NBME 29

6 Exam Section 1: Item 6 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 6. A study is done to determine the efficacy of St. John's wort (Hypericum perforatum) for treatment of endogenous depression. One hundred patients with mild depression are randomly assigned to either a treatment group (900 mg of St. John's wort daily) or a placebo group. Scores on a Hamilton Depression Rating Scale are recorded for each subject before treatment and 4 weeks after the start of the study. Which of the following best describes this study design? A) Case-control B) Cohort C) Controlled trial D) Crossover E) Cross-sectional

C. A controlled trial is a type of prospective, interventional study design wherein patients are assigned to receive a particular intervention. The intervention may be compared against placebo therapy or against the current standard of care (control group), depending on the study design and the disease being studied. The patients are randomly assigned to receive the intervention or placebo, and both the patients and investigators are blinded to the assignment (double-blinded). Controlled trials are the standard for medical research, and they generate more powerful evidence than other forms of observational or descriptive studies. In this case, the intervention is St. John's wort, and it is being compared with a placebo. A clinical trial typically utilizes a randomized, controlled design. There are four phases to clinical trials. In phase I, a small number of healthy volunteers are tested to determine if the drug is safe in healthy individuals. In phase II, a small number of diseased volunteers are tested to determine if the drug is effective in the disease population. In phase III, a large number of patients are assigned to receive either the drug or placebo to determine if the drug is better than the placebo or current standard of care. In phase IV, or post-market surveillance, patients who take the drug once it is approved are observed for any unpredicted adverse effects. Adverse results in a phase IV trial can result in a drug being removed from the market. Incorrect Answers: A, B, D, and E. A case-control study (Choice A) investigates an association between an exposure and an outcome. In this study design, a group of patients with the disease (cases) are identified. A group of patients without the disease (controls) are matched on baseline characteristics to the cases. Exposure data for the two groups is collected, and these data are compared to determine association with the outcome (disease) in question. An odds ratio may be calculated to compare exposures between groups. A cohort study (Choice B) identifies a group of patients with or without an exposure 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. A crossover study (Choice D) is a type of controlled trial in which patients begin the study in either the intervention or placebo group and then cross over to the other group at a predetermined time point. For example, if this study had a crossover design, patients receiving St. John's wort would be followed until a given point in time, then switched to the placebo and followed to a second point in time. A cross-sectional study (Choice E) seeks to identify the prevalence of a condition at a particular point in time. An example of a cross-sectional study would be a single survey of a population inquiring whether patients have depression and concurrently inquiring about their use of St. John's wort. 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 point in time. Causation cannot be determined from a cross-sectional study, only correlation. Educational Objective: A randomized, blinded, controlled trial is the standard for medical research. Subjects are randomly assigned to receive either the intervention or placebo, and their outcomes are closely measured. These studies generate the most compelling evidence that a given intervention has an effect on a given outcome, and they compare that effect to either a placebo or the current standard of care. Previous Next Score Report Lab Values Calculator Help Pause

197 Exam Section 4: Item 48 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 48. A 6-day-old female newborn is brought to the physician by her parents because of a 3-day history of difficulty feeding, vomiting, and progressive lethargy. Her parents have noticed that her urine has a burnt-sugar odor. Physical examination shows lethargy and hypotonia. Deficiency of which of the following is the most likely cause of these findings in this patient? A) Acid maltase B) Branched-chain a-keto acid dehydrogenase C) Carnitine palmitoyltransferase D) Debranching enzyme E) Glucuronyl transferase

B. Branched-chain a-keto acid dehydrogenase deficiency resulting in maple syrup urine disease is the most likely cause of this patient's lethargy, hypotonia, and abnormal urine. Branched-chain amino acids include isoleucine, leucine, and valine, and they are important in physiologic processes such as gluconeogenesis and cholesterol synthesis. To break down branched-chain amino acids, they must first be converted to a-ketoacids by branched-chain aminotransferases (BCAT) with specific enzymes for each amino acid. This is followed by decarboxylation by branched-chain a-keto acid dehydrogenase complex (BCKDC). These amino acids are ultimately converted to acetoacetate, acetyl-CoA, and succinyl-CoA, respectively. Maple syrup urine disease occurs as a result of deficiency in BCKDC and leads to increased circulating concentrations of these branched-chain amino acids and their keto-acids. The transport of neutral amino acids into the central nervous system is required for synthesis of neurotransmitters, and increased concentrations of leucine affect this process. Additionally, increased concentrations of isoleucine impart a characteristic smell to the urine. Incorrect Answers: A, C, D, and E. Acid maltase (Choice A) deficiency causes Pompe disease, also known as glycogen storage disease, type II. Pompe disease leads to cardiomegaly, cardiomyopathy, hepatomegaly, and hypotonia. Carnitine palmitoyltransferase (Choice C) deficiency causes disorders of long-chain fatty acid oxidation. Carnitine palmitoyltransferases catalyze the bond between palmitate and carnitine, which permits the long-chain fatty acid palmitate to translocate between the cytoplasm and the mitochondrial matrix where B-oxidation (breakdown of the fatty acid) occurs. Deficiency results in cardiomyopathy, myopathy, and liver failure. Debranching enzyme (Choice D) deficiency causes glycogen storage disease, type III, also known as Cori disease. This leads to accumulation of abnormally structured glycogen and ineffective breakdown. Symptoms include skeletal- and cardiomyopathy. Glucuronyl transferase (Choice E) deficiency results in Gilbert syndrome, a benign condition characterized by episodic jaundice throughout life during periods of stress or infection. Glucuronyl transferase conjugates bilirubin to glucuronic acid which makes bilirubin water soluble so that it can be excreted in bile. Educational Objective: Branched-chain a-keto acid dehydrogenase deficiency causes maple syrup urine disease, which is characterized by a buildup of the branched-chain amino acids leucine, isoleucine, and valine. Increased concentrations of isoleucine give the urine its characteristic maple urine smell, and increased concentrations of leucine account for many of the neurologic complications. Previous Next Score Report Lab Values Calculator Help Pause

106 Exam Section 3: Item 6 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 6. A 20-year-old man with intellectual disability is brought to the physician for a routine examination. Physical examination shows a long face, prominent ears, and large testes. His younger brother and a maternal uncle have similar features. Analysis of the patient's DNA shows 800 CGG repeated sequences (N<60) in the 5' untranslated region of the FMR1 gene. This region of expanded CGG trinucleotide repeats is heavily methylated. Which of the following is the most likely effect of these expanded nucleotide repeats on transcription of FMR1 MRNA? A) Alteration of MRNA splicing B) Decreased transcription C) Enhancement of mRNA degradation D) Incorporation of CGG repeats into mRNA E) Increased binding of RNA polymerase

