NBME 30

1 Exam Section 1: Item 1 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 1. Shortly after delivery, a full-term male newborn is found to have black hair with a white forelock. His mother, a brunette, also has a white forelock and wears hearing aids. Physical examination shows heterochromia of irides. Otoacoustic emissions testing and brain stem auditory evoked responses show bilateral sensorineural hearing loss. Which of the following is the most likely cause of the findings in this patient? A) Abnormal neural crest development B) Abnormality of connexins C) Deficiency of homogentisic acid oxidase activity D) Deficiency of tyrosinase activity E) Failure of internalization of melanin granules by keratinocytes F) Failure of melanosome transportation along dendrites

A. Abnormal neural crest development leads to Waardenburg syndrome. Waardenburg syndrome is a syndrome of patchy depigmentation of the skin, hair, irises, and cochlear dysfunction that primarily illustrates an autosomal dominant inheritance pattern. Because of genetic mutations of genes encoding transcription factors, neural crest cells do not properly differentiate into melanoblasts (melanocyte precursors), or melanoblasts do not migrate to their appropriate location. Patients typically have a white forelock and eyelashes, depigmented skin patches, iridic heterochromia, and sensorineural deafness. The eyes may also be laterally displaced. The clinical diagnosis may be confirmed with genetic testing. Treatment includes audiologic evaluation and genetic consultation. Incorrect Answers: B, C, D, E, and F. An abnormality of connexins (Choice B) would lead to abnormal formation of the plasma membrane channels of diverse cell types. Different combinations of sensorineural hearing loss, ichthyosis, alopecia, and peripheral neuropathy may occur. Depigmentation would be atypical. Deficiency of homogentisic acid oxidase activity (Choice C) would lead to decreased metabolism of the amino acids phenylalanine and tyrosine, which instead degrade into homogentisic acid. Homogentisic acid accumulates in the skin and joints, causing increased pigmentation and arthritis, respectively. Depigmentation would be atypical. Deficiency of tyrosinase activity (Choice D) occurs in oculocutaneous albinism, which presents with uniformly hypopigmented hair and skin (versus the patchy depigmentation of Waardenburg syndrome) and eye abnormalities (eg, iris hypopigmentation, refractive errors, nystagmus). In tyrosinase deficiency, melanocytes are unable to synthesize melanin from the amino acid tyrosine. Iridic heterochromia and sensorineural deafness would be atypical. Failure of internalization of melanin granules by keratinocytes (Choice E) as well as the failure of melanosome transportation along dendrites (Choice F) would lead to decreased pigmentation of keratinocytes, the primary cell type of the epidermis. These abnormalities would likely lead to uniform depigmentation (versus the patchy depigmentation of Waardenburg syndrome), and iridic heterochromia and sensorineural hearing loss would be atypical. Educational Objective: Waardenburg syndrome is a syndrome of patchy depigmentation of the skin, hair, irises, and cochlear dysfunction that results from a defect in the differentiation of neural crest cells into melanocytes. Patients typically present with a white forelock and eyelashes, depigmented skin patches, iridic heterochromia, and sensorineural hearing loss. %3D Next Score Report Lab Values Calculator Help Pause

98 Exam Section 2: Item 48 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 48. A 25-year-old man who immigrated to the USA from Pakistan 2 months ago comes to the physician for a follow-up examination. He was diagnosed with hepatitis C at the age of 5 years but has had no symptoms or clinical signs of this condition. Physical examination shows no abnormalities. Serum studies done today show an AST activity of 68 U/L, ALT activity of 100 U/L, and hepatitis C virus RNA of 1.5 million IU/mL. A photomicrograph of the liver biopsy specimen is shown. Which of the following characteristic features of virus-induced histopathology is seen in this patient? A) Apoptosis B) Coagulation necrosis C) Endoplasmic reticulum proliferation D) Immune-mediated cell lysis E) Mallory hyaline inclusion

A. Apoptosis is demonstrated in the photomicrograph provided, which shows a councilman body (eosinophilic cytoplasm) in the lower left quadrant of the slide. Councilman bodies represent hepatocytes in a state of apoptosis or necrosis, and their distinct appearance histologically contrasts them against normal surrounding cells and architecture. They possess a small nuclear fragment but no distinct nucleus and an intensely eosinophilic cytoplasm. They can be identified both in acute and chronic viral hepatitis, although in chronic hepatitis they may be associated with features of chronic injury, including spotty necrosis, multinucleated hepatocytes, and Kupffer cell hyperplasia. More severe forms of hepatic injury, including bridging and confluent necrosis, are less common in chronic hepatitis C. Incorrect Answers: B, C, D, and E. Coagulation necrosis (Choice B), in contrast to apoptosis, is cell death that occurs as a result of ischemic injury and involves distinctly different cellular pathways. It is histologically characterized by absent nuclei but preserved cell architecture that may persist for several days, followed by cellular destruction via heterolysis. Endoplasmic reticulum (ER) proliferation (Choice C) may occur in cellular ischemia because of cessation of the ATP dependent sodium-potassium transporter in the plasma membrane; increased quantity of ER may also occur in states of high protein synthesis (eg, in plasma cells), and in cases of considerable or repeated exposure to toxic substances degraded by the smooth ER (eg, in hepatocytes). This is not seen on this photomicrograph. Immune-mediated cell lysis (Choice D) describes several distinct processes, but commonly involves the binding of an antibody to the surface of cells, leading either to complement fixation with activation of the classical complement cascade or destruction by cytotoxic T or natural killer cells. Mallory hyaline inclusions (Choice E) are eosinophilic, spiral structures made of cytokeratin that are found in ballooning hepatocytes. They are seen frequently in alcoholic hepatitis and cirrhosis. They are not apparent on this slide. Educational Objective: Councilman bodies are eosinophilic hepatocytes with a nuclear fragment that represent cells undergoing apoptosis. They are frequently seen in both acute and chronic hepatitis. %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 32-year-old man begins to laugh while eating dinner with his friends. A small particle of food irritates his larynx and provokes him to cough. Which of the following best describes the position of his vocal cords throughout this sequence? ITTT Immediately After Laryngeal Irritation While Swallowing While Coughing A) Closed closed open B) Closed оpen closed C) Closed оpen open D) Оpen closed closed E) Оpen closed оpen OF) Оpen open closed

A. Coughing can be a voluntary or reflexive action that protects the airway from aspiration, foreign bodies, and environmental irritants, and clears bronchial secretions and debris. The forceful expulsion of air requires coordinated mechanics between the respiratory muscles and laryngeal structures. While swallowing, the vocal cords are closed to prevent inadvertent introduction of food into the airway and lungs. Immediately after laryngeal irritation, the vocal cords and epiglottis remain closed while the respiratory muscles contract to generate increased pressure in the airways. While coughing, the vocal cords open to allow forceful exhalation of air to remove the irritant. Respiratory muscle weakness and neuromuscular disorders that interfere with the coordinated mechanics of cough are risk factors for aspiration and associated pneumonia. Incorrect Answers: B, C, D, E, and F. Choices D, E, and F are incorrect as the epiglottis is inverted and the vocal cords are closed during swallowing to protect the airway and lungs. Choices B, C, and F are incorrect as the vocal cords remain closed as the cough reflex arc is initiated by laryngeal irritation in order to generate pressure with the trapped air in the lungs. Choices B, D, and F are incorrect as the vocal cords must open during the cough to allow air to escape the lungs and remove the offending irritant. Educational Objective: Coughing is a protective mechanism for clearing material from the airways. To generate high positive pressure for expelling air from the lungs, the vocal cords close as the respiratory muscles contract, trapping air in the lungs. Opening of the vocal cords then allows a forceful expulsion of air to occur. %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 28-year-old woman delivers a full-term newborn who has generally flattened facial features, including a broad, flat nose, and bilateral clubfoot deformity with tibial torsion. Ultrasonography at 18 weeks' gestation showed oligohydramnios and small dysplastic fetal kidneys with scattered cysts. Which of the following general classifications best describes the extrarenal physical features? A) Deformation B) Disruption C) Malformation D) Multifactorial E) Teratogenic effect

A. Deformation describes a disruption in organ morphogenesis that occurs after the embryonic period. It arises because of an extrinsic force that limits normal development. In this example, renal dysplasia has resulted in minimal fetal urine production, and therefore limited production of amniotic fluid (oligo- or anhydramnios) as fetal urine is a major contributor. Without the production of amniotic fluid to distribute force evenly over the surface of the developing fetus, deformities will result. Fetuses with renal agenesis or dysplasia typically exhibit compressed, flattened facial features, limb dysplasia, and pulmonary hypoplasia caused by inadequate amniotic fluid-induced lung expansion. This is known as the Potter sequence and is an example of an extrinsic influence on organ morphogenesis from an inadequate intrauterine environment. Incorrect Answers: B, C, D, and E. Disruption (Choice B) defines an error in organ morphogenesis that arises from the breakdown of an organ or body region that was developmentally normal. The classic example of disruption is the presence of amniotic bands that constrict, compress, or encircle the developing fetus, resulting in morphogenic abnormalities involving, for example, the limbs, penis, fingers, or toes. Malformation (Choice C) defines an intrinsic disruption in organ development, which generally occurs earlier in the embryonic period. Multifactorial (Choice D) defines the presence of many influences that result in organ morphogenic errors, an example of which could be a fetus exposed to alcohol in utero with resultant abnormal facial development that also experienced disruption from amniotic bands later in utero. Teratogenic effect (Choice E) describes the result of embryonic exposure to an external toxic substance (often a medication or illicit drug) that results in malformation. Examples include the maternal use of ACE inhibitors, which impair fetal renal development, lithium, which is classically associated with an atrialized right ventricle, and thalidomide, which is known to cause limb deformities. Educational Objective: Deformation describes a disruption in organ morphogenesis that occurs after the embryonic period. It arises because of an extrinsic force that limits normal organ development. %3D Previous Next Score Report Lab Values Calculator Help Pause

88 Exam Section 2: Item 38 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 38. A 62-year-old man is brought to the physician by his wife because of increasing confusion during the past 6 months. His wife says that he has become lost twice in the past month when going to work, even though he has been going to the same office for 12 years. She adds that he often has difficulty finding objects such as his glasses and keys, sometimes cannot recall his grandchildren's names, and has become very critical of her cooking, which he used to enjoy. When asked, he can name only the current president and none of the candidates for an upcoming presidential election, although he and his wife watch the television news together each night. Neurologic examination shows no motor or sensory abnormalities. His Mini-Mental State Examination score is 19/30. Treatment with a cholinesterase inhibitor is most likely to improve this patient's memory because of its ability to target synaptic connections between which of the following structures? A) The basal forebrain and neurons in the cerebral cortex B) The dentate nucleus and the thalamus C) The fornix and neurons in the mammillary bodies D) The substantia nigra and the neurons of the globus pallidus E) The thalamus and neurons in layer 4 of the cerebral cortex

A. Dementia, Alzheimer type, results from decreased cholinergic signaling in the cerebral cortex and basal forebrain. Cholinergic neurons (which primarily originate from the locus coeruleus and basal forebrain) project widely to areas including the cortex, mediating attention, learning, and memory. Alzheimer dementia is the most prevalent type of dementia, presenting with progressive cognitive decline that begins with short-term memory impairment, progresses to apraxia and language abnormalities, and culminates in behavioral and personality changes preventing the patient from performing basic activities of daily living. Acetylcholinesterase normally hydrolyzes acetylcholine in synaptic clefts. Donepezil and other cholinesterase inhibitors non-competitively and reversibly inhibit acetylcholinesterase activity and thereby increase the amount of synaptic acetylcholine available for neurotransmission. Though these medications may slow the rate of cognitive decline and modestly improve functionality, they are not curative. Incorrect Answers: B, C, D, and E. The dentate nucleus and thalamus (Choice B) mediate output from the cerebellum. These structures are not typically affected in Alzheimer dementia. The fornix and neurons in the mammillary bodies (Choice C) constitute output pathways from the hippocampus and mediate episodic memory. Lesions may result in amnesia. Patients with Alzheimer dementia demonstrate hippocampal atrophy and decreased cholinergic signaling in the hippocampus, but the fornix and mammillary bodies are less commonly affected. The substantia nigra and the neurons of the globus pallidus (Choice D) together modulate voluntary movement. These brain regions are affected in Parkinson disease, not Alzheimer dementia. The thalamus and neurons in layer 4 of the cerebral cortex (Choice E) mediate the transmission of sensory information to the primary sensory cortices. Sensory domains are not significantly affected in Alzheimer dementia. Educational Objective: Dementia, Alzheimer type, results from decreased cholinergic signaling in the cortex and basal forebrain. Alzheimer dementia is the most prevalent type of dementia, presenting with progressive cognitive decline that culminates in the inability to perform basic activities of daily living. %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. The graph shows the rate of breakdown of skeletal muscle glycogen by phosphorylase in the presence and absence of AMP. Based on the data, which of the following best describes the action of AMP? Vmax +AMP A) Allosteric activator No effector Vmax B) Allosteric inhibitor C) Catabolite activator D) Catabolite repressor [S] E) Covalent modifier of the enzyme F) Covalent modifier of phosphorylase kinase

A. Enzymes are kinetically categorized by their Michaelis constant, Km, which inversely reflects their affinity for their substrate, and by Vmax, the maximum catalysis rate of the enzyme. In other words, as Km decreases, affinity for the substrate increases. Km is defined as the substrate concentration at which the enzyme reaches one-half of its Vmax. In this case, the kinetic analysis of the patient's glycogen phosphorylase shows that the concentration of substrate [S] at one-half of Vmax decreases in the presence of AMP. Therefore, AMP must increase the affinity of the enzyme for its substrate, a phenomenon that classically results from a conformational change in the active site of the enzyme where substrate binds. Allosteric activators influence enzymes in exactly this manner - inducing conformational change at the active site which in turn increases binding affinity for the substrate. This is reflected by the decreased Km. AMP, a breakdown product of adenosine triphosphate (ATP), is generated in states of fasting. Glycogen, a storage form of glucose in liver and muscle, is broken down by glycogen phosphorylase to liberate free glucose molecules when glucose is depleted in fasting. AMP, when present in excess, signals a state of fasting. It upregulates glycogen degradation through glycogen phosphorylase by acting as an allosteric activator of the enzyme. Incorrect Answers: B, C, D, E, and F. Allosteric inhibitors (Choice B) modify enzymes in a manner that their affinity for their target substrate is reduced at the active site, not increased. Allosteric inhibitors would be reflected in enzyme kinetic analysis by an increase in the Km, indicating that a greater concentration of substrate would be required to elicit the same binding. Catabolite activator (Choice C) and catabolite repressor (Choice D) describe phenomena in which transcription of genes is affected; this was first described in the lac operon. When given the option between using glucose or lactose as a substrate, glucose was preferred, and synthesis of enzymes related to lactose metabolism was repressed. By contrast, in an absence of glucose, synthesis of lactose-metabolizing enzymes was promoted. Covalent modifier of the enzyme (Choice E) and covalent modifier of phosphorylase kinase (Choice F) suggest modification of the enzyme by phosphorylation or dephosphorylation. Covalent modification can affect conformation at the active site, however, AMP is not acting as a phosphate donor or recipient in this case; it binds directly to induce conformational change. Educational Objective: Allosteric activators influence enzymes by inducing conformational change at the active site, which increases binding affinity for the substrate. This is reflected by a decreased Km. AMP, a breakdown product of adenosine triphosphate, is generated in states of fasting and thus upregulates glycogen phosphorylase to liberate stored glucose and supply the cells with a substrate for glycolysis when none is nutritionally available. %3D Previous Next Score Report Lab Values Calculator Help Pause Initial velocity V.

58 Exam Section 2: Item 8 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 8. One day after being admitted to the hospital for treatment of peritonitis, a 42-year-old man has progressive shortness of breath and moderate distress. Pulse oximetry on room air shows an oxygen saturation of 53%. Bilateral basilar crackles are heard on auscultation. Acute lung injury is suspected. The patient is intubated and mechanically ventilated with 90% oxygen at a lung-protective tidal volume of 550 mL and respirations of 16/min. Positive end-expiratory pressure also should be implemented to prevent which of the following complications? A) Absorption atelectasis B) Alveolar overdistention C) Oxygen toxicity D) Pleural exudate formation E) Pulmonary hypertension F) Tension pneumothorax

A. Mechanical ventilation is indicated for respiratory failure. 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. This helps keep alveoli stented open and able to participate in gas exchange. The patient is suffering acute lung injury and has impaired gas exchange caused by fluid filling the alveoli, resulting in absorption atelectasis. Oxygenation can be improved by implementing positive end-expiratory pressure to resist alveolar collapse and atelectasis. Incorrect Answers: B, C, D, E, and F. Alveolar overdistention (Choice B) is a potential complication of mechanical ventilation from excessive alveolar pressure. Large tidal volumes and shear stress from alveoli collapsing and opening can result in inflammatory cytokine release and secondary ventilator- induced lung injury. Lung-protective tidal volume strategies reduce the risk for alveolar overdistension. Oxygen toxicity (Choice C) is a potential complication of mechanical ventilation with a high inspired fraction of oxygen. Cell damage occurs secondary to reactive oxygen species, which can worsen lung injury. Pleural exudate formation (Choice D) is not prevented by positive end-expiratory pressure. Underlying causes for exudative pleural effusion include infection, malignancy, and autoimmune disorders. Pulmonary hypertension (Choice E) has numerous potential causes. It may be secondary to conditions such as left heart disease, chronic hypoxemic lung disease, and connective tissue disorders. Positive end-expiratory pressure is not used to prevent pulmonary hypertension apart from reducing transient hypoxic vasoconstriction. Tension pneumothorax (Choice F) is a potential complication of mechanical ventilation, especially with large tidal volumes or high peak pressures that can cause alveolar sac rupture and air communication into the pleural space. Lung-protective tidal volume strategies and close monitoring of peak inspiratory pressure reduce the risk for tension pneumothorax formation. Educational Objective: Oxygenation of mechanically ventilated patients in the setting of acute lung injury is challenging because of reduced lung compliance and absorptive atelectasis from fluid in the alveolar space. Positive end-expiratory pressure can reduce the amount of alveolar collapse and atelectasis. %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. A 66-year-old man is brought to the emergency department 30 minutes after the sudden onset of substernal chest pain that radiates to the neck and left arm; the pain is associated with weakness, nausea, and profuse sweating. He was jogging during the onset of pain. He has a 5-year history of hypertension treated with indapamide. His pulse is 90/min, and blood pressure is 150/90 mm Hg. Cardiac examination shows an S, Treatment with sublingual nitroglycerin resolves his pain within 2 minutes. An ECG shows a 2-mm ST-segment elevation in the anterior leads. Treatment with a fibrinolytic drug is beneficial to this patient because of which of the following mechanisms of action? A) Catalyzing the formation of plasmin B) Catalyzing the platelet Ilb/llla receptor complex C) Inhibition of ADP-induced platelet-fibrinogen binding D) Inhibition of conversion of fibrinogen to fibrin E) Inhibition of the synthesis of thromboxane A2 F) Inhibition of vitamin K-dependent coagulation factor synthesis

A. Myocardial infarction classically presents with acute-onset chest pain, often radiating to the neck, jaw, or arm, along with shortness of breath, nausea, and/or lightheadedness. Risk factors include a history of smoking, diabetes mellitus, hypertension, hyperlipidemia, and/or obesity. If the infarction results in left ventricular dysfunction, patients may experience cardiogenic pulmonary edema resulting in shortness of breath, tachypnea, hypoxia, audible rales, and S3 or S4gallops. Diagnosis is made using clinical history, analysis of cardiac biomarkers, and an ECG, which may show ST-segment elevation or depression, inverted T-waves, or Q-waves. Treatment requires the use of antiplatelet agents (eg, aspirin, clopidogrel) plus anticoagulants (eg, heparin), pain control, and revascularization through angioplasty, thrombolysis, or coronary artery bypass grafting. For patients who cannot receive percutaneous coronary intervention in a timely manner, thrombolytics are indicated. Thrombolytics such as tissue plasminogen activator function by catalyzing the formation of plasmin from plasminogen. Plasmin is a serine protease that cleaves fibrin clots. Incorrect Answers: B, C, D, E, and F. Catalyzing the platelet Ilb/Illa receptor complex (Choice B) is not the mechanism of action of thrombolytics. The platelet Ilb/Illa receptor binds to fibrinogen during the process of platelet aggregation. Inhibition of ADP-induced platelet-fibrinogen binding (Choice C) occurs with ADP receptor blockers (eg, clopidogrel) that inhibit expression of the Ilb/llla receptor and with monoclonal antibodies (eg, abciximab) that directly bind the receptor. Inhibition of conversion of fibrinogen to fibrin (Choice D) is the mechanism of heparin anticoagulants and direct thrombin inhibitors such as lepirudin and bivalirudin. Thrombin catalyzes the conversion of fibrinogen to insoluble fibrin polymers for clot formation. Heparin increases the binding affinity of antithrombin III, which inactivates thrombin. Direct thrombin inhibitors directly inhibit thrombin activity. Inhibition of the synthesis of thromboxane A, (Choice E) occurs with cyclooxygenase inhibitors, such as aspirin, which prevent the conversion of arachidonic acid to thromboxane A, Thromboxane A, is involved in platelet activation and its decreased synthesis results in diminished platelet activity. Inhibition of vitamin K-dependent coagulation factor synthesis (Choice F) occurs with warfarin because of the inhibition of vitamin K epoxide reductase. This enzyme is necessary for recycling vitamin K, an essential cofactor in the synthesis of coagulation factors in the liver. Educational Objective: Myocardial infarction carries a high risk for morbidity and mortality. It is most commonly caused by atherosclerotic plaque rupture in the coronary arteries with thrombosis formation. Antithrombotic agents are a mainstay of therapy. For patients who cannot receive percutaneous coronary intervention or coronary artery bypass grafting in a timely manner, thrombolytics are indicated. %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 7. A 12-year-old boy with seizure disorder is brought to the physician by his parents for a follow-up examination. His mother has two neurofibromas, and his brother and maternal uncle each have numerous café au lait spots. Physical examination shows numerous flat, irregular pigmented skin lesions. He is individual IV-2 in the pedigree shown. Which of the following is the most likely pattern of inheritance of this family's disorder? | || A) Autosomal dominant with variable expressivity B) Autosomal dominant with variable penetrance II C) Autosomal recessive with high heterozygote frequency D) Autosomal recessive with low heterozygote frequency IV E) X-linked dominant O F) X-linked recessive O Unaffected female Unaffected male Affected female Affected male

A. Neurofibromatosis 1 is an autosomal dominant neurocutaneous disorder that typically results from an inherited mutation of the NF1 tumor suppressor gene on chromosome 17. Neurofibromatosis 1 is inherited with complete penetrance and variable expressivity, signifying that all family members who inherit the mutation manifest the disease but may demonstrate different phenotypic presentations. Rarely, the mutation in the NF1 gene will occur sporadically instead. The disorder presents with cutaneous neurofibromas (benign neoplasms derived from neural crest cells), café au lait spots (flat, irregular pigmented lesions), pigmented iris hamartomas (Lisch nodules), optic gliomas (which may lead to seizures), and/or pheochromocytomas. The diagnosis is typically based on physical examination findings, but genetic testing may be utilized to confirm ambiguous cases. Treatment includes surgical debulking of symptomatic tumors, management of seizures and other neurologic complications, and genetic counseling. Incorrect Answers: B, C, D, E, and F. Autosomal dominant disorders with variable penetrance (Choice B) typically result in less than half of the progeny of an affected parent manifesting the disease, as not all offspring with the mutation will manifest the disease. This patient's pedigree instead shows that slightly more than half of the offspring of affected parents manifest the disease, which is more consistent with complete penetrance. Autosomal recessive inheritance with either high or low heterozygote frequency (Choices C and D) would necessitate that both parents of the affected offspring either have the disease or are heterozygous carriers. A high heterozygote frequency signifies a low homozygote frequency (ie, a low frequency of offspring with two mutated alleles) and therefore a low frequency of disease in the offspring. A low heterozygote frequency is conversely associated with a higher frequency of disease in the offspring. In this patient's pedigree, the disease is present in every generation. In autosomal recessive diseases, typically only a single generation is affected, although the risk for disease is increased in consanguineous families. X-linked dominant or recessive disorders (Choices E and F) manifesting in male offspring can only be transmitted maternally. In this patient's pedigree, the male of generation three receives the mutated allele from his father. In X-linked dominant disorders, both males and females possessing the mutated allele manifest the disease, while in X-linked recessive disorders, only males with the mutated allele manifest the disease. Educational Objective: Neurofibromatosis 1 is an autosomal dominant neurocutaneous disorder resulting from an inherited mutation of the NF1 tumor suppressor gene, which is transmitted with complete penetrance and variable expressivity. The disorder presents with cutaneous neurofibromas (benign neoplasms derived from neural crest cells), café au lait spots, pigmented iris hamartomas (Lisch nodules), optic gliomas, and/or pheochromocytomas. %3D Previous Next Score Report Lab Values Calculator Help Pause 2. 2.

72 Exam Section 2: Item 22 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 22. An investigator is conducting a study of the effect of salt intake on renal function. The diet of an experimental animal is modified to contain only 25% of normal salt content. Which of the following is most likely to be increased in this experimental animal? A) Epithelial sodium channel transcription B) Extracellular fluid C) Plasma atrial natriuretic peptide concentration D) Serum potassium concentration E) Urine pH

A. The nephron is characterized by an apical epithelial membrane that faces the tubular lumen and a basal layer that faces the interstitium. This epithelial membrane, whether in the convoluted tubules, loops, or collecting duct is the site of reabsorption of solute or water into the blood. The collecting duct is the terminal segment of the nephron and reabsorbs sodium and water while secreting potassium and protons, primarily through epithelial sodium channels and aquaporins. It is the site of action of aldosterone. Increased dietary salt intake suppresses aldosterone, while decreased dietary salt intake increases aldosterone activity. Aldosterone increases epithelial sodium channel transcription and activity, resulting in increased sodium reabsorption. Incorrect Answers: B, C, D, and E. Increased dietary sodium can lead to fluid accumulation in extracellular tissue. Decreased dietary salt would be unlikely to increase extracellular fluid (Choice B). Plasma atrial natriuretic peptide (ANP) concentration (Choice C) would likely be decreased with low salt intake. ANP is a hormone secreted by atrial cardiac myocytes in response to increased hydrostatic pressure. It results in diuresis, natriuresis, and decreased blood pressure. Decreased dietary salt intake generally decreases systolic blood pressure in hypertensive individuals, whereas increased dietary salt intake would be more likely to increase blood pressure and therefore plasma ANP concentration. Decreased dietary salt would increase aldosterone activity, which increases epithelial sodium channel activity and apical potassium secretion, leading to decreased, not increased, serum potassium concentration (Choice D). Aldosterone causes increased H+-ATPase activity, resulting in H* secretion into the urine. This would decrease, not increase, urine pH (Choice E). Educational Objective: Increased dietary salt intake suppresses aldosterone, while decreased dietary salt intake increases aldosterone activity. Aldosterone increases epithelial sodium channel transcription and activity, resulting in increased sodium reabsorption. %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. The graph shows the mean steady state plasma concentration of Drug X as a function of time in a subject who has normal weight (solid curve) and a subject who is obese (dashed curve). Both subjects take 10 mg of Drug X daily. Which of the following best describes the disposition of Drug X in the obese patient compared with the patient who has normal weight? A) Greater volume of distribution B) Higher clearance C) Lower bioavailability D) Slower absorption rate E) Shorter half-life 1 3 4 6 7 Time (days)

A. Volume of distribution is the theoretical volume in which a drug exists given its plasma concentration following a dose of medication. There are three main compartments into which a medication can distribute: the intravascular compartment, the interstitial compartment, and the intracellular compartment (including fat). With an increased volume of distribution, the same dose of medication will lead to a smaller plasma concentration (ie, it will distribute more widely into tissue), and it will take longer for the medication to reach a steady state within the plasma. Large molecules or protein-bound molecules generally remain within the vasculature and quickly reach a steady state concentration given the relatively low volume of the intravascular compartment. Smaller hydrophilic molecules tend to move into the interstitial space, which increases their volume of distribution and increases the amount of time it takes for the medication to reach a steady state concentration. Small lipophilic molecules, however, exhibit the largest volume of distribution as they are commonly taken up by adipocytes, which greatly increases the volume in which they are distributed. Therefore, it takes these molecules longer to reach a steady state concentration, especially in obese patients in which the volume of distribution is even greater given the presence of increased fat stores, as seen in this graph. Incorrect Answers: B, C, D, and E. Higher clearance (Choice B) of a medication indicates that more medication is removed from the body in a given period of time. Differences in clearance are typically because of changes in hepatic or renal function and are commonly displayed in a graph that shows the decline in drug concentration. An increase in clearance with the same volume of distribution would require a higher maintenance dose to reach steady state. Lower bioavailability (Choice C) is described as a decreased absorption of the medication, lowering the amount of medication that is available to act in the body. It commonly refers to oral medications, as they are subject to first pass metabolism in the liver. Bioavailability is not subject to alterations based on body fat percentage, and a decrease in bioavailability would require an increase in maintenance dosing to reach a similar steady state concentration. Slower absorption rate (Choice D) in the setting of similar volumes of distribution and clearance would result in an increased time to reach steady state concentration. However, the absorption rate of oral medications is not typically affected by body weight. Shorter half-life (Choice E) is a decrease in the amount of time that it requires for a drug's concentration in plasma to decrease by 50%. Half-life is indirectly related to clearance, such that it decreases with increases in plasma clearance. Conversely, it is directly proportional to the volume of distribution. A decreased half-life would decrease the time necessary for the medication to reach steady state concentration, unlike this scenario. Educational Objective: The volume of distribution is the theoretical volume in which a drug exists based on its plasma concentration. It is altered by changes in volume of the intravascular, interstitial, and intracellular compartments. Increases in volume of distribution are exhibited by a slower rate of increase in plasma drug concentration and a prolonged period before reaching steady state concentration. %3D Previous Next Score Report Lab Values Calculator Help Pause Plasma [drug X] N.

77 Exam Section 2: Item 27 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 27. A61-year-old man has had three episodes of painless gross hematuria during the past week. The left flank is mildly tender to palpation. Urinalysis shows blood but no WBCS, bacteria, or cellular casts. He says to the physician, "Tell me like it is. Do I have cancer?" Which of the following is the most appropriate response? A) "Cancer is one of the possible causes of blood in your urine. We will need to perform additional tests to determine the cause." B) "Hematuria can be caused by many things. Trust me. I will let you know if we find anything worrisome." C) "It is too soon to tell. Don't worry about that for now." D) "Yes, cancer looks likely, but many effective treatments are now available." E) "Yes, I'm afraid this usually is due to cancer. Would you like to have someone with you while we discuss it?"

