next Score Report Lab Values Calculator Help Pause

129 Exam Section 3: Item 29 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 29. A 24-year-old man who is comatose is admitted to the hospital because of a drug overdose and pneumonia. On admission, he is intubated and mechanically ventilated. While receiving a tidal volume of 500 mL, the patient's end-expiratory pressure is +5 cm H,O, and his end-inspiratory airway pressure is +25 cm H,0. An esophageal balloon is inserted, and an end-inspiratory pleural pressure of +20 cm H,0 is measured. Which of the following best represents this patient's respiratory system compliance? A) 0.01 cm H,O/mL B) 0.04 cm H,O /mL OC) 0.05 cm H,O/mL D) 20 mL/cm H,0 E) 25 mL/cm H,0 F) 100 mL/cm H,0

E. Compliance is the change in volume of a system per change in pressure. In the respiratory system, the compliance of the chest wall and the lungs are both considered to determine total system compliance. In a ventilated patient, the total compliance can be calculated by dividing the tidal volume by the change in pressure needed to deliver the breath (eg, the difference between end-inspiratory pressure and end-expiratory pressure). In this case, the compliance is equal to the tidal volume of 500 mL divided by the difference between end-inspiratory and end-expiratory pressure (25 cm H20 - 5 cm H20 = 20 cm H20), which equals 25 mL/cm H20. Complications of positive-pressure mechanical ventilation include barotrauma (pressure-induced injury). The pressure experienced by the lung tissue is the transpulmonary pressure, which is the difference between the alveolar pressure and the pleural pressure. The alveolar pressure is determined by the ventilator. The pleural pressure is challenging to measure directly and is typically estimated by utilizing an esophageal balloon. Incorrect Answers: A, B, C, D, and F. Choices A, B, and C are incorrect as they represent the inverse relationship between the change in volume and the change in pressure (pressure/volume = H2O/mL) in describing compliance, which should be the change in volume divided by the change in pressure. Choice D is obtained by dividing the tidal volume by the end-inspiratory pressure of 25 cm H20. This does not represent the change in pressure of the system during the respiratory cycle, which must account for the end-expiratory pressure. Choice F is obtained by using the difference between the end-inspiratory pressure and the pleural pressure as the change in pressure of the system. This value represents the transpulmonary pressure at the end of inspiration, which determines the risk of suffering barotrauma, but does not provide information about the total respiratory system compliance. Educational Objective: Compliance is the measure of the change in volume of a system in response to a change in pressure. The respiratory system compliance, in the setting of mechanical ventilation, can be determined by dividing the tidal volume delivered by the difference between the end-expiratory and end-inspiratory pressures. %3D Previous Next Score Report Lab Values Calculator Help Pause

132 Exam Section 3: Item 32 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 32. A 67-year-old woman has had decreased pain and temperature sensation in the left lower extremity for 1 week. The most likely cause is damage to which of the following labeled sites in the cross section of the spinal cord? Left JA B Right H. F A) B) C) D) E) F) G) H) 1) J)

E. Damage to the right-sided spinothalamic tract (label E) in this cervical spinal cord section (as indicated by the ovoid shape and predominance of white matter) would cause left-sided impairment of pain and temperature sensation. The spinothalamic tract originates from the sensory nerve ending, travels to the dorsal root ganglion where its cell body is located, enters the spinal cord, and synapses with the second-order neuron in the ipsilateral dorsal horn of the spinal cord. The second-order neuron then decussates at the anterior white commissure of the spinal cord (ascending two levels during its decussation), and then ascends within the contralateral spinothalamic tract to the ventral posterolateral nucleus of the thalamus. The third-order neuron then ascends to the primary sensory cortex. An injury to the spinothalamic tract in the spinal cord results in contralateral neurologic deficits as the second-order neuron has already decussated from the side of the first-order neuron sensory nerve ending. The spinothalamic tract is organized somatotopically in the spinal cord such that the lateral region controls pain and temperature and the anterior region controls crude touch and pressure. Incorrect Answers: A, B, C, D, F, G, H, I, and J. Labels A and J represent the fasciculi gracilis of the dorsal column-medial lemniscus tract, while labels B and I represent the fasciculi cuneatus of the dorsal column-medial lemniscus tract. The dorsal column-medial lemniscus tract originates from the sensory nerve ending and ascends ipsilaterally within the fasciculus gracilis or cuneatus until it synapses with the second-order neuron in the ipsilateral nucleus gracilis or cuneatus, respectively. The second-order neuron then decussates in the medulla and synapses in the thalamus. The third-order neuron then ascends to terminate in the primary sensory cortex. Dorsal column-medial lemniscus tract lesions within the cervical spinal cord will lead to ipsilateral deficits in pressure, vibration, fine touch, and proprioception. The fasciculi gracilis contain ascending sensory neurons for the lower body, while the fasciculi cuneatus contain ascending sensory neurons for the upper body. Labels C and H represent the lateral corticospinal tracts. Upper motor neurons of the lateral corticospinal tract originate in the primary motor cortex, descend ipsilaterally through the internal capsule, crus cerebri of the midbrain, ventral pons, and medullary pyramids, where they then decussate in the caudal medulla, and then descend contralaterally in the lateral corticospinal tract of the spinal cord to synapse with the contralateral lower motor neuron. Lateral corticospinal tract lesions within the cervical spinal cord will lead to an ipsilateral upper motor neuron pattern of motor weakness, including hyperreflexia and increased muscle tone. Labels D and G represent the anterior horns of the spinal cord, which contain the cell bodies of lower motor neurons. Lesions of the anterior horn would result in an ipsilateral lower motor neuron pattern of skeletal muscle weakness including hyporeflexia and decreased muscle tone. Label F represents the left-sided spinothalamic tract. Lesions of this area would result in impairment in pain and temperature sensation in the right lower extremity. Educational Objective: Spinal cord lesions of the lateral spinothalamic tract cause contralateral deficits in pain and temperature, whereas lesions of the anterior spinothalamic tract cause contralateral deficits in crude touch and pressure. Alternatively, spinal cord lesions of the corticospinal and dorsal column-medial lemniscus tracts produce ipsilateral neurologic deficits. Previous Next Score Report Lab Values Calculator Help Pause E.