B. Decreased transcription as a result of hypermethylation of the FMR1 gene on the X-chromosome is the most common genetic abnormality resulting in fragile X syndrome. Expansion of the trinucleotide sequence CGG in the 5' untranslated region (UTR) of the FMR1 gene results in hypermethylation of the gene. Methylation is achieved via the action of DNA methyltransferases that preferentially transfer a methyl group to cytosine residues. Areas with high numbers of CGG repeats are at risk for hypermethylation. In the case of fragile X syndrome, hypermethylation renders the FMR1 gene silent by preventing recruitment of transcription factors and stopping RNA polymerase from binding. This gene repression results in an absence of the FMR1 protein, which is important in neural synaptic plasticity. Common clinical manifestations include postpubertal macro-orchidism, a large jaw and face, large and everted ears, congenital heart defects, and intellectual disability. Incorrect Answers: A, C, D, and E. Alteration of MRNA splicing (Choice A) commonly occurs when there are mutations in introns or exons, resulting either in exclusion of essential exons from the final MRNA transcript or abnormal inclusion of introns. This frequently results in an abnormal MRNA that is either prematurely degraded or translated into an abnormal, nonfunctional protein. This is known to occur in some patients with B-thalassemia and phenylketonuria. Enhancement of MRNA degradation (Choice C) can occur when there is failure to polyadenylate the tail of an MRNA transcript, a necessary step in pre-mRNA processing that stabilizes the transcript and prevents premature degradation. Mutations in the polyadenylation site are known to account for some cases of B-thalassemia. Incorporation of CGG repeats into MRNA (Choice D) is not a feature of fragile X syndrome; CGG repeats result in excessive methylation and repression of the gene, but they are not incorporated into the MRNA transcript. Patients with between fifty and two hundred CGG repeats in the FMR1 gene do not exhibit hypermethylation and continue to express a normal FMR1 gene product. This indicates that CGG repeats lead to excessive methylation and suppression of transcription, not to the creation of an abnormal protein. Increased binding of RNA polymerase (Choice E) is not correct as this would result in increased expression of the FMR1 gene, not suppression of transcription as occurs in fragile X syndrome. Educational Objective: Fragile X syndrome is caused by expansion of CGG repeats in the FMR1 gene leading to hypermethylation and repression of gene transcription. Loss of the FMR1 protein leads to intellectual disability, macro-orchidism, and characteristic physical features including a large jaw and ears. Previous Next Score Report Lab Values Calculator Help Pause

199 Exam Section 4: Item 50 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 50. A previously healthy 55-year-old man comes to the physician 2 months after he noticed a lump in his neck. His vital signs are within normal limits. Physical examination shows a 2.5-cm, firm, cervical lymph node on the lower right. A photomicrograph of a biopsy specimen of the node stained with hematoxylin and eosin is shown. Flow cytometry of lymphoid cells from the node yields the following phenotype: CD3 CD19 CD10 3% 70% 70% 70% 0% Карра Lambda Which of the following is the most likely diagnosis? A) Diffuse large B-cell lymphoma B) Follicular lymphoma C) Nodal marginal zone lymphoma D) Precursor B-cell lymphoblastic lymphoma E) Small lymphocytic lymphoma

B. Follicular lymphoma is the most likely diagnosis in this patient. Follicular lymphoma is an indolent B-lymphocyte malignancy that arises from lymph node germinal centers, and it is common for patients to have symptoms for months without significant disease progression. As the malignant cells are B lymphocytes, their immunophenotype is consistent with B lymphocytes. They express CD19, CD10, and surface immunoglobulins. The presence of kappa light chains in the absence of lambda light chains indicates a clonal population. The presence of CD3 in 3% of cells evaluated most likely indicates the presence of normal T lymphocytes in the lymph node and does not indicate that the follicular lymphoma cells express CD3. Further evaluation would show an absence of CD5 and CD43; cytoplasmic presence of BCL-2 is also seen in many low-grade follicular lymphomas. Treatment can sometimes be deferred because follicular lymphoma is not a particularly aggressive malignancy, but clear indications for treatment include bulky nodal disease; B symptoms such as fevers, weight loss, and night sweats; effusions; and cytopenias indicating diffuse bone marrow involvement. Incorrect Answers: A, C, D, and E. Diffuse large B-cell lymphoma (DLBCL) (Choice A) is an aggressive variety of B-lymphocyte lymphoma and has a different immunophenotype than follicular lymphoma. These cells will express B- lymphocyte markers including CD19 and CD20, but they will also express CD45, unlike follicular lymphoma. There is also typically high expression of BCL2 and BCL6. Immediate treatment for DLBCL is recommended. Nodal marginal zone lymphoma (Choice C) typically presents with diffuse nodal disease, although the clinical characteristics are widely variable. The immunophenotype shows cells that stain positive for CD20 but are CD10-negative. Precursor B-cell lymphoblastic lymphoma (Choice D) is a rare malignancy that develops from malignant precursor B lymphocytes. Diffuse marrow involvement is common and complete effacement of nodal architecture is seen. Cells are positive for CD10 and CD19. Genetic analysis may show translocations including t(9,22) or 11q23 mutations. Small lymphocytic lymphoma (Choice E) is the nodal equivalent to chronic lymphocytic leukemia (CLL). These are malignancies of mature B lymphocytes and stain accordingly, with positive CD19 and CD20, but they also express CD5. Educational Objective: Follicular lymphoma is an indolent B-lymphocyte lymphoma that typically presents with a slowly enlarging lymph node. Immunophenotype shows cells with kappa or lambda restriction. There is positive staining for CD10 and CD19, but the absence of CD5 and CD45, which sets follicular lymphoma apart from other B-lymphocyte lymphomas. %3D Previous Next Score Report Lab Values Calculator Help Pause

125 Exam Section 3: Item 25 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 25. A 55-year-old man comes to the physician because of pain and stiffness of his right hand and wrist for 3 months. The pain is exacerbated by hand movement and is relieved by rest. He was a professional boxer for 15 years. Physical examination shows moderate swelling over the dorsum of the right wrist. Examination of fluid aspirated from the wrist shows: Appearance Leukocyte count Neutrophils Glucose Clear, yellowish 250/mm3 (Ns200) 5% (N<25%) 101 mg/dL (N=80-100) Which of the following is the most likely cause of this patient's condition? A) Acute gouty arthritis B) Osteoarthritis C) Rheumatoid arthritis D) Septic arthritis E) Normal age-related changes in joints

B. Osteoarthritis is a noninflammatory arthropathy that occurs secondary to deterioration of articular cartilage. It typically occurs in older persons and may affect any joint. It is commonly associated with the fingers and wrists because of repetitive use; in this patient, it is related to previous trauma from boxing. The hips and knees are commonly affected because of chronic weight-bearing, especially in obese patients. Osteoarthritis is not associated with infectious or inflammatory symptoms and is generally insidious in onset. Patients typically complain of chronic pain in a joint that is worse with movement or exertion and that improves with rest, often with limited range of motion because of underlying joint space narrowing and pain. Fever, chills, erythema over a joint, traumatic injury, or extra-articular manifestations (eg, rash, uveitis, urethritis) suggest an alternate, inflammatory or infectious cause. Imaging may show osteophytosis, subchondral cystic or sclerotic changes, and joint space narrowing. Arthrocentesis typically shows leukocytes of less than 2,000/mm3 with no additional abnormalities, as in this case. Treatment includes rest, physical therapy, nonsteroidal anti-inflammatory drugs (NSAIDS), intra-articular injections (eg, cortisone), and joint replacement surgery for severe or refractory cases. Incorrect Answers: A, C, D, and E. Acute gouty arthritis (Choice A) is a monoarticular arthropathy resulting from an intra-articular inflammatory reaction to the precipitation of monosodium urate crystals. It typically presents with atraumatic joint pain, erythema, and swelling, which can recur in patients who are under- or untreated. The most common joint involved is the first metatarsophalangeal, but other joints include the knees and elbows. Arthrocentesis generally shows 10,000 to 50,000 leukocytes/mm3 with a mixture of neutrophils and lymphocytes, and needle-like, birefringent crystals are seen on microscopy. Treatment consists NSAIDS, colchicine, and/or corticosteroids for acute exacerbations. Rheumatoid arthritis (Choice C) is an autoimmune inflammatory arthropathy that results in symmetric involvement of multiple joints, most commonly the metacarpophalangeal and proximal interphalangeal joints of the hands, demonstrating articular erosions and progressive joint space narrowing. Ulnar deviation of the fingers and boutonniere and swan neck deformities commonly occur. Arthrocentesis, if performed, generally shows 10,000 to 50,000 leukocytes/mm3 Septic arthritis (Choice D) is generally acute and painful, and it results from bacterial infection of the synovial joint space, most commonly by S. aureus. The affected joint demonstrates pain, effusion, and erythema. Arthrocentesis is characteristically purulent and shows greater than 50,000 leukocytes/mm3 Normal age-related changes in joints (Choice E) include joint stiffening, decreased flexibility, and thinned cartilage. Swelling, as experienced by this patient, is more consistent with osteoarthritis given his history of boxing. Educational Objective: Osteoarthritis is a noninflammatory arthropathy that occurs secondary to deterioration of articular cartilage. It is commonly associated with the fingers and wrists because of repetitive use and the hips and knees because of weight-bearing and obesity. Arthrocentesis typically shows leukocytes of less than 2,000/mm3 with no additional abnormalities. Previous Next Score Report Lab Values Calculator Help Pause

