NBME 26

40 ---------- Exam Section 1: Item 41 of 50 National, Board of Medical Examiners Comprehensive Basic Science Self-Assessment 41. An investigator is studying patients with West Nile virus infection. During a 5-year period, data are collected on 25 patients diagnosed with this condition as confirmed by testing at the Centers for Disease Control and Prevention. Demographic information on the patients is reported (age, gender, and ethnicity), as well as information about likely sources of infection. Which of the following best describes this study design? A) Case series B) Case-control study C) Cohort study D) Correlational study E) Cross-sectional study

A. A case series is a descriptive study design in which a number of consecutive or nonconsecutive cases of a disease and/or treatment are described in detail, with information about exposure, demographics, and comorbidities. Case series do not imply a cause-and- effect relationship. They do not test a hypothesis nor are they randomized. They are useful in characterizing the natural history of a disease or response to treatment. They are also useful in describing rare diseases, as small amounts of patients may not permit conduction of larger case-control, cohort, or randomized trials with sufficient power. Incorrect Answers: B, C, D, and E. A case-control study (Choice B) investigates an association between exposure and an outcome. In this study design, a group of patients with the disease (cases) are identified. A group of patients without the disease (controls) are matched on baseline characteristics to the cases. Exposure data for the two groups is collected, and these data are compared to determine association with the outcome (disease) in question. An odds ratio may be calculated to compare exposures between groups. A cohort study (Choice C) identifies a group of patients and follows them over time to identify whether an exposure is associated with an outcome of interest. Cohort studies may be retrospective or prospective in design. In a prospective design, the hypothesis and analysis protocols are established prior to the start of the study period. In a retrospective design the hypothesis or question is designed after the study time period has passed. Ăn example of a prospective cohort study would be following a group of 1,000 smokers for a time period of 10 years and identifying the proportion of these patients who develop pancreatic cancer to identify the relative risk of pancreatic cancer in smokers. Correlational studies (Choice D) are descriptive of research methods that do not imply causal relationships. Correlational studies can be subject to errors of interpretation. For example, a researcher may find a correlation between coffee drinking and pancreatic cancer. There may be a confounding relationship between this correlation. Coffee drinkers may be more likely to be smokers, and smoking may have a causal relationship with pancreatic cancer instead. Thus, the study may be subject to confounding and causality should not be inferred. A cross-sectional study (Choice E) seeks to identify the prevalence of the condition at a particular point in time. An example of a cross-sectional study would be a single survey of a population inquiring whether patients have coronary artery disease and concurrently inquiring about activity levels and diet. Thus, the risk factor and the outcomes are measured simultaneously. The study does not follow patients over time. All information is collected at a single time point. Educational Objective: A case series includes a small number of patients and seeks to describe the course of a disease or treatment. It does not imply a causal relationship. It is observational in nature and is useful in describing a disease or treatment with limited numbers of patients.

84 ---------- Exam Section 2: Item 34 of 50 National Board of Medical Examiners ement 34. A 63-year-old woman comes to the physician because of a 5-day history of shortness of breath and swollen legs. Her respirations are 25/min and labored, and blood pressure is 130/50 mm Hg. She has a large subclavian arteriovenous fistula caused by a stab wound to the left supraclavicular area 15 years ago. Physical examination shows 2+ edema of the lower extremities. Which of the following findings is most likely in this patient? O A) Decreased arterial oxygen saturation O B) Decreased mixed venous oxygen saturation O C) Decreased stroke volume D) Increased resting cardiac output E) Increased systemic vascular resistance

D. Increased resting cardiac output is an expected finding in this patient with clinical signs and symptoms of high-output heart failure in the setting of a large subclavian arteriovenous fistula. While most cases of heart failure are divided into heart failure with preserved (HFPEF) or reduced (HFREF) ejection fraction, high-output cardiac failure is a unique variety of cardiac failure. Causes of high-output cardiac failure include endocrine abnormalities such as hyperthyroidism, chronic lung disease, sepsis, cirrhosis, and large arteriovenous (AV) fistulas as seen in this patient. AV fistulas are either congenital or acquired and result in an abnormal connection between a high-pressure system (arterial) and a low-pressure system (venous), without the benefit of the stepwise reduction in pressure that occurs with blood transmission through the peripheral arterioles and capillaries. AV fistulas effectively bypass the arteriolar and capillary systems leading to significant shunting of blood directly from the arterial to the venous system. This results in increased venous return to the right side of the heart, with a subsequent increase in heart rate and stroke volume. As cardiac output (CO) is defined by the product of stroke volume and heart rate, this results in increased CO. Eventually, due to the persistent increased myocardial oxygen demand of the heart in the setting of high-output, patients will decompensate and present with signs and symptoms of heart failure, including dyspnea, increased jugular venous pressure, and peripheral edema. Incorrect Answers: A, B, C, and E. Decreased arterial oxygen saturation (Choice A) is not a feature of AV fistulas. Arterial oxygen saturation will be normal. Decreased mixed venous oxygen saturation (Choice B) is not a feature of AV fistulas. Highly oxygenated arterial blood passes directly into the venous circulation via the AV fistula, so the mixed venous oxygen saturation will be higher than expected. Decreased stroke volume (Choice C) is not found in cardiac failure from AV fistulas. As a result of increased preload and stretch on the right ventricle, the stroke volume will be higher than normal. Increased systemic vascular resistance (SVR) (Choice E) is not an expected finding in high-output cardiac failure from an AV fistula. SVR is largely maintained by the action of the arterioles and bypassing this part of the arterial circuit via an AV fistula results in low SVR. Educational Objective: Large arteriovenous (AV) fistulas can result in high-output cardiac failure. The abnormal connection between an artery and vein results in decreased systemic vascular resistance and an increased venous return to the right heart, with a consequent increase in stroke volume and cardiac output.

47 ---------- Exam Section 1: Item 48 of 50 National, Board of Medical Examiners Comprehensive Basic Science Self-Assessment 48. A 23-year-old woman is brought to the emergency department because of shortness of breath for 2 weeks. Her respirations are 28/min. Physical examination shows no other abnormalities. Laboratory studies show: Serum Na+ K+ 135 mEq/L 4.0 mEq/L 110 mEq/L 15 mEq/L CI- HCO3- Arterial pH on room air 7.25 Urine Sodium Chlorine 20 mEq/L 30 mEq/L 15 mEq/L Potassium Which of the following is the most likely diagnosis? O A) Alcoholic ketoacidosis B) Crohn disease C) Diabetic ketoacidosis D) Lactic acidosis E) Renal tubular acidosis F) Salicylate poisoning

E. Renal tubular acidosis (RTA) results in a hyperchloremic, non-anion gap metabolic acidosis from the dysregulation of the renal tubules causing the loss of bicarbonate or inability to excrete hydrogen ions. Metabolic acidosis can present with a compensatory respiratory alkalosis, manifesting as tachypnea and shortness of breath. Types of RTA vary in etiology and pathophysiology, but all result in a non-anion gap metabolic acidosis. RTAS are not due to the accumulation of an unmeasured anion and do not present with an increased anion gap. This patient demonstrates a low bicarbonate level, with a normal anion gap of 12, consistent with a non-anion gap acidosis. Distal RTA (type 1) results from a defect in the renal secretion of H*, which results in the lack of HCO3 generation, ultimately leading to a metabolic acidosis. It is characterized by a urine pH greater than 5.5. Some common causes of type 1 RTA include analgesic nephropathy, congenital obstructive urinary tract anomalies, medications, and autoimmune disease. Incorrect Answers: A, B, C, D, and F. Alcoholic ketoacidosis (Choice A) refers to a starvation state induced by excess alcohol intake that results in increased ketone production and an anion gap metabolic acidosis due to alterations in the ratio of NAD+ to NADH. Diabetic ketoacidosis (Choice C) typically occurs in patients with type 1 diabetes mellitus as a result of increased insulin requirements during times of physiologic stress (eg, infection) or insulin nonadherence. Diabetic ketoacidosis results in a metabolic acidosis with an increased anion gap. Lactic acidosis (Choice D) can result from any hypoperfusion state, hypoxia, deficiency in the citric acid cycle, impaired oxidative phosphorylation, or from vitamin deficiencies (eg, thiamine), and is associated with an increased anion gap metabolic acidosis. Salicylate poisoning (Choice F) initially results in a respiratory alkalosis from the stimulation of respiratory centers causing hyperventilation. Subsequently, it results in the development of a delayed anion gap metabolic acidosis secondary to its decoupling effects on the electron transport chain, resulting in increased anaerobic metabolism and the production of lactate and ketoacids. Educational Objective: There three types of renal tubular acidosis (RTA), which vary in etiology and pathophysiology, but all result in a hyperchloremic, non-anion gap metabolic acidosis from underlying dysregulation of the renal tubules.

7. A 65-year-old man who has chronic lymphocytic leukemia has the sudden onset of fatigue and shortness of breath. Laboratory studies show: Hemoglobin 6.5 g/dL Hematocrit 19% Leukocyte count 50,000/mm3 Platelet count 170,000/mm3 Reticulocyte count 8% Serum bilirubin 4 mg/dL Total Conjugated 0.5 mg/dL A peripheral blood smear is shown. Which of the following is the most likely cause of anemia? A) Aplastic anemia B) Autoimmune hemolysis C) Hereditary spherocytosis D) Microangiopathic hemolysis E) Monoclonal gammopathy

B. Autoimmune hemolysis explains this patient's anemia, unconjugated hyperbilirubinemia, reticulocytosis (large blue-colored erythrocytes), and spherocytes on the peripheral smear. Autoimmune hemolytic anemia (AIHA) occurs due to the 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 demonstrate normocytic anemia, an increased reticulocyte count indicating hematopoiesis, and unconjugated hyperbilirubinemia from hemoglobin released from lysed erythrocytes. Diagnosis is made by laboratory analysis, a compatible clinical history, and a 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-lgG, and an 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, a positive test. Chronic lymphocytic leukemia (CLL) predisposes to the development of warm AIHA because malignant CLL cells produce autoantibodies. Incorrect Answers: A, C, D, and E. Aplastic anemia (Choice A) results from bone marrow destruction of erythrocyte precursors and can be a primary autoimmune process, secondary to viral infections, medications, myelotoxin exposures (eg, heavy metals), or acquired clonal abnormalities. It manifests with anemia and inappropriately low reticulocyte count. Hereditary spherocytosis (Choice C) is caused by mutations in spectrin or ankyrin cytoskeletal anchoring genes that leads to fragile erythrocyte membranes and poor distensibility. Patients present with chronic hemolytic anemia (and family history of anemia) and on peripheral smear. While this patient does have spherocytes, his clinical history of CLL makes AIHA much more likely. Microangiopathic hemolysis (MAHA) (Choice D) presents with anemia, increased serum-free hemoglobin and lactate dehydrogenase, decreased haptoglobin, and increased unconjugated bilirubin, however, schistocytes (fragmented RBCS) would be seen on peripheral smear. MAHA often results from an infectious or inflammatory insult damaging endothelium. Subsequent microthrombi form and erythrocyte destruction occur from shearing across them. Monoclonal gammopathy (Choice E) is seen in multiple myeloma, Waldenstrom macroglobulinemia, and monoclonal gammopathy of unknown significance. While anemia can occur, it is most commonly a result of bone marrow infiltration by malignant cells (eg, myeloma). Peripheral smear may show RBC aggregation in rouleaux formation but would not demonstrate spherocytes or reticulocytes. 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 (Coombs), unconjugated hyperbilirubinemia, low haptoglobin, and the presence of spherocytes and reticulocytes on the peripheral smear.

8. A 47-year-old man with a history of rheumatic valvular disease comes to the physician because of chest pain and difficulty breathing for the past 2 days. An ECG shows an acute myocardial infarction. Examination shows a systolic murmur. Pressure tracings from the aorta left ventricle (LV), and left atrium (LA) are shown. Which of the following is the most likely diagnosis? A) Aortic obstruction B) Aortic regurgitation C) Left ventricular aneurysm D) Mitral obstruction E) Rupture of the chordae tendineae

E. Rupture of the chordae tendineae with development of mitral regurgitation accounts for the findings depicted in the graph. The diagram depicts changes in pressure within the aorta, LV, and LA over a single cardiac cycle. Normally the LV will undergo isovolumetric contraction with the expulsion of blood through the aortic valve once the pressure inside the LV exceeds the aortic pressure. When the pressure in the aorta (diastolic pressure) exceeds or equals the LV pressure, the aortic valve closes and the LV is able to relax, corresponding to a rapid decrease in pressure. This will allow for the ventricle to fill during diastole. During ventricular systole, the LA should demonstrate only a minimal increase in pressure because a competent mitral valve prevents regurgitation of blood back into the LA. However, if the mitral valve is incompetent (mitral regurgitation), blood from the LV will enter the LA through the mitral valve. This is depicted in the diagram as an increase in pressure in the LA during the LV systole, which is abnormal. Following a myocardial infarction, a common cause of mitral valve regurgitation is rupture of a papillary muscle and associated chordae tendineae, which typically anchor the leaflets of the mitral valve to the ventricles and prevent regurgitation during ventricular systole. When ruptured, there is no support structure to keep the mitral valve leaflets closed, so the increased LV pressure forces them back into the LA with consequent development of mitral regurgitation. Depending upon the severity, this can result in acutely decompensated heart failure, and surgical repair may be required. Incorrect Answers: A, B, C, and D. Aortic obstruction (Choice A) from aortic stenosis means that the LV must generate a much higher pressure to overcome the stenotic valve and effectively eject blood into the aorta. This would be depicted by an LV pressure tracing that exceeds (taller than) the aortic pressure tracing during systole. Aortic regurgitation (Choice B) is the result of an incompetent aortic valve. After the closure of the aortic valve, blood will continue to flow backward into the LV, which would appear on the graph as a reduced and protracted decrease in pressure on the LV tracing after systole. There will also be a decrease in aortic diastolic pressure because of the regurgitation. Left ventricular aneurysm (Choice C) results in paradoxical outward motion of the myocardium during systole with reduced stroke volume. It would not cause increased left atrial pressure during systole, as in mitral regurgitation. Mitral obstruction (Choice D) from mitral stenosis results in an abnormally increased LA pressure due to the impaired flow of blood from the LA into the LV. Additionally, the LV stroke volume may also be reduced by impaired diastolic filling. Educational Objective: Rupture of the chordae tendineae and papillary muscle is a potential complication of myocardial infarction and results in acute mitral regurgitation. On a pressure-time graph, this manifests as an increased left atrial pressure during ventricular systole from regurgitant blood. Pressure (mm Hg)

93 ---------- Exam Section 2: Item 43 of 50 National, Board pf Medical Examiners ancive Basic Saance S ement 43. A 15-year-old girl is brought to the physician because of a 3-day history of fever, sore throat, and malaise. Her temperature is 39.2°C (102.6°F). Physical examination shows diffuse pharyngeal erythema, moderately enlarged tonsils, and tender anterior and posterior cervical lymphadenopathy. A complete blood count shows: 19,500/mm3 (N=3500- 10,500) Leukocyte count Segmented neutrophils 30% Bands 7% 2% 25% Eosinophils Lymphocytes Lymphocytes, atypical Monocytes 30% 6% Incubation of this patient's serum with sheep erythrocytes results in agglutination. The atypical lymphocytes in this patient are most likely which of the following cell types? O A) B lymphocytes O B) CD4+ T lymphocytes C) CD8+ T lymphocytes D) FOXP3-expressing regulatory T lymphocytes O E) Natural killer cells

C. Infectious mononucleosis is a viral illness caused often by Epstein-Barr virus (EBV). EBV is transmitted through respiratory secretions and saliva, causing the illness to be commonly acquired by teenagers and young adults. EBV infects B lymphocytes through CD21. While B lymphocytes are the cell type infected, atypical lymphocytes seen on peripheral blood smear are actually cytotoxic, CD8+ T lymphocytes, which are reacting to the viral infection. As a result, patients often have leukocytosis with a lymphocytic predominance. Cytomegalovirus (CMV) is another cause of infectious mononucleosis and an acute HIV infection may present with a similar syndrome. Clinically, the syndrome commonly presents with fever, lymphadenopathy (typically involving the posterior cervical lymph nodes), and hepatosplenomegaly along with pharyngitis. Patients with infectious mononucleosis will typically test positive on the Monospot test, which detects heterophile antibodies through the agglutination of sheep or horse erythrocytes. Treatment is supportive and patients should be counseled to avoid any contact sports until the splenomegaly has resolved due to the risk of splenic rupture. If amoxicillin is administered for treatment of a misdiagnosed pharyngitis thought to be from Streptococcus pyogenes, patients may develop a generalized maculopapular rash. Incorrect Answers: B, D, and E. While the EBV virus infects B lymphocytes through CD21 (Choice A), these are not the cell type that appear atypical on peripheral blood smear. T lymphocytes can be broken into three groups. CD8+ T lymphocytes are cytotoxic and recognize major histocompatibility complex (MHC)-I on the surface of antigen presenting cells. They are responsible for the direct killing of infected cells. CD4+ T lymphocytes (Choice B) recognize MHC-II on the surface of antigen presenting cells. They assist B lymphocytes in making antibodies and create a cytokine milieu, which recruits macrophages and leukocytes to the site of infection. Regulatory T lymphocytes (Choice D), regulate the adaptive immune response to preserve self-tolerance and prevent autoimmunity. Neither CD4+ T lymphocytes nor regulatory T lymphocytes take on an atypical appearance during EBV infection. Natural killer cells (Choice E) are a component of the innate immune system. They directly cause the apoptosis of virally infected and neoplastic cells through the use of perforin and granzymes. They are activated by nonspecific activation signals such as the absence of MHC-I or the presence of pathogen associated molecular proteins. They do not take on an atypical appearance during EBV infection. Educational Objective: Infectious mononucleosis is a viral illness caused often by infection with Epstein-Barr virus (EBV). While EBV infects B lymphocytes through CD21, the atypical lymphocytes seen on peripheral blood smear are cytotoxic CD8+ T lymphocytes reacting to the viral infection.

38 ---------- Exam Section 1: Item 39 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 39. An investigator is studying a laboratory strain of HIV-1. During the experiment, point mutations are introduced in the long terminal repeat region of the virus. Infection of a human T-lymphocyte line with the mutant virus results in markedly decreased viral replication compared with infection with the wild-type virus. A mutation most likely directly affected which of the following phases of the viral life cycle? A) Attachment of the virus to the cell surface B) Budding of the virus from the cell membrane C) Fusion of the viral envelope and cell membrane D) Integration of the provirus into the host cell genome E) RNA splicing to produce an mRNA encoding the envelope glycoproteins

D. Long terminal repeats (LTRS) are found at each of the 5' and 3' ends of retroviral DNA and transposons. In the case of retroviral DNA, they are formed by reverse transcription, which is required for the replication and virulence of HIV. LTRS are used to integrate the viral genome with that of the host, and in this case, integration of the provirus into the host cell genome is most directly affected by mutations within the LTRS. The LTR contains regulators of gene expression. The integrated HIV provirus 5' LTR contains the promoter sequence for transcription, while the 3' end contains the polyadenylation sequence. The viral integrase enzyme catalyzes the integration of proviral DNA with host DNA. Once viral DNA is integrated, the cell is permanently infected and carries the genome for the remainder of its life. Integrase inhibitors prevent the action of this enzyme, limiting the degree of reverse transcribed DNA that can fuse with the host genome. Incorrect Answers: A, B, C, and E. Attachment of the virus to the cell surface (Choice A) involves glycoprotein (eg, gp120), CD4, CXCR4, and CCR5 surface proteins. Mutations in host proteins may confer immunity. Maraviroc, an entry inhibitor, blocks HIV attachment to CCR5. Budding of the virus from the cell membrane (Choice B) is the final step in the release of newly synthesized HIV virions. Proteases are involved in the final packaging and release of virions, and protease inhibitors prevent successful completion of these steps. Fusion of the viral envelope and cell membrane (Choice C) is required for viral penetration into the cytoplasm of the host cell, and involves interactions between surface glycoproteins (eg, gp120, gp41) and host membrane proteins (eg, CD4). Enfuvirtide inhibits the binding of gp41 to prevent the creation of entry pores in the cell membrane. RNA splicing to produce an mRNA encoding envelope glycoproteins (Choice E) would be affected by mutations at splice sites, which typically occur in the exon and intron segments of transcribed mRNA, not in the LTRS. Educational Objective: Long terminal repeats (LTRS) are found at each of the 5' and 3' ends of retroviral DNA and transposons and are used to integrate the viral genome into the host genome. Mutations in these regions may prevent integration.

79 ---------- Exam Section 2: Item 29 of 50 National, Board pf Medical Examiners ment * 29. A 15-year-old girl is brought to the physician by her mother because of a 1-year history of increasingly severe abdominal pain that occurs periodically, but not regularly. Physical examination shows abdominal tenderness that is most acute around the umbilicus. The physician suspects ectopic endometrial tissue, and an operation is scheduled. During the operation, an 8-cm piece of tissue is removed from the ileal region on the antimesenteric side of the bowel approximately 90 cm proximal to the ileocecal junction. The patient's pain resolves postoperatively. The ectopic tissue was most likely located in which of the following structures? O A) Ascending colon B) Falciform ligament C) lleal polyp O D) Meckel diverticulum O E) Vermiform appendix

D. Meckel diverticulum results from persistence of the vitelline duct and can demonstrate ectopic, functional gastric, pancreatic, or, rarely, endometrial tissue. It can present with acute or chronic hematochezia or melena. It is often painless but may also present with abdominal pain. It typically presents in childhood, most commonly before age 2 years, although because of its indolent nature, it may not be diag length. In children, it may serve as a potential lead-point for intussusception. A technetium 99m pertechnetate scan demonstrates uptake in the right lower quadrant, corresponding with ectopic gastric mucosa within the Meckel diverticulum. Treatment of symptomatic cases is through surgical resection. sed until later in life. It is co only located two feet proximal to the ileocecal valve and is approxir ately two inches in Incorrect Answers: A, B, C, and E. The ascending colon (Choice A) is a common site of colon polyps. The ascending colon is distal to the ileocecal valve and is not consistent with the location of the excised tissue. The falciform ligament (Choice B) is a peritoneal fold that divides the liver; it is anatomically distant from the described location of resection in this case. lleal polyps (Choice C) are less common than their counterparts in the colon, and may be associated with hereditary polyposis syndromes, such as Peutz-Jeghers syndrome or familial adenomatous polyposis. The length of the removed tissue and its presentation with intermittent abdominal pain is more suggestive of a Meckel diverticulum. The vermiform appendix (Choice E) is the site of acute appendicitis, which presents with periumbilical pain that migrates to the right lower quadrant, fever, anorexia, nausea, vomiting, diarrhea, and leukocytosis. Educational Objective: Meckel diverticulum results from persistence of the vitelline duct and presents as an approximately two-inch diverticulum located two feet proximal to the ileocecal valve. It can cause abdominal pain, hematochezia, or melena. Meckel diverticulum frequently harbors ectopic tissue, which is most commonly gastric or pancreatic tissue. Rarely, Meckel diverticulum may contain ectopic endometrial tissue.

67 ---------- Exam Section 2: Item 17 of 50 Natjonal, Board of Medical Examinersment 17. To investigate the association between cellular telephone use and the development of brain tumors, 782 patients with brain tumors are compared with 799 age- and sex-matched patients who do not have a malignancy. The results of the study find that there was no increase in the odds ratio for cellular telephone use in patients who had brain cancer compared with those who did not have cancer. Which of the following best describes this study design? A) Case-control study O B) Case series O C) Cross-sectional study O D) Prospective cohort study O E) Randomized clinical trial O F) Retrospective cohort study

A. A case-control study investigates an association between exposure and an outcome. In this study design, a group of patients with the disease (cases) are identified. A group of patients without the disease (controls) are matched on baseline characteristics to the cases. Exposure data for the two groups is collected, and these data are compared to determine association with the outcome (disease) in question. An odds ratio may be calculated to compare exposures between groups. Incorrect Answers: B, C, D, E, and F. A case series (Choice B) is a descriptive study design in which a number of consecutive or nonconsecutive cases of a disease and/or treatment are described in detail, with information about exposure, demographics, and comorbidities. Case series do not imply a cause-and-effect relationship. They do not test a hypothesis nor are they randomized. They are useful in characterizing the natural history of a disease or response to treatment. They are also useful in describing rare diseases, as small amounts of patients may not permit conduction of larger case-control, cohort, or randomized trials with sufficient power. A cross-sectional study (Choice C) seeks to identify the prevalence of the condition at a particular point in time. An example of a cross-sectional study would be a single survey of a population inquiring whether patients have coronary artery disease and concurrently inquiring about activity levels and diet. Thus, the risk factor and the outcomes are measured simultaneously. The study does not follow patients over time. All information is collected at a single time point. A prospective cohort study (Choice D) identifies a group of patients and follows them over time seeking to identify whether an exposure is associated with an outcome of interest. In a prospective design, the hypothesis and analysis protocols are established prior to the start of the study period. An example of a prospective cohort study would be following a group of 1,000 smokers for a time period of 10 years and identifying the proportion of these patients who develop pancreatic cancer to identify the risk of pancreatic cancer in smokers. A randomized clinical trial (Choice E) is an experimental study design. Patients are randomly allocated to two or more interventional arms or control arms, and these patients are followed over time to evaluate an outcome of interest. Randomized design minimizes opportunity for bias; thus, a randomized interventional study can be used to imply causation. Common examples of randomized trials include therapeutic comparisons between a new drug and the previous standard of care. A retrospective cohort study (Choice F) identifies a group of patients and reviews them historically in previous time, seeking to identify whether an exposure is associated with an outcome of interest. In a retrospective design, the hypothesis or question is designed after the study time period has passed. Educational Objective: A case-control study compares individuals with a particular disease or outcome (cases) with a matched group selected from the same population (controls). This comparison allows for the identification of unique exposures in the case group.

20. A healthy 35-year-old woman who wishes to become a regular platelet donor comes to a blood donation center. Blood is obtained by venous phlebotomy into sodium citrate and processed to yield platelet-rich plasma (PRP) for platelet function studies. She has no personal or family history of bleeding disorders. Aliquots of the PRP are subsequently analyzed using a light-transmittance platelet aggregometer. Adding a 5 µM solution of which of the following reagents to the PRP is most likely to cause a rapid and irreversible aggregation pattern in these studies? A) Adenosine diphosphate B) Collagen monomer C) Immune globulin D) Norepinephrine E) Prostacyclin (PG|2)

A. Adenosine diphosphate (ADP) addition to PRP will result in rapid and irreversible aggregation. ADP is critical in normal hemostasis. It acts via the receptors P2Y1 and P2Y12 on the surface of platelets. Binding to these receptors results in G-protein coupled receptor signaling, with consequent alteration of platelet shape and activation of GPII6/Illa via the PI3K signaling pathway, which results in rapid platelet aggregation. The P2Y12 receptor is the target of clopidogrel, an anti-platelet medication used commonly for the treatment of acute coronary syndrome following percutaneous coronary intervention. Incorrect Answers: B, C, D, and E. Collagen monomers (Choice B) are essential building blocks of collagen superstructures that function to help adhere platelets at the site of clots, but they do not have a role in the process of rapid and irreversible platelet aggregation. Immune globulin (Choice C) is given in the intravenous form (IVIG) for various indications. It has been shown to cause platelet aggregation and thrombosis via the interaction of platelets with the Fc portion of IgG, but this reaction is considered a potential side effect of IVIG and does not occur reliably. Norepinephrine (Choice D) may induce platelet aggregation by binding to alpha-adrenergic receptors on the surface of platelets with a subsequent increased risk of microthrombi formation, but this is not a reliable effect. Prostacyclin (PG|2) (Choice E) is produced through the cyclooxygenase pathway. It inhibits the activation and aggregation of platelets and it plays a role in vasodilation. Epoprostenol is a synthetic prostacyclin that can be used in the treatment of pulmonary arterial hypertension. Educational Objective: ADP rapidly and irreversibly causes platelet aggregation through its action on the P2Y1 and P2Y12 receptors.

83 ---------- Exam Section 2: Item 33 of 50 Natjonal, Board pf Medical Examinersment andive Rasic Saance 33. A 2-year-old boy is brought to the physician because of a 6-month history of failure to thrive. Cardiac examination shows a grade 4/6 systolic murmur caused by increased pulmonic flow, followed by a fixed, widely split S2. Echocardiography shows hypertrophy of the right atrium, right ventricle, and pulmonary arteries. This patient most likely has which of the following congenital cardiac anomalies? A) Atrial septal defect O B) Patent ductus arteriosus O G) Persistent truncus arteriesus O D) Tetralogy of Fallot O E) Transposition of the great arteries O F}VentriculaF-septal defect

A. Atrial septal defect is a common congenital malformation of the interatrial septum. The most common type is an ostium secundum defect, although ostium primum defects are commonly associated with trisomy 21. The atrial septal defect results in a left-to-right shunt with abnormal flow of blood from the left atrium to the right atrium, resulting in relative volume overload of the right atrium and ventricle. This increased stroke volume of the right ventricle results in delayed closure of the pulmonic valve, which presents as a fixed, split S2, and low-grade physiologic ejection murmur on cardiac auscultation. The increased right heart volumes also result in a prominent right ventricular impulse on physical examination and may present an increased risk for the development of a right bundle branch block. Chest radiographs characteristically demonstrate increased caliber of the main pulmonary artery and increased pulmonary vascular markings. If the atrial septal defect remains uncorrected, it can result in the development of Eisenmenger syndrome secondary to prolonged pulmonary vasculature remodeling resulting in pulmonary arterial hypertension and shunt reversal leading to cyanosis. Incorrect Answers: B, C, D, E, and F. Patent ductus arteriosus (PDA) (Choice B) is a persistent conduit between the aorta and the pulmonary artery that has failed to obliterate after birth. It results in a continuous, machine-like murmur best heard in the left second intercostal space, radiating to the clavicle. Persistent truncus arteriosus (Choice results from failure of the aorticopulmonary septum to divide the aorta and pulmonary trunk. Most patients present with a concomitant ventricular septal defect and cyanosis noted in the neonatal period. Tetralogy of Fallot (Choice D) consists of pulmonary infundibular stenosis, overriding aorta, ventricular septal defect, and right ventricular hypertrophy, resulting in a right-to-left shunt with cyanosis first noted in infancy. The murmur heard with tetralogy of Fallot relates to the underlying ventricular septal defect characterized by a holosystolic murmur best heard in the left lower sternal border. Transposition of the great arteries (Choice E) results in aberrant reversal of the normal anatomical association of the aorta and pulmonary artery to the cardiac chambers secondary to failure of the aorticopulmonary septum to spiral. This results in separate systemic and pulmonary circulations, with the aorta leaving the right ventricle and the pulmonary trunk leaving the left ventricle. This is incompatible with life unless an additional shunt allowing admixture of blood is present (eg, PDA, ventricular septal defect). Ventricular septal defect (Choice F) refers to a defect in the interventricular septum, most commonly the membranous portion. It is characterized by a holosystolic murmur best heard in the left lower sternal border. Educational Objective: A fixed, widely split S2 is characteristic of an atrial septal defect due to increased blood flow through the pulmonic valve. Severe defects can result in pulmonary hypertension and development of Eisenmenger syndrome over time, with reversal of the left-to-right shunt.

41 ---------- Exam Section 1: Item 42 of 50 National, Board of Medical Examiners Comprehensive Basic Science Self-Assessment 42. A previously healthy 20-year-old woman comes to the emergency department because of a 3-day history of fever, shaking chills, headaches, fatigue, and joint and muscle pain. She recently spent the summer working as a lifeguard on Long Island, New York. She has never traveled outside the USA. She underwent splenectomy for injuries sustained in a motor vehicle collision at the age of 6 years. Her temperature is 39.1°C (102.4°F). Physical examination shows no other abnormalities. A peripheral blood smear shows small intraerythrocytic rings; the result of a polymerase chain reaction test for Plasmodium species is negative. Which of the following is the most likely causal organism? A) Babesia microti O B) Bartonella bacilliformis C) Bordetella bronchiseptica D) Borrelia burgdorferi E) Brucella melitensis

A. Babesiosis is a disease presenting similar to malaria that is caused by the protozoa Babesia microti. It is predominantly found in the northeastern United States and transmitted via the Ixodes tick. It classically presents with fever, chills, myalgias, weakness, headache, and an associated hemolytic anemia. Asplenia is a risk factor for severe disease. Diagnosis is made by direct visualization of organisms on a Giemsa-stained peripheral blood smear, which may appear in ring forms or in a tetrad formation resembling a cross within erythrocytes. Treatment is with atovaquone plus azithromycin. Incorrect Answers: B, C, D, and E. Bartonella bacilliformis (Choice B) is a Gram-negative, coccobacillus found in Peru, Ecuador, Colombia, and some parts of southern Florida that causes bartonellosis (Carrión disease) or Oroya fever in the acute phase of infection and verruga peruana in the chronic phase. It is transmitted by sandflies. Bordetella bronchiseptica (Choice C) is a Gram-negative bacillus and is a common cause of respiratory disease in dogs and cats, but rarely causes disease in humans. Borrelia burgdorferi (Choice D) is a spirochete common in the northeastern United States that causes Lyme disease. It is transmitted by the Ixodes tick. Babesia microti may be co-transmitted with Borrelia burgdorferi in this manner. Brucella melitensis (Choice E) is one of the species of Gram-negative coccobacilli that cause brucellosis, also known as undulant fever. The other species that infect humans are B. abortus, B. canis, and B. suis. Brucellosis is commonly contracted by the consumption of unpasteurized milk. Educational Objective: Babesiosis is a malaria-like disease caused by the protozoa Babesia microti. It can be recognized by the presence of intraerythrocytic ring forms or a tetrad formation on a peripheral blood smear.