A. When patients express the desire to be informed about serious diagnoses, physicians should inform patients directly and compassionately. This physician should inform the patient that his hematuria may reflect an underlying cancer but that the diagnosis is not yet confirmed. Once the patient has been informed that this diagnosis is a possibility, the physician should validate the patient's emotions by listening receptively, using reflective statements, inferring emotions from the expressed content, and/or normalizing the patient's experience. The physician should also deliver education and explain the next steps using small pieces of information. Incorrect Answers: B, C, D, and E. Informing the patient that his diagnosis is unclear (Choices B and C) would avoid the patient's question about whether he has cancer. The physician should respond that cancer is a possibility. Answering the patient's question about whether he has cancer with a definitive "yes" (Choices D and E) would be premature. Further discussion and treatment planning should occur only after the cancer diagnosis is definitively made. Educational Objective: When patients express the desire to be informed about serious diagnoses, physicians should inform patients directly and compassionately about possible or confirmed diagnoses. The physician should also deliver education and explain the next steps using small pieces of information. %3D Previous Next Score Report Lab Values Calculator Help Pause

40 Exam Section 1: Item 40 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 40. An 18-month-old boy has pinched, wrinkled facies and total loss of subcutaneous fat. He is alert. His height is at the 50th percentile and his weight is at the 2nd percentile for his age. The liver is not enlarged, and there is no edema. Which of the following is the most likely diagnosis? A) Kwashiorkor B) Marasmus C) Pellagra D) Rickets E) Scurvy

B. Acute protein-energy malnutrition is characterized clinically by a decrease in mid-upper arm circumference, weight for length, or body mass index. Unlike in chronic malnutrition, children maintain their linear growth because the weight is more quickly impacted than the height. Acute malnutrition can be divided into two groups based on the presence or absence of edema. Marasmus, or wasting syndrome, occurs when the overall caloric intake is deficient, but no specific vitamins or minerals are missing. Wasting of muscle mass and consumption of body fat stores is characteristic. It is not associated with edema. Affected children appear emaciated and weak and may be irritable. Autonomic functions may suffer, causing bradycardia, hypotension, and hypothermia. Redundant skin folds, as seen in this case, arise because of loss of subcutaneous fat. In contrast, kwashiorkor is also a form of protein-energy malnutrition but is accompanied by edema. The edema begins in dependent areas but, with continued malnutrition, progresses to involve the presacral, genitalia, and preorbital areas. Children with either marasmus or kwashiorkor malnutrition have complications in nearly every organ system of the body including immune dysfunction, increased cortisol concentrations, decreased intestinal motility and absorptive capacity, and a tenuous fluid balance predisposing to acute heart failure. Incorrect Answers: A, C, D, and E. Kwashiorkor (Choice A), also known as edematous malnutrition, is differentiated from marasmus by the presence of edema. Skin lesions and fatty liver, both absent in this case, are typically present in kwashiorkor. Pellagra (Choice C) is caused by an acquired deficiency of niacin (nicotinic acid, vitamin B3). It is characterized by diarrhea, dementia, and dermatitis. Signs of protein-energy malnutrition are not present. Pellagra-like symptoms can also be seen in Hartnup disease, an autosomal recessive disorder caused by decreased gastrointestinal absorption of tryptophan, the precursor to nicotinic acid. Rickets (Choice D), or childhood vitamin D deficiency, presents with multiple skeletal deformities including frontal bossing, nodules at the costochondral junctions of the anterior chest, and delayed closure of the growth plates because of impaired bone mineralization. Children may demonstrate bowing of the tibia and femur. Children will have hypoplastic teeth and multiple caries as mineralization of dentin is impaired. Vitamin D deficiency in children commonly comes from low exposure to UV radiation and low dietary vitamin D intake (particularly from low vitamin D in breast milk). Signs of protein-energy malnutrition are not present. Deficiency in vitamin C leads to scurvy (Choice E), 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. Educational Objective: Acute protein-energy malnutrition includes a non-edematous type (marasmus) and an edematous type (kwashiorkor). Both are characterized clinically by wasting of muscle mass and consumption of body fat stores leading to emaciation. Nearly every organ system of the body is affected by this severe malnourishment. %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 37. A51-year-old man develops diaphoresis, tachycardia, and a blood pressure of 155/100 mm Hg 24 hours after undergoing an abdominal operation. Two hours later, he has a generalized tonic-clonic seizure. Which of the following is most likely responsible for these adverse effects in this patient? A) Acute renal failure B) Alcohol withdrawal C) Anaphylactic reaction D) Narcotic pain medication E) Sepsis from a gram-positive organism

B. Alcohol withdrawal typically leads to sympathetic overdrive and, in severe cases, generalized seizures. Chronic alcohol use causes changes in the expression of different subunits of the N-methyl-d-aspartate (NMDA) and gamma-aminobutyric acid (GABA) receptors. As a result of these adaptations, the discontinuation of alcohol leads to sympathetic overdrive. Initially, this hyperexcitability results in tremors, anxiety, diaphoresis, hypertension, tachycardia, and nausea. Within hours of the last drink, alcoholic hallucinosis (auditory or visual hallucinations without confusion or autonomic instability) and seizures (caused by cortical hyperactivity) can occur. Delirium tremens, the most concerning and life-threatening complication of alcohol withdrawal, involves severe confusion, disorientation, fluctuations in consciousness, agitation, visual hallucinations, and autonomic instability (fluctuations in pulse and blood pressure with hyperthermia). Incorrect Answers: A, C, D, and E. Acute renal failure (Choice A) can lead to oliguria and may be associated with hyperkalemia, metabolic acidosis, and uremia. None of these changes typically cause seizures. Additionally, sympathetic overactivation would be uncommon. An anaphylactic reaction (Choice C) is an acute, life-threatening reaction to an allergen that leads to systemic mast cell degranulation. Anaphylactic reactions typically present with hypotension caused by increased vascular permeability from histamine. Furthermore, a seizure would be atypical in the presentation of an anaphylactic reaction. Narcotic pain medications (Choice D) cause symptoms of central nervous system depression including sedation, respiratory depression, bradycardia, and hypotension as opposed to this patient's tachycardia and hypertension. Sepsis from a gram-positive organism (Choice E) typically causes hypotension. Additionally, gram-positive or gram-negative bacterial infections may occur in the setting of an abdominal wound; however, these infections do not typically develop until several days postoperatively. A seizure would be atypical. Educational Objective: Alcohol withdrawal is associated with sympathetic overdrive and initially presents with tremors, anxiety, diaphoresis, hypertension, and tachycardia. In severe alcohol withdrawal, seizures and delirium tremens may occur. %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. A 22-year-old man is brought to the physician because of a 1-week history of bleeding gums and bruising of his legs and a 3-month history of weakness and fatigue. His temperature is 37°C (98.6°F). Physical examination shows pallor, boggy gingival hypertrophy, and several petechial lesions and ecchymoses scattered over the lower extremities. The spleen tip is palpated 3 cm below the left costal margin. Laboratory studies show: Hemoglobin Hematocrit 8.2 g/dL 24% Leukocyte count Platelet count 32,000/mm3 23,000/mm3 Microscopic examination of a peripheral blood smear shows 42% blast forms. Which of the following is the most likely diagnosis? A) Acute lymphocytic leukemia B) Acute myelogenous leukemia C) Chronic lymphocytic leukemia D) Chronic myelogenous leukemia E) Hairy cell leukemia

B. Acute myelogenous leukemia (AML) is the most likely diagnosis in this patient with pancytopenia, gingival hypertrophy, and 42% blasts on peripheral smear. AML is a neoplasm arising from cells within the myeloid lineage, which is distinct from the lymphoid lineage. There are myriad mutations that can be present in malignant myeloid cells and the types of mutations confer different risk profiles. Patients with poor risk mutations undergo stem cell transplant if they are able to achieve remission after induction and consolidation chemotherapy with cytarabine and idarubicin. Common presenting symptoms include fatigue, malaise, fevers, and weight loss. Gingival hypertrophy is commonly seen in a particular subtype of AML known as acute promyelocytic leukemia and is a result of infiltration of the gums with malignant cells. Leukemia cutis, a rash that represents infiltration of the skin by malignant cells, can also occasionally be seen. Other findings include petechiae or purpura from thrombocytopenia and anemia from diffuse marrow infiltration by blasts. Induction therapy for AML is considered urgent and is usually done with cytarabine and idarubicin, which is followed by consolidation therapy. Patients with poor risk mutations invariably go on to stem cell transplant if they can achieve remission initially. Incorrect Answers: A, C, D, and E. Acute lymphocytic leukemia (ALL) (Choice A) is primarily a malignancy of childhood but is occasionally seen in adults. Cytopenias are common and peripheral smear also shows blasts as in this patient, although these blasts will be of the lymphoid lineage and appear different from myeloid cells. Gingival hypertrophy is not generally associated with ALL. Chronic lymphocytic leukemia (Choice C) is a malignancy of mature, differentiated B lymphocytes. Common findings include lymphocytosis, and peripheral smear classically shows numerous lymphocytes with large blue nuclei and scant cytoplasm, many of which may appear ruptured. Blasts are not seen on peripheral smear as the malignant cells are mature. Treatment with ibrutinib is initiated if patients have profound cytopenias or disabling symptoms such as malaise and bulky lymphadenopathy. Chronic myelogenous leukemia (Choice D) is defined by the presence of the Philadelphia (Ph) chromosome, which is created by a translocation between chromosomes 9 and 22 resulting in constitutive activation of the ABL1 tyrosine kinase. Peripheral smear will show a diverse set of immature or partially mature cells of the granulocytic cell lineage with basophilia. Hairy cell leukemia (Choice E) is typically a disease of older adults and is diagnosed in the setting of pancytopenia, lymphocytes with cytoplasmic projections on peripheral smear, and positive tartrate resistant acid phosphatase (TRAP). Malaise, abdominal distention, early satiety, and pain from splenomegaly are common presenting complaints. Treatment is reserved for symptomatic patients. Educational Objective: AML is an aggressive malignancy of myeloid cells and commonly presents in adults with pancytopenia and circulating blasts on peripheral smear. Gingival hypertrophy is not an uncommon finding and represents infiltration of the gingiva by malignant cells. Depending upon the types of mutations present, patients are divided into good, intermediate, or poor risk subcategories. Those with poor risk disease invariably go on to stem cell transplant after induction and consolidation chemotherapy. %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 13-year-old girl is brought to the physician by her mother because of a mass in her armpit for 3 days. She has a history of several similar masses since early childhood that have resolved with treatment. She also has had several episodes of pneumonia, rhinitis, and perianal abscesses since birth. Her temperature is 37°C (98.6°F). Physical examination shows a 2-cm, raised, red, tender, fluctuant mass in the right axilla. There is mild hepatosplenomegaly. A complete blood count shows a mildly increased leukocyte count. A chest x-ray shows bilateral interstitial infiltrates. A photomicrograph of lung tissue obtained on biopsy is shown. A nitroblue tetrazolium test shows deficient reduction by granulocytes. The axillary mass is incised and drained, and culture of the fluid grows Staphylococcus aureus. Which of the following is the most likely diagnosis? A) Chédiak-Higashi syndrome B) Chronic granulomatous disease C) Myeloperoxidase deficiency D) Neutrophil-specific granule deficiency E) Wiskott-Aldrich syndrome

B. Chronic granulomatous disease (CGD) is the most likely diagnosis in this patient with a Staphylococcus aureus abscess, history of prior similar infections, and an abnormal nitroblue tetrazolium test. CGD is defined by deficiency of the NADPH oxidase complex, which is essential for normal neutrophil intracellular killing of pathogens. 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 incapable of neutralizing catalase-positive bacteria, which are capable of neutralizing their own hydrogen peroxide, thus leaving the host cells without the substrate necessary to complete the respiratory burst. Diagnosis is made by an abnormal dihydrorhodamine test or a nitroblue tetrazolium reduction test. In this latter test, normal phagocytes use the action of NADPH to reduce nitroblue, which leads to a color change from yellow to blue. Patients with CGD will not demonstrate color change. Recurrent pneumonia is the most common presenting infection in patients with CGD, and the most common infecting bacteria include Staphylococcus species, Aspergillus species, Burkholderia cepacia, and Nocardia species. Patients with CGD are also at risk for fungal infections, especially Aspergillus species. Incorrect Answers: A, C, D, and E. presentingChédiak-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. Myeloperoxidase deficiency (Choice C) is a common 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 D) results from 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. Patients have a normal nitroblue tetrazolium test. Wiskott-Aldrich syndrome (Choice E) is caused by a mutation in the WAS gene on the X-chromosome that encodes a protein essential for the actin cytoskeleton rearrangement that occurs during interactions between T lymphocytes, antigen-presenting cells, and B lymphocytes; this can lead 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. This presents as recurrent pyogenic infections with catalase-positive organisms such as S. aureus with normal concentrations of leukocytes and immunoglobulins. %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 60-year-old man is being evaluated for episodes of carpopedal spasm. Laboratory studies of serum show: 6.2 mg/dL 4.0 g/dL 8.6 mg/dL 200 pg/mL (N=10-65 pg/mL) Ca2+ Albumin Phosphorus Parathyroid hormone Which of the following conditions would most likely account for these findings? A) Acute tubular necrosis B) Chronic renal failure C) Dehydration D) Rapidly progressive glomerulonephritis E) Renal cell carcinoma

B. Chronic renal failure can have variable cause, such as with chronic prerenal disease in the setting of heart failure or cirrhosis with decreased renal perfusion, intrinsic renal disease such as nephrosclerosis or atherosclerotic renal artery disease, chronic nephritic or nephrotic glomerular disease, or chronic postrenal obstructive disease. Chronic kidney disease is more common in older patients after years of underlying kidney injury. Active vitamin D (1,25-dihydroxycholecalciferol) is synthesized through cutaneous, hepatic, and renal routes, with the final conversion of 25-hydroxycholecalciferol to active 1,25-dihydroxycholecalciferol in the kidney via 1-alpha hydroxylase. Patients with chronic kidney disease have diminished capacity for native vitamin D synthesis because of underlying renal parenchymal damage. The decreased production of active vitamin D results in impaired gastrointestinal uptake of dietary calcium, leading to hypocalcemia. This, in turn, leads to increased serum PTH and secondary hyperparathyroidism. Chronic renal failure is also characterized by the inability of the kidney to excrete phosphate, resulting in hyperphosphatemia. Hyperphosphatemia and hypocalcemia result in the upregulation of PTH, which increases bony turnover in an attempt to raise serum calcium. 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 common on urinalysis. It would not cause hypocalcemia. Dehydration (Choice C) can result in prerenal acute kidney injury but would not lead to the hypocalcemia and secondary hyperparathyroidism seen in chronic renal failure. Rapidly progressive glomerulonephritis (Choice D) is a nephritic syndrome characterized by acute renal failure associated with hematuria, red blood cell casts, and hypertension. It does not cause hypocalcemia directly. Renal cell carcinoma (RCC) (Choice E) 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. Hypercalcemia can result from bony metastasis. Educational Objective: Secondary hyperparathyroidism typically presents with increased PTH, hypocalcemia, and hyperphosphatemia. It results from chronic renal failure because of the inability of the kidney to excrete phosphate and to produce active vitamin D, which results in hypocalcemia. Hyperphosphatemia and hypocalcemia result in the upregulation of PTH, which increases bony turnover in an attempt to raise serum calcium. %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 4-year-old boy has had a viral upper respiratory tract infection for the past 3 days. Clearance of the virus is most dependent on which of the following? A) Activation of macrophages B) Class I MHC-viral peptide complex presentation C) Class II MHC-viral peptide complex presentation D) Phagocytosis of viral particles by CD8+ T lymphocytes E) Production of memory B lymphocytes F) Proliferation of memory T lymphocytes G) Proliferation of plasma cells

B. Class I MHC-viral peptide complex presentation with subsequent activation of CD8+ T lymphocytes will most likely contribute to clearance of the virus. Viruses are obligate intracellular pathogens and must enter a cell to proliferate. Viral particles that are released from infected cells are taken up by professional antigen presenting cells (APCS) such as dendritic cells. These large proteins are shuttled to the proteasome, which generates smaller fragments of protein that are in turn shuttled to the endoplasmic reticulum (ER). In the ER, these peptides are attached to class I MHC molecules before being transferred to the Golgi apparatus and then to the cytoplasmic membrane where they are expressed. Viral particles may also be presented on class I MHC molecules on the surface of infected cells, although these cells often lack the costimulatory capabilities of APCS. CD8+ T lymphocytes possess unique T-cell receptors (TCRS) that recognize particular viral antigens, and binding of the TCR to the class I MHC-antigen complex activates the CD8+ T lymphocytes. In turn, they release cytokines such as interferon-y and tumor necrosis factor; they also interact with cell surface Fas and FasL to induce apoptosis of infected cells. They can release granzyme and perforin, which cause membrane pore formation and lead to apoptosis of infected cells. Incorrect Answers: A, C, D, E, F, and G. Activation of macrophages (Choice A) occurs primarily through IL-12/IFN-y signaling. Activated macrophages can become histiocytes and may form granulomas in the case of some infections (eg, Mycobacterium tuberculosis). Class II MHC-viral peptide complex presentation (Choice C) activates CD4+ T lymphocytes, which assist B lymphocytes in making antibodies, and recruit macrophages, CD8+ T lymphocytes, and leukocytes to the site of activation. While they do contribute to viral clearance, they are incapable of directly killing infected cells, unlike Class I MHC-ČD8+T lymphocyte interactions. Phagocytosis of viral particles by CD8+ T lymphocytes (Choice D) is incorrect. Viral particles are phagocytosed by APCS and their peptide fragments are presented via class II MHC molecules on the cell surface. Production of memory B lymphocytes (Choice E) occurs during the course of a viral infection after B lymphocytes class switch and differentiate into plasma cells (Choice G), which secrete large quantities of immunoglobulin. While a critical part of humoral immunity, the production of specific antibodies takes time and does not contribute to early clearance of a virus unless specific memory B lymphocytes and plasma cells already exist. Proliferation of memory T lymphocytes (Choice F) is a feature of viral infections, although it occurs later in the course and does not contribute to early viral clearance upon initial infection unless the patient is re-exposed to the same virus. Memory T lymphocytes reside in peripheral tissues; upon re-exposure rapidly differentiate into effector cells such as CD8+ T lymphocytes. Educational Objective: Early viral clearance is predicated upon the actions of cytotoxic CD8+ T lymphocytes, which recognize viral antigen complexed to class I MHC molecules presented on the surface of APCS and virally infected cells. Binding results in T lymphocyte activation with the subsequent release of inflammatory cytokines to attract other immune cells, release of granzyme and perforin to form membrane pores in infected cells, and initiation of apoptosis in infected cells. %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 26-year-old man comes to the physician because of fever, cough, chest pain, and malaise for 2 weeks. He moved to central California 6 months ago. A complete blood count shows mild eosinophilia. A chest x-ray shows patchy bronchopneumonia. Culture of the sputum grows a mold with the morphology shown in the photomicrograph, with the arrow indicating the infectious particle. Which of the following is the most likely causal organism? A) Actinomyces israelii B) Coccidioides immitis C) Histoplasma capsulatum D) Legionella pneumophila E) Mycobacterium tuberculosis OF) Nocardia brasiliensis G) Staphylococcus aureus H) Streptococcus pneumoniae

B. Coccidioides immitis is an endemic fungus of the Southwestern United States and central valley of California that typically causes coccidioidomycosis, a self-limited respiratory illness. Signs and symptoms include fever, fatigue, cough, arthralgia, and myalgia. Patients may also present with erythema nodosum. Coccidioidomycosis can potentially present with disseminated disease, especially in immunocompromised patients, and cause infections of the skin, bone, and central nervous system. Silver stain of sputum or tissue biopsy demonstrates large spherules containing endospores. Diagnosis can be facilitated with enzyme-linked immunosorbent assay (ELISA) testing and be confirmed with polymerase chain reaction. Treatment is with oral or intravenous antifungals, including agents from the azole or polyene classes. Incorrect Answers: A, C, D, E, F, G, and H. Actinomyces israelii (Choice A) is a Gram-positive, anaerobic bacillus that forms branching filaments and yellow granules. It typically causes abscesses of the head and neck, often with draining fistulous tracts. Histoplasma capsulatum (Choice C) is a fungus native to the Ohio river and Mississippi river valleys that can cause pneumonia and is spread by the droppings of birds or bats. It is identifiable as oval yeasts within macrophages. Legionella pneumophila (Choice D) is a Gram-negative bacillus that is transmitted primarily through aerosols from water sources. It causes Legionnaire disease, characterized by fever, pneumonia, and gastrointestinal symptoms. It can be identified by positive staining with silver stains and its fastidious nature, requiring culture on charcoal yeast extract with supplemental iron and cysteine. Mycobacterium tuberculosis (Choice E) is an acid-fast bacillus that causes tuberculosis. It is identifiable by its thick, waxy capsule, positive acid-fast staining, and by the presence of caseating granulomas with multinucleated giant cells. Nocardia brasiliensis (Choice F) is a Gram-positive, aerobic, acid fast bacillus that forms branching filaments. It can cause pneumonia, as well as cutaneous and central nervous system infections in immunocompromised patients. Staphylococcus aureus (Choice G) is a Gram-positive coccus that is a common cause of skin and soft tissue infections in adults. It is identifiable as Gram-positive cocci in clusters. Streptococcus pneumonia (Choice H) is a Gram-positive diplococcus that is a common cause of community-acquired pneumonia infections in adults. It is identifiable as Gram-positive cocci in chains or pairs. Educational Objective: Coccidiomycosis is a fungal infection endemic to the Southwestern United States and to the central valley of California that typically presents as a self-limited respiratory illness. Signs and symptoms typically include fever, fatigue, cough, arthralgia, and myalgia. Disseminated disease is possible, especially in immunocompromised patients. 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 study is conducted to assess the effectiveness of a registry for identifying patients 50 years old and older eligible for colon cancer screening. Medical practices are selected based on equal and proportional representation of both genders and socioeconomic and ethnic groups, including those who have medical insurance and those who do not. Half of the practices will use the registry for a 15-year period and the other half will provide usual care without the registry during the same period. Results show that the colon cancer mortality rate is decreased in the practices using the registry, and the researchers recommend use of a patient registry to decrease the number of deaths due to colon cancer. Which of the following study characteristics most directly supports the researchers' recommendation? A) Accuracy B) External validity C) Face validity D) Precision E) Reliability

B. External validity determines whether an experiment can be generalized and its conclusions applied to groups beyond those considered in the study population. In other words, will the conclusions hold for different populations at different times? In this example, a registry that identifies patients over age 50, but then broadly selects practices for inclusion that have representation of gender, socioeconomic and ethnic breadth, and wide financial and insurance status will likely capture a reasonable sample of the population as a whole presenting to medical practices. The study period was also quite long, 15 years, which is likely sufficiently long to capture outcomes related to colorectal carcinoma, a condition with a long latency period. Given that the study showed reduced mortality rate through the use of the registry, the researchers make the recommendation to use it broadly, a recommendation based on proximate similarity of practices that will use it. In other words, the study results should apply to many medical practices beyond those purely with the characteristics of those included. Incorrect Answers: A, C, D, and E. Accuracy (Choice A) describes the absence of error or bias and reflects how correct the test measurements are as compared to actuality. In other words, if a patient actually weighs 50 kg, and the test measures the patient's weight, will the test show a weight of 50 kg? If so, the test may be considered accurate. Face validity (Choice C) is a relatively weak form of a validity assessment that subjectively asks whether the test or experiment measures what it is supposed to measure. In other words, is the test suitable to its aims? This experiment appears to be valid at face value - it is designed to measure whether a registry will reduce the mortality from colorectal carcinoma and compares its use over a long time across medical practices. Precision (Choice D) describes the reproducibility of results. In other words, how reliable is the test? More precise measurements suffer from less variation in measurement. Decreasing random error in a test increases precision. If a patient actually weighs 50 kg, and the test measures the patient's weight, will the test show a weight of 50 kg each time? If so, the test may be considered precise. Reliability (Choice E) is a related concept to precision that describes the consistency and reproducibility of results. A reliable test will yield the same, precise result each Educational Objective: External validity determines whether an experiment can be generalized, and its conclusions applied to groups beyond those considered in the study population. %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. During an experiment, a solution of mixed fatty acids is injected into the duodenum of an experimental animal. Under these conditions, the clearance rate of an intravenous glucose load from the circulation is doubled. In contrast, an injection of an equal volume amount of 0.9% saline into the duodenum has much less effect on the plasma clearance rate of glucose. These findings are most likely caused by the secretion of which of the following hormones? A) Gastrin B) Glucose-dependent insulinotropic peptide C) Motilin D) Secretin E) Somatostatin

B. Glucose-dependent insulinotropic peptide (GIP) is secreted by K cells in the duodenum and jejunum and functions to decrease gastric acid production and stimulate insulin release from the pancreas. Its secretion is promoted by the presence of fatty acids, amino acids, and intestinal glucose. While serum glucose also stimulates insulin secretion by the pancreas, the effect of intraluminal glucose on GIP and subsequent insulin secretion leads to increased concentrations of insulin compared to parenteral glucose administration. Insulin promotes peripheral tissue uptake of glucose, glycolysis, glycogen synthesis, protein synthesis, and fatty acid synthesis, resulting in decreased glucose concentration in the serum. Incorrect Answers: A, C, D, and E. Gastrin (Choice A) is produced by G cells in the gastric antrum and stimulates parietal cells within the gastric body to produce hydrochloric acid. Gastric acid has no effect on serum glucose concentration. Motilin (Choice C) is secreted by the small intestine and stimulates intestinal peristalsis. Motilin receptors are targeted by erythromycin and metoclopramide, used therapeutically in gastroparesis. Secretin (Choice D) is produced by duodenal S cells. It promotes the release of bicarbonate-rich pancreatic secretions and bile and inhibits gastric acid production. Somatostatin (Choice E) is a regulatory peptide secreted by D cells of the pancreas and gastrointestinal mucosa that inhibits gastric acid and pepsinogen secretion, gallbladder contraction, and insulin and glucagon release. Somatostatin would have an indirect effect on glucose through counterregulatory action of both insulin and glucagon. Educational Objective: Glucose-dependent insulinotropic peptide (GIP) is secreted by K cells in the duodenum and jejunum, and it functions to decrease gastric acid production and stimulate insulin release from the pancreas. Insulin promotes the peripheral tissue uptake of glucose and hepatic glucose storage, resulting in a decreased glucose concentration in the serum. %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. A 27-year-old man with AIDS comes to the physician because of a 6-week history of progressive memory loss, imbalance, clumsiness, and difficulty finding words. The symptoms began with an inability to concentrate 3 months ago. He is not adherent to combination antiretroviral therapy and recently recovered from Pneumocystis jirovecii (formerly P. carinii) pneumonia. An MRI shows diffuse, poorly defined, bilateral hyperintensities within cerebral white matter. A stereotactic biopsy specimen shown in the photomicrograph depicts cells scattered predominantly along the perivascular spaces. Which of the following is the most likely diagnosis? A) Cryptosporidiosis B) HIV encephalopathy C) HIV protease inhibitor toxicity O D) Lymphoma E) P. jirovecii infection

B. HIV encephalopathy is the most likely diagnosis in this patient with AIDS, progressive memory loss, and motor deficits who does not adhere to antiretroviral therapy. HIV encephalopathy is a diagnosis of exclusion in patients with HIV, and other causes of encephalopathy such as toxoplasmosis, meningitis, encephalitis, bacterial abscess, progressive multifocal leukoencephalopathy (PML), or primary central nervous system (CNS) lymphoma should be ruled out. The classic triad of HIV encephalopathy involves movement disorders, psychomotor impairment, and memory deficits. While it can occasionally be confused with PML, patients with PML tend to have a more rapid disease progression, focal deficits, and different findings on MRI. In HIV encephalopathy, the MRI demonstrates multiple, symmetric, and poorly demarcated T2 hyperintense lesions scattered in the subcortical white matter. Brain biopsy characteristically demonstrates microglial nodules with multinucleated giant cells as in this case. Incorrect Answers: A, C, D, and E. Cryptosporidiosis (Choice A) is caused by Cryptosporidium parvum that presents with severe diarrhea in patients with AIDS. Symptoms include fever, weight loss, symptoms of dehydration and orthostasis, severe, watery diarrhea, cramping abdominal pain, nausea, and vomiting. Disseminated infection can involve the lungs and liver but encephalopathy or encephalitis does not occur. HIV protease inhibitor toxicity (Choice C) can present acutely with nausea and vomiting, but long-term complications commonly include lipodystrophy and increased cardiovascular risk. They do not cause encephalopathy. Lymphoma (Choice D) in patients with AIDS can take on many forms, but primary CNS lymphoma is common. It often presents with seizures, lethargy, subacute memory loss, and headache. Physical examination may show neurologic deficits stemming from structural disruption caused by the location of the mass. MRI will usually show a dominant cortical mass. Multiple lesions distributed throughout the cortex would be atypical. P. jirovecii (Choice E) is an opportunistic, yeast-like fungal organism that can cause pneumonia in immunocompromised patients. Chest imaging typically shows diffuse, bilateral infiltrates, often prominently about the hila. Methenamine silver or toluidine blue selectively stain the cyst walls and are used to confirm the diagnosis. Educational Objective: HIV encephalopathy is typically a diagnosis of exclusion. Patients commonly present with subacute to chronic memory and language impairment and psychomotor deficits. MRI shows symmetrically distributed, poorly defined, T2 hyperintense lesions in the subcortical white matter. Brain biopsy shows multinucleated giant cells and microglial nodules. %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 23-year-old woman comes to the physician because of a 1-week history of intermittent episodes of fever and chills, followed by a rash. She has had four operations to correct scoliosis, in addition to undergoing a tonsillectomy, an appendectomy, removal of a lipoma from her scalp, and dermabrasion to correct acne scarring. She was admitted to the hospital 1 year ago for 2 weeks because of nausea and vomiting of unexplained origin. Her temperature is 37°C (98.6°F). Physical examination shows approximately 12 evenly spaced punctate marks in a linear pattern on the abdomen and upper and lower extremities. There are no marks on the face or back. Which of the following is the most likely diagnosis? A) Acute myelogenous leukemia B) Factitious disorder C) Immune thrombocytopenic purpura D) Lyme disease E) Syphilis

B. In factitious disorder, patients consciously produce symptoms (ie, purposely injure 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. Early losses, neglect, or abuse may disrupt patient identity and relationships, and adopting the sick role may represent an attempt to cope with stress and stabilize their identity and relationships. This young patient's extensive history of elective surgeries and hospital admissions illuminate a pattern of care seeking, and linearly spaced punctate marks are consistent with self-inflicted puncture wounds rather than a known disease. Treatment includes regular follow-up with one physician who oversees management (as opposed to symptom-triggered appointments) and possible psychotherapy. Incorrect Answers: A, C, D, and E. Acute myelogenous leukemia (Choice A) is a hematologic cancer resulting from the malignant transformation of myeloid precursor cells. Proliferated myeloid precursor cells may deposit in soft tissues and present as nodules or plaques. Punctate lesions would be atypical. Immune thrombocytopenic purpura (Choice C) is an acquired thrombocytopenia resulting from autoantibodies against platelet antigens. Patients may present with a petechial or purpuric rash in dependent body regions (eg, feet, legs, and hands) rather than punctate marks on the abdomen and legs. Lyme disease (Choice D) is a tick-borne illness that may initially present with erythema migrans, a lesion that forms around the site of the tick bite. The lesion may develop central clearing. Multiple punctate lesions would be inconsistent with Lyme disease. Patients with syphilis (Choice E), a sexually transmitted spirochete infection, may demonstrate a diffuse macular or papular rash that involves the trunk and extremities (including the palms and soles) during secondary syphilis or may present with gummatous disease during tertiary syphilis, which can manifest as singular or multiple ulcerative or granulomatous lesions. Punctate marks in a linear pattern would be atypical. Educational Objective: In factitious disorder, patients consciously produce symptoms (ie, purposely injure themselves) for primary gain. A history of multiple procedures and subjective or objective findings that are inconsistent with known diseases suggest factitious disorder. %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. A 28-year-old woman has had fatigue and heavy menstrual periods during the past year. She is a vegetarian who eats eggs and dairy products. Her diet contains little fruit. Which of the following erythrocyte types is most likely to be present on a peripheral blood smear from this patient? OA) Macrocytes B) Microcytes C) Ovalocytes D) Spherocytes E) Stippled cells

B. Microcytes are most likely to be present in this patient's peripheral blood smear, which is suggestive of iron deficiency anemia (IDA) in the setting of heavy menstrual periods and low dietary iron intake (vegetarianism). Erythrocytes in the setting of IDA are pale (hypochromic) and small in size (microcytic) but normal in shape. Iron is required for the synthesis of heme, which is a necessary component of the hemoglobin molecule, and thus, of erythrocytes. It functions to shuttle oxygen to and from peripheral tissues. In individuals who do not have an adequate intake of dietary iron in the form of heme obtained from animal meat, especially when other causes of ongoing blood loss such as heavy menstruation are present, deficiency can develop. IDA may also develop as the result of chronic blood loss from colorectal bleeding, as a result of malabsorption syndromes (eg, celiac disease), and in patients who have undergone gastric bypass surgery. Erythrocytes on the peripheral blood smear are hypochromic and microcytic as a result of deficient hemoglobin concentration. It is hypothesized that erythrocytes are microcytic as a result of continuing erythrocyte division in order to reach an adequate hemoglobin concentration; because hemoglobin stores are inadequate, cell division continues beyond what would normally occur and causes the cells to be smaller than normal. Treatment for this patient would include management of her menstrual bleeding and oral iron supplementation. Incorrect Answers: A, C, D, and E. Macrocytes (Choice A) refer to larger than normal erythrocytes and are seen in the setting of folate or vitamin B 12 deficiency. This patient has chronic blood loss most likely leading to iron deficiency and is likely to have adequate stores of folate and vitamin B 12 which are consumed in high amounts in most vegetarian diets. Ovalocytes (Choice C) are oval shaped erythrocytes also known as elliptocytes. While they can be seen in IDA, they are more commonly seen in conditions such as myelofibrosis, thalassemia, megaloblastic anemia, and myelodysplastic syndrome. Spherocytes (Choice D) are small, dark, round cells with a high concentration of hemoglobin that are most commonly seen in hereditary spherocytosis or in cases of autoimmune hemolytic anemia. They are not seen in IDA. Stippled cells (Choice E) suggest the presence of precipitated ribosomes and ribosomal RNA within an erythrocyte. Classically, coarse basophilic stippling is associated with lead poisoning. Educational Objective: Patients who follow a strict vegetarian diet and have clinically significant blood loss from menstruation are at risk for iron deficiency anemia as iron is required for the normal synthesis of heme. IDA presents with hypochromic, microcytic erythrocytes on peripheral blood smear. Treatment includes control of the underlying bleeding and oral iron supplementation. %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 70-year-old woman comes to the physician for a routine pelvic examination. During speculum examination of the vagina and cervix, the Valsalva maneuver causes a bulge of the anterior vaginal wall. Which of the following is the most likely cause of this finding? A) Bartholin gland cyst B) Cystocele C) Obturator hernia D) Rectocele E) Uterine prolapse