196 Exam Section 4: Item 47 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 47. A 44-year-old woman with a history of breast cancer comes to the physician for an initial examination. She says, "I'm concerned about my future. Can you give me any advice?" She recently moved to the city because her husband changed jobs. She underwent a mastectomy for inflammatory breast cancer, and she completed a course of chemotherapy 2 months ago. Physical examination shows an absent left breast and a well-healed scar on the left side of the chest. There are no palpable lymph nodes. Which of the following initial responses by the physician is most appropriate? A) "I recommend that you attend a breast cancer support group to learn from other women who have gone through a similar experience." B) "I understand you've been treated by other physicians. Tell me what you know about your disease." C) "I would like you to review these pamphlets by the American Cancer Society, and we can discuss them at your next appointment." D) "The survival statistics for persons with the type of cancer you have are not favorable." E) "Unfortunately, you are not out of the woods yet. You may have a recurrence of cancer in the next several months."

B. When patients express concerns about an existing medical condition, physicians should initially ask open-ended questions to assess patients' understanding of their condition. An open-ended question may encourage this patient to elaborate on her specific concerns. The physician can then tailor further discussion to address the patient's knowledge gaps and specific concerns. If a patient expresses distress about their situation, the physician should also validate the patient's emotions (eg, restate the patient's words or infer the underlying emotion). In this situation, the physician should first clarify the patient's specific concerns and emotions. Incorrect Answers: A, C, D, and E. Before providing recommendations (Choices A and C), the physician should ask an open-ended question to assess the patient's understanding and specific concerns. Though it's possible this patient could benefit from a support group, the physician should first confirm this with the patient since she may already have other supports (Choice À). The patient may already know the basics of breast cancer and may not benefit from pamphlets (Choice C). Providing education about the patient's poor prognosis (Choices D and E) would be premature given the patient's reassuring examination and may startle the patient. Any conversation about the patient's prognosis should be approached cautiously after assessing the patient's understanding of her prognosis. Educational Objective: When a patient expresses concern about an existing medical condition, the physician should initially ask open-ended questions to assess the patient's understanding of their condition. The physician can then tailor further discussion to address the patient's knowledge gaps and specific concerns. %D Previous Next Score Report Lab Values Calculator Help Pause

134 Exam Section 3: Item 34 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 34. A 45-year-old woman is brought to the emergency department because of the sudden onset of severe abdominal pain. Examination shows an irregularly irregular pulse, a grade 3/6, apical diastolic murmur, and an opening snap. An x-ray of the abdomen shows distention of the small intestine, the ascending colon, and a portion of the transverse colon. Which of the following is the most likely underlying cause of these findings? A) Atherosclerosis B) Embolization C) Hemorrhage D) Plaque rupture E) Thrombosis

B. The patient is likely suffering acute mesenteric ischemia caused by occlusion of a mesenteric vessel from embolization of a thrombus. Mesenteric ischemia presents with acute, severe abdominal pain and may be associated with nausea, vomiting, and bloody diarrhea, commonly in a patient with atrial fibrillation, splanchnic atherosclerosis with plaque rupture and arterial occlusion, or a hypercoagulable state. Acute mesenteric ischemia is life-threatening because of the potential complications of bowel necrosis including sepsis, peritonitis, and perforation. The superior mesenteric artery provides blood supply from the duodenum to the proximal two-thirds of the transverse colon and is the most commonly affected mesenteric vessel by thromboembolic events. The patient's abnormal distension of bowel correlating to the superior mesenteric artery distribution on abdominal imaging suggests that this artery has become acutely occluded. This patient likely has underlying atrial fibrillation given the irregularly irregular pulse with a murmur suggestive of mitral stenosis. Mitral stenosis can result in left atrial enlargement and a predisposition for atrial fibrillation. Uncoordinated contraction of the atrium results in blood pooling, which promotes thrombus formation. A left atrial thrombus can potentially embolize to the arterial circulation and cause systemic embolic events such as stroke, myocardial infarction, mesenteric ischemia, or acute limb ischemia. Patients with atrial fibrillation should be initiated on therapeutic anticoagulation to prevent thromboembolic events. Incorrect Answers: A, C, D, and E. Atherosclerosis (Choice A) is a risk factor for chronic mesenteric ischemia because of narrowing of the mesenteric arteries and diminished perfusion to the gut. It is classically associated with postprandial abdominal pain and weight loss. Atherosclerosis involving the mesenteric vessels or the aorta may lead to plaque rupture (Choice D) and thrombosis (Choice E), resulting in acute mesenteric ischemia with a similar presentation. However, embolization is more likely in this young patient with atrial fibrillation. Nonatherosclerotic thrombosis is unlikely in the absence of an arterial thrombophilia, such as antiphospholipid antibody syndrome. Hemorrhage (Choice C) can result in acute mesenteric ischemia secondary to hypovolemic shock. This typically affects watershed areas at the periphery of the arterial supply (eg, between the vascular territories of the superior and inferior mesenteric arteries) such as the splenic flexure of the colon and the rectosigmoid junction. This patient with an underlying atrial arrhythmia, lack of other symptoms, and involvement of the entire superior mesenteric artery bowel distribution more likely has suffered an embolic event. Educational Objective: Atrial fibrillation is a risk factor for arterial thromboembolic disease such as stroke, myocardial infarction, acute mesenteric ischemia, and acute limb ischemia. The superior mesenteric artery is the most commonly affected mesenteric vessel by thromboembolic events. Previous Next Score Report Lab Values Calculator Help Pause

109 Exam Section 3: Item 9 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment メ which of the following? 9. For aerobic catabolism of compounds with equivalent numbers of carbon atoms, the greatest amount of energy in the form of ATP is obtained from compounds with the largest number of A) Aldehyde groups B) Amino groups C) Hydrogen atoms D) Keto groups E) Oxygen atoms