15. A previously healthy 19-year-old woman comes to student health services because of a 3-day history of vulvar itching and vaginal discharge. She has a new sexual partner and uses no contraception. Speculum examination shows a profuse yellow-gray discharge in the vagina. The pH of the discharge is 6. Microscopic examination of the discharge shows clue cells. Addition of KOH to the discharge produces a strong amine odor. Which of the following is the most likely cause? A) Bacterial vaginosis B) Bacteroides fragilis C) Candida albicans D) Haemophilus ducreyi E) Herpes simplex virus F) HIV G) Human papillomavirus H) Pneumocystis jirovecii (formerly P. carinii) I) Treponema pallidum J) Trichomonas vaginalis

A. Bacterial vaginosis is a common gynecologic condition characterized by a shift of vaginal flora and overgrowth of particular bacterial species, most commonly Gardnerella vaginalis. G. vaginalis is a Gram-variable, facultative anaerobe. Bacterial vaginosis presents with gray, thỉn, malodorous vaginal discharge and may cause vulvovaginal pruritus. Vaginal pH is typically above 4.5, and a fishy odor is detected upon KOH testing. The identification of clue cells on wet mount preparation is diagnostic. Metronidazole is the first-line therapy. Bacterial vaginosis is not considered a sexually transmitted infection and treatment of asymptomatic sexual partners is not recommended. Incorrect Answers: B, C, D, E, F, G, H, I, and J. Bacteroides fragilis (Choice B) is an anaerobic, Gram-negative bacillus and a normal component of the gastrointestinal microbiota. It only results in infections when displaced from the colon such as following surgery, rupture, or trauma, where it has the potential to cause bacteremia, intra-abdominal infections, peritonitis, and subcutaneous abscesses. Candida albicans (Choice C) is a fungus that also forms part of the normal vaginal flora. Overgrowth of Candida species may produce vulvovaginitis. Candidal vulvovaginitis, in contrast to bacterial vaginosis, produces pruritus and a white discharge. Vaginal pH is normal, and hyphae may be visualized on a wet mount. Haemophilus ducreyi (Choice D) is a Gram-negative organism that causes the sexually transmitted infection of chancroid. Chancroid presents with painful genital ulceration and inguinal lymphadenopathy. Herpes simplex virus (Choice E) is an enveloped, double-stranded DNA virus that causes sexually transmitted infections among other syndromes (eg, encephalitis, herpes labialis, herpetic whitlow). Symptoms of genital herpes include pain, burning, and a vesicular rash. HIV (Choice F) is an enveloped, single-stranded RNA virus of the retrovirus family that causes acquired immunodeficiency syndrome. The acute viral syndrome resembles an influenza-like illness that transitions into a prolonged dormant period, followed by the onset of immunodeficiency and opportunistic infections. Human papillomavirus (HPV) (Choice G) is a non-enveloped, double-stranded DNA virus of the papillomaviridae family that causes genital warts. Infection with HPV is a significant and preventable risk factor for cervical, penile, anal, and oropharyngeal cancer. Pneumocystis jirovecii (formerly P. carinii) (Choice H) is an opportunistic fungal organism that can cause pneumonia in immunocompromised patients. Pneumonia secondary to P. jirovecii has become rare following the widespread use of prophylaxis with trimethoprim-sulfamethoxazole in vulnerable patients. Treponema pallidum (Choice I) is the causative spirochete bacterium of syphilis, which presents in multiple stages with varying symptoms, including primary with a painless chancre, secondary with fever, diffuse maculopapular rash, lymphadenopathy, and condylomata lata, and tertiary with tabes dorsalis, aortitis, and gummas. Trichomonas vaginalis (Choice J) is an anaerobic, flagellated protozoan that causes vaginal trichomoniasis. Trichomoniasis presents with pruritus, dysuria, dyspareunia, cervical erythema, and abnormal discharge that is described as green, frothy, and malodorous. Vaginal pH is often above 4.5 and motile trichomonads can be visualized on wet mount preparation. Educational Objective: Bacterial vaginosis is caused by the vaginal overgrowth of Gardnerella vaginalis, a Gram-variable, facultative anaerobe. It presents with gray, thin, malodorous vaginal discharge, a vaginal pH greater than 4.5, a fishy odor upon KOH testing, and clue cells on microscopy.

42 ---------- Exam Section 1: Item 43 of 50 National, Board of Medical Examiners Comprehensive Basic Science Self-Assessment 43. A female newborn delivered at term develops jaundice, hepatomegaly, and rash shortly after delivery. Pregnancy was complicated by a 2-week febrile illness during the first trimester. The newborn is at the 25th percentile for length, 30th percentile for weight, and 20th percentile for head circumference. One week later, a CT scan of the head shows intracranial calcifications. Two days later, she becomes increasingly lethargic and develops respiratory distress and seizures. She has a 2-year-old sister who attends day care. Which of the following is the most likely causal virus? A) Cytomegalovirus B) Herpes simplex virus C) Measles virus D) Mumps virus E) Varicella-zoster virus

A. Congenital infection with cytomegalovirus (CMV), also known as human herpesvirus 5 (HHV-5), 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. Symptoms of maternal primary infection typically include fever, myalgias, and laboratory evidence of atypical lymphocytosis, often with abnormal liver enzymes. In an immunocompromised host, CMV can also present with a variety of atypical syndromes including colitis, encephalitis, pneumonia, esophagitis, or retinitis. Histopathologic examination of infected cells demonstrates prominent inclusion bodies. Unlike other viruses in the HHV family, CMV is not susceptible to guanosine analogs like acyclovir and valacyclovir as the virus does not possess the thymidine kinase necessary to activate this class of medications. Incorrect Answers: B, C, D, and E. Herpes simplex virus (Choice B) infection can occur via the placenta or through direct inoculation as the neonate passes through the birth canal in a mother with active lesions. Infection with herpes simplex virus can cause both cutaneous herpetic lesions and systemic manifestations like meningoencephalitis. It is not a cause of intracranial calcifications. Measles virus (Choice C) may be acquired via placental transmission or at the time of delivery. Clinical manifestations include a prodrome stage of fever, malaise, and anorexia. This is followed by the classic triad of cough, coryza, and conjunctivitis. An erythematous rash beginning on the face and spreading cephalocaudally is also characteristic. Neonates with measles infections may present with a spectrum of disease ranging from mild to severe. Mumps virus (Choice D) infection is caused by a paramyxovirus and is more likely to be transmitted through respiratory droplets and direct contact than through the placenta. Clinical manifestations include fever, myalgias, and fatigue, which are then followed by parotid gland swelling. Mumps orchitis, swelling of the testes, is a potential complication in males. Varicella-zoster virus (Choice E) may be transmitted vertically through the placenta during pregnancy but is more likely to be transmitted after delivery from mother to neonate. Neonates with varicella infection present with fevers and a vesicular rash with individual lesions classically in different stages of healing. Neonatal varicella has a mortality rate of up to 30%. Educational Objective: Congenital infection with cytomegalovirus (CMV) results in hearing loss, seizures, petechial rash, intracranial calcifications, organomegaly, and lymphadenopathy. The virus is transmitted via the placenta from the mother.

91 ---------- Exam Section 2: Item 41 of 50 National Board of Medical Examiners andive Rasic Science ment * 41. A 75-year-old woman with non-small cell carcinoma of the lung is brought to the physician because of a 2-day history of lethargy and cognitive impairment. One month ago, she completed a 1-year regimen of chemotherapy and cranial irradiation. One week ago, she received the diagnosis of major depressive disorder and began therapy with a selective serotonin reuptake inhibitor. She speaks with a blunted affect. Physical examination shows no abnormalities. Her Mini-Mental State Examination score is 20/30. Serum studies show a sodium concentration of 122 mEg/L. The patient appears confused and falls asleep during the examination. Which of the following is the most likely diagnosis? O A) Delirium O B) Dementia, Alzheimer type O C) Dementia caused by Pick disease O D) Serotonin syndrome O E) Vascular dementia

A. Delirium is an acute confusional state typically associated with acute medical illness in older patients. The pathogenesis of delirium is poorly understood but may be related to deficient acetylcholine signaling in the brain. Delirium presents with acute disturbances in awareness, attention, and baseline cognition that fluctuate in severity over the course of the day and appear related to a substance or acute medical condition. The cognitive changes can include memory or language deficits, disorientation, or perceptual disturbances (eg, visual hallucinations or paranoia). This patient demonstrates clear disturbances in awareness and cognition, as well as severe hyponatremia (likely due to syndrome of inappropriate antidiuretic hormone secretion associated with brain irradiation and selective serotonin reuptake inhibitor initiation). Delirium prevention with cognitive stimulation and avoidance of cognition-impairing medications is key. Delirium is managed by addressing the underlying medical problem and supportive management such as frequent reorientation and promoting sleep only at night to re-establish the appropriate circadian rhythm. Antipsychotic medication can manage symptoms of agitation or distressing hallucinations but does not shorten the duration of delirium. Incorrect Answers: B, C, D, and E. Dementia, Alzheimer type (Choice B) or caused by Pick disease (Choice C) would present with insidious cognitive decline instead of acute confusion. Fluctuations in the severity of cognitive deficits over the course of the day would be atypical, unless a concomitant delirium was also present. Baseline dementia is a risk factor for developing acute delirium. Pick disease (also known as frontotemporal dementia) typically develops when patients are in their 50s and features personality changes in addition to cognitive decline. Serotonin syndrome (Choice D), the result of serotonin toxicity (eg, from serotonergic medications such as selective serotonin reuptake inhibitors), can lead to the triad of acute confusion, autonomic hyperactivity (hyperthermia, tachycardia, hypertension), and neuromuscular excitability (tremor, myoclonus, hyperreflexia). Due to serotonin receptors in the gastrointestinal tract, vomiting and diarrhea can also occur. In this patient with severe hyponatremia and no evidence of neuromuscular excitability or autonomic instability, delirium is more likely. Vascular dementia (Choice E) features stepwise cognitive decline following cerebral infarction (stroke). This patient does not demonstrate focal neurological deficits indicative of a current or prior stroke, and fluctuations in the severity of cognitive deficits over the course of the day would be atypical for vascular dementia. Educational Objective: Delirium is an acute confusional state typically associated with acute medical illness in older patients. Delirium presents with acute disturbances in awareness, attention, and baseline cognition that fluctuate in severity over the course of the day.

39 ---------- Exam Section 1: Item 40 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 40. A 52-year-old woman undergoes surgical excision of a large multinodular goiter. During the operation, the superior thyroid artery is used as a landmark to avoid damage to a nerve proximal to that vessel. Which of the following best describes this nerve? A) External branch of the superior laryngeal nerve B) Inferior root of the ansa cervicalis C) Nerve to the mylohyoid D) Phrenic nerve E) Transverse cervical nerve

A. During thyroid surgery, there are a number of imp laryngeal nerve arises from the vagus nerve and courses along the superior thyroid artery, supplying innervation to the cricothyroid muscles. Injury to this nerve will cause alterations in the pitch of the voice. The recurrent laryngeal nerve is also important for vocalization, as it provides motor innervation for all of the intrinsic muscles of the larynx except for the cricothyroid muscle. It provides the sensory innervation for the laryngeal structures inferior to the vocal cords. The recurrent laryngeal nerves additionally have unique anatomy as they branch from the vagus nerves distally in the thorax. The left recurrent laryngeal nerve loops inferior and posterior to the aortic arch, while the right recurrent laryngeal nerve loops inferior and posterior to the right subclavian artery. Both of these nerves then progress cranially, rather than caudally. ant structures at risk. These structures include the parathyroid glands, the superior and inferior thyroid arteries, the recurrent laryngeal nerve, and the external branch of the superior laryngeal nerve. The superior Incorrect Answers: B, C, D, and E. The ansa cervicalis (Choice B) is a loop of nerves in the anterior neck that originate from the spinal nerve roots of C1-C3. The inferior root is formed by the fibers of C2 and C3 and provides branches to the inferior portion of the omohyoid, sternothyroid, and sternohyoid muscles. The nerve to the mylohyoid (Choice C) arises from the mandibular division of the trigeminal nerve and supplies motor innervation for the mylohyoid muscle and the anterior belly of the digastric muscle. The phrenic nerve (Choice D) arises from the cervical roots C3-C5 and provides motor innervation to the diaphragm. It also provides sensory innervation to the mediastinal pleura and the pericardium. Phrenic nerve injury can occur during cardiothoracic surgery or thoracic or cervical trauma and can lead to diaphragmatic paralysis. The transverse cervical nerve (Choice E) arises from the C1 and C2 cervical nerve roots. It provides sensory innervation to the skin on the anterior neck. It is anatomically distant from the thyroid. Educational Objective: The superior laryngeal nerve and the recurrent laryngeal nerve have unique anatomic relationships. It is important that these nerves are not injured during anterior neck surgery (eg, thyroidectomy) as this will leave the patient with laryngeal muscle paralysis, compromising phonation and airway protection.

65 ---------- Exam Section 2: Item 15 of 50 Natjonal, Board of Medical Examinersment ancive Rasic Saance 15. Histone acetyltransferases catalyze the acetylation of lysine residues in the amino-terminal tails of histones. Which of the following is the most likely effect of this covalent modification on chromatin structure? O A) Decreases the affinity of histones for DNA O B) Decreases the nucleosome content of the nucleus O C) Increases the affinity of histones for DNA O D) Increases the nucleosome content of the nucleus O E) Removes histone H1 from DNA O F) Removes histone H4 from DNA

A. Histone acetyltransferases decrease the affinity of histones for DNA by adding acetyl groups to histones lysine residues. Histones are proteinaceous cores, rich in lysine and arginine (positively charged amino acids). Without modification (eg, acetylation), the positively charged histones attract negatively charged DŇA, which loops around it, forming a chromatin nucleosome. Condensed heterochromatin is transcriptionally inactive. Acetylation of a histone is important in gene transcription as it permits relaxation of a DNA-histone complex by reducing the molecular affinity between DNA and histone. When DNA is tightly bound to a histone, gene transcription does not occur, since binding sites for RNA polymerase |I (which transcribes messenger RNA) will be occupied by the presence of the histone. When the affinity between the histone and DNA is reduced, the unpacked (euchromatin) DNA can be bound by transcriptional enzymes, promoting gene expression. Incorrect Answers: B, C, D, E, and F. Decreasing the nucleosome content of the nucleus (Choice B) implies dissolution of the nucleosome structure, a discrete length of DNA (usually 126 base pairs, or nearly two turns of DNA) that is wrapped around a set of eight histones (H2A, H2B, H3, and H4) forming a histone octamer. During transcription or DNA replication for cell division, the nucleosome is transiently disassembled, but then rapidly reassembled. Conversely, increasing the nucleosome content of the nucleus (Choice D) implies the formation of more nucleosomes. Adding additional nucleosomes to existing DNA would indicate winding the chromatin tighter, but the tight regulation of DNA folding prevents this from occurring. The addition of an extra chromosome, however, such as in trisomy 21, would technically increase the nucleosome content of the nucleus. Increasing the affinity of histones for DNA (Choice C) is accomplished by the enzyme histone deacetylase. It results in an increased affinity of the histone for DNA, causing the formation of heterochromatin. Removing histone H1 from DNA (Choice E) is not correct because this occurs via the action of histone chaperone proteins. Removing H1, which normally links nucleosomes together, allows for DNA unwinding. Removing histone H4 from DNA (Choice F) is not correct as histone acetyltransferase only modifies existing histones. Educational Objective: Histone acetyltransferase serves to add acetyl groups to histone lysine residues, causing a decreased affinity of histone for DNA. This results in relaxation of the DNA-histone complex (euchromatin), which allows for gene transcription.

95 ---------- Exam Section 2: Item 45 of 50 Natjonal, Board of Medical Examinersment andive Rasic Science * 45. A 10-year-old boy with spinal muscular atrophy is brought to the emergency department because of a 2-week history of nausea and vomiting. He also has had decreased appetite during this period. One week prior to his illness, his entire family had similar symptoms which improved after a few days. Physical examination shows decreased skin turgor, dry mucous membranes, and diffuse muscle weakness and atrophy. Laboratory studies show: Serum 140 mEq/L (N=138- 145) 4 mEq/L (N=3.4-4.7) 105 mEq/L (N=95- 105) 18 mEq/L (N=22-28) 18 mg/dL (N=5-18) 30 mg/dL (N=60-100) Na+ K+ CI- HCO3- Urea nitrogen Glucose Urine 5.0 (N=4.6-8.0) 1.020 (N=1.003- 1.030) pH Specific gravity Glucose none Protein trace Ketones 4+ RBC O/hpf O/hpf WBC Which of the following biochemical cycles in the liver is most likely providing this patient's brain with an energy source at this time? O A) Alanine-glucose cycle O B) Carnitine-acylcarnitine cycle C) Lactic acid cycle O D) Tricarboxylic acid cycle O E) Urea cycle

B. Spinal muscular atrophy is a group of autosomal recessive disorders that lead to the degeneration of motor neurons and result in progressive hypotonia and areflexia. Patients may exhibit difficulty with mobility and frequently experience respiratory distress and difficulty feeding. The brain primarily utilizes glucose for energy, and during the fasting state is highly dependent on glycogenolysis and hepatic gluconeogenesis. In the initial fasting state, glycogen is broken down in the liver to assist in the maintenance of blood glucose levels. Glycogen stores are typically depleted after 1 day of fasting. Following this, blood glucose is maintained through the breakdown of muscle and fat to provide substrates for hepatic gluconeogenesis. After approximately 3 days of fasting, fatty acids become the primary source for energy through the production of ketone bodies. This patient's laboratory values, including profound hypoglycemia, decreased bicarbonate, and ketonuria, are suggestive of starvation ketoacidosis in the setting of acute illness and prolonged fasting from persistent nausea and vomiting. Carnitine and acylcarnitine are components of the carnitine shuttle and serve to transport acyl-CoA into the mitochondrial matrix to allow for the beta-oxidation of fatty acids. During the initial step of the shuttle, carnitine palmitoyltransferase I transfers the acyl group from fatty acyl-CoA to carnitine, forming acylcarnitine. Acylcarnitine is then transported across the inner mitochondrial membrane by carnitine-acylcarnitine translocase, a step which is susceptible to negative feedback by increased malonyl-CoA. In the final step, carnitine and fatty acyl-CoA are liberated by carnitine palmitoyltransferase I| on the inner mitochondrial membrane. This allows beta-oxidation to proceed within the mitochondrial matrix, which will support the continued production of ketone bodies to provide the brain with an energy source during prolonged fasting. Incorrect Answers: A, C, D, and E. The alanine-glucose cycle (Choice A), also known as the Cahill cycle, utilizes alanine aminotransferase (ALT) to facilitate the transfer of alanine from skeletal muscle to the liver, where it can be used for gluconeogenesis, while also allowing skeletal muscle to rid itself of nitrogenous waste. The lactic acid cycle (Choice C), also known as the Cori cycle, transports lactate produced by skeletal muscle to the liver, where it can be used for gluconeogenesis. The lactic acid cycle is primarily active in the context of muscular exercise, rather than during the fasting state. Ketoacidosis secondary to fasting is a more likely etiology of metabolic acidosis than lactic acidosis in this patient. Tricarboxylic acid cycle (Choice D) utilizes acetyl-CoA to produce NADH, FADH2, and GTP. The citric acid cycle relies on the glycolytic production of pyruvate and is dependent on the supply of glucose. This is not the primary source of energy for the brain in a prolonged fasting state. The urea cycle (Choice E) aids in the excretion of nitrogenous waste by converting ammonia to urea. The urea cycle does not produce glucose or other sources of energy for the brain. Educational Objective: During a prolonged fasting state (greater than 3 days), the brain is dependent upon ketone bodies as its primary source of energy. Beta-oxidation, which requires the carnitine-acylcarnitine cycle to transfer acyl groups from the cytoplasm to the matrix, allows for the production of ketone bodies. This manifests on laboratory evaluation with hypoglycemia, metabolic acidosis, and ketonuria.

44 ---------- Exam Section 1: Item 45 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 45. A 70-year-old man with poorly controlled hypertension is brought to the emergency department 30 minutes after the sudden onset of weakness of the left side of his face and body. He is alert and able to follow commands. His pulse is 80/min, and blood pressure is 190/90 mm Hg. Physical examination shows mild flattening of the nasolabial fold on the left side and weakness and hyperreflexia of the left upper and lower extremities. There are no sensory deficits, but he has mild dysarthria. Six months later, histologic examination of the area of the injury is most likely to show proliferation of which of the following cell types? A) Astrocytes B) Ependymal cells C) Microglial cells O D) Neurons E) Oligodendrocytes

A. In the aftermath of a cerebral infarction (stroke), several histologic changes occur that culminate in a glial scar composed of proliferated astrocytes. One to 2 days after an ischemic event, neurons shrink and release lysosomal enzymes that lead to liquefactive necrosis. Three to 5 days after the event, macrophages (called microglial cells in the central nervous system) transiently proliferate and phagocytose dead neurons. Within weeks, astrocytes migrate to the site of injury (as opposed to fibroblasts in the peripheral nervous system) and begin to proliferate. After 2 weeks, the ischemic area becomes occupied by a permanent glial scar, which is composed of a dense network of neuroglial cells (astrocytes) and fluid-filled cystic spaces. Incorrect Answers: B, C, D, and E. Ependymal cells (Choice B) form the permeable lining of the ventricles that separates the CSF system and brain parenchyma. They do not migrate and are not involved in post-stroke histologic changes. Microglial cells (Choice C), derived from precursor monocytes, proliferate and phagocytose dead neurons days after an ischemic event and may remain in the region weeks after the initial injury. Microglial cells would not be present 6 months after the event. Neurons (Choice D) undergo liquefactive necrosis after a stroke. They are permanently differentiated cells, similar to cardiac myocytes, and are unable to re-enter the cell cycle for regeneration. Therefore, few neurons would be seen on histologic examination after 6 months. Oligodendrocytes (Choice E) are glial cells that produce myelin in the central nervous system. These cells are damaged in strokes. They slowly regenerate over months but would not be expected to multiply to the degree that scar-forming astrocytes proliferate. Educational Objective: After a ce bral infarction, neurons undergo liquefactive necrosis and are replaced by a glial scar consisting of proliferated astrocytes. e glial scar forms approximately 2 the event.

26 ---------- Exam Section 1: Item 27 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 27. A 32-year-old woman comes to the physician 1 day after she noticed a large red lesion on her back. One week ago, she returned from vacation in Cape Cod. After evaluation, a diagnosis of erythema chronicum migrans is made. Prior to prescribing an antibiotic, the physician asks the patient if she could possibly be pregnant because an adverse effect of the antibiotic active against the causal organism is incorporation into fetal bones and teeth, causing yellow discoloration. The most likely mechanism of this drug is inhibition of which of the following? O A) Attachment of tRNA to ribosomes B} Dihydrofelate reductase C) DNA repair by DNA gyrase Đ} Electron tranAspert E) Transpeptidation of peptidoglycan

A. Lyme disease is a tick-borne disease caused by Borrelia burgdorferi endemic in the Northeastern United States. In its early stages, it causes erythema chronicum migrans, which is an expanding rash that may have a central ring-like clearing. It can also be associated with a flu-like illness. In the early stages of Lyme disease, it can be treated with doxycycline, which acts by preventing the attachment of aminoacyl-tRNA to ribosomes. Side effects of tetracycline antibiotics include gastrointestinal distress, photosensitivity, and in children, discoloration of teeth and inhibition of bone growth, which makes it contraindicated in pregnancy. Later stages of Lyme disease can be characterized by carditis, atrioventricular heart block, facial nerve palsy (which is often bilateral), migratory arthralgias, and if left untreated, encephalopathy and chronic arthritis. Later stages of Lyme disease that affect the heart or central nervous system are generally treated with parenteral ceftriaxone. Incorrect Answers: B, C, D, and E. Dihydrofolate reductase (Choice B) is an enzyme used in the synthesis of purines and thymidine, precursors of nucleic acids. Inhibition of dihydrofolate reductase by trimethoprim prevents folate synthesis and is bacteriostatic. It is avoided in the first trimester of pregnancy but has no role in treating Lyme disease. DNA gyrase (Choice C) functions in DNA synthesis and repair by relaxing supercoils during replication and transcription. Fluoroquinolones inhibit prokaryotic DNA gyrase (also known as topoisomerase), side effects of which include cartilage damage, tendonitis, or tendon rupture. Inhibitors of electron transport (Choice D) such as antimycin A inhibit the proton gradient and prevent normal synthesis of ATP. Doxycycline does not inhibit electron transport. Transpeptidation of peptidoglycan (Choice E) provides strength to wall. Monobactams inhibit transpeptidation by inhibiting penicillin-binding protein 3. They have no effect on Borrelia burgdorferi, a spirochete. Educational Objective: Lyme disease is a tick-borne disease caused by Borrelia burgdorferi and characteristically causes erythema chronicum migrans, an expanding rash that may have a central ring-like clearing. Early Lyme disease is treated with doxycycline, which prevents the attachment of aminoacyl-TRNA to ribosomes. Doxycycline can cause gastrointestinal distress, photosensitivity, and in children, discoloration of teeth and inhibition of bone growth, which makes it contraindicated in pregnancy.

9. After being given an infusion of mannitol (400 mM), a healthy person is most likely to have which of the following changes in plasma osmolality and plasma ADH (vasopressin) concentration? Plasma osmolality (mOsmol/L) Control before infusion A) B) C) D) E)

A. Mannitol, an osmotically active carbohydrate monomer, raises serum osmolality when given as an infusion. Mannitol, when infused, is a hyperosmolar solution. Following infusion, the net serum osmolality will be increased as mannitol molecules remain within the serum. In turn, the increased osmolality of serum will draw fluid into the intravascular space from the interstitium. This principle is used therapeutically in the treatment of acute increases in intracranial and intraocular pressure, for example in cases of acute cerebral edema, impending brain herniation, or glaucoma. Osmoreceptors detect the change in osmolality and trigger the production of antidiuretic hormone (ADH) by the hypothalamus with the release by the posterior pituitary gland. ADH, in turn, increases the insertion of aquaporin channels in the collecting duct of the nephron. Free water is reclaimed by the aquaporins, which dilutes the serum and in turn normalizes osmolality. In this graph, person A has serum studies that reflect increased osmolality and increased production of ADH, which serum changes following the administration of mannitol. Incorrect Answers: B, C, D, and E. Choice B reflects increased osmolality without associated increased serum ADH. This set of values may be obtained in the immediate moments following a rapid bolus of mannitol, but ADH levels would quickly increase to compensate for increased serum osmolality. Choice C reflects increased osmolality and decreased ADH, which is not consistent with an appropriate physiologic response to increased osmolality. Choice D reflects decreased osmolality, not increased as would occur with the administration of mannitol. In the case of physiologically decreased serum osmolality, as in psychogenic polydipsia, low serum ADH would be expected as a normal physiologic response. Choice E reflects decreased serum osmolality and increased serum ADH, which would be consistent with a diagnosis of syndrome of inappropriate ADH, a frequent cause of euvolemic, hypoosmolar hyponatremia. Educational Objective: Mannitol, an osmotically active carbohydrate monomer, raises serum osmolality when given as an infusion. Osmoreceptors detect the change in osmolality and trigger the production of antidiuretic hormone by the hypothalamus.

16. A 9-year-old boy is brought to the physician because he told his teacher that his fingers felt "funny" after he swung on the monkey bars 2 hours ago. Neurologic examination shows numbness and tingling in the ring and small fingers of the left hand. The physician concludes that the ulnar nerve may have been stretched while he was hanging from the bars. Which of the following other functions should be checked to assess the integrity of the ulnar nerve in the left limb? A) Abduction of the index, middle, ring, and small fingers B) Abduction of the thumb C) Extension of the index, middle, ring, and small fingers D) Extension of the thumb E) Opposition of the thumb

A. Peripheral nerve injuries of the upper extremity are commonly caused by compression, direct trauma, or nerve traction. In this patient, hanging from the monkey bars placed tension on the nerves that originate from the lower brachial plexus. Roots C8 and T1 contribute to the ulnar nerve. Injury to this nerve is causing the patient's distal motor and sensory symptoms. The ulnar nerve innervates the intrinsic muscles of the hand (eg, lumbricals, interossei, hypothenar, and palmaris brevis), and muscles of the forearm (eg, flexor carpi ulnaris and the ulnar portion of flexor digitorum profundus). It also provides sensation to the volar aspects of the fourth and fifth digit. Injury to this nerve will cause weakness in abduction and adduction of the index, middle, ring, and small fingers, as the dorsal interossei and the palmar interossei control these movements, respectively. To isolate ulnar nerve motor function, the patient's abduction and adduction of the digits can be evaluated with resistive strength testing. Sensation can be tested along the volar aspect of the fourth and fifth fingers. Incorrect Answers: B, C, D, and E. Abduction of the thumb (Choice B) is primarily controlled by abductor pollicis brevis and abductor pollicis longus. Abductor pollicis brevis is innervated by the recurrent branch of the median nerve which can be injured during carpal tunnel surgery. The abductor pollicis longus is innervated by the posterior interosseous nerve, a branch of the radial nerve. This nerve can be injured in a humerus fracture. Extension of the index, middle, ring, and small fingers (Choice C) is provided by the extensor digitorum communis muscle, which is innervated by the radial nerve. Extension of the thumb (Choice D) is controlled primarily by extensor pollicis brevis and extensor pollicis longus, both of which are innervated by the posterior interosseous nerve. Opposition of the thumb (Choice E) is driven by opponens pollicis, which is innervated by the recurrent branch of the median nerve. Educational Objective: The ulnar nerve innervates the majority of the intrinsic muscles of the hand. It may be injured when traction is placed on the brachial plexus. Extensors of the digits and wrist, as well as the abductors of the thumb, are innervated by the radial nerve. The median nerve innervates the majority of the flexors of the wrist and fingers as well as some intrinsic muscles of the thenar eminence.

77 ---------- Exam Section 2: Item 27 of 50 National Board of Medical Examiners nent V 27. A 52-year-old man who is in the hospital with malignant hypertension, heart failure, and a mild renal impairment is being treated with captopril, nitroprusside, and furosemide. After 48 hours of therapy, his blood pressure is controlled, but he develops metabolic acidosis, lethargy, and difficulty breathing. Which of the following substances most likely caused these adverse effects in this patient? A) Cyanide O B) Iron O C) Lead O D) Nitric oxide E) Selenium

A. Sodium 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 guanosine monophosphate (CGMP) pathway. A byproduct of sodium nitroprusside is cyanide, and in rare cases, infusion of nitroprusside can cause cyanide toxicity. Cyanide poisoning occurs due to its action of inhibiting the electron transport chain, leading to metabolic lactic acidosis. Cyanide inhibits the mitochondrial enzyme cytochrome c oxidase, preventing the synthesis of adenosine triphosphate, leading to lactic acidosis. The presentation of cyanide poisoning is characterized by tachypnea due to compensatory respiratory drive, hypotension, confusion, lethargy, and bradycardia. The management of cyanide poisoning includes supportive care with intravenous fluids and vasopressors, sodium thiosulfate which increases the conversion of cyanide to thiocyanate for renal excretion, amyl nitrites to induce methemoglobinemia which binds with cyanide and prevents it from acting on the electron transport chain, and hydroxocobalamin which combines with cyanide to create nontoxic cyanocobalamin. Incorrect Answers: B, C, D, and E. Iron toxicity (Choice B) is characterized by gastrointestinal irritation, and later, anion gap metabolic acidosis, hepatic toxicity, acute kidney injury, coagulopathy, and fulminant hepatic failure. Due to the severity of toxicity, early management may include whole bowel irrigation with polyethylene glycol or gastric lavage. Lead toxicity (Choice C) can occur for example from workplace exposure, lead paint in a residence, or contaminated drinking water. Symptoms of lead exposure can be variable and nonspecific, including abdominal pain, constipation, anorexia, myalgias and arthralgias, fatigue, headache, difficulty concentrating, and microcytic anemia. Nitric oxide (Choice D) is the expected byproduct of sodium nitroprusside that leads to vascular smooth muscle relaxation and vasodilation, which is helpful in the management of hypertensive emergencies. It does not cause metabolic acidosis or difficulty breathing. Selenium (Choice E) is a naturally occurring mineral that is harmless at normal doses, but in acute or chronic excess, can cause nausea, vomiting, hair loss, fatigue, and irritability. It is not associated with metabolic acidosis or sodium nitroprusside administration. Educational Objective: Sodium nitroprusside breaks down to release nitric oxide, which leads to vascular smooth muscle relaxation and vasodilation via a CGMP pathway. A potential adverse byproduct of nitroprusside includes cyanide, which causes uncoupling of the electron transport chain, leading to lactic acidosis, compensatory respiratory alkalosis, hypotension, lethargy, and bradycardia.

50 ---------- Exam Section 2: Item 1 of 50 National, Board of Medical Examiners Comprehensive Basic Science Self-Assessment 1. A 35-year-old woman and her 35-year-old husband come to the physician for genetic counseling after their son is diagnosed with a rare metabolic disease. Assuming Hardy-Weinberg equilibrium, the physician tells the parents about the incidence and carrier frequency of this disorder. Which of the following is most likely to disturb the Hardy-Weinberg equilibrium of this disorder? A) Appreciable rate of gene mutation B) Random matings in the population C) Relatively large population D) No selection against a certain genotype E) No significant immigrant population

A. The Hardy-Weinberg principal, known as Hardy-Weinberg equilibrium, proposes that allele frequencies will remain constant across generations in the absence of evolutionary change. This model of population genetic stability requires a number of assumptions, including that natural selection does not occur, that there are no random mutations, that no movement in or out of the population occurs in equilibrium, and that mating is completely random. With these assumptions, the frequencies of alleles can be calculated as they will remain constant over time. If deviations from these assumptions occur, the allele frequencies within the population will change over time. The Hardy-Weinberg equilibrium equation is: p< + 2pq + q² = 1, where p is the frequency of the dominant allele and q is the frequency of the recessive allele. 2pq is the probability of heterozygosity, p- is the probability of dominant homozygote, and q- is the probability of being a recessive homozygote. If there is an appreciable rate of gene mutation occurring within the population, over time the allele frequency will change. ASsuming equivalent mutation rates in each gene, eventually all allele frequencies would change. Gene mutation prevents the stability or equilibrium of genotypes over time. %3D Incorrect Answers: B, C, D, and E. Random matings in the population (Choice B) are required for Hardy-Weinberg equilibrium. Nonrandom mating promotes imbalances in the equilibrium with resultant selection for or against certain alleles. A relatively large population (Choice C) is not necessary for Hardy-Weinberg equilibrium. However, small population sizes with small numbers of particular alleles are more subject to genetic drift. No selection against a certain genotype (Choice D) is required for Hardy-Weinberg equilibrium. If a certain gene is selected for, that is, the organism with a specific genotype is more likely to breed and pass along the allele, then the prevalence of that particular genotype will increase. Conversely, if a certain allele is selected against, the frequency of that allele will decrease over time. Immigration transfers genetic material from one population to another. An influx of new genetic material would disrupt the Hardy-Weinberg equilibrium. Conversely, no significant immigration (Choice E) into the population would promote Hardy-Weinberg equilibrium. Educational Objective: Hardy-Weinberg equilibrium models the frequency of alleles within a population from generation to generation, which considers alleles to be in stable equilibrium over time. Violation of its assumptions will cause deviation from this ideal behavior and variation in the prevalence of alleles over time.