106 Exam Section 3: Item 6 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 6. A 28-year-old woman at 32 weeks' gestation comes to the physician because of a 4-day history of fever and back pain. She says that during this time she also has been crying frequently. Her temperature is 38°C (100.4°F). Physical examination shows costophrenic angle tenderness. The photomicrograph shown represents her disease. Which of the following mechanisms is the most likely cause? A) Chorioamnionitis B) Endometritis C) Glomerulonephritis D) Hematogenous infection E) Obstructive uropathy F) Pelvic inflammatory disease

E. Fever, chills, and flank pain are classic findings in the diagnosis of pyelonephritis, an infection of the kidney that commonly results from retrograde spread of a urinary tract infection. Urinary tract infections are much more common in women due to the shorter urethra and favorable regional environment for bacterial growth. Pregnant patients are at higher risk for urinary tract infections including pyelonephritis due to the stagnation of urine from progesterone-induced decreased ureteral tone and motility, along with mechanical compression of the ureters at the pelvic brim, bladder, and ureteral orifices by the gravid uterus, resulting in obstructive uropathy. If untreated, bacteria can travel up the ureter to infect the kidney, resulting in pyelonephritis. Pyelonephritis typically presents with dysuria, flank pain, and associated systemic symptoms such as fever, rigors, nausea, vomiting, myalgias, arthralgias, and fatigue. On examination, patients with pyelonephritis typically demonstrate high fevers, tachycardia, leukocytosis, and costovertebral angle tenderness. Urinalysis in pyelonephritis often shows >10 leukocytes per high power field, may show red cells as evidence of mucosal inflammation, and may show white blood cell casts. Bacteria are generally seen on microscopy, with Gram- negative rods being the most common pathogen. Incorrect Answers: A, B, C, D, and F. Chorioamnionitis (Choice A) is a bacterial infection of the fetal membranes and commonly occurs in the setting of premature or prolonged rupture of membranes. Maternal fever and leukocytosis are characteristic, and the uterus may be tender to palpation. Endometritis (Choice B) is an acute, typically polymicrobial, infection of the uterine endometrium involving a mixture of aerobes and anaerobes from the genital tract. During labor and delivery, microbes from the genital tract can enter the uterine cavity and result in infection. Patients typically present with fever, severe tenderness of the uterine fundus, and mucopurulent vaginal discharge. Glomerulonephritis (Choice C) 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. It would not typically cause ureteral obstruction. Hematogenous infection (Choice D) can result from pyelonephritis or other infections that spread into the bloodstream, causing sepsis. Patients with sepsis would typically present with tachycardia, systemic symptoms, and may be ill-appearing. Pelvic inflammatory disease (Choice F) can be caused by sexually transmitted infections such as Chlamydia trachomatis and Neisseria gonorrhoeae. It typically presents with fever, mucopurulent cervical discharge, and cervical or adnexal tenderness. Adnexal tenderness may be present in cases of tubo-ovarian abscess. Educational Objective: Pyelonephritis typically presents with fevers, nausea, vomiting, and flank pain with associated costovertebral angle tenderness on physical examination. This most commonly occurs from ascending urinary tract infections. Pregnant patients are at a higher risk for pyelonephritis due to the stagnation of urine from progesterone-induced decreased ureteral tone and motility, along with mechanical compression of the ureters at the pelvic brim, bladder, and ureteral orifices by the gravid uterus. Previous Next Score Report Lab Values Calculator Help Pause