C. Catabolism of compounds is a mechanism by which cells obtain energy for use in other cellular processes. Eukaryotic cells catabolize molecules such as glucose, glycogen, amino acids, alcohols, and ketones for use as energy substrates. Catabolism often requires the creation of intermediate molecules from these precursor substrates prior to direct use as energy sources. An example of this is the synthesis of hydrated nicotinamide adenine dinucleotide (NADH) from NAD+ which is created during the catabolism of glucose, alcohols, and ketones. NADH, in turn, donates protons and electrons to the electron transport chain (ETC) within the mitochondria, which creates a proton gradient across the inner mitochondrial membrane that supplies the energy to synthesize ATP. ATP can then be used to supply energy to maintain transmembrane gradients, activate or deactivate enzymes, facilitate muscle contraction, or participate in hundreds of other intracellular energy-dependent processes. The reduced NAD+ molecule, NADH, transports energy stored as protons (hydrogen atoms). Similarly, flavin adenine dinucleotide (FAD) transports hydrogens as hydrated FAD, abbreviated FADH, Both of these compounds donate electrons to the ETC to ultimately generate ATP via ATP synthase, powered by a hydrogen gradient across the mitochondrial membrane. The more reduced a substrate is, the more it can reduce FAD or NAD to FADH, or NADH. Therefore, heavily hydrogenated substrates supply the greatest amount of energy. Incorrect Answers: A, B, D, and E. Aldehyde groups (Choice A) and keto groups (Choice D) are already partially oxidized. Each of these groups contains a double bond to an oxygen atom. A terminal aldehyde group may be bound to one additional hydrogen atom, which would permit oxidation to a carboxylic acid and the creation of a reduced FAD or NAD+ conjugate. However, the amount of reduced conjugate created is less than if the group began as a hydroxyl. Amino groups (Choice B) may, depending on their primary, secondary, tertiary, or quaternary status, be able to undergo some degree of oxidation. However they generally maintain a lone pair of electrons instead of existing in a fully reduced form. Oxygen atoms (Choice E) are not able to participate in any reaction donating hydrogen, and therefore, they do not generate FADH, or NADH. Educational Objective: Heavily hydrogenated molecules can undergo oxidation to create FADH, and NADH for use in the generation of ATP. Any partially oxidized group cannot contribute as much hydrogen as a completely reduced group, and therefore, cannot generate as much ATP. Previous Next Score Report Lab Values Calculator Help Pause

124 Exam Section 3: Item 24 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 24. A 63-year-old man from southern China has had a bloody nasal discharge for the past 5 months. Physical examination shows an enlarged, fixed lymph node in the upper posterior cervical triangle of the left neck. Examination of tissue obtained on needle biopsy of the lymph node shows neoplastic cells that are positive for keratin and Epstein-Barr nuclear antigen. Which of the following is the most likely diagnosis? A) Burkitt lymphoma B) Kaposi sarcoma C) Nasopharyngeal carcinoma D) Sinonasal carcinoma E) Squamous cell carcinoma

C. Epstein-Barr virus (EBV) is a herpesvirus (human herpesvirus 4), which can cause the acute illness infectious mononucleosis. It is also associated with numerous malignancies such as nasopharyngeal carcinoma, Burkitt lymphoma, primary central nervous system lymphoma, gastric carcinoma, oral hairy leukoplakia in patients with HIV co-infection, and lymphoproliferative diseases. Nasopharyngeal carcinoma typically arises from keratinizing squamous cells of the nasopharyngeal epithelium. Risk factors include tobacco use, alcohol use, consumption of foods containing nitrosamine preservatives, and EBV infection in patients of Asian descent. Early symptoms include epistaxis and/or unilateral nasal obstruction. Metastatic disease typically presents with enlarging cervical lymph nodes, otitis media, and nasal obstruction. Additional symptoms depend on the extent and location of tumor spread. Treatment includes surgical resection, radiotherapy, and/or chemotherapy. Incorrect Answers: A, B, D, and E. The endemic form of Burkitt lymphoma (Choice A) is associated with EBV infection. The majority of these cases occur in children in the African equatorial belt and classically present with a mass in the jaw and/or kidneys. Histologic features include a diffuse infiltrate of monomorphic lymphoid cells with large numbers of macrophages. Kaposi sarcoma (Choice B) is a neoplasm of lymphatic endothelial cells and is associated with human herpesvirus 8 in AlIDS or states of immunosuppression (eg, post-transplantation). The lesions are heterogenous and may be either red or purple, flat or raised; they commonly occur on the face, oral mucosa, legs, and torso. Lesions may also arise in the gastrointestinal tract and the lungs. Histologic characteristics include spindle-shaped cells that form clefts. Sinonasal carcinoma (Choice D) is a type of neoplasm that may arise from the epithelium of the nasal cavity or paranasal sinuses. It may present with epistaxis, nasal discharge, and partial nasal obstruction. It is associated with human papilloma virus (HPV), but not with EBV. While nasopharyngeal carcinoma is a subtype of keratinizing squamous cell carcinoma (Choice E), the association with EBV is specific to nasopharyngeal carcinoma. Educational Objective: Epstein-Barr virus (EBV) is associated with numerous malignancies. Nasopharyngeal carcinoma is a squamous cell neoplasm strongly associated with EBV, especially in endemic regions of Asia. Previous Next Score Report Lab Values Calculator Help Pause

114 Exam Section 3: Item 14 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 14. A 1-month-old boy with X-linked severe combined immunodeficiency is enrolled in a gene therapy trial. Bone marrow stem cells are extracted from the patient and are treated ex vivo with a retrovirus expressing the wild-type gene. Although the patient initially responds well, 30 months after treatment he develops a form of T-lymphocyte leukemia. Molecular studies show that the T lymphocytes are monoclonal. Which of the following viral processes is the most likely cause of the leukemia in this patient? A) Deactivation of the p53 tumor suppressor gene B) Generation of the Philadelphia (Ph1) chromosome C) Integration adjacent to an oncogene D) Phosphorylation of pRB E) Reverse transcription of PTEN

C. Integration of the wild type gene adjacent to an oncogene most likely accounts for this patient's new diagnosis of T-lymphocyte leukemia. The use of retroviruses offers a promising tool for integrating functional genes into the cells of patients with monogenic disorders. In these instances, the functional gene is engineered to fit within a segment of viral RNA. Following entry into the target cell-in this case, T-lymphocytes-the viral RNA is reverse transcribed into DNA. Through the use of viral integrase proteins, the functional gene can then be integrated into the host genome, which allows for the functional gene to be expressed and to be passed onto daughter cells during mitosis. In prior cases of gene therapy for SCID that resulted in leukemia, analysis of the T-lymphocyte clones showed that the wild-type gene was commonly inserted in the transcription start site of the LMO2 oncogene. Subsequent expression of the oncogene likely accounts for the development of leukemia in this patient. Incorrect Answers: A, B, D, and E. Deactivation of the p53 tumor suppressor gene (Choice A) is a common occurrence in many cancers but occurs in only a fraction of acute T-lymphocyte leukemias. It is more common in relapsed leukemia or therapy-related leukemia where additional mutations have developed over time. Inactivation of a single tumor suppressor gene is often insufficient for the development of cancer and typically requires at least one other mutation. Generation of the Philadelphia (Ph1) chromosome (Choice B) is an uncommon occurrence in T-lymphocyteleukemia, although it does occur. The Philadelphia chromosome is created by a translocation between chromosomes 9 and 22, leading to constitutive activation of the ABL1 tyrosine kinase. It is most commonly found in chronic myelogenous leukemia (CML). Phosphorylation of pRB (Choice D) leads to inactivation of this tumor suppressor gene and allows cells to enter the cell cycle. Like other tumor suppressor genes, a second genomic mutation is often required for the development of cancer. Inactivation of this protein is most commonly associated with retinoblastoma, cholangiocarcinoma, bladder and lung cancer. Inactivation is less common in T-lymphocyte leukemia and is not consistent with the mechanism of oncogenesis in this patient. Reverse transcription of PTEN (Choice E) would decrease the incidence of malignancy if expressed, as it is a tumor suppressor gene, not an oncogene. Educational Objective: Gene therapy for monogenic diseases using retroviruses encoding wild-type genes can occasionally result in the insertion of these genes next to oncogenes, which can predispose to the development of T-lymphocyte leukemia. Previous Next Score Report Lab Values Calculator Help Pause