B. Normal pelvic support is caused by the action of several pelvic muscles and ligaments of the pelvic floor, including the levator muscles, urogenital diaphragm, perineal body, endopelvic fascia, uterosacral and cardinal ligaments, anal sphincter, and urethral sphincter. Defects in this pelvic support may result in urogenital prolapse, which is caused by the loss of connective tissue and muscular support of the uterus, paravaginal tissue, bladder wall, urethra, or distal rectum. Risk factors for pelvic support defects include collagen and connective tissue disorders, previous vaginal delivery, menopause, prior pelvic surgery, and increased intra-abdominal pressure (eg, obesity and chronic constipation). Signs and symptoms include a vaginal mass, vaginal pressure, perineal discomfort, and urinary or fecal retention or incontinence. In the case of a cystocele, the bladder prolapses into the anterior vaginal wall. It is associated with stress urinary incontinence, as well as urinary retention. Physical examination will show a bulge of the anterior vaginal wall, as in this patient. Management for a symptomatic cystocele may include the insertion of a vaginal pessary, which is a mechanical device that provides pelvic support, along with pelvic floor exercises or surgical correction. Incorrect Answers: A, C, D, andE. A Bartholin gland cyst (Choice A) appears as a swollen, fluctuant mass at the posterior labium majora caused by obstruction of Bartholin gland outflow. If left untreated, a cyst can progress to an abscess, with the development of erythema and tenderness to palpation. It does not cause a mass on the anterior vaginal wall. An obturator hernia (Choice C) is a herniation of the abdominal contents through the obturator foramen, along with the obturator vessels and nerve. It presents with abdominal pain and a potentially palpable mass in the groin. It is more common in patients with increased intra-abdominal pressure, as well as elderly multiparous women. A rectocele (Choice D) is caused by weakness of the pelvic floor, with the rectum prolapsing into the posterior wall of the vagina, rather than the anterior wall, as in this patient. Symptoms include constipation and a palpable bulge of the posterior vaginal wall. Uterine prolapse (Choice E) is related to pelvic floor weakness and presents with prolapse of the uterus into the vaginal canal. This would be seen on speculum examination with lowering of the cervix toward the vaginal opening with a Valsalva maneuver, rather than an anterior vaginal wall bulge. Educational Objective: A cystocele is a form of pelvic organ prolapse caused by weakness of the pelvic floor musculature, commonly occurring in the setting of connective tissue disorders, previous vaginal delivery, menopause, prior pelvic surgery, and increased intra-abdominal pressure. A cystocele presents most commonly with urinary retention or incontinence and an anterior vaginal wall bulge on physical examination. %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. A 54-year-old man comes to the physician because of a 2-week history of burning epigastric pain. He is recently divorced. His diet mainly consists of fast food. He smokes cigars and drinks one to two 12-ounce cans of beer on the weekends. He was treated for Helicobacter pylori infection 1 year ago. He uses high doses (600 mg daily) of over-the-counter ibuprofen for chronic joint pain. Abdominal examination shows tenderness to deep palpation of the epigastric area. A peptic ulcer is suspected. Which of the following is the strongest predisposing risk factor for this patient's condition? A) High-fat diet B) Ibuprofen use C) Ingestion of alcohol D) Recent psychosocial stressor E) Smoking cigars

B. Peptic ulcer disease describes the presence of ulcers in the stomach or duodenum, which classically present with worsening abdominal pain related to consumption (gastric) or lack of consumption (duodenal) of food. The most common cause of peptic ulcer disease is infection with the bacterium Helicobacter pylori, which accounts for nearly all duodenal ulcers. Gastric ulcers can be caused by prolonged or excess usage of nonsteroidal anti-inflammatory drugs (NSAIDS), such as ibuprofen, which inhibit cyclooxygenase in the gastrointestinal tract, leading to a reduction of prostaglandin secretion and decreased protection of the gastric mucosa. Persistent inflammation related to a peptic ulcer can result in complications including fibrosis, stricture, hemorrhage, and perforation. Proton pump inhibitor therapy is first line in the management of gastroesophageal reflux and peptic ulcer disease along with eradication of H. pylori and cessation of inciting factors (eg, smoking, alcohol, NSAIDS). Incorrect Answers: A, C, D, and E. High-fat diet (Choice A) can be related to biliary colic or hepatobiliary pathology. Biliary colic is characterized by epigastric or right upper quadrant pain after ingestion of a fatty meal promoting gallbladder contraction. Ingestion of alcohol (Choice C) in excessive amounts can cause damage to the gastric mucosal barrier. This patient drinks a moderate amount of alcohol occasionally, making it unlikely to be the strongest cause of his peptic ulcer disease. Recent psychosocial stressor (Choice D) has been associated to a limited extent with increased risk for peptic ulcer disease. However, NSAID use has a stronger association with the development of peptic ulcer disease. Smoking cigars (Choice E) has been associated with peptic ulcer disease. Risk for peptic ulcer disease increases accordingly with increasing pack-years of smoking. This patient smokes cigars intermittently; it is not the strongest predisposing risk factor. Educational Objective: Peptic ulcer disease presents with epigastric or left upper quadrant abdominal pain in association with meals. It is caused by infection from H. pylori, or from inciting factors that compromise the gastric mucosal barrier such as NSAIDS, smoking, and alcohol use. %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. A 25-year-old man is brought to the emergency department because of a 2-hour history of nausea, vomiting, severe abdominal pain, and shortness of breath. His respirations are 32/min. Abdominal examination shows guarding and rigidity. An x-ray of the chest and upper abdomen with the patient seated upright is shown. Based on these findings, this patient most likely has which of the following? A) Diaphragmatic hernia B) Perforated duodenal ulcer C) Rectovesical fistula D) Tension pneumothorax E) Traumatic injury to the liver

B. Perforation of a duodenal ulcer commonly presents with severe, diffuse abdominal pain, nausea, vomiting, fever, tachycardia, and septic shock. Bowel contents in the peritoneal space cause rapid infection in the form of peritonitis, marked by abdominal rigidity, rebound tenderness, and guarding. An upright chest x-ray may show air under the diaphragm (pneumoperitoneum) caused by leakage of bowel contents into the peritoneal cavity, as in this patient. Perforated viscus (and resultant peritonitis) constitutes a surgical emergency requiring exploratory laparotomy to identify and repair any perforation, lavage the peritoneum and prevent or control sepsis, manage exsanguination if vessels are involved, and resect any necrotic bowel. Broad spectrum antibiotics are also indicated. Incorrect Answers: A, C, D, and E. Diaphragmatic hernia (Choice A) describes the presence of abdominal contents within the thorax, and can present with chest pain, shortness of breath, and bowel sounds within the lung fields. It most commonly occurs in the left hemithorax and may be visualized as intestinal loops on radiographs. Rectovesical fistula (Choice C) is a condition in which there is an abnormal connection between the rectum and the urinary bladder. This can lead to complications of urine leakage through the rectum and urinary tract infections plus pneumaturia. It would not cause pneumoperitoneum. Tension pneumothorax (Choice D) occurs when air accumulates within the pleural cavity to an extent that results in compression and shifting of the mediastinal structures. Such structural compromise results in impaired venous return leading to diminished cardiac output. It typically presents as severe respiratory distress, jugular venous distention, tracheal deviation, diminished breath sounds, hyperresonance to percussion, and hemithorax hyperinflation. Vital signs disclose tachypnea, tachycardia, and hypotension. Traumatic injury to the liver (Choice E), such as a liver laceration, could present with abdominal pain and peritonitis as a result of blood causing irritation and inflammation the peritoneum. However, it would not cause pneumoperitoneum. Educational Objective: Perforation of a duodenal ulcer commonly presents with severe, diffuse abdominal pain, nausea, vomiting, fever, tachycardia, and septic shock. An upright chest x-ray may show air under the diaphragm (pneumoperitoneum) caused by leakage of bowel contents into the peritoneal cavity. %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 33-year-old woman comes to the physician because of a 3-day history of nausea and light-headedness. Her last menstrual period was 5 weeks ago. She is apprehensive. Physical examination shows no abnormalities. Her serum B-human chorionic gonadotropin concentration is increased. At this stage in the pregnancy, progesterone is most likely produced within which of the following structures? A) Corpus albicans B) Corpus luteum C) Pituitary gland D) Placental cytotrophoblast E) Placental syncytiotrophoblast

B. Pregnancy is suspected when there is a missed or delayed menstrual period. During a normal menstrual period, follicle-stimulating hormone and luteinizing hormone (FSH and LH, respectively) concentrations increase and stimulate the developing follicle. The follicle produces estrogen, which leads to proliferation of the endometrium in preparation for implantation of a fertilized ovum. As estrogen rises, a surge occurs, which in turn stimulates a surge in LH that causes ovulation. Immediately following ovulation, the corpus luteum forms. The corpus luteum secretes progesterone to maintain the endometrial lining. If no implantation occurs, the corpus luteum degrades to the corpus albicans and estrogen and progesterone concentrations decrease, causing menstruation. However, during pregnancy, the placenta develops from embryo implantation. The placenta then begins to secrete human chorionic gonadotropin, which acts to maintain the corpus luteum and its secretion of progesterone, which is necessary for maintenance of the pregnancy. As the placenta develops, it becomes primarily responsible for progesterone production around seven to ten weeks of gestation. As this patient is at five weeks of gestation, her progesterone production is still primarily performed by the corpus luteum. Incorrect Answers: A, C, D, andE. The corpus albicans (Choice A) is the degraded corpus luteum that develops because of the absence of embryo implantation. It does not secrete hormones. The pituitary gland (Choice C) is responsible for secreting FSH and LH, as well as prolactin and thyroid-stimulating hormone. While FSH and LH play a role in the development of an ovarian follicle, the pituitary gland does not directly produce progesterone. Placental cytotrophoblast (Choice D) is the inner layer of the chorion and is vital for the implantation of an embryo. It does not secrete hormones. Placental syncytiotrophoblast (Choice E) is the outer layer of the chorionic villi and is responsible for secreting human chorionic gonadotropin, which maintains the corpus luteum and its production of progesterone. It does not directly produce progesterone. Educational Objective: During pregnancy, progesterone is produced by the corpus luteum for the first seven to ten weeks of gestation secondary to stimulation by human chorionic gonadotropin from the placental syncytiotrophoblast, after which time it is primarily produced by the placenta. %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. A study is done to determine the relationship between use of oral contraceptives and cervical cancer. Study subjects include 50,000 women who are using oral contraceptives and 50,000 women who have had a tubal ligation. After 2 years of follow-up, the rate of in situ cervical cancer is 18 per 10,000 in the oral contraceptive group and 3 per 10,000 (p < 0.05) in the tubal ligation group. Which of the following is the estimated relative risk of cervical cancer among women who have had a tubal ligation compared with women who use oral contraceptives? A) 3 - 50,000 = 0.00006 B) 3 18 = 0.17 C) 18 - 3 = 6 D) 18 - 3 +10 = 17.7 E) 18 - 3 = 15 F) 18 + 3 = 21 G) Indeterminable from the data given

B. Relative risk (RR) describes the difference in likelihood of the occurrence of a particular disease outcome between two groups of patients with or without a particular exposure. In this case, the outcome of cervical cancer in situ is compared between two exposures, the use of oral contraceptives or the use of tubal ligation for contraception. Calculations of relative risk are commonly performed in cohort studies. RR is calculated by dividing the fraction of patients with a positive exposure and who developed disease (a) amongst all patients who were exposed which includes those exposed who did not develop disease (b), (a + b), by the fraction of patients with a negative exposure and who developed disease (c) amongst all patients who were not exposed which includes those who did not develop disease (d), (c + d). RR thus equals (a / (a + b) / (c/ (c + d)). RR values greater than 1.0 indicate an increased risk for developing disease in association with the exposure, whereas values less than 1.0 indicate a reduced risk for developing disease, and RR equal to 1.0 indicates that the disease outcome and the exposure are not related. In this case, RR of cervical cancer in situ among women who have had a tubal ligation compared with women who use oral contraceptives is calculated as (3/ 10,000) - (18/10,000) = 3 18 = 0.17. %3D Incorrect Answers: A, C, D, E, F, and G. 3 - 50,000 = 0.00006 (Choice A) is an incorrect computation of the total fraction of women with a tubal ligation who developed cervical cancer in situ, as the data are presented per 10,000 patients. Therefore, if the fraction of women with a tubal ligation who developed cervical cancer in situ were to be calculated, it would be 3/10,000 persons or 15/50,000 in absolute numbers. 18 3 = 6 (Choice C) computes the RR of developing of cervical cancer in situ among women who use oral contraceptives compared with women who have had a tubal ligation. 18 - 3 +10 = 17.7 (Choice D) is not a meaningful computation in the context of the presented data and reflects a misinterpretation or a mathematical error. | 18 - 3 = 15 (Choice E), if reflecting the numbers of cervical carcinoma in situ per 10,000 patients, would reflect the attributable risk to oral contraceptive use. %3D 18 + 3 = 21 (Choice F), if reflecting the numbers of cervical carcinoma in situ per 10,000 patients, would reflect the total risk within the overall study population. Indeterminable from the data given (Choice G) is incorrect, as the relative risk can be computed from the data provided. Educational Objective: Relative risk (RR) describes the difference in likelihood of the occurrence of a particular disease outcome between two groups of patients with or without a particular exposure. RR is calculated by dividing the fraction of patients with a positive exposure and who developed disease (a) amongst all patients who were exposed including those exposed who did not develop disease (b), (a + b), by the fraction of patients with a negative exposure and who developed disease (c) amongst all patients who were not exposed which includes those who did not develop disease (d), (C + d). RR thus equals (a / (a + b))/ (c/ (c+ d)). %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 study is conducted to assess the accuracy of a new rapid test to detect a virulent bacterial infection. This infection has an 80% mortality rate if it is not identified early in its course; however, prompt administration of antibiotics decreases the mortality rate to less than 5%. The risks of this antibiotic therapy are minimal. A total of 10,000 participants are enrolled and undergo assessment with the new test. The graph shows the distribution of infected and noninfected participants according to the results of the test. Which of the following labeled points is most appropriate for use as the optimal diagnostic cut point for results of this test? Not Infected infected A B C D E A) B) C) D) E)

B. The cut point of the test should be set such that all persons with the infection result as positive so that there are few false negatives. The threshold should be set to maximize sensitivity, which equals the true positive test results divided by the sum of true positive and false negative test results. Therefore, as sensitivity approaches 1.0, all patients who have the disease will be detected by the test (there will be no false negatives). In this example, high sensitivity is critical, as the disease has an 80% mortality rate if not detected and treated early. Therefore, a priority should be placed on sensitivity when deciding the cut point threshold for positive and negative results. Specificity is calculated by the true negative test results divided by the sum of true negative and false positive results. Increasing sensitivity comes at the expense of specificity; by setting the cut point to include all persons with the disease, many false positives will be introduced as persons without the disease have results above the cut point. Patients testing negative on a sensitive test can be safely considered disease free, whereas patients who test positive should receive additional diagnostic evaluation or treatment. In this example, a bimodal distribution of patients with and without disease is presented. All patients without disease are included in the area beneath the not-infected curve, while all patients with disease are included in the area beneath the infected curve. There is overlap between the two, suggesting that some patients will test equivalently on the test but may or may not have the disease. Setting the cut point to point B would permit all patients within the area under the infected curve to be identified as positive, which maximizes sensitivity, while minimizing the number of false positives, and therefore optimizing specificity while prioritizing sensitivity. Incorrect Answers: A, C, D, and E. Choice A would permit 100% sensitivity, as all patients within the area under the infected curve would test positive, however, in comparison with point B (which also identifies all infected patients), point A would include a larger fraction of false positives. This would reduce the specificity of the test without further increasing sensitivity, which would worsen its performance. Choices C, D, and E all miss a substantial fraction of infected patients, and would all have a sensitivity less than 100%, making them suboptimal for use in ruling out a significantly fatal disease. Educational Objective: The cut point of a test should be set to optimize sensitivity, specificity, or both depending on the clinical utility of the test. A sensitive test should be employed when ruling out a significantly mortal or morbid diagnosis. II Previous Next Score Report Lab Values Calculator Help Pause Number of participants

36 Exam Section 1: Item 36 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 36. An investigator is studying Po, in an experimental animal. It is found that Po, in the renal vein is relatively high compared with venous Po, from most other organs. Which of the following best explains this finding? A) Bicarbonate reabsorption causes a Bohr shift of the oxyhemoglobin dissociation curve of the renal vein B) Blood flow per gram of tissue is greater in kidneys than in other organs OC) Erythropoietin synthesized in the kidneys increases local hematocrit and Po2 D) Fenestrations in glomerular capillaries promote the convective transport of O, from Bowman space to the efferent arteriole E) Most of the active transport to form urine in the kidney occurs in the renal podocytes

B. The kidneys receive approximately one-fourth to one-fifth of total cardiac output, and because of the small size of the kidneys, they receive among the largest blood flow per gram of tissue. Blood flow functions to provide oxygen and nutrients. In the kidneys, an additional function of blood flow beyond native supply of the organ parenchyma is to contribute to filtration of plasma to create ultrafiltrate and urine. Thus, the blood flow in the kidney is in excess of what is needed to meet its metabolic demands. This results in a small difference in arterial and venous Po, and Po, in the renal vein is relatively high compared to other organs that depend on blood flow to meet local metabolic demand. Incorrect Answers: A, C, D, and E. Shifting of the oxyhemoglobin dissociation curve (Choice A) describes hemoglobin's sigmoidal affinity for oxygen depending on the partial pressure of oxygen. The oxyhemoglobin dissociation curve can shift and is affected by H+ and the partial pressure of CO, not bicarbonate. Other factors that can affect the oxyhemoglobin dissociation curve include exercise, altitude, and temperature. Erythropoietin is a hormone produced by kidneys that stimulates proliferation of erythrocytic progenitors in the bone marrow. It does not increase local hematocrit and Po, (Choice C). Fenestrations in glomerular capillaries do not promote the convective transport of O, from Bowman space to the efferent arteriole (Choice D). Fenestrations in the glomerulus function for the filtration of plasma solutes and water to create ultrafiltrate in the Bowman space. Most of the active transport to form urine in the kidney does not occur in the renal podocytes (Choice E). The renal podocytes are cells in the Bowman space, which is the first site of filtration. Filtration between the glomerulus and Bowman space is secondary to hydrostatic and oncotic pressures, not as a result of active transport. Educational Objective: The kidney's relatively low metabolic demands compared with high blood flow results in a small difference in renal arterial and venous Po, The Po, in the renal vein is relatively high compared to other organs that depend on blood flow to meet local metabolic demands. %3D 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 25-year-old man comes to the physician because of abdominal pain and vomiting for 1 day. He has had greasy yellow plaques on his skin and recurrent abdominal pain following ingestion of fatty meals since infancy. Physical examination shows xanthomas over the trunk and lipemia retinalis. There is tenderness over the epigastrium, and bowel sounds are decreased. Laboratory studies show leukocytosis and an increased serum amylase activity. The plasma appears milky. Which of the following is most likely increased in this patient's serum? A) Apo A-II B) Chylomicrons C) Free cholesterol D) HDL-cholesterol E) LDL-cholesterol

B. The patient is presenting with characteristic features of familial hyperchylomicronemia, a genetic lipid disorder caused by lipoprotein lipase deficiency. Lipoprotein lipase catalyzes the conversion of triglycerides carried by circulating chylomicrons into fatty acids in extrahepatic tissue. The fatty acids can then be absorbed by cells for use in metabolism. Lipoprotein lipase deficiency is marked by increased serum concentrations of chylomicrons and triglycerides. Patients typically present with abdominal pain following ingestion of fatty meals and steatosis. Physical examination may disclose eruptive xanthomas and lipemia retinalis (cream-colored discoloration of the retinal vessels associated with hyperchylomicronemia). Build-up of chylomicrons and triglycerides in the serum results in a milky appearance and increased amylase activity. Incorrect Answers: A, C, D, and E. Apo A-II (Choice A) is a protein that is found in HDL-cholesterol and is not involved in chylomicron transport and processing. Apo C-Il is a component of VLDL-cholesterol and chylomicrons, and it activates lipoprotein lipase. Deficiency of Apo C-Il can also result in familial hyperchylomicronemia. Free cholesterol (Choice C) concentrations are not typically increased in hyperchylomicronemia, and the condition is not associated with an increased risk for atherosclerosis, as with other lipid disorders. HDL-cholesterol (Choice D) is composed of the apolipoproteins ApoA-1, ApoC-II, and ApoE. HDL concentrations are not increased in hyperchylomicronemia. LDL-cholesterol (Choice E) is primarily composed of the apolipoprotein ApoB-100, which mediates binding to the LDL-receptor on hepatocytes. LDL concentrations are increased in familial hypercholesterolemia and combined hyperlipidemia. Educational Objective: A deficiency in lipoprotein lipase results in familial hyperchylomicronemia caused by the inability of endothelial cells to catalyze the breakdown of triglycerides from circulating chylomicrons. Patients present with abdominal pain following ingestion of fatty meals. %3D Previous Next Score Report Lab Values Calculator Help Pause

5 Exam Section 1: Item 5 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 5. A 76-year-old woman comes to the emergency department because of a 6-hour history of moderate right low back and flank pain. Vital signs are within normal limits. Physical examination shows mild vertebromediastinal recess (costovertebral angle) tenderness. Urinalysis shows no blood. A CT scan of the pelvis shows hydronephrosis with obstruction of the right ureter caused by external compression at the pelvic brim from a vascular structure. This patient most likely has an aneurysm of which of the following arteries? A) Abdominal aorta B) Common iliac C) Femoral D) Inferior mesenteric E) Renal

B. The ureter courses from the renal pelvis to the urinary bladder. It initially descends through the lower abdomen anterior and medial to the psoas major. It then enters the pelvis and crosses immediately anterior to the bifurcation of the common iliac artery, before continuing its descent anterior to the internal iliac artery. The ureters are commonly constricted at three locations: at the ureteropelvic junction, at the crossing point over the common iliac artery, and at the vesicoureteral junction at the entrance to the bladder. The ureter may be constricted by a vascular aneurysm of the common iliac artery and, additionally, renal calculi tend to lodge at these three locations. Ureteral obstruction leads to hydroureteronephrosis and possible postrenal kidney injury. Incorrect Answers: A, C, D, and E. The abdominal aorta (Choice A) is located medial to the ureters, just anterolateral to the vertebral bodies. It bifurcates into the two common iliac arteries and does not come in direct proximity to either ureter. Abdominal aortic aneurysms are usually not of sufficient size to cause ureteral obstruction. The femoral artery (Choice C) is found in the thigh as a continuation of the external iliac artery. It does not come in proximity to the ureters. Femoral arterial aneurysms occur but are uncommon. The inferior mesenteric artery (Choice D) originates from the abdominal aorta near the level of the third lumbar vertebra and provides arterial supply to the distal colon. As it is a midline structure, it does not come in close proximity to the ureters. The renal artery (Choice E) supplies blood to the kidney and the proximal portion of the ureter. While aneurysms of the renal artery may form, they are rare and are unlikely to obstruct the ureter. Educational Objective: The ureter is commonly constricted at three sites: at the ureteropelvic junction, at the crossing point over the common iliac artery, and at the vesicoureteral junction at the entrance to the bladder. The ureter may be constricted by a vascular aneurysm of the common iliac artery and, additionally, renal calculi tend to lodge at these three locations. %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 previously healthy 18-year-old man is brought to the emergency department by his mother 45 minutes after she found him crouched in his closet, saying he had to hide from the intruders "from that other place." There were no intruders in the house. He has no history of alcohol or drug use. He appears afraid of the physician and sits far away from her. During the interview, he says he has heard two male voices in his head for the past 2 weeks. Physical examination and laboratory studies show no abnormalities. Which of the following is the most likely diagnosis? A) Bipolar disorder B) Brief psychotic disorder C) Delusional disorder D) Schizoaffective disorder E) Schizophrenia

B. This patient meets the diagnostic criteria for brief psychotic disorder. Brief psychotic disorder is characterized by the acute onset of one or more psychotic symptoms (eg, delusions, hallucinations, disorganized speech, disorganized behavior) lasting less than one month. This patient is experiencing a paranoid delusion, disorganized behavior (crouching in his closet despite the absence of an intruder), and auditory hallucinations. Risk factors for brief psychotic disorder include stressful life events and personality disorders. Treatment may include hospitalization and antipsychotic treatment depending on symptom severity and functional impairment, though some symptoms and the presentation overall may resolve without medications. Incorrect Answers: A, C, D, and E. Bipolar disorder (Choice A) is characterized by discrete episodes of depression and mania, which can sometimes be accompanied by psychotic symptoms such as hallucinations and delusions. This patient demonstrates psychotic symptoms without evidence of mood symptoms, which makes a primary psychotic disorder such as brief psychotic disorder more likely. Delusional disorder (Choice C) features one or more delusions for a month or longer without other psychotic symptoms. This patient does not meet the duration criterion for delusional disorder and demonstrates disorganized behavior and auditory hallucinations, which are atypical of delusional disorder. Patients with chronic psychotic disorders such as schizoaffective disorder (Choice D) or schizophrenia (Choice E) demonstrate at least two of the following five symptoms: delusions, hallucinations, disorganized speech, disorganized behavior, or negative symptoms (eg, flat affect, apathy, alogia). Many patients also illustrate prodromal periods of strange behavior and decreased functioning. Patients with schizoaffective disorder experience these psychotic symptoms in the absence of mood symptoms though also experience prominent depressive or manic symptoms. The total duration of symptoms must exceed six months to meet diagnostic criteria. This patient does not meet the duration criterion and did not demonstrate prodromal symptoms, making brief psychotic disorder the most appropriate diagnosis. Educational Objective: Brief psychotic disorder is characterized by at least one acute psychotic symptom (eg, delusions, hallucinations, disorganized speech, disorganized behavior) lasting less than one month. %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. An 83-year-old woman is brought to the physician by her daughter to discuss the results of a complete dementia work-up. The patient has had mild memory impairment for 8 months. She takes no medications. Vital signs are normal. Her Mini-Mental State Examination score is 23/30. A rapid plasma reagin is 1:4, and a microhemagglutination assay for Treponema pallidum is positive. Which of the following is the best next step for the physician? A) Discussion of the diagnosis with the daughter privately B) Discussion of the diagnosis with the patient privately C) Disregarding the results since the patient is too old for treatment D) Repeated tests E) Lumbar puncture

B. This patient's syphilis diagnosis should be discussed with her privately. All patients, including those with dementia, should be assumed to possess decisional capacity to make medical decisions unless the physician determines the patient does not possess decisional capacity. This physician should first discuss this sensitive diagnosis with the patient privately and assess her decisional capacity surrounding the diagnosis. Patients who are deemed to lack decisional capacity should still be informed of the diagnosis, which may give the patient a partial understanding of their situation and relieve potential distress about the diagnostic uncertainty. The next-of-kin or designated medical decision maker of a patient without decisional capacity should also be informed after the diagnosis is discussed with the patient. Neurosyphilis refers to the spread of syphilis to the central nervous system. Neurosyphilis can manifest as meningitis, a progressive dementia syndrome known as general paresis, or disease of the posterior columns of the spinal cord known as tabes dorsalis. Diagnosis involves confirming the diagnosis of syphilis with both treponemal and nontreponemal (eg, rapid plasma reagin and VDRL) serum testing followed by examining the cerebrospinal fluid (CSF) and performing CSF treponemal or nontreponemal tests. Treatment of neurosyphilis is crucial to prevent neurologic progression and requires the administration of penicillin G. Incorrect Answers: A, C, D, and E. Rather than discuss this sensitive diagnosis with the patient's daughter (Choice A), the physician should honor this patient's autonomy by first discussing the diagnosis with the patient. After this discussion, if the physician believes that the patient lacks decisional capacity and the daughter is the next-of-kin or a designated medical decision-maker, then the physician should discuss the diagnosis and management plan with the daughter. If the patient possesses decisional capacity and the daughter wishes to be informed of her mother's medical situation, the physician should ask the patient's permission to share the diagnosis with her daughter or encourage the patient to share the diagnosis with her daughter herself. There is no age limit for syphilis treatment (Choice C). This patient with neurosyphilis should be treated with antibiotics to prevent progression of the neurologic manifestations. Repeated tests are unnecessary (Choices D). Since both a treponemal and nontreponemal serum test were positive, this patient has a confirmed syphilis diagnosis. Though cerebrospinal fluid testing from a lumbar puncture confirms the diagnosis of neurosyphilis (Choice E), the physician first needs to discuss the syphilis diagnosis with the patient and determine the patient's decisional capacity. The physician can then obtain the patient's (or next-of-kin's) informed consent for the lumbar puncture. Educational Objective: Neurosyphilis, or the spread of syphilis to the central nervous system, may present with a progressive dementia. All patients, including those with dementia, should be assumed to possess decisional capacity to make medical decisions unless the physician determines the patient does not possess decisional capacity. Physicians should therefore discuss new diagnoses, such as syphilis, with patients privately before discussing the diagnosis with family members or proceeding with further management steps. %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. The local health department is investigating an outbreak of acute hepatitis A in two city districts by reviewing the medical records at a health center in each district. The investigators plan to analyze the clinical and epidemiologic characteristics of patients who tested positive for hepatitis A serum IgM antibody compared with characteristics of those who tested negative for the antibody. Which of the following best describes this study design? A) Case series B) Case-control study C) Prospective study D) Randomized trial E) Retrospective cohort

B. This study compares one group of patients with an outcome under study (seropositive for Hepatitis A IgM antibody) (cases) against a second matched group without that outcome (controls) and identifies the associated exposure within each group. This study design is known as a case-control study. Case-control studies can be conducted in a prospective or retrospective manner but are always observational studies. By grouping patients by outcome and comparing differences in the odds of exposure, case-control studies can detect associations between exposure and outcome, such as exposure to certain risk factors and outcome of contracting hepatitis A as in this study. This is described statistically as an odds ratio (OR). Two unrelated variables will have an OR of 1.0, whereas positive association between an exposure and an outcome will have an OR greater than 1.0 and negatively associated variables will have an OR less than 1.0. Case-control studies are therefore capable of establishing association between exposure and outcome, but they do not establish causality. Incorrect Answers: A, C, D, and E. A case series study (Choice A) is a descriptive study that describes the history, possible exposures, and clinical findings of a group of patients with a similar diagnosis. Case series are non-analytic studies. They do not test a hypothesis and do not generally contain a control group. A prospective study (Choice C) is one in which one group of patients experiences an intervention or exposure and an associated control group does not. The two groups are followed and the desired outcome tracked. A randomized trial (Choice D) is a stringent type of prospective study wherein patients are randomly assigned to receive a particular intervention. The intervention may be compared against placebo therapy or against standard therapy, depending on the study design. When combined with blinding procedures, prospective, randomized control studies form the standard for medical research and generate strong evidence as compared to observational or descriptive studies. Retrospective cohort studies (Choice E) examine a group of patients with a known exposure and determine the risk for developing a given disease outcome within the exposed group compared to a control group lacking the known exposure. Cohort studies may be prospective or retrospective. Cohort studies differ from case-control studies in that patients are grouped according to exposure status in the former design and are grouped according to disease outcome status in the latter design. Cohort studies calculate relative risk (RR). Educational Objective: Case-control studies group patients according to disease outcome status and analyze the odds of exposure to a particular hazard. The appropriate statistical measure of case-control studies is the odds ratio (OR). %3D Previous Next Score Report Lab Values Calculator Help Pause

68 Exam Section 2: Item 18 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 18. A 27-year-old woman delivers monozygotic twins at 34 weeks' gestation. The larger twin has a hematocrit of 68%; the smaller twin is pale and has a hematocrit of 25%. Which of the following is the most likely explanation for these findings? A) Amniotic fluid leak across intervening membranes B) Artery-to-artery chorionic surface anastomoses C) Chronic abruptio placentae D) Funisitis E) Knotting of the umbilical cords F) Oligohydramnios

B. Twin-twin transfusion syndrome (TTTS) and twin anemia polycythemia sequence (TAPS) are complications of monochorionic twin gestation. TTTS occurs because of the formation of arteriovenous anastomoses in the chorion of the placenta that allow blood to pass from one fetus to the other. Less commonly, it can also involve the formation of artery-to-artery chorionic surface anastomoses. It typically presents on prenatal ultrasound with unequal amniotic fluid indices between the two amniotic sacs. It can also present with anemia of one fetus and polycythemia of the other fetus when chronic, which is referred to as TAPS. Monochorionic twin gestations are typically monitored with serial ultrasounds to watch for the development of these conditions, as they have a high morbidity and mortality. Ultrasound findings also include discrepancies in nuchal translucency and crown-rump length, and abnormal ductus venosus flow. Inequalities in amniotic fluid distribution are caused by relative hypovolemia of one fetus, with resultant activation of the renin-angiotensin-aldosterone system and consequent oliguria. In contrast, the hypervolemia of the other twin causes release of atrial natriuretic peptide, which results in diuresis and relative increases in the amniotic fluid index. Complications of this syndrome also include congenital anatomic abnormalities, hydrops fetalis, heart failure, and growth restriction. Options for management include laser ablation of the anastomotic vessels, amnioreduction, and/or selective fetal reduction. Incorrect Answers: A, C, D, E, and F. Amniotic fluid leak across intervening membranes (Choice A) could lead to oligohydramnios in one fetus and polyhydramnios in the other fetus if the movement of fluid was unidirectional. However, movement of amniotic fluid from one fetus to another would not cause discordant hematocrit values in the newborns. Chronic abruptio placentae (Choice C) presents with intermittent vaginal bleeding, oligohydramnios caused by placental insufficiency, and fetal growth restriction. As monozygotic twins share a placenta, chronic abruptio placentae would be expected to affect both fetuses similarly. Funisitis (Choice D) is an infection of the umbilical cord that occurs in the setting of chorioamnionitis. Chorioamnionitis is a bacterial infection of the fetal membranes that most commonly occurs with premature or prolonged rupture of membranes. Funisitis would not cause an alteration in hematocrit concentrations in the newborns. Knotting of the umbilical cord (Choice E) can lead to fetal hypoxemia caused by cord compression during fetal descent through the vaginal canal. However, knotting of the umbilical cords is relatively rare and would likely only occur in monochorionic monoamniotic twins. It would not affect hematocrit concentrations in either fetus. Oligohydramnios (Choice F) presents with a low amniotic fluid index and can result from premature rupture of membranes, postdate pregnancies, or abnormalities of the fetal genitourinary tract, as well as TTTS. Complications include intrauterine growth restriction but do not include anemia or polycythemia. Educational Objective: Twin-twin transfusion syndrome is caused by arteriovenous anastomoses in the chorion of monochorionic twins with resultant movement of red blood cells and plasma from one fetus to the other. Complications include twin anemia polycythemia sequence, discordant amniotic fluid indices, congenital anatomic abnormalities, hydrops fetalis, heart failure, and intrauterine growth restriction. %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 12-year-old girl is brought to the emergency department 15 minutes after she accidentally sliced her left palm with a knife. Physical examination shows a 2-cm laceration over the left palm. The wound is cleansed and sutured. One week later, the sutures are removed. At this time, which of the following factors is most instrumental in the migration of fibroblasts into the area of wound healing? A) Collagen B) Fibronectin C) Нераrin D) Immunoglobulin O E) Plasminogen