4. A male newborn is found to have a defect in anchoring fibrils. Which of the following skin findings is most likely in this patient? A) Blisters B) Easy bruising C) Eczematous rash D) Inability to sweat E) Thickened skin F) Ulcers

A. The hemidesmosome is an intricate complex of proteins whose ultimate function is to anchor the basal keratinocytes of the epidermis to the dermis at the dermal-epidermal junction. Any impairment of the hemidesmosome will cause the basal keratinocytes to separate from the dermis, causing a blister to form. Because the hemidesmosomes of neighboring skin are still intact, these will be tense blisters. Anchoring fibrils are made of type VII collagen and are a component of the hemidesmosome. A mutation or antibody to collagen type VII, as seen in epidermolysis bullosa, will cause blistering to occur. Bullous pemphigoid is another disease that affects the hemidesmosome. In contrast, pemphigus Vulgaris is caused by antibodies to desmosomes, the protein complex which maintains cell to cell adhesion in the epidermis. Because the target of pemphigus Vulgaris is more superficial, in the epidermis, those blisters will be fragile and flaccid. Incorrect Answers: B, C, D, E, and F. Easy bruising (Choice B) and petechiae may be seen in Vitamin C deficiency or scurvy. Vitamin C is necessary to produce collagen, as it is a cofactor in the hydroxylation of proline and lysine residues, which is a key step in the conversion of preprocollagen to procollagen. Eczematous rashes (Choice C) may be seen in many genetic disorders including atopic dermatitis, autosomal dominant hyper-IgE syndrome (Job syndrome), and Wiskott-Aldrich syndrome. While acute eczematous reactions can have small vesicles due to edema within the epidermis, blisters are not seen. Itchy, erythematous patches are a classic finding. Inability to sweat (Choice D) is seen in a group of inherited disorders called ectodermal dysplasias. In these disorders, ectoderm-derived structures including the sweat glands, hair, teeth, and nails are abnormal. The anchoring fibrils are intact, and blisters are not typically seen. Thickened skin (Choice E) is a feature of a group of disorders called keratodermas, which can be inherited or acquired. The thick, scaly skin in these disorders is due to thickening of the stratum corneum, the top layer of the epidermis. The hemidesmosome is not affected and blisters are not seen. An ulcer (Choice F) is a defect in the epidermis that leaves the dermis exposed. Ulcers can be a feature of autoimmune diseases such as systemic lupus erythematosus and Behçet disease, however, they are not due to a defect in anchoring fibrils or the hemidesmosome. Educational Objective: A defect in any component of the hemidesmosome, including the anchoring fibrils, will result in blister formation. Epidermolysis bullosa is characterized by antibodies against or a mutation of the anchoring fibrils.

18. A 35-year-old man is brought to the emergency department because of multiple injuries sustained in a motor vehicle collision 45 minutes ago. He is lethargic and confused. The right pupil is sluggishly reactive to light and is 1 mm larger than the left pupil. There is mild left hemiparesis. Further examination shows a fracture of the right femur, ruptured spleen, and hemorrhagic pleural effusion. A CT scan of the head shows a subdural hematoma on the right with mass effect and midline shift. He is intubated and mechanically hyperventilated. Which of the following is the most likely rationale for hyperventilation in this patient? A) Cause cerebral vasoconstriction B) Decrease pulmonary atelectasis C) Increase cerebral blood flow Đ} Increase intracranial pressure E) Increase oxygenation

A. The intracranial compartment is a fixed compartment surrounded by a rigid skull containing brain parenchyma, cerebrospinal fluid, and intravascular blood. Pathologic lesions that take up space in the intracranial compartment such as tumors, abscesses, and extravascular blood such as subdural or epidural hematoma may displace normal intracranial structures, causing mass effect and midline shift. Cerebral blood flow (CBF) supplies intravascular blood in the intracranial space. Hypocapnia induces cerebral vasoconstriction, decreasing the cerebral blood flow and the volume of intravascular, intracranial blood. Therapeutic mechanical hyperventilation induces hypocapnia and has been shown to reduce intracranial pressure in the acute setting. This method for intracranial pressure reduction is not a long-term solution; rather, it is a temporizing measure prior to definitive treatment such as a craniotomy. Incorrect Answers: B, C, D, and E. Hyperventilation may decrease pulmonary atelectasis (Choice B), however, a decrease in pulmonary atelectasis does not correlate with increased ventilation and hypocapnia directly. Decreasing atelectasis may be accomplished through incentive spirometry, or in the case of a mechanically-ventilated patient, through increased positive end-expiratory pressure (PEEP). Increased cerebral blood flow (Choice C) and increased intracranial pressure (Choice D) would not be helpful in this patient with a subdural hematoma displacing intracranial structures. The intracranial compartment is rigid, and increased blood flow or intracranial pressure would cause further mass effects in this patient. Hyperventilation decreases cerebral blood flow and intracranial pressure through inducing cerebral vasoconstriction. Hyperventilation has little effect on increasing oxygenation (Choice E). The oxygenation of a mechanically ventilated patient may be influenced by the fraction of inspired oxygen (FiO2) or by PEEP. Educational Objective: The intracranial compartment is a fixed, rigid space, and any lesions that occupy space in it may displace normal brain parenchyma, causing midline shift and mass effect. Therapeutic hyperventilation and resulting hypocapnia cause cerebral vasoconstriction, which results in decreased cerebral blood flow and intracranial pressure.

52 ---------- Exam Section 2: Item 3 of 50 National, Board of Medical Examiners Comprehensive Basic Science Self-Assessment 3. A 68-year-old man comes to the physician's office because of the sudden onset of loss of vision in the left eye. There is no pain. He has a 15-year history of well controlled type 2 diabetes mellitus and hypertension. He had a myocardial infarction 2 years ago. Temperature is 36.7°C (98.1°F), pulse is 72/min, respirations are 16/min, and blood pressure is 200/100 mm Hg. Funduscopic examination shows a pale, white retina with retinal artery narrowing and decreased filling of the retinal veins. Which of the following is the most likely cause of the loss of vision? A) Malignant hypertension B) Retinal artery embolism C) Retinal detachment D) Retinal vein thrombosis E) Vitreous hemorrhage secondary to neovascularization

B. Central retinal artery occlusion (CRAO) presents with sudden, monocular vision loss, which is usually painless. Vision loss is typically profound, with severely reduced visual acuity, and occurs over seconds. There may be a history of preceding amaurosis fugax. Physical examination shows a relative afferent pupillary defect, a pale fundus with a cherry red spot, and narrowing of the retinal arterioles with arteriolar boxcarring. The most common cause of CRAO is a retinal artery embolism from a carotid artery atheroma or from a cardiac valvular vegetation. Embolic disease may also manifest with Hollenhorst plaques, which are smaller, refractile deposits that can be visualized at the bifurcations of retinal arterioles. CRAO is considered a form of embolic stroke affecting the eye, and suspicion of CRAO should therefore prompt an immediate stroke risk factor evaluation. ČRÃO may also occur secondary to giant cell arteritis. There is no proven effective treatment for CRAO and the prognosis for visual recovery is poor. Patients who have been affected by CRAO are at risk of secondary ocular neovascularization and glaucoma due to persistent retinal ischemia. Incorrect Answers: A, C, D and E. Malignant hypertension (Choice A) presents with optic disc edema, retinal arteriolar narrowing with arteriovenous nicking and silver wiring, retinal edema, cotton wool spots, and flame hemorrhages. It is typically a symmetric, bilateral process. Retinal detachment (Choice C) presents with painless, monocular vision loss but is typically preceded by photopsia and bursts of floaters. The detachment is identifiable on dilated fundus examination as a distinct region of retinal elevation and is not usually associated with retinal whitening, arteriolar narrowing, or reduced venous filling. Retinal vein thrombosis (Choice D), presents with painless, monocular vision loss, and retinal venous dilatation and tortuosity, retinal hemorrhages, and nerve fiber layer infarcts (eg, cotton wool spots) on fundoscopic examination. Retinal vein thrombosis may affect a single branch of the retinal veins, known as a branch retinal vein occlusion (BRVO) or may affect the central retinal vein (CRVO). Affected patients in both cases are at increased risk of subsequent neovascularization due to retinal ischemia. Vitreous hemorrhage secondary to neovascularization (Choice E) presents with painless, monocular vision loss; patients may report seeing floaters and visual haze. Examination generally reveals hemorrhage in the vitreous cavity or in the subhyaloid space. Although this patient has type 2 diabetes mellitus, which is a risk factor for neovascularization and vitreous hemorrhage, the contralateral eye will usually reveal a similar degree of retinopathy. Educational Objective: Central retinal artery occlusion (CRAO) presents with sudden, painless, monocular vision loss, relative afferent pupillary defect, retinal whitening with a cherry red spot, and retinal arteriolar narrowing with boxcarring. The most common cause of CRAO is a retinal artery embolism from a carotid artery atheroma or a cardiac valvular vegetation. CRAO is considered a form of embolic stroke affecting the eye, and suspicion of CRAO should therefore prompt an immediate stroke risk factor evaluation.

58 ---------- Exam Section 2: Item 9 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 9. A 45-year-old man comes to the physician because of progressively worsening, constant pain in his left thigh over the past 3 months. He is a long-distance runner. An x-ray of the femur shows thickening of the diaphysis and disruption of the cortex with focal areas of increased calcification. A glistening mass containing several cysts is surgically excised. A photomicrograph of tissue from the mass is shown. Which of the following is the most likely diagnosis? A) Atypical stress fracture B) Chondrosarcoma C) Enchondroma O D) Ewing sarcoma E) Giant cell carcinoma O F) Multilobulated bone cyst G) Osteosarcoma

B. Chondrosarcoma is a malignant neoplasm that arises from the mesenchymal cells that produce cartilage. It commonly occurs in the axial skeleton, particularly in the pelvis, and often in patients in the fifth decade of life. It is a slow-growing tumor that is initially asymptomatic, permitting extensive growth before detection. Pelvic chondrosarcoma will often present with pelvic pain and compression of adjacent structures such as the lumbosacral plexus, leading to neurologic symptoms in the ipsilateral lower extremity. Chondrosarcoma can be primary or arise secondary to

59 ---------- Exam Section 2: Item 9 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment me diapnysis ana disrupuon or tne cortex WILN Tocal areas onincreased caiciication. Agiistening mass containng severa cysts is surgicany excIsed. A pnotomicrograpn or ussue Trom the mass is shown. Which of the following is the most likely diagnosis? A) Atypical stress fracture B) Chondrosarcoma C) Enchondroma D) Ewing sarcoma E) Giant cell carcinoma F) Multilobulated bone cyst G) Osteosarcoma

B. Chondrosarcoma is a malignant neoplasm that arises from the mesenchymal cells that produce cartilage. It commonly occurs in the axial skeleton, particularly in the pelvis, and often in patients in the fifth decade of life. It is a slow-growing tumor that is initially asymptomatic, permitting extensive growth before detection. Pelvic chondrosarcoma will often present with pelvic pain and compression of adjacent structures such as the lumbosacral plexus, leading to neurologic symptoms in the ipsilateral lower extremity. Chondrosarcoma can be primary or arise secondary to

33 ---------- Exam Section 1: Item 34 of 50 National, Board of Medical Examiners Comprehensive Basic Science Self-Assessment 34. A 54-year-old woman with a 25-year history of rheumatoid arthritis that is refractory to therapy is admitted to the hospital for a joint replacement operation. Preoperative laboratory studies show: Hemoglobin Mean corpuscular volume Leukocyte count Platelet count 10.5 g/dL 79 μm3 8900/mm3 230,000/mm 3 Serum Iron Transferrin 40 μg/dL 220 mg/dL (N=200-400) A photomicrograph of a Prussian blue iron-stained smear of bone marrow aspirate is shown. Which of the following is the most likely cause of the anemia? A) Autoimmune hemolysis B) Chronic disease C) Iron deficiency O D) Marrow replacement by neoplastic tissue O E) Myelodysplasia O F) B-Thalassemia minor

B. Chronic disease (anemia of chronic disease [ACD]) accounts for this patient's anemia. ACD is found in many chronic inflammatory, infectious, and malignant disease states including rheumatoid arthritis, lymphoma, and tuberculosis. This patient's rheumatoid arthritis with microcytic anemia, normal iron levels, and bone marrow smear with Prussian blue-iron staining revealing increased iron storage are consistent with this diagnosis. Anemia is often categorized into macrocytic, normocytic, and microcytic etiologies. Other causes of microcytic anemia include thalassemia, iron deficiency anemia, and sideroblastic anemia. To determine the specific cause of microcytic anemia (which guides therapy), diagnostic evaluation is necessary and includes red blood cell indices (eg, red cell distribution width, mean cell volume, mean corpuscular hemoglobin concentration) and iron studies (eg, transferrin, serum iron level, total iron binding capacity, ferritin). In ACD, iron studies will show normal or low serum iron level, and normal or increased ferritin level. While ACD usually causes microcytosis, it can present with normal mean corpuscular volume when early. While not routinely performed, a bone marrow aspirate with Prussian blue staining for iron will reveal increased deposits of iron in the bone marrow, as shown in the photomicrograph. The pathophysiology may relate to upregulated hepcidin, resulting in reduced intestinal absorption of iron and reduced release of iron from macrophages. While patients may not be iron deficient, their iron stores are unavailable for hematopoiesis, leading to anemia. Incorrect Answers: A, C, D, E, and F. Autoimmune hemolysis (Choice A) occurs as a result of antibody production against circulating red blood cell surface antigens, leading to phagocytosis or activation of complement, both of which lead to destruction of the red cell in the spleen or intravascular space. Diagnosis is made by a positive direct antibody test (Coombs). Laboratory analysis will also reveal unconjugated hyperbilirubinemia and low haptoglobin. Iron deficiency anemia (Choice C) commonly results from occult blood loss through the gastrointestinal tract or via menstruation. Bone marrow biopsy, if performed, would demonstrate an absence of iron on Prussian blue staining. Serum ferritin level would be low. In men or post-menopausal women with iron deficiency anemia, sites of potential blood loss should be evaluated with endoscopy and colonoscopy to assess for gastrointestinal causes (eg, colorectal carcinoma), along with a pelvic examination in women to assess for abnormal uterine bleeding. Marrow replacement by neoplastic tissue (Choice D) would present with systemic signs of malignancy such as fever, weight loss, and malaise, and cytopenia on laboratory analysis as normal bone marrow is displaced by malignant cells. Bone marrow aspirate would reveal malignant cells rather than a normal Prussian blue stain. Myelodysplasia (Choice E) is frequently encountered in older individuals or in those who previously received cytotoxic chemotherapy. Findings on bone marrow aspirate include abnormal appearing cells of multiple lineages. This does not fit with the image shown or the patient's presentation or marrow biopsy. B-Thalassemia minor (Choice F) is caused by mutation of one of the two paired B-globin chains. Patients will have an abnormal hemoglobin electrophoresis but are typically asymptomatic or only mildly symptomatic. Peripheral smear demonstrates microcytosis

34 ---------- Exam Section 1: Item 35 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 35. A newborn is delivered at 40 weeks' gestation to a 20-year-old woman after an uncomplicated pregnancy. Physical examination shows male-like external genitalia, with clitoral hypertrophy and labial fusion. The karyotype is 46,XX. The serum concentration of cortisol is decreased and serum concentrations of 17-hydroxyprogesterone, androstenedione, and testosterone are increased. Which of the following is the most likely cause of the findings in this newborn? AJAndrogen receptor mutation B) Defect in adrenal steroid biosynthesis 67 Deficiency ofenzyme activity in-gonadal steroidhermone biosynthesis Đ} Increased anti-müllerian hormone secretion E) Translocation of the SRY gene to one of the X chromosomes

B. Congenital adrenal hyperplasia (CAH) is a defect in adrenal steroid biosynthesis, particularly of cortisol, with variable effects on mineralocorticoids and androgens. The most common form of CAH is 21-hydroxylase deficiency, which presents with salt wasting in a newborn or precocious puberty in a child, as well as virilization of infants with an XX karyotype (eg, clitoral hypertrophy, labial fusion), due to increased levels of androgens, as seen in this patient. Due to the enzyme deficiency, levels of 17-hydroxyprogesterone, androstenedione, and testosterone are increased, while aldosterone and cortisol are decreased, leading to hypotension and hyperkalemia. Treatment consists of exogenous glucocorticoid and mineralocorticoid supplementation. Incorrect Answers: A, C, D, and E. Androgen receptor mutations (Choice A) would lead to androgen insensitivity in genotypic males, which presents as normal-appearing female external genitalia, decreased or absent body hair, and lack of a uterus and ovaries. Patients will have functioning testes and testosterone levels will be increased. However, as this biological female is presenting with virilization, androgen receptor mutations are an unlikely diagnosis. Deficiency of enzyme activity in gonadal steroid hormone biosynthesis (Choice C) would lead to decreased levels of estradiol, progesterone, and testosterone, which is not seen in this patient. Increased anti-müllerian hormone secretion (Choice D) or müllerian inhibitory factor would lead to the absence of development of the female internal reproductive organs in this patient but would not cause abnormalities in cortisol levels. Translocation of the SRY gene to one of the X chromosomes (Choice E) would cause a fetus with an XX karyotype to develop as a biologic male, including external genitalia and testes. Hormone levels would be normal, making this an unlikely diagnosis in this patient. Educational Objective: Congenital adrenal hyperplasia results from a defect in adrenal steroid biosynthesis. Its most common form, 21-hydroxylase deficiency, presents with salt-wasting in infants or precocious puberty in children. It also causes the virilization of XX fetuses, elevated levels of 17-hydroxyprogesterone, hypotension, and hyperkalemia. Treatment includes the exogenous administration of mineralocorticoids and glucocorticoids.

82 ---------- Exam Section 2: Item 32 of 50 Natjonal, Board of Medical Examinersment asive Basic Science 32. A 52-year-old man with chronic renal failure receives a cadaveric kidney transplant. Postoperatively, he is given cyclosporine for immunosuppressive therapy. Six weeks after the operation, he develops hypoxemia with an arterial Po2 of 40 mm Hg. A photomicrograph of a transbronchial biopsy specimen is shown. Which of the following pathophysiologic mechanisms best explains this patient's hypoxemia? A) Alveolar exudation due to Cryptococcus neoformans infection O B) Cytomegalovirus pneumonitis with diffuse alveolar damage O C) Neutrophilic alveolar consolidation due to pneumococcal pneumonia O D) Pneumocystis jirovecii (formerly P. carinii) infection with alveolar exudation E) Vascular invasion by Aspergillus with pulmonary infarction

B. Cytomegalovirus (CMV), also known as human herpesvirus-5 (HHV-5), is an opportunistic infection that commonly occurs in immunocompromised patients in the setting of solid-organ or allogeneic stem cell transplantation, severe ulcerative colitis, or HIV/AIDS. CMV pneumonitis with diffuse alveolar damage is one of the many presentations that can occur in patients taking immunosuppressants to prevent organ rejection following solid organ transplant. It typically occurs in the first 3 to 6 months following transplantation. Radiologic imaging features are nonspecific but may demonstrate an interstitial pneumonitis pattern with the potential for infiltrative consolidations and ground-glass opacities to develop. The classic histologic findings of CMV pneumonitis are infected cells with prominent basophilic nuclear inclusions, as seen in this patient's lung biopsy. Other manifestations of CMV infection associated with immunosuppression include systemic infectious mononucleosis, retinitis, esophagitis, colitis, and encephalitis. Incorrect Answers: A, C, D, and E. Alveolar exudation due to Cryptococcus neoformans infection (Choice A) may be seen in pulmonary cryptococcosis. Cryptococcus neoformans is an encapsulated yeast and opportunistic pathogen in immunocompromised patients, especially patients with HIV/AIDS with low CD4 counts. Disseminated disease can also cause cryptococcal meningitis and encephalitis. Histology reveals fungal cells with narrow-based budding and a bright red inner capsule on staining with mucicarmine. Neutrophilic alveolar consolidation due to pneumococcal pneumonia (Choice C) is seen in lung infections due to Streptococcus pneumoniae. Pneumonia classically presents with fever, dyspnea, productive cough, and pleuritic chest pain. Gram-positive diplococci would be seen on sputum cultures or bronchoalveolar lavage. Pneumocystis jirovecii (formerly P. carinii) infection with alveolar exudation (Choice D) is present in pneumocystis pneumonia. Chest radiologic imaging typically reveals diffuse, bilateral infiltrates often occurring prominently about the bilateral hila. Methenamine silver or toluidine blue stains selectively stain the cyst walls and are used to confirm the diagnosis. Vascular invasion by Aspergillus with pulmonary infarction (Choice E) is a feature of angioinvasive aspergillosis. Immunosuppression is a risk factor for this disease. Patients present with persistent fever, cough, hemoptysis, and chest pain. Vascular invasion can result in distal pulmonary infarctions. Histologic findings include invasive hyphae in the lung parenchyma with extension into the vasculature. Educational Objective: Solid-organ transplantation and immunosuppression to prevent organ rejection is frequently complicated by opportunistic pathogens. CMV is a common cause of disease in these patients and can present with pneumonitis with diffuse alveolar damage, colitis, retinitis, esophagitis, and encephalitis. Infected cells typically demonstrate a large, atypical appearance and feature the presence of basophilic nuclear inclusions.

25 ---------- Exam Section 1: Item 26 of 50 National, Board of Medical Examiners Comprehensive Basic Science Self-Assessment 26. A health care screening for diabetes mellitus is conducted at a weekend market with subjects including both men and women ranging in age from 25 to 95 years. Blood samples are obtained, and the blood glucose concentrations are determined. The results are shown in the table: Patient Glucose (mg/dL) 70 120 120 140 110 95 75 145 105 10 85 85 12 90 120 100 190 13 14 15 Which of the following best represents the median blood glucose concentration for this group (in mg/dL)? A) 95 B) 105 C) 110 D) 120 E) 125

B. Measures of central tendency within distributions of data include the mean, median, and mode. The median is the middle value of listed data points ordered from least to greatest. If the dataset contains an even number of values, the median is considered the mean of the middle two values. The mode is the most frequently occurring value within a dataset. The mean is calculated by the sum of the values, divided by the total number of values, and is the most affected by the presence of extreme or outlier data points. In this case, ordering the fifteen values of patient blood glucose (in mg/dL) in order yields the series 70, 75, 85, 85, 90, 95, 100, 105, 110, 120, 120, 120, 140, 145, 190. The middle value in the set is 105 mg/dL, which is the median of this dataset. Incorrect Answers: A, C, D, and E. 95 mg/dL (Choice A) and 125 mg/dL (Choice E), are neither the mean, median, nor mode of the dataset. 110 mg/dL (Choice C), is the mean of the dataset and is computed by the sum of all data points divided by the total number of data points. The sum is equal to 1,650 mg/dL. Dividing by 15 total data points yields a mean of 110 mg/dL. 120 mg/dL (Choice D), is the most frequently occurring value in the set, and is therefore the mode. Educational Objective: Measures of central tendency within distributions of data include the mean, median, and mode. The median is the middle value of listed data points ordered from least to greatest. If the dataset contains an even number of values, the median is considered the mean of the middle two values. 123 4 567∞ o =N345

51 ---------- Exam Section 2: Item 2 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 2. A40-year-old man has moderate edema. A 24-hour urine collection contains 15 g of protein (NS150 mg/24 h). Which of the following is the most likely mechanism of the edema? A) Decreased capillary hydrostatic pressure O B) Decreased plasma oncotic pressure C) Increased plasma oncotic pressure D) Decreased permeability of postcapillary venules E) Increased permeability of postcapillary venules

B. Flow into the interstitial space is dependent on interactions between capillary hydrostatic pressure, interstitial fluid hydrostatic pressure, plasma colloid oncotic pressure, and interstitial fluid oncotic pressure. If capillary hydrostatic pressure is high relative to interstitial fluid hydrostatic pressure, fluid will flow down a pressure gradient into the interstitial space. If the opposite is true (interstitial fluid pressure is higher than capillary pressure), fluid will flow into the capillaries. If plasma fluid oncotic pressure is high relative to interstitial fluid oncotic pressure, fluid will be drawn back into the capillaries. Conversely, if interstitial fluid oncotic pressure exceeds plasma fluid oncotic pressure, fluid will move into the interstitial space. Typically, excess fluid in the interstitial space is removed by the lymphatic system. However, when this system is overwhelmed, edema develops. Therefore, this patient has developed peripheral edema due to decreased plasma oncotic pressure secondary to the loss of protein through proteinuria. Incorrect Answers: A, C, D, and E. Decreased capillary hydrostatic pressure (Choice A) and increased plasma oncotic pressure (Choice C) would draw fluid into the capillaries from the interstitial space, decreasing edema. Decreased or increased permeability of postcapillary venules (Choices D and E) would affect edema, as fluid would be prevented from moving into or out of the interstitial space despite the influence of plasma or interstitial fluid hydrostatic and oncotic pressures. Any influence of venule permeability is a less likely underlying mechanism given this patient's profound proteinuria and subsequent decreased plasma oncotic pressure. Educational Objective: Movement of fluid between capillaries and the interstitial space is dependent on the hydrostatic and oncotic pressures of both compartments. High capillary hydrostatic pressure and low plasma oncotic pressure will increase the movement of fluid out of the capillaries and into the interstitial space. Conversely, high interstitial fluid hydrostatic pressure and low interstitial fluid oncotic pressure will drive fluid back into the capillaries and away from the interstitial space.

2. Free purine and pyrimidine bases are reutilized in normal metabolism. In children with Lesch-Nyhan syndrome who have intellectual disability, poor muscle coordination, and self-mutilation tendencies, there is a defect in the salvage of which of the following pairs of bases? A) Adenine and thymine B) Guanine and hypoxanthine C) Guanine and uric acid D) Uracil and cytosine E) Xanthine and hypoxanthine

B. Lesch-Nyhan syndrome presents with intellectual disability, aggressive behavior, self-mutilation, gout, and dystonia. The disorder is due to inactivating mutations of hypoxanthine-guanine phosphoribosyltransferase (HGPRT), a key enzyme in the purine salvage pathway, and is inherited in an X-linked recessive fashion. HGPRT catalyzes the conversion of guanine to guanosine monophosphate and hypoxanthine to inosine monophosphate. Patients with deficient activity of HGPRT are unable to salvage guanine and hypoxanthine and develop resultant increased levels of xanthine and uric acid. Hyperuricemia in Lesch-Nyhan syndrome is treated with xanthine oxidase inhibitors, such as allopurinol or febuxostat, in order to reduce the synthesis of uric acid. Incorrect Answers: A, C, D, and E. Adenine and thymine (Choice A) are purine and pyrimidine bases, respectively. Purine and pyrimidine salvage are handled through two distinct pathways that are not commonly involved in a single disease process. Guanine and uric acid (Choice C) accumulation may occur as part of Lesch-Nyhan syndrome, however, the accumulation of uric acid is also secondary to the accumulation of hypoxanthine. Choice B more accurately describes defective salvage of guanine and hypoxanthine as the fundamental effect of HGPRT dysfunction. The accumulation of uric acid is secondary. Uracil and cytosine (Choice D) are pyrimidine nucleotides. Pyrimidine salvage is not affected by mutations of HGPRT. Defects of xanthine and hypoxanthine (Choice E) metabolism may result from defects in HGPRT. However, HGPRT dysfunction results in impaired hypoxanthine salvage with resultant excessive production of xanthine, rather than impaired xanthine salvage. Educational Objective: Lesch-Nyhan syndrome presents with intellectual disability, aggressive behavior, self-mutilation, gout, and dystonia. Mutations in HGPRT cause this disorder by impairing salvage of the purines guanine and hypoxanthine, which leads to increased levels of xanthine and uric acid. Hyperuricemia in Lesch-Nyhan syndrome is treated with xanthine oxidase inhibitors, such as allopurinol or febuxostat, in order to reduce synthesis of uric acid.

49 ---------- Exam Section 1: Item 50 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 50. An investigator conducts a phase 1 clinical trial to test the efficacy of a systemic antagonist to lymphocyte function-associated antigen 1 (LFA-1). The drug is approved for topical administration to treat keratoconjunctivitis sicca in patients with multiple sclerosis. During the trial, it is found that these patients are at an increased risk for bacterial infection as a result of this drug. The most likely explanation for this risk is drug-induced blockade of which of the following segmented neutrophil activities? A) Activation of phagocytosis B) Adhesion to the endothelium C) Diapedesis through the capillary wall D) Migration to the site of injury E) Rolling at the vessel periphery

B. Neutrophil recruitment to sites of infection is an essential component of the innate immune response. Neutrophils directly eliminate pathogens by generation of reactive oxygen species and phagocytosis, recruitment of macrophages, and creation of a pro- inflammatory environment. The stages of neutrophil recruitment are margination, rolling, adhesion, extravasation, and chemotaxis/migration. Lymphocyte function-associated antigen 1 (LFA-1) is an integrin protein involved in neutrophil adhesion to the endothelium. LFA-1 antagonism leads to decreased neutrophil adhesion and an impaired innate immune response. The genetic disorder leukocyte adhesion deficiency type 1 is caused by a defect in LFA-1. Incorrect Answers: A, C, D, and E. Activation of phagocytosis (Choice A) is achieved by numerous inflammatory mediators, including bacterial proteins such as lipopolysaccharides, peptidoglycan, and teichoic acids, as well as complement proteins, inflammatory cytokines, and prostaglandins. Integrins are not involved in this process. Diapedesis through the capillary wall (Choice C), also called extravasation, is the process by which neutrophils move from the vascular compartment into the extracellular space. It is mediated by the integrin platelet/endothelial cell adhesion molecule 1 (PECAM-1), not LFA-1. Migration to the site of injury (Choice D), also called chemotaxis, is the movement of inflammatory cells through the extracellular space to the site of injury. It is mediated by numerous chemotactic agents, including IL-8, C5a, leukotriene B4, 5-HETE, and formyl- methionyl peptides. Rolling at the vessel periphery (Choice E) occurs following margination of the neutrophils to the periphery of the blood vessel and prior to neutrophil adhesion to the endothelial wall. It is mediated by selectins and glycoproteins (eg, Sialyl-Lewis X antigen). Educational Objective: The recruitment of neutrophils from the peripheral circulation to sites of tissue injury and infection involves the distinct stages of margination, rolling, adhesion, extravasation, and chemotaxis/migration. Lymphocyte function-associated antigen 1 (LFA-1) is involved in the adhesion of neutrophils to the endothelium.

13. A 64-year-old woman develops pain and stiffness of the proximal interphalangeal joints and the right knee. The pain is made worse by activity and is relieved by rest. X-rays of the knee show a narrowed joint space with radiodense subchondral bone and cyst formation. Knee fluid aspirate shows: Appearance Leukocyte count Neutrophils Glucose clear, yellowish 250/mm3 (N<200) 5% (N<25%) 101 mg/dL (N=80-100) Which of the following is the most likely cause of her condition? A) Acute gouty arthritis B) Osteoarthritis C) Rheumatoid arthritis D) Septic arthritis E) Normal age-related changes in joints

B. Osteoarthritis is the most common disease of joints in humans. It affects 60% to 80% of adults over the age of 65. It is characterized by global degeneration of a joint with fraying, fibrillation, and degradation of cartilage, thickening of the subchondral plate with sclerotic bone, and subchondral cysts due to infiltration of synovial fluid into the bone. On radiographs, the disease is demonstrated by joint space narrowing, articular sclerosis, radiolucent subchondral cysts, and sharpened corners of the joints with new bony protrusions (osteophytosis). If occurring before age 65 years, patients may have a history of prior traumatic injury to a joint that incited an inflammatory cascade and accelerated the onset of osteoarthritis. This presentation is known as posttraumatic arthritis. Joint aspiration is characterized by normal-appearing aspirate on Gram stain and culture. However, this synovial fluid typically contains cartilage and bone degradation particles as well as inflammatory cytokines and metalloproteases that are not detectable on traditional aspiration. Osteoarthritis may occur concomitantly with inflammatory or septic arthritis. The diagnosis of osteoarthritis does not preclude the existence or occurrence of more aggressive arthropathies. Incorrect Answers: A, C, D, and E. Acute gouty arthritis (Choice A) is an inflammatory arthropathy incited by hyperuricemia, with the deposition of uric acid crystals inside the joint and surrounding tissues. This type of arthritis presents following consumption of a purine-rich meal or alcohol (which competes with renal excretion of uric acid). A knee aspirate would show negatively birefringent needle-shaped crystals and a leukocyte count of typically less than 50,000 cells/mm°. Rheumatoid arthritis (Choice C) is a polyarticular inflammatory arthritis that is characterized by increased serum levels of rheumatoid factors. It most commonly affects the metacarpophalangeal and proximal interphalangeal joints and improves with use during the day. Increased serum antibodies against cyclic citrullinated peptides are specific for rheumatoid arthritis. Septic arthritis (Choice D) is most commonly bacterial and presents with acute joint swelling, erythema, pain, and tenderness with a range of motion. Aspirate typically reveals bacteria on Gram stain, most often Gram-positive organisms, and leukocytes of greater than 50,000 cells/mm3 with a neutrophilic predominance. Normal age-related changes in joints (Choice E) include increased stiffness and decreased strength and water content of cartilage. The thickness of the subchondral bone plate decreases as does the capacity of joint tissues to heal and regrow. Normal joint aging does not cause pain or joint space narrowing. Educational Objective: Osteoarthritis is the most common disease of joints in humans. It affects 60% to 80% of adults over the age of 65. It is characterized by global degeneration of a joint with fraying, fibrillation, and degradation of cartilage, thickening of the subchondral plate with sclerotic bone, and subchondral cysts due to infiltration of synovial fluid into the bone. Arthrocentesis typically demonstrates few leukocytes, normal appearance, and normal glucose.