175 Exam Section 4: Item 25 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 400 25. A 72-year-old woman comes to the physician because of diffuse muscle pain and weakness for 6 months. The muscle pain is exacerbated by activity. Physical examination shows proximal muscle weakness and tenderness over the surface of both shins. The dot on the nomogram shown indicates the relationship between her serum calcium and parathyroid hormone concentrations. Which of the following conditions is the most likely cause of this patient's symptoms? 200 - A) Hypoparathyroidism 100 B) Metastatic breast cancer 60 Normal C) Osteomalacia 10 OD) Osteoporosis O E) Primary hyperparathyroidism ト 8 9 10 11 12 13 14 Calcium (mg/dL)

C. Osteomalacia is characterized by diffuse bone pain and weakness, generally caused by vitamin D, calcium, or phosphorus deficiency or malabsorption. Inadequate bone mineralization results, which presents in children with rickets and in adults with diffuse myalgias, muscle weakness, bone pain, and fragility fractures. Vitamin D deficiency is the most common cause; thus, patients living in locations distant from the equator with limited sunlight exposure or those who have dietary deficiency are most likely to be affected. Muscle weakness occurs because of calcium deficiency. In this patient's case, her serum parathyroid hormone (PTH) is increased, but her serum calcium concentration remains below the lower limit of normal. This suggests an appropriate increase in PTH in an attempt to compensate for hypocalcemia. However, insufficient calcium intake, insufficient vitamin D, or an inability to absorb calcium persists. Osteolysis occurs to maintain serum calcium concentration, with resultant bone pain, fragility, and demineralization. Treatment begins with vitamin D and calcium supplementation. Incorrect Answers: A, B, D, and E. Hypoparathyroidism (Choice A) may result from surgical removal of the parathyroid glands, infiltrative disease, inflammatory or autoimmune destruction, or malignancy. Hypocalcemia with decreased PTH would be expected on laboratory analysis. Metastatic breast cancer (Choice B) classically presents with hypercalcemia from bone destruction. Breast cancer can present with mixed osteoblastic and osteoclastic lesions and bone pain. In this case, the patient's decreased serum calcium makes metastatic breast cancer less likely. Osteoporosis (Choice D) occurs because of the loss of bone mass, most commonly appearing in elderly women of European descent. The condition results from the interplay of three possible mechanisms: failure to achieve peak bone density, increased bone resorption, and decreased new bone formation. The condition is often asymptomatic until a traumatic injury results in a fracture, most commonly involving the hip, distal radius, or vertebrae. Classically, serum calcium, PTH, vitamin D, and phosphate are within normal limits. Primary hyperparathyroidism (Choice E) presents with hypercalcemia, hypophosphatemia, and increased PTH. It occurs because of parathyroid adenoma, hyperplasia, or carcinoma. Educational Objective: Osteomalacia is characterized by bone demineralization and often results from vitamin D deficiency. Hypocalcemia may be observed on laboratory analysis, with resultant appropriate increases in serum PTH as a means of compensation via feedback regulation. Previous Next Score Report Lab Values Calculator Help Pause Serum parathyroid hormone (pg/mL) 88

149 Exam Section 3: Item 49 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 49. A 35-year-old woman comes to the physician because of a 2-day history of painful blisters on her upper and lower lips. She has a 10-year history of similar symptoms interspersed with periods of remission. A photograph of the lesions is shown. Which of the following neuronal processes was most likely involved in the initial establishment of the viral infection in this patient? A) Action potential B) Bulk axoplasmic flow C) Protein synthesis D) Retrograde axon transport E) Synaptic transmission

D. This patient's symptoms of recurrent, painful blisters on the lips are consistent with infection by herpes simplex virus in the form of herpes labialis. Infection typically begins with the formation of vesicles, which lyse and progress to shallow, painful ulcers with an erythematous border. HSV-1 and HSV-2 are both members of the herpesvirus family of double-stranded DNA enveloped viruses. Herpes labialis is most often caused by HSV-1 but can be caused by HSV-2. In contrast, genital herpes is most often caused by HSV-2 but can also be caused by HSV-1. Co-infection with both strains is possible. HSV infects the cells of the epidermis and dermis and then travels proximally to the sensory ganglia. The viral particles utilize retrograde axon transport, or movement toward the cell body, to travel in retrograde fashion from the dendrites of the neuron proximally through the axon, and finally to the sensory ganglia. In the sensory ganglia, HSV establishes a reservoir of virus, which is protected from the immune system. It remains latent in the sensory ganglia until reactivation, when the virus then migrates distally to cause painful vesicles and erosions on the lips, sometimes with associated cervical lymphadenopathy. Systemic manifestations are also possible, including viral meningitis and encephalitis. The diagnosis may be confirmed by a polymerase chain reaction test, but performing a Tzanck smear is also an efficient method of diagnosis. On Tzanck smear, herpes infected multinucleated giant cells may be seen. Treatment for herpetic infections involves drugs that inhibit viral DNA polymerase, classically by guanosine analogs such as acyclovir, valacyclovir, and famciclovir. Incorrect Answers: A, B, C, and E. Action potential (Choice A) refers to the electrical signal propagated down the neuron. It is created by the rising and falling of the membrane potential (depolarization and repolarization), which ultimately results in the release of neurotransmitters at the neuronal synapse. This does not play a role in HSV infection. Bulk axoplasmic flow (Choice B) is the process by which intracellular organelles such as mitochondria or vesicles are moved from the cell body of a neuron through the cytoplasm of the neuron (also called the axoplasm). Bulk flow occurs when the entirety of axon contents are moved toward or away from the cell body. Protein synthesis (Choice C) occurs when a virus infects a cell and then utilizes the host cell's existing machinery to make viral proteins. This process is seen in HSV infection but is not the neuronal process by which the virus establishes a reservoir in the sensory ganglion. Synaptic transmission (Choice E) is the process by which a neuron communicates with a second neuron or target cell across a synapse. HSV infection does not require movement from one neuron to another; rather, it moves in a retrograde fashion within the initial neuron it infected until it reaches the sensory root ganglion. Educational Objective: Herpes labialis is most often caused by HSV-1 or HSV-2, which infect the cells of the epidermis and dermis and then travel via retrograde axonal transport to the sensory ganglia, where they remain latent. Upon reactivation, the virus migrates distally to cause painful vesicles and erosions on the lips, sometimes with associated cervical lymphadenopathy. Previous Next Score Report Lab Values Calculator Help Pause