B. Wound healing occurs via a staged process. In early wound healing, platelet aggregation and platelet plug formation occur to achieve hemostasis. In the subsequent 1 to 7 days, neutrophils and macrophages infiltrate the area and release growth factors and cytokines that stimulate fibroblast proliferation. Fibronectin is essential for fibroblast migration by providing a pathway for migration during wound healing. Fibroblasts bind to peptide sequences within fibronectin, which guide them to the site of healing. Granulation tissue forms as collagen is deposited into the area by fibroblasts and neovascularization begins to occur. During this time, wound edges contract from the action of myofibroblasts. Epidermal cells migrate across the newly deposited collagen matrix to reconstitute normal skin appearance. In the following weeks and months, scar formation and remodeling occur via metalloproteinase-mediated collagen breakdown. Incorrect Answers: A, C, D, and E. Collagen (Choice A) is synthesized by fibroblasts and contributes to scar formation in the process of wound healing. Dermal collagen fibers increase as a result of scarring. While the majority of collagen fibers in healthy skin are type I collagen, scarring is initially created by type III collagen. Heparin (Choice C) is a common anticoagulant that potentiates the action of antithrombin III to inhibit multiple coagulation factors. It does not relate to fibroblast migration in wound healing. Immunoglobulins (Choice D) are found on B lymphocyte membranes or are secreted into the serum by plasma cells to recognize antigens and activate the immune system in response to a pathogen. Plasminogen (Choice E) is converted to plasmin, which subsequently degrades fibrin clots leading to clot dissolution. Plasminogen is made in the liver and its role is to degrade and prevent clot formation, not guide fibroblasts. Educational Objective: Fibronectin is essential for fibroblast migration and provides a pathway for migration during wound healing. Fibroblasts bind to peptide sequences within fibronectin, which guide them to the site of healing. %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 19. A 55-year-old homeless man comes to the emergency department because of a 2-month history of intermittent fever, cough productive of blood-tinged sputum, night sweats, and fatigue; he has had a 3.2-kg (7-lb) weight loss due to decreased appetite during this period. On arrival, he is cachectic and has a chronic cough. His temperature is 38°C (100.4°F). Rhonchi are heard over the right upper lobe.. A chest x-ray shows areas of cavitation in the right upper lobe. Which of the following is the most likely cause of tissue injury in this patient's condition? A) Antigen-antibody complex deposition B) Arthus reaction C) Delayed-type hypersensitivity D) Pseudomembrane formation E) Wheal and flare reaction

C. Delayed-type hypersensitivity reaction, also known as a type IV hypersensitivity reaction, most likely accounts for this patient's lung injury in the setting of pulmonary tuberculosis. Type IV hypersensitivity is characterized by a cell-mediated response which involves maturation of antigen specific CD4+ or CD8+ T lymphocytes, which in this case, are antigens specific to Mycobacterium tuberculosis (MTB). When the antigen is encountered, CD4+ T lymphocytes release cytokines leading to inflammation and macrophage activation, while CD8+ T lymphocytes directly kill cells expressing the antigen. Activated macrophages become histiocytes and form granulomas, which in the case of MTB, may be caseating granulomas. Local inflammation at the site of MTB infection is an attempt to encapsulate and/or destroy MTB, but the local immune response results in substantial damage to the surrounding lung parenchyma. Incorrect Answers: A, B, D, and E. Antigen-antibody complex deposition (Choice A), also called a Type lII hypersensitivity reaction, is a result of immune complex (antigen-antibody) deposition in tissues resulting in a local inflammatory response and complement fixation with subsequent damage to tissues. Common diseases in which this plays a role include polyarteritis nodosa, systemic lupus erythematosus, and serum sickness. Arthus reaction (Choice B) is a localized type III (antigen-antibody complex) hypersensitivity reaction whereby complexes that fix complement deposit in blood vessels and cause local necrosis. It occasionally occurs with hepatitis B or tetanus vaccines, although this is uncommon. Pseudomembrane formation (Choice D) occurs in infection with Clostridium difficile, a bacterium that causes colitis in patients with recent exposure to antibiotics. Colonoscopy classically shows pseudomembranes on the surface of colonic mucosa. Pseudomembrane formation is not a characteristic of MTB infection. Wheal and flare reaction (Choice E) indicates the presence of mast cell degranulation and typically manifests as hives. The cause may be related to medication effects, heat or cold exposure, allergies, or may be idiopathic. Educational Objective: MTB infection induces a type IV (delayed-type) hypersensitivity reaction in which MTB antigens induce maturation of CD4+ and CD8+ T lymphocytes specific to that antigen. It results in a local inflammatory response, with direct cytotoxic killing of infected cells by CD8+T lymphocytes and stimulation of macrophage activation with subsequent granuloma formation. %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 3-month-old girl is brought to the physician 2 days after the mother felt masses in the infant's groin while bathing her. Physical examination shows bilateral inguinal hernias and a shallow vagina that ends blindly. Serum studies show increased concentrations of luteinizing hormone, follicle-stimulating hormone, testosterone, and dihydrotestosterone. During repair of the hernias, gonads are found. A biopsy specimen of the gonads shows bilateral testes. The karyotype is 46,XY. Ultrasonography shows no uterus or fallopian tubes. Which of the following is the most likely cause of the findings in this patient? A) Decreased 21-hydroxylase activity B) Decreased 5a-reductase activity C) Defective androgen receptor D) Failure of testis to secrete anti-müllerian hormone E) Mutation of the sex-determining region Y gene (SRY)

C. Androgen insensitivity syndrome is caused by a defect in the androgen receptor complex resulting in a genotypic XY male to develop external female or ambiguous genitalia and female secondary sexual characteristics. Testes are present and produce testosterone normally, but absence of a functioning androgen receptor prevents hormone binding and thereby prevents the 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. Results of laboratory studies show increased concentrations of testosterone, estrogen, and luteinizing hormone. Menses will not occur because of the lack of cycled progesterone and estrogen, and the lack of a functional uterus with endometrial lining. Incorrect Answers: A, B, D, and E. Decreased 21-hydroxylase activity (Choice A) occurs in the setting of congenital adrenal hyperplasia, of which the most common form is 21-hydroxylase deficiency. Lack of this enzyme prevents production of aldosterone and cortisol, and results in excessive androgen production. Genetically female patients present with hypoaldosteronism as well as virilization during infancy, and genetically male patients present with precocious puberty in childhood. Decreased 5a-reductase activity (Choice B) results in the insufficient conversion of testosterone to dihydrotestosterone (DHT), leading to decreased concentrations of DHT and impaired virilization of the male urogenital tract. Individuals with 5a-reductase deficiency exhibit either female or ambiguous external genitalia, although demonstrate normal male internal genitalia because of normal concentrations of testosterone. Increased concentrations of DHT, as in this infant, would not be found. Failure of testis to secrete anti-müllerian hormone (Choice D) results in failure of suppression of the development of the paramesonephric (Müllerian) duct. Patients with an XY karyotype appear externally male but have undescended testes and an incompletely developed uterus. Mutation of the sex-determining region Y gene (SRY) (Choice E) would cause an XY male to develop as a phenotypic female since the non-functional gene would result in the development of ovaries from the undifferentiated gonads, not testes. This would then consequently result in the development of both normal internal and external female genitalia. Educational Objective: Androgen insensitivity syndrome occurs because of a defect in the androgen receptor, and results in an XY genotypic male developing external female or ambiguous genitalia and female secondary sexual characteristics. %3D 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 34-year-old woman comes to the physician because of a 2-week history of easy fatigability. She is pale. Physical examination shows scleral icterus and mild jaundice. Laboratory studies show: 9.2 g/dL 10% of red cells Hemoglobin Reticulocyte count Serum Total bilirubin Direct bilirubin 3.0 mg/dL 0.3 mg/dL Which of the following is most likely responsible for the development of anemia in this patient? A) Chronic disease OB) Hemoglobinopathy C) Hemolysis OD) Hypersplenism E) Ineffective erythropoiesis

C. Autoimmune hemolysis explains this patient's anemia, unconjugated hyperbilirubinemia, and reticulocytosis. Autoimmune hemolytic anemia (AIHA) occurs because of production of antibodies targeting circulating red blood cell (RBC) surface antigens, marking them for removal in the reticuloendothelial system (RES) or fixing complement leading to intravascular hemolysis. IgG antibodies are active at the physiologic temperature of the human body (called warm AIHA); these do not routinely activate complement. Instead, they mark the cells for removal. Phagocytosis of the IgG antibody along with a component of the erythrocyte membrane creates sphere-shaped RBCS seen on peripheral smear. In contrast, IgM antibodies bind and fix complement at lower temperatures and result in intravascular hemolysis (called cold AIHA). IgG antibodies are known as warm antibodies, and IgM antibodies are known as cold antibodies. Regardless of the type of antibody, laboratory findings show normo- or microcytic anemia, an increased reticulocyte count consistent with increased hematopoiesis, and unconjugated hyperbilirubinemia from hemoglobin released from lysed erythrocytes. Diagnosis is made by laboratory analysis, a compatible clinical history, and positive direct antibody test (DAT, Coombs). In the DAT, the patient's erythrocytes are washed free of plasma, and incubated with Coombs reagent, an anti-IgG and anti-complement antibody. If an autoantibody or complement is bound to the surface of the erythrocytes, the Coombs reagent will bind to it and cause agglutination. Incorrect Answers: A, B, D, and E. Chronic disease (Choice A) such as chronic infection or autoimmune disease can cause anemia. The mechanism involves increased hepcidin concentrations that decrease iron absorption and increase iron storage in the bone marrow, preventing use of iron by erythrocyte precursors, thereby resulting in anemia. As hematopoiesis is ineffective, the reticulocyte count will be decreased. Increased bilirubin is not consistent with this diagnosis. Hemoglobinopathy (Choice B) describes numerous conditions including the thalassemia and sickle cell disease among others. These conditions cause anemia because of deficient production of normal adult hemoglobin or production of abnormal hemoglobin. Reticulocytosis and increased bilirubin concentrations can be seen if hemolysis is present, but these conditions are chronic and present at birth. Hypersplenism (Choice D) can result from infection, myeloproliferative neoplasms, or cirrhosis with portal hypertension. Erythrocytes passing through an enlarged spleen may have slower transit times and may be removed prematurely from the circulation, but hypersplenism does not cause hemolysis or manifest with an increased bilirubin. Ineffective erythropoiesis (Choice E) is a feature of many diseases including iron deficiency anemia, anemia of chronic disease, and bone marrow malignancies. Patients with these conditions demonstrate a low reticulocyte count. Educational Objective: AIHA results from the binding of an autoantibody to RBC surface antigens, leading to removal via the RES or to intravascular hemolysis. Diagnosis is made by a positive DAT, unconjugated hyperbilirubinemia, and reticulocytes on the peripheral smear. %3D Previous Next Score Report Lab Values Calculator Help Pause

49 Exam Section 1: Item 49 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 49. A 1-day-old female newborn with aniridia undergoes genetic testing. Her father and several other paternal family members also have aniridia. Chromosomal microarray analysis of the patient shows no gain or loss of gene dosage throughout the genome, including the chromosome 11p15 region that contains the PAX6 commonly mutated in patients with aniridia. Routine chromosomal analysis shows a balanced translocation between chromosomes 1 and 11, with the chromosomal breakpoint on chromosome 11 being 11p15. The same balanced translocation is also found in all of the affected family members but not in the unaffected family members. However, DNA sequencing of the PAX6 located at 11p15 shows no mutation in any of the exons of PAX6 in the patient. Which of the following is the most likely cause of this patient's condition? A) Creation of a new stop codon B) Deletion of an intron C) Disruption of a regulatory sequence D) Skipping of an exon E) Transition of a base

C. Balanced translocations occur when a segment of a chromosome has switched places with another broken chromosomal segment. In this patient scenario, a segment of chromosome 1 has been relocated and joined onto chromosome 15 just as a segment of chromosome 15 has been joined onto chromosome 1 in place of the translocated segment. Frequently, chromosomal translocations carry no significant consequence unless the break point affects a regulatory sequence or section of an exon that results in reduced or absent translation of the encoded protein. In this case, chromosomal analysis showed that the exons of PAX6, a highly conserved gene that encodes the iris, demonstrated no mutations. Therefore, the translocation must affect a promoter, enhancer, silencer, or similar regulatory sequence that prevents appropriate expression of the gene itself. Aniridia is an autosomal dominant mutation; homozygosity at this gene locus is often severe and fatal. This patient inherited one copy of the abnormal chromosome from her father and a normal copy from the mother, leading to the observed phenotype. Regulatory sequences in DNA include promoters, the sites bound by RNA polymerase and transcription factors, enhancers, sites bound by transcription factors that generally upregulate transcription, and silencers, regions bound by repressor proteins. Incorrect Answers: A, B, D, and E. Creation of a new stop codon (Choice A) would result in cessation of ribosomal translation of the protein prior to completion. A stop codon mutation would likely be detected in the chromosomal analysis of the patient's PAX6 gene, which was reportedly found to be without mutation. Deletion of an intron (Choice B) would generally have no effect on the phenotype, as introns are non-translated intervening regions of DNA between exons that are spliced out during pre-messenger RNA processing. Skipping of an exon (Choice D) would result in the synthesis of an incomplete or misfolded protein, lacking peptides encoded by the skipped exon. In the case of a translocation, from the point of translocation and downstream, the translated protein would differ entirely from the protein produced as an alternate gene would be translated instead based on the regulatory elements governing the transposed segment. Transition of a base (Choice E) defines a type of point mutation where a like-kind mutation occurs within a nucleotide (eg, purine for purine, pyrimidine for pyrimidine). Point mutations can be silent (the new nucleotide codes for the same base), missense (the nucleotide codes for an alternate and sometimes chemically different base), or nonsense (the point mutation generates an early stop codon). Transition mutations occur within a DNA sequence. Educational Objective: Translocation may result in failure to transcribe RNA because of disruption of a regulatory element (eg, promoter) within a gene. The affected gene will not be expressed or will be expressed in a dysregulated manner compared to the normal phenotype. %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. A 57-year-old man dies 5 days after a stroke. He had progressively severe hypertension over the past 2 years but refused treatment. Blood pressure just prior to the stroke was 220/110 mm Hg. Which of the following is the most likely histologic finding in his kidneys at autopsy? A) Amyloidosis B) Chronic pyelonephritis C) Hyperplastic arteriolitis D) Nodular glomerulosclerosis E) Renal papillary necrosis

C. Chronic hypertension is associated with vascular changes in the small arteries and arterioles called arteriosclerosis, marked by thickening of the vessel walls and loss of elasticity. There are two primary types, which are hyaline arteriosclerosis and hyperplastic arteriosclerosis (also called hyperplastic arteriolitis). Hyaline arteriosclerosis is characterized by protein deposition in the vessel walls. Hyperplastic arteriosclerosis is characterized by concentric thickening of the vessel wall with a laminar appearance caused by the proliferation of smooth muscle cells and is associated with severe, chronic hypertension, as in this patient. Incorrect Answers: A, B, D, and E. Amyloidosis (Choice A) is an infiltrative disorder caused by the deposition of abnormal proteins in tissue. The kidneys are commonly involved, with abnormal protein deposits in the mesangium that display an apple-green birefringence with Congo red dye on histologic analysis. Renal impairment and nephrotic syndrome may develop. It is not caused by chronic hypertension. Chronic pyelonephritis (Choice B) may develop from recurrent infections of the genitourinary tract with reflux into the renal pelvis, which may occur in the setting of obstructive uropathy, nephrolithiasis, and vesicoureteral reflux. The kidneys display atrophy, calyceal deformities, and fibrosis of the renal parenchyma. Nodular glomerulosclerosis (Choice D) is associated with diabetic nephropathy and amyloidosis. It may progress to nephrotic syndrome. Renal papillary necrosis (Choice E) may result from severe ischemic injury to the kidney. Risk factors include sickle cell disease, obstructive nephropathy, nonsteroidal anti-inflammatory analgesic use, diabetes mellitus, and severe pyelonephritis. Educational Objective: Chronic hypertension can result in arteriosclerosis, which commonly affects the renal arterioles. Histologic changes associated with hyperplastic arteriosclerosis include laminar thickening of the vessel wall caused by smooth muscle proliferation. %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 2-month-old girl is brought to the physician for a follow-up examination 2 days after a complete blood count was found to be abnormal. Her blood group is A, Rh-positive. She was born at term to a healthy 23-year-old woman whose blood group is O, Rh- positive. Birth weight was 3890 g (8 lb 9 oz). Pregnancy and delivery were uncomplicated. The patient has been active and feeding wellI. Her temperature is 37.6°C (99.7°F), pulse is 120/min, and respirations are 30/min. Physical examination shows no abnormalities. Laboratory studies done since birth are shown. Patient Age 1 Week 15.7 49 24 Hours 16.9 1 Month Hemoglobin (g/dL) (N=11-15) Hematocrit (%) (N=28-45) Leukocyte count (/mm3) (N=5000-19,500) Segmented neutrophils Eosinophils Basophils Lymphocytes Monocytes Platelet count (/mm3) (N=150,000-400,000) 12.7 39 51 17,200 4% 4% 15,600 3% 4% 12,900 5% 2% 1% 77% 4% 69% 19% 2% 73% 18% 198,000 15% 185,000 205,000 Which of the following is the most likely diagnosis? A) Alloimmune hemolytic disease of the newborn B) Congenital cytomegalovirus infection C) Congenital neutropenia D) DiGeorge syndrome E) Severe combined immunodeficiency

C. Congenital neutropenia refers to the presence of neutropenia at or around the time of birth and is the most likely diagnosis in this case. Neutropenia is defined by an absolute neutrophil count (ANC) of less than 1500/mm3, although most patients with congenital neutropenia have an ANC less than 1000/mm3, at which point the risk for bacterial infection increases. The underlying cause most commonly includes mutations that affect the production of myeloid cells. Bone marrow biopsy, if performed, may show decreased cellularity with cells that are arrested at the promyelocyte or myelocyte stage. Many patients will remain largely asymptomatic. In more severe cases with lower neutrophil counts, patients may develop stomatitis and recurrent mucocutaneous ulcerations, often with abdominal pain from gastrointestinal ulceration. Neutropenia is found in many conditions present at birth including Chédiak-Higashi syndrome, X-linked agammaglobulinemia, and von Gierke disease, and should be distinguished from these by mutational analysis and evaluation for concomitant physical findings. Incorrect Answers: A, B, D, and E. Alloimmune hemolytic disease of the newborn (Choice A) occurs in patients who are Rh-positive and are born to mothers who are Rh-negative but were previously sensitized either by a prior birth or blood transfusion to make anti-Rh antibodies. Both the patient and her mother are Rh-positive, making this diagnosis unlikely. Congenital cytomegalovirus infection (Choice B) causes hearing loss, seizures, a petechial rash, and intracranial calcifications in the neonate when acquired in utero. Hepatomegaly, splenomegaly, and lymphadenopathy are also commonly seen. The virus is transmitted via the placenta from the mother, who often contracts the virus after interacting closely with young children. DiGeorge syndrome (Choice D) 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. Severe combined immunodeficiency (Choice E) can result from numerous mutations but commonly presents with recurrent bacterial, viral, and fungal infections caused by severe impairment in both B- and T-lymphocyte function. Educational Objective: Congenital neutropenia is defined as neutropenia that is present at or around the time of birth, is not related to a secondary cause, and persists. In cases of mild neutropenia, symptoms are less common, although patients may eventually develop stomatitis and gastrointestinal ulceration. In severe cases, there is a substantially heightened risk for invasive bacterial infection. %3D Previous Next Score Report Lab Values Calculator Help Pause

66 Exam Section 2: Item 16 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 16. The high therapeutic index displayed by acyclovir and its derivatives in treating herpesvirus infections is due to the fact that the drug must be activated by a viral enzyme. Which of the following enzymes is most likely being described? A) DNA ligase O B) Guanylyltransferase C) Thymidine kinase D) Uracil DNA glycosylase E) Viral protease

C. Herpetic infections are treated with drugs that inhibit viral DNA polymerase, classically by guanosine analogs such as acyclovir, valacyclovir, and famciclovir. Prior to exerting their antiviral effects, guanosine analogs must be phosphorylated by the viral enzyme thymidine kinase. Phosphorylation is the process by which a phosphate group is added to the molecule. These molecules 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 Ēpstein-Barr virus, and not effective against cytomegalovirus. Development of a mutation in the viral thymidine kinase enzyme would prevent drug phosphorylation and confer resistance to guanosine analog medications. Incorrect Answers: A, B, D, and E. DNA ligase (Choice A) is an enzyme related to DNA replication and modification. It joins Okazaki fragments together by catalyzing the synthesis of the phosphodiester bond within double-stranded DNA. It does not interact with guanosine analogs. Guanylyltransferase (Choice B) is an enzyme which transfers a guanosine monophosphate group to another molecule, usually pyrophosphate, in the process of MRNA formation. While the class of medications used to treat herpetic infections are guanosine analogs, they are activated by phosphorylation, not via transfer of or to a larger group. Uracil DNA glycosylase (Choice D) is a component of the base excision repair pathway and removes uracil groups from DNA, which form from deamination of cytosine residues. It is not involved in the activation of guanosine analogs. Viral proteases (Choice E) cleave the initial polypeptides produced by the translation of viral RNA into smaller, functional parts. Protease inhibitors are used in the treatment of HIV and include darunavir, indinavir, ritonavir, and saquinavir. Educational Objective: Acyclovir, famciclovir, and valacyclovir are guanosine analogs that inhibit viral DNA polymerase and are used to treat herpes simplex virus infections. Guanosine analogs require phosphorylation by thymidine kinase to be activated. %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 36-year-old man with a 4-year history of HIV infection comes to the physician for a follow-up examination. Current medications include lopinavir-ritonavir and lamivudine (3TC)-zidovudine (AZT). His CD4+ T-lymphocyte count is 410/mm3 (N2500); 6 months ago, it was 720/mm3 Reverse transcriptase polymerase chain reaction test results show that his remaining CD4+ T lymphocytes are positive for CCR5. Based on the expression of CCR5, which of the following cell types in this patient is most likely also infected with HIV? A) B lymphocytes B) Eosinophils C) Macrophages D) Mast cells E) Natural killer cells

C. Macrophages are the other cell type in addition to CD4+ T lymphocytes that are most likely also infected with HIV as they possess CD4 and CCR5 on their cell surface. To enter the cell, HIV must first bind to CD4 via the HIV gp120 and gp41 subunits. This binding induces a conformational change in the HIV envelope allowing for binding to a coreceptor such as CCR5 or CXCR4. CCR5 is expressed on the surfaces of antigen presenting cells such as macrophages and dendritic cells, in addition to T lymphocytes. Following this, the HIV envelope fuses with the cellular membrane and expels its contents into the host cell. The HIV RNA genome is subsequently reverse transcribed into DNA and inserted into the host cell genome via HIV integrase, followed by transcription of HIV genes and translation into HIV proteins. These proteins are packaged and cleaved to form the mature HIV virus. Following infection with HIV, macrophages may also may serve to perpetuate infection for several reasons: they are present in high numbers at sites of viral entry, are resistant to the cytotoxic effects of HIV infection allowing for HIV persistence and replication, and have a remarkably long life span with the ability to survive for months to years in peripheral tissues. They also distribute widely throughout the body, including the central nervous system. Incorrect Answers: A, B, D, and E. B lymphocytes (Choice A), eosinophils (Choice B), mast cells (Choice D), and natural killer cells (Choice E) do not express both CD4 and CCR5 on their surface, thus HIV is unable to enter and infect these cell types. Educational Objective: HIV binds to CD4 and a coreceptor such as CCR5 to enter CD4+ T lymphocytes. Macrophages also possess CD4 and CCR5, which allows for HIV viral entry. Infection can persist given the long half-life of macrophages, their resistance to the cytotoxic effects of HIV, and their wide dissemination within tissues. %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. A 67-year-old white woman undergoes punch biopsy of an urticarial lesion on her back. When the biopsy specimen container is received in the laboratory, it is determined that it contains two specimens instead of one. Review of the additional specimens received from the physician's office indicates that specimens from two separate patients were inadvertently shipped in the same container. The other specimen is from the back of a dark-skinned, 58-year-old Nigerian woman. An increase in which of the following in the specimen from the Nigerian patient is most likely to differentiate these two specimens? A) Number of melanocytes B) Number of melanophages C) Number of melanosomes D) Size of melanocytes

C. Melanin is a pigment made within a specialized cellular organelle called a melanosome found in melanocytes, which are dendritic cell derivatives that reside in the basal layer of the epidermis. Melanocytes are derived from neural crest cells, which originate at the dorsal neural tube. During embryologic development, they migrate from dorsal to ventral, and then to the epidermis. The initial substrate in the creation of melanin is the amino acid tyrosine. After producing melanin from tyrosine, the melanosomes are transferred to the surrounding keratinocytes. The variation in skin tones seen in different ethnic groups is based on the number of the melanosomes and the distribution of melanin within them in the keratinocytes, not the number of melanocytes themselves. In patients with darker skin, the melanosomes are more numerous and more densely packed with melanin. In patients with lighter skin, the melanin is distributed with less density, and there are fewer melanosomes. Melanin production is regulated in part by melanocyte-stimulating hormone (MSH). MSH is a byproduct of proopiomelanocortin (POMC) which also produces adrenocorticotropic hormone (ACTH). In adrenal insufficiency, stimulation of ACTH production by POMC simultaneously produces MSH, leading to diffuse hyperpigmentation in affected patients. Incorrect Answers: A, B, and D. Neither the number of melanocytes (Choice A) nor their size (Choice D) differs between patients with different skin types. Increased number, size, and nuclear/cytoplasmic ratio of melanocytes would be seen in melanoma, a neoplasm of melanocytes. Lesions concerning for melanoma are characterized clinically by asymmetry, irregular-appearing borders, variable coloration, a diameter greater than 6 mm, and rapid evolution in characteristics. Melanophages are macrophages which have taken up melanin and reside in the dermis. Increased number of melanophages (Choice B) would not distinguish two patients with different skin types from each other. Melanophages are not routinely present in healthy skin; rather, they develop when an inflammatory process has destroyed the basal layer of the epidermis, causing melanocytes and keratinocytes to release melanin. Educational Objective: Melanin is a pigment made by melanocytes within a specialized cellular organelle called a melanosome. The number of melanosomes and density of melanin are what provide variation in skin tone. %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. A 14-year-old boy is brought to the physician by his parents because of increasingly severe left knee pain during the past week. The pain is exacerbated by running or kneeling. He plays multiple sports and is currently in the middle of soccer season. There has been no recent trauma to the area or related sports injuries. Physical examination shows full range of motion of both lower extremities. There is a discrete area of swelling just below the left patella over the proximal portion of the tibia. Palpation of the area produces pain. Pain is also reproduced when he does a full squat. Examination of the right knee shows no abnormalities. Which of the following is the most likely cause of this patient's condition? A) Chondromalacia B) Ligamental tear C) Osgood-Schlatter disease D) Osteochondritis E) Stress fracture

C. Osgood-Schlatter disease refers to osteochondrosis or traction apophysitis of the tibial tubercle that typically occurs in adolescent, athletic children. Repetitive tension via the patellar tendon transmits to the tibial tubercle, presenting as pain reported during activities that increase the stress on the tubercle, such as kneeling, squatting, kicking, or similar activities that increase the extensor force transmitted by the quadriceps. Patients localize pain to the anterior aspect of the proximal tibia and knee. Physical examination typically discloses tenderness over an enlarged tibial tubercle. The condition is diagnosed clinically; x-rays, if obtained, may show increased lucency in the area of the tibial tubercle. Treatment is through rest, cryotherapy, and non-steroidal anti-inflammatory medications, as the condition resolves with time and unloading. Incorrect Answers: A, B, D, and E. Chondromalacia (Choice A) as related to the knee refers to deterioration of cartilage along the posterior aspect of the patella. It is common in young athletes, especially runners, and presents with knee pain that is worse with bending. It can be differentiated from Osgood-Schlatter disease by the location of pain, whereas chondromalacia will not present with pain at the tibial tuberosity. Ligamental tear (Choice B) is a broad term that in the context of the knee could describe damage to any of the collateral ligaments, such as the anterior or posterior cruciate or the medial or lateral collateral ligaments. Tear or rupture of these ligaments presents with pain which is worse when placing the affected ligament under strain, tenderness in the area of the injury, and laxity when evoking the motion restricted by the ligament itself. Osteochondritis (Choice D) describes the inflammation of bone or cartilage within a joint; it may also include the dissection of the cartilaginous layer away from adjacent bone. It is a cause of pain within, not adjacent to, a joint. Stress fracture (Choice E) is a repetitive use injury that results in microtrauma and small discontinuities in bone. Stress fractures are common in the lower extremities but would be less likely than Osgood-Schlatter disease to present with focal pain or tenderness over the tibial tubercle. Educational Objective: Apophysitis of the tibial tubercle, known as Osgood-Schlatter disease, is a common cause of knee pain at the insertion of the patellar tendon in young athletes. It presents with focal pain and tenderness that is worse with the application of tension to the tibial tubercle. %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. A transgenic animal containing a targeted mutation in the gene coding for macrophage colony-stimulating factor is prepared. Which of the following generalized skeletal abnormalities is expected in this animal? A) Achondroplasia B) Osteomalacia C) Osteopetrosis D) Osteoporosis E) Skeletal agenesis

C. Osteopetrosis is a rare, inherited condition in which an imbalance in the function of osteoclasts and osteoblasts arises, leading to excess bone mineral deposition with resulting dense, abnormally mineralized bones. Consequences of imbalanced bone density in osteopetrosis include early mineralization and fusion of the physis with stunted growth, cranial nerve impingement with resultant blindness, deafness, or loss of cranial nerve function, bone pain, craniosynostosis, scoliosis, hypocalcemia, anemia, and immunodeficiency. In contrast, conditions of low bone density such as osteoporosis generally present with fragility fractures. There are many genetic causes of osteopetrosis, with the final common pathway being the failure of the osteoclast to appropriately resorb bone. Genetic mutations may include channelopathies (CLCN7), receptor activator of nuclear factor kappa-B ligand (RANKL), the RANK receptor, or carbonic anhydrase (CÁ2) deficiencies. The net effect of any such defect is impaired resorption by the osteoclast. Osteoclasts originate from the monocyte-macrophage lineage (a hematopoietic line); therefore, mutations that inactivate macrophage colony-stimulating factor would prevent differentiation of these cells. RANKL is also required for their differentiation and activation. By contrast, osteoblasts arise from mesenchymal stem cells. Osteoclasts are large, multinucleated cells found in pits at the surface of bone, where they produce matrix metalloproteases, cathepsins, and hydrogen ions, which dissolve bone matrix. Incorrect Answers: A, B, D, and E. Achondroplasia (Choice A) results in impaired growth of long bones caused by an autosomal dominant mutation in the fibroblast growth factor receptor 3 (FGFR3) gene. Affected patients have short extremities and a torso of normal length. Intelligence is generally unaffected, but patients may experience sleep apnea caused by craniofacial obstructive effects. Homozygous dominance is generally fatal by infancy, at the latest. Osteomalacia (Choice B) is a condition of imbalanced bone mineral deposition that results in inappropriately soft bones, generally arising from dietary vitamin D deficiency or impaired calcium resorption. Osteoporosis (Choice D) defines a state of low bone mineral density and is common among postmenopausal women. The etiology is multifactorial with genetic predisposition and environmental influences leading to the phenotype. Skeletal agenesis (Choice E) describes a heterogeneous group of conditions in which all or part of a bone fails to develop. This may be primary, as in fibular hemimelia, or related to multiple malformations, as seen in caudal regression syndrome. Educational Objective: Osteopetrosis describes a disease of excessive mineralization of bone, leading to abnormally dense bone. It results from failure of osteoclastic resorption. The osteoclast arises from the monocyte-macrophage lineage, and defects in signaling of macrophage colony-stimulating factor may play a role in the pathogenesis. %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 40-year-old man is evaluated because his skin is extremely sensitive to sunlight. Exposure to the sun causes the formation of vesicles and blisters on the skin, which frequently take weeks to heal. He is diagnosed with a disorder caused by the increased synthesis of compounds in the skin that are subject to excitation by visible light. Which of the following biochemical pathways is most likely defective in this patient? A) Bile acid synthesis B) Bilirubin degradation C) Heme synthesis D) Melanin synthesis E) Riboflavin metabolism