109 Exam Section 3: Item 9 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 9. A 98-year-old woman dies of respiratory failure. At autopsy, gross examination shows a small heart with a brown discoloration. Histologic examination of the cardiac tissue by light microscopy shows light brown pigment within the cardiac myocytes, particularly in the perinuclear region. This material does not stain positively for iron. Which of the following best describes this pigment? A) Anthracotic pigment B) Free fatty acids C) Hemosiderin D) Laminin E) Lipofuscin

E. Lipofuscin is a yellow-brown pigment that is deposited with time and indicates wear on the tissues. It is associated with normal aging and an autopsy of an aged individual, as in this case, will demonstrate lipofuscin deposits in many organs, including the heart, colon, liver, and kidney. Lipofuscin is formed by oxidation of the organellar membranes, which have been autophagocytosed. Another brown pigment that may be seen in tissues is hemosiderin. Hemosiderin is pigment from iron that deposits in tissues after hemorrhage or erythrocyte extravasation. In hemochromatosis, a condition of increased iron deposition, hemosiderin is found on liver biopsy. Hemosiderin and lipofuscin can be differentiated from each other by iron staining. Hemosiderin stains positively for iron while lipofuscin does not. Incorrect Answers: A, B, C, and D. Anthracotic pigment (Choice A) is secreted by pigment-producing fungi. It also presents as carbonaceous deposits in the lung in pulmonary anthracosis. Pigment-producing fungi are rare but induce infections like phaeohyphomycosis and chromoblastomycosis. These organisms are endemic in soil and cause superficial cutaneous infections. Free fatty acids (Choice B) may appear yellow when they are deposited together in the skin, such as in xanthelasma. However, they would not demonstrate any yellow or brown coloration on light microscopy. Hemosiderin (Choice C) is an iron deposited in the tissues and thus will stain positively for iron on histopathologic examination. It would not be expected to be present within cardiac myocytes outside of hemochromatosis. Laminin (Choice D) is a protein found in cell membranes and hemidesmosomes. Lipofuscin is formed by oxidation of organellar membranes, which may include laminin, however laminin alone does not appear yellow-brown on light microscopy. Educational Objective: Lipofuscin is a yellow-brown pigment that indicates wear on the tissues and is present in most organs on autopsy of an aged individual. It is derived from organellar membranes, not iron. %3D Previous Next Score Report Lab Values Calculator Help Pause

187 Exam Section 4: Item 37 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 37. A 32-year-old man comes to the physician because of a 6-month history of progressive drooping of the eyelids and facial weakness. Muscle weakness is exacerbated by use and relieved by rest. If a neoplasm is the cause of these findings, which of the following is the most likely location in this patient? A) Femur B) Liver C) Lymph nodes D) Meninges E) Thymus

E. Myasthenia gravis is associated with thymomas. Myasthenia gravis is an autoimmune disorder of neuromuscular transmission that has a bimodal age of onset, peaking in both young adults (teens to 20s) and older adults (50s to 60s). Patients with myasthenia gravis possess antibodies that block or destroy nicotinic acetylcholine receptors (NACHRS), which interferes with neuromuscular junction signaling. Many patients with myasthenia gravis demonstrate thymic hyperplasia or a thymoma. In myasthenia gravis, the thymus, which is the site of T-lymphocyte maturation, may demonstrate abnormalities in the presentation of the NACHR as an antigen to helper T lymphocytes. This abnormal antigen presentation is postulated to result in the autoimmune attack targeting NACHRS on skeletal muscle cells. Patients present with progressively worsening skeletal muscle weakness at rest and fatigability on activation in focal or generalized muscle groups. The ocular, bulbar, and neck muscles are commonly affected. Most concerningly, respiratory muscle weakness can lead to respiratory insufficiency, and in an acute flare, these patients may require mechanical ventilation. Management includes symptomatic treatments such as acetylcholinesterase inhibitors along with immune-modulating therapies. In patients with or without a thymoma, complete thymectomy may be curative. Incorrect Answers: A, B, C, and D. Neoplasms of the femur, liver, lymph nodes, and meninges (Choices A, B, C, and D) have not been implicated in the pathogenesis of myasthenia gravis. Though T lymphocytes are involved in the autoimmune etiology of myasthenia gravis, thymomas rather than lymphomas contribute to abnormal antigen presentation. Educational Objective: In patients with myasthenia gravis, a thymoma promotes the autoimmune attack of nAChRs on skeletal muscle cells, leading to focal or generalized muscle weakness and fatigability. Complete thymectomy may be curative. %3D Previous Next Score Report Lab Values Calculator Help Pause