154 Exam Section 4: Item 4 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 4. The kinetics of a regulatory enzyme are shown in the figure as curve A. The kinetics in the presence of an allosteric modifier are represented by curve B. Which of the following best describes the effect of the modifier on the activity of this enzyme? A A) Completely inhibits enzyme activity B) Decreases Km C) Decreases V, max D) Increases Km [S] E) Increases V max F) Has no effect on enzyme activity

D. Enzymes are kinetically categorized by their Michaelis-Menten constant, Km, which inversely reflects their affinity for the substrate, and by Vmax, the maximum catalysis rate of the enzyme. In other words, as Km increases, affinity for the substrate decreases. Km is defined as the substrate concentration at which the enzyme reaches one-half of its Vmax In the pictured graph, the presence of an allosteric modifier shifts the enzymatic curve to the right while the shape of the curve remains unchanged. The X-axis in this instance represents the concentration of substrate, so shifting the graph to the right means that a greater concentration of substrate is required to achieve the same enzymatic velocity, likely because the allosteric modifier has decreased the enzyme's affinity for its substrate. This is characterized by an increase in Km- Incorrect Answers: A, B, C, E, and F. Completely inhibits enzyme activity (Choice A) is not correct because the shape of the enzymatic curve has not changed, only shifted to the right, implying a decreased affinity for its substrate. Complete inhibition would be represented by a flat horizontal line. Decreases Km (Choice B) is not correct. This would represent an instance in which the allosteric modifier increased the affinity of the enzyme for its substrate and would cause the curve to shift to the left. Decreases Vmax (Choice C) or increases Vmax (Choice E) are not correct choices. Changes in Vmax Would appear as changes in the overall height of curve B on the Y-axis. The Vmax of these two curves is the same. Has no effect on enzyme activity (Choice F) is incorrect; if this were the case, the graphs would overlap. Educational Objective: In the presence of an allosteric modifier, the affinity of an enzyme for its substrate may be decreased, which shifts its kinetic curve to the right. This correlates with an increase in the Km of the enzyme. Previous Next Score Report Lab Values Calculator Help Pause Velocity

148 Exam Section 3: Item 48 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 48. An 8-year-old boy is admitted to the hospital because of severe flank pain and blood in his urine. He has had intellectual developmental disorder since early infancy. Examination shows choreoathetosis and self-mutilation of the lips and fingers. His serum uric acid concentration is increased. Urinalysis shows numerous uric acid crystals and erythrocytes. The most likely cause of the hyperuricemia is deficient activity of which of the following enzymes? A) õ-Aminolevulinate synthase B) Glycine-oxaloacetic transaminase C) Glycine synthase D) Hypoxanthine-guanine phosphoribosyl transferase E) Serine transhydroxymethylase F) Xanthine oxidase

D. Hypoxanthine-guanine phosphoribosyl transferase (HGPRT) inactivating mutations are the most likely cause of hyperuricemia in this patient with Lesch-Nyhan syndrome. HGPRT is 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 concentrations of xanthine and uric acid. Clinical findings include intellectual disability, aggressive behavior, self-mutilation, gout, and dystonia. Hyperuricemia in Lesch-Nyhan syndrome is treated with xanthine oxidase inhibitors such as allopurinol or febuxostat in order to decrease the synthesis of uric acid. Incorrect Answers: A, B, C, E, and F. Ö-Aminolevulinate synthase (Choice A) deficiency causes X-linked sideroblastic anemia. This enzyme catalyzes the synthesis of õ-Aminolevulinic acid, which is a precursor molecule required for synthesis of heme. Glycine-oxaloacetic transaminase (Choice B) is not the enzyme implicated in Lesch-Nyhan syndrome. Glycine synthase (Choice C) is composed of four distinct proteins: T-, P-, L-, and H-protein. They catalyze both the formation of glycine and the reverse reactions that metabolize glycine. Mutations can increase glycine concentrations and cause encephalopathy. Serine transhydroxymethylase (Choice E) catalyzes conversion of serine to glycine and tetrahydrofolate to 5,10-methylenetetrahydrofolate. Deficiency occurs in Smith-Magenis syndrome and is characterized by developmental delay, abnormal facial features, and bizarre behavior. Xanthine oxidase (Choice F) is an enzyme that converts xanthine to uric acid, which is soluble and excreted in the urine. Deficiency causes increased concentrations of xanthine that can precipitate to form renal stones. Educational Objective: Lesch-Nyhan syndrome presents with intellectual disability, aggressive behavior, self-mutilation, gout, and dystonia. Mutations in HGPRT cause this disorder by impairing the salvage of guanine and hypoxanthine, which leads to increased concentrations of xanthine and uric acid. Hyperuricemia in Lesch-Nyhan syndrome is treated with xanthine oxidase inhibitors such as allopurinol or febuxostat in order to decrease the synthesis of uric acid. %3D Previous Next Score Report Lab Values Calculator Help Pause

144 Exam Section 3: Item 44 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 44. An experiment is performed in which an isolated segment of skeletal muscle undergoes isotonic contraction. Which of the following would determine the maximum rate of shortening? A) Amount of muscle phosphocreatine B) Amplitude of the action potentials C) Frequency of the action potentials D) Rate of cross-bridge recycling E) Rate of Na-K*ATPase activity

D. Isotonic contraction refers to contraction under constant force or tension, with resultant change in the length of the muscle fiber. This is in contrast to isometric contraction, which is the generation of force (contraction) without a change in length. Contractions can be concentric, wherein the muscle shortens while generating force, or eccentric, wherein the muscle lengthens under tension. The action of shortening within a muscle is governed by interactions between actin and myosin. In response to a depolarization of the motor end plate, calcium enters the myocyte, which then triggers additional calcium release from the sarcoplasmic reticulum. Calcium in turn triggers conformational change in the troponin-tropomyosin complex, exposing myosin binding sites. From here, the cross-bridge contraction cycle occurs. Myosin, bound by ADP and inorganic phosphate, binds its active site on actin. The power stroke bends the myosin head, shifting the parallel position of actin with respect to myosin, shortening the sarcomere. ADP and inorganic phosphate dissociate in this step. ATP binding releases the myosin head back to its cocked position and detaches the cross-bridge. With hydrolysis of the ĀTP, the myosin head forms a new cross bridge with a downstream actin molecule. This continued cycling results in shortening of the muscle fiber. Thus, the rate of cross-bridge recycling determines the maximum rate at which the muscle can shorten. In contrast to skeletal muscle, vascular smooth muscle has a requirement for constant basal tension to maintain vascular tone. Actin-myosin cross bridging constantly cycles at a slow rate, which facilitates sustained contraction. Skeletal muscle generally cycles much more rapidly. Incorrect Answers: A, B, C, and E. The amount of muscle phosphocreatine (Choice A) is important during intense muscle contractions in which ATP is rapidly depleted as creatine acts as a phosphate donor to regenerate ATP from ADP. It does not affect the rate of shortening as directly as the rate of cross-bridge cycling. However, in prolonged, sustained, intense contraction, ATP stores may be depleted without a phosphocreatine reservoir. Amplitude of the action potentials (Choice B) and frequency of the action potentials (Choice C) determine the number of muscle cell depolarizations and, therefore, the number of contractions. Amplitude of an action potential is independent of the endplate potential that produced it; action potentials are all-or-nothing events that spread across the myocyte membrane. Frequency of action potentials does affect the regulation of tension; the increase in force seen with increased action potential frequency yields a summative response from motor units that had not yet been recruited. This affects the force of contraction, not the rate of cycling. Rate of Nat-K* ATPase activity (Choice E) affects the generation and maintenance of membrane potentials because of ionic concentrations of sodium, potassium, and calcium. Inhibition of this channel increases the concentration of intracellular sodium, which directly increases the concentration of intracellular calcium through sodium-calcium exchange transporters. Increased cytoplasmic calcium can increase the force of contraction, but not the rate of shortening. Educational Objective: The rate of shortening within a muscle is governed by interactions between actin and myosin via cross-bridge contraction cycling. The more rapid the rate of cross-bridge recycling, the more rapid the rate of shortening. Previous Next Score Report Lab Values Calculator Help Pause