C. Porphyria cutanea tarda is characterized by severe cutaneous photosensitivity with blistering and hyperpigmentation after exposure to sunlight and is the most common of the porphyrias. It is caused by decreased activity of uroporphyrinogen decarboxylase, an enzyme used to in the production of heme. The initial substrates for heme are glycine and succinyl-CoA. Their conversion to heme begins in the mitochondria with a rate limiting step catalyzed by aminolevulinate synthase. A series of additional steps then occurs which take place in the cellular cytoplasm. In one of these intermediate steps, uroporphyrinogen decarboxylase catalyzes the conversion of uroporphyrinogen III to coproporphyrinogen III. Not only will a deficiency in uroporphyrinogen decarboxylase prevent correct heme synthesis, it will also cause uroporphyrinogen III to accumulate. Accumulated uroporphyrinogen IIl is then deposited in the skin. Upon exposure to light of wavelength 400nm, the molecule enters an excited state and releases photons which in turn create reactive oxygen species within the skin. These reactive oxygen species damage the basement membrane, lipids, and proteins nearby resulting in dermoepidermal separation and blister formation. Incorrect Answers: A, B, D, and E. Bile acid synthesis (Choice A) uses cholesterol as the initial substrate. Cholesterol 7a-hydroxylase is the rate-limiting step in the creation of bile acids. Porphyrins are not intermediate byproducts of this pathway. Impaired bile acid production will lead to decreased intestinal absorption of lipids and fat-soluble vitamins, but not photosensitivity and blister formation. Bilirubin forms as a result of the degradation of heme. Bilirubin degradation (Choice B) is the process by which bilirubin is first conjugated in the liver to become water soluble and then excreted into the gut via bile. It is then degraded by gut bacteria to urobilinogen. Urobilinogen is either converted to stercobilin and excreted in the feces or is reabsorbed through the small intestine. If reabsorbed, 10% is excreted by the kidneys as urobilin and 90% progresses to the liver where it is recycled. No enzyme deficiency in this pathway causes porphyrins to accumulate. Melanin synthesis (Choice D) takes place in melanocytes, located in the basal layer of the epidermis. Tyrosine is the initial substrate of melanin and tyrosinase catalyzes the rate limiting step. In patients with impaired melanin synthesis, such as in oculocutaneous albinism, photosensitivity is severe and blistering sunburns may form. However, this is not because compounds are being excited by visible light, but rather because there is no melanin to protect from the harmful effects of ultraviolet radiation on keratinocytes. Impaired riboflavin, or vitamin B, metabolism (Choice E) does not generate porphyrins. Metabolism of riboflavin creates FAD and FMN, two cofactors used in reduction-oxidation reactions such as the citric acid cycle. Riboflavin deficiency presents with lip and oral commissure fissuring (cheilosis) and corneal vascularization. Educational Objective: Porphyria cutanea tarda, the most common of the porphyrias, is characterized by severe cutaneous photosensitivity with blistering and hyperpigmentation after exposure to sunlight. It is caused by a decreased activity of uroporphyrinogen decarboxylase, an enzyme used in the production of heme. %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. An investigator is studying the adverse effects of a proteasome inhibitor, bortezomib, on immune responses. Bortezomib is administered to a group of patients with relapsed multiple myeloma, and immune responses are observed. Which of the following immunologic processes is primarily affected by this drug? A) Activation of the complement cascade B) Activation of perforin C) Presentation of antigens to CD8+ T lymphocytes D) Secretion of histamine by mast cells E) Secretion of interleukin-1 (IL-1) by macrophages

C. Presentation of antigens to CD8+ T lymphocytes would be affected by the use of the proteasome inhibitor bortezomib. Proteasomes are large proteases with numerous domains that are present in both the cytoplasm and the nucleus. They are made of a and B subunits and have an extensive role in cell cycle regulation, but also in immune function. The caspase, chymotrypsin, and trypsin-like protease capabilities of this enzyme complex are also critical to their role as antigen processing centers; intracellular peptides are processed in the proteasome for presentation on class I major histocompatibility complex (MHC). Antigen-bound class I MHC molecules are recognized by cytotoxic CD8+ T lymphocytes, so proteasome inhibitors blunt the response of CD8+ T lymphocytes by altering antigen presentation on the surface of MHC I molecules. Incorrect Answers: A, B, D, and E. Activation of the complement cascade (Choice A) is not a feature of proteasome inhibitors. The complement cascade is activated by the classical, alternative, or lectin pathways. Activation leads to opsonization of invading microbes, direct microbial killing via formation of the membrane attack complex (MAC), and proinflammatory signaling (eg, C5a mediation of neutrophil chemotaxis). Activation of perforin (Choice B) occurs as a result of activated natural killer cells, which are immune cells that primarily respond to an absence of MHC I molecules on the surface of cells. Perforins, along with granzyme, are proapoptotic agents and induce cellular death. Proteasome inhibitors interfere with antigen presentation on MHC I molecules but do not affect the presence of MHČ I molecules. Secretion of histamine by mast cells (Choice D) occurs as a result of mast cell degranulation, a process that primarily occurs during allergic reactions but can also occur in the setting of mast cell activation syndrome. This is not affected by proteasome inhibitors. Secretion of interleukin-1 (IL-1) by macrophages (Choice E) has numerous effects, including the promotion of fever, vasodilation, and stimulation of endothelial cells to express adhesion proteins for leukocyte recruitment. Secretion is not affected by proteasome inhibitors. Educational Objective: Proteasome inhibitors prevent antigen processing and presentation on class I MHC molecules, thereby blunting CD8+ cytotoxic T lymphocyte activation. While this is an effective strategy to treat multiple myeloma, it also makes patients more prone to infections, especially viral infections. %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. The patient whose cardiac function is illustrated most likely has which of the following? Normal A) Arteriovenous malformation B) Cardiac tamponade C) Congestive heart failure Patient D) Cor pulmonale E) Restrictive cardiomyopathy Left ventricular end-diastolic volume

C. The Frank-Starling mechanism describes the phenomena by which cardiac output is dependent on the amount of cardiomyocyte fiber stretch prior to contraction, as represented by the left ventricular end-diastolic volume. A greater pre-contraction stretch results in a greater force of contraction (to a point), and the relationship is demonstrated by Frank-Starling curves. A given Frank-Starling curve applies for constant afterload and inotropy. Changes in afterload and/or inotropy shift the curve up or down. This patient has a Frank-Starling curve that is shifted down, indicating that for a given preload, there is reduced cardiac output relative to normal. This may occur in decreased inotropic states such as congestive heart failure, with the administration of negative inotropes, or in the setting of increased afterload. The curve shifts up in positive inotropic states and/or with decreased afterload. Incorrect Answers: A, B, D, and E. Arteriovenous malformation (Choice A) results in low-resistance, high-volume flow of blood from the arterial to the venous system with greatly increased venous return. The increase in preload causes a greater distension in the cardiomyocyte fibers at the end of diastole, which results in increased cardiac output per the Frank-Starling relationship. Cardiac tamponade (Choice B) result in decreased ventricular filling because of compression of the heart by fluid in the pericardium. In the absence of other factors affecting afterload or cardiac contractility, the Frank-Starling curve would not be depressed. Cor pulmonale (Choice D) describes right ventricular failure resulting from chronic pulmonary hypertension. Left ventricular contractility and afterload are not affected, and the Frank-Starling curve for the left ventricle would not shift. Restrictive cardiomyopathy (Choice E) results in decreased compliance of the ventricular wall, with diastolic dysfunction and preserved left ventricular systolic function. Educational Objective: The Frank-Starling mechanism describes the relationship between cardiac output and left ventricular end-diastolic volume (preload). Cardiac output can be described by a family of curves that shift depending on afterload and cardiac contractility. %3D Previous Next Score Report Lab Values Calculator Help Pause Cardiac output

35 Exam Section 1: Item 35 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 35. A 22-year-old primigravid woman at 20 weeks' gestation comes to the physician because of a 4-day history of dull pelvic pain that radiates to the right side of her labia majora. She says, "It feels like something is pulling." She is concerned that she is going into labor. Physical examination shows no dilation of the cervix. Stretching of which of the following ligaments is the most likely cause of this patient's pain? A) Broad ligament B) Mesosalpinx C) Round ligament D) Suspensory ligament of the ovary E) Uterosacral ligament

C. The round ligament of the uterus is a fibromuscular band of tissue that extends from the superolateral part of the uterus where the fallopian tubes insert, passes through the inguinal canal, and ends in the labia majora. It is a remnant of the gubernaculum, an embryonic structure which assists in the descent of the gonads. In pregnancy, as the uterus increases in size, the round ligament becomes stretched. Increased concentration of the hormone relaxin during pregnancy also contributes to its stretching. Relaxin allows for the ligaments of the pelvis and hip girdle to stretch so the birth canal can accommodate the fetus during delivery. The combination of increased elasticity and increased size of the uterus strains the round ligament and commonly leads to pain during pregnancy. Sudden movements or vigorous activity may precipitate this pain. There is no association of round ligament pain with preterm labor, but other causes of abdominal pain during pregnancy should be ruled out before attributing pain to the round ligament. No treatment is necessary, as the pain is self-limited and typically mild. Incorrect Answers: A, B, D, and E. The uterus is supported by the broad ligament (Choice A) which attaches to the superior aspect of the uterus and is a layer of peritoneum. The mesometrium, mesosalpinx (Choice B), and mesovarium are named portions of the broad ligament. While these structures are all contiguous, the mesometrium covers the uterus and creates the majority of the broad ligament, the mesosalpinx covers the fallopian tubes, and the mesovarium covers the ovaries. Stretching of this ligament during pregnancy does not cause labial pain. The suspensory ligament of the ovary (Choice D) extends laterally from the ovary to the wall of the pelvis and runs adjacent to the ovarian artery and vein. Its size does not significantly change in pregnancy. The uterosacral ligament (Choice E) holds the uterus in place in the pelvis. It is a paired structure which connects the uterus to the sacrum at the level of the cervix. They may also be called the rectouterine ligaments or sacrocervical ligaments. It is not the cause of this patient's pain. Educational Objective: The round ligament of the uterus is a fibromuscular band of tissue that extends from the superolateral part of the uterus where the fallopian tubes insert, passes through the inguinal canal, and ends in the labia majora. Stretching of the round ligament commonly leads to pain during pregnancy. %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 20-year-old woman comes to the office because of a 1-month history of a mass in her neck. She has a 16-year history of cerebellar ataxia and progressive scleral telangiectatic lesions. Her brother has a similar condition. Physical examination shows a large, firm, nonmoveable lymph node palpable in the left anterior side of the neck. A biopsy specimen of the node shows evidence of Hodgkin lymphoma. Which of the following genetic mechanisms is the most likely cause of this patient's lymphoma? A) Activation of a tumor suppressor gene B) Deletion of the BCL gene C) DNA repair defect D) Initiation of somatic hypermutation

C. This patient with ataxia telangiectasia possesses a genetic mutation of the ATM gene leading to a DNA repair defect. The ATM gene product delays the progression of cells with DNA damage through the cell cycle, which allows the cell to repair the DNA. When the ATM gene product is defective, somatic mutations accumulate, leading to the dysfunction of diverse proteins and an increased risk for malignancy. Patients present with cerebellar atrophy and progressive ataxia, telangiectatic lesions of the eyes and exposed areas of the skin, and other movement abnormalities such as dystonia, abnormal eye movements (eg, saccades), and decreased immunoglobulin concentrations (IgA, IgG, and IgE). Conditions associated with ataxia telangiectasia include pulmonary disease (eg, bronchiectasis from recurrent sinopulmonary infections) and hematologic malignancies such as Hodgkin lymphoma. Ataxia telangiectasia demonstrates an autosomal recessive inheritance pattern. Management requires the surveillance and treatment of disease complications. Incorrect Answers: A, B, and D. Activation of a tumor suppressor gene (Choice A) such as P53 would arrest the cell cycle. Activating tumor suppressor genes would decrease cellular proliferation and would not result in malignancy. Deletion of the BCL gene (Choice B) may result in hematologic deficiencies such as anemia. The BCL gene is an oncogene that normally inhibits apoptosis and may promote malignancy (eg, follicular lymphoma) when activated. Initiation of somatic hypermutation (Choice D) occurs during normal immunoglobulin class switching. Somatic hypermutation is not directly associated with malignancy. Educational Objective: Ataxia telangiectasia is an autosomal recessive disorder resulting from a DNA repair defect. Consequently, somatic mutations accumulate, leading to protein dysfunction and an increased susceptibility to hematologic malignancy such as Hodgkin lymphoma. Patients typically present with progressive ataxia, telangiectasias, abnormal eye movements, and decreased immunoglobulin concentrations (leading to recurrent sinopulmonary infections). %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. Human gene fragments in appropriate vectors may be introduced into bacterial cells by transformation. Which of the following is most likely to be used to transfer cloned human genetic material to bacteria? A) DNA complexed with bacterial histones O B) Human chromosomal fragments C) Naked DNA D) Purified euchromatin E) Purified heterochromatin

C. Transformation describes the ability of bacteria to uptake and incorporate exogenous genetic material (naked DNA) from the environment. Following lysis of a cell DNA may exist extracellularly; a bacterium may uptake such material through the cell membrane. Transformation is one of three methods of horizontal gene transfer by which bacteria can acquire novel genes which may confer a survival benefit. Notably, Streptococcus pneumoniae and Haemophilus influenzae type B have demonstrated acquisition of genetic material by transformation. In order for transformation to occur, a bacterium must demonstrate competence (the ability to uptake such material). Competence occurs in states of stress such as starvation. Beyond transformation, horizontal gene transfer may also occur via conjugation (in which two cells in direct contact, typically involving a sex pilus), exchange material, and transduction (transfer of genetic material via a viral vector such as a bacteriophage). Incorrect Answers: A, B, D, and E. DNA complexed with bacterial histones (Choice A), human chromosomal fragments (Choice B), purified euchromatin (Choice D), and purified heterochromatin (Choice E) all describe fragments of DNA associated with additional protein sizes. The process of transformation requires naked DNA, which describes DNA that is not associated with proteins, lipids, or molecules that may shield it. exes and of variable Educational Objective: Transformation describes the ability of bacteria to uptake and incorporate genetic material (naked DNA) that is not associated with proteins or lipids from the environment. This mechanism may promote genetic diversity and acquisition of survival advantage. %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 28-year-old woman has had restlessness, exercise intolerance, palpitations, diarrhea, and excessive sweating for the past 3 days. She had an upper respiratory tract infection 3 weeks ago. The thyroid gland is diffusely enlarged and tender. Serum total thyroxine (T) and total triiodothyronine (T3) concentrations and free T, index are increased. Serum thyroid-stimulating hormone concentration and radioactive iodine uptake by the thyroid gland are both decreased. Which of the following is the most likely cause of her condition? A) Autoimmune (Hashimoto) thyroiditis B) Diffuse toxic goiter (Graves disease) C) Euthyroid sick syndrome D) Subacute thyroiditis E) Toxic adenoma F) Toxic multinodular goiter

D. This patient's presenting findings are highly suggestive of subacute granulomatous thyroiditis, also known as de Quervain thyroiditis. Subacute granulomatous thyroiditis is a self-limited inflammatory condition of the thyroid gland that often follows an acute viral illness. Presenting findings are typically suggestive of clinical hyperthyroidism and may include diaphoresis, palpitations, neck pain, dysphagia, fever, tachycardia, increased serum concentrations of T3 and T4 and a decreased serum concentration of thyroid- stimulating hormone. Patients may eventually become hypothyroid or euthyroid following the acute inflammatory, hyperthyroid phase. Subacute granulomatous thyroiditis is distinguished from other forms of hyperthyroidism by the presence of a painful, tender thyroid, which is highly suggestive of the diagnosis. It is also characterized by the absence of iodine uptake on a radionuclide scan, and by the presence of multinucleated giant cells on fine needle aspiration. Incorrect Answers: A, B, C, E, and F. Autoimmune (Hashimoto) thyroiditis (Choice A) 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. While patients in the acute stage of the disease may present with symptoms of hyperthyroidism, the thyroid is not typically painful or tender. 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. The thyroid is not typically painful or tender. Euthyroid sick syndrome (Choice C) occurs in the setting of acute systemic illness and is characterized by decreased concentrations of thyroid hormone and a euthyroid clinical appearance. Euthyroid sick syndrome does not present with an enlarged or tender thyroid. Toxic adenomas (Choice E) are a type of hyperfunctioning thyroid follicular adenoma that secrete abnormally high concentrations of thyroid hormone and present with symptoms of hyperthyroidism. They typically present as a solitary nodule, which avidly takes up iodine on radionuclide scintigraphy, and are not typically tender. Toxic multinodular goiter (Choice F) is a common cause of hyperthyroidism and is secondary to thyroid hormone-secreting nodules that function independent of thyroid-stimulating hormone. Patients present with symptoms of hyperthyroidism, but the thyroid is not typically painful or tender. Educational Objective: Subacute granulomatous thyroiditis, also known as de Quervain thyroiditis, is a self-limited thyroiditis that often follows an acute viral illness and that presents with symptoms of hyperthyroidism. A painful, tender thyroid is highly suggestive of the diagnosis. %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 31. The table reports survival of patients who had an operation for a particular form of cancer: Number of Patients at the Number of Patients Who Percent of Patients Surviving This Interval 80 Interval Beginning of the Interval Died During the Interval 500 0-1 year 1-2 years 400 2-3 years 350 3-4 years 315 100 50 35 31 87.5 90 90 Which of the following is the probability of a patient surviving 4 years? A) (.875 - .80) x 90 x .90 B).875 C) .90 D) .80 x .875 x 90 x .90 E) (.875 - .80) x .90

D. A survival analysis is used to convey the fraction of a group of patients who are alive at a given time point after an intervention has been made. Graphically, this is often done with a Kaplan-Meier survival curve, the data for which is shown in the table. For this particular cancer the time interval of 0 to 1 year is accompanied by a 20% death rate, or 80% survival rate. This means that if 100 patients were given the same diagnosis and treated with the same operation, 80 of them would be living at one year. After surviving to 1 year, there is then a likelihood of 87.5% that they would survive to 2 years, and so on. This analysis allows prognostication for a patient newly diagnosed with this cancer when that patient's outcome is yet unknown. The probability that a patient with a new diagnosis who undergoes treatment survives until 4 years is the combined probability of surviving until 1 year, then 2 years, then 3 years, and finally 4 years. This is represented by 0.80 x 0.875 × 0.90 x 0.90. Incorrect Answers: A, B, C, and E. Calculating the 4-year survival rate does not require calculating the difference between two probabilities. Neither (.875 - 80) x .90 x .90 (Choice A) nor (.875 - .80) x .90 (Choice E) will correctly calculate the combined probability of survival through all four years. .875 (Choice B) is the survival rate of patients between the 1st and 2nd year after diagnosis. It does not take into account the probability of the first, third or fourth years. Likewise, .90 (Choice C) is the survival rate of patients between either the 2 nd and 3rd year or 3rd and 4th year after diagnosis. The probability of surviving from diagnosis to the end of the fourth year is the product of the probabilities of surviving each year. Educational Objective: Kaplan-Meier estimates are used to provide prognostic data to patients with new diagnoses. The survival rate is the percentage of patients who survive a given time interval. The probability that a patient with a new diagnosis who undergoes treatment survives until 4 years is the combined probability of surviving until 1 year, then 2 years, then 3 years, and finally 4 years. %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 20-year-old man is brought to the physician because of a 4-hour history of abdominal pain, nausea, and vomiting. He says that he had been drinking ethanol heavily all weekend, and he took three doses of acetaminophen within 2 hours after the onset of a severe headache Monday morning. This patient is at increased risk for liver injury because of which of the following actions of ethanol? A) Activation of IgE-mediated mast cell degranulation B) Decreased acetaminophen clearance via glucuronidation C) Increased bioavailability of acetaminophen D) Induction of cytochrome P450 enzymes that activate acetaminophen to a hepatotoxic metabolite E) Metabolic acidosis due to an increased NADH:NAD+ ratio

D. Acetaminophen poisoning is more common in patients with underlying risk factors, including exposure to alcohol and hepatotoxic drugs. When taken at therapeutic doses, acetaminophen is safely metabolized through phase II conjugations, including glucuronidation and sulfation. In the setting of acetaminophen overdose, saturation of phase II metabolic pathways leads to excess acetaminophen metabolized by CYP-mediated reactions to N-acetyl-p-benzoquinoneimine (NAPQI), which has strong oxidizing properties and is directly hepatotoxic. Chronic alcohol use increases the risk for hepatotoxicity in patients who take high doses of acetaminophen. Ethanol leads to the induction of P450 enzymes that result in the increased production of NAPQI when high doses of acetaminophen are ingested. The antioxidant molecule glutathione conjugates NAPQI, allowing it to be safely excreted, and depletion of glutathione is a hallmark of acetaminophen toxicity. Acetaminophen toxicity is treated by repleting hepatic stores of glutathione through intravenous or oral administration of N-acetylcysteine. Incorrect Answers: A, B, C, and E. Activation of IgE-mediated mast cell degranulation (Choice A) is the mechanism of action of type I hypersensitivity reactions and underlies many allergic reactions. Alcohol does not increase hepatic sensitivity to acetaminophen through an allergic mechanism. Decreased acetaminophen clearance via glucuronidation (Choice B) does not occur as a result of alcohol exposure. Rather than affecting the capacity of the liver to perform glucuronidation reactions, alcohol increases the capacity of the liver to produce NAPQI through the induction of P450 enzymes. Increased bioavailability of acetaminophen (Choice C) does not occur. Acetaminophen has high oral bioavailability, which is not altered by the ingestion of alcohol. Metabolic acidosis due to an increased NADH:NAD+ ratio (Choice E) may occur as a result of alcohol ingestion but does not impact the metabolism of acetaminophen. The metabolism of ethanol to acetaldehyde by alcohol dehydrogenase converts NAD+ to NADH, leading to decreased NAD+ and increased NADH. Because of the LeChatelier principle, the excess NADH is responsible for an increased activity of numerous metabolic pathways that utilize NADH as a cofactor, including the production of lactic acid and the inhibition of allosteric enzymes within glycolysis and the tricarboxylic acid cycle. However, decreased NAD+ and increased NADH do not play a direct role in acetaminophen-mediated hepatotoxicity. Educational Objective: Alcohol leads to the induction of P450 enzymes that convert acetaminophen to N-acetyl-p-benzoquinoneimine (NAPQI), a toxic metabolite that causes hepatotoxicity in the setting of acetaminophen overdose. %3D Previous Next Score Report Lab Values Calculator Help Pause

23 Exam Section 1: Item 23 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 23. A 35-year-old man comes to the physician because of a 3-year history of an enlarging nose, coarsening of his facial features, muscle weakness, and increased hand and foot size. Physical examination shows a large fleshy nose and prognathism. Fasting serum studies show increased insulin-like growth factor-l, glucose, and triglyceride concentrations. Increased serum growth hormone concentrations do not decrease after the administration of oral glucose. Compared with a healthy patient, which of the following best describes the metabolic changes in this patient? Muscle Glucose Adipose Lipolysis Неpatic Gluconeogenesis Uptake O A) ↑ ↑ ↑ B) ↑ ↑ C) ↑ D) ↑ ↑ E) ↑ F)

D. Acromegaly is caused by a growth hormone (GH) secreting pituitary adenoma that stimulates the excessive production of insulin-like growth factor-1 (IGF-1) by the liver. IGF-1 interacts with its receptor, a tyrosine kinase-based receptor, that stimulates cell growth, proliferation, and growth of the axial and appendicular skeleton. Acromegaly occurs after the closure of growth plates, and in this context excessive IGF-1 leads to expansion of flat bones and tissues. Excessive GH and IGF-1 leads to deranged glucose homeostasis by increasing peripheral insulin resistance, impairing muscle and adipose uptake of glucose, increasing adipose lipolysis, and increasing hepatic gluconeogenesis. Consequently, up to half of patients with acromegaly develop diabetes mellitus. Patients with acromegaly also exhibit increased rates of hypertriglyceridemia. Incorrect Answers: A, B, C, E, and F. Choices A, B, and C reflect states of increased muscle glucose uptake, which would occur in the setting of increased peripheral insulin concentrations or receptor sensitivity. Excessive concentrations of GH and IGF-1 lead to reduced peripheral insulin sensitivity. Choices B, C, and F reflect states of decreased lipolysis, whereas deficient insulin signaling will lead to increased lipolysis. Choices C, E, and F reflect states of decreased hepatic gluconeogenesis. GH directly stimulates gluconeogenesis, which is increased in acromegaly. Educational Objective: Acromegaly is associated with excessive GH and IGF-1 concentrations that lead to deranged glucose homeostasis by increasing peripheral insulin resistance, impairing muscle and adipose uptake of glucose, and increasing lipolysis and hepatic gluconeogenesis. Patients with acromegaly experience increased rates of diabetes mellitus and hypertriglyceridemia. %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 45-year-old woman comes to the physician because of a 2-month history of fatigue, nausea, and generalized bone pain. She has had a 2-kg (4.4-lb) weight loss during this period because of loss of appetite. She has a 4-month history of renal insufficiency. Her blood pressure is 170/100 mm Hg. Physical examination shows pallor and 2+ pitting edema of the feet. Laboratory studies show anemia. Her serum urea nitrogen concentration is 55 mg/dL, and serum creatinine concentration is 4 mg/dL. Bone x-rays show widened osteoid seams and subperiosteal erosions. Which of the following additional sets of serum findings is most likely in this patient? 1,25- Dihydroxycholecalciferol Calcium Phosphate Parathyroid Hormone O A) ↑ ↑ ↑ B) ↑ ↑ ↑ C) ↑ D) ↑ E) ↑ F) G)

D. Chronic renal disease is a cause of secondary hyperparathyroidism. Secondary hyperparathyroidism typically presents with increased parathyroid hormone (PTH), hypocalcemia, and hyperphosphatemia. It results from chronic renal failure caused by the inability of the kidney to excrete phosphate, reabsorb calcium, and produce active vitamin D, as the final conversion of 25-hydroxycholecalciferol to active 1,25-dihydroxycholecalciferol occurs in the kidney via 1-a hydroxylase. Hypocalcemia is exacerbated by the decreased intestinal absorption of calcium in secondary to the deficiency of active vitamin D. Hyperphosphatemia and hypocalcemia result in the upregulation of PTH, which increases bony turnover to raise serum calcium and promotes renal excretion of phosphate. Unregulated, this process results in renal osteodystrophy, the breakdown of bone caused by excessive PTH stimulus of osteoclasts. This is clinically manifest with widened osteoid seams and subperiosteal erosions, as seen in this patient. Incorrect Answers: A, B, C, E, F, and G. While parathyroid hormone and phosphate are increased in chronic kidney disease (Choice A), calcium and 1,25-dihydroxycholecalciferol are decreased, not increased. Active vitamin D is decreased because the kidney parenchyma is the site of its activation and is lost in chronic disease. Hypocalcemia, not hypercalcemia, occurs as a result of the combination of decreased calcium reabsorption, lack of activation of vitamin D, and decreased phosphate excretion. In chronic renal disease, secondary hyperparathyroidism, not hypoparathyroidism (Choices B, C, E, and G) occurs. This is in response to the hypocalcemia that is perpetuated by decreased calcium reabsorption, lack of activation of vitamin D, and decreased phosphate excretion. Parathyroid hormone is increased, calcium is decreased, and active vitamin D is decreased in chronic renal disease, but phosphate is increased, not decreased (Choice F). Patients with kidney failure typically develop an increased serum phosphate because of impaired excretion of phosphate from tubular dysfunction, which does not necessarily relate to increased intraluminal phosphate in the gastrointestinal tract. Educational Objective: Secondary hyperparathyroidism as a result of chronic kidney disease typically presents with increased PTH, hypocalcemia, and hyperphosphatemia. Serum concentration of 1,25-dihydroxycholecalciferol is also decreased. Renal osteodystrophy may ensue. %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 man who is a firefighter comes to the physician because of a 3-month history of persistent cough and repeated lung infections. He has smoked 2 packs of cigarettes daily for 30 years. His respirations are 16/min. Physical examination shows dry cough and decreased lung sounds. A chest x-ray shows no abnormalities. Histologic examination of this patient's respiratory epithelium is most likely to show which of the following types of metaplasia? A) Columnar B) Connective tissue C) Neuroendocrine D) Squamous E) Transitional

D. Chronic tobacco use is associated with numerous detrimental changes to the respiratory system. Tobacco smoke directly disrupts the ability of the airway to clear foreign material by resulting in increased mucus production and secretion by the respiratory epithelium, decreased activity of airway cilia, and inhibition of alveolar macrophage function. Chronic irritation from tobacco exposure also results in squamous metaplasia of the respiratory epithelium, which is normally primarily composed of pseudostratified columnar cells in the trachea and bronchi, and cuboidal cells in the bronchioles. Šquamous metaplasia is a preneoplastic lesion that has the potential to transform into squamous cell carcinoma, which is the second most common type of primary lung cancer following adenocarcinoma. Features associated with squamous cell carcinoma of the lung include pulmonary cavitations, central location, and hypercalcemia caused by paraneoplastic parathyroid hormone-related peptide (PTHRP) production. Incorrect Answers: A, B, C, and E. Columnar (Choice A) metaplasia occurs with Barrett esophagus because of chronic chemical irritation from gastric contents. The normal squamous epithelium of the esophagus transforms into columnar epithelium and is associated with an increased risk for esophageal adenocarcinoma. Connective tissue (Choice B) metaplasia refers to abnormal formation of mesenchymal cells (bone, cartilage, adipose) in tissue. An example is myositis ossificans, the formation of bone in muscle tissue that may occur following trauma. Neuroendocrine (Choice C) tumors of the lung, such as small cell lung cancer, result from hyperplasia and malignant transformation of normal neuroendocrine cells residing in the lung, rather than metaplasia in response to an irritant. Transitional (Choice E) epithelium is located in the genitourinary tract and undergoes metaplasia to squamous epithelium in response to injury, which may be a precursor to bladder cancer. Educational Objective: Metaplasia is the replacement of one differentiated cell type by another in response to normal maturation signals or abnormal irritants. The presence of metaplasia can be a risk factor for developing tissue-specific neoplasms. Metaplasia of squamous cells in the respiratory epithelium is a risk factor for squamous cell carcinoma of the lung. %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. A physician prescribes a recently marketed drug for 20 patients. After several months, five patients develop increased serum AST and ALT activities and show clinical symptoms of hepatotoxicity. The physician discontinues the drug for all patients and reports the adverse effects to the FDA. The physician has participated in which of the following phases of clinical drug testing? A) Phase 1 B) Phase 2 C) Phase 3 D) Phase 4

D. Clinical trials are an important aspect of the development of any new pharmaceutical or intervention. They involve the experimental study on human subjects and compare a new treatment against a placebo or an alternative treatment. The FDA regulates trials for newly developed drugs, and a series of phases must be undertaken to demonstrate safety, efficacy, comparability to standard of care, and long-term adverse effects. In this example a drug that has been recently marketed is prescribed to 20 patients, 25% of whom show signs of hepatotoxicity after several months. The action of reporting this to the FDA describes the physician participating in Phase 4 of the drug trial, which is characterized by post-distribution and post-prescription surveillance following treatment approval. Long-term adverse effects and rare adverse effects or outcomes are often identified this way, and failure of a drug during Phase 4 can result in discontinuation/withdrawal of the drug from the markets and loss of approval by the FDA. Incorrect Answers: A, B, and C. Phase 1 (Choice A) is the initial phase of a clinical trial. In this phase, a small number of volunteers receive the treatment, and data is collected to gauge safety and toxicity, adverse effects, interactions, pharmacodynamics, and pharmacokinetics. The volunteers in this stage are healthy. Failure at this stage limits further progress of the trial, as the intervention is deemed unsafe for human use. Phase 2 (Choice B) involves the trial of the intervention on patients who have the disease to be treated. As in Phase 1, a small number of volunteers are involved in this phase. Efficacy is assessed in this phase along with dosing. Continued study of safety and adverse effects occurs. Phase 3 (Choice C) involves the conduction of a larger, randomized controlled trial once safety and efficacy have been established as in Phases 1 and 2. In this phase, patients are assigned, ideally via a randomized, blinded process, to the treatment or placebo (or standard of care) arm of a trial. In this manner, whether the new intervention is equivalent to, better than, or worse than the control is assessed. Approval at this stage can lead to marketing of the drug and commercial use. Educational Objective: Clinical trials occur in phases. Phase 1 assesses safety in healthy volunteers. Phase 2 assesses efficacy in a small number of sick volunteers. Phase 3 is generally a large, randomized trial comparing the new intervention to placebo or the previous standard, and Phase 4 describes post-marketing surveillance for long-term and rare effects. %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. A 43-year-old man comes to the physician because of a 1-month history of chronic sinus congestion, occasional bleeding from the nose, and intermittent shortness of breath with exertion. He has had no fever or chills but has felt very "run down." His temperature is 37.4°C (99.3°F), pulse is 88/min, respirations are 16/min, and blood pressure is 160/100 mm Hg. Physical examination shows bilateral maxillary and frontal sinus tenderness to palpation and poor transillumination. Isolated crackles are heard in the left mid lung. Cardiac examination shows no abnormalities. Laboratory studies show: 9.6 g/dL 28% Hemoglobin Hematocrit Serum Na+ K+ CI- 132 mEq/L 5 mEq/L 100 mEq/L 21 mEq/L 48 mg/dL 2.4 mg/dL negative positive negative HCO,- Urea nitrogen Creatinine Antinuclear antibody Cytoplasmic antineutrophil cytoplasmic antibody (c-ANCA) Anti-glomerular basement membrane antibody Urine pH Specific gravity Blood Protein 6.4 1.015 3+ 1+ A chest x-ray shows an infiltrate in the lower segment of the left upper lobe of the lung with normal cardiac silhouette and no significant mediastinal adenopathy. A CT scan of the sinuses shows opacities bilaterally. Which of the following patterns of inflammatory vascular injury most likely underlies the pathophysiology of this patient's disease? A) Eosinophilic vasculitis B) Leukocytoclastic vasculitis O C) Lymphomatoid granulomatosis D) Necrotizing granulomatous vasculitis E) Temporal arteritis