155 Exam Section 4: Item 5 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 5. A 10-year-old boy is brought to the physician by his parents because of increased urinary frequency and progressive fatigue during the past 3 weeks. Physical examination shows no abnormalities. His fasting serum glucose concentration is 350 mg/dL. Urinalysis shows 4+ glucose. Which of the following mechanisms in the proximal tubule is the most likely cause of the glycosuria in this patient? A) Decreased sodium diffusion across the apical membrane B) Inhibition of the sodium/glucose cotransporter-1 (SGLT-1) carrier protein C) Saturation of the glucose transporter protein-1 (GLUT-1) carrier protein D) Saturation of glucose transporter protein-2 (GLUT-2) carrier protein E) Saturation of the sodium/glucose cotransporter-1 (SGLT-1) carrier protein

E. Saturation of the sodium/glucose cotransporter-1 (SGLT-1) carrier protein in the renal proximal tubule accounts for this patient's glycosuria, likely in the setting of a new diagnosis of type I diabetes mellitus. Autoimmune destruction of pancreatic islet cells causes profound hyperglycemia. Most cellular glucose transport proteins require the presence of insulin to function; thus, insulin deficiency leads to high serum levels of glucose but low intracellular stores of glucose. Circulating glucose is filtered through the glomerulus and resorbed from filtrate in the tubules via the action of SGLT-1 and SGLT-2, which transport glucose across the apical membrane in an adenosine triphosphate (ATP)-dependent manner. SGLT-1 accounts for only about 10% of glucose resorption while SGLT-2 accounts for about 90%. These receptors have a maximum rate at which they can resorb glucose in filtrate, and when blood glucose concentration exceed approximately 180 mg/dL, the rate of glucose filtration exceeds the ability of the transporter proteins to resorb it. This results in glycosuria. Incorrect Answers: A, B, C, and D Decreased sodium diffusion across the apical membrane (Choice A) would not result in glycosuria, although hyperglycemia can indirectly affect total body sodium balance. Sodium is passively reabsorbed along a concentration gradient along with water and chloride and is also resorbed via the action of SGLT-1 and a sodium-hydrogen antiporter. Inhibition of the sodium/glucose cotransporter-1 (SGLT-1) carrier protein (Choice B) does not account for glycosuria in hyperglycemia. SGLT-1 is expressed on the luminal brush border in enterocytes where it absorbs glucose from the lumen and is expressed in the renal proximal tubules where it contributes to glucose resorption. Inhibition of SGLT-2, not SGLT-1, is a characteristic of a class of medications used to treat type |l diabetes mellitus and includes medications such as empagliflozin and canagliflozin. These medications are not used to treat type I diabetes mellitus, which occurs as a result of insulin deficiency. Saturation of the glucose transporter protein-1 (GLUT-1) carrier protein (Choice C) is not the mechanism by which glycosuria occurs in patients with hyperglycemia. The GLUT-1 transporter is most active on the surface of erythrocytes and acts to bring glucose into the cytoplasm where it is phosphorylated to glucose-6-phosphate. It is also expressed in endothelial cells and is not the primary glucose transport protein in the nephron. Saturation of glucose transporter protein-2 (GLUT-2) carrier protein (Choice D) occurs after large oral glucose loads. GLUT-2 is expressed in the liver, pancreas, kidneys, and gastrointestinal tract. It functions as a glucose-sensing receptor in the liver and pancreas but plays less of a role of glucose transport in the nephron. Educational Objective: Filtered glucose is reclaimed by the nephron through the action of sodium-glucose cotransporter proteins. When serum glucose concentrations exceed approximately 180 mg/dL, these transporters begin to saturate and some filtered glucose cannot be recovered from the tubular filtrate, resulting in glycosuria. %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

167 Exam Section 4: Item 17 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 17. Which of the following characteristics of the glucocorticoid receptor is the most likely cause of the persistence of the pharmacologic effects of dexamethasone after the drug is eliminated from the body? A) Catalyzes formation of stable tyrosine-phosphate bonds O B) Competes with B-arrestin for binding to adrenergic receptors OC) Couples to G proteins with low intrinsic GTPase activity D) Depletes cellular stores of glutathione E) Induces proteins that remain functional after the drug is eliminated