76 ---------- Exam Section 2: Item 26 of 50 National Board of Medical Examiners ansive Basic Saance amont Y 26. A 32-year-old woman comes to the physician because of a 3-month history of headaches. Two weeks ago, she moved to an area women's shelter because her husband had been hitting her. She takes no medications. Physical examination shows multiple bruises in different stages of healing over the face, neck, and trunk. There is circumferential bruising on both sides of the neck. Which of the following statements is most appropriate to gather pertinent information from this patient? O A) "Did you bring your records from the emergency room with you?" B) "I see some bruising on your neck. How did that happen?" C) "Please demonstrate to me how this neck bruising occurred." %3D O D) "This looks like you were choked. Is that true?" E) "What sort of device did your husband use on your neck?"

B. Physicians should ask direct, open-ended questions about physical examination findings indicative of abuse. Physicians should start by summarizing their objective observations and end with a question about the findings. To protect patients' well-being, physicians have an obligation to investigate evidence of abuse. Additionally, abuse may cause physical or psychological injury that may require diagnosis and treatment. Physicians should treat the immediate symptoms, which may manifest as medical or psychiatric symptoms, and continue to provide regular care to monitor for long-term consequences of abuse. Physicians are mandated by many states to report abuse of children and vulnerable adults to protective agencies. Incorrect Answers: A, C, D, and E. Asking about emergency room records (Choice A) would avoid the topic of ongoing abuse and be unlikely to inform the physician about the patient's current social situation. Physicians should directly ask about concerning physical examination findings. Physicians should not ask patients to reenact possible abuse (Choice C), which may retraumatize patients. Patients should be invited to discuss abuse using open-ended questions instead. The physician should not assume that the bruises came from the husband's abuse (Choices D and E) or that the patient is willing to discuss this sensitive topic. The patient may feel intruded upon or defensive, which would preclude effective discussion. Asking an open-ended question based on an objective observation would likely increase the patient's comfort level with a discussion about the abuse. Educational Objective: Physicians should ask direct, open-ended questions about physical examination findings indicative of abuse without verbalizing assumptions about the etiology of the examination findings.

70 ---------- ivA Medical Examinersment Exam Section 2: Item 20 of 50 National, Board 20. A 55-year-old woman comes to the physician because of a 3-year history of weakness, fatigue, decreased appetite, and constipation. She says that she did not receive medical treatment for these symptoms because she did not have health insurance until recently. She was admitted to the hospital 1 year ago because of renal calculi, but she could not afford to pay for follow-up care. X-rays at that time were indicative of osteitis fibrosa cystica. Vital signs are within normal limits. Physical examination shows bony tenderness and muscle strength of 4/5 with decreased tone. Serum studies show an increased calcium concentration and a decreased phosphorus concentration. Which of the following is the most likely diagnosis? O A) Adult T-lymphocyte leukemia B) Chief cell adenoma of the parathyroid gland O C) Familial (benign) hypocalciuric hypercalcemia O D) Medullary carcinoma of the thyroid gland O E) Multiple myeloma

B. Primary hyperparathyroidism results in hypercalcemia and hypophosphatemia secondary to the aberrant increased production of parathyroid hormone (PTH). The most common cause of primary hyperparathyroidism is a chief cell adenoma of the parathyroid gland. PTH is produced by parathyroid chief cells and stimulates 1a-hydro phosphate. PTH also plays a role in the regulation of calcium and phosphate by stimulating osteoclastic bone reabsorption and distal convoluted tubular calcium reabsorption and phosphate excretion in the kidney. Common symptoms of primary hyperparathyroidism are secondary to hypercalcemia and include recurrent nephrolithiasis, bone pain from osseous resorption (osteitis fibrosa cystica), polyuria, dehydration, constipation, and psychiatric disturbances. kylase in the kidn which leads to the conversion of 25-hydroxycholecalciferol to 1,25-dihydroxycholecalciferol (active vitamin D). Vitamin Ď acts to increase the intestinal sorpt of calcium and Incorrect Answers: A, C, D, and E. Adult T-lymphocyte leukemia (Choice A) is a rare neoplasm of T lymphocytes that results as a sequela of infection with human T-lymphotropic virus type 1 (HTLV-1). While this disease produces lytic bone lesions and hypercalcemia, it is also accompanied by other findings including generalized lymphadenopathy, hepatosplenomegaly, and immunodeficiency. Familial (benign) hypocalciuric hypercalcemia (Choice C) arises from an abnormal calcium sensing receptor on parathyroid cells. As a result, PTH is not suppressed by normal or increased serum calcium. This leads to a mild hypercalcemia and decrease urine calcium concentration in the setting of increased PTH due to the normal mechanism of action of PTH on osteoclasts and renal tubules. Symptomatic hypercalcemia and renal calculi are uncommon in this condition. Medullary carcinoma of the thyroid gland (Choice D) can result in the excessive production of calcitonin by the parafollicular cells. Symptoms of hypocalcemia, rather than hypercalcemia, may be present. Medullary thyroid carcinoma is associated with multiple endocrine neoplasia (MEN) types 2Á and 2B. Multiple myeloma (Choice E) is a plasma cell neoplasia that presents with multiple lytic bone lesions in the skull, pelvis, or vertebrae, and may present with hypercalcemia secondary to bone destruction. Symptoms also include anemia, renal failure, frequent infections, and diffuse skeletal pain. The absence of characteristic radiographic findings makes this diagnosis unlikely. Educational Objective: Parathyroid chief cell adenoma is the most common cause of primary hyperparathyroidism, and presents with symptoms of hypercalcemia, including recurrent nephrolithiasis, bone pain from osseous resorption, polyuria, dehydration, constipation, and psychiatric disturbances.

14. Which of the following labeled depolarizations is associated with the smallest stroke volume? A) В) C) D) E)

B. The electrical complex labeled B in the rhythm strip is characteristic of a premature ventricular contraction (PVC) and is associated with a lower stroke volume when compared to a normal sinus beat. Normal cardiac conduction starts with an electrical impulse in the sinoatrial (SA) node located in the right atrium, which is depicted on the rhythm strip as the "P wave." The electrical impulse is conducted throughout the atria, causing coordinated contraction of the right and left atria, with a consequent active filling of the right and left ventricles. The electrical impulse also reaches the atrioventricular (AV) node located in the interatrial septum. Conduction is temporarily halted in the AV node to allow for complete active and passive filling of the ventricles, followed by conduction of the electrical impulse down the His-Purkinje system in the interventricular septum, then to the right and left bundle, and finally to the myocardial cells of the ventricles. This results in contraction of the right and left ventricles (QRS complex) with the expulsion of blood into the pulmonary and systemic circulations, respectively. The ventricles subsequently repolarize (T wave) and relax to allow filling for the next cardiac cycle. Stroke volume is defined as the difference between the end-diastolic-volume and the end-systolic-volume. To achieve an ideal stroke volume, diastolic filling time must be sufficiently long to allow the ventricles to fill completely before contraction occurs. This is achieved by delaying action potential conduction in the AV node. A premature ventricular contraction, which is shown by the letter "B," indicates premature ventricular systole. When this occurs, diastolic filling time is reduced, so the amount of blood in the ventricles prior to ventricular systole is less than would exist had filling time been sufficient. This results in a reduced stroke volume. Incorrect Answers: A, C, D, and E. Each of these answers (Choices A, C, D, and E) depicts normal sinus beats. There is a P-wave followed by a QRS complex and a T wave. Only Choice A lacks a T wave because of the superimposed, subsequent PVC. Normal cardiac conduction allows for adequate filling of the ventricles before systole. Educational Objective: Stroke volume is the difference between the ventricular end-diastolic volume and the end-systolic volume. PVCS shortens diastolic filling times leading to lower diastolic volumes with a consequent reduction in stroke volume during ventricular systole.

63 ---------- Exam Section 2: Item 13 of 50 National Board of Medical Examiners aiveB A lance S ment * 13. A 32-year-old woman comes to the physician because she is concerned about fertility. Menses occur at regular 28-day intervals. Physical examination shows no abnormalities. An endometrial biopsy specimen is obtained from the uterine wall. Histologic sections show a well-developed stratum basale, intact spiral arteries, abundant endometrial stroma, and long, straight uterine glands with no secretions in their lumens. Based on this description, which of the following labeled options best corresponds to the point in the patient's menstrual cycle at which the biopsy was done? A B Menses Ovulation O A) B) C) O D)

B. The first half of the menstrual cycle, the follicular phase, begins with menses. After menses, follicle-stimulating hormone and luteinizing hormone (FSH and LH, respectively) levels begin to increase, which stimulates the development of a follicle. The follicle produces estrogen, which leads to the proliferation of the endometrium. This proliferative endometrium consists of highly active endometrial epithelial cells, stromal cells, straight tubular endometrial glands, and extension of the uterine spiral arteries, as seen in this patient. As estrogen rises, a surge occurs, which in turn stimulates a surge in LH that causes ovulation 14 days prior to menses. Immediately following ovulation, the luteal (secretory) phase begins as the corpus luteum forms. The corpus luteum secretes progesterone to produce and maintain a secretory endometrial lining, which consists of tortuous glands full of secretions, full length spiral arteries, and edematous stromal cells. If no implantation occurs, the corpus luteum degrades to the corpus albicans, causing estrogen and progesterone levels to decrease, resulting in menstruation. The loss of progesterone causes the endometrial lining to slough off due to constriction of the spiral arteries, and the follicular phase begins again. Incorrect Answers: A, C, and D. Immediately after menses (Choice A), the endometrium would be much less developed, as the entirety of the stratum functionalis (eg, the superficial layer) would have been sloughed off during menses. During menses, the spiral arteries retract into the stratum basalis and the glands are partially lost. Well-developed spiral arteries and uterine glands, as well as abundant stroma, would not be expected immediately after menses. The secretory phase (Choices C and D) consists of the time from ovulation to menstruation, which is typically 14 days in length. The endometrium undergoes differentiation during this time secondary to progesterone stimulation, with increases in the tortuosity of the uterine glands and amou proliferative phase, prior to ovulation. of secretions, further development of the stroma cell layer, and extension of spiral art to cover the entire thicknes the ometrium. The histologic findings in this biopsy are more consistent with those found in the late Educational Objective: Proliferative endometrium is present during the follicular phase of menstruation prior to ovulation. Histologically, it demonstrates straight uterine glands, extension of the spiral arteries, and abundant stromal cells.

28 ---------- Exam Section 1: Item 29 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 29. A 40-year-old woman comes to the physician for a follow-up examination 1 week after having a second increased blood pressure measurement. Her pulse is 72/min, and blood pressure is 168/104 mm Hg in both arms while sitting. Physical examination shows no other abnormalities. Treatment with lisinopril and hydrochlorothiazide is initiated. One month later, her blood pressure is 120/80 mm Hg. Which of the following sets of hormonal changes is most likely in this patient? Angiotensin I Angiotensin II Bradykinin A) ↑ no change B) ↑ ↑ C) D) ↑ ↑ E)

B. Thiazide diuretics, such as hydrochlorothiazide (HCTZ), and ACE inhibitors, such as lisinopril, are first-line therapies for the treatment of hypertension. ACE inhibitors target the renin-angiotensin-aldosterone system which is one of the primary regulators of blood pressure. The system is activated by the release of renin from the renal juxtaglomerular apparatus in response to low-flow states in the afferent arterioles or through the distal convoluted tubule (decreased afferent arteriole pressure and decreased delivery of Na+ and Cl- ions to the macula densa both stimulate renin release). Renin acts on angiotensinogen via proteolytic cleavage to generate angiotensin I. Angiotensin I is then converted to angiotensin II by ACE, which is found in vascular endothelium (particularly in the lungs). Angiotensin II has several important effects, including potent vasoconstriction leading to increased systemic vascular resistance, increased Na+ reabsorption in the renal tubules, aldosterone release by the adrenal cortex leading to increased Na+ and fluid retention, vasopressin release from the posterior pituitary leading to increased fluid retention, enhanced adrenergic function through inhibition of norepinephrine reuptake in sympathetic nerve endings, and stimulation of hypothalamic thirst centers. All of these effects serve to increase blood pressure. ACE inhibitors decrease blood pressure by blocking the conversion of angiotensin I to angiotensin II. This results in increased angiotensin I, decreased angiotensin II, and increased bradykinin levels, as ACE is also responsible for the degradation of bradykinin (a potent vasodilator). Thiazide diuretics act on the distal convoluted tubule to decrease Na+ and Cl- reabsorption and do not directly affect the renin-angiotensin-aldosterone pathway. Incorrect Answers: A, C, D, and E. Choices A and C are incorrect as inhibition of ACE affects the enzymatic degradation of bradykinin. Bradykinin levels are increased with use of ACE inhibitors. In addition to vasodilation, bradykinin is also a protussive agent, and persistent cough is one of the main reasons ACE inhibitors are discontinued. Angioedema is also a common occurrence with ACE inhibitors due to the effects of bradykinin and can be life-threatening when the airway is involved. Choices D and E are incorrect as levels of the ACE substrate angiotensin I increases when the enzyme is inhibited, while levels of the enzymatic product angiotensin II decrease. Educational Objective: ACE inhibitors block the conversion of angiotensin I to angiotensin II, with the downstream effect of decreasing blood pressure. ACE is also involved in degrading bradykinin.

37 ---------- Exam Section 1: Item 38 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 38. A 55-year-old woman who is a physician is admitted to the hospital because of a fractured femur sustained in a motor vehicle collision. Two days after admission, she develops tachycardia, restlessness, diaphoresis, and anxiety. She says that "vague shapes" are coming out of the walls. Which of the following is the most likely cause of this patient's current condition? A) Acute stress disorder B) Alcohol withdrawal C) Bipolar disorder D) Histrionic personality disorder E) Panic disorder

B. This patient is likely developing delirium tremens, a serious complication of alcohol withdrawal. Alcohol increases the central nervous system activity of gamma-aminobutyric acid (GABA), the main inhibitory neurotransmitter. Abrupt discontinuation of alcohol therefore leads to sympathetic excess. Uncomplicated alcohol withdrawal typically presents with tremors, anxiety, restlessness, headache, nausea, and diaphoresis, which can occur hours after the cessation of alcohol use. Delirium tremens, the most concerning and life-threatening complication of alcohol withdrawal, involves severe confusion and disorientation, fluctuations in consciousness, agitation, visual hallucinations, and autonomic instability (fluctuations in heart rate and blood pressure with hyperthermia). As in this patient, delirium tremens can begin 2 to 5 days after the patient's last alcoholic drink. Patients with delirium tremens may die from associated dysrhythmias, seizures, or respiratory failure. Delirium tremens is managed supportively with benzodiazepines to address agitation and prevent the symptoms of withdrawal, fluid repletion, nutritional supplementation, and frequent assessment, including vital sign checks. Other complications of alcohol withdrawal include alcoholic hallucinosis (auditory or visual hallucinations without confusion or autonomic instability). Incorrect Answers: A, C, D, and E. Acute stress disorder (Choice A) refers to symptoms of intrusive thoughts, hyperarousal, dissociation, avoidance, and/or depressed mood within 1 month of a traumatic event such as a life-threatening motor vehicle collision. This patient's tachycardia and diaphoresis could represent symptoms of hyperarousal, and hallucinations do occasionally occur during trauma-related dissociative experiences. However, hallucinations occur less commonly in acute stress disorder than in delirium tremens, and the acute onset of this patient's symptoms 48 hours after hospital admission is concerning for delirium tremens. Bipolar disorder (Choice C) is characterized by discrete episodes of depression and mania, which can sometimes be accompanied by psychotic symptoms such as hallucinations and delusions. Tachycardia and diaphoresis would be uncommon, and primary psychiatric disorders are more frequently associated with auditory, rather than visual, hallucinations. Histrionic personality disorder (Choice D), a cluster B personality disorder, is characterized by theatrical, superficial expressions of emotion that unconsciously serve to garner attention from others to fulfill emotional needs. Episodic tachycardia, diaphoresis, anxiety, and visual hallucinations would be atypical. Panic disorder (Choice E) features panic attacks that occur unexpectedly and are associated with changes in behavior to avoid panic attack triggers. Though panic attacks can include restlessness, tachycardia, and diaphoresis, visual hallucinations would be atypical. Additionally, the timing of this patient's symptoms 48 hours after hospital admission is instead classic for delirium tremens. Educational Objective: Delirium tremens is a life-threatening complication of alcohol withdrawal that presents with severe confusion, agitation, visual hallucinations, and autonomic instability. Delirium tremens typically begins 2 to 5 days after the patient's last alcoholic drink.

45 ---------- Exam Section 1: Item 46 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 46. A 35-year-old man comes to the physician because of a 6-month history of burning abdominal pain that occurs 1 to 2 hours after he eats. He also has had black stools for 2 days. Use of over-the-counter antacids and histamine-2 (H2)-receptor blocking agents has not been effective in relieving his symptoms. He is sweating profusely and has light-headedness when he stands. His blood pressure is 105/70 mm Hg while sitting. Physical examination shows epigastric tenderness. A CT scan of the abdomen shows a 1-cm mass in the pancreas. Immunohistochemical labeling of neoplastic cells in a biopsy specimen is most likely to involve the use of antibodies directed to which of the following substances? A) Amylase B) Gastrin C) Glucagon D) Human pancreatic polypeptide E) Insulin F) Lipase G) Serotonin H) Somatostatin I) Vasoactive intestinal polypeptide

B. This patient's findings of postprandial abdominal pain, melena, diaphoresis, and orthostatic hypotension in the setting of a pancreatic lesion are suggestive of a peptic ulcer secondary to Zollinger-Ellison syndrome (ZES). ZES stems from a pancreatic or duodenal gastrin-secreting tumor. Ġastrin is typically produced by gastric G cells and stimulates parietal cells to produce hydrochloric acid. Excessive production of acid in the setting of ZES leads to recurrent, chronic duodenal or jejunal ulcers, which can present with abdominal pain, diarrhea secondary to malabsorption, and possible hematemesis, melena, or hematochezia. Immunohistochemical labeling of neoplastic cells using an anti-gastrin antibody reveals positive staining. Pancreatic gastrinoma and ZÉS may present as a component of multiple endocrine neoplasia type 1 (MEN1). Incorrect Answers: A, C, D, E, F, G, H, and I. Amylase (Choice A) assists in the digestion of carbohydrates and is secreted in saliva and from the exocrine pancreas. Increased levels of amylase may indicate sialadenitis or pancreatitis. Glucagon (Choice C) is produced by pancreatic alpha cells and primarily works to increase serum glucose. Positive glucagon staining on histology is characteristic of a glucagonoma, which presents with mild diabetes mellitus, necrolytic migratory erythema, and weight loss. Human pancreatic polypeptide (Choice D) regulates pancreatic endocrine and exocrine secretions. Its overexpression is nonspecific and may be observed in a variety of different hormonally active pancreatic tumors. Insulin (Choice E) is produced by pancreatic beta cells and primarily works to lower serum glucose. Positive staining for insulin on histology is characteristic of an insulinoma, a rare neoplasm that produces insulin. Patients present with lethargy, weakness, fatigue, diplopia, and hypoglycemia. Symptoms are exacerbated by fasting or exercise. Lipase (Choice F) is a pancreatic enzyme that functions to digest triglycerides. Some pancreatic acinar cell carcinomas may demonstrate staining for lipase. Serotonin (Choice G) is typically produced by carcinoid tumors. Carcinoid tumors of the gastrointestinal tract are neuroendocrine tumors that present with weight loss, flushing, diarrhea, bronchospasm, right sided cardiac valvular disease, gastrointestinal bleeding, and abdominal pain. Somatostatin (Choice H) is produced by a somatostatinoma, an uncommon neuroendocrine malignancy that presents with weight loss, mild diabetes mellitus, steatorrhea due to limited cholecystokinin release, and achlorhydria from the inhibition of gastrin. Vasoactive intestinal polypeptide (VIP) (Choice l) can be produced by VIPomas. VIPoma is a rare tumor often associated with MEN1 that presents with profound watery diarrhea, electrolyte disturbances (eg, hypokalemia, hypercalcemia), achlorhydria, alkalosis, flushing, and vasodilation. Educational Objective: Zollinger-Ellison syndrome stems from a gastrin-secreting tumor located typically within the pancreas or duodenum that results in excessive production of acid by the gastric parietal cells. Patients develop recurrent, chronic duodenal or jejunal ulcers, which can present with abdominal pain, diarrhea secondary to malabsorption, and possible hematemesis, melena, or hematochezia. Immunohistochemical labeling of neoplastic cells will reveal positive staining for gastrin.

10. A 37-year-old man comes to the physician because of a 1-week history of nausea, vomiting, and abdominal pain. He ate raw oysters while vacationing in Mexico 2 weeks ago. He has smoked 1 pack of cigarettes daily for the past 15 years but has now lost the desire to smoke. Physical examination shows scleral icterus and slight enlargement and tenderness of the liver. Which of the following pathogens is most likely responsible for the development of these symptoms in this patient? A) Epstein-Barr virus B) Hepatitis A virus C) Hepatitis B virus D) Hepatitis C virus E) Vibrio vulnificus

B. This patient's presenting findings of nausea, vomiting, abdominal pain, icterus, hepatomegaly and right upper quadrant tenderness two weeks after eating raw shellfish is consistent with an acute hepatitis A virus (HAV) infection. HAV is a non-enveloped, single-stranded RNA virus that is spread through fecal-oral transmission. It is commonly acquired through ingestion of poorly cooked, improperly handled, or raw foods, including shellfish. HAV has an incubation period of several weeks. Presenting signs and symptoms include fever, nausea, vomiting, poor appetite, abdominal pain, jaundice, and hepatomegaly, although many patients are asymptomatic. Laboratory findings include transaminitis, increased alkaline phosphatase, hyperbilirubinemia, and bilirubinuria. The diagnosis is confirmed by the serologic detection of anti-HAV IgM antibodies. HAV infection is typically a self-limited illness and does not progress to chronic infection. Incorrect Answers: A, C, D, and E. Epstein-Barr virus (Choice A) causes infectious mononucleosis, which presents with fever, hepatosplenomegaly, pharyngitis, and posterior cervical lymphadenopathy. It is transmitted primarily through saliva and respiratory secretions and is not transmitted through uncooked food. Hepatitis B (Choice C) and C virus (Choice D) are transmitted through bodily fluids. Hepatitis B is commonly transmitted through blood, sexual contact, or perinatally. Hepatitis C is commonly transmitted through blood and is especially common in the setting of intravenous drug use. The patient's history is suggestive of oral transmission through raw shellfish. Vibrio vulnificus (Choice E) is a Gram-negative, primarily water-borne bacterium that causes sepsis or bacteremia and is associated with the consumption of raw shellfish. V. vulnificus sepsis presents acutely after the ingestion of shellfish with fever, hypotension, shock, bullous skin lesions, gastrointestinal hemorrhage, and coagulopathy, which may progress to disseminated intravascular coagulation. It is distinguished from HAV infection by its rapid onset, absence of jaundice, and generally severe clinical presentation. Educational Objective: Hepatitis A virus (HAV) causes a self-limited infection that presents with nausea, vomiting, abdominal pain, icterus, hepatomegaly, and right upper quadrant tenderness that evolves over several weeks following the ingestion of poorly cooked, improperly handled, or raw foods, including shellfish. HAV is diagnosed by the serologic detection of anti-HAV IgM antibodies.

22 ---------- Exam Section 1: Item 23 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment Normal AB Type 2 Diabetes Mellitus Fasting Serum Glucose 23. During an investigational study, fasting serum glucose concentrations are collected from a population of healthy individuals and a population of patients with type 2 diabetes mellitus. The results are shown in the graph. Switching from cut point A to cut point B would most likely result in which of the following changes in test characteristics regarding identification of patients with type 2 diabetes mellitus? A) Decreased specificity B) Improved sensitivity C) Increase in false-negatives D) Increase in false-positives E) No change

C. Any population of patients will include those who have the disease and those who do not, and overlap may exist between these groups when considering testing for nonbinary diagnoses. For example, type 2 diabetes mellitus is understood to be an insensitivity of peripheral tissues to insulin, marked by diminished insulin uptake and increased blood glucose. It is impossible to specifically measure the degree to which cellular receptors are resistant to insulin, therefore, alternative diagnostic criteria are used reflecting increased blood glucose and its effects. As of 2020, the American Diabetes Association criteria include two fasting serum blood glucose levels greater than or equal to 126 mg/dL, a random serum blood glucose level greater than or equal to 200 mg/dL with signs and symptoms of diabetes, a hemoglobin-A1c level of greater than or equal to 6.5%, or a 2-hour oral glucose tolerance test with increased blood sugar beyond 200 mg/dL. Not all patients with a fasting serum blood glucose greater than or equal to 126 mg/dL will have diabetes mellitus. For example, confounding factors such as chronic steroid use may result in increased blood glucose without impaired cellular uptake. These patients may receive positive tests, but do not have the underlying condition. Therefore, the set points for diagnostic tests will result in some true positives (patients who have the disease and who the tests identifies as having the disease), true negatives (patients who do not have the disease who the test identifies as not having the disease), false positives (patients who do not have the disease who the test identifies as having the disease), and false negatives (patients who have the disease who the test identifies as not having the disease). The set point will determine the fraction of patients who are true or false positive and true or false negative and should be chosen to optimize the sensitivity and specificity of the test. Point A would result in the majority of persons with type 2 diabetes mellitus being identified as positive; therefore, the number of false negatives would be İow, and the number of true positives would be high. Moving the cut point from point A to point B (raising the threshold) would result in fewer people with the disease being identified, thereby increasing the number of false negatives and reducing the number of false positives, which would decrease sensitivity and increase specificity. Incorrect Answers: A, B, D, and E. Decreased specificity (Choice A) and improved sensitivity (Choice B) would result from an increase in false positives (Choice D) and a decrease in false negatives, which would occur if the cut point were moved from B to A. Moving the cut point from A to B would decrease the number of false positives and increase the number of false negatives. Specificity is calculated as true negatives / (true negatives + false positives). Sensitivity is calculated as true positives / (true positives + false negatives). No change (Choice E) is incorrect as adjusting the cut point alters the fraction of true and false positives and negatives, which affects the sensitivity and specificity of the test. Educational Objective: Raising the threshold for a positive test to diagnose patients with a disease decreases sensitivity and increases specificity of the test by increasing the number of false negatives and decreasing the number of false positives.

35 ---------- Exam Section 1: Item 36 of 50 National, Board of Medical Examiners Comprehensive Basic Science Self-Assessment 36. A 40-year-old woman with chronic paraplegia caused by multiple sclerosis is brought to the physician because of severe, debilitating muscle cramps for the past 3 weeks. Treatment with baclofen resolves her muscle cramps. Which of the following receptors most likely decreased the muscle spasticity in this patient? A) a2Adrenoreceptor B) B2Adrenoreceptor C) y-Aminobutyric acid B receptor D) Calcium-sensing receptor E) Ryanodine receptor

C. Baclofen is a y-aminobutyric acid (GABA) analogue that acts as an agonist at GABA-B receptors. GABA-B receptors are metabotropic G-protein coupled receptors that increase efflux of potassium and thus hyperpolarize skeletal muscle cells and decrease action potential frequency. As such, baclofen is a first-line treatment for muscle spasticity. GABA-B receptors are also found in the central nervous system (CNS); therefore, baclofen can cause sedation, especially when used in conjunction with other CNS depressants. Incorrect Answers: A, B, D, and E. a2-Adrenoreceptors (Choice A) are G-protein coupled receptors localized to the CNS that decrease sympathetic outflow when activated. Examples of a2-adrenergic agonists include clonidine and guanfacine, which treat attention-deficit/hyperactivity disorder (ADHD) and hypertensive urgency. Tizanidine is an a2-adrenergic agonist that is used as a muscle relaxant. B2-Adrenoreceptors (Choice B) are G-protein coupled receptors involved in smooth muscle relaxation. Albuterol, a bronchodilator used in asthma, is one example of a B2-Adrenergic agonist. B2-Adrenergic agonists do not act on skeletal muscle, though tremors can be seen as a consequence of increased sympathetic tone. Calcium-sensing receptors (Choice D) are G-protein coupled receptors located in the renal tubule and parathyroid gland that sense the extracellular concentration of calcium ions. These receptors influence calcium homeostasis by regulating renal calcium reabsorption and parathyroid hormone release. Ryanodine receptors (Choice E) are calcium channels located on the sarcoplasmic reticulum membranes in skeletal muscle. These channels mediate the intracellular release of calcium during the excitation-contraction coupling essential to muscle contraction. Dantrolene antagonizes ryanodine receptors to treat muscle spasticity and malignant hyperthermia. Educational Objective: Baclofen is a GABA-B receptor agonist, which hyperpolarizes muscle cells and relieves muscle spasticity.

55 ---------- Exam Section 2: Item 6 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 6. An investigator obtains a serum sample from an individual previously immunized with a vaccine that elicits immunity to Clostridium tetani. Administration of this serum to a guinea pig prior to challenge with a potentially lethal dose of C. tetani is most likely to mediate protection to the challenge dose through binding of antibody to which of the following products of the bacterial cells? A) Capsular polysaccharide B) Cell wall polysaccharide C) Exotoxin D) Fimbria E) Flagellum F) Peptidoglycan

C. Clostridium tetani is the bacterial cause of tetanus, resulting in the development of spastic paralysis following bacterial inoculation via a penetrating traumatic wound. Physical examination findings include trismus, risus sardonicus, opisthotonos, and rigid abdominal muscles. In neonates, a foul-smelling and erythematous umbilical stump that was cut at birth using non-sterile technique is associated with C. tetani infection. The major virulence factor is tetanospasmin, which is a protease exotoxin that cleaves the soluble NSF- attachment protein receptor (SNARE) complex in Renshaw cells of the spinal cord, preventing the release of inhibitory GABA and glycine neurotransmitters by impairing the fusion of vesicles. It is carried from the site of inoculation to the spinal cord via retrograde axonal transport. Infection can be prevented by immunization with a tetanus toxoid vaccine. The tetanus toxoid is a modified exotoxin that is no longer virulent but still stimulates an antigenic response in the host. The antibodies produced after immunization target the exotoxin of C. tetani, disabling its function and preventing the onset of spastic paralysis. Incorrect Answers: A, B, D, E, and F. Capsular polysaccharide (Choice A) targets are used to make vaccines for pneumococcal disease, meningococcal disease, and Salmonella typhi. Clostridium tetani does not have a capsule. Cell wall polysaccharide (Choice B) is a bacterial virulence factor that forms a glycocalyx projecting from the bacterial cell wall. A glycocalyx protects bacteria from phagocytosis and aides in the adherence to other bacterial and inert materials. Fimbria (Choice D) is a virulence factor that aides in the adherence and colonization of bacteria to host cells. It is composed of glycoproteins and is a common virulence factor seen in Gram-negative bacteria, especially E. coli. Flagellum (Choice E) is a virulence factor that increases bacterial mobility and chemotaxis. It is composed of the protein flagellin. A flagellum is not a virulence factor present in the bacteria C. tetani. Peptidoglycan (Choice F) is a polymer of sugars and amino acids that provides stability to the bacterial cell wall. B-lactam antibiotics inhibit the transpeptidase cross-linking of peptidoglycan in the cell wall. Educational Objective: Toxoid vaccines are composed of exotoxins that have been chemically modified to remove their virulence. They stimulate the host production of antibodies that target the virulent toxin, disabling its function. These toxin antibodies would prevent the onset of spastic paralysis if inoculated with C. tetani.