D. Granulomatosis with polyangiitis is a systemic necrotizing granulomatous vasculitis that affects small and medium vessels. It is associated with c-ANCA and primarily affects the sinopulmonary tract, kidneys, and skin. Patients typically present with constitutional symptoms such as fever, weight loss, and fatigue, with additional symptoms based on specific organ involvement. Sinopulmonary manifestations include recurrent sinusitis, otitis media, mastoiditis, nasal ulcerations with epistaxis, hemoptysis, cough, and dyspnea. The kidneys are typically involved, and patients often present with hematuria with an increased risk for developing rapidly progressive glomerulonephritis and renal failure. Skin lesions are common, especially lower extremity purpura. Incorrect Answers: A, B, C, and E. Eosinophilic vasculitis (Choice A) is associated with allergic granulomatous angiitis (Churg-Strauss syndrome), which is a small vessel vasculitis that may present with asthma, nasal inflammation, and lung, gastrointestinal, cardiac, or kidney involvement. Pathology shows a granulomatous, necrotizing vasculitis and eosinophilia. Leukocytoclastic vasculitis (Choice B) refers to small vessel vasculitis that primarily involves the skin. It affects women more than men, and typically presents with a painful, burning rash as well as systemic symptoms of fever, weight loss, fatigue, and myalgias. Palpable purpura is commonly noted on physical examination. Lymphomatoid granulomatosis (Choice C) is a rare lymphoproliferative disorder that most commonly affects the lungs. Patients may also develop a patchy rash and subcutaneous nodules. In this patient, sinopulmonary involvement is more suggestive of granulomatosis with polyangiitis. Temporal arteritis (Choice E) is an autoimmune vasculitis that affects large and medium sized arteries. It is more common in women. Patients classically present with constitutional symptoms, jaw claudication, and headache. Complications include the permanent loss of vision. Educational Objective: Vasculitis may be categorized by whether small, medium, or large vessels are affected. Granulomatosis with polyangiitis is an ANCA-positive necrotizing granulomatous vasculitis of small and medium vessels that is characterized by sinopulmonary, renal, and skin involvement. %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 25-year-old man comes to the physician with his wife because they have been unable to conceive for 2 years. Previous examination of the wife showed no abnormalities. Examination of the patient shows no abnormalities. The patient's laboratory studies show autoantibodies to sperm. Dysfunction of which of the following cell types is the most likely cause of these findings? A) Leydig cells B) Primary spermatocytes C) Secondary spermatocytes D) Sertoli cells E) Type A spermatogonia

D. In normal male anatomy, the Sertoli cells lining the seminiferous tubules create a blood-testis barrier through tight junctions which allows the seminiferous tubule to exist as an immune-protected site. The spermatozoa developing within the seminiferous tubule contain genetic material that may be foreign to the body as a result of recombination during meiosis. Thus, when this barrier is intact, it prevents potential novel antigens from leaving the seminiferous tubule and encountering the host immune system. If the barrier is disrupted, as may occur in testicular trauma, the spermatozoa can elicit an immune response that may include production of autoantibodies against the gametes. Sperm autoantibodies are a theoretical cause of male infertility. Other causes of male infertility include Klinefelter syndrome (XXY karyotype causing dysgenesis of the seminiferous tubules and testicular atrophy), increased testicular temperature (seen in cryptorchidism or varicocele), or gonadotropin-releasing hormone (GnRH) deficiency. Incorrect Answers: A, B, C, and E. Spermatogenesis takes place in the seminiferous tubules of the testes. Pulsatile GNRH produced by the hypothalamus causes the pituitary to secrete follicle stimulating hormone (FSH) and luteinizing hormone (LH) which drive spermatogenesis. LH acts on the Leydig cells (Choice A), located adjacent to but outside of the seminiferous tubules, and causes them to secrete testosterone, which is necessary to create the proper hormonal environment within the seminiferous tubule required for spermatogenesis. Dysfunction of Leydig cells would not damage the blood-testis barrier leading to antibody formation. Lining the seminiferous tubules are spermatogonia, which generate the primary spermatocytes (Choice B) that then mature into secondary spermatocytes (Choice C) and will ultimately become spermatozoa. So long as the blood-testis barrier is intact, a defect in either primary or secondary spermatocytes will not induce antibody formation. Spermatogonia lining the seminiferous tubules are either type A (Choice E) or type B. Type A spermatogonia undergo active mitosis and divide to produce type B spermatogonia. The type B spermatogonia progress to become primary spermatocytes. The type A spermatogonia do not mature, themselves, but continue to divide. Educational Objective: Without the benefit of the blood-testis barrier, the gametes may elicit an immune reaction and production of autoantibodies. This is a theoretical cause of male infertility. Other causes of male infertility include Klinefelter syndrome, increased testicular temperature, and gonadotropin-releasing hormone deficiency. %3D Previous Next Score Report Lab Values Calculator Help Pause

95 Exam Section 2: Item 45 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 45. A 54-year-old woman comes to the physician because of a 10-day history of episodes of severe stabbing pain of her left cheek and jaw. She says that the pain occurs suddenly and can be precipitated by chewing, speaking, or brushing her teeth; it lasts several seconds and occurs 20 to 30 times daily. She adds, "It feels like lightning is striking my cheek." Touching the left cheek reproduces her symptoms. Which of the following is the most likely cause of this patient's symptoms? A) Compression of the trigeminal nerve by an acoustic neuroma (vestibular schwannoma) B) Inflammation of the temporomandibular joint C) Inflammation of the trigeminal nerve by a multiple sclerosis plaque D) Microvascular compression of the trigeminal nerve E) Recurrent transient ischemia in the vertebrobasilar system

D. Trigeminal neuralgia is a neuropathic pain disorder that presents with severe shooting or burning pain within the innervated regions of the sensory branches of cranial nerve V (the trigeminal nerve). Cranial nerve V divides into three sensory branches, V, to V3, which innervate the forehead (V,), the skin overlying the zygomatic and maxillary bone (V2), and the skin overlying the mandible (V). The pathophysiology of the pain is believed to be related to compression of the trigeminal nerve by adjacent blood vessels or bone as it exits the skull base. Symptoms may be triggered by light touch, shaving, swallowing, or may occur without any precipitant. The pain is generally described as severe. Treatment includes carbamazepine or surgical decompression. Incorrect Answers: A, B, C, and E. Compression of the trigeminal nerve by an acoustic neuroma (vestibular schwannoma) (Choice A) would be unlikely, as acoustic neuromas generally occur in association with cranial nerve VIII and generally cause sensorineural deafness, vertigo, and disequilibrium. Pain would be an unusual presentation, as would a lesion large enough to also affect cranial nerve V. Inflammation of the temporomandibular joint (Choice B) would cause pain with chewing and be specifically located in the area of the temporomandibular joint. Patients may also complain of locking of the jaw. Inflammation of the trigeminal nerve by a multiple sclerosis plaque (Choice C) is a known cause of trigeminal neuralgia, however this patient demonstrates no symptoms of multiple sclerosis such as neurologic dysfunction affecting multiple body systems over time (eg, urinary incontinence, weakness, or optic neuritis). Recurrent transient ischemia in the vertebrobasilar system (Choice E) presents with functional loss affecting the brainstem and cerebellum, such as vertigo, ataxia, nystagmus, nausea, or vomiting. Pain is not characteristic. Educational Objective: Trigeminal neuralgia is a neuropathic pain disorder that presents with severe shooting or burning pain within the innervated regions of the sensory branches of cranial nerve V (the trigeminal nerve). The pathophysiology of the pain is believed to be related to compression of the trigeminal nerve by adjacent blood vessels or bone as it exits the skull base. %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. A3-month-old girl is brought to the physician by her mother for a well-child examination. Her mother says that the infant can lift her head when lying prone. The patient is eating well and recently started sleeping about 6 uninterrupted hours through the night. The mother adds that the baby does not yet smile in response to human faces and has not started to coo or gurgle with attention. The patient is at the 25th percentile for length and weight and 10th percentile for head circumference. Physical examination shows no other abnormalities. Which of the following best describes this infant's development? Motor Social Verbal and Cognitive A) Normal normal normal B) Normal normal delayed C) Normal delayed normal D) Normal delayed delayed OE) Delayed normal normal F) Delayed normal delayed G) Delayed delayed normal H) Delayed delayed delayed

D. In the first 3 months of life, infants are expected to demonstrate the gross motor skills of holding their head up and pushing their body up with their arms when lying prone. Expected fine motor skills include grasping and shaking toys and bringing their hands to their mouth. Most 3-month-olds demonstrate a social smile. Verbally, 3-month-old infants are expected to coo, babble, and imitate vowel sounds. This patient does not meet social or verbal/cognitive milestones. Screening for developmental delays alerts physicians to the potential need for further medical evaluation and/or early referral to developmental specialists. Early intervention has been demonstrated to improve outcomes. Incorrect Answers: A, B, C, E, F, G, and H. This patient, who does not smile or coo yet, does not meet the social or verbal/cognitive milestones for a 3-month-old infant (Choices A, B, C, E, F, G, and H). This baby should be further evaluated for causes of developmental delay. Educational Objective: By age 3 months, infants are typically able to hold their head up, grasp toys, smile socially, and babble or coo. %3D Previous Next Score Report Lab Values Calculator Help Pause

65 Exam Section 2: Item 15 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 15. A 48-year-old man comes to the physician because he is concerned about a pigmented lesion that has enlarged and changed color during the past 3 months. Physical examination shows a 2-cm oval lesion. Histologic examination of an excisional biopsy specimen shows melanoma. If a biopsy of the lesion had been done prior to the malignant transformation, which of the following would have been the most likely diagnosis? A) Acanthosis nigricans B) Actinic keratosis C) Freckle D) Nevocellular nevus 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. It is a neoplastic proliferation of melanocytes and can develop within a giant congenital nevus (nevocellular nevus). Compared to acquired nevi, congenital nevi tend to be larger, darker in color, thicker, and more likely to contain hair. These nevi should be monitored by the patient for development of any features concerning for malignant degeneration. Malignant melanoma has the ability to rapidly invade and metastasize, which carries a poor prognosis when diagnosed late. Subtypes include superficial spreading, nodular, lentigo maligna, and acral lentiginous. Any lesion with features suggestive of malignant melanoma should be surgically excised with negative margins and pathologically examined for the depth of dermal invasion. Incorrect Answers: A, B, C, 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 mellitus. It is neither premalignant 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. They present as chronic rough, scaly patches of skin in areas of prolonged sun exposure (eg, face, ears, hands). Freckle (Choice C), or ephelis, is a benign, light tan macule which darkens with sun exposure. It is caused by an increase in melanin production and is neither premalignant nor malignant. 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 because of the keratin produced by the epidermis rather than melanin. Educational Objective: Melanoma should be suspected when lesions demonstrate asymmetry, border irregularity, multiple colors, diameter greater than 6 mm, or changing features. It may arise from a preexisting congenital nevus (nevocellular nevus). %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 LDS 1 cm 26. A 65-year-old woman dies 6 months after the onset of severe headaches. Her brain as seen at autopsy is shown in the photograph. Which of the following is the most likely cell of origin of this neoplasm? A) Astrocyte O B) Endothelial OC) Melanocyte D) Meningeal E) Squamous epithelial

D. Meningiomas are the most common primary brain tumors in adults and arise from the meninges (the protective coverings of the brain). This patient's meningioma is pictured along the inferior aspect of the forebrain within the anterior interhemispheric fissure between the frontal lobes, resulting in significant mass effect and compression of the adjacent frontal lobes and olfactory bulbs. Patients may be asymptomatic, or experience seizures or focal neurologic deficits related to the tumor location. Meningiomas are typically benign but can be malignant. Malignant meningiomas may grow rapidly, leading to symptoms of increased intracranial pressure (postural headache, nausea, papilledema), or may metastasize. Meningiomas appear as well-circumscribed, extra-axial (outside of the brain parenchyma) masses that compress the parenchyma, as in this patient. Treatment may include surgical resection to alleviate the tumoral mass effect if significant. Incorrect Answers: A, B, C, and E. Astrocytes (Choice A) proliferate to form glial tumors in adults, for example, glioblastoma. These highly malignant tumors appear as intra-axial masses in the cerebral hemispheres, which typically demonstrate areas of necrosis and can spread to the opposite hemisphere across the corpus callosum. Endothelial cells (Choice B) may proliferate to form hemangioblastomas.Hemangioblastomas typically arise in the posterior fossa as intra-axial tumors. Tumors arising from melanocytes (melanomas) (Choice C) or squamous epithelial cells (squamous cell carcinomas) (Choice E) are primarily extracranial and would not compress the brain. Brain metastases from these malignancies would most likely be intra- axial rather than extra-axial. Educational Objective: Meningiomas are the most common primary brain tumors in adults. Malignant meningiomas may grow rapidly and cause symptoms of increased intracranial pressure such as headache. Meningiomas appear as well-circumscribed, extra- axial masses that compress the adjacent brain parenchyma. %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 4-year-old girl has acute lymphoblastic leukemia that has not responded to aggressive treatment. The physician recommends palliative care to the parents and suggests that they talk to their daughter about her impending death and answer any questions that she may have. The parents ask the physician to help them respond to her questions. When speaking with the child, the parents should anticipate that she will most likely understand death as which of the following? A) She will blame God for her illness and death B) She will comprehend that all living things die C) She will have no understanding of death D) She will view death as temporary and reversible

D. Most children between the ages of 3 and 5 years view death as temporary and reversible. Partly related to a lack of experience of death, they also do not understand that all living things inevitably die. Children of this age range who are dying may believe dying is a punishment from their parents, signifying an inability to abstract. However, these gaps in understanding can be filled by adults explaining death to children and by children's experiences of death. It is suggested that the parents of dying children help their child understand the situation. Incorrect Answers: A, B, and C. This patient is unlikely to blame God for her illness and death (Choice A) as it is too abstract a concept for a child this age. The patient is more likely to blame her parents. She may, depending on her parents' religious beliefs, believe that death represents going to Heaven without an abstract understanding of Heaven. Starting at age five, children begin to understand that death is inevitable and irreversible (Choice B). At this age, children may continue to demonstrate an incomplete understanding of death. Media images of death (eg, ghosts) may be part of this incomplete understanding. At age ten, children typically understand that death is universal, irreversible, and renders people inanimate (versus believing in ghosts). Only infants have no understanding of death (Choice C). Toddlers typically begin to become aware of the concept of death over time. Educational Objectives: Most children between the ages of 3 and 5 years view death as temporary and reversible. Further, preschool age children are frequently unaware of the inevitability of death. %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 3-month-old boy is brought to the physician because of yellow eyes and skin and weakness since birth. Physical examination shows jaundice, large fontanels, a flat midfacial area, hypotonia, and hepatomegaly. Serum studies show: Total bilirubin (mainly direct) AST ALT increased increased increased increased Very-long-chain fatty acids A liver biopsy specimen shows foamy, lipid-filled hepatocytes, necrosis, and absence of a specific organelle. This organelle is most likely which of the following? O A) Golgi complex B) Lysosomes C) Mitochondria D) Peroxisomes E) Smooth endoplasmic reticulum

D. Zellweger syndrome is caused by a genetic mutation that leads to an absence of peroxisomes. Peroxisomes synthesize cholesterol and bile acids, as well as cell membrane substrates. As peroxisomes are also responsible for metabolizing very long chain fatty acids (VLCFA), phytanic acid, pipecolic acid, pristanic acid, hydrogen peroxide, and ethanol, their absence leads to an accumulation of these products. Zellweger syndrome classically presents with seizures, intellectual disability, hypotonia, hepatomegaly, jaundice, kidney disease, cataracts, hearing loss, and craniofacial abnormalities. Physical examination will show an absence of reflexes, and results of laboratory studies will show increased concentrations of VLCFA, hepatic transaminases, and direct bilirubin, as observed in this patient. There is no effective treatment, and life expectancy is less than six months of age. Incorrect Answers: A, B, C, andE. The golgi complex (Choice A) sorts, modifies, and transports proteins within the cell and to the cell membrane for export. Deficiency of the golgi complex in hepatocytes would likely lead to cell death, although it would not result in an increased concentration of VLCĒAS. Lysosomes (Choice B) are responsible for the degradation of a variety of substrates, including those brought into the cell through endocytosis and materials of the cell itself (eg, autophagy). Deficiencies in lysosomal enzymes lead to lysosomal storage disorders, which present broadly with neurocognitive decline, skeletal abnormalities, and craniofacial abnormalities. Mitochondria (Choice C) are responsible for producing energy for the cell in the form of adenosine triphosphate. Disorders of mitochondrial structure or function include MELAS (mitochondrial encephalomyopathy, lactic acidosis, and stroke-like episodes) syndrome and myoclonic epilepsy with ragged red fibers. Complete absence in the hepatocytes would lead to cell death and liver failure. The smooth endoplasmic reticulum (Choice E) is responsible for synthesizing lipids and steroids, as well as detoxifying toxins. While not present in erythrocytes, their absence in hepatocytes would likely lead to cell death and liver failure, although it would not result in an increased concentration of VLČFAS. Educational Objective: Zellweger syndrome is caused by a genetic absence of peroxisomes, and presents classically with seizures, intellectual disability, hypotonia, hepatomegaly, kidney disease, cataracts, hearing loss, and craniofacial abnormalities. Results of laboratory studies will show increased VLCFAS, hepatic transaminases, and direct bilirubin. %3D Previous Next Score Report Lab Values Calculator Help Pause

82 Exam Section 2: Item 32 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 32. A 45-year-old man is admitted to the hospital because of congestive heart failure. Two first-degree relatives died in their 40s with dilated cardiomyopathy and cirrhosis. Physical examination shows cardiac enlargement and generalized hyperpigmentation. His serum glucose concentration is 320 mg/dL. Histologic examination of endomyocardial tissue is most likely to show an excess of which of the following? A) Amyloid B) a-Antitrypsin C) Cerebroside D) Copper E) Eosinophils F) Iron OG) Lysosomal glycogen H) Mucopolysaccharide

F. Iron deposition will most likely be appreciated on histologic examination of endomyocardial tissue in this patient with a probable diagnosis of hemochromatosis. Hemochromatosis may be acquired or inherited secondary to mutations in the HFE gene, leading to abnormally increased intestinal absorption of iron. This results in accumulation of iron in the body, increased serum iron, and increased ferritin. Iron can accumulate in several organs, including the liver, pancreas, skin, heart, and joints. Because of 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 gonadal atrophy. Hemochromatosis, when acquired, may occur in the setting of transfusion-dependent anemias such as thalassemia, but family history of similar symptoms makes hereditary hemochromatosis more likely in this patient. Diagnostic studies may include biopsy, which commonly shows iron deposition on Prussian blue stain. Treatment involves serial phlebotomy and management of secondary disorders such as heart failure and diabetes mellitus. Incorrect Answers: A, B, C, D, E, G, and H. Amyloid (Choice A) deposition is seen in patients with amyloidosis, a disorder in which low molecular weight proteins deposit in tissues and cause organ dysfunction. These proteins are often immunoglobulin light chains. Amyloidosis can cause restrictive cardiomyopathy. Cirrhosis may occur but bronzing of the skin and diabetes mellitus are more consistent with hemochromatosis. a-Antitrypsin (Choice B) deficiency results in liver disease and emphysema, especially in individuals who smoke. Diabetes mellitus and hyperpigmentation are less common. Cerebroside (Choice C) deposition is seen in metachromatic leukodystrophy, which is caused most commonly by deficiency of arylsulfatase A, resulting in abnormal accumulation of cerebroside sulfate in the central and peripheral nervous systems. Copper (Choice D) deposition occurs in Wilson disease. Patients present with liver dysfunction and neuropsychological impairment (eg, parkinsonism, ataxia, tremors, dystonia, dementia, hallucinations, and/or personality changes). Eosinophils (Choice E) can be found in organs in numerous disease states. One examplle is eosinophilic esophagitis, which presents commonly with food impaction. Biopsy shows eosinophils infiltrating the esophageal mucosa. Eosinophils also play a primary role in various hypereosinophilic syndromes, related to the site of eosinophil deposition including myocarditis or colitis. Lysosomal glycogen (Choice G) deposits are found in glycogen storage diseases. One example is Pompe disease, which leads to accumulation of glycogen in the lysosomes with findings of cardiomegaly, cardiomyopathy, hepatomegaly, and hypotonia. Mucopolysaccharide (Choice H) deposition is a feature of mucopolysaccharidoses. One example is mucopolysaccharidosis I (Hurler syndrome), which results from deficiency of the enzyme a-L-iduronidase, with subsequent lysosomal accumulation of glycosaminoglycans (eg, heparin sulfate and dermatan sulfate). It is an autosomal recessive disease and presents with physical features such as coarse facial features with a large, elongated face, widely spaced orbits, corneal opacification, and hepatosplenomegaly. Educational Objective: Hemochromatosis presents with liver failure, diabetes mellitus, arthritis, heart failure, darkening of the skin, and gonadal atrophy secondary to excess total body iron. Biopsy commonly shows excess iron on Prussian blue stain. Inherited hemochromatosis is the result of abnormally increased intestinal absorption of iron, while acquired forms are often the result of chronic blood transfusions. %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. The graph shows diastolic and end-systolic relationships of the left ventricle. The solid line shows the control pressure-volume loop in a single cardiac cycle. If a healthy person is given nitroprusside and reflex responses are blocked, which of the following labeled end-systolic pressure points is most likely? 200 A •B 100- E• C D 100 300 400 200 Left ventricular volume (mL) A) B) C) D) E)

D. Nitroprusside is an intravenous, titratable vasodilator that can be used for the treatment of hypertensive emergencies. It breaks down in circulation to release nitric oxide, which in turn activates guanylate cyclase in vascular smooth muscle to result in vascular smooth muscle relaxation and vasodilation via a cyclic GMP pathway. It typically induces a reflex tachycardia. Sodium nitroprusside preferentially dilates arterial vessels over venous vessels, resulting in a decrease in afterload without significant change in preload. An isolated decrease in afterload results in a shortened pressure-volume loop (less left ventricular pressure is needed to open the aortic valve) and a decreased end-systolic left ventricular volume (because of an increased stroke volume and ejection fraction). The associated end-systolic pressure point on this graph is represented by point D. Incorrect Answers: A, B, C, and E. Choice A represents a point that may be observed if afterload is increased and cardiac contractility is increased. The left ventricular pressure needed to open the aortic valve is increased, and the increased contractility maintains the stroke volume to result in the same end-systolic left ventricular volume as the control. Choice B represents the end-systolic pressure point associated with an isolated increase in afterload. The left ventricular pressure needed to open the aortic valve is increased and the stroke volume is reduced, resulting in an increased end-systolic left ventricular volume. Choice C represents an end-systolic pressure point that may be seen in the case of decreased afterload and decreased cardiac contractility, with reduced end-systolic left ventricular pressure and a reduced stroke volume resulting in increased end-systolic left ventricular volume. Choice E represents a point that may occur with an isolated increase in cardiac contractility. The afterload and end-systolic left ventricular pressure remain the same, and an increase in stroke volume results in a lower end-systolic left ventricular volume. Educational Objective: Pressure-volume loops provide information about cardiac mechanics. The shape and position of the loop is dependent on preload, afterload, and cardiac contractility. A decrease in afterload results in a shorter loop that is shifted to the left, as lower ventricular pressure is needed to open the aortic valve and a greater volume of blood is ejected. %D Previous Next Score Report Lab Values Calculator Help Pause Left ventricular pressure (mm Hg)

80 Exam Section 2: Item 30 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 30. A 12-year-old girl is brought to the physician by her mother after the dentist found many unerupted and supernumerary teeth. Physical examination shows frontal bossing, hypertelorism, and retained deciduous teeth. When prompted, the patient is able to place her shoulders together anteriorly. Serum studies show a decreased alkaline phosphatase activity. A chest x-ray shows hypoplasia of the clavicles. Skull x-rays show open sutures and multiple Wormian bones. Genetic testing shows a mutation in the CBFA1 gene. Which of the following cell types is directly affected by the mutation in this patient? A) Chondroblasts B) Chondroclasts C) Chondrocytes D) Osteoblasts E) Osteoclasts F) Osteocytes

D. Osteoblast differentiation is likely impaired in this patient with a physical presentation concerning for cleidocranial dysplasia (CCD). CCD is inherited either de novo or in an autosomal dominant manner and presents with bony dysplasia. Diminution or absence of the clavicle permits abnormal movement of the shoulders anteriorly as seen in this patient. Additional symptoms include impaired skull development, often involving the frontal or parietal bones, the presence of Wormian (intrasutural) bones, supernumerary teeth, hypoplasia of the maxilla, frontal bossing, hypertelorism, and delayed ossification of the pubic symphysis. CBFA1 (core binding factor activity 1) gene may be mutated. This gene codes for a transcription factor involved in osteoblast differentiation. Membranous bone, often found along the midline, is most commonly affected. The life expectancy of affected patients is generally normal, as is neurologic development. Incorrect Answers: A, B, C, E, and F. Chondroblasts (Choice A), chondroclasts (Choice B), and chondrocytes (Choice C) are the primary cell types involved in the synthesis (-blast) and breakdown (-clast) of cartilage. In pathology, chondroblastoma can result from the oversynthesis of cartilage from chondroblast stimulation. They are not involved in the pathophysiology of CCD, which results from impaired osteoblast differentiation as opposed to imbalanced synthesis or breakdown of cartilage. Osteoclasts (Choice E) are the primary cell type involved in the degradation of bone. Excess osteoclast activation results in loss of bone mineral density and weakened bone, as seen in multiple myeloma and

74 Exam Section 2: Item 24 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 24. Two days after admission to the hospital because of a sickle cell disease crisis, a 24-year-old man suddenly develops blood in his urine and flank pain. Laboratory studies at the time of admission showed that his hematocrit was decreased to 11% from his normal baseline of 25%. Urinalysis shows gross blood, rare WBCS, and no WBC casts. Which of the following is the most likely cause of the hematuria in this patient? OA) Glomerulonephritis B) Nephrolithiasis C) Prostatitis D) Renal papillary necrosis E) Transitional cell carcinoma

D. Renal papillary necrosis (RPN) 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 medications (NSAIDS). In sickle cell disease, renal papillary necrosis is common and occurs secondary to renal medullary ischemia and infarction. It typically presents with hematuria and acute flank pain. It can also be complicated by urinary tract obstruction and/or infection secondary to necrotic tissue sloughing into the renal collecting system and ureters. It characteristically presents with gross hematuria and proteinuria on urinalysis. Incorrect Answers: A, B, C, and E. Glomerulonephritis (Choice A) refers to a variety of glomerular diseases, including nephritic and nephrotic syndromes. Nephritic syndromes typically present with acute renal failure with associated hematuria, red blood cell urine casts, and hypertension. Nephrotic syndrome typically presents with excessive proteinuria (greater than 3g per day) hyperlipidemia, hypoalbuminemia, and edema. It would not typically cause ureteral obstruction. Nephrolithiasis (Choice B) typically presents with unilateral flank pain that is colicky and sometimes radiates to the groin with associated hematuria. In a patient with sickle cell crisis, renal papillary necrosis is a more likely cause of flank pain and hematuria. Prostatitis (Choice C) can present with dysuria, frequency, and urinary urgency similar to a urinary tract infection. On examination, a tender, boggy, enlarged prostate can be palpated and treatment is with antibiotics. Urinalysis would show WBCs and would be unlikely to indicate hematuria. Transitional cell carcinoma (Choice E) is the most common tumor of the urinary tract system and typically presents with painless gross hematuria. It is associated with smoking, aniline dyes, and cyclophosphamide. Educational Objective: Renal papillary necrosis (RPN) 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 NSAIDS. It typically presents with gross hematuria and proteinuria on urinalysis. %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 46-year-old woman comes to the physician because of a 2-year history of progressive shortness of breath while supine; she now requires two pillows to help her sleep. She also has had a cough with occasional production of rusty-colored sputum during this period. Her respirations are 16/min. Bilateral crackles and a grade 4/6 mid-diastolic murmur are heard. Echocardiography shows left atrial dilation with mitral stenosis. No other valvular lesions are noted. The photograph shown is characteristic of her disease process. Which of the following is the most likely cause of these findings? A) Congenital rubella B) Coronary artery atherosclerosis C) Illicit intravenous drug use D) Recurrent group A streptococcal infections E) Systemic sclerosis (scleroderma) with CREST syndrome

D. Rheumatic heart disease is a complication of acute rheumatic fever caused by group A streptococcal infection resulting in progressive inflammatory damage to the heart valves with fibrosis and calcification. It commonly causes insufficiency in the acute phase and stenosis later in life, predominantly affecting the mitral valve. Rheumatic fever typically presents with joint inflammation, pancarditis, subcutaneous inflammatory nodules, erythema marginatum, and Sydenham chorea. Mitral stenosis presents with a diastolic murmur radiating to the axilla. If severe enough, it can result in left atrial enlargement, cardiogenic pulmonary edema, and arrhythmias such as atrial fibrillation and flutter. Incorrect Answers: A, B, C, and E. Congenital rubella (Choice A) syndrome is caused by fetal rubella infection during pregnancy, which results in severe malformations in the fetus. It is diagnosed in the neonatal period and classically presents with bilateral hearing loss, congenital heart defects, a violaceous rash, and hepatosplenomegaly. Coronary artery atherosclerosis (Choice B) may result in chronic ischemic changes to the myocardium and risk for papillary muscle rupture in the setting of acute tissue infarction. It does not cause chronic inflammation and calcification of the mitral valve. Illicit intravenous drug use (Choice C) is a risk factor for infective endocarditis caused by the introduction of microbes into the venous circulation. The tricuspid valve is most commonly affected. Vegetations may form on the valve, and patients may present with fever, dyspnea, tricuspid regurgitation, and septic pulmonary emboli. Systemic sclerosis (scleroderma) with CREST syndrome (Choice E) is a subtype of scleroderma characterized by fibrotic skin changes of the distal extremities and face; it is associated with calcinosis cutis, Raynaud phenomenon, esophageal dysmotility, sclerodactyly, and telangiectasias. Educational Objective: Calcification of the mitral valve can result in mitral stenosis, classically heard as an opening snap, followed by a diastolic rumble loudest over the cardiac apex. Mitral stenosis commonly occurs as a long-term complication of rheumatic heart disease caused by group A streptococcal infections. II 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. A 53-year-old man comes to the physician because of a 6-day history of shortness of breath, cough, and pleuritic chest pain. His temperature is 38.9°C (102°F), and respirations are 35/min. Sputum is purulent and rust colored. Physical examination shows decreased breath sounds, and crackles are heard at the left base. A Gram stain of sputum shows gram-positive diplococci. A chest x-ray shows left-sided lobar consolidation. Which of the following anatomic structures most likely allowed rapid spread of organisms between alveoli to involve the entire left lobe in this patient? A) Alveolar capillaries B) Germinal centers C) Lymphatic capillaries D) Pores of Kohn E) Vascular sinusoids

D. The pores of Kohn form connections between alveoli and are present in normal lung tissue. They are composed, at least in part, of type Il alveolar cells and allow for the passage of air, fluid, phagocytes, and in the setting of pneumonia, bacteria between adjacent alveoli. By allowing equilibration between adjacent alveoli, the pores of Kohn aid in normal oxygenation and in the prevention of atelectasis. However, infectious organisms and inflammation may also spread between adjacent alveoli through these apertures. Incorrect Answers: A, B, C, and E. Alveolar capillaries (Choice A) surround each alveolus and are crucial for gas exchange. However, they do not provide a direct connection between adjacent alveoli, provided that the alveolar and capillary endothelia remain intact. Germinal centers (Choice B) allow for the development of B lymphocytes and are important in mounting an adaptive immune response to pulmonary pathogens. They do not form connections between alveoli. Lymphatic capillaries (Choice C) allow for the drainage of interstitial fluid from the lung parenchyma to the lymphatic ducts and eventually to the heart. These passages do not form direct anastomoses between adjacent alveoli. Vascular sinusoids (Choice E) are present in hepatic tissue and create anastomoses between the portal and systemic circulation, allowing for the exchange of nutrients with hepatocytes. They are not present in pulmonary tissue. Educational Objective: The pores of Kohn form connections between alveoli and are present in normal lung tissue. They allow for the passage of air, fluid, phagocytes, and, in the setting of pneumonia, bacteria between adjacent alveoli. %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 66-year-old man with type 2 diabetes mellitus and hypertension is brought to the emergency department 30 minutes after the sudden onset of left eyelid drooping, double vision, and mild weakness of the right hand and leg. His pulse is 88/min and regular, and blood pressure is 159/99 mm Hg. Examination of the head shows a substantially droopy left eyelid, and in primary gaze, the left eye is exotropic and somewhat lower than the right. He has slowed finger movements on the right, a pronator drift with the right hand, and mild hyperreflexia on the right. Which of the following is the most likely site and diagnosis of the patient's lesion? A) Angular gyrus (Gerstmann syndrome) B) Dorsolateral thalamus (Dejerine-Roussy syndrome) C) Lower medulla (Wallenberg syndrome) D) Lower midbrain (Weber syndrome) E) Upper spinal cord (Horner syndrome)