113 Exam Section 3: Item 13 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 13. A 50-year-old man with schizophrenia is admitted to the hospital because of chest pain. He develops acute fever, lethargy, and muscular rigidity 5 days later. He has been receiving haloperidol treatment since admission. His temperature is 41°C (105.8°F), pulse is 120/min, and blood pressure is 140/80 mm Hg. He is disoriented and agitated and has diffusely increased muscle tone. Serum creatine kinase activity is 1500 U/L. Which of the following is the most likely diagnosis? A) Catatonia B) Dystonia C) Malignant hyperthermia D) Neuroleptic malignant syndrome E) Tardive dyskinesia

D. Neuroleptic malignant syndrome (NMS) is most likely in this patient who has received high-potency antipsychotic medication. NMS is a life-threatening condition associated with the antagonism of dopamine; it presents with altered mental status, muscle rigidity, hyperthermia, and autonomic dysfunction. Common causative agents are antipsychotics (especially high-potency antipsychotics such as haloperidol) and antidopaminergic nausea medications. Rigidity results from dopamine antagonism in the nigrostriatal pathway, while autonomic dysfunction and hyperthermia may result from dopamine antagonism in the hypothalamus. Characteristic laboratory abnormalities include leukocytosis and increased serum creatine kinase, lactate dehydrogenase, and alkaline phosphatase activity (caused by prolonged muscle rigidity and consequent myolysis). Treatment is supportive and includes discontinuation of the offending medication, hydration, cooling blankets, and benzodiazepines if needed for agitation. Dantrolene is potentially beneficial. Incorrect Answers: A, B, C, and E. Catatonia (Choice A) is a neurobehavioral syndrome marked by the inability to move or speak or the inability to stop moving or speaking repetitively that typically arises from an exacerbation of a psychiatric disorder. In a severe form of catatonia called malignant catatonia, patients additionally demonstrate autonomic instability, which can present similarly to NMS. However, this patient's lack of preceding psychiatric symptoms or stereotyped movements make malignant catatonia less likely. Acute dystonia (Choice B), an adverse reaction from antipsychotics, refers to involuntary contractions of major muscle groups and can present as torticollis, oculogyric crisis, or opisthotonus. Confusion and autonomic dysfunction would be atypical. Malignant hyperthermia (Choice C) is a hypermetabolic state caused by exposure to an anesthetic in genetically vulnerable patients that features muscle rigidity, rhabdomyolysis, increased creatine kinase, hyperthermia, and hypercapnia. This patient has no history of anesthetic administration. Tardive dyskinesia (Choice E) is a syndrome of involuntary movements (eg, lip smacking, choreoathetoid movements of the tongue) that results from the chronic use of antipsychotics. Acute- onset muscle rigidity would be atypical. Autonomic dysfunction and confusion are also inconsistent with tardive dyskinesia. Educational Objective: Neuroleptic malignant syndrome is a life-threatening condition associated with the antagonism of dopamine that presents with altered mental status, muscle rigidity, hyperthermia, and autonomic dysfunction. Characteristic laboratory abnormalities include leukocytosis and increased creatine kinase, alkaline phosphatase, and lactate dehydrogenase activity. Previous Next Score Report Lab Values Calculator Help Pause

116 Exam Section 3: Item 16 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 16. A 24-year-old nulligravid woman comes to the office because of a 6-month history of chronic headache and vision difficulties. She also recently had several episodes of milky discharge from her breasts. Menses occur at irregular 50- to 90-day intervals; her last menstrual period was 6 months ago. She takes an oral contraceptive continuously and does not have withdrawal bleeding. Pulse is 80/min, respirations are 14/min, and blood pressure is 110/78 mm Hg. Ophthalmologic examination shows bitemporal hemianopia. Physical examination shows galactorrhea. Urine pregnancy test result is negative. MRI of the brain shows a 2-cm mass. Which of the following labeled structures in the normal MRI of the brain is the most likely location of the mass in this patient? C E A) B) C) D) E)

D. This patient's bitemporal hemianopia is most likely the result of optic chiasm compression from a pituitary adenoma, which is often associated with galactorrhea and amenorrhea in women secondary to the overproduction of prolactin. The pituitary gland (Choice D) is located in the sella turcica. A pituitary adenoma is a benign brain tumor that can present with mass effect symptoms such as headache and bitemporal hemianopsia (loss of bilateral temporal visual fields). Bitemporal hemianopsia occurs because of mass effect of the tumor on the optic chiasm, which contains crossing fibers from bilateral nasal optic nerves (containing visual information from the temporal eye fields). Pituitary adenomas can also cause symptoms from hyperpituitarism or hypopituitarism depending on whether the tumor is functionally secreting hormones. The most common functional pituitary adenoma is a prolactinoma; other types of pituitary adenomas secrete follicle- stimulating hormone (FSH), luteinizing hormone (LH), adrenocorticotropic hormone, TSH, or growth hormone. Increased prolactin concentrations lead to galactorrhea and suppress gonadotropin- releasing hormone, which results in decreased FSH and LH production and manifests as amenorrhea. Incorrect Answers: A, B, C, and E. Choice A identifies the hypothalamus, which is responsible for body homeostasis by secreting hormones that regulate functions such as body temperature, hunger, and sleep. The most common hypothalamic tumors are gliomas or metastases from extracranial primary malignancies. Choice B identifies the midbrain, and Choice E identifies the pons. Both brain stem regions contain structures that mediate diverse functions such as sensory and motor control, descending pain modulation, cranial nerve function, and consciousness. Midbrain and pontine masses are rare and would not cause bilateral hemianopia or secrete prolactin. Choice C identifies the thalamus, which primarily serves as a sensory input and relay nucleus. Thalamic masses are rare and would not cause bilateral hemianopia from mass effect or secrete prolactin. Educational Objective: Pituitary adenomas can cause mass effect symptoms such as headache and bitemporal hemianopsia from compression of the adjacent optic chiasm. Additionally, they may be functional or nonfunctional in nature. When functional, this can result in the hypersecretion of hormones, most commonly prolactin, which typically manifests with amenorrhea and galactorrhea in females. Previous Next Score Report Lab Values Calculator Help Pause

133 Exam Section 3: Item 33 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 33. A 50-year-old woman has azotemia. Renal ultrasonography shows bilateral hydroureters and hydronephrosis. The most likely cause of these findings is primary carcinoma of which of the following? A) Colon B) Kidney C) Ovary D) Ureter E) Uterine cervix