90 ---------- Exam Section 2: Item 40 of 50 Natjonal, Board of Medical Examinersmen' * 40. A 28-year-old white woman comes to the physician for a follow-up examination. Two days ago, she was discharged from the hospital after being treated for a hypertensive crisis with seizure. She has no other history of major medical illness. Current medications include a B-adrenergic blocker, dihydropyridine calcium channel blocker, and a centrally acting agent stimulating presynaptic a2 receptors. Her temperature is 37°C (98.6°F), pulse is 54/min, and blood pressure is 182/100 mm Hg. Physical examination shows trace ankle edema. Cardiac examination shows an S4 gallop. A soft abdominal bruit is heard left of the umbilicus. Abdominal ultrasonography shows normal-sized kidneys and a normal liver, spleen, pancreas, and gallbladder; a small left ovarian cyst is noted. Laboratory studies show that complete blood count, serum electrolyte concentrations, and renal and liver function test results are within the reference ranges. Which of the following is the most likely cause of this patient's secondary hypertension? O AJChronie kidney disease O B}Goarctation of the aorta O C) Fibromuscular dysplasia of the left renal artery D) Hyperthyroidism O E} Primary-hyperaldosteronism

C. Fibromuscular dysplasia is a non-inflammatory and non-atherosclerotic angiopathy of medium-sized arteries (eg, renal, carotid), that results in multifocal fibrous and muscular thickening of the arterial wall. The resultant stenosis causes secondary hypertension due to abnormal stimulation of the juxtaglomerular apparatus from low afferent blood flow leading to excessive production of renin and angiotensin. It can lead to severe hypertension in an otherwise healthy, young patient with no medical comorbidities or laboratory abnormalities. Examination findings often include signs of left ventricular hypertrophy such as an S4 gallop and a renal artery bruit auscultated lateral to the umbilicus on the affected side. Angiography may reveal a bead-like appearance of the renal artery. Treatment involves angioplasty or stenting of the stenosed renal artery to improve flow. ACE inhibitors can be considered for unilateral stenosis but can lead to acute renal failure in the setting of significant bilateral renal artery stenosis. Incorrect Answers: A, B, D, and E. Chronic kidney disease (Choice A) can have variable etiology, 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. Coarctation of the aorta (Choice B) often presents in childhood with hypertension in an upper extremity and a systolic murmur. It is associated with weak lower extremity pulses and differentially lower blood pressure in the lower extremities. Hyperthyroidism (Choice D) may cause hypertension, but would be associated with other symptoms of hyperthyroidism, including heat intolerance, sweating, hair loss, warm, flushed skin, amenorrhea, diarrhea, palpitations, and weight loss. Vital signs may also reveal hyperthermia, tachycardia, and tachypnea. Primary hyperaldosteronism (Choice E) may cause hypertension but would be associated with hypokalemia and metabolic alkalosis. This would be unlikely in a patient with normal serum electrolytes. Educational Objective: Fibromuscular dysplasia is a non-inflammatory and non-atherosclerotic angiopathy of medium-sized arteries that can lead to renal stenosis and secondary hypertension due to excessive production of renin and angiotensin. It commonly presents in a young patient with no medical comorbidities or abnormal laboratory findings.

48 ---------- Exam Section 1: Item 49 of 50 National, Board of Medical Examiners Comprehensive Basic Science Self-Assessment 49. An investigator conducts a clinical study of 300 patients with stages III and IV colon carcinoma. It is found that over time, metastases show an increasing potential for growth and resistance to chemotherapy. Which of the following mechanisms of the cell cycle best explains this observation? A) Decreased expression of cyclin B during G 2/M transition B) Decreased phosphorylation of RB1 protein during G1-S transition C) Genomic instability during S, G2 and M phases D) Overexpression of p15, p16, and p18 during G1 phase E) Overexpression of p21, p27, and p57 proteins during all phases

C. Genomic instability during S, G2, and M phases explains the increased potential for growth and resistance to chemotherapy agents used to treat stage III and IV colon carcinoma. The two primary forms of genomic instability that predispose to the formation of malignant cells are microsatellite and chromosome instability. In the former, the failure of mismatch repair pathways (MMR) from mutations in MLH1, MSH2, APC, and other genes, results in the propagation of aberrant DNA and accumulation of mutations that predispose to malignant transformation. This is the mechanism of oncogenesis found in familial cancer syndromes such as Lynch syndrome or familial adenomatous polyposis (FAP). In the latter instance of chromosome instability, specific activating mutations of oncogenes or loss of function mutations of tumor suppressor genes results in oncogenic potential. This pathway accounts for the majority of sporadic colon cancers. Activating mutations of oncogenes such as KRAS and inactivating mutations of tumor suppressor genes such as P53 predispose cells to abnormal growth through abnormal entry into the cell cycle, which results in cellular proliferation. Within the cell cycle, errors of replication further predispose to malignant growth and may account for resistance to chemotherapeutic agents. The normal cell cycle in somatic cells involves several stages: G1, Š, G2, and M phase. Go phase denotes cell cycle arrest. During G1 phase, the cellular contents, with the exception of the chromosomes, are replicated. In S phase, chromosomal duplication occurs. In G2 phase, the fidelity of replication is checked. M phase involves the attachment of chromosomes to spindles followed by their separation and division into two identical cells, with the stages of M phase denoted as prophase, metaphase, anaphase, and telophase. Derangement at any point during mitosis can predispose to the propagation of an abnormal and unstable genome, with the possible development of resistance to chemotherapy agents. Incorrect Answers: A, B, D, and E. Decreased expression of cyclin B during G2/M transition (Choice A) paradoxically reduces cell proliferation. Depleted cyclin B levels mean the cyclin B/CDK1 complex cannot form, which prevents cells from entering M phase. Decreased phosphorylation of RB1 protein during G1-S transition (Choice B) is not the correct answer. RB1 is a tumor suppressor gene and phosphorylation results in its inhibition, thereby allowing the cell to enter the cell cycle. Reduced phosphorylation would prevent entry into the cell cycle and would inhibit growth. Overexpression of p15, p16, and p18 during G1 phase (Choice D) would result in the reduction of cellular proliferation as these genes are all tumor suppressors. Overexpression of p21, p27, and p57 proteins during all phases (Choice E) of the cell cycle would result in decreased cellular proliferation as all of these encode cyclin-dependent kinases that halt progression of the cell cycle when upregulated. Educational Objective: There are two primary mechanisms of genomic instability in the pathogenesis of colon cancer: microsatellite instability (15% of cases) and chromosome instability (85% of cases). Chromosome instability results initially in activating mutations of oncogenes and/or loss of function mutations of tumor suppressor genes, resulting in the consequent aberrant entry into the cell cycle. Once cells enter mitosis, genomic instability occurs during chromosome replication (S phase), fidelity checks of the chromosomes (G2 phase), and during separation of the chromosomes (M phase), which results in the accumulation of mutations, potentially causing increased proliferation and resistance to chemotherapy.

96 ---------- Exam Section 2: Item 46 of 50 National Board of Medical Examiners ment 46. A 45-year-old man comes to the physician because of a 6-month history of progressive shortness of breath with exertion. He has a history of frequent nosebleeds since adolescence. He is 178 cm (5 ft 10 in) tall and weighs 79 kg (175 lb); BMI is 25 kg/m2. Physical examination shows the findings in the photographs. Inhaled albuterol does not improve his symptoms. Which of the following is the most likely cause of this patient's condition? O A) Atrial septal defect O B) Hypertension o C) Pulmonary arteriovenous shunting O D) Reactive airway disease O E) Thromboembolism

C. Hereditary hemorrhagic telangiectasia (HHT) is an autosomal dominant disorder of the vasculature that is characterized by arteriovenous malformations (AVMS) and telangiectasias, which are small vascular malformations which appear as bright red, blanching lesions of the skin and mucous membranes (as seen in the oral photograph). There are several gene mutations associated with the disorder that result in abnormal formation of capillaries and the manifestations of the disease. Patients classically present with recurrent epistaxis, hematuria, and gastrointestinal bleeding. Cerebral AVMS may cause hemorrhagic stroke. Pulmonary AVMS can lead to embolic stroke as venous blood (and a potential embolus) is able to bypass the pulmonary capillary bed and enter the systemic arterial circulation. Pulmonary AVMS also result in pulmonary arteriovenous shunting with subsequent chronic hypoxemia, which can lead to hypertrophic osteoarthropathy (digital clubbing) as seen in the photograph of the fingers. HHT is associated with pulmonary arterial hypertension and iron deficiency anemia secondary to chronic blood loss. Incorrect Answers: A, B, D, and E. Atrial septal defect (Choice A) is a common congenital malformation of the interatrial septum that results in a left-to-right shunt with abnormal flow of blood from the left atrium to the right atrium. If the atrial septal defect remains uncorrected, it can result in the development of Eisenmenger syndrome secondary to prolonged pulmonary vasculature remodeling resulting in pulmonary arterial hypertension and shunt reversal leading to cyanosis. It is not associated with recurrent bleeding or mucocutaneous lesions. Hypertension (Choice B) is a leading cause of cardiovascular mortality. Chronic hypertension can result in left ventricular hypertrophy and ultimately left-sided heart failure. Left-sided heart failure is the most common cause of right-sided heart failure. Hypertension is not associated with mucocutaneous lesions. Reactive airway disease (Choice D) is a term used to describe conditions which result in airway sensitivity to physical, chemical, or pharmacologic stimuli (eg, asthma, COPD). Chronic, undiagnosed reactive airway disease may result in hypertrophic osteoarthropathy. Symptoms would be expected to improve with albuterol. Thromboembolism (Choice E) is a complication of Behçet syndrome, which is an autoimmune disorder characterized by small and large vessel vasculitis. It often presents with fever, weight loss, arthritis, enthesitis, and thrombosis or aneurysm secondary to vascular inflammation. Physical examination may reveal recurrent oral aphthous ulcers, genital ulcers, and various skin lesions. The mucocutaneous lesions in this case are more consistent with a diagnosis of HHT. Educational Objective: Hereditary hemorrhagic telangiectasia is an inherited disorder of vascular formation that results in arteriovenous malformations and telangiectasias throughout the body. It is characterized by recurrent bleeding. Pulmonary arteriovenous shunting can result in chronic hypoxemia.

22. An 18-month-old boy in West Africa with perinatally acquired HIV infection is exposed to a child in his village with a fever and an erythematous, maculopapular rash. Three weeks later, the 18-month-old boy develops giant cell pneumonia, but there is no evidence of a rash. Which of the following viruses is the most likely cause of the pneumonia? A) Coxsackievirus B) Influenza virus C) Measles virus D) Rubella virus E) West Nile virus

C. Measles (rubeola) virus is a highly contagious paramyxovirus that causes the disease measles, which is associated with acute febrile illness and potentially severe sequelae. It is more common in children than adults. It characteristically presents with prodromal fever, cough, coryza, conjunctivitis, and a confluent maculopapular rash that starts at the head/neck and spreads to the trunk, excluding the palms/soles. Physical examination may reveal bright red macules with a bluish-white center on the buccal mucosa and lymphadenopathy. The virus replicates in epithelial cells of the respiratory tract and lymph nodes causing lymphadenitis. Virulence factors include the F (fusion) protein which aids in viral fusion to host cells and can result in the formation of multinucleated giant cells. Hemagglutinin (HA) protein is involved in viral binding to host cells. Treatment is supportive. Vitamin A deficiency is associated with severe disease and complications, and supplementation should be provided, especially in children who are malnourished. Complications include subacute sclerosing panencephalitis (a neurodegenerative disease associated with intellectual regression, personality changes, motor deterioration, seizures, and premature death), encephalitis, and giant cell pneumonia. Giant cell pneumonia is rare, except in patients who are immunosuppressed. It is often fatal and characterized by the presence of inclusion bodies and multinucleated giant cells in the respiratory epithelium. Incorrect Answers: A, B, D, and E. Coxsackievirus (Choice A) is an RNA enterovirus which is transmitted via oral secretions or feces. Coxsackievirus type A causes hand, foot, and mouth disease, and herpangina. Coxsackievirus type B may cause myocarditis and pericarditis. Influenza virus (Choice B) is an orthomyxovirus transmitted via respiratory droplets. Symptoms include high fever, chills, myalgias, malaise, headaches, nausea, and vomiting. Pneumonia may develop. Influenza types A and B are responsible for seasonal pandemics. Complications include secondary bacterial infection of the respiratory tract or lungs and myocarditis. Rubella virus (Choice D) is a togavirus transmitted via respiratory secretions that causes the disease rubella. Symptoms include fever, arthralgias, and lymphadenopathy (most often postauricular) with a confluent rash that spreads from the head to the trunk and then centrifugally to the extremities. Complications include vertical transmission to the fetus in women who are pregnant (resulting in congenital rubella syndrome), thrombocytopenic purpura, and Guillain-Barré syndrome. West Nile virus (Choice E) is a flavivirus transmitted via the bite of an infected mosquito and may cause West Nile fever with progression to meningoencephalitis. West Nile fever presents with headache, myalgias, nausea, vomiting, and a non-specific maculopapular rash distributed symmetrically on the extremities and trunk. Meningoencephalitis is associated with acute flaccid paralysis, myoclonus, ataxia, and seizures. Complications include rhabdomyolysis and Guillain-Barré syndrome. Educational Objective: Giant cell pneumonia is a severe complication of measles that is rare in patients who are not immunosuppressed. Histology reveals inclusion-bodies and multinucleated giant cells in the respiratory epithelium.

57 ---------- Exam Section 2: Item 8 of 50 National, Board of Medical Examiners Comprehensive Basic Science Self-Assessment 8. A 56-year-old man dies 1 month after the onset of difficulty with short-term memory and anxiety. He underwent surgical excision of small cell carcinoma 1 year ago. Degenerative changes are most likely to be concentrated at which of the following labeled sites on the normal MRI of the head? A C A) B) D) E) F)

C. Memory and emotional regulation are mediated by the limbic system, which includes the amygdala, mammillary bodies, stria medullaris, hippocampus, cingulate gyrus, prefrontal cortex, parahippocampal gyrus, ventral tegmentum, and multiple deep nuclei. Label "C" in the axial T2-weighted MRI image shown identifies the mesial temporal lobe, where the hippocampus and parahippocampal gyrus are located. The hippocampus is particularly sensitive to radiation, and degenerative changes or damage of the hippocampus commonly result in short-term memory loss. Chemotherapy has also been shown to injure hippocampal synapses. This patient's symptoms of short-term memory loss, anxiety, and a history of malignancy suggest damage to the limbic system with likely involvement of the hippocampus and mesial temporal lobe. Mesial temporal lobe atrophy is also associated with Alzheimer disease and frontotemporal dementia. Incorrect Answers: A, B, D, E, and F. Choice A identifies the eye and superior orbit, in which rectus muscles and the globe are visible. These structures are involved in vision and do not play a role in memory or emotional regulation. Choice B identifies the region of the frontal pole, inferior frontal gyrus, straight gyrus, olfactory gyrus, and olfactory sulcus. Olfactory sensations are closely related to emotional processing and memory. However, this region is less likely than the mesial temporal lobe to result in short-term memory loss. Choice D identifies the right substantia nigra. The substantia nigra is a nucleus for dopaminergic cells and is critical for the modulation of movement in coordination with the basal ganglia. Pathology affecting this area is associated with Parkinson disease. Choice E identifies the superior cerebellar vermis. Chronic alcoholism is associated with atrophy of this region. Atrophy and lesions of this area are associated with truncal ataxia and gait disturbances. Choice F identifies the right occipital lobe, associated with visual processing. Lesions in this area would typically present with contralateral homonymous hemianopsia and would be unlikely to affect memory or emotional processing. Educational Obiective: Memorv and emotional reaulation are mediated bv the limbic sustem which includes the hinnocamnus narahinnocamnal avrus and amvadala located in the mesial temporal lobe Cancer-related radiation and chemotherany commonly

53 ---------- Exam Section 2: Item 4 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 4. A 22-year-old woman who has B-thalassemia has a B-hemoglobin allele that carries a single base substitution (A → G) at the splice acceptor site of the second intron. Which of the following events in expression of the B-hemoglobin gene is most likely to be impaired by this mutation? A) Binding of nucleosomal histones to the transcript B) Cleavage of the transcript by RNA ligase C) Modification of the transcript by small nuclear ribonucleoproteins D) Recognition of the transcript by signal recognition particle E) Transport of the transcript out of the nucleus

C. Modification of the transcript by small nuclear ribonucleoproteins (snRNPs) is most likely to be affected by a single base substitution at the splice acceptor site of the second intron of the B-hemoglobin allele. Transcription is the process by which pre-messenger RNA (MRNA) is made from DNÅ, but pre-mRNA must undergo post-transcriptional modification to create mature mRÑA, including the removal of introns and connection of exons. Splice sites exist at the end of an intron and the beginning of an exon, and are recognized by snRŇPs, which are RNA-protein complexes that combine to form a spliceosome. These snRNPs remove the introns, and exons are subsequently spliced together. A mutation in the splice acceptor site can result in the inclusion of introns into the final mRÑA, or exclusion of exons, thereby rendering the MRNA and its translated protein non-functional. In the instance of B-thalassemia, a single abnormal B-hemoglobin gene would result in B-thalassemia minor. Incorrect Answers: A, B, D, and E. Binding of nucleosomal histones to the transcript (Choice A) would not be affected by a mutation of the splice site. Histones are proteins with positively charged amino acids that bind to negatively charged DNA. A complex of histones is called a nucleosome. Unwinding of the nucleosome is required for transcription and occurs via the action of histone acetylase. Cleavage of the transcript by RNA ligase (Choice B) is not affected by mutations in the splice acceptor site. RNA ligases are a class of enzymes that join RNA strands to one another. Recognition of the transcript by signal recognition particle (SRP) (Choice D) occurs after transcription has already occurred. The SRP binds new peptides as they are being synthesized by the ribosome and results in a transient reduction in the pace of translation to facilitate early initiation of protein translocation. Transport of the transcript out of the nucleus (Choice E) occurs via a mechanism that involves assimilation of MRNA into ribonucleoprotein particles (RNP) with the subsequent recruitment of specific mRNA exporter proteins. These export proteins help move mRNA out of the nucleus and into the cytoplasm. This would not be affected by a mutation in the splice acceptor site. Educational Objective: The splice acceptor site resides between an intron and exon on pre-mRNA strands and functions to recruit snRNPs that remove introns, with subsequent connection of exons via the spliceosome, forming fully mature MRNA. Mutations in the acceptor site lead to abnormal inclusion of introns or exclusion of exons and ultimately result in an abnormal, nonfunctional protein.

69 ---------- Exam Section 2: Item 19 of 50 National, Board pf Medical Examiners mont 19. A newborn delivered at 28 weeks' gestation is in severe respiratory distress. The immature alveoli of this newborn's lungs have a diminished ability to serve as sites of effective gas exchange. An increase in which of the following parameters best explains this finding? O AJAlveolar radii O B} Lung compliance C) Lung elastic recoil O D) Pleural pressure O E}Surfactant secretion

C. Neonatal respiratory distress syndrome presents with dyspnea and tachypnea during the first few hours of life, most often occurring in premature neonates. Physical examination reveals evidence of increased respiratory effort with nasal flaring, expiratory grunting, and intercostal retractions. This typically occurs secondary to an insufficient quantity of pulmonary surfactant production related to immature type Il pneumocytes, which occurs in the setting of premature delivery or the decreased stimulation of surfactant production from mature pneumocytes (eg, cesarean delivery with reduced fetal cortisol production in the absence of vaginal compressive stress). Pulmonary surfactant is a phospholipid mixture that lines the alveoli and reduces surface tension. Surface tension is directly proportional to the collapsing pressure experienced by the alveolus. Impaired surfactant production leads to alveolar collapse with two important consequences. The first consequence is a substantially reduced area for gas exchange resulting in hypoxemia, while the secondary consequence is decreased lung compliance, requiring greater force to expand the lungs. Incorrect Answers: A, B, D, and E. An increase in alveolar radii (Choice A) would result in increased lung compliance and decreased lung elastic recoil. This can be observed with the law of Laplace, which describes the pressure on a system (in this case, alveolar collapsing pressure) as a function of surface tension divided by radius. As the radius of the alveolus increases, the collapsing pressure decreases. Lung compliance (Choice B) is decreased in neonatal respiratory distress syndrome. Compliance is the change in volume of a system in response to a change in pressure. The increased lung elastic recoil from collapsed alveoli leads to a decreased lung compliance, as it takes a greater change in pressure to expand the lungs than if the alveoli were not collapsed. Pleural pressure (Choice D) is decreased in neonatal respiratory distress syndrome relative to healthy lungs. A greater negative pressure is required to expand the lungs due to the decreased lung compliance during inspiration, and less positive pleural pressure is needed for expiration to occur given the increased lung elastic recoil. Surfactant secretion (Choice E) is decreased, not increased, in neonatal respiratory distress syndrome. The decreased production surfactant by type II pneumocytes is the underlying pathophysiologic process of neonatal respiratory distress syndrome. Educational Objective: Neonatal respiratory distress syndrome is characterized by the reduced production of pulmonary surfactant. Surfactant is critical for reducing alveolar surface tension and collapsing pressure; without it, alveolar collapse results in impaired gas exchange, decreased lung compliance, and increased lung elastic recoil.

23 ---------- Exam Section 1: Item 24 of 50 National, Board of Medical Examiners Comprehensive Basic Science Self-Assessment 24. A 30-year-old man develops urinary incontinence 2 weeks after successful treatment of a fracture of the left pelvis that was sustained at work. He was pinned against a loading dock by a truck moving in reverse. Physical examination shows a distended bladder. Cystometrography shows absence of micturition reflexes. After the bladder fills to capacity, overflow of urine occurs through the urethra a few drops at a time. This patient most likely sustained additional injury to which of the following during his initial accident? O A) External urinary sphincter B) Hypogastric nerve C) Pelvic nerves D) Pudendal nerves E) Skeletal motor fibers

C. Overflow incontinence is characterized by chronic urinary retention and a chronically distended bladder. When intravesical pressure exceeds outlet resistance, incontinence results as urine flows. It may be caused by physical barriers such as tumors or urethral stricture, neurologic lesions, or as a side effect of some medications (eg, anticholinergics). After pelvic trauma, the pelvic nerve plexus can be damaged. Pelvic parasympathetic nerves in the pelvic plexus function to excite and contract the detrusor muscle of the bladder via muscarinic acetylcholine receptors while sympathetic nerves mediate relaxation of the internal urethral sphincter via a1-adrenergic receptors, leading to normal urination. Damage to the pelvic nerve plexus can affect both bladder contraction and urethral sphincter relaxation, leading to overflow incontinence. Overflow incontinence is treated with managing inciting conditions, timed voiding, or placement of a urethral catheter. Incorrect Answers: A, B, D, and E. The external urinary sphincter (Choice A) is innervated by the pudendal nerve (Choice D). The pudendal nerve supplies sensory neurons to the external genitalia along with somatic skeletal motor fibers (Choice E) to pelvic muscles, including the external urethral sphincter. The pudendal nerve does not carry any parasympathetic fibers. Motor fibers in the pudendal nerve act on the external urethral sphincter via acetylcholine release on nicotinic receptors, causing contraction and urinary retention. Damage to these nerves would result in loss of voluntary control over voiding. The hypogastric nerve (Choice B) contains sympathetic fibers that release norepinephrine and result in relaxation of the detrusor muscle and contraction of the urethral smooth muscle to prevent voiding. Damage to these fibers would lead to total incontinence. Educational Objective: Pelvic parasympathetic nerves in the pelvic nerve plexus function to excite and contract the bladder via muscarinic acetylcholine receptors while sympathetic nerves mediate relaxation of the internal urethral sphincter via a1-adrenergic receptors, leading to normal urination. Damage to the pelvic nerve plexus can affect both bladder contraction and internal urethral sphincter relaxation, leading to overflow incontinence.

54 ---------- Exam Section 2: Item 5 of 50 National, Board of Medical Examiners Comprehensive Basic Science Self-Assessment 5. A 22-year-old woman comes to the physician because of a 6-month history of frequent headaches and fatigue. She tells the physician that she often overeats when she is anxious and has been making herself vomit and using laxatives for the past 2 years. She is 168 cm (5 ft 6 in) tall and weighs 59 kg (130 lb); BMI is 21 kg/m2. Physical examination will most likely show which of the following conditions? A) Arthritis B) Decreased body hair C) Dental caries D) Galactorrhea E) Pituitary adenoma F) Tachycardia O G) Thyroid gland enlargement

C. Patients with eating disorders such as anorexia nervosa and bulimia nervosa who purge by vomiting may demonstrate dental caries on physical examination. This patient likely has bulimia nervosa (purging type). Bulimia nervosa (purging type) involves cycles of uncontrollable eating and compensatory behaviors such as vomiting, laxative, or diuretic overuse that occur at least once a week over 3 months or more. Unlike patients with anorexia nervosa, patients with bulimia nervosa typically have a normal BMI. Patients with the binging/purging type of anorexia or bulimia nervosa can demonstrate dental caries from gastric hydrochloric acid erosion of enamel, parotid gland swelling, and scars on the knuckles secondary to abrasions from the incisors when inducing vomiting. The loss of gastric hydrochloric acid leads to hypochloremia and metabolic alkalosis. In severe cases, signs of hypovolemia such as tachycardia and hypotension may be present. Treatment of bulimia nervosa is through a combined medical and psychiatric approach and involves correcting fluid and electrolyte derangements alongside behavioral and pharmacologic therapy. Incorrect Answers: A, B, D, E, F, and G. Arthritis (Choice A) is not a known eating disorder complication. Decreased bone mineral density and an increased risk for stress fractures are possible musculoskeletal sequelae of eating disorders, which are related to malnutrition and estrogen deficiency. Decreased body hair (Choice B) is a common physical examination finding in anorexia nervosa resulting from decreased protein stores. The severe malnutrition and associated examination findings in anorexia nervosa are less common in bulimia nervosa. Galactorrhea (Choice D) refers to abnormal, non-physiologic discharge of milk from the nipple. It typically results from hyperprolactinemia, which can occur due to medications (eg, dopamine antagonists), prolactinomas, and hypothyroidism (increased thyrotropin- releasing hormone stimulates prolactin release). Patients with bulimia nervosa do not possess any of these risk factors for galactorrhea. Pituitary adenoma (Choice E) may present with headaches, bitemporal hemianopsia (from optic chiasm compression), and hyper- or hypopituitarism. Bulimia nervosa may present with headaches and amenorrhea (though amenorrhea is more commonly seen in anorexia nervosa). Bulimia nervosa and pituitary adenoma do not typically co-occur. Tachycardia (Choice F) may occur in severe cases of the purging type of anorexia or bulimia nervosa due to hypovolemia or torsades de pointes from hypokalemia. However, dental caries are much more common and can occur in less severe cases. Thyroid gland enlargement (Choice G) would not be expected in patients with bulimia nervosa. The binging phase of bulimia nervosa may stimulate thyroid activity, but thyroid enlargement would be atypical. Educational Objective: Bulimia nervosa (purging type) involves cycles of uncontrollable eating and compensatory behaviors such as vomiting or overuse of laxatives or diuretics. Patients with the binging/purging type of bulimia nervosa can demonstrate dental caries from gastric hydrochloric acid erosion of enamel.

78 ---------- Exam Section 2: Item 28 of 50 National, Board pf Medical Examiners aiveB A ance ement 28. A 78-year-old man comes to the physician with his daughter because of low back pain for 3 months. He underwent a bilateral orchiectomy 2 years ago for prostate cancer. He speaks French fluently, but he knows only a few English words. His daughter offers to serve as an interpreter. After thanking the daughter for her offer, it is most appropriate for the physician to state which of the following? O A) "Have you interpreted for your father before in a medical capacity?" O B) "I believe it would be awkward for a daughter to interpret for her father." C) "It is best that I use a professional French interpreter." O D) "Please ask your father if he's ready to begin the examination." E) "That would be fine. I'll just ask you to step out of the room for the physical examination."

C. Professional interpreters can provide high-quality translations that maintain patient safety. Credentialed interpreters are trained in medical knowledge and vocabulary, ethics, and cultural sensitivity. Relying on a family member for interpreting increases the risk of miscommunication about medical or culturally relevant details and can jeopardize informed consent for treatment and patient safety. Incorrect Answers: A, B, D, and E. Having past experience translating for a family member (Choice A) is suboptimal compared to the rigorous training professional interpreters receive. Irrespective of interpersonal awkwardness or the physician's personal beliefs (Choice B), family members should not act as interpreters for patients (Choices D and E). The nuances of a medical situation may be misunderstood by an untrained interpreter and therefore miscommunicated to the patient. The only situation in which interpretation by a family member is acceptable is an emergency when no other option exists. Educational Objective: Physicians should utilize professional interpreters instead of family members to translate for patients. Professional interpreters are trained to accurately communicate in medical settings and are therefore crucial for patient safety and informed consent to treatment.

61 ---------- Exam Section 2: Item 11 of 50 Natjonal, Board of Medical Examinersment 11. A bronchial smooth muscle preparation is maintained in a tissue bath. Bioactive substances can be applied to the smooth muscle, and contraction and relaxation of the smooth muscle can be measured. The results of administration of thromboxane A2 (TXA2) and prostaglandin E2 (PGE2) are shown. Which of the following is the most likely mechanism of the effect of application of the PGE2? PGE, O A) Activation of phospholipase A2 TXA, O B) Closing K+ channels C) Increase in CAMP O D) Increase in inositol 1,4,5-trisphosphate E) Inhibition of 5-lipoxygenase

C. Smooth muscle cell contraction occurs when myosin light chains are phosphorylated, leading to cross-bridge formations between myosin heads and actin filaments. This reaction is performed by myosin light chain kinase (MLCK). This process is opposed by myosin light chain phosphatase (MLCP), which dephosphorylates the myosin light chains resulting in decreased muscle tone. Prostaglandin E2 (PGE2) stimulates adenylyl cyclase in smooth muscle cells, which leads to an increase in CAMP. CAMP inhibits the activity of myosin light chain kinase, resulting in decreased phosphorylation of myosin light chains, which leads to decreased smooth muscle tension. Incorrect Answers: A, B, D, and E. Activation of phospholipase A2 (Choice A) occurs in proinflammatory states. Phospholipase A2 catalyzes the conversion of phospholipids into arachidonic acid, which is then modified into eicosanoids (such as thromboxane A2) to mediate inflammatory effects. Closing K* channels (Choice B) is not a direct action of PGE2. K* channels are ubiquitous on cell membranes and important in regulating action potentials and membrane depolarization/repolarization. Increase in inositol 1,4,5-triphosphate (Choice D) causes increased smooth muscle tension. Inositol 1,4,5-triphosphate (IP3) is a signaling molecule that increases intracellular calcium release from the sarcoplasmic reticulum of muscle cells. Calcium binds to calmodulin, and the calcium-calmodulin complex activates MLCK, leading to increased muscle contraction. Inhibition of 5-lipoxygenase (Choice E) decreases the production of leukotrienes from arachidonic acid. 5-lipooxygenase inhibitors are commonly used in the treatment of asthma. PGE2 does not affect the activity of 5-lipoxygenase. Educational Objective: Prostaglandin E2 is a potent bronchodilator that acts by stimulating adenylyl cyclase in smooth muscle cells to increase the intracellular concentration of CAMP. CAMP leads to smooth muscle relaxation by inhibiting myosin light chain kinase. Smooth muscle tension Decrease | Increase

17. A 22-year-old man is brought to the emergency department 20 minutes after sustaining a knife wound to the right side of his chest in a fight at a local bar. On arrival, he is short of breath. His pulse is 108/min, respirations are 30/min, and blood pressure is 100/60 mm Hg. Physical examination will show that the trachea has deviated to the left. Further physical examination of the traumatized side is most likely to show which of the following pulmonary findings? A) Crackles B) Egophony C) Hyperresonant percussion D) Increased fremitus E) Stridor F) Wheezing G) Whispered pectoriloquy

C. Tension pneumothorax can be a complication of penetrating thoracic trauma and occurs when air is able to enter the pleural space but is prevented from exiting. Increasing amounts of trapped air in the pleural space compress the lung parenchyma, creating an inability for the lung to expand and causing it to collapse. As the pneumothorax progresses and continues to increase in size, it can result in tracheal and/or mediastinal deviation away from the affected hemithorax. It can also result in decreased venous return to the heart, which manifests with hypotension and obstructive shock. Ďullness in percussion occurs over solid structures such as bones or pleural effusions. Tension pneumothorax is characterized by absent breath sounds and hyperresonant percussion due to the increased air in the intrathoracic space displacing normal lung parenchyma. Incorrect Answers; A, B, D, E, F, and G. Lesions that cause increased consolidation of the lung parenchyma, such as pneumonia, result in rales or crackles (Choice A), egophony (Choice B), dullness to percussion, and increased fremitus (Choice D) and whispered pectoriloquy (Choice G). Rales or crackles refer to an abnormal rattling sound with respiration, which may be seen with pneumonia, but is more common in the setting of pulmonary edema. Dullness to percussion as well as decreased fremitus occurs when there is atelectasis or a pleural effusion present. Egophony occurs when there is an increased resonance of voice sounds heard when auscultating the lungs with a stethoscope, which is due to a change in the frequency of sound passing through consolidated lung tissue. Fremitus refers to vibration and tremors that can be palpated on the chest during patient phonation, which increases lung consolidation and decreases when there is a pleural effusion that is separating the lung tissue from the chest wall. Whispered pectoriloquy refers to an increased sound of whispers auscultated with a stethoscope over an area of lung consolidation. In a tension pneumothorax, these findings would not be present. Stridor (Choice E) is a high-pitched sound caused by turbulent airflow through obstructed or narrowed upper airway structures, such as the pharynx or trachea. It can be caused by illnesses such as croup or epiglottitis, or by structural lesions such as a laryngeal tumor, foreign body, tracheomalacia, or laryngospasm. Wheezing (Choice F) is a continuous low-pitched whistling sound on auscultation that occurs when there is a narrowing of small airways, such as in asthma or chronic obstructive pulmonary disease. It would not likely be present in a patient with tension pneumothorax. Educational Objective: Tension pneumothorax occurs when air is able to enter the pleural space but is prevented from exiting. This leads to an increasing amount of air trapped in the pleural space with the subsequent collapse of the lung parenchyma and mass effect on the mediastinum. It is characterized by respiratory distress, tracheal deviation, hypotension, decreased breath sounds, and hyperresonant percussion on physical examination.