D. This patient likely has a lesion of the lower midbrain that affects the cerebral peduncle and the oculomotor nerve, which causes contralateral hemiparesis and ipsilateral oculomotor palsy, known as Weber syndrome. The cerebral peduncle refers to the anterior portion of the midbrain and includes the crus cerebri, which contains the corticospinal tract and is supplied by paramedian branches of the posterior cerebral artery. Upper motor neurons of the corticospinal tract originate in the primary motor cortex, descend ipsilaterally through the internal capsule and midbrain (within the crus cerebri), decussate in the caudal medulla, and then descend contralaterally in the spinal cord to synapse with the contralateral lower motor neuron. A brainstem lesion of the corticospinal tract leads to contralateral weakness in an upper motor neuron pattern of dysfunction (eg, spastic paralysis, pronator drift, and hyperreflexia). The oculomotor nerve (cranial nerve III) is located anteriorly between the two cerebral peduncles and may also be affected by strokes of the paramedian branches of the posterior cerebral artery, leading to ipsilateral dysfunction of the extraocular muscles and consequent diplopia along with ptosis and a deficit in pupillary constriction. CVAS occur because of ischemic or hemorrhagic loss of blood supply to the brain. Approximately 80-85% of CVAS are ischemic, commonly arising from thromboembolic disease (eg, middle cerebral artery occlusion from a thrombus), whereas 15-20% of CVAS are hemorrhagic and caused by blood vessel rupture (eg, hypertension-related intraparenchymal hemorrhage from a perforating artery). Risk factors for CVAS include smoking, hypertension, diabetes mellitus, carotid or intracranial atherosclerotic disease, history of hypercoagulability, atrial fibrillation, and advanced age. Classically, CVAS manifest as a neurologic deficit related to the affected part of the brain. Incorrect Answer: A, B, C, and E. Patients with angular gyrus (Choice A) damage may demonstrate agraphia, acalculia, finger agnosia, and left-right disorientation. This constellation of symptoms is known as Gerstmann syndrome. The dorsolateral thalamus (Choice B) connects with the limbic system and may mediate the emotional aspects of pain. As such, lesions of this brain area may lead to post-stroke intractable pain, known as thalamic pain syndrome or Dejerine-Roussy syndrome. Lesions of the lower medulla (Choice C) may affect the nuclei of multiple cranial nerves, the spinal trigeminal nucleus, solitary nucleus, vestibular nuclei, and nucleus ambiguus; the sympathetic, spinothalamic, and spinocerebellar tracts may also be affected. Consequently, lesions of the lower medulla typically lead to ipsilateral facial numbness, impaired taste sensation, vertigo with nystagmus, dysphonia, dysarthria, dysphagia, Horner syndrome, pain and temperature sensation deficits in the contralateral arms and legs, and/or ipsilateral ataxia. This constellation of symptoms is known as Wallenberg syndrome. Lesions of the upper spinal cord (Choice E) may interrupt the sympathetic nerve supply to the ipsilateral face and eye, which results in Horner syndrome. Horner syndrome classically presents with ipsilateral ptosis, miosis, and anhidrosis. Educational Objective: In Weber syndrome, a stroke of the paramedian branches of the posterior cerebral artery affect the cerebral peduncle, containing the corticospinal tract, and the neighboring oculomotor nerve. Patients typically present with ipsilateral oculomotor nerve palsy (diplopia, ptosis, primary gaze palsy, pupillary constriction deficits) and contralateral hemiparesis with an upper motor neuron pattern of dysfunction. %3D Previous Next Score Report Lab Values Calculator Help Pause

38 Exam Section 1: Item 38 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 38. A 29-year-old woman comes to the physician because of a 6-day history of rash, joint pain, and fatigue. Her temperature is 37.4°C (99.4°F), and blood pressure is 150/90 mm Hg. Physical examination shows a raised, erythematous, blotchy malar rash and mild peripheral edema. There is tenderness and swelling of several joints. Laboratory studies show: Leukocyte count Segmented neutrophils Eosinophils Lymphocytes Monocytes Platelet count 3200/mm3 62% 3% 29% 6% 74,000/mm3 2.8 g/dL positive positive Serum albumin Antinuclear antibody Antibody to double-stranded DNA Urine Protein RBC casts 3+ 2+ Which of the following serologic studies is most likely to produce a false-positive result in this patient? A) Cold agglutinins B) Hepatitis B surface antigen O C) Heterophile antibody D) p24 antigen E) Rapid plasma reagin

E. A rapid plasma reagin false positive result in this patient would be expected. This young woman with malar rash, arthritis, leukopenia, thrombocytopenia, positive antinuclear antibody (ANA), positive anti double-stranded DNA antibody (anti-dsDNA), and an active urine sediment meets criteria for a diagnosis of systemic lupus erythematosus (SLE). Many patients with SLE also possess one or more antiphospholipid antibodies, including anticardiolipin antibody, lupus anticoagulant, or anti-beta-2-glycoprotein-1 antibody, which predisposes to the development of venous or arterial thrombosis and recurrent pregnancy loss. Additionally, these antibodies are known to interfere with the rapid plasma reagin (RPR) test for syphilis. The RPR tests for the presence of antibodies against a cardiolipin-cholesterol-lecithin antigen, which are called reagin antibodies. In patients who possess an anti-cardiolipin antibody, the RPR will be falsely positive. Follow up tests to evaluate for infection with syphilis should involve a treponemal test such as Treponema pallidum particle agglutination assay (TPA) or fluorescent treponemal antibody absorption (FTA-ABS) test. Incorrect Answers: A, B, C, and D. Cold agglutinins (Choice A) are circulating IgM antibodies against erythrocyte antigens that, upon binding to the erythrocyte antigen, cause agglutination and extravascular hemolysis. These antibodies are active at colder temperature, so agglutination occurs primarily in the extremities. This is a rare cause of autoimmune hemolytic anemia in patients with SLE. Hepatitis B surface antigen (HBSA9) (Choice B) may persist for several weeks after hepatitis B vaccination, but a false positive HBSAg would not be expected in SLE. Heterophile antibodies (Choice C) are antibodies that react to antigens from unrelated species such as sheep or horse erythrocyte antigens. The heterophile antibody test is commonly used to diagnose mononucleosis from Epstein-Barr virus. While false positive tests have been reported in patients with SLE, they are far less common than a false positive RPR. p24 antigen (Choice D) is found on the surface of HIV and is a component of the screening test for HIV. False positive results are known to occur and should be followed up with confirmatory testing, typically with an HIV-1/2 antibody differentiation assay and/or HIV RNĀ level. Educational Objective: Patients with SLE also commonly have antiphospholipid antibodies. The RPR test for syphilis evaluates for the presence of antibodies against specific syphilis antigens to which anticardiolipin antibodies react, resulting in a false positive RPR. Previous Next Score Report Lab Values Calculator Help Pause

27 Exam Section 1: Item 27 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 27. A study is conducted in which the right renal artery of an experimental animal is constricted to decrease arcuate artery pressure by 20 mm Hg. Measurement of inulin clearance shows that renal artery constriction has no effect on glomerular filtration rate (GFR). Which of the following best explains the maintenance of a constant GFR in this experiment? A) Decreased glomerular filtration coefficient B) Decreased glomerular hydrostatic pressure C) Decreased renal blood flow D) Increased afferent arteriolar resistance E) Increased efferent arteriolar resistance F) Increased renal blood flow

E. Autoregulation is an essential process for maintaining perfusion of end organs despite variations in systemic blood pressure and takes place in numerous organs throughout the body. Renal autoregulation is adapted to produce a relatively constant GFR despite changes in glomerular perfusion and is mediated by the delivery of filtered sodium to the macula densa in the early distal convoluted tubule. In renal artery stenosis, there is decreased glomerular perfusion. Reductions in glomerular perfusion result in decreased filtration and decreased sodium delivery to the macula densa. The macula densa compensates for this by promoting activation of the renin-angiotensin system by stimulating the juxtaglomerular apparatus to secrete renin, which subsequently causes an increased angiotensin Il-mediated constriction of the glomerular efferent arteriole. This causes increased efferent resistance and induces an increased intraglomerular pressure to maintain GFR. Systemic hypertension can also be counteracted by renal autoregulation through a similar but opposite mechanism. As with all autoregulatory systems, renal autoregulation is only able to maintain glomerular filtration within a particular range of systemic blood pressures and can be overwhelmed by either profound hyper- or hypotension. Incorrect Answers: A, B, C, D, and F. Decreased glomerular filtration coefficient (Choice A) is not a mechanism of renal autoregulation. The glomerular filtration coefficient is an inherent property of the renal capillary and is not freely modified in response to changes in blood pressure. Some diseases that result in glomerular capillary damage, such as diabetes mellitus, may decrease the filtration coefficient. Decreased glomerular hydrostatic pressure (Choice B) would result in decreased glomerular filtration and exacerbate the hypofiltration of sodium that underlies the mechanism of renal autoregulation. Decreased renal blood flow (Choice C) is the direct result of the experimental constriction of the renal artery. This answer choice describes the effects of the experimental intervention rather than the physiologic mechanisms that compensate for this intervention. Increased afferent arteriolar resistance (Choice D) would also result in decreased glomerular filtration and exacerbate the hypofiltration of sodium that underlies the mechanism of renal autoregulation. Increased renal blood flow (Choice F) would not be observed following experimental constriction of the renal artery. Autoregulation does not alter renal blood flow itself, but rather maintains GFR despite changes in renal blood flow. Educational Objective: Renal autoregulation is mediated by the delivery of filtered sodium to the macula densa. Hypoperfusion of the glomerulus leads to decreased sodium filtration, and activation of the renin-angiotensin system by the macula densa, which causes angiotensin II-mediated constriction of the glomerular efferent arteriole and a consequent increase in intraglomerular pressure and GFR. %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. A 45-year-old woman comes to the physician because of a 3-week history of pain over her lateral left hip that is most severe when she awakens in the morning. She is unable to sleep on her left side due to the pain. Range of motion of the left hip is full with no pain on internal or external rotation. There is tenderness to palpation over the lateral hip. Sensation is intact. Inflammation of which of the following structures is the most likely cause of this patient's symptoms? A) Hip joint B) Inguinal ligament C) Lateral femoral cutaneous nerve D) Sacroiliac joint E) Trochanteric bursa

E. Bursae are thin, sac-like, fluid-filled structures that reduce friction and permit motion between layers of connective tissue. Bursae are frequently implicated in pathology and are often associated with pain and swelling, such as the subacromial bursa, which can cause pain in the shoulder, the trochanteric bursa, which can cause pain over the lateral aspect of the hip (in this case), and the olecranon bursa, which can cause pain and swelling posterior to the elbow. When inflamed or injured, fluid can collect inside the potential space of the bursa causing swelling, which can be appreciated on examination as a fluctuant, tender, circumscribed mass. Trochanteric bursitis presents with pain to palpation over the affected bursa and swelling and erythema in the affected area, but generally without limited range of motion or pain in the hip joint itself. If erythema and warmth are present, these findings can signify infection of the bursa (septic bursitis), which should be treated with antibiotics. In this patient's case, an intact sensory examination, full range of motion of the joint, and focal tenderness in the area of a bursa implicates trochanteric bursitis as the likely diagnosis. Incorrect Answers: A, B, C, and D. Hip joint (Choice A) pain can arise from pathologies such as osteoarthritis, avascular necrosis, synovitis, or septic arthritis. Pathology within the joint would present with pain and tenderness with range of motion of the joint, whereas this patient has normal range of motion and pain focally over the lateral hip. Inguinal ligament (Choice B) inflammation is uncommon, however pain in the inguinal region is a frequent presenting complaint. Etiologies of such pain include but are not limited to hernias (inguinal or femoral), ligament and muscle strain and sprain, neuropathic pain, stress fractures, referred pain, urinary tract infections, renal calculi, and appendicitis. This pain is located in the anterior groin, not in the lateral hip. Lateral femoral cutaneous nerve (Choice C) is a branch of the lumbosacral plexus and supplies sensation to the lateral thigh. It can become impinged on its course leading to meralgia paresthetica, a condition of pain and paresthesia in its territory. Sacroiliac joint (Choice D) pain is experienced in the inferior back and buttocks and may arise from athletic activity or inflammatory disease. It generally does not refer to the lateral hip. Educational Objective: Bursae are thin, sac-like, fluid-filled structures that reduce friction and permit motion between layers of connective tissue. Trochanteric bursitis presents with pain and swelling of the lateral hip over the greater trochanter. When erythematous and swollen, infectious bursitis may be present. %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. An 88-year-old man loses consciousness when his neck is palpated during a routine health maintenance examination. He has a 40-year history of hypertension that has been well controlled with hydrochlorothiazide and a 20-year history of degenerative osteoarthritis, for which he occasionally takes anti-inflammatory drugs. Which of the following is the most likely cause of the syncope? A) Complete heart block B) Laryngospasm C) Peripheral arteriole constriction D) Peripheral arteriole dilation E) Sinus bradyarrhythmia F) Ventricular tachyarrhythmia

E. Carotid sinus syndrome is characterized by an exaggerated response to carotid baroreceptor stimulation, with resultant sinus bradycardia, hypotension, and syncope. Baroreceptors located at the carotid bifurcations and the aortic arch are important in regulating cardiac output and maintaining adequate perfusion pressures to the brain and organs of the upper body. Baroreceptors are sensitive to mechanical pressure; normally, an increase in blood pressure stimulates baroreceptor activity, which activates a reflex arc resulting in parasympathetic innervation of the sinoatrial and atrioventricular nodes and inhibition of sympathetic vascular tone. This results in decreased pulse and peripheral vascular resistance. Mechanical compression of the baroreceptors from external palpation can stimulate this reflex, which may be exacerbated in the setting of volume depletion, such as the presented case where the patient is taking a thiazide diuretic. Incorrect Answers: A, B, C, D, and F. Complete heart block (Choice A), also called third-degree atrioventricular block, occurs when no sinus impulses are transmitted through the AV conduction pathway, resulting in complete dissociation between the P waves and the QRS complexes. Patients usually require a pacemaker. The temporal association with neck palpation in this patient is more suggestive of carotid sinus syndrome and sinus bradyarrhythmia. Laryngospasm (Choice B) is an exaggerated protective airway reflex to prevent aspiration that can result in partial or complete airway obstruction. It typically presents with acute dyspnea, increased work of breathing, and may result in hypoxemia-induced bradycardia. Peripheral arteriole constriction (Choice C) occurs with increased sympathetic activity and signaling at vascular adrenergic receptor sites, resulting in an increase in blood pressure. It does not lead to syncope. Peripheral arteriole dilation (Choice D) does result from carotid sinus stimulation. Patients experience orthostatic hypotension with presyncope symptoms such as lightheadedness and tunnel vision. Ventricular tachyarrhythmia (Choice F) includes ventricular tachycardia, ventricular fibrillation, and ventricular premature beats. Ventricular tachyarrhythmias are potentially life-threatening because of the potential for the sudden loss of adequate cardiac output. Educational Objective: The cardiovascular system must adjust to postural changes to maintain adequate perfusion pressure to the head and upper body. Baroreceptors in the carotid sinus and aortic arch detect changes in pressure and stimulate the body's response. Carotid baroreceptor hypersensitivity can provoke sinus bradycardia and syncope. %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. A 70-year-old man comes to the physician because of a 3-day history of shortness of breath and right-sided chest pain. His respirations are 25/min. Chest x-rays show a right pleural effusion and an associated ipsilateral, scallop-shaped pleural density. Examination during an open lung biopsy shows a thick, firm, white pleural tumor that ensheathes the right lung. Which of the following is the strongest predisposing risk factor for this patient's condition? A) Family history of lung cancer B) Radiation therapy C) Raising pigeons D) Smoking cigarettes E) Working in a shipyard

E. Chronic cough, shortness of breath, and pleural effusion plus a calcified pleural plaque on x-ray is consistent with a diagnosis of malignant mesothelioma, which can be confirmed by biopsy. Development of malignant mesothelioma is strongly associated with pulmonary asbestosis. A history of working in a shipyard is a known risk factor for exposure to asbestos, along with roofing and plumbing. Asbestosis is a progressive fibrotic disease that manifests decades after exposure. Long-term complications include interstitial fibrosis, chronic respiratory failure, and malignancy (especially malignant mesothelioma and bronchogenic carcinoma). Malignant mesothelioma typically presents with an exudative pleural effusion caused by the malignant pleural disease. Treatment includes chemotherapy, surgical resection, and radiation. Incorrect Answers: A, B, C, and D. Family history of lung cancer (Choice A), radiation therapy (Choice B), and smoking cigarettes (Choice D) are risk factors for developing primary bronchogenic carcinoma. Asbestosis is also a risk factor for bronchogenic carcinoma, although it also has a strong association with the development of malignant mesothelioma. Raising pigeons (Choice C) is a risk factor for chronic hypersensitivity pneumonitis, an inflammatory condition of the lungs in response to environmental irritants. The condition may progress to pulmonary interstitial fibrosis, and the chronic inflammation increases the risk for developing primary bronchogenic carcinoma. Educational Objective: Asbestosis is a chronic progressive pulmonary disorder associated with interstitial fibrosis and an increased risk for both primary bronchogenic carcinoma and malignant mesothelioma. Occupational risk factors for asbestos exposure include shipbuilding, roofing, and plumbing. %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. A 27-year-old man who is a weightlifter comes to the physician for a routine examination. He has been taking a synthetic androgen for the past year to increase his muscle mass. Examination shows increased scalp hair loss and mild testicular atrophy. Which of the following is most likely responsible for the decreased size of this patient's testicles? A) Compression of the testes by overdeveloped pelvic muscles B) Downregulation of pituitary luteinizing hormone receptors C) Downregulation of testicular follicle-stimulating hormone receptors D) Increased secretion of inhibin E) Increased negative feedback on gonadotropin secretion F) Increased sexual activity

E. Excessive circulating androgen concentrations in the setting of the exogenous use of synthetic androgens results in increased negative feedback on gonadotropin secretion. Testosterone provides negative feedback upon gonadotropin releasing hormone (GNRH) neurons in the hypothalamus, with subsequent decreases in the release of luteinizing hormone (LH) and of follicle-stimulating hormone (FSH) by the anterior pituitary. Downregulation of LH secretion results in decreased stimulus for the testicular production of testosterone, which leads to testicular atrophy. Testicular atrophy may commonly occur in patients who are taking anabolic steroids or in patients who are taking testosterone replacement therapy. Incorrect Answers: A, B, C, D, and F. Compression of the testes by overdeveloped pelvic muscles (Choice A) would not occur. The testes are located within the scrotum, outside of the pelvis, and are not vulnerable to compression by pelvic muscle hypertrophy. Downregulation of pituitary luteinizing hormone receptors (Choice B) would not cause testicular atrophy. The LH receptor is responsible for the transduction of LH-dependent signaling that causes the synthesis of androgens and is primarily located in gonadal tissue. Downregulation of LH receptors in gonadal tissue will lead to testicular atrophy. LH receptor is not expressed in substantial quantities in extragonadal tissues. Downregulation of testicular follicle-stimulating hormone receptors (Choice C) leads to decreased spermatogenesis. In the testis, FSH receptors are primarily expressed on Sertoli cells. Testicular atrophy is primarily mediated by the inhibition of LH signaling, rather than that of FSH. Increased secretion of inhibin (Choice D) is performed by Sertoli cells and provides negative feedback upon the anterior pituitary to decrease FSH secretion. Increased secretion of inhibin would not result in changes in testosterone production by the testis and would not lead to testicular atrophy. Increased sexual activity (Choice F) does not lead to atrophy or hypertrophy of the testes. Educational Objective: Testicular atrophy is primarily caused by decreased testicular testosterone synthesis as a result of the absence of stimulation by luteinizing hormone. This commonly occurs in the setting of anabolic steroid use or testosterone replacement therapy, as these exogenous androgens provide excessive negative feedback upon the hypothalamus, thereby causing the decreased production of GNRH and, subsequently, LH. %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 34-year-old woman comes to the physician because of a 10-year history of constipation. She usually has fewer than three bowel movements weekly. She often has bloating and cramping, and she has to strain and apply perineal pressure to defecate. She says that she does not have problems with sleep, appetite, or energy. Palpation of the abdomen produces diffuse discomfort; bowel sounds are present. Which of the following is the most likely cause of this patient's gastrointestinal symptoms? A) Colon polyps B) Congenital megacolon (Hirschsprung disease) C) Factitious disorder D) Gluten enteropathy E) Irritable bowel syndrome F) Major depressive disorder

E. Irritable bowel syndrome (IBS) is characterized by recurrent, intermittent abdominal pain and alteration of bowel habits as either a change in stool frequency or consistency. The abdominal pain may be either exacerbated or relieved by defecation and can vary widely in location and character. IBS commonly presents with intermittent episodes of diarrhea and/or constipation, often which alternate. When diarrhea does occur, it usually does so during waking hours which serves to differentiate it from other causes of diarrhea. It is most common in middle-aged women and the pathogenesis is unknown. It is not caused by a structural abnormality, and physical examination will show nonspecific abdominal discomfort rather than localized pain. Stress management may improve bowel habits in patients with irritable bowel syndrome, which is often associated with concomitant anxiety. If medication is needed, dicyclomine is an antispasmodic and diphenoxylate slows gut motility. Both are used in the management of irritable bowel syndrome. Incorrect Answers: A, B, C, D, and F. Colon polyps (Choice A) present in a variety of subtypes, from non-neoplastic polyps (eg, hamartomatous, mucosal, inflammatory, hyperplastic) to potentially malignant polyps (adenomatous, serrated). Polyps are typically asymptomatic and are only recognized after screening colonoscopy. They are not a common cause of chronic constipation. Inflammatory polyps may be seen in inflammatory bowel disease, but not irritable bowel syndrome. Congenital megacolon (Hirschsprung disease) (Choice B), or intestinal aganglionosis, is caused by the congenital absence of the distal portion of the myenteric plexus, a part of the enteric nervous system located between the inner and outer layers of the muscularis externa. This often leads to a failure to pass stool within the first few days of life and constipation thereafter. An explosive expulsion of feces on rectal examination is a classic clinical finding. It does not present in adulthood. In factitious disorder (Choice C), 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 does not appear to be causing her constipation for primary gain. Gluten enteropathy (Choice D), or Celiac disease, is an immunologic intolerance to the protein gliadin. Celiac disease typically presents with abdominal discomfort, bloating, nausea, and diarrhea, not constipation, following exposure to gluten, and may be associated with weight loss and dermatitis herpetiformis. Major depressive disorder (Choice F) presents with depressed mood, anhedonia, insomnia, weight loss, fatigue, impairments in concentration, guilt, psychomotor slowing, and suicidal ideation, none of which is patient is experiencing. Irritable bowel syndrome is more likely to be seen in conjunction with anxiety than depression. Educational Objective: Irritable bowel syndrome (IBS) is characterized by recurrent, intermittent abdominal pain and alteration of bowel habits as either a change in stool frequency or consistency. The condition is chronic and in most patients the severity of symptoms varies over time. %3D 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 38-year-old woman with asthma comes to the physician for advice concerning contraception. She is sexually active with one partner, and they use condoms inconsistently. Current medications include inhalational fluticasone and albuterol. She has smoked one-half pack of cigarettes daily for 20 years and drinks two alcoholic beverages daily. She does not use illicit drugs. Her maternal grandmother developed breast cancer at the age of 58 years. She is 157 cm (5 ft 2 in) tall and weighs 82 kg (180 lb); BMI is 33 kg/m2 Her vital signs are within normal limits. Physical examination shows no other abnormalities. This patient should be advised to avoid the use of an oral contraceptive because of which of the following historical factors? A) Age B) Alcohol use C) Asthma D) Family history E) Obesity F) Tobacco use

F. Combined oral contraceptive pills (OCPS) contain a combination of estrogen and progesterone. They are primarily used for contraceptive purposes but can also be used in a variety of gynecologic disorders, such as polycystic ovarian syndrome (PCOS), menorrhagia, and endometriosis. A sustained release of progestin and estrogen prevents ovulation and causes thickening of the cervical mucus along with thinning of the endometrial layer by inhibiting the release of gonadotropin-releasing hormone, follicle- stimulating hormone, and luteinizing hormone. OCPS are typically administered as a daily pill for three weeks, followed by a daily placebo pill for one week, during which time withdrawal bleeding occurs. They are associated with increased hypercoagulability and are therefore contraindicated in patients with a history of deep venous thrombosis, pulmonary embolism, stroke, or myocardial infarction. They are also contraindicated in patients who are 35 years of age or older and concomitantly smoke 15 or more cigarettes per day, along with patients diagnosed with hypertension, migraine with aura, severe cirrhosis, and breast cancer. Common adverse effects include nausea, breast tenderness, irregular menstrual periods, and a mildly delayed return to fertility following cessation. OCPS are contraindicated in this patient caused by her tobacco use. Incorrect Answers: A, B, C, D, and E. Age (Choice A) can increase the risks associated with OCPS, as there is an increased baseline risk for cardiovascular disease. However, this patient is only 38 years old, and age alone is not a contraindication to OCP use without smoking history. Alcohol use (Choice B) can preclude the use of OCPS if severe and associated with consequent cirrhosis, as they are generally contraindicated in severe decompensated cirrhosis. However, alcohol use alone would not be an indication to avoid OCPS. Asthma (Choice C) commonly presents with shortness of breath, prolonged cough, and bronchospasm. OCPS do not cause alterations in respiratory mechanics and are not contraindicated in asthmatic patients. Family history of breast cancer (Choice D) could raise concern for the use of estrogen-containing oral contraceptives, as a strong family history of breast cancer could indicate a possible BRCA1 or 2 mutation that would increase the patient's risk for developing breast cancer in the setting of exogenous estrogen administration. However, this patient's grandmother developed breast cancer at 58, making a BRCA1 or 2 mutation unlikely. Obesity (Choice E) independently increases the risk for cardiovascular disease, stroke, and thromboembolism. When administered to an obese patient, OCPS do increase the relative risk for thromboembolism. However, the absolute risk is less than the risk for thrombosis in an active smoker taking oral contraceptive pills. Educational Objective: Combined oral contraceptive pills contain estrogen and progesterone to promote anovulation and prevent pregnancy, as well as to treat PCOS, menorrhagia, and endometriosis. They should be avoided in patients with a history of deep venous thrombosis, pulmonary embolism, stroke, myocardial infarction, hypertension, migraine with aura, severe cirrhosis, and current breast cancer. They are also contraindicated in patients 35 years or older who concomitantly smoke cigarettes. %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 16-year-old boy is brought to the emergency department 30 minutes after he dove into a 3-foot-deep swimming pool at night. On examination, he is unable to move the right upper and lower extremities. The most likely cause of the movement deficits in this patient is damage to which of the following labeled regions in the photograph of a cross section of the spinal cord shown? A BC D Right Left A) B) C) D) E) F) G) H)

E. Label E represents the right lateral corticospinal tract in this cervical spinal cord section (as indicated by the ovoid shape and predominance of white matter). Upper motor neurons of the corticospinal tract originate in the primary motor cortex, descend ipsilaterally through the internal capsule and midbrain, decussate in the caudal medulla, and then descend contralaterally in the spinal cord to synapse with the contralateral lower motor neuron. Since the decussation occurs superior to the cervical spinal cord, lesions involving the lateral corticospinal tract within the cervical spinal cord will lead to ipsilateral motor weakness and an upper motor neuron pattern of dysfunction (eg, increased muscle tone, hyperreflexia, Babinski reflex). Traumatic spinal cord injuries can occur through diverse mechanisms such as cord compression by a vertebral column fracture. Given this patient's hemiparesis of the right upper and lower extremities, the location of the injury was likely at the C5 level or higher. Incorrect Answers: A, B, C, D, F, G, and H. Labels A and D represent the fasciculi cuneatus of the dorsal column-medial lemniscus tract, while labels B and C represent the fasciculi gracilis of the dorsal column-medial lemniscus tract. The dorsal column-medial lemniscus tract decussates in the medulla. Dorsal column-medial lemniscus tract lesions within the cervical spinal cord lead to ipsilateral deficits in pressure, vibration, fine touch, and proprioception. The fasciculi gracilis contain these sensory fibers for the lower body, while the fasciculi cuneatus contain these sensory fibers for the upper body. Label F represents the left lateral corticospinal tract. Lesions in this location would result in an upper motor neuron pattern of dysfunction in the left upper and/or lower extremities. Labels G and H represent the spinothalamic tracts, which carry fibers that mediate pain, temperature, crude touch, and pressure sensation. Lesions of the spinothalamic tracts in the cervical spinal cord would lead to deficits in these sensory domains in the contralateral extremities. This occurs because of the decussation of the tract across the anterior white commissure immediately after the peripheral sensory nerve synapses with the second order neuron in the ipsilateral dorsal horn of the spinal cord. Educational Objective: Upper motor neurons of the corticospinal tract decussate in the caudal medulla. Lesions of the lateral corticospinal tract in the cervical spinal cord lead to an upper motor neuron pattern of dysfunction in the ipsilateral extremities. II 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. A 56-year-old woman comes to the physician because of a 3-week history of increasing weakness of her arms and legs. She has smoked 1 pack of cigarettes daily for 30 years. Physical examination shows pronounced weakness of the hip girdle muscles and lesser weakness of the shoulder girdle muscles, both of which improve with repetitive testing. A chest x-ray shows a hilar lung mass. Electromyography is compatible with a malfunction of the neuromuscular junction. Impairment of which of the following is the most likely cause of these findings? A) Acetylcholinesterase B) Binding of acetylcholine on the postsynaptic membrane C) Direct depolarization of muscle fibers by Ca2+ D) Postsynaptic membrane potential E) Presynaptic release of acetylcholine

E. Lambert-Eaton myasthenic syndrome (LEMS) is an uncommon neurologic condition that affects neuromuscular junction transmission. Normally, action potentials lead to depolarization in the terminal bouton of the axon. Voltage-gated calcium channels consequently open and allow calcium influx, which triggers the exocytosis of neurotransmitter-filled vesicles into the synaptic cleft. Patients with LEMS possess antibodies to voltage-gated calcium channels on presynaptic lower motor neurons, which decreases calcium influx and acetylcholine release to postsynaptic skeletal muscle cells. This patient with a smoking history and central lung mass likely has a small cell lung cancer that is producing autoantibodies to the voltage-gated calcium channel, causing a paraneoplastic LEMS. Patients consequently demonstrate progressive proximal muscle weakness with depressed or absent deep tendon reflexes. The muscle weakness typically improves with repetitive motion and exercise as small amounts of acetylcholine accumulate in the synaptic cleft over time, increasing the frequency of postsynaptic action potentials in skeletal muscle cells. Autonomic symptoms such as decreased salivation and consequent dry mouth may also occur. The diagnosis may be confirmed with nerve conduction studies and electromyography with exercise testing. Removal of the antibody-producing malignancy is crucial for treatment. Incorrect Answers: A, B, C, and D. Acetylcholinesterase (Choice A) inhibition leads to decreased degradation of synaptic acetylcholine and is the mechanism of medications (eg, pyridostigmine) utilized for myasthenia gravis. This mechanism typically improves muscle strength and would not be expected to lead to the progressive muscle weakness demonstrated in LEMS. Impairment of the binding of acetylcholine on the postsynaptic membrane (Choice B) is the cause of myasthenia gravis. Myasthenia gravis is an autoimmune disorder of neuromuscular transmission that presents with muscle weakness and fatigability (versus the postexercise facilitation demonstrated in LEMS). Additionally, myasthenia gravis is more likely to affect the distal extremity and ocular musculature than LEMS. Direct depolarization of muscle fibers by Ca2+ (Choice C) is not a known mechanism of neuromuscular junction transmission. Postsynaptic depolarization is mediated by sodium influx via nicotinic acetylcholine receptors. The postsynaptic membrane potential (Choice D) may be decreased in LEMS secondary to the decreased presynaptic release of acetylcholine. Educational Objective: Patients with Lambert-Eaton myasthenic syndrome (LEMS) possess antibodies to voltage-gated calcium channels on presynaptic lower motor neurons, which decreases calcium influx and acetylcholine release to postsynaptic skeletal muscle cells. In some patients, malignancies (eg, small cell lung cancer) trigger production of these antibodies, resulting in paraneoplastic LEMS. Patients present with progressive proximal muscle weakness that improves with repetitive motion and exercise. %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 70-year-old man with metastatic prostate cancer has been taking leuprolide for the past 3 months. Which of the following best describes this patient's current serum luteinizing hormone (LH) and testosterone concentrations compared with concentrations before treatment? LH Testosterone A) ↑ ↑ B) ↑ C) No change ↑ D) no change E)

E. Leuprolide is a gonadotropin-releasing hormone (GNRH) analog. If given in a pulsatile fashion, mimicking the physiologic secretion of GNRH, it will act as a GNRH receptor agonist and increase follicle-stimulating hormone (FSH) and luteinizing hormone (LH). When leuprolide is initially started in a patient, the hypothalamus reacts as if it were an agonist and there is a transient rise in LH and FSH for the first week of treatment. However, with continued use in a non-pulsatile fashion, as in this case, it will act as a GNRH receptor antagonist and subsequently decrease FSH and LH. By decreasing the production of LH and the stimulation of Leydig cells, leuprolide indirectly lowers testosterone. Prostate cancer is a hormonally sensitive cancer; androgens play a critical role in its growth. Thus, by inhibiting testosterone production, this driver of cancer growth is removed, and sensitive tumors begin to shrink. This approach to treating prostate cancer is termed medical castration. Some tumors may become castration-resistant, in which mutations develop that allow the cancer to continue growing without hormonal stimulation. Incorrect Answers: A, B, C, and D. While there is a transient increase in LH and testosterone (Choice A) for the first week of treatment, after three months of treatment leuprolide will be exerting GNRH antagonist effects. This will lead to decreased LH and testosterone concentrations. Because LH stimulates the Leydig cells to produce testosterone, when LH concentration is decreased, testosterone concentrations will also decrease (Choice B). A medication which continuously increases testosterone would not be appropriate for use in prostate cancer as it is an androgen-dependent malignancy and would continue to grow under androgen stimulation. LH decreases rather than showing no change (Choice C) when leuprolide is administered given its antagonistic effects on the GNRH receptors when given in a continuous fashion. Likewise, testosterone decreases rather than showing no change (Choice D) in response to the decreased stimulation of the Leydig cells. Educational Objective: Leuprolide is used in prostate cancer as a form of medical castration. Administration of leuprolide in a continuous fashion decreases LH release by the pituitary and subsequently decreases the production of testosterone by the Leydig cells. %3D 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. A 25-year-old man is brought to the emergency department because of a 2-hour history of nausea, vomiting, abdominal cramping, and difficulty passing flatus. Examination of the face shows the lesions in the photograph; similar lesions are seen on the fingers and toes. Laboratory studies show hypochromic microcytic anemia. Test of the stool for occult blood is positive. Which of the following is the most likely diagnosis? A) Cowden disease B) Gardner syndrome C) Muir-Torre syndrome D) Neurofibromatosis E) Peutz-Jeghers syndrome