E. Steroid hormones, such as dexamethasone, bind and activate intracellular receptors and is able to traverse the cell membrane due to their lipophilic nature. Other steroid hormones include estrogen, progesterone, testosterone, and aldosterone. All of these hormones are derived from cholesterol as a common precursor. Steroid hormone receptors are intracellular and reside either in the cytoplasm or the nucleus and regulate gene expression. Because of the effects mediated through gene regulation, the onset of action of steroid hormones occurs over the course of hours to days. Intracellular steroid receptors have a number of different protein domains. The DNA binding domain classically has a zinc finger motif as a tertiary structure. This domain has affinity with the regulatory (eg, promoter, silencer) regions of DNA. The altered protein expression induced by steroid hormones may persist even after the hormone itself is eliminated because the synthesized peptides may have a half-life distinctly longer than that of the drug itself. Incorrect Answers: A, B, C, and D. Formation of stable tyrosine-phosphate bonds (Choice A) is a feature of tyrosine kinase receptors, such as the insulin receptor. Once activated by homodimerization, the cytoplasmic portions of the receptor engage in autophosphorylation of specific tyrosine residues, which creates binding sites for downstream signaling effectors. Competition with B-arrestin for binding to adrenergic receptors (Choice B) is a feature of G proteins. Adrenergic receptors are G protein-coupled receptors and interact with G proteins once activated. B-arrestins compete to bind with and inactivate the adrenergic receptor in order to attenuate the signaling response. Coupling to G proteins with low intrinsic GTPase activity (Choice C) is also a feature of G protein-coupled receptors. Depletion of cellular stores of glutathione (Choice D) is characteristic of acetaminophen toxicity and is not a result of treatment with dexamethasone. Educational Objective: Steroid hormones, such as dexamethasone, bind and activate intracellular receptors that function as transcription factors. Activation of the steroid hormone receptor induces alterations in protein expression that persist after the hormone itself is eliminated. %3D Previous Next Score Report Lab Values Calculator Help Pause

101 Exam Section 3: Item 1 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 1. A 65-year-old man comes to the physician because of fever and a worsening cough productive of approximately % cup of blood-tinged sputum daily. He has smoked 12 packs of cigarettes daily for 45 years. His temperature is 38.4°C (101.2°F), respirations are 20/min, and blood pressure is 140/90 mm Hg. Physical examination shows bilateral clubbing of the digits. Pulmonary examination shows egophony, whispered pectoriloquy, and dullness to percussion in the area overlying the 2nd and 3rd ribs on the right anteriorly, just inferior to the right clavicle. Which of the following structures is the most likely site of an obstructing carcinoma? A) Main carina B) Right lower lobe bronchus C) Right main bronchus D) Right middle lobe bronchus E) Right upper lobe bronchus

E. The gross lobar anatomy of the lung contains three lobes of the right lung (upper, middle, and lower) and two lobes of the left lung (upper and lower). The lingula is a distinct projection of the upper lobe of the left lung and is homologous to the middle lobe of the right lung. The right upper lobe lies deep to the anterior 2nd and 3rd ribs on the right side of the thorax just inferior to the clavicle, and the right upper lobe bronchus is the most likely site of this patient's obstructing carcinoma. Risk factors for all major types of lung cancer include tobacco use, secondhand smoke, asbestos or radon exposure, and a family history of lung cancer. Lung cancer, in general, typically presents with cough, unintentional weight loss, hemoptysis, chest pain, dyspnea, and hoarseness. Occasionally, wheezing, focal rhonchi, or hypertrophic osteoarthropathy may be noted on examination. A mass that occludes a segment of the airway increases the risk for post-obstructive pneumonia. On physical examination, this manifests with egophony, whispered pectoriloquy, increased tactile fremitus, and dullness to percussion in the region of the pneumonia, which in this patient, correlates with the right upper lobe. Definitive diagnosis is made by chest imaging and biopsy. Prognosis is a function of the cancer type along with grading and staging of the disease. It is often detected once metastatic, at which point the prognosis is poor. Incorrect Answers: A, B, C, and D. The main carina (Choice A) marks the divergence of the trachea into the right and left mainstem bronchi. It is typically located at the level of the fourth thoracic vertebral body. An obstructing mass in this location may result in complete airway obstruction of both lungs. The right lower lobe bronchus (Choice B) branches from the bronchus intermedius and supplies the right lower lobe, which occupies most of the inferior and posterior aspects of the right hemithorax. A post-obstructive pneumonia involving the right lower lobe would be best appreciated by the presence of pneumonia exam findings along the anterior-inferior or posterior chest wall. The right main bronchus (Choice C) branches from the trachea at the carina and lies at the level of the fifth thoracic vertebral body along the right parasternal border. An obstructing mass of the right main bronchus may result in complete right lung collapse. The right middle lobe bronchus (Choice D) branches from the bronchus intermedius and supplies the right middle lobe, which forms a triangular wedge in the anterior-medial right hemithorax. Educational Objective: Post-obstructive pneumonia is a common complication of masses that occlude the large and medium-sized airways. Localization of the pneumonia and potential underlying may be possible on physical examination, although definitive diagnosis is achieved with chest imaging followed by a biopsy of the mass. %3D Previous Next Score Report Lab Values Calculator Help Pause