85 ---------- Exam Section 2: Item 35 of 50 Natjonal, Board of Medical Examinersment ansive Rasic Saance 35. A study is conducted to assess 32 patients in a community of 1000 who have developed drug-resistant tuberculosis during a 1-year period. These patients are removed from the community for treatment. Assuming that the risk for infection and susceptibility to the disease is constant, which of the following best represents the number of individuals most likely to develop subsequent drug-resistant tuberculosis during the next year? O A) 27 O B) 29 C) 31 O D) 32 E) 33

C. The incidence rate of disease in a population equals the number of new cases of the disease divided by the number of persons at risk for contracting the disease. Incidence is a measure of risk and is the probability of a given condition occurring within a specified period of time. It is often expressed as a single number of new cases but is more appropriately expressed as a fraction of a population. Importantly, when individuals are added to or are removed from a population, the denominator must also change to reflect the new number of persons at risk for contracting the condition. The incidence rate is mathematically expressed as (the number of new cases) / (the number of persons at risk). In this study, 32 patients contracted drug-resistant tuberculosis over a 1-year period of time. 1,000 persons were at risk over that same period of time. The incidence rate of contracting drug-resistant tuberculosis is therefore 3.2% (32/1000). In the following year, once the 32 affected persons were removed from the community, with risk of infection and susceptibility remaining constant, 968 persons would be at risk (1000-32). The number of individuals likely to develop drug-resistant tuberculosis in the following year is therefore computed by: Incidence rate = number of new cases/number of persons at risk = 3.2% = number of new cases/968 individuals =3.2%*968 individuals = 30.97 individuals, or approximately 31 individuals. Incorrect Answers: A, B, D, and E. 27 (Choice A) and 29 (Choice B) would be potentially correct if a larger number of persons than 32 were removed from the population at risk. 32 (Choice D) reflects the number of persons who would be expected to contract drug-resistant tuberculosis if the population at-risk remained 1,000 individuals. The number of persons added to or removed from a population must be considered. 33 (Choice E) reflects the number of persons who would be expected to contract drug-resistant tuberculosis if the population at-risk contained 1,032 patients, that is, if the number of people supposed to be removed from the at-risk population were instead mistakenly added. This could be arrived at as follows: Incidence rate = number of new cases/number of persons at risk = 3.2% = number of new cases/1,032 individuals = 3.2%*1,032 individuals = 33.024 individuals, or approximately 33 individuals. Educational Objective: The incidence rate of disease in a population equals the number of new cases of the disease divided by the number of persons at risk for contracting the disease. Incidence is a measure of risk and is the probability of a given condition occurring within a specified period of time. Importantly, when individuals are added to or are removed from a population, the denominator must also change to reflect the new number of persons at risk for contracting the condition.

12. A previously healthy 35-year-old woman develops hypoxemia 35 minutes after ingesting a near-lethal dose of barbiturates. She has not aspirated. Which of the following sets of arterial blood gas values (in mm Hg) is most likely in this patient? Po2 Pco2 (A-a)02 A) B) C) D) E)

D. Barbiturates are central nervous system depressants which act on the chloride pore of the GABA receptor, increasing the time the receptor is opened. Symptoms of overdose include encephalopathy and respiratory depression. Respiratory depression leads to hypoventilation as a decreased volume of air is moved to the alveolar-capillary interface for gas exchange. Hypoventilation can be identified by decreased PO2, increased CO-2, and a normal alveolar-arterial (A-a) gradient on arterial blood gas analysis. The alveolar-arterial (A-a) gradient, also written as (A-a)O2, is the difference between the partial pressure of oxygen in the alveoli (PAO 2) and the arterial blood (PaO2). The PaO2 is measured directly on the blood gas analysis. The PAO2 must be estimated using the alveolar gas equation, which is a function of the fraction of inspired oxygen (FiO2), the partial pressure of carbon dioxide (PaCO2), and the respiratory quotient (RQ), which relates to the volume of CO2 produced by tissue to the volume of O2 consumed. If there is no disruption of the alveolar-capillary interface or impaired perfusion to ventilated lung, the PAO2 and PaO2 will equalize as oxygen diffuses readily across the membrane. A normal A-a gradient of~10 mm Hg exists based on the physiologic shunt, which is a volume of blood that does not interact with ventilated lungs. This occurs due to a normal ventilation-perfusion mismatch that exists due to gravity (increased ventilation in the apices of the lung, increased perfusion in the bases) as well as circulation that bypasses the alveolar-capillary interface entirely (eg, bronchial arteries to pulmonary veins, smallest cardiac veins that drain directly into the left ventricle). This deoxygenated blood mixes with freshly oxygenated blood returning from the lungs, leading to a decreased PaO2 relative to the PAO¬2. Hypoventilation results in decreased gas exchange, with no disruption of diffusion or perfusion. A normal A-a gradient is expected. Incorrect Answers: A, B, C, and E. Choices A, B, and E are incorrect as they reveal an increased A-a gradient. The causes of hypoxemia can be divided into those that affect the A-a gradient and those that do not. A normal A-a gradient is seen in decreased environmental FiO2 (eg, high altitude) and hypoventilation. An increased A-a gradient is seen in ventilation-perfusion (V/Q) mismatch (ventilation of lung that is not perfused, eg, pulmonary embolism), right-to-left shunt (perfusion in regions of the lung that are not ventilated, eg, airway obstruction), and diffusion limitation (destruction or impairment of the alveolar-capillary interface, eg, interstitial lung disease, pulmonary edema). These conditions disrupt normal gas exchange and increase the gradient between oxygen pressure in the alveoli and the arterial blood. Choice C is incorrect as it represents a respiratory alkalosis (decreased PCO2). This is caused by hyperventilation, which may be secondary to central nervous system lesions, hypoxemia, a primary lung process, or iatrogenic causes. Central lesions include traumatic brain injury, stroke, hyperthyroidism, medications (eg, salicylates), anxiety, pain, and fear. Primary lung processes include pneumothorax, pulmonary embolism, pneumonia, and acute asthma exacerbation. latrogenic hyperventilation is primarily due to over ventilation of intubated patients, which may be intentional in cases of acute elevations in intracranial pressure. In this case, the patient has respiratory depression which causes CO¬2 retention and respiratory acidosis. Educational Objective: Respiratory depression results in hypoventilation, which is characterized by hypoxemia, respiratory acidosis (due to CO2 retention), and a normal alveolar-arterial gradient on arterial blood gas analysis. An increased A-a gradient is seen in conditions that cause V/Q mismatch, shunt, and/or diffusion limitation.

29 ---------- Exam Section 1: Item 30 of 50 National, Board of Medical Examiners Comprehensive Basic Science Self-Assessment 30. A 55-year-old woman with systemic lupus erythematosus has an increased serum C-reactive protein (CRP) concentration. Which of the following is the most likely source of the increased serum CRP concentration in this patient? O A) Bone marrow O B) Brain C) Kidney D) Liver E) Lymph node F) Skin

D. C-reactive protein (CRP) is an acute phase reactant protein that is synthesized in the liver in response to the pro-inflammatory cytokine interleukin-6. It forms an important element of the innate immune response. CRP functions as an opsonin and aids in the targeting of cells for phagocytosis and in the fixation of complement. Clinically, serum measurement of CRP is used as an indicator of active inflammation. Serum CRP is a non-specific inflammatory marker and can indicate inflammation due to infectious, rheumatologic, or malignant etiologies. Incorrect Answers: A, B, C, E, and F. Bone marrow (Choice A) is the site of hematopoiesis and of the production of immune cells. B lymphocyte maturation occurs in the bone marrow. CRP is not synthesized in the bone marrow. Brain (Choice B) is an immune-privileged site and does not play a role in the synthesis of acute phase reactants. Kidney (Choice C) is the site of production of 1,25-dihydroxycholecalciferol, or vitamin D. There is evidence that vitamin D plays a role in regulation of the immune response. CRP is not synthesized in the kidney. Lymph nodes (Choice E) are the site of mature B and T cells and function as the site of antigen presentation and activation of the acquired immune response. ) are Skin (Choice F) has a principal function in immunity, in addition to thermoregulation and fluid balance. Antigen-presenting cells within skin identify antigens and present them to other immune cells, which may produce a host of cytokines including interleukin-6, which triggers the synthesis of CRP by the liver. Educational Objective: CRP is an acute phase reactant protein that is synthesized in the liver. CRP functions as an opsonin and aids in the targeting of cells for phagocytosis and in the fixation of complement. Serum CRP serves as a non-specific inflammatory marker and can be useful in the diagnosis and monitoring of a variety of inflammatory diseases.

94 ---------- Exam Section 2: Item 44 of 50 Natjonal, Board of Medical Examinersment anive Rasic Science S V 44. A 75-year-old man with alcoholism is brought to the emergency department 30 minutes after he began vomiting blood. He has a history of portal hypertension. Abdominal examination shows spider angiomata. Which of the following is the most likely cause of this patient's hematemesis? O AJAnastomosis between the superior and inferior mesenteric veins O B) Aortoesophageal fistula O C) Inflammation of the distal esophagus D) Retrograde blood flow between tributaries of the left gastric to the azygos veins O E}Shunting ef blood from the portal paraumbilical veins to the superier epigastric veins

D. Cirrhosis can present with edema, ascites, increased bilirubin, jaundice, spider angiomata, and sequelae of portal hypertension such as esophageal varices, splenomegaly, caput medusae, and rectal varices. It typically occurs in patients with preceding conditions such as alcohol abuse or chronic hepatitis. In the setting of portal hypertension, there is the potential for retrograde blood flow between tributaries of the left gastric and azygos veins (portosystemic anastomosis). This results in the formation of esophageal varices, which are dilated submucosal blood vessels. Acutely bleeding esophageal varices typically present with hematemesis, and treatment requires supportive therapy with intravenous fluids and transfusions, along with attempts to stop the bleeding through endoscopic guided sclerosis or banding of the esophageal varices. A transjugular intrahepatic portal shunt can be used in the prevention of variceal bleeding by creating a shunt between the portal vein and hepatic vein, which decreases portal pressure by shunting blood to the systemic circulation, bypassing the liver. Incorrect Answers: A, B, C, and E. Anastomosis between the superior and inferior mesenteric veins (Choice A) is a normal anatomic variant. The inferior mesenteric vein normally drains into the splenic vein, which in turn combines with the superior mesenteric vein to form the portal vein. An aortoesophageal fistula (Choice B) is a life-threatening abnormal communication between the aorta and esophagus, leading to gastrointestinal bleeding and hematemesis. It can be caused by a thoracic aortic aneurysm, foreign body such as an aortic graft, esophageal malignancy, or postoperative complication. It causes massive upper gastrointestinal hemorrhage and is often fatal. It is not associated with portal hypertension. Inflammation of the distal esophagus (Choice C) such as with chronic reflux esophagitis can lead to tissue metaplasia and portends an increased risk of esophageal adenocarcinoma. It would be an unlikely cause of large volume hematemesis. Shunting of blood from the portal paraumbilical veins to the superior epigastric veins (Choice E) of the anterior abdominal wall is also a potential consequence of portal hypertension. This presents as engorged superficial abdominal wall veins referred to as caput medusae. These do not present with hematemesis. Educational Objective: Portal hypertension is a consequence of cirrhosis and can lead to the formation of esophageal varices secondary to retrograde flow through the portosystemic anastomosis between the left gastric and azygos veins. Varices are dilated submucosal blood vessels prone to bleeding and may cause massive gastrointestinal bleeding and hematemesis.

60 ---------- Exam Section 2: Item 10 of 50 National, Board pf Medical Examiners ment V 10. A 34-year-old man has a herniated lumbar intervertebral disc. Laminectomy and removal of the herniated fragment are scheduled. Which of the following labeled sites on the normal lumbar axial CT scan shown is the most appropriate location for surgical entrance to the neural canal? FRONT A C O A) O B) D) E)

D. Disc herniations occur when the nucleus pulposus herniates through the outer fibrocartilaginous ring of the intervertebral disc, the annulus fibrosus. Disc herniations can occur from acute trauma, but often have an underlying degenerative component. Degeneration of intervertebral discs is characterized by decreased water content and flexibility of the spinal column. When the nucleus pulposus herniates through the annulus, it can cause direct compression of neural elements as well as irritation, cytokine release, and inflammatory cell infiltration in the area of the affected nerve root. Often, acute disc herniation occurs laterally in the spinal canal due to the integrity of the posterior longitudinal ligament. This can cause acute radiculopathy, which classically presents with pain in a dermatomal distribution, decreased sensation, and weakness in the corresponding myotome. Typically, the affected nerve root is that which exits the spinal canal one level below the disc herniation, because the nerve root is compressed as it traverses distally prior to exiting the neural foramen. Compression of the distal aspect of the cord, known as conus medullaris syndrome, will present with bilateral upper motor neuron type (eg, hyperreflexia, spastic paresis) lower extremity weakness, perineal numbness, and urinary retention leading to overflow incontinence. Cauda equina syndrome is the compression of the bundle of nerve roots that dwell in the spinal canal distal to the conus medullaris. Compression of the cauda equina can cause similar symptoms to conus medullaris syndrome, however, deficits are more likely to be asymmetric as the bilateral roots are seldom compressed in identical fashion, and it typically results in lower motor neuron type weakness (eg, hyporeflexia). Along with anti-inflammatory medications, surgical decompression may be indicated. During a disc decompression, portions of the posterior elements of the lumbar vertebrae are removed. Commonly, the spinal lamina (the bony arch connecting the pedicles, transverse, and spinous processes) is excised to permit access to the herniated disc. Incorrect Answers: A, B, C, and E. The vertebral body (Choice A) is the anterior portion of the spinal column. The intervertebral disc sits between each vertebral body and allows for shock absorption as well as relative motion between the vertebral bodies. Injuries to the vertebral body include compression fractures in osteoporosis and burst fractures. The pedicle (Choice B) is a stout bony pillar that connects the vertebral body to the posterior elements of the spinal column. Nerve roots exit above and below these pedicles through the neural foramina. The transverse processes (Choice C) of the vertebrae serve as origination points for the psoas muscles, as well as longissimus and multifidus muscles. The spinous process (Choice E) is the posterior-most bony element of the vertebrae. The interspinous ligaments attach between these processes, serving to stabilize and resist forward flexion of the spine. These processes serve as an origin for the latissimus dorsi muscles. Educational Objective: Injury to the spinal column involving bony, ligament, or disc disruption can lead to compression of neural elements. This presents as acute neurologic symptoms such as pain, weakness, and change in sensation in a dermatomal and myotomal distribution. The lamina serves as the most appropriate location for surgical entrance to the spinal canal to access the herniated disc.

32 ---------- Exam Section 1: Item 33 of 50 National, Board of Medical Examiners Comprehensive Basic Science Self-Assessment 33. A 45-year-old homeless man is brought to the emergency department by police 30 minutes after he was found unconscious. His breath and clothes smell of alcohol. His temperature is 36.8°C (98.2°F), pulse is 68/min, respirations are 14/min, and blood pressure is 110/55 mm Hg. Physical examination shows bronzed skin and spider angiomata on the chest. Laboratory studies show: 10 g/dL 30% Hemoglobin Hematocrit Mean corpuscular volume Leukocyte count Segmented neutrophils Lymphocytes Monocytes Platelet count 110 um3 9000/mm3 70% 20% 10% 160,000/mm3 Serum 200 ng/mL 500 pg/mL (N=160-950) 20 ng/mL (N=166-640) Ferritin Vitamin B12 (cobalamin) RBC folate A peripheral blood smear shows occasional hypersegmented neutrophils and 3+ oval macrocytes. Serum studies are most likely to show which of the following sets of additional findings? Methylmalonic Acid Homocysteine A) Increased increased B) Increased normal C) Increased decreased D) Normal increased E) Normal normal F) Normal decreased G) Decreased increased H) Decreased normal 1) Decreased decreased

D. Folic acid, or vitamin Bg, is converted in the body to tetrahydrofolic acid which functions as a coenzyme in the synthesis of nucleic acids. Folate is contained in leafy vegetables and primarily absorbed in the jejunum. Folate supplementation is recommended in pregnancy to decrease the risk of neural tube defects. It is also commonly warranted in patients with rapid cellular turnover, such as acute leukemia and sickle cell disease, to offset the consumption of folic acid in cell division. Folate deficiency is often seen in patients with malnutrition, alcoholism, and patients taking anti-folate medications (eg, phenytoin, methotrexate). Megaloblastic anemia occurs in the setting of impaired DNA synthesis, most commonly related to folate or vitamin B12 (cobalamin) deficiency and is characterized by erythrocyte macrocytosis and hypersegmented neutrophils. Laboratory evaluation reveals anemia, increased mean cell volume, and normal white cell and platelet indices. The presence of hypersegmented neutrophils on peripheral smear is a characteristic finding in megaloblastic anemia. Since folate and vitamin B12 (cobalamin) are needed for the conversion of homocysteine to methionine, deficiency is associated with elevated homocysteine levels. However, vitamin B12 (cobalamin) also acts as a cofactor for the conversion of methylmalonyl-CoA to succinyl-CoA, and its deficiency is associated with increased methylmalonic acid levels. Therefore, folate deficiency would result in increased homocysteine levels but normal methylmalonic acid levels. Incorrect Answers: A, B, C, E, F, G, H, and I. Increased methylmalonic acid levels (Choices A, B, C, and D) would result from vitamin B12 (cobalamin) deficiency, which would also cause increased homocysteine levels. Vitamin B12 (cobalamin) deficiency presents with macrocytic, megaloblastic anemia, but also with neurologic symptoms such as subacute combined degeneration of the spinal cord. Normal (Choice E) or decreased (Choice F) homocysteine levels in the setting of normal methylmalonic acid levels would not be expected in the setting of folate deficiency, which would typically cause an increased homocysteine level. Folate deficiency does not cause decreased methylmalonic acid levels (Choices G, H, and I), as it does not play a role in the metabolism of methylmalonic acid. Educational Objective: Folate deficiency is often seen in patients with malnutrition, alcoholism, or high cell-turnover conditions such as acute leukemia and sickle cell disease. It is associated with megaloblastic anemia, hypersegmented neutrophils, erythrocyte O O OO

86 ---------- Exam Section 2: Item 36 of 50 Natjonal, Board pf Medical Examinersment Sive Basic Science 36. An 81-year-old woman is admitted to the hospital because of a massive pulmonary embolism from a deep venous thrombosis. Her platelet count is 160,000/mm3. Appropriate pharmacotherapy is started. One week later, her platelet count is 55,000/mm3. The thrombocytopenia was most likely caused by a drug with which of the following mechanisms of action? O A) Activates tissue plasminogen O B) Interferes with the carboxylation of coagulation factors O C) Irreversibly inactivates cyclooxygenase D) Potentiates the action of antithrombin III E) Selectively inhibits factor Xa

D. Heparin results in immediate anticoagulation through binding and potentiating the action of anti-thrombin III, causing the inhibition of thrombin and factor Xa. It is commonly used for anticoagulation in the setting of deep vein thrombosis, pulmonary embolism, and acute coronary syndrome. Heparin-induced thrombocytopenia (HIT) is a potential complication of heparin therapy. HIT is characterized by thrombocytopenia (a decrease of greater than 50% from baseline) that typically occurs between 5 to 10 days following heparin exposure, along with evidence of arterial or venous thrombosis. Complications include organ ischemia, skin necrosis, and limb gangrene. HIT occurs secondary to the development of IgG antibodies against the complex of heparin and platelet factor 4. This leads to the activation of platelets, which can cause thrombosis and consumption of platelets. This abnormal activation yields progressively decreasing platelet counts, which typically nadir at greater than 50% from their initial level. Treatment requires cessation of all heparin- related products (eg, unfractionated heparin and low molecular weight heparin), and the initiation of a direct thrombin inhibitor or direct Xa inhibitor to ensure adequate anticoagulation without worsening the thrombocytopenia. Incorrect Answers: A, B, C, and E. Tissue plasminogen activator (TPA) activates tissue plasminogen (Choice A). TPA is a thrombolytic that cleaves plasminogen to form active plasmin. Plasmin degrades the cross-links between fibrin leading to the dissolution of blood clots. The major adverse effect of tPA is bleeding. Warfarin is an anticoagulant that interferes with the carboxylation of coagulation factors (Choice B). This occurs through the inhibition of vitamin K epoxide reductase, which is necessary for recycling vitamin K, an essential cofactor in the synthesis of coagulation factors in the liver. Warfarin may introduce a transient hypercoagulable state when initiated, which increases the risk for skin necrosis. Typically, warfarin is begun with the concomitant administration of heparin to avoid this complication. Aspirin is an antiplatelet agent that irreversibly inactivates cyclooxygenase (Choice C), which prevents the conversion of arachidonic acid to thromboxane A2. Thromboxane A2 is involved in platelet activation and its decreased synthesis results in diminished platelet activation. Complications of aspirin use include bleeding, Reye syndrome, and a mixed respiratory alkalosis-metabolic acidosis (in the setting of salicylate toxicity). Novel (or direct) oral anticoagulants include apixaban, rivaroxaban, and edoxaban, which selectively inhibit factor Xa (Choice E). The other major novel oral anticoagulant is dabigatran, which is a direct thrombin inhibitor. The major adverse effect of the direct oral anticoagulants is bleeding. Educational Objective: Heparin is a common anticoagulant that potentiates the action of anti-thrombin III to inhibit multiple coagulation factors. A potential complication of heparin therapy is HIT, which is an antibody-mediated reaction to heparin-platelet factor 4 complexes that results in the abnormal activation of platelets, with subsequent thrombocytopenia and arterial or venous thromboembolic disease.

87 ---------- Exam Section 2: Item 37 of 50 Natjonal, Board of Medical Examinersment andive Rasic Saance 37. A 68-year-old woman comes to the physician because of a 1-year history of severe abdominal pain after meals; she also has had an unintentional 9-kg (20-lb) weight loss during this period. The pain is relieved when she decreases the amount of food that she eats. She has a history of atherosclerosis and underwent triple coronary artery bypass grafting 2 years ago. Physical examination shows a soft, nontender abdomen and an abdominal bruit. Pedal pulses are diminished. The most likely cause of this patient's symptoms is stenosis of which of the following arteries? O A) Greater pancreatic о в) Нерatic O C) Right gastric D) Superior mesenteric O E) Supraduodenal

D. Mesenteric ischemia occurs when there is a reduction in blood flow to the small intestine. It is either acute or chronic, in which acute mesenteric ischemia occurs due to an embolic or thrombotic arterial obstruction, and chronic mesenteric ischemia is caused by chronic atherosclerotic stenosis of the celiac or mesenteric arteries, with the superior mesenteric artery the most commonly affected. Chronic mesenteric ischemia is characterized by recurring episodes of abdominal pain after eating, which is hypothe secondary to mismatch between intestinal blood flow and metabolic demands, or secondary to small intestinal hypoperfusion when blood is shunted to the stomach during meals. Symptoms can be progressive and lead to the avoidance of eating and weight loss. Clinicians should have a high index of suspicion for chronic mesenteric ischemia especially in patients who exhibit other sequelae of atherosclerotic disease, including peripheral arterial disease with diminished pedal pulses or coronary artery disease. ed to be Incorrect Answers: A, B, C, and E. The greater pancreatic artery (Choice A) is a branch of the splenic artery that supplies the pancreatic tail and body. The common hepatic artery (Choice B) supplies the liver, pylorus, duodenum, pancreas, and gallbladder. It branches into the proper hepatic artery, gastroduodenal artery, and the right gastric artery (Choice C), which supplies the lesser curvature of the stomach. The supraduodenal artery (Choice E) is a branch of the gastroduodenal artery that supplies the first and second parts of the duodenum. None of these arteries supply the small intestine and would be unlikely to contribute to the symptoms of chronic mesenteric ischemia. Educational Objective: Mesenteric ischemia occurs when there is a reduction in blood flow to the small intestine. Chronic mesenteric ischemia is caused by chronic atherosclerotic disease of the celiac or mesenteric arteries, with the superior mesenteric artery the most commonly affected, and is characterized by recurring episodes of abdominal pain after eating, avoidance of food, and weight loss.

81 ---------- iv Medical Examinersment Exam Section 2: Item 31 of 50 National Board * 31. A strain of Escherichia coli produces a temperature-sensitive tryptophan synthase and therefore requires exogenously supplied tryptophan for growth at 42°C but not at 30°C. Which of the following mutations is most likely responsible for this phenotype? O A) Deletion O B) Frameshift C) Insertion O D) Missense O E) Nonsense

D. Missense mutations are a type of point mutation in which the change of a single base pair forms a codon that results in substitution of a single, different amino acid from that usually encoded. Substitution of a single amino acid may result in a wide range of changes in protein function depending on the position of the substitution. The change in amino acid may result in no or minimal change to protein function or may lead to non-functional or misfolded proteins. The described change in the mutant tryptophan synthase, in which the protein retains normal function except under particular temperature conditions, is a minor change in protein function that is most consistent with a missense mutation. Other types of mutation typically result in more severe changes in protein function. Incorrect Answers: A, B, C, and E. Deletion (Choice A) of a portion of the genetic sequence results in shifts of the reading frame if the deletion does not involve a number of base pairs that is divisible by three. Deletion mutations typically result in more severe alterations of protein function (generally loss of function) compared to missense mutations. Frameshift mutations (Choice B) result from changes to the reading frame after insertion or deletion of a number of base pairs that is not divisible by three. Frameshift mutations typically result in severe loss of function. Insertion mutations (Choice C) involve the addition of base pairs to the genetic sequence and may result in shift of the reading frame if the insertion does not involve a number of base pairs that is divisible by three. Insertion mutations typically result in more severe alterations of protein function compared to missense mutations. Nonsense mutations (Choice E) are point mutations that result in premature stop codons and protein truncation. Nonsense mutations typically result in loss of protein function. Educational Objective: Missense mutations are a type of point mutation in which the change of a single base pair forms a codon that results in substitution of a single, different amino acid from that usually encoded. Substitution of a single amino acid may result in a wide range of changes in protein function depending on the position of the substitution.

11. An 8-year-old boy is brought to the physician by his parents because of disruptive behavior. They say, "He is easily distracted, constantly interrupts us while we are talking, and seems to be always moving. His teacher says he is always talking with his neighbors, has trouble completing tasks, and refuses to wait his turn when playing games." Physical examination shows no abnormalities. If drug therapy is indicated, administration of a drug with which of the following mechanisms of action is most appropriate? A) Antagonism at B-adrenergic receptors B) Blockade of voltage-gated Na+ channels C) Enhanced action of y-aminobutyric acid (GABA) at GABAA receptors D) Increased release of dopamine and norepinephrine E) Selective inhibition of uptake of serotonin at nerve terminals

D. The first-line treatment for attention-deficit/hyperactivity disorder (ADHD) is stimulant medication such as amphetamine salts, which increase the presynaptic release of dopamine and norepinephrine. ADHD presents with chronic symptoms of hyperactivity/impulsivity and/or inattention that occur in more than one setting and impair academic, social, or emotional function. Children with predominant hyperactivity symptoms are unable to sit still and may have difficulty taking turns (postulated to result from a dysregulated reward pathway), whereas children with predominant inattentive symptoms tend to daydream, process information slowly, and demonstrate difficulty in completing tasks (likely related to norepinephrine dysregulation). This patient is experiencing symptoms of both inattention and hyperactivity. Combination treatment with medications and psychotherapy may be the most effective option, though medication monotherapy is also appropriate for children older than 6 years. Stimulants, which include amphetamine salts and methylphenidate, are first-line agents. Both types of stimulants increase synaptic dopamine and norepinephrine, which improve reward processing and attention, respectively. Incorrect Answers: A, B, C, and E. Antagonism at ß-adrenergic receptors (Choice A) is the mechanism of B-adrenergic blocker medications. Some B-adrenergic blocker medications, such as propranolol, are centrally acting and can be used for performance anxiety or traumatic brain injury-related agitation. These medications are unhelpful in ADHD. Blockade of voltage-gated Na" channels (Choice B) is the mechanism utilized by several mood stabilizers and antiepileptics including carbamazepine and lamotrigine. These medications are used for bipolar disorder, not ADHD. Enhanced action of y-aminobutyric acid (GABA) at GABAA receptors (Choice C) is the mechanism of sedative agents such as benzodiazepines and barbiturates. Benzodiazepines can be used for panic disorder. However, these medications are not useful for ADHD symptoms. Selective inhibition of uptake of serotonin at nerve terminals (Choice E) is the mechanism of first-line treatments for major depressive disorder, which belong to the class of selective serotonin reuptake inhibitors. These medications do not target ADHD. Educational Objective: The first-line treatment for attention-deficit/hyperactivity disorder is stimulant medication such as amphetamine salts and methylphenidate. These medications increase presynaptic release of dopamine and norepinephrine, which improve hyperactivity and inattention, respectively.

6. A 62-year-old man is being evaluated for rectal bleeding. An x-ray of the gastrointestinal tract with contrast material is shown. Which of the following is the most likely explanation for the feathery appearance in the portion of the gastrointestinal tract indicated by X when compared with the portion indicated by Y? A) Absence of circular muscle B) Fewer villi C) Greater bowel motility D) Greater mucosal surface area E) Less circular and longitudinal smooth muscle

D. The jejunum (X) can be distinguished from the ileum (Y) by several characteristic features. The inner mucosal folds, or plicae circulares, are more prominent, more numerous, and taller in the jejunum compared to the ileum. This leads to an increased mucosal surface area and an associated feathered appearance after the administration of oral contrast material. The jejunum also demonstrates a larger caliber, thicker muscular walls, longer vasa rectae, and fewer arcades. The ileum is notable for the presence of Peyer patches and mucosal lymphoid follicles, which can be identified histologically. Incorrect Answers: A, B, C, and E. Absence of circular muscle (Choice A) is not a feature of either the jejunum or the ileum. The muscularis externa, which is organized into circular and longitudinal layers, extends through all segments of the small intestine. Fewer villi (Choice B) is not a feature of the jejunum. Both the jejunum and the ileum contain villi. However, the plicae circulares are much more prominent and numerous in the jejunum, leading to an increased overall surface area and an associated increased number of mucosal villi compared to the ileum. Greater bowel motility (Choice C) is not a distinguishing characteristic of the jejunum as compared to the ileum. Bowel motility is controlled by peristaltic contractions and migrating motor complexes that are under the control of the enteric and central nervous systems. Less circular and longitudinal smooth muscle (Choice E) is a feature of the ileum (Y) rather than the jejunum (X). The jejunum has a thicker muscular wall in comparison to the ileum, but its feathered appearance with oral contrast material is primarily due to its prominent plicae circulares and increased mucosal surface area. Educational Objective: The jejunum possesses several characteristic features, including prominent plicae circulares, which lends the jejunum an increased mucosal surface area and a feathered appearance with oral contrast material. The jejunum also demonstrates a larger caliber, thicker muscular walls, longer vasa rectae, and fewer arcades.

75 ---------- Exam Section 2: Item 25 of 50 National, Board pf Medical Examiners diveB A ance nent * 25. In a healthy 25-year-old man with a normal diet and fluid intake, which of the following substances filtered by the kidney has the greatest fraction excreted in urine? O A) Glucose B) Magnesium C) Sodium D) Urea E) Water

D. The fractional excretion of a solute measures the percentage of the original filtered amount that is present in the excreted urine. The calculations are made using the plasma and urine concentrations of the solute in question. The fractional excretion of a molecule is more reliable than the urine concentration of that molecule as urine concentration can vary with the amount of water present in the urine. Out of the molecules listed, urea has the highest fractional excretion. Urea is created by the urea cycle as a way to excrete nitrogen and prevent the accumulation of ammonia, which is toxic. Urea is filtered by the glomerulus and then reabsorbed in the proximal convoluted tubule at a rate that is lower than that of water. It is also an integral component of the medullary countercurrent system, allowing for the reabsorption of water in the loop of Henle. Approximately 40% of the initially filtered urea is found in the final excreted urine. Incorrect Answers: A, B, C, and E. Glucose (Choice A) has a fractional excretion of nearly 0% in healthy, nondiabetic patients with serum glucose levels less than 180 mg/dL. The proximal convoluted tubule is the primary site of glucose reabsorption. In diabetes mellitus, the filtered load of glucose exceeds the resorptive capacity of the proximal convoluted tubule and glucosuria ensues. Under physiologic conditions, the fractional excretion of magnesium (Choice B) is approximately 3% to 5%. Unlike most other solutes, the site of reabsorption of magnesium is the thick ascending loop of Henle, not the proximal convoluted tubule. The fractional excretion of sodium (Choice C) differs based on volume status but is usually less than 1%. When dehydration is present, increased sodium is reabsorbed to create an osmotic gradient for water to follow. In a well-hydrated patient, there is less reabsorption of sodium, leading to an increased fractional excretion. The fractional excretion of sodium is often used clinically to determine the cause of an acute renal injury. The specific gravity of urine is the measurement more often used to determine the amount of water (Choice E) in urine. Fractional excretion applies to solutes rather than water, which is a solvent. Educational Objective: Fractional excretion is the percentage of a filtered solute that is excreted in the urine. Urea has a fractional excretion of approximately 40%, which is much higher than that of magnesium (3%-5%), sodium (less than 1%), and glucose (0%) under physiologic conditions.

80 ---------- Exam Section 2: Item 30 of 50 National, Board pf Medical Examiners ent 30. A 14-year-old girl is brought to the physician by her mother because of episodes of increasingly severe facial blemishes during the past 6 months. She says that she does not eat high-fat foods or use makeup. Use of an astringent soap has not resolved her symptoms. Menarche occurred at the age of 12 years. A photograph of her face is shown. Which of the following best describes the pathologic mechanism of this patient's condition? O A) Acceleration of the proliferation of cells in the dermis by increased estrogen production O B) Colonization of apocrine sweat glands by fungi O C) Decreased blood flow to skin capillaries by overproduction of elastic fibers D) Follicular epidermal hyperproliferation with excess production of sebum E) Thickening of the basement membrane in response to increased pubertal serum estrogen concentrations

D. The initial lesion of acne vulgaris is a comedo, a hair follicle that has been blocked by keratin debris. The inciting event in comedo formation is hyperproliferation of the epidermis and abnormal keratinization. This is compounded by androgenic stimulation of sebaceous glands associated with the hair follicle, which are together called a pilosebaceous unit, leading to increased sebum production. This increased sebum provides a substrate for bacterial overgrowth of normal skin flora including Štaphylococcus epidermidis and Propionibacterium acnes. With the accumulation of keratin debris, increased sebum production, and bacterial overgrowth, the comedo becomes inflamed and ruptures, causing a papule or cyst to form. Clinically, areas with increased sebaceous activity including the face, upper back, and chest are prone to acne. It is a common disorder that affects the vast majority of adolescents. Treatments are aimed at reducing comedo formation (retinoids), decreasing sebum production (anti-androgens), and mitigating bacterial overgrowth (antibiotics). Incorrect Answers: A, B, C, and E. Androgens, not estrogen (Choice A), contribute to acne vulgaris. While the sebaceous glands reside in the dermis, it is not proliferation of the sebaceous cells but increased production of their secretions that contribute to acne. Bacterial colonization of the pilosebaceous unit, not colonization of the apocrine glands by fungi (Choice B) leads to acne vulgaris. Apocrine glands are located in the axillae, areolae, and anogenital regions. These areas are not typically affected by acne vulgaris.Dermal collagen fibers, rather than elastic fibers (Choice C), increase as a result of scarring. While the majority of collagen fibers in healthy skin are type I collagen, scar is initially created by type III collagen. The basement membrane (Choice E) is not thickened by estrogen nor does it play a role in the development of acne vulgaris. Educational Objective: Comedo formation, the key inciting event in acne vulgaris, is initiated by proliferation and abnormal keratinization of the hair follicle epidermis. Excess production of sebum then provides a hospitable environment for bacterial overgrowth. Acne treatments are aimed at addressing these aberrancies.