E. Peutz-Jeghers syndrome is an autosomal dominant polyposis in which hamartomatous polyps occur in the colon and gastrointestinal (GI) tract, and pigmented macules are found in or around the mouth, lips, hands, and genitalia. It is associated with increased risk for breast and Gl tract cancers (eg, colorectal, stomach, small bowel, pancreatic). Hamartomatous polyps are characterized by disorganized growth of tissue similar to normal native tissue. When solitary, they are usually benign and do not carry a significant risk for malignant transformation, however when syndromic they are associated with increased risk in gastrointestinal tract cancers. Patients with Peutz-Jeghers syndrome may have intestinal obstruction or intussusception as their presenting symptom, such as in this patient. Polyps can serve as a lead point for intussusception or obstruct the gastrointestinal lumen. Polyps can also cause chronic occult bleeding, leading to a hypochromic microcytic anemia. Incorrect Answers: A, B, C, and D. Cowden disease (Choice A) is an autosomal dominant polyposis associated with hamartomatous polyps in the GI tract. It is associated with mucocutaneous neuromas, oral papillomas, and cutaneous trichilemmomas. Gardner syndrome (Choice B) is a familial adenomatous polyposis characterized by the presence of thousands of adenomatous polyps arising after puberty. It is associated with osteomas, soft tissue tumors, supernumerary teeth, and hypertrophy of the retinal pigment epithelium. Muir-Torre syndrome (Choice C) is a form of hereditary nonpolyposis colorectal cancer. Affected individuals are prone to developing colon, genitourinary, or skin cancers. Cutaneous manifestations can include keratoacanthomas and sebaceous gland tumors. Neurofibromatosis (Choice D) type 1 is a heritable neurocutaneous disorder that presents with cutaneous neurofibromas (benign neoplasms derived from neural crest cells), café au lait spots, pigmented iris hamartomas (Lisch nodules), optic gliomas, and pheochromocytomas. This autosomal dominant disorder results from a de novo or inherited mutated NF1 tumor suppressor gene. Educational Objective: Peutz-Jeghers syndrome is an autosomal dominant syndrome that is characterized by hamartomatous polyps in the colon and Gl tract, and pigmented macules in the mouth, lips, hands, and genitalia. Hamartomatous polyps in the GI tract can present with intussusception, bowel obstruction, and hypochromic microcytic anemia. 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 1 cm 23. A 56-year-old woman comes to the physician because of cough productive of blood-tinged sputum for 8 days. She also has had fatigue, loss of appetite, and an unintentional 6.8-kg (15-lb) weight loss during the past 2 months. She has a 5-year history of chronic cough and shortness of breath that she had attributed to smoking 2 packs of cigarettes daily for 42 years. She appears ill. Physical examination shows dullness to percussion over the right lower lung lobe. A chest x-ray shows a mass in the right upper lung lobe. The gross and microscopic appearances of a lung from a patient with a similar condition are shown. Which of the following is the most likely type of neoplasm in this patient? A) Adenocarcinoma B) Giant cell carcinoma C) Large cell undifferentiated carcinoma D) Small cell carcinoma E) Squamous cell carcinoma

E. Squamous cell carcinoma of the Ilung, which typically presents as a central lesion, is the second most common type of primary lung cancer following adenocarcinoma, which more often presents as a peripheral lesion. Risk factors for all major types of lung cancer include tobacco use, secondhand smoke, asbestos, and radon exposure, and a family history of lung cancer. Features associated with squamous cell carcinoma of the lung include pulmonary cavitations, central location, and hypercalcemia caused by paraneoplastic parathyroid hormone-related peptide (PTHPP) production. Histologic characteristics include polygonal cells with intercellular bridges, eosinophilic cytoplasm, keratin pearls (as seen in the photomicrograph), and extensive necrosis. Lung cancer, in general, typically presents with cough, unintentional weight loss, hemoptysis, chest pain, dyspnea, and hoarseness; occasionally, wheezing, focal rhonchi, or hypertrophic osteoarthropathy may be noted on examination. Obstruction of the airways can lead to recurrent postobstructive pneumonia. Diagnosis is made by chest imaging and examination of a biopsy specimen. Prognosis is a function of the cancer type along with grading and staging of the disease. It is often detected once metastatic, at which point the prognosis is poor. Incorrect Answers: A, B, C, and D. Adenocarcinoma (Choice A) of the lung is the most common overall primary lung cancer and the most common among nonsmokers. It typically presents as a chronic consolidation in the periphery of the lung rather than centrally. A glandular pattern is classically seen on histology with mucin-positive staining. Giant cell carcinoma (Choice B) is a rare carcinoma of the lung that contains pleomorphic giant, multinucleated cells on histology. It most commonly involves the upper lobes and the lung periphery. Large cell undifferentiated carcinoma (Choice C) is a neoplasm of epithelial cells that lacks glandular, squamous, or neuroendocrine characteristics on histology. It most often presents as a peripheral lesion. Small cell carcinoma (Choice D) is also centrally located and associated with tobacco use. It is a neoplasm of neuroendocrine cells and is associated with numerous paraneoplastic syndromes, including Cushing syndrome, syndrome of inappropriate antidiuretic hormone, Lambert-Eaton myasthenic syndrome caused by presynaptic calcium channel antibody production, paraneoplastic limbic encephalitis, and subacute cerebellar degeneration. Histologic features include small, dark blue tumor cells lacking nucleoli with a high nuclear to cytoplasm ratio. Educational Objective: Centrally located primary lung cancers include squamous cell carcinoma of the lung and small cell carcinoma of the lung. Squamous cell carcinoma is the more common subtype and is associated with polygonal cells with intercellular bridges and keratin pearls on histology. 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. In an animal model of ischemic myocardial injury, a branch of the coronary artery is clamped. Which of the following morphologic changes is most likely to occur first? O A) Defects in the plasma membrane B) Myelin figures O C) Nuclear pyknosis D) Rupture of lysosomes E) Swelling of the endoplasmic reticulum

E. Swelling of the endoplasmic reticulum (ER) is most likely to be the first morphologic change that will occur after clamping a branch of the coronary artery. Myocardial cells have a high metabolic demand, and complete cessation of blood flow to the myocardium leads to myocardial ischemia. If blood flow is not restored, myocardial ischemia progresses to myocardial infarction, at which point cardiomyocyte apoptosis occurs. One of the first cellular changes that occurs after cessation of blood flow is swelling of the cell and its organelles, including the ER. Decreased oxygen impairs the ability of the cell to generate ÁTP, which is used to pump sodium out of the cell and potassium into the cell via the action of the Na -K+-ATPase in the plasma membrane. Cessation of this pump results in intracellular accumulation of sodium, thereby shifting the osmotic gradient and leading to cellular swelling. As protein synthesis is an energy intensive process, ischemia can also alter the folding of cellular proteins. These proteins accumulate in the ER. 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. If blood flow is restored, these changes can be reversed, and cardiomyocyte apoptosis is avoided. Incorrect Answers: A, B, C, and D. Defects in the plasma membrane (Choice A) leading to increased plasma membrane permeability is a feature of early coagulative necrosis and indicates irreversible cellular damage that is the hallmark of myocardial infarction. Similarly, nuclear pyknosis (Choice C) describes the condensation of nuclear chromatin that occurs as the cell is undergoing apoptosis, while the rupture of lysosomes (Choice D) is also a feature of irreversible cell injury. Myelin figures (Choice B) describe cytoplasmic structures that resemble in appearance the myelin fibers associated with neurons, although they are unrelated. Myelin figures are aggregates of phospholipids from damaged mitochondria and ER. They can be seen in both reversible and irreversible cellular injury and are a later finding than swelling of the ER. Educational Objective: Myocardial ischemia from cessation of coronary blood flow describes a stepwise process of cellular injury. One of the first manifestations of cardiomyocyte injury is swelling of the cell and its organelles, which occurs as the intracellular accumulation of sodium leads to an alteration in the osmotic gradient. If blood flow is not restored, apoptosis eventually occurs. %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. An investigator is conducting a study of the effects of a newly discovered spider venom on neuromuscular transmission. An isolated nerve muscle preparation is bathed in a solution containing the venom; there is a decrease in the end-plate potential amplitude following stimulation of the nerve. The presence of the venom does not change the amplitude of the nerve action potential or the muscle potential change in response to the exogenous application of acetylcholine to the neuromuscular junction. Blockade of which of the following by the venom best explains the decreased end-plate potential amplitude in this study? A) Chloride conductance in the nerve terminal B) Function of acetylcholinesterase C) Inactivation of voltage-gated sodium channels in the muscle D) Insertion of acetylcholine receptors into the muscle membrane E) Presynaptic, voltage-gated calcium channels

E. The blockade of presynaptic, voltage-gated calcium channels would lead to decreased end-plate potential amplitude. Normally, action potentials lead to depolarization in the terminal bouton of the axon. Presynaptic, voltage-gated calcium channels consequently open and allow calcium influx, which triggers the exocytosis of acetylcholine-filled vesicles into the synaptic cleft. Acetylcholine diffuses across the synaptic cleft and binds nicotinic acetylcholine receptors (NÁCHRS) on the skeletal muscle cell membrane. The bound NACHRS allow sodium influx (and a lesser degree of potassium efflux), leading to depolarization of the postsynaptic membrane potential. Once the threshold membrane potential is reached, voltage-gated sodium channels open and the depolarization propagates down the postsynaptic membrane, representing the muscle action potential. Each NACHR is associated with its own end-plate potential, or voltage change, that results from the binding of acetylcholine. The summation of the end-plate potentials represents the membrane potential. Decreased synaptic acetylcholine concentrations result in decreased end-plate potential amplitudes. Incorrect Answers: A, B, C, and D. Blockade of chloride conductance in the nerve terminal (Choice A) would lead to decreased chloride influx and a more positive presynaptic membrane potential. Consequently, more voltage-gated calcium channels would open, leading to calcium influx and increased release of acetylcholine vesicles. The amplitude of end-plate potentials would increase. Blockade of the function of acetylcholinesterase (Choice B) leads to decreased degradation of synaptic acetylcholine and is the mechanism of medications (eg, pyridostigmine) utilized for myasthenia gravis. An increased synaptic acetylcholine concentration would increase the amplitude of end-plate potentials. Blockade of the inactivation of voltage-gated sodium channels in the muscle (Choice C) would not affect end-plate potential amplitudes, which are instead influenced by synaptic acetylcholine concentrations. Blockade of sodium channel inactivation would result in an increased amplitude of muscle action potentials. Blockade of the insertion of acetylcholine receptors into the muscle membrane (Choice D) would not decrease the end-plate potential amplitude, which is instead influenced by synaptic acetylcholine concentration. Fewer acetylcholine receptors would lead to a decreased muscle membrane potential. Educational Objective: When acetylcholine binds postsynaptic nicotinic acetylcholine receptors, each receptor causes its own postsynaptic voltage change, which is called an end-plate potential. The summation of the end-plate potentials represents the membrane potential. The synaptic acetylcholine concentration influences the amplitude of end-plate potentials. %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 45-year-old woman is brought to the emergency department 30 minutes after the sudden onset of left-sided facial drooping and left arm and leg weakness. One week ago, she returned from a business trip that involved a 6-hour airplane flight. She has no history of major medical illness. She does not smoke. Her only medications are an oral contraceptive and occasional acetaminophen for headache. Her pulse is 88/min and regular, and blood pressure is 160/90 mm Hg. Physical examination shows left- sided facial paralysis. Examination of the left lower extremity shows swelling, tenderness, and localized erythema over the calf. Muscle strength is 3/5 in the left upper and lower extremities. Cardiac examination shows no abnormalities. Echocardiography is most likely to show which of the following findings in this patient? A) Aortic stenosis B) Endocarditis C) Hypertrophy of the interventricular septum D) Mitral valve prolapse E) Patent foramen ovale

E. This patient with a deep venous thrombosis (DVT) most likely possesses a patent foramen ovale, allowing a thromboembolism to bypass the pulmonary circulation and cause a thromboembolic cerebrovascular accident (CVA, or stroke). DVT typically presents with lower extremity edema, erythema, warmth, and pain at rest along with calf pain on dorsiflexion of the foot (Homan sign). Blood clots typically occur in patients who possess some or all of the risk factors of the Virchow triad: stasis (long flight, postoperative immobility), hypercoagulability (oral contraceptive use, hypercoagulable disorders), and endothelial damage (smoking, hypertension, atherosclerosis, injury). DVT may lead to pulmonary embolism if the clot embolizes to the right heart and subsequently to the pulmonary circulation. In patients with a patent foramen ovale (a congenital defect between the right and left atria), the clot may travel from the right atrium to the left atrium and then to the systemic circulation. This patient's clot likely embolized to the right internal carotid artery and then the right middle cerebral artery (MCA), occluding blood flow to the right primary motor cortex resulting in left-sided weakness. CVAS occur because of ischemic or hemorrhagic loss of blood supply to the brain. Approximately 80 to 85% of CVAS are ischemic, commonly arising from thromboembolic disease, whereas 15 to 20% of CVAS are hemorrhagic as a result of blood vessel rupture (eg, hypertension-related intraparenchymal hemorrhage from a perforating artery). Classically, CVAS manifest as a neurologic deficit related to the affected part of the brain. Incorrect Answers: A, B, C, and D. Aortic stenosis (Choice A), hypertrophy of the interventricular septum (Choice C), and mitral valve prolapse (Choice D) are structural abnormalities of the heart that lead to abnormal cardiac outflow, which may result in volume overload and left atrial enlargement over time. Hemostasis (blood pooling) in the enlarged left atrium increases the risk for thromboembolic stroke. However, this patient's physical examination shows a DVT as the source of the thromboembolism. The classic example of this scenario is left atrial stasis in the setting of atrial fibrillation from mitral stenosis, left ventricular hypertrophy, or chronically increased afterload (eg, hypertension, aortic stenosis). Endocarditis (Choice B) refers to an infection of the endocardium, which may feature valvular vegetations. These vegetations may embolize to the brain. However, this patient's physical examination shows a DVT as the source of the occlusion instead. This patient lacks physical examination findings concerning for endocarditis (eg, splinter hemorrhages, Osler nodes, Janeway lesions). Educational Objective: Patients with the risk factors of Virchow triad (stasis, hypercoagulability, and/or endothelial damage) may develop deep venous thrombosis (DVT). A DVT can embolize to the pulmonary circulation, resulting in pulmonary embolism. However, in patients with a patent foramen ovale, thromboembolism from DVT may bypass the pulmonary circulation and cause a thromboembolic stroke. %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 21-year-old woman who is a college student comes to the physician because of a severe left-sided headache, nausea, and vomiting for 1 hour. Her symptoms began with flashing lights in both eyes while she was studying for final exams. Physical examination shows no abnormalities. The most appropriate treatment is a drug that acts on a receptor for which of the following endogenous mediators? A) B-Endorphin B) Histamine C) Neuropeptide Y D) Nitric oxide E) Serotonin

E. This patient with a migraine headache may be treated with triptan medications, which are serotonin agonists. Migraine is a relatively common condition that presents with recurrent primary headaches. Migraine headaches are typically severe, unilateral, throbbing, and associated with nausea, photophobia, and phonophobia. Some patients additionally experience an aura before or during the headache, which may present with vision loss followed by the perception of scintillating shapes or lines and/or paresthesias. Migraine triggers include stress, menstruation, weather changes, fasting, and alcohol. Migraines are postulated to arise from deficient serotonin inhibition of descending pain pathways. Acute migraine treatment includes triptans (serotonin agonists), dopamine antagonists, antiemetics, acetaminophen, and/or nonsteroidal anti-inflammatory drugs (NSAIDS). For patients with disabling or frequent migraines, prophylactic treatment is also recommended and includes B-adrenergic blockers, antidepressants (such as tricyclic antidepressants and serotonin-norepinephrine reuptake inhibitors), and anticonvulsants. Incorrect Answers: A, B, C, and D B-Endorphin (Choice A) is an endogenous opioid that is released during intense exercise and stress. Opioid medications mimic endogenous B-endorphin but are not recommended for migraines because of the risk for opioid overuse and the potentially resultant worsening of headaches. Histamine (Choice B) release causes vasodilation and increased vascular permeability and contributes to environmental allergies. Antihistaminergic medications (eg, diphenhydramine, cetirizine) are utilized to address allergies and are used off-label to treat anxiety and insomnia. Antihistaminergic medication does not necessarily play a role in acute migraine treatment, however the anticholinergic and sedating effects of first-generation antihistamines can counteract adverse effects of dopamine antagonists (eg, akathisia, dystonia) and improve patient comfort. Neuropeptide Y (Choice C) is an orexigenic peptide neurotransmitter that modulates peripheral sympathetic transmission. Neuropeptide Y is not involved in migraine pathogenesis or treatment. Nitric oxide (Choice D) is a signaling molecule that modulates synaptic plasticity and promotes vasodilation. Nitric oxide is not involved in migraine pathogenesis or treatment. Educational Objective: Migraines are typically unilateral, pulsating headaches that are associated with nausea, photophobia, and phonophobia and may also present with visual or sensory auras. Acute migraine treatment includes triptans, which are serotonin agonists. %3D 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. During a study, a healthy 35-year-old woman and a 35-year-old woman with Crohn disease are given an oral solution of radiolabeled lactulose. The appearance of radioactivity in their urine is monitored over the next 24 hours. Urinary excretion of radioactivity in the patient with Crohn disease is much greater than that of the healthy subject. This finding is most likely due to inflammation-associated changes in the integrity of which of the following epithelial structures? A) Adherens junctions (zonulae adherentes) B) Basement membrane C) Desmosomes (maculae adherentes) D) Gap junctions E) Tight junctions (zonulae occludentes)

E. Tight junctions (zonulae occludentes) are a type of epithelial cell connection composed of proteins called claudins and occludins. Tight junctions are located between two cells, toward the apical surface, and create a watertight seal to prevent the paracellular movement of solutes or water through an epithelial or endothelial lining. Inflammation of an epithelial lining causes it to become more permeable to substances via impairment of tight junctions. As seen in this patient with Crohn disease, inflammation of the wall of the small intestine impairs tight junctions and allows substances which would not normally be absorbed by the enterocytes to be transmitted paracellularly and then diffuse into the interstitial blood vessels. Another example of this same phenomenon is seen in an exudative pleural effusion. Inflammation from pneumonia, for example, impairs tight junctions and allows for solutes to move paracellularly from the intravascular space or alveoli into the pleural space. The lack of tight junctions explains the movement of large molecules which could not normally diffuse across cells. Incorrect Answers: A, B, C, and D. Adherens junctions (zonulae adherentes) (Choice A) are epithelial cell junctions which connect the actin cytoskeletons of adjacent cells to each other. A defect in adherens junctions, such as when E-cadherin expression is lost in breast cancer, allows for cells to metastasize. Basement membrane (Choice B) is a barrier which separates an epithelial lining from the interstitial tissue surrounding it. For example, the basement membrane separates the epidermis from the dermis in the skin. The basement membrane is made of type IV collagen, laminin, heparan sulfate, and fibronectin among other molecules. Type IV collagen is defective in Alport syndrome, causing dysfunction of the glomerular basement membrane. Desmosomes (maculae adherentes) (Choice C) are a second protein complex which maintain cell-to-cell adhesion in the epidermis by connecting the cytokeratin skeleton of two adjacent cells. Pemphigus vulgaris is caused by antibodies to desmosomes and presents with fragile, flaccid blisters. Gap junctions (Choice D) are connections between the cytoplasm of two adjacent cells that allow for the movement of small molecules in either direction. They are made of channel proteins called connexins. They play a role in allowing a chemical or electrical signal to transfer between cells, such as cardiac myocytes. Educational Objective: Epithelial cell connections include tight junctions, adherens junctions, desmosomes, and gap junctions. Tight junctions are responsible for creating a tight seal to prevent the paracellular movement of solutes or water through an epithelial or endothelial lining. Inflammation often impairs the efficacy of tight junctions. %3D 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 00 9. A 4-year-old boy is brought to the physician by his mother because of a 3-month history of hyperactivity and a decreased attention span. Physical examination shows no abnormalities. Laboratory studies show: 10.5 g/dL (N=11-15) 27% (N=28%-45%) 24 pg/cell (N=25.4-34.6) 76 um3 (N=77-98) Hemoglobin Hematocrit Mean corpuscular hemoglobin Mean corpuscular volume A photomicrograph of a peripheral blood smear is shown. Which of the following is the most likely underlying cause of these findings? A) Decreased B-globin synthesis B) Decreased glucose 6-phosphate dehydrogenase activity C) Dietary deficiency of B vitamins D) Dietary deficiency of iron E) Functional asplenia F) Inhibition of ferrochelatase G) a-Spectrin mutation

F. Although lead is generally no longer used in the manufacture of paint and gasoline, children can continue to be exposed to lead through ingestion of paint chips and dust in older homes or from contaminated drinking water. Lead poisoning is characterized by basophilic stippling on peripheral blood smear (shown here) and ringed sideroblasts within the bone marrow. Lead poisoning typically affects the central and peripheral nervous systems, the heme synthesis pathway, and gastrointestinal and kidney function. This patient demonstrates neurobehavioral symptoms which may be permanent if occurring in youth. Children are susceptible to the toxic effects of lead because of an immature central nervous system and lead passage across the blood brain barrier. Acute encephalopathy, hearing loss, and peripheral neuropathy are possible neurologic complications. Clinical manifestations of lead poisoning include renal tubular dysfunction, interstitial nephritis, vomiting, abdominal pain, constipation, and anemia. Lead inhibits the enzymes ferrochelatase and õ-ALA dehydratase, resulting in decreased heme synthesis, increased serum õ-ALA, urinary coproporphyrin, and erythrocyte zinc protoporphyrin. The diagnosis is suspected on exposure and symptoms and is confirmed with an increased serum lead concentration. Management of lead poisoning depends on the blood lead concentration and presence of symptoms, and includes chelation with dimercaprol, ethylene diamine tetraacetate, or succimer. Sources of lead should be investigated and eliminated. Incorrect Answers: A, B, C, D, E, and G. Decreased B-globin synthesis (Choice A) is the cause of B-thalassemia. B-Thalassemia causes microcytic anemia, but peripheral blood smear shows microcytosis and target cells with normal iron concentrations. Patients will have an abnormal hemoglobin electrophoresis. It would not explain hyperactivity or decreased attention span. Decreased glucose 6-phosphate dehydrogenase activity (Choice B) is a potential cause of normocytic, hemolytic anemia. Basophilic stippling is not a consistent smear finding, and cells missing sections of their membrane are more commonly seen. Dietary deficiency of B vitamins (Choice C) including B6 (pyridoxine), Bg (folic acid), and B12 (cobalamin), are responsible for causing anemia. Pyridoxine causes sideroblastic anemia because of impaired hemoglobin synthesis and excess iron but would not explain the patient's hyperactivity or decreased attention span. Folic acid and cobalamin deficiencies, in contrast, cause megaloblastic anemia which would have an increased mean corpuscular volume. Dietary deficiency of iron (Choice D) is a cause of iron deficiency anemia. This is a microcytic anemia which demonstrates increased total iron binding capacity and decreased ferritin. Fatigue, conjunctival pallor, pica, and koilonychia, but not hyperactivity or decreased attention span, are features of iron deficiency. Functional asplenia (Choice E) or hyposplenia leads to retention within red blood cells of basophilic nuclear remnants, or Howell-Jolly bodies. These are typically removed from red blood cells by splenic macrophages. This patient's peripheral blood smear demonstrates basophilic stippling, not Howell-Jolly bodies. The two can be distinguished by their number, as Howell-Jolly bodies are usually solitary. a-Spectrin mutation (Choice G) is a cause of hereditary spherocytosis, which causes hemolytic anemia because of the increased osmotic fragility of the affected red blood cells. Peripheral blood smear would demonstrate spherocytes rather than basophilic stippling. Educational Objective: Lead inhibits the enzymes ferrochelatase and õ-ALA dehydratase, resulting in decreased heme synthesis. Clinical manifestations of lead poisoning include encephalopathy, cognitive dysfunction, peripheral nephropathy, hearing loss, renal tubular dysfunction, interstitial nephritis, vomiting, abdominal pain, constipation, and anemia. %D 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 12-year-old boy is brought to the physician by his parents because of bed-wetting. He urinates 10 to 15 times each day. He says that he is constantly thirsty and drinks large quantities of liquid throughout the day. Urinalysis shows a specific gravity of 1.003. A 24-hour urine collection shows a creatinine clearance of 140 mL/min, with a total urine volume of 7600 mL and urine osmolality of 230 mOsmol/kg H,O. A tumor affecting which of the following hypothalamic nuclei is most likely involved in producing this patient's symptoms? A) Anterior B) Arcuate C) Dorsomedial D) Lateral E) Posterior F) Supraoptic G) Ventromedial

F. Central diabetes insipidus may result from a tumor involving the hypothalamic paraventricular and supraoptic nuclei and is characterized by the inadequate secretion of antidiuretic hormone (ADH, also known as vasopressin) from the posterior pituitary. Normally, increased plasma osmolality will be sensed by osmoreceptors and result in the increased production of vasopressin in the hypothalamic paraventricular and supraoptic nuclei. Vasopressin is subsequently released by the posterior pituitary gland. ADH binds V2 receptors of the late distal convoluted tubule and the collecting duct principal cells, which results in the insertion of aquaporin channels in the luminal surface and increases free water absorption. This process is typically utilized to maintain tight regulation of the serum osmolality. Inadequate production of ADH by the hypothalamus and release by the posterior pituitary gland may occur in the setting of various pathologies, such as pituitary tumors, trauma, tuberculosis, or intracranial surgery. This results in a decreased ability to absorb adequate amounts of free water within the collecting duct, which results in dilute urine and increased serum osmolality. Patients typically present with extreme thirst, production of large volumes of dilute urine, and altered mental status. Treatment requires the administration of desmopressin, an analog of vasopressin (ADH), which results in the reabsorption of free water via aquaporin insertion in the collecting duct. Incorrect Answers: A, B, C, D, E, and G. The anterior hypothalamic nucleus (Choice A) is involved in thermoregulation, specifically in cooling. The arcuate hypothalamic nucleus (Choice B) contains complex regulatory functions that are involved in appetite stimulation. The dorsomedial hypothalamic nucleus (Choice C) is involved in the regulation of feeding and circadian rhythm. The lateral hypothalamic nucleus (Choice D) is important in promoting appetite under the stimulation of ghrelin. The posterior hypothalamic nucleus (Choice E) is involved in thermoregulation, specifically in heating. The ventromedial hypothalamic nucleus (Choice G) is important in promoting satiety under the stimulation of leptin. Educational Objective: Central diabetes insipidus may result from a tumor involving the hypothalamic paraventricular and supraoptic nuclei and is characterized by inadequate secretion of antidiuretic hormone. Lack of antidiuretic hormone leads to decreased free water reabsorption from collecting ducts, resulting in an inability to concentrate urine and an abnormally increased serum osmolality. %3D 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 female newborn is born at 39 weeks' gestation following an uneventful pregnancy. The Apgar scores are 10 at 1 and 5 minutes, respectively. The newborn is at the 50th percentile for length and 25th percentile for weight. She has normal morphologic features. The parents are told she has an enzyme defect that, if left untreated, will result in failure to attain early developmental milestones, microcephaly, hyperactivity, seizures, intellectual disability, and a mousy odor of the skin and urine. The patient's diet needs to contain which of the following to provide substrate for catecholamine production? A) Dihydroxyphenylalanine B) Dopamine C) Epinephrine D) Norepinephrine E) Phenylalanine F) Tyrosine

F. Tyrosine supplements are required for catecholamine production in this patient with phenylketonuria (PKU). PKU results from a deficiency in the enzyme phenylalanine hydroxylase (PAH) that converts phenylalanine to tyrosine. Deficiency results in accumulation of phenylalanine and its metabolites phenylacetate and phenyllactate. Rarely, mutations in enzymes that either produce or recycle tetrahydrobiopterin (BH , which is a cofactor for PAH, can have similar clinical manifestations as classic PKÚ. Phenylalanine is a precursor to tyrosine, and tyrosine is used to synthesize catecholamines such as dopamine, epinephrine, and norepinephrine. Tyrosine is converted to dihydroxyphenylalanine by tyrosine hydroxylase, followed by conversion to dopamine by DOPA-decarboxylase. Dopamine can subsequently be converted to norepinephrine, and methylation of norepinephrine results in formation of epinephrine. Patients with PKU from PAH deficiency require supplementation with tyrosine to bypass the dysfunctional enzymatic step of converting phenylalanine to tyrosine. Incorrect Answers: A, B, C, D, and E. Dihydroxyphenylalanine (Choice A), abbreviated as DOPA, is a precursor to dopamine and is formed via the action of tyrosine hydroxylase. This enzymatic pathway is functional in patients with PKU so supplementation with DOPA is not required. Similarly, dopamine (Choice B) is created from DOPA by the enzyme DOPA decarboxylase, while epinephrine (Choice C) and norepinephrine (Choice D) are made from dopamine by separate enzymes, none of which are deficient in PKU. Supplementation of tyrosine is sufficient to allow for normal synthesis of all of these catecholamines. Phenylalanine (Choice E) would worsen the symptoms of PKU as the primary cause of this condition is the inability to convert phenylalanine to tyrosine. Educational Objective: PKU commonly results from deficiencies in the enzyme phenylalanine hydroxylase, which converts phenylalanine to tyrosine. Tyrosine, in turn, is required for synthesis of the catecholamines dopamine, norepinephrine, and epinephrine. Supplementation with tyrosine effectively bypasses the defective enzyme and allows for the normal synthesis of catecholamines. %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 50-year-old man is brought to the emergency department because of a 2-hour history of vomiting blood. He is in acute distress. His temperature is 37.5°C (99.5°F), pulse is 110/min, respirations are 18/min, and blood pressure is 80/50 mm Hg. Despite appropriate treatment, he dies shortly thereafter, and an autopsy is done. The gross and microscopic appearances of a section of the esophagus are shown. Which of the following best explains these autopsy findings? A) Angiosarcoma B) Candidiasis C) Cytomegalovirus infection SPECIMEN DATE D) Esophageal carcinoma E) Gastric metaplasia OF) Hemangioma G) Miliary tuberculosis H) Portal hypertension

H. The microscopic findings taken from the esophageal sample demonstrate dilation of the submucosal venous plexus. This correlates with the dilated vessels seen on the gross specimen. The inferior esophageal veins drain into the left gastric vein and then into the portal vein. They are part of the portal venous system, a series of interconnected veins that drain blood from the colon, small intestines, spleen, liver, stomach, and inferior esophagus. While this blood eventually makes its way to the systemic circulation via the inferior vena cava, it must first pass through the liver. In cirrhosis, obliteration of the hepatic sinusoids through progressive fibrosis increases the resistance to blood flow through the liver, which is transmitted to the portal venous system causing portal hypertension (PH). Increased venous pressure is transmitted to all the veins which drain into the portal vein, including the inferior esophageal veins, resulting in esophageal varices after they are engorged. In addition to esophageal varices, manifestations of PH include gastric varices, caput medusae, hemorrhoids, splenomegaly, and ascites. The risk for acute gastrointestinal bleeding from esophageal varices is high. It presents with large volume hematemesis followed by hemodynamic collapse if untreated. Incorrect Answers: A, B, C, D, E, F, and G. Angiosarcoma (Choice A) and hemangioma (Choice F) are both vascular neoplasms caused by proliferation of endothelial cells. Angiosarcoma is a malignant vascular neoplasm which is characterized by large, atypical endothelial cells and vascular slit-like spaces. In contrast, hemangioma is a proliferation of benign endothelial cells creating duplicated vascular channels. This histopathology does not demonstrate cellular atypia and the number of vascular channels is normal, only they are dilated. Esophageal candidiasis (Choice B) is seen in HIV patients and presents with white, friable patches on the mucosa of the esophagus. It is commonly accompanied by oral candidiasis. It causes dysphagia, but not large volume hematemesis and hemodynamic instability. Cytomegalovirus (CMV) infection (Choice C) is an opportunistic infection commonly occurring in immunocompromised patients in the setting of solid-organ or allogeneic bone marrow transplantation or HIV/AIDS. When causing esophagitis, linear ulcers are characteristic. Large volume hematemesis is not a feature of CMV esophagitis. Gastric metaplasia (Choice E) is a complication of longstanding gastritis. Persistent mucosal irritation and inflammation leads to replacement of the normal gastric epithelium with columnar, intestinal epithelium. Untreated, this can increase the risk for developing gastric carcinoma. Similarly, intestinal metaplasia of the distal esophagus secondary to longstanding gastroesophageal reflux (Barrett esophagus) increases the risk for developing esophageal carcinoma (Choice D). These are less likely a cause of large volume hematemesis than esophageal varices. Miliary tuberculosis (Choice G) is caused by hematologic dissemination of Mycobacterium tuberculosis. The esophagus is an unlikely site of miliary tuberculosis. Miliary tuberculosis commonly involves the lymphatic system, bones, liver, and central nervous system and is usually accompanied by fever and night sweats. Educational Objective: Esophageal varices appear microscopically as dilated submucosal veins. They are caused by portal hypertension in which the resistance generated by a cirrhotic liver is conveyed through the portal venous system and portosystemic shunts. Esophageal varices are at a high risk for bleeding leading to hematemesis and hemodynamic instability. Previous Next Score Report Lab Values Calculator Help Pause