E. End-stage kidney disease with azotemia is a potential presenting symptom of advanced cervical cancer; local invasion and the subsequent development of bilateral ureteral obstruction causes associated postrenal azotemia. Cervical squamous cell carcinoma begins from the transformation zone or squamocolumnar junction of the cervix, which demarcates the ectocervix and its squamous epithelium from the endocervix and its columnar epithelium. This region is the most likely initial site of cervical cancer development. It is the target region of a Pap smear, a screening test for cervical neoplasia. Risk factors for cervical cancer include infection with human papillomavirus (HPV) high risk strains 16, 18, 31, or 33, early age at onset of sexual activity, tobacco use, oral contraceptive use, immunosuppression, and infection with HIV or other sexually transmitted infections. Červical cancer classically presents with vaginal bleeding with or without intercourse. As it progresses, it may locally invade the urinary bladder, or the rectum if advanced, and it can result in obstruction of bilateral distal ureters near their insertion into the urinary bladder. The obstruction results in consequent bilateral hydroureteronephrosis, which presents as postrenal azotemia because the inability to appropriately drain urine into the bladder can cause progressively worsening kidney failure if untreated. If distant spread occurs, the lung is a common site of metastatic disease. Incorrect Answers: A, B, C, and D. Colon carcinoma (Choice A) most commonly occurs in the rectosigmoid region of the large bowel, which can result in partial obstruction of the bowel lumen and a change in stool caliber. Lesions in the ascending colon more commonly present with chronic, microscopic blood loss and iron deficiency anemia. Colon cancer commonly metastasizes to the liver, and it would be unlikely to cause hydroureteronephrosis. Kidney carcinoma (Choice B) most commonly presents with a unilateral renal mass, flank pain, and hematuria. It does not present with bilateral hydroureteronephrosis as the lesion would be located proximal to the renal collecting system. Ovarian carcinoma (Choice C) classically presents indolently in an older, postmenopausal woman with abdominal distention and weight loss. Physical examination may show a palpable adnexal mass. It is less likely to cause bilateral hydroureteronephrosis than cervical carcinoma. Ureteral carcinoma (Choice D) can present with hematuria and flank pain. If large enough to obstruct the ureteral lumen, it would cause unilateral hydroureteronephrosis, rather than bilateral as in this patient. Educational Objective: When locally advanced, cervical cancer can obstruct bilateral distal ureters near their insertion into the urinary bladder, which causes bilateral hydroureteronephrosis. This presents as postrenal azotemia because of the inability to appropriately drain urine into the bladder, which can cause progressively worsening kidney failure if not treated. Previous Next Score Report Lab Values Calculator Help Pause

162 Exam Section 4: Item 12 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 12. A 30-year-old woman comes to the physician for evaluation of recurrent postprandial retrosternal burning occurring three times weekly during the past 6 months. The symptoms occur only during the day and are noticeably more severe after large meals, especially if she eats fried foods. Initially, the symptoms responded well to over-the-counter antacids, but in the past 2 months, the antacids have failed to control her symptoms. She had similar symptoms 5 years ago during pregnancy that also responded well to antacids and then resolved after the delivery of her child. She has had no difficulty swallowing, weight loss, or gastrointestinal bleeding. She is sleeping well. She takes no other medications and does not smoke cigarettes or drink alcohol. Physical examination shows no abnormalities. Which of the following is the most likely cause of her condition? A) Gastrin-induced inhibition of antral contractility B) Gastrin-induced inhibition of lower esophageal sphincter tone C) Somatostatin-induced stimulation of gastric acid secretion D) Vagal inhibition of esophageal contractility E) Vagal stimulation of lower esophageal sphincter relaxation

E. Gastroesophageal reflux disease (GERD) presents with chest discomfort, often in association with consumption of a large meal or trigger foods such as an acidic beverage, fatty foods, coffee, chocolate, or tomatoes. It is often worse with supine positioning. It occurs because of lower esophageal sphincter relaxation or incompetence, which may be secondary to vagal stimulation. Relaxation permits stomach acid to reflux into the esophagus. As stomach acid refluxes into the esophagus, it causes mucosal irritation and esophagitis, which the patient experiences as chest discomfort. Over time, metaplasia can occur, leading to Barrett esophagus, a state of pre-malignancy marked by intestinal metaplasia of the distal esophagus. Untreated, it can lead to esophageal adenocarcinoma. Other complications of untreated GERD include esophageal strictures. First-line therapy for GERD is an empiric trial of proton pump inhibitors or H, receptor blockers. Nissen fundoplication is a surgical treatment for GERD caused by incompetence of the lower esophageal sphincter and involves plication of the gastric fundus around the lower esophagus. Incorrect Answers: A, B, C, and D. Gastrin is secreted by G cells in the antrum of the stomach and the duodenum and stimulates gastric acid secretion and gastric motility. It does not inhibit gastric antral contractility (Choice A) or inhibit lower esophageal sphincter tone (Choice B). Somatostatin is secreted by D cells in the pancreas and gastrointestinal mucosa and functions to inhibit secretion of various hormones and enzymes that function in digestion. It decreases gastric acid secretion (Choice C). The vagus nerve, via parasympathetic fibers, stimulates contractions and peristalsis in the muscle of the esophagus. It does not inhibit esophageal contractility (Choice D). Educational Objective: GERD presents with chest discomfort, often in association with consumption of a large meal or trigger foods such as an acidic beverage, fatty foods, coffee, chocolate, or tomatoes. Parasympathetic fibers from the vagal nerve promote relaxation of the lower esophageal sphincter and can contribute to reflux. Previous Next Score Report Lab Values Calculator Help Pause

187 Exam Section 4: Item 38 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 38. An 18-year-old woman comes to the physician because of irregular menstrual periods since menarche at the age of 13 years. Menses occur at 60- to 120-day intervals. She takes no medications. She does not smoke cigarettes, drink alcohol, or use illicit drugs. She is sexually active with one partner, and they use condoms consistently. She is 157 cm (5 ft 2 in) tall and weighs 75 kg (165 lb); BMI is 30 kg/m2 Physical examination shows facial hirsutism. Pelvic examination shows no cervical motion tenderness or masses. Which of the following is the most likely diagnosis? A) Adrenal insufficiency B) Congenital adrenal hyperplasia C) Hyperaldosteronism OD) Hypothyroidism E) Polycystic ovarian syndrome

E. Polycystic ovarian syndrome (PCOS) presents with irregular menses, infertility, and hirsutism, often in patients with an increased body weight. It requires two of the following for diagnosis: oligo/anovulation, polycystic ovaries on ultrasound, and clinical or biochemical signs of hyperandrogenism (eg, hirsutism). It is also associated with signs of insulin resistance, such as acanthosis nigricans and increased blood glucose concentration. PCOS is associated with an abnormal hypothalamic hormonal feedback response, which leads to increased luteinizing hormone (LH) in comparison with follicle stimulating hormone (FSH), as well as increased androgen (eg, testosterone) production from the theca lutein cells. Treatment includes weight loss, combined oral contraceptives for menstrual regulation, spironolactone or ketoconazole for hirsutism, and management of diabetes mellitus if present. Clomiphene stimulates increased ovulation and can be utilized for individuals desiring pregnancy. If the patient is diagnosed with diabetes mellitus, oral antihyperglycemic medications or insulin may also be necessary. A potential complication of long- standing PCOS is endometrial carcinoma caused by the prolonged exposure of the endometrium to unopposed estrogen. Incorrect