56 ---------- Exam Section 2: Item 7 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 7. A 7-year-old boy is about to undergo an appendectomy. An intravenous catheter needs to be inserted, but the patient is extremely fearful of being stuck with a needle while awake. The most appropriate anesthesia administered by mask to anesthetize this patient quickly would have which of the following characteristics? A} High blood-selubility B} High-cerebrespinal-fluid-solubility C) High lipid solubility D) Low blood solubility E) Low lipid solubility

D. The rate of induction, or onset of anesthesia, with volatile anesthetics relies on blood solubility, cardiac output, and minute ventilation. Induction is primarily dependent on the partial pressure of inhaled volatile anesthetic present in the alveoli. Solubility in this case can be equated with the blood gas partition coefficient, which describes the relative division between two phases that is typically taken by an anesthetic. Low blood solubility increases the speed of induction, as it allows the alveolar concentration to quickly come into equilibrium with the concentration of anesthetic in the blood, preventing further anesthetic from being removed by the blood. Similarly, low cardiac output decreases the amount of blood delivered to the alveoli, and the anesthetic will more quickly equilibrate with the blood that is present. Increases in minute ventilation also speed induction time by delivering more anesthetic to the alveoli, leading to a more rapid increase in the gas concentration. The converse of these statements is true, and induction speed decreases with high blood solubility, high cardiac output, and low minute ventilation. Therefore, choosing a volatile anesthetic agent with low blood solubility will lead to rapid induction of anesthesia. Incorrect Answers: A, B, C, and E. High blood solubility (Choice A) will delay the speed of induction, as it increases the amount of time it takes for the concentration of gas in the alveoli to reach equilibrium with the concentration of anesthetic in the blood. High cerebrospinal fluid solubility (Choice B) would also increase the time required for induction, as anesthetic would be rapidly taken up into the cerebrospinal fluid from the bloodstream, increasing the time it takes for alveolar gas concentration to equilibrate with blood anesthetic concentration. Neither high lipid solubility (Choice C) or low lipid solubility (Choice E) would affect induction time. Potency of volatile anesthetic is dependent on lipid solubility, with those medications that are highly lipid soluble being much more potent. Educational Objective: Induction time of anesthesia with volatile anesthetics depends on the blood solubility of the anesthetic, cardiac output, and minute ventilation. Low blood solubility decreased cardiac output, and high minute ventilation will increase the speed of induction. Conversely, high blood solubility, increased cardiac output, and low minute ventilation will decrease the spe of induction.

92 ---------- Exam Section 2: Item 42 of 50 National Board of Medical Examiners nent 42. An 18-year-old man is brought to the emergency department 30 minutes after sustaining injuries in a motor vehicle collision. He has abdominal pain. His pulse is 120/min, and blood pressure is 80/60 mm Hg. Physical examination shows multiple contusions over the trunk and abdominal tenderness. A chest x-ray shows multiple rib fractures. A CT scan of the abdomen is shown. Which of the following organs is most likely injured in this patient? O A) Bladder O B) Liver O C) Pancreas D) Spleen O E) Stomach

D. The spleen is one of the most common organs damaged during blunt abdominal trauma, sustaining injuries such as laceration and rupture. The CT image demonstrates the upper abdomen, with a normal-appearing liver in the right upper quadrant and a complex high- grade laceration and rupture of the spleen in the left upper quadrant surrounded by extravasated blood. Splenic rupture can cause large amounts of blood to leak into the peritoneal space, which can result in hemodynamic instability characterized by tachycardia and hypotension. Patients with splenic injuries may have left upper quadrant tenderness, peritonitis from blood products irritating the peritoneum, or referred pain to the left shoulder secondary to diaphragmatic irritation. In hemodynamically stable patients, minor splenic injuries may be treated nonoperatively with observation and reassessment. In unstable patients, exploratory laparotomy is required for repair of the injury or splenectomy, depending on the nature and severity of the injury. Incorrect Answers: A, B, C, and E. The bladder (Choice A) is not shown in this CT image. Blunt abdominal trauma to the lower abdomen or pelvis can cause bladder rupture and urine leakage into the abdominopelvic cavity. Bladder injuries are commonly associated with pelvic fractures. Patients with bladder injuries may present with gross hematuria, suprapubic tenderness, difficulty urinating, or peritonitis from irritation by intraperitoneal urine. Liver (Choice B) injuries in blunt abdominal trauma may cause hemoperitoneum, resulting in hemodynamic instability. However, the CT image reveals a normal appearing liver in the right upper quadrant without any evidence of injury. Pancreatic (Choice C) injuries in blunt trauma are less common. Clinical evaluation is nonspecific for pancreatic injury, and may present with abdominal tenderness, abdominal wall ecchymosis, emesis, or may be asymptomatic. Diagnosis can be made with CT imaging to assess for discontinuity of the pancreatic parenchyma and duct, but the pancreas is not shown in this image. The stomach (Choice E) is located in the left upper quadrant adjacent to the spleen, but rupture or perforation of the stomach following blunt trauma is uncommon. Findings on CT scan suggesting stomach or other gastrointestinal injury include pneumoperitoneum and intra-abdominal fluid in the absence of a solid organ injury. There is intra-abdominal fluid seen on the image, but it is related to blood products from the splenic rupture and not due to a stomach injury. Educational Objective: The spleen is one of the most commonly injured organs in the setting of blunt abdominal trauma. It can present with abdominal tenderness and if severe, hemodynamic instability from large volume hemoperitoneum. Treatment is supportive for stable patients but requires exploratory laparotomy with repair or resection of the spleen for unstable patients.

68 ---------- Exam Section 2: Item 18 of 50 Natjonal, Board of Medical Examinersment Cell Membrane Phospholipids 18. The illustration shows the arachidonic cascade that is activated when skin is exposed to poison ivy. Within this cascade, which of the following is the major mechanism of analogs of corticosteroids? Phospholipase A, Arachidonic Acid A) Acceleration of the metabolism of arachidonic acid Cyclooxygenase Lipoxygenase O B) Inhibition of receptors mediating selected responses to prostaglandins and leukotrienes Prostaglandins Leukotrienes O C) Nonselective inhibition of cyclooxygenase and lipoxygenase Tissue receptors Tissue receptors O D) Regulation of phospholipase A2 E) Selective inhibition of lipoxygenase

D. Topical corticosteroids, such as triamcinolone, betamethasone, and clobetasol, are the treatment of choice for allergic contact dermatitis. These medications are analogues of the corticosteroids produced by the zona fasciculata of the adrenal gland. While corticosteroids decrease levels of virtually all cytokines, in the arachidonic cascade they specifically inhibit phospholipase A2. They also inhibit NF-KB, a positive regulator of cyclooxygenase (COX) 2 and thus indirectly inhibit COX2. Allergic contact dermatitis is a type IV hypersensitivity reaction caused by exposure of the skin to either an irritant or allergen causing an eczematous rash to occur in the distribution of the exposure. The role of topical corticosteroids in decreasing cytokines is helpful because in this type of hypersensitivity, sensitized helper T cells rely on cytokines to cause inflammation and attract macrophages. By decreasing production of cytokines, the inflammatory response and therefore the degree of symptom severity is significantly reduced. Incorrect Answers: A, B, C, and E. Acceleration of the metabolism of arachidonic acid (Choice A) would increase the rate of production of prostaglandins and leukotrienes. Prostaglandins and leukotrienes increase neutrophil chemotaxis and decrease vascular tone, respectively, which both facilitate the inflammatory response to an insult. This would cause worsening of inflammation, not improvement. Inhibition of receptors mediating selected responses to prostaglandins and leukotrienes (Choice B) is a mechanism used by several medications, such as montelukast. Montelukast is an example of a leukotriene receptor antagonist that is selective for leukotrienes C4, D4, and E4. This medication is used to prevent smooth muscle contraction and airway edema in asthma. However, inhibition of these or prostaglandin receptors would not decrease the cytokine milieu contributing to the type IV hypersensitivity reaction to poison ivy. Aspirin and non-steroidal anti-inflammatory drugs (NSAIDS), excluding celecoxib, are nonselective inhibitors of COX 1 and 2 (Choice C). Zileuton is a selective inhibitor of lipoxygenase (Choice E). Both of these targets reduce inflammation and the downstream effects of prostaglandins and leukotrienes. However, steroids act upstream in the pathway when compared with COX and lipoxygenase inhibitors. Educational Objective: Topical corticosteroids are first-line therapy for allergic contact dermatitis. They decrease the cytokine milieu by inhibiting phospholipase A2, thus preventing helper T cells from recruiting additional inflammatory cells to the site of exposure.

30 ---------- Exam Section 1: Item 31 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 31. A 45-year-old man comes to the physician because of a 2-month history of a dull pain on the left side of his scrotum. Examination of the left side of the scrotum shows a bluish appearance of the skin and a palpable mass that feels like a bag of worms. Which of the following veins is most likely obstructed in this patient? A) Inferior mesenteric vein B) Inferior vena cava C) Inferior vesical vein D) Internal iliac vein E) Left renal vein

E. A varicocele refers to dilated veins of the pampiniform plexus due to elevated venous pressure. Varicoceles are commonly located on the left side due to increased resistance to venous flow as the left gonadal vein drains into the left renal vein. On the right side, the right gonadal vein drains directly into the inferior vena cava and is less likely to become obstructed. An obstruction of the left renal vein, such as may be caused by local invasion of a renal tumor, causes pressure to mount behind the obstruction which is transmitted through the left gonadal vein to the pampiniform plexus surrounding the testicle. The dilation of the pampiniform plexus causes a bluish appearance of the overlying skin. Palpation of the dilated plexus demonstrates varicosities, which are often described as having a vermiform (worm-like) feeling. Unlike a hydrocele, which is caused by failure of the processus vaginalis to close leading to accumulation of fluid around the testicle, a varicocele will not transilluminate with application of light to the scrotum. Infertility is a potential complication of varicoceles, as the dilated veins result in increased temperature of the testicle, which impairs spermatogenesis. Treatment for varicoceles may require surgical embolization or ligation to manage infertility. Incorrect Answers: A, B, C, and D. The inferior mesenteric vein (Choice A) drains the large intestine into the splenic vein, which in turn merges with the superior mesenteric vein to form the hepatic portal vein. Obstruction of the inferior mesenteric vein would cause venous congestion of the large intestine and superior rectum, not the pampiniform plexus in the scrotum. The inferior vena cava (Choice B) is the large vessel formed by the merging of the bilateral common iliac veins which carries blood from the abdomen, pelvis, and lower extremities back to the heart. Its wide lumen is less likely to become completely obstructed than that of a smaller vein. If occlusion did occur, widespread venous congestion of the bilateral lower extremities, pelvic organs, and gastrointestinal tract would occur. The inferior vesical vein (Choice C) drains blood from the urinary bladder. Its obstruction would not lead to dilation of the pampiniform plexus. The internal iliac vein (Choice D) and its tributaries drain blood from the pelvic organs and pelvic wall into the common iliac veins. The gor dal veins are not tributaries of the internal iliac vein and thus obstruction would not lead to varicocele formation. Educational Objective: The testes are drained by the gonadal veins, which drain into the left renal vein on the left side and directly into the inferior vena cava on the right side. Obstruction of or increased pressure within the left gonadal vein or left renal vein leads to a left-sided varicocele.

89 ---------- Exam Section 2: Item 39 of 50 Natjonal, Board of Medical Examinersent 39. A 76-year-old woman comes to the physician because of a 2-day history of abdominal pain, nausea, and vomiting. Her temperature is 38.9°C (102°F), pulse is 88/min, respirations are 26/min, and blood pressure is 117/79 mm Hg. Abdominal examination shows tenderness of the left lower quadrant. Laboratory studies show a leukocyte count of 22,500/mm3 with 8% bands. A CT scan of the abdomen shows an abscess in the left lower quadrant. The patient undergoes drainage of the abscess and a partial sigmoidectomy for a ruptured diverticulum. Neutrophil activation in this patient is directed, in part, by bacterial lipopolysaccharides. As a result, these lipopolysaccharides most likely bind to which of the following neutrophil receptor types? O A) Cytokine O B) G protein-linked O C) lon channel-linked D) Nuclear E) Toll-like

E. Acute diverticulitis presents due to infection or inflammation of existing colonic diverticula, which are small outpouchings of the bowel wall, most commonly located in the sigmoid colon. Diverticulitis presents with acute left lower quadrant abdominal pain, sometimes accompanied by lipopolysaccharide binding to toll-like receptors on the neutrophil surface. Toll-like receptors bind to pathogen-associated molecular patterns (PAMPS) such as bacterial lipopolysaccharides and activate the NF-KB pathway. NF-KB is a transcription factor that induces the expression of several proinflammatory genes, promoting the production of cytokines, chemokines, and other inflammatory mediators. diarrhea, fever, and tenderness to palpation in the left lower quadrant. Complications of acute diverticulitis include abscess or fistula formation, and perforation (rupture). The innate immune response to acute diverticulitis includes neutrophil activation Incorrect Answers: A, B, C, and D. Cytokine (Choice A) receptor binding modulates the response of neutrophils to infection. Proinflammatory cytokines include interleukin-1 (IL-1), 6, 8, 12 and 18, interferons, and tumor necrosis factor. Anti-inflammatory cytokines include IL-4, 10, 11, and 13. Lipopolysaccharides bind to toll-like receptors on neutrophils, not cytokine receptors. G protein-linked (Choice B) receptors are involved in neutrophil chemotaxis. Ligands include leukotriene B4, platelet-activating factor, and complement fragment C5a. lon channel-linked (Choice C) receptors are involved in modulating several stages of neutrophil recruitment. They do not bind with pathogen-associated molecular patterns. Nuclear (Choice D) receptors are ligand-related transcription factors that control gene expression in response to external signals. Lipopolysaccharides bind to extracellular toll-like receptors on neutrophils, not intracellular nuclear receptors. Educational Objective: Neutrophil activation as part of the innate immune response to infection involves binding of pathogen-associated molecular patterns to toll-like receptors on the neutrophil surface, leading to NF-KB activation and production of proinflammatory mediators.

31 ---------- Exam Section 1: Item 32 of 50 National, Board of Medical Examiners Comprehensive Basic Science Self-Assessment 32. A 55-year-old man comes to the physician because of a 2-month history of decreased appetite; he also has had a 9-kg (20-lb) weight loss and an intermittent rash during this period. Physical examination shows necrolytic migratory erythema over the axillae and groin. Laboratory studies show a serum glucose concentration of 280 mg/dL and plasma glucagon concentration of 1500 pg/mL (N=20-100). A CT scan of the abdomen shows a mass at the head of the pancreas. Which of the following processes is most likely occurring in this patient's liver? A) Decreased activity of carnitine acetyltransferase O B}Glycogen synthesis C) Increased activity of acetyl-CoA carboxylase D) Inhibition of ketogenesis E) Partial oxidation of fatty acids

E. Glucagonoma is a rare functional malignancy that secretes glucagon, which raises blood glucose. Occasionally, glucagonoma can present as a component of multiple endocrine neoplasia. Symptoms are nonspecific and include mild diabetes mellitus, paraneoplastic rash (eg, necrolytic migratory erythema), and weight loss. Glucagon acts on hepatocytes via a CAMP pathway activating protein kinase A that ends in the activation of glycogen phosphorylase, with release of glucose monomers into serum. When a glucagonoma is present, glucagon is continuously secreted, unlike in the physiologic state. When the glycogen stores are exhausted, the body then relies on the breakdown of fat through beta oxidation of fatty acids to provide energy. Beta oxidation occurs in the mitochondria of the cell. In each cycle of beta oxidation, two carbon atoms are cleaved to form acetyl-ČOA, which leaves the mitochondria via the carnitine shuttle. In addition to a molecule of acetyl-CoA, which then enters the citric acid cycle, a molecule of NADH is garnered with each cycle. The ability to generate energy from the body's fat stores during a period of starvation is critical to maintaining the function of the vital organs, such as the brain. Incorrect Answers: A, B, C, and D. Carnitine acetyltransferase (Choice A) is a transmembrane protein found on the surface of mitochondria in liver, muscle, and brain, and catalyzes the bond between acyl-CoA and carnitine, which permits the long-chain fatty acid to translocate into the mitochondrial matrix where beta-oxidation (breakdown of the fatty acid) occurs. Its activity is increased in the presence of glucagon, not decreased. Glycogen synthesis (Choice B) occurs when there is a surplus of glucose in order to store energy for future use. In contrast, a glucagonoma secretes glucagon which signals a state of starvation to the liver, causing glycogen phosphorylase to be activated which results in the degradation of glycogen stores for maintenance of serum glucose concentration. Acetyl-CoA carboxylase (Choice C) regulates the metabolism of fatty acids and, when active, adds a carboxy group to acetyl-CoA to make malonyl-CoA. This is a key ingredient in the formation of new fatty acids. If the activity of this enzyme is increased, fatty acids will be created rather than broken down. This would be seen in the fed state rather than the starvation state mimicked by increased glucagon levels. Ketogenesis (Choice D) is the process by which ketone production occurs through fatty acid and amino acid degradation. Ketogenesis occurs in the starvation state in order to supply the brain, heart, and skeletal muscle with adequate nutrition. In the starvation state mimicked by increased glucagon levels, ketogenesis is increased, not inhibited. Educational Objective: Oxidation of fatty acids in the mitochondrial matrix occurs during a state of starvation, or a state with increased glucagon levels, in order to maintain nutrition when glycogen stores have been depleted. OO O OO

27 ---------- Exam Section 1: Item 28 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 28. A female newborn is delivered vaginally at 37 weeks' gestation after an uncomplicated pregnancy, labor, and delivery. Physical examination shows a narrow cranium that is elongated in the anteroposterior dimension. Premature closure of which of the following during embryogenesis is the most likely cause of this anomaly? A) Bregma B) Coronal suture C) Lambda D) Lambdoid suture E) Sagittal suture

E. Craniosynostosis refers to the premature ossification that occurs between the plates of the cranial bones leading to deformity of the skull. Most infants with this condition are asymptomatic. Genetic etiologies may involve abnormalities in expression of FGF and TGF. Normally, the skull is composed of five independent bones. These include the two frontal bones, two parietal bones, and the occipital bone. The fontanelles are cartilaginous gaps that exist between these bony plates that, along with the cartilaginous sutures, allow for movement, growth, and compression of the cranial bones particularly during childbirth when the skull passes through the pelvis and vaginal canal. Closure of one or more of these regions prematurely leads to abnormal growth and fusion of the skulI. The resulting deformity can be both cosmetic (abnormal head shape) and pathologic (limiting normal central nervous system growth). A number of patterns of deformity occur depending on which sutures prematurely fuse. In this patient, fusion of the sagittal suture arrests the medial-lateral growth of the skull. This leads to continued anterior-posterior growth in disproportion to medial-lateral growth. A narrow, elongated anteroposterior cranium results, known as scaphocephaly. Incorrect Answers: A, B, C, and D. The bregma (Choice A) is the point at which the sagittal suture and the coronal suture meet. It is the anterior counterpart to the lambda. Premature closure of the coronal suture (Choice B) leads to brachycephaly, which is a short, broad, flattened skull. This results in arrest of the anterior-posterior growth of the skull. Lambda (Choice C) is the landmark where the sagittal suture intersects the lambdoid sutures, that is, the point at which the two parietal bones meet the occipital bone. A unilateral lambdoid suture (Choice D) craniosynostosis will lead to a twisted appearance of the skull due to asymmetric growth. Educational Objective: Craniosynostosis results from the pren constriction of the central nervous system. ture fusion of cranial bone sutures, leading to asymmetric growth in the areas of remaining, unfused sutures. Skull shape deformity results, which is both cosmetic and physiologically problematic due to

Comprehensive Basic Science Self-Assessment 5. A 65-year-old woman has ascites. Which of the following additional findings indicates a diagnosis of constrictive pericarditis rather than cirrhosis? A) Edema of the lower extremities B) Esophageal varices C) Hypoalbuminemia D) Hyponatremia E) Increased jugular venous pressure F) Splenomegaly

E. Increased jugular venous pressure (JVP) is an expected finding in constrictive pericarditis (CP) and would not be found in patients with volume overload secondary to cirrhosis. CP describes a pathologic state whereby the pericardium, which encases the entirety of the heart and the origins of the great vessels, loses its elasticity. This can occur in patients with viral infections, connective tissue disease, tuberculosis, or as a result of cardiac surgery or radiation. During the normal cardiac cycle, increased venous return to the right atrium (RA) and right ventricle (RV) during inspiration leads to transient expansion of the RV with slight bowing of the interventricular septum into the left ventricle (LV). This increased RV preload does not impair LV filling as the pericardial sac expands to accommodate the increased RV volume. In constrictive pericarditis, the pericardial sac loses its elasticity. When venous return to the right heart increases, the pericardial sac is unable to expand, which exacerbates movement of the interventricular septum into the LV. This impairs diastolic filling of the LV and reduces cardiac output. LV diastolic filling is further reduced by a reduction in preload from the pulmonary veins. The constricted pericardium does not respond to normal changes in intrathoracic pressure during inspiration, but the pulmonary venous system, which lies outside of the pericardium, experiences a normal drop in pressure during inspiration. This difference creates an abnormal pressure gradient that reduces LV preload and leads to reduced cardiac output. The RV and LV develop interventricular dependence whereby increased pressure in each ventricle begins to affect the other ventricle. CP ultimately results in equalization of pressures in all four chambers, with clinical evidence of right heart failure exhibited by an increased JVP on physical examination, along with peripheral edema, and occasionally congestive hepatopathy or ascites. While cirrhosis can lead to hypervolemia, this is often manifested by peripheral edema, pulmonary edema, pleural effusions, or ascites, rarely increased JVP pressure, unless concomitant right heart failure is also present. Incorrect Answers: A, B, C, D, and F. Edema of the lower extremities (Choice A) is seen frequently in cirrhosis and is caused by hypoalbuminemia and systemic vasodilation with capillary leak. In CP, extremity edema is secondary to right heart failure. Esophageal varices (Choice B) are a consequence of portal hypertension and associated retrograde flow across the portosystemic venous anastomosis of the lower esophagus. This causes dilation of the submucosal veins in the esophagus (varices), which are a risk factor for gastrointestinal bleeding. It is encountered in cirrhosis but is not a feature of CP. Hypoalbuminemia (Choice C) is seen in cirrhosis but not in CP. Albumin is synthesized in the liver, which is impaired in patients with cirrhosis. Hyponatremia (Choice D) is common in cirrhosis with ascites. Portal hypertension results in systemic vasodilation and decreased effective circulating volume. This activates the renin-angiotensin-aldosterone system and induces the release of antidiuretic hormone. The net effect is an increase in resorption of free water resulting in hyponatremia. Splenomegaly (Choice F) in cirrhosis is a consequence of portal hypertension and is typically not seen in CP. Educational Objective: Constrictive pericarditis (CP) results from the loss of the normal elasticity of the pericardium, which causes interventricular interdependence and equalization of pressures throughout the heart with impaired diastolic filling. Increased jugular venous pressure (JVP) is often seen on physical examination and is a result of elevated filling pressures and right heart failure.

43 ---------- Exam Section 1: Item 44 of 50 National, Board of Medical Examiners Comprehensive Basic Science Self-Assessment 44. A 62-year-old man undergoes a physical examination on his arrival at the penitentiary to serve his prison sentence. A routine PPD skin test shows a 14-mm area of induration at 48 hours. A chest x-ray is shown. Reciprocal activation of the two immune cell types involved in the pathogenesis of this patient's pulmonary lesions is mediated by interferon-gamma and which of the following cytokines? O A) Interleukin-4 (IL-4) B) IL-5 C) IL-7 D) IL-10 E) IL-12

E. Interleukin-12 (IL-12) and interferon-gamma activate the two primary cells responsible for granuloma formation in this patient with Mycobacterium tuberculosis (MTB) infection. IL-12 is an inflammatory cytokine secreted by macrophages that are infected with MTB, an intracellular pathogen. IL-12 acts on both natural killer cells (NK cells) and cytotoxic T lymphocytes, causing them to release interferon-gamma, which has several roles including the promotion of Th1 T-cell differentiation, the augmentation of antigen presentation by macrophages, and the differentiation of macrophages into multinucleated giant cells. Collectively, this activating pathway of IL-12 and interferon-gamma allow for granuloma formation, a histologic hallmark of infection with MTB, isolating the invasive pathogen. Contrary to other diseases that result in granuloma formation, MTB infection often results in caseating (necrotic) granulomas. Incorrect Answers: A, B, C, and D. Interleukin-4 (Choice A) induces differentiation of helper T cells to Th2 cells, not Th1 cells. It is important in the process of activating macrophages to assist with wound repair and fibrosis, not granuloma formation. IL-5 (Choice B) is secreted by Th2 helper T cells and mast cells. It activates B lymphocytes to secrete IgA. IL-7 (Choice C) is secreted by numerous types of cells and is important in the maturation of both B and T lymphocytes but does not play a role in granuloma formation. IL-10 (Choice D) is secreted by Th2 and regulatory T cells. It has myriad effects but overall decreases the inflammatory response. It downregulates the expression of MHC class II antigens and Th1 cytokines. Additionally, it inhibits the activity of activated macrophages, which would decrease granuloma formation. cational Objective: MTB infects macrophages and results in granuloma formation within the lungs via the actions of IL- and interferon-gamma on macrophages, Th1 helper T cells, and NK cells. OO O O 0

64 ---------- Exam Section 2: Item 14 of 50 Natjonal, Board of Medical Examinersment encive Rasic S ance V 14. An 18-month-old boy is brought to the emergency department because of lethargy for 3 hours. He has not eaten well for the past 24 hours. He had cardiorespiratory arrest associated with hypoglycemia after an episode of diarrhea 10 months ago. On arrival, he is unresponsive. His temperature is 37°C (98.6°F), pulse is 140/min, respirations are 25/min, and blood pressure is 100/60 mm Hg. Physical examination shows hepatomegaly. Laboratory studies show: Serum Glucose Ketones Carnitine Urine Ketones decreased decreased decreased decreased Dicarboxylic acids present Following intravenous administration of glucose, he becomes responsive. Medium-chain triglycerides are administered daily for 6 months. At a follow-up examination, his serum glucose concentration is within the reference range. A deficiency of which of the following enzymes is the most likely cause of these findings? O A) Glucose 6-phosphatase O B) HMG-CoA lyase O C) Hormone-sensitive lipase D) Lipoprotein lipase E) Long-chain acyl-CoA dehydrogenase F) Phosphoenolpyruvate carboxykinase

E. Long-chain acyl-CoA dehydrogenase is an enzyme involved in the beta-oxidation of long-chain fatty acids. Deficiency of this enzyme is an autosomal recessive disorder and results in the accumulation of long-chain fatty acids in the mitochondrial matrix. It is characterized by two forms: an early-onset form that results in severe, life-threatening cardiomyopathy, and a late-onset form that is characterized by episodes of symptomatic hypoglycemia. Signs and symptoms may include irritability, lethargy, hepatomegaly (due to excess fat accumulation), and hypoglycemia. Symptomatic episodes can be triggered by illness, fasting, stress, or exercise. Hypoglycemia occurs without the production of ketones due to impaired fatty acid metabolism and ketone body production since there is deficient production of acetyl-COA from long-chain fatty acids. Immediate treatment involves correction of the patient's hypoglycemia and long-term management involves preventing acute episodes of hypoglycemia and dietary supplementation with medium-chain fatty acids, which will utilize medium-chain acyl-CoA dehydrogenase for beta-oxidation, circumventing the patient's enzymatic deficiency for long-chain fatty acid beta-oxidation. Incorrect Answers: A, B, C, D, and F. Glucose 6-phosphatase (Choice A) is an enzyme located in the liver that functions in gluconeogenesis and converts glucose-6-phosphate to glucose for its release into the blood. Impaired gluconeogenesis can lead to severe hypoglycemia during periods of fasting. Glycogen storage disease type I (von Gierke disease) is secondary to glucose 6-phosphatase deficiency and is characterized by excess glycogen in the liver, fasting hypoglycemia, increased levels of triglycerides, uric acid, and hepatomegaly. HMG-COA lyase (Choice B) is an enzyme that converts HMG-CoA into acetoacetate in the mitochondria and is key in ketogenesis during periods of fasting. Deficient ketogenesis can present with vomiting, diarrhea, dehydration, lethargy, and hypotonia. Hormone-sensitive lipase (Choice C) functions in the metabolic degradation of stored triglycerides in adipocytes, resulting in the release of free fatty acids into the bloodstream. It is inhibited by insulin during the fed-state. Lipoprotein lipase (Choice D) functions in the degradation of triglycerides and release of free fatty acids in both chylomicrons and very low-density lipoprotein (VLDL). Ďeficiency can lead to accumulation of triglycerides in the body with resultant xanthomata, hepatosplenomegaly, and pancreatitis. Phosphoenolpyruvate carboxykinase (Choice F) is an enzyme that catalyzes the irreversible conversion of oxaloacetate to phosphoenolpyruvate, a key regulatory step in gluconeogenesis. Deficiency can lead to hypoglycemia, failure to thrive, hepatomegaly, and lactic acidosis. Treatment includes a diet with simple carbohydrates, especially during periods of fasting, illness, or exercise. Educational Objective: Long-chain acyl-CoA dehydrogenase deficiency is an autosomal recessive disorder of fatty acid metabolism. Signs and symptoms include irritability, lethargy, hepatomegaly (due to excess fat accumulation), and hypoglycemia. Episodes can be triggered by illness, fasting, stress, or exercise. Immediate treatment involves correction of hypoglycemia and long-term management involves preventing acute episodes of hypoglycemia and dietary supplementation with medium-chain fatty acids.

21. A 35-year-old woman, gravida 3, para 2, develops a massive hemorrhage after the vaginal delivery of a healthy female newborn at 38 weeks gestation. She underwent a cesarean delivery 2 years ago because of cephalopelvic disproportion. A hysterectomy is required at this time to control the bleeding. The gross and microscopic appearances of the uterus are shown. Which of the following is the most likely cause of the postpartum hemorrhage in this patient? A) Abruptio placentae B) Ectopic pregnancy C) Endometriosis D) Leiomyomata uteri E) Placenta accreta F) Placental site trophoblastic tumor

E. Placenta accreta refers to the abnormal attachment of the placenta to the myometrium, rather than just the endometrium. This can be seen in the gross specimen and histology presented in this case. It most commonly presents with postpartum hemorrhage due to retained products of conception but is also commonly recognized on routine prenatal ultrasound examination. Risk factors include prior cesarean delivery or uterine surgery, increased maternal age, placenta previa, and multiparity. Management of known placenta accreta consists of elective cesarean delivery as early as 34 weeks, potentially followed by a hysterectomy depending on the extent of the accreta and retained products of conception. Incorrect answers: A, B, C, D, and F. Abruptio placentae (Choice A) presents with vaginal bleeding, severe uterine pain, and tetanic contractions, typically in the third trimester prior to delivery. It commonly results from lower abdominal trauma, smoking, or cocaine use. It is a less likely cause of postpartum hemorrhage in this patient with prior uterine instrumentation. Ectopic pregnancy (Choice B) occurs when the fertilized embryo implants in a location outside of the uterine cavity. Rupture of an ectopic pregnancy can cause vaginal bleeding and abdominal pain. An ectopic pregnancy is an unlikely cause of postpartum hemorrhage in this patient, as she delivered a healthy term baby. Endometriosis (Choice C) is the development of endometrial tissue outside of the uterine cavity. It can cause abnormal vaginal bleeding, pelvic pain, dyspareunia, and pain with defecation (dyschezia), but does not cause postpartum hemorrhage. Leiomyomata uteri (fibroids) (Choice D) are benign tumors of the uterus, which can present with abnormal uterine bleeding and pelvic pain. However, they are more common in older women and are associated with an irregularly shaped uterus on physical examination. They do not typically cause postpartum hemorrhage. Plaental site trophoblastic tumor (Choice F) a form of gestational trophoblastic disease, which also includes invasive moles and choriocarcinoma. It may develop following a pregnancy, abortion, or molar pregnancy. It is a typical cause of acute postpartum hemorrhage. Educational Objective: Placenta accreta is the abnormal attachment of the placenta directly to the myometrium, which most commonly presents with retained products of conception and postpartum hemorrhage. Risk factors include prior uterine instrumentation (including cesarean delivery), increased maternal age, placenta previa, and multiparity. Given the high risk of morbidity and mortality, elective surgical delivery at 34 to 36 weeks is recommended.