126 Exam Section 3: Item 26 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 26. A 13-year-old boy is scheduled to receive chemotherapy for a leukemia that has the histologic features of malignant lymphocytes. This neoplasm is further typed for cell surface and intracellular markers specific for lymphocyte subsets. The neoplastic cells do not express the following markers: CD4, CD8, surface IgM, surface IgG, cytoplasmic IgM and p-heavy chain, cytoplasmic IgG, and y-heavy chain. The leukemic cells express class I MHC molecules and show rearrangement of the T-lymphocyte receptor B-chain gene D and J segments. Which of the following is the normal counterpart of these malignant lymphocytes? A) Activated cytolytic effector T lymphocytes in the circulation B) Mature IgM-secreting B lymphocytes in the lymph node C) Mature immunoglobulin-secreting plasma cells in the lymph node D) Pre-B lymphocyte progenitor of mature B lymphocytes in the bone marrow E) T-lymphocyte thymocytes localized to the thymic cortex

E. The laboratory analysis of this acute lymphoblastic leukemia is most suggestive of malignant transformation of normal T-lymphocyte thymocytes localized to the thymic cortex. Leukemia may arise from the lymphoid or myeloid cell lines. Acute lymphoblastic leukemia is composed of lymphoid progenitor cells and is the most common hematologic malignancy of childhood. Cell markers are used to identify the underlying cell type. Normal T-lymphocyte development begins in the bone marrow with differentiation of hematopoietic stem cells into T-lymphocyte precursors. The precursors then migrate to the thymic cortex where they undergo rearrangement of the T-lymphocyte receptor B-chain gene D and J segments. I lymphocytes that recognize class I MHC molecules undergo positive selection and proliferate. The T lymphocytes at this stage are double negative (do not express CD4 or CD8). Following positive selection, T lymphocytes undergo rearrangement of the a-chain gene and enter the double positive stage with expression of CD4 and CD8. Double positive cells then differentiate into single positive CD4+ or CD8+ cells via interaction with either class II MHC or class I MHC molecules, respectively. The single positive cells migrate to the thymic medulla to undergo negative selection. T lymphocytes with a high affinity for self-antigens will undergo apoptosis to ensure self-tolerance. T lymphocytes that survive negative selection will enter the circulation as mature T lymphocytes. Incorrect Answers: A, B, C, and D. Activated cytolytic effector T lymphocytes in the circulation (Choice A) is incorrect as these cells will express CD8, which binds to class MHC on target cells. Effector T lymphocytes eliminate neoplastic cells and virus-infected cells by releasing cytotoxic granules and inducing apoptosis. Mature IgM-secreting B lymphocytes in the lymph node (Choice B) express surface IgM, cytoplasmic IgM, and p-heavy chain cell markers. IgM-secreting B lymphocytes are one of the first cells activated in the primary antibody response to a foreign antigen. Pre-B lymphocyte progenitors of mature B lymphocytes in the bone marrow (Choice D) also initially express IgM and u-heavy chains before progressing to the next stage of development. Mature immunoglobulin-secreting plasma cells in the lymph node (Choice C) express surface IgG and y-heavy chains. IgG is the primary class of immunoglobulin present in the circulation, which effects the secondary antibody response to a foreign antigen. Educational Objective: T lymphocytes undergo selection and maturation in the thymus prior to entering the circulation. Prior to positive selection for recognition of class I MHC molecules, T lymphocytes do not express CD4 or CD8. After positive selection, they express both before differentiating into single positive CD4+ or CD8+ cells. %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

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

139 Exam Section 3: Item 39 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 39. A67-year-old woman with acute myelogenous leukemia develops fever and chills 6 days after undergoing induction therapy. Her temperature is 38.3°C (100.9°F), and blood pressure is 110/60 mm Hg. Physical examination shows scattered petechiae over the upper and lower extremities and trunk. Laboratory studies show a leukocyte count of 100/mm3 (98% lymphocytes and 2% monocytes) and platelet count of 4000/mm3. Blood cultures grow Pseudomonas aeruginosa. A deficiency of which of the following is the most likely predisposing factor for development of bacteremia in this patient? A) Complement B) Immunoglobulin C) Lymphocytes D) Macrophages E) Platelets F) Segmented neutrophils

F. Acute myelogenous leukemia is an aggressive bone marrow malignancy that primarily affects adults. It is characterized by the presence of circulating blasts of the myeloid lineage. Patients are classified into different prognostic tiers based upon the presence or absence of specific mutations. The goal of initial therapy is to induce cancer remission, generally with cytarabine and daunorubicin. Complications of both the leukemia itself and induction therapy include pancytopenia because of the suppression of normal bone marrow. Patients may present with fatigue and dyspnea secondary to anemia, bleeding and petechiae secondary to thrombocytopenia, and infections due to neutropenia. Neutropenic fever is a life-threatening medical condition that refers to an acute infection and fever presenting in the setting of a deficiency of segmented neutrophils, defined as an absolute neutrophil count less than 500/mm3. Patients may not display typical signs of infection due to a severely impaired immune system. Patie broad-spectrum antibiotic therapy that includes activity against Pseudomonas aeruginosa. require hospitalization and Incorrect Answers: A, B, C, D, and E. Complement (Choice A) deficiencies are not typically associated with acute myelogenous leukemia as complement is synthesized in the liver. Complement deficiencies are associated with recurrent neisserial infections. Immunoglobulin (Choice B) and lymphocyte (Choice C) deficiencies are associated with congenital disorders of lymphocytes such as X-linked agammaglobulinemia, selective IgA deficiency, common variable immunodeficiency, severe combined immunodeficiency, and thymic aplasia. Lymphocytes and macrophages (Choice D) are also suppressed in the setting of cytotoxic chemotherapy, but an impaired initial innate immune response with neutropenia is the major predisposing factor for bacteremia. Platelet (Choice E) deficiencies result in bleeding complications, which may manifest as bruises, petechiae, mucosal bleeding from the gingiva, epistaxis, and/or more severe gastrointestinal hemorrhage or menorrhagia. Educational Objective: Complications of acute myelogenous leukemia are related to the impaired production of normal cell lines. This is further complicated by the administration of cytotoxic chemotherapy, which suppresses normal bone marrow function. Neutropenic fever is a medical emergency and is defined as the presence of a fever in the setting of an absolute neutrophil count less than 500/mm3. %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

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

137 Exam Section 3: Item 37 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 37. A physician is sad because he has to inform a patient of recent test results that indicate progression of carcinoma to the terminal phase. When the patient sees the physician's face, he begins to cry and says, "It's bad news, isn't it?" Which of the following responses by the physician is most appropriate? A) "How have you been since the last time I saw you?" B) "Let's talk about the positive aspects first." C) "Look on the bright side of things." D) "Tell me how you are feeling." E) "There are other people who have it a lot worse than you." F) "Yes, it is." G) "You've had several years better off than many others with this disease."

F. In patients who express the desire to be informed of a terminal diagnosis, physicians should deliver bad news directly and compassionately. This physician should answer the patient's direct question. Once the patient has been informed of the diagnosis, the physician should validate the patient's emotions by listening receptively, using reflective statements, inferring emotions from the expressed content, and normalizing the patient's experience. The physician should also deliver education and explain the next steps using small pieces of information. Incorrect Answers: A, B, C, D, E, and G. Small talk such as asking how the patient has been (Choice A) would constitute an avoidant approach to answering the patient's question and eliciting the patient's emotions. The physician should instead focus on the most emotionally salient topic, which is the bad news. The physician can then support the patient's emotions. Framing the news positively (Choices B, C, E, and G) would avoid answering the question and may invalidate this patient's natural emotional response. Encouraging words would minimize the situation and probably feel like platitudes to the patient given the gravity of the news. Allowing the patient to feel sadness and validating these emotions will help the patient process the news and ultimately move forward to make end-of-life preparations. Inviting the patient to share how he is feeling (Choice D) would be appropriate later in the conversation so that the physician can support the patient's emotions. First, the physician should answer the patient's question about the news. Educational Objective: For patients who desire to be informed of a terminal diagnosis, physicians should deliver bad news directly. Once the patient has been informed, the physician should validate the patient's emotions. %3D Previous Next Score Report Lab Values Calculator Help Pause

189 Exam Section 4: Item 39 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 39. A 65-year-old man comes to the physician for a follow-up examination. He has a 15-year history of poorly controlled type 2 diabetes mellitus, resulting in multiple peripheral neuropathies. Neurologic examination shows wasting of the interosseous muscles of the left hand and inability to abduct the fingers of this hand; plantar flexion of the right foot is absent. The function of which of the following pairs of nerves is most likely impaired in this patient? A) Median and common fibular (peroneal) B) Median and tibial C) Radial and common fibular (peroneal) D) Radial and tibial E) Ulnar and common fibular (peroneal) F) Ulnar and tibial