72 ---------- Exam Section 2: Item 22 of 50 Natjonal, Board of Medical Examinersment. 22. Receptors for all of the members of the steroid hormone family share which of the following features? O A) Calcium-mediated intracellular effects O B) GTP-binding proteins O C) Mitochondrial membrane association O D) Plasma membrane association E) Zinc-containing DNA-binding domains

E. Steroid hormones act on intracellular receptors and have the ability to traverse the cell membrane due to their lipophilic nature. Some of these hormones include cortisol, estrogen and progesterone, testosterone, and aldosterone. All of these molecules are derived from cholesterol as a common precursor. Steroid hormone receptors are intracellular and reside either in the cytoplasm or the nucleus and regulate gene expression. Because of the effects mediated through gene regulation, the onset of action of steroid hormones occurs over the course of hours to days. Intracellular steroid receptors have a number of different protein domains. The DNA-binding domain classically has a zinc finger motif as a tertiary structure. This domain has affinity with the regulatory (eg, promoter, silencer) regions of DNA. Incorrect Answers: A, B, C, and D. Activation of Go-protein coupled receptors leads to an increase in calcium-mediated intracellular effects (Choice A). The signaling cascade of Gg includes activation of phospholipase C, which cleaves membrane bound phospholipids, leading to the formation of inositol triphosphate (IP3) and diacylglycerol (DAG). This results in the subsequent increase in intracellular calcium concentration, which then activates protein kinase C and/or smooth muscle contraction. GTP-binding proteins (Choice B) are also known as G proteins which are components of many transmembrane receptors. G proteins enter an activated state when they are bound to GTP molecules. Once the GTP molecule is cleaved to GDP, the G protein becomes inactive. G proteins are commonly involved in signaling pathways of peptide hormones. Mitochondrial membrane association (Choice C) is a description of pathways of cellular apoptosis. During activation, caspases will cleave proteins causing permeability of the cell membrane. In addition, caspase 8 associates with the mitochondrial membrane and alters its permeability. Plasma membrane association (Choice D) is a component of many signal transduction pathways. G-protein coupled receptors have a transmembrane domain, which in turn connects to an intracellular G protein. Signaling molecules with receptor tyrosine kinases, such as insulin, are also closely associated with the cell membrane. Educational Objective: Steroid hormones function within the nucleus by regulating gene expression. Steroid receptor DNA-binding domains contain zinc motifs.

19. A 52-year-old woman with a long-standing history of generalized anxiety disorder comes to the physician for a follow-up examination. She has been taking an anxiolytic drug for 12 years, but she is not currently under a psychiatrist's care because she says she is feeling better and that she prefers this physician to psychiatrists. She has hyperlipidemia, hypertension, and mild type 2 diabetes mellitus treated with pharmacotherapy. Following a physical examination and laboratory studies, the physician makes adjustments to the patient's medication regimen. Two days later, the physician receives 12 calls from the patient for clarification of treatment recommendations; this pattern has occurred frequently in the past. In particular, the patient fixates on the potential toxicities of her medications. She often asks questions that have specifically been answered during the recent office visit. Which of the following is the best approach to manage this patient's needs? A) Advise the patient that constant questioning of recommendations suggests a lack of trust and that she should seek care from a provider in whom she has confidence B) Advise staff to ignore the patient's calls if they are too frequent C) Contact one of the patient's family members to engage his or her help in enforcing more responsible use of the physician's time D) Insist that the patient obtain routine psychiatric care E) Set conditions and establish limits for phone calls, and set up a regular schedule of office visits

E. The physician should manage this anxious patient's frequent phone calls by establishing limits and setting a regular appointment schedule. The physician-patient relationship has professional boundaries (also known as a therapeutic frame) that both parties should respect. These boundaries include specific meeting times and relationship expectations. For instance, a physician should abstain from dominating the conversation with details about their personal life since their role is to be attentive and therapeutic to the patient. When a patient transgresses these boundaries, such as making many phone calls outside of scheduled appointments, both parties will benefit from the reestablishment of expectations. Many patients who are anxious are destabilized by poor interpersonal boundaries and will respond well to structure. Incorrect Answers: A, B, C, and D. Referring the patient elsewhere (Choices A and D) without attempting to problem-solve with the patient would be premature. Boundary transgressions occur regularly in physician-patient relationships and should not warrant immediate referral, especially if the transgressions are related to an anxiety disorder the physician is managing. Due to this patient's acute anxiety and attachment to this physician, terminating the relationship would likely lead to significant distress. Using other people to mediate the interpersonal relationship (Choices B and C) would prevent the direct communication and boundary setting that is necessary in this physician-patient relationship. One therapeutic role of the physician is to demonstrate that they can handle challenges and difficult feelings that arise, which can assuage the patient's anxiety. Educational Objective: The physician-patient relationship benefits from boundaries. When a patient transgresses these boundaries, the physician should utilize direct communication to reestablish expectations.

46 ---------- Exam Section 1: Item 47 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 47. A 56-year-old man is brought to the emergency department 1 hour after having a generalized tonic-clonic seizure during his morning run. Vital signs are within normal limits. MRIS of the brain show a mass in the lateral aspect of the precentral gyrus of the frontal lobe. Based on the MRI findings, which of the following additional deficits is most likely in this patient? A) Atrophy of the left upper extremity B) Decreased proprioception in the right upper extremity C) Decreased sensation in the left lower extremity D) Deviation of the tongue to the right E) Weakness of the left lower area of the face F) Weakness of the right lower extremity

E. The precentral gyrus of the frontal lobe contains the primary motor cortex that controls voluntary movements of the contralateral body. Medially, the precentral gyrus contains neurons that control the lower extremity. Laterally, the precentral gyrus controls the contralateral face. The contralateral arm is controlled by neurons between the two areas. This distribution of control is known as somatotopy and permits localization of physical examination findings to a particular brain region. In this case, a lesion in the lateral aspect of the right precentral gyrus would disrupt motor neurons controlling the left face, resulting in weakness, facial droop, and an upper motor neuron pattern of dysfunction (eg, hyperreflexia). Importantly, the forehead would be spared, as the facial nerve (cranial nerve VII) nucleus that controls forehead musculature is dually innervated by upper motor neurons from bilateral precentral gyri. Because of this, the patient would be expected to symmetrically raise his eyebrows. Alternatively, lesions involving the lower motor neurons of facial expression (eg, facial nerve inflammation in Bell palsy) affect the forehead and lower face together. Incorrect Answers: A, B, C, D, and F. Atrophy of the left upper extremity (Choice A) may be associated with an upper motor neuron lesion located in the right precentral gyrus that would be located medially relative to the location of this patient's mass. The contralateral arm is innervated by neurons located in the parasagittal region of the precentral gyrus. As well, atrophy of the left upper extremity may be associated with lower motor neuron pathology in the ipsilateral spinal cord, spinal nerves, or peripheral nerves. Decreased proprioception in the right upper extremity (Choice B) or decreased sensation in the left lower extremity (Choice C) could be associated with lesions of the contralateral thalamus, tracts leading to the postcentral gyrus, or within the postcentral gyrus, as these areas are associated with position sense and touch sensation. Peripheral lesions associated with decreased proprioception or touch could involve the ipsilateral sensory neurons, dorsal root ganglia, or the dorsal column-medial lemniscus pathway within the spinal cord. Deviation of the tongue to the right (Choice D) is associated with loss of motor control from the right cranial nerve XII, which innervates most of the intrinsic and extrinsic muscles of the tongue. Absent tone on the right-side results in deviation toward the right. The upper motor neurons for the tongue exist within the lateral contralateral precentral gyrus, but facial weakness would likely also be present. Weakness of the right lower extremity (Choice F) may be associated with an upper motor neuron lesion in the left medial precentral gyrus in the region of the paracentral lobule. As well, right lower extremity weakness can be secondary to pathology involving the ipsilateral lumbar spinal cord, cauda equina, or peripheral nerves. Educational Objective: The precentral gyrus of the frontal lobe contains the primary motor cortex controlling the voluntary movement of the contralateral body. Laterally, the precentral gyrus controls the contralateral face. Pathology affecting the upper motor neuron innervation to the contralateral face typically presents with only lower face weakness as the facial nerve nucleus that controls forehead musculature is dually innervated by upper motor neurons from bilateral precentral gyri.

74 ---------- Exam Section 2: Item 24 of 50 Natjonal, Board of Medical Examinersent V 24. A 49-year-old man comes to the physician because of a 1-month history of lower abdominal pain and increased urge to defecate. Capability to prevent spontaneous defecation is intact. He sustained nondisplaced multiple fractures of the pelvis 2 months ago in a motor vehicle collision. He initially was on bed rest with pelvic support, progressed to non-weight-bearing crutch walking, and now is partial weight bearing with crutches. Medications now include naprosyn three times daily. Physical examination shows an anal wink. Neurologic examination shows intact rectal sphincter tone. This patient is able to sustain cortical control over defecation via which of the following nerves? O A) Conus medullaris O B) Hypogastric nerves O C) Myenteric plexus D) Pelvic nerves E) Pudendal nerves

E. The pudendal nerve is formed from the sacral plexus by the S2, S3, and S4 nerve roots. The nerve carries both sensory and motor fibers and innervates the genitalia and the pelvic floor musculature, including the levator ani muscle and the external anal and urethral sphincters. Pudendal neuropathy can occur in Alcock canal syndrome with pelvic trauma or surgery, and in childbirth. The pudendal nerve courses just medially to the ischial tuberosity, making this an important landmark for regional anesthetic blocks used in vaginal delivery or anorectal surgery. Given this patient's history of pelvic trauma, there is likely disruption involving the complex interactions between sympathetic and parasympathetic nerves that provide autonomic control over urination, ejaculation, and defecation. However, his ability to maintain voluntary (cortical) control over defecation demonstrates an intact neural pathway involving the pudendal nerve, a somatic nerve, even if mild injury to this nerve was sustained. often due to impingement of the pudendal erve between the sacrotuberous or sacrospinous ligaments. This pathology can occur in cyclists, due to prolonged compression in this area, patients Incorrect Answers: A, B, C, and D. The conus medullaris (Choice A) is the distal termination of the spinal cord. Compression of the conus will lead to saddle anesthesia, overflow incontinence, and weakness or paralysis of the bilateral lower extremities. The hypogastric nerves (Choice B) arise from roots of T11-L4 and provide sympathetic innervation to portions of reproductive organs. They are responsible for emission of semen into the posterior urethra which is the first stage of male ejaculation. The myenteric plexus (Choice C) is part of the enteric nervous system located between inner and outer layers of the muscularis externa. Absence of this plexus is responsible for Hirschsprung disease, which is a congenital failure in the distal development of the myenteric plexus, leading to large bowel obstruction. The pelvic nerves (Choice D) references pelvic splanchnic nerves, which form from the sacral and lumbosacral plexus. These nerves supply autonomic innervation to the rectum, bladder, and the genitalia. Injury to these nerves causes a loss of function of bladder and bowel, however they are under visceral, not somatic, control. Educational Objective: The pudendal nerve provides somatic sensory and motor innervation to the genitals and the pelvic floor musculature, including the levator ani muscle and external anal sphincter. Injury to this nerve can occur with pelvic trauma or chronic compression.

88 ---------- Exam Section 2: Item 38 of 50 Natjonal, Board of Medical Examinersment. 38. A 52-year-old woman with type 2 diabetes mellitus and hypercholesterolemia comes to the physician for a follow-up examination. Since her diagnosis 14 months ago, she has been exercising regularly, eating healthier foods, and taking omega-3 fatty acid supplements. She has had a 9-kg (20-lb) weight loss during this period. She is 168 cm (5 ft 6 in) tall and now weighs 62 kg (137 Ib); BMI is 22 kg/m2. Physical examination shows no abnormalities. Serum studies show decreased glucose and cholesterol concentrations since her last visit 2 months ago. She says, "I'm glad everything is fine, but I am worried that I won't be able to stay on this regimen and that my health will get worse." Which of the following best describes this patient's stage of change? O A) Precontemplation O B) Contemplation O C) Preparation D) Action E) Maintenance

E. This patient has already made significant behavioral changes, resulting in improved health parameters, but is concerned she may revert to her previous unhealthy diet and exercise habits. This scenario is consistent with the maintenance stage of behavioral change. The physician and patient may collabora a health-related behavior such as treatment adherence, substance use, diet, or exercise habits. In sequential order, the stages of behavioral change are precontemplation, contemplation, preparation, action, maintenance, and termination. Physicians aim to move patients through these stages over time with an interview technique called motivational interviewing. Motivational interviewing involves using open-ended, non-judgmental questions to help the patient explore their reasons for wanting to change or maintain the habit. vely strategize about how to avoid relapse, which may involve bolstering skills for coping with temptation or joining a diabetes support group. The stages of behavioral change are used to define a tient's readiness to change Incorrect Answers: A, B, C, and D. Patients in the precontemplation stage (Choice A) are not interested in changing their habit and may not see the benefit in making a change. This patient is aware of the benefits of exercise and a healthy diet and has already changed her habits, which signifies she has surpassed the precontemplation stage. Patients in the contemplation stage (Choice B) are typically aware of the benefits of the behavior change but may be ambivalent or indecisive about change. This patient has already improved her diet and exercise habits, which signifies that she is beyond the contemplation stage. Patients in the preparation stage (Choice C) have committed to making a change and are ready to discuss strategies and resources to help them make the change. For example, a patient may research information about a low-carbohydrate diet to improve his eating habits. This patient has already changed her habits and has therefore surpassed the preparation stage. In the action stage (Choice D), patients start to take steps to change their behavior. For example, a patient may begin a low-carbohydrate diet to improve his unhealthy eating habits. This patient has established a new routine that involves healthier diet and exercise habits, signifying that she is beyond the action stage. Educational Objective: In the maintenance stage of behavioral change, patients have sustained a behavioral change. Physicians and patients may strategize about how to prevent relapse.

62 ---------- Exam Section 2: Item 12 of 50 Natjonal, Board of Medical Examinersment 12. A 25-year-old man comes to the physician because of sores on his penis for 2 days. Physical examination shows vesicular and ulcerative lesions on the penis. Drug therapy is initiated. One week later, his symptoms resolve. Which of the following enzymes is required to activate the most commonly prescribed drug? O A) ATPase O B) Cytochrome P450 O C) Protease O D) Reverse transcriptase E) Thymidine kinase

E. This patient's vesicular and ulcerative lesions on the penis are likely caused by herpes simplex virus in the form of genital herpes. Treatment for herpetic infections involves drugs that inhibit viral DNA polymerase, classically by guanosine analogs such as acyclovir, valacyclovir, and famciclovir. Prior to exerting their antiviral effects, guanosine analogs must be phosphorylated by the viral enzyme thymidine kinase. 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 Epstein-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, C, and D. ATPases (Choice A) are a heterogenous group of enzymes that hydrolyze ATP into ADP and a molecule of inorganic phosphate. Guanosine analogs require phosphorylation, not dephosphorylation, in order to be active. Cytochrome P450 (Choice B) enzymes are found in the liver and play a key role in the metabolism of many drugs. In most cases, this process alters a drug from its active form to an inactive metabolite. However, some drugs are converted from their prodrug form into an active metabolite by the P450 enzymes, such as clopidogrel, but guanosine analogs are not among them. Proteases (Choice C) 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. Reverse transcriptase (Choice D) transcribes viral DNA from viral mRNA for incorporation into host DNA. This enzyme is inhibited by both nucleoside reverse transcriptase inhibitors, such as abacavir and didanosine, and non-nucleoside reverse transcriptase inhibitors, such as efavirenz and nevirapine. These medications are used to treat HIV. 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.

36 ---------- Exam Section 1: Item 37 of 50 National Board of Medical Examiners Comprehensive Basic Science Self-Assessment 37. A 20-month-old girl is brought to the physician by her mother because the mother is concerned about her daughter's development and behavior. The girl is able to walk and run and has a vocabulary of at least 100 words. However, when she attends playdates with other children, she prefers to play by herself and will strike another child on the head if that child attempts to join her in playing with a particular toy. Which of the following best explains this patient's actions during play? A) Attention-deficit/hyperactivity disorder B) Conduct disorder C) Early sign of autistic disorder D) Oppositional defiant disorder E) Normal behavior for age

E. This toddler demonstrates normal behavior for age with age-appropriate gross motor, language, and social functioning. Children are expected to be able to walk up stairs by age 18 months and run by age 2 years, and the typically developing 18-month-old has a vocabulary of approximately 100 words. Social skills may include pointing to show others something interesting, playing pretend, and showing affection to familiar others. Cooperative play is not expected to develop until age 4 years, so this child's inclination not to share her toy is typical. Hitting and aggression are relatively common in toddlers. Typical fine motor skills at this age include stacking several blocks. Incorrect Answers: A, B, C, and D. Attention-deficit/hyperactivity disorder (Choice A) is a childhood disorder that presents with hyperactivity, impulsivity, and/or inattention. Though this child impulsively hits others, this behavior is common for her age. This child does not demonstrate hyperactivity or inattention. Conduct disorder (Choice B) refers to persistent emotional and behavioral problems such as purposeful aggression toward people and animals in children and adolescents. The behavioral problems can also include lying and serious rule violations. This child's aggression is borne out of age-appropriate impulsivity rather than intentional bullying. Early signs of autistic disorder (Choice C) may include deficits in social and language skills, sensory overstimulation, and frequent tantrums. This patient has age-appropriate social and language skills. Oppositional defiant disorder (Choice D) manifests as a persistent pattern of angry mood, defiant behavior, and vindictiveness that causes distress and impaired functioning in the child or their caretakers. This child's aggression is borne out of age-appropriate impulsivity rather than defiance. Educational Objective: Developing 20-month old children can typically walk up stairs (and may be able to run), stack several blocks, use 100 or more vocabulary words, and play pretend. Aggressive, impulsive behavior is not uncommon, and cooperative play does not typically develop until the age of 4 years. OO O

3. A 42-year-old man is struck by a motor vehicle. His only injury is a closed fracture of the proximal tibia. Initial neurovascular examination shows no deficits. Twenty-four hours later, he has increased leg pain and paresthesias in the dorsal space between his first and second toes. The patient begins to pass dark red urine and becomes oliguric. Urinalysis is positive for blood but no erythrocytes are seen on microscopic examination. Which of the following acute disorders is the most likely cause of the renal failure? A) Glomerulonephritis B) Hemolytic-uremic syndrome C) Interstitial nephritis D) Nephrotic syndrome E) Tubular necrosis

E. Tibial fractures present a high risk for compartment syndrome. The fracture results in blood vessel injury and muscle injury, inflammation, and edema. Because the fascia containing the anterior compartment of the leg does not stretch, bleeding and swelling can cause increased pressure in the compartment. This increased pressure in turn inhibits venous drainage, further increasing pressure in the compartment. Eventually, the nerve supply and associated arteries are compromised, leading to the classic signs and symptoms of compartment syndrome. Signs and symptoms of compartment syndrome include pain out of proportion to examination findings, pain with passive movement of the muscles, paresthesia, pallor, pulselessness, and paralysis. Compromised blood supply deprives muscle and tissue of oxygen and glucose, leading to tissue ischemia and necrosis. Muscle necrosis leads to rhabdomyolysis, myoglobinuria, and acute renal failure. Evaluation of rhabdomyolysis reveals red or brown urine and urinalysis is typically positive for blood due to the presence of myoglobinuria without microscopic evidence of red blood cells. A complication of rhabdomyolysis is acute kidney injury from acute tubular necrosis secondary to the release of nephrotoxic myoglobin and nonprotein heme pigments. Acute tubular necrosis typically occurs following an ischemic or nephrotoxic insult to the kidneys, which results in loss of the tubular epithelium. Granular, muddy brown casts are common on urinalysis. Compartment syndrome is treated by immediate fasciotomy to decrease compartment pressure and support tissue perfusion. Incorrect Answers: A, B, C, and D. Glomerulonephritis (Choice A) refers to a variety of glomerular diseases, including nephritic and nephrotic syndromes. Nephritic syndromes typically present with acute renal failure associated with hematuria, red blood cell urine casts, and hypertension. Nephrotic syndrome typically presents with excessive proteinuria (greater than 3 g/day), hyperlipidemia, hypoalbuminemia, and edema. Hemolytic-uremic syndrome (Choice B) is classically associated with infection from Escherichia coli 0157:H7, a Gram-negative rod. It often presents in children with bloody diarrhea, microangiopathic hemolytic anemia, thrombocytopenia, and renal failure. It is not associated with trauma or compartment syndrome. Interstitial nephritis (Choice C) is usually caused by a hypersensitivity reaction to offending drugs (eg, nonsteroidal anti-inflammatory medications, diuretics, and sulfonamides) and is characterized by a rash, acute kidney injury, and eosinophiluria. Nephrotic syndrome (Choice D) is characterized by proteinuria (greater than 3 g/day), edema, hypoalbuminemia, and hyperlipidemia. Complications include hypercoagulability due to urinary loss of antithrombin-3 and infection due to urinary loss of gamma globulins. Educational Objective: Tibial fractures can be complicated by compartment syndrome due to increased compartment pressure in the anterior leg. Compartment syndrome leading to muscle necrosis and rhabdomyolysis can be complicated by acute tubular necrosis due to the nephrotoxicity of myoglobin and its byproducts.

t disease) O D) Sarcoidosis E) Tuberculosis

E. Tuberculous disease of the spine, also known as Pott disease, is a spinal osteomyelitis or osteodiscitis caused by Mycobacterium tuberculosis. M. tuberculosis is a slow-growing organism transmitted via respiratory secretions. The slow growth and evasion of immune defenses result in the disease presenting with primary, latent, and reactivation patterns. In the primary infection, patients present with subacute fevers, weight loss, night sweats, cough, and malaise. Latent tuberculosis can occur as the organism proliferates within macrophages and is maintained within caseating granulomas. Although tuberculosis is primarily a disease of the respiratory tract, it can have a number of extrapulmonary manifestations. These include meningitis, dermatologic tuberculosis, and vertebral discitis or osteomyelitis. In this case, the vertebral body and intervertebral disc are involved. Radiographs or CT/MRI imaging will demonstrate localized destruction of bony anatomy, potentially with compression of the spinal nerve roots or spinal cord. Because of its airborne transmission pattern, patients with tuberculosis must be isolated to prevent transmission to other persons. Tuberculosis remains prevalent in homeless, incarcerated, and immunocompromised patients. Active disease is treated primarily with a combination of rifampin, isoniazid, pyrazinamide, and ethambutol. Incorrect Answers: A, B, C, and D. Hyperparathyroidism (Choice A) leads to bony resorption and hypercalcemia through increased production of parathyroid hormone. Primary hyperparathyroidism results from functionally active parathyroid adenoma, hyperplasia, or carcinoma. Secondary hyperparathyroidism is commonly due to chronic hyperphosphatemia and hypocalcemia in the setting of end-stage renal disease. As parathyroid hormone activates osteoclasts to maintain calcium levels, bone resorption results. This does not cause focal destruction of vertebrae or intervertebral discs. Multiple myeloma (Choice B) is a neoplasm of proliferating plasma cells in the bone marrow, triggering local osteoclast activation with resultant lytic bone lesions. Classically, multiple myeloma presents with hypercalcemia, renal failure, and anemia, in addition to bone lesions.Osteitis deformans (Choice C), also known as

71 ---------- Exam Section 2: Item 21 of 50 National Board of Medical Examiners mont 21. A 3-year-old girl is brought to the physician because of a 2-week history of easy bruisability. She has had frequent, large, foul-smelling stools for 3 months. She was found to have cystic fibrosis at the age of 18 months. She is below the 3rd percentile for height and weight. Physical examination shows several ecchymotic areas over the extremities. This patient most likely has a deficiency of which of the following? O A) Antithrombin III O B) Platelets O C) Protein C O D) Vitamin B12 (cobalamin) O E) Vitamin C F) Vitamin K O G) von Willebrand factor

F. Cystic fibrosis is an autosomal recessive disorder due to a defect in the CFTR gene, leading to a deficiency in a chloride channel that secretes chloride in the lungs and gastrointestinal tract and reabsorbs chloride in sweat glands. This abnormal chloride transport causes decreased chloride and water secretion and increased water reabsorption, leading to abnormally viscous mucous in the lungs and gastrointestinal tract. The thick mucus in the gastrointestinal tract contributes to pancreatic insufficiency, steatorrhea, and malabsorption, leading to a deficiency of fat-soluble vitamins A, D, E, and K. Vitamin K plays a critical role in the synthesis of hepatic coagulation proteins; it is oxidized in the liver during carboxylation of glutamic acid residues on coagulation factors II, VII, IX, X, and proteins Č and S. A deficiency in vitamin K leads to coagulopathy with elevated prothrombin time (PT) and activated partial thromboplastin time (APTT) and is characterized by easy bruising. Incorrect Answers: A, B, C, D, E, and G. Antithrombin Il (Choice A) deficiency can be inherited or acquired secondary to nephrotic syndrome. Antithrombin III functions to inhibit coagulation factors II, VII, IX, X, XI, and XII. Deficiency leads to hypercoagulability. Platelets (Choice B) function in primary hemostasis in which they adhere to areas of endothelial damage and form a temporary platelet plug to stop bleeding prior to the activation of the coagulation cascade. Qualitative or quantitative platelet defects are all characterized by increased bleeding time, and generally present with prolonged mucous membrane bleeding (eg, epistaxis, menorrhagia), petechiae, or purpura. Protein C (Choice C) inactivates factors Va and VIlla in the coagulation cascade, and deficiency leads to hypercoagulability, not easy bruising. Vitamin B12 (cobalamin) (Choice D) is classically characterized by erythrocyte macrocytosis and hypersegmented neutrophils. It is most often seen in patients with malnutrition, alcoholism, pernicious anemia, or Crohn disease. Vitamin C (Choice E) is an essential nutrient that is found in fruits and vegetables and is necessary for collagen synthesis, iron absorption, immune function, and conversion of dopamine to norepinephrine. Deficiency causes scurvy, characterized by swollen gums, bruising, poor wound healing, petechiae, perifollicular and subperiosteal hemorrhages, and short, fragile, curly hair. Von Willebrand factor (Choice G) deficiency is one of the most common hereditary bleeding disorders and is due to a quantitative or qualitative abnormality of von Willebrand factor, which binds platelets and subendothelial collagen in primary hemostasis. It causes a mild bleeding disorder with elevated bleeding time. Educational Objective: Cystic fibrosis results in pancreatic insufficiency, which leads to the gastrointestinal malabsorption and deficiency of fat-soluble vitamins A, D, E, and K. Vitamin K plays a critical role in the synthesis of hepatic coagulation proteins, and a deficiency in vitamin K leads to coagulopathy with elevated PT and aPTT.

66 ---------- Exam Section 2: Item 16 of 50 Natjonal, Board of Medical Ęxaminersont * 16. A 57-year-old woman comes to the physician because of chronic shortness of breath and swollen legs for 2 months. Her respirations are 12/min. Physical examination shows edema of the lower extremities. Auscultation of the chest shows each cardiac systole associated with a thrusting impulse palpable along the left sternal border, most prominently in the third, fourth, and fifth intercostal spaces. Which of the following disorders is the most likely cause of this cardiac finding? O A) Aneurysm of the ascending aorta O B) Left atrial hypertrophy O C) Left ventricular hypertrophy D) Pericardial effusion O E) Right atrial hypertrophy F) Right ventricular hypertrophy

F. The physical examination findings of right ventricular hypertrophy may include a parasternal heave (also called a precordial heave), which is an impulse observed or palpated along the left sternal border. The right ventricle is the most anterior of the four heart chambers and lies deep to the third, fourth, and fifth intercostal spaces when in its usual anatomic location. Right ventricular hypertrophy is a consequence of increased right ventricle pressure, which most commonly occurs secondary to chronic lung disease resulting in increased pulmonary vascular resistance and resultant pulmonary hypertension. Chronic lung disease is a risk factor for the development of pulmonary hypertension due to diffuse hypoxia-induced vasoconstriction and pulmonary vascular remodeling. When an area of lung has low oxygen concentration, reflexive contraction of vascular smooth muscle diverts blood flow to a region with increased ventilation. This is called ventilation-perfusion matching. Chronic, severe lung disease leads to widespread pulmonary vasoconstriction, and the increased resistance to flow results in increased right ventricle pressures and development of right ventricular hypertrophy. Other etiologies of right ventricular hypertrophy include primary pulmonary hypertension, severe tricuspid regurgitation, high altitude, left to right cardiac shunts, atrial or ventricular septal defects, anomalous pulmonary venous return, Eisenmenger syndrome, pulmonic valve stenosis, and cardiomyopathy. Patients often present with signs and symptoms of right-sided heart failure, including dyspnea, dependent edema, jugular venous distention, and hepatic congestion. Incorrect Answers: A, B, C, D, and E. Aneurysm of the ascending aorta (Choice A) is a dilatation of the three layers of the aortic wall proximal to the aortic arch. They are often asymptomatic unless ruptured or infected. Physical examination may reveal an aortic murmur or bruit. While abdominal aortic aneurysms are often palpable on physical examination, thoracic lesions are not. Left atrial hypertrophy (Choice B) can result from mitral stenosis, which presents with a murmur classically heard as an opening snap followed by a diastolic rumble, loudest over the cardiac apex with radiation to the axilla. Severe left atrial hypertrophy may present with a discernible parasternal heave, though this is much rarer than right ventricular hypertrophy. Left ventricular hypertrophy (Choice C) is a common disorder that may be caused by chronic systemic hypertension, aortic stenosis, or hypertrophic cardiomyopathy. It is associated with an S3 or S4 gallop, a crescendo-decrescendo systolic murmur (if aortic stenosis is present) and may rarely cause a parasternal heave. Right ventricular hypertrophy is a much more common cause due to its anterior location. Pericardial effusion (Choice D) refers to the presence of an abnormal amount of fluid in the pericardium. If severe enough, it can potentially progress to cardiac tamponade with compression of the cardiac chambers and obstructive shock. It typically causes the equalization of pressures amongst the four heart chambers. Physical examination findings include Beck triad when in tamponade (hypotension, distended jugular veins, and distant heart sounds). Right atrial hypertrophy (Choice E) is often a result of tricuspid regurgitation, tricuspid stenosis, or right ventricular hypertrophy, though a parasternal heave is less likely to be felt as the right atrium is retrosternal and has much weaker contractile force than the right ventricle. Educational Objective: A parasternal heave on physical examination is most commonly associated with right ventricular hypertrophy, which is the most anterior chamber of the heart. Right ventricular hypertrophy results from pulmonary hypertension, which is frequently a complication of chronic lung disease due to hypoxia-induced pulmonary vasoconstriction and vascular remodeling.

24 ---------- Exam Section 1: Item 25 of 50 National, Board of Medical Examiners Comprehensive Basic Science Self-Assessment 25. A 1-year-old boy is brought to the physician by his parents because of a 4-week history of progressive generalized tonic-clonic seizures and a strange odor to his urine. He has a history of delayed development. He was adopted from an orphanage in Russia at the age of 6 months. Physical examination shows fair skin and blond hair. His phenylalanine hydroxylase gene is homozygous for a point mutation (GT AT) in intron 12 of the affected gene that causes skipping of exon 12. Which of the following is the most likely explanation for exon skipping in this patient's affected gene? A) Alternative polyadenylation site B) Deletion of 5' untranslated region C) Expansion of trinucleotide repeat D) Gene duplication E) Nonhomologous recombination F) Nonsense mutation G) RNA splice error H) X inactivation

G. After DNA is transcribed into pre-messenger RNA (pre-mRNA), processing occurs to form the finished mRNA product. One element of this processing is the splicing out of introns, non-coding portions of the genetic material which do not contribute to the final protein, leaving behind only exons. From a single pre-mRNA, several unique proteins may be made if splicing variations occur. Correct splicing is dependent on the spliceosome recognizing a splice site at the start of an intron, which typically consists of base pairs G-U-A-A-G. A point mutation in DNA changing a guanine to an adenine, as in this case, may mask a splice site and either cause an intron to remain in the final mRNA product erroneously or cause a larger region of pre-mRNA to be spliced, leading to the removal of an exon. In this case, incorrect removal of exon 12 has created a defective version of phenylalanine hydroxylase and led to the condition phenylketonuria. Patients with phenylketonuria do not have the ability to convert phenylalanine to tyrosine. Accumulation of phenylalanine leads to physical manifestations including a musty odor, impaired neurologic development, and seizures. Incorrect Answers: A, B, C, D, E, F, and H. Polyadenylation of the 3' end and capping of the 5' end with non-translated nucleotides are two additional processes completed during pre-MRNA processing. An alternative polyadenylation site (Choice A) would lead to an addition of numerous adenine residues. Deletion of the 5' untranslated region (Choice B) would remove a series of nucleotides at the 5' end of the MRNA. Neither would cause the substitution at a splice site seen in this case. Expansion of a trinucleotide repeat (Choice C) occurs when a series of three nucleotides in repetition is replicated incorrectly and too many repeats are included in the replicated product. Conditions caused by trinucleotide repeat expansions include fragile X syndrome, Huntington disease, Friedrich ataxia, and myotonic dystrophy. Gene duplication (Choice D) occurs when a region of DNA containing a gene is erroneously duplicated. The nucleotide sequence of the gene remains intact, but there are now two copies of the same gene on the same chromosome. Nonhomologous recombination (Choice E) is a repair mechanism by which double-stranded DNA breaks are repaired. The break is repaired without a need for a homologous strand to be used as a template, giving it the name nonhomologous recombination. This process is error-prone and tends to add or delete genetic information rather than causing a single nucleotide substitution, as in this case. A nonsense mutation (Choice F) occurs when a nucleotide is substituted in the replication or transcription process leading to an early stop codon to form. This leads to a truncated protein, which is usually nonfunctional. In this case, a single base pair substitution led to the masking of a splice site rather than an early stop codon. In females with two X chromosomes, X inactivation (Choice H) occurs randomly in each cell early in development. One of the two X chromosomes is chosen to be epigenetically silenced in that cell and its progeny. In an individual heterozygous for a mutated allele on the X chromosome, the cell lines which randomly inactivate the X chromosome containing the wildtype allele will then express the dysfunctional protein. This is a mechanism of mosaicism in females. Educational Objective: Before a strand of mRNA is mature, splicing of the introns must occur, which requires specific splice sites to be recognized by the spliceosome. If a splice site is masked, an intron may be inappropriately retained, or an exon may be erroneously removed.