F. Long-standing diabetes mellitus can lead to the development of diabetic neuropathy. Chronic hyperglycemia leads to glycosylation of axonal proteins along with sorbitol-related osmotic damage to the neuron, with resulting progressive sensory and motor neuropathy. Initially, the condition mostly affects distal nerves, that is, the small nerves of the feet and fingers. Over time, the neuropathy progresses proximally, leading to pain and decreased sensation in the extremities. The associated decreased motor innervation can lead to muscle wasting, while decreased sensation may lead to skin and joint damage that fails to heal as the patient fails to decrease pressure on the injured region as protective sensation is limited. This patient presents with the inability to abduct the fingers of his hand and plantar flex his right foot. The interosseous muscles of the hand abduct and adduct the fingers and are innervated by the ulnar nerve. Plantar flexion of the foot is controlled by the muscles of the posterior compartment such as the gastrocnemius and soleus, which are innervated by the tibial nerve. Incorrect Answers: A, B, C, D, and E. The median nerve (Choice A and B) provides motor innervation to the majority of the flexors of the forearm and the thenar eminence. In the anterior forearm, the only muscles that are not innervated by the median nerve are the ulnar half of flexor digitorum profundus and flexor carpi ulnaris, which are innervated by the ulnar nerve. The median nerve also supplies sensory innervation to the palmar radial side of the hand including the thumb, index, middle, and half of the ring finger. The tibial nerve (Choices B and D) provides innervation to the posterior compartment of the lower leg and the intrinsic muscles of the foot. It provides sensory innervation to the bottom of the foot via the medial and lateral plantar nerves. However, the median nerve and radial nerve do not correlate with this patient's lesion. The radial nerve (Choices C and D) provides sensory innervation to the posterior distal part of the forearm. It also innervates all of the extensors of the elbow, wrist, and hand. The ulnar nerve (Choice E) innervates most of the intrinsic muscles of the hand except for the lateral two lumbrical muscles and those of the thenar eminence. It also provides sensory innervation to the medial aspect of the forearm, small finger, and ulnar aspect of the ring finger. The common fibular nerve (peroneal) (Choices A and C) provides motor innervation to the anterior and lateral compartments of the lower leg and provides sensation to the lateral lower leg and dorsum of the foot. Educational Objective: Long-standing diabetes mellitus can lead to the development of neuropathy. Muscle wasting, sensory loss, and dysfunction in specific dermatomal and myotomal distributions can allow for identification of specific nerve involvement. %3D Previous Next Score Report Lab Values Calculator Help Pause

170 Exam Section 4: Item 20 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment V 20. A 19-year-old woman comes to the physician 6 hours after the sudden onset of left-sided, cramping abdominal pain. She does not have nausea. She also has a 3-week history of pain with sexual intercourse, and she is sexually active with three male partners. Her last menses started 5 days ago. Current medication is an oral contraceptive. Her temperature is 38°C (100.4°F). Physical examination shows a normal cervix with no discharge. Bimanual examination shows tenderness and fullness on the left side compared with the right, and cervical motion tenderness. A urine pregnancy test result is negative. Which of the following structures is most likely affected by this patient's current condition? O A) Appendix B) Bladder C) Endometrium D) Kidney E) Peritoneum F) Uterine (fallopian) tube

F. Neisseria gonorhoeae and Chlamydia trachomatis can both cause vaginitis, cervicitis, and pelvic inflammatory disease (PID). Vaginitis typically presents with mucopurulent discharge and pruritus. Cervicitis presents with similar discharge, but also with an erythematous, friable cervix and cervical motion tenderness on examination. When infection spreads to the fallopian tubes and ovaries, PID can result, which presents with cervical motion tenderness, purulent cervical discharge, uterine and adnexal tenderness, and systemic signs and symptoms such as fever, fatigue, nausea, and myalgias. If left untreated, PID can become complicated, with tubo-ovarian or intra-abdominal abscess presenting with focal tenderness and a fluctuant mass. Treatment of PID involves antibiotics such as cephalosporins and tetracyclines. Risk factors for PID and sexually transmitted infections include adolescence, multiple sexual partners, and not using barrier contraception such as condoms. Incorrect Answers: A, B, C, D, and E. Acute appendicitis (Choice A) presents with right lower quadrant or migratory periumbilical abdominal pain, nausea, anorexia, fever, leukocytosis, and tenderness to palpation. Rupture or perforation of appendicitis can lead to inflammation of the peritoneum (Choice E), which presents with abdominal guarding, rebound tenderness, and rigidity. Cervical motion tenderness and left sided abdominal pain is more likely due to PID involving the fallopian tube. Bladder (Choice B) and urinary tract infections can present with urinary frequency, urgency, and dysuria. Urinalysis for typical urinary tract infections reveals >10 leukocytes per high power field, may show red cells as evidence of mucosal inflammation, and may show white blood cell casts if infection has ascended to pyelonephritis. Bacteria are generally seen on microscopy. Ascending urinary tract infections can involve the kidney (Choice D) resulting in pyelonephritis. Pyelonephritis typically presents with fevers, nausea, vomiting, and flank pain with associated costovertebral angle tenderness on physical examination. Endometritis is an acute, typically polymicrobial infection of the uterine endometrium (Choice C) involving a mixture of aerobes and anaerobes from the genital tract. Patients typically present with fever, severe tenderness of the uterine fundus, and mucopurulent vaginal discharge. Educational Objective: PID secondary to N. gonorrhoeae or C. trachomatis involves the fallopian tubes and can present with cervical motion tenderness, purulent cervical discharge, uterine and adnexal tenderness, and systemic symptoms such as fever. %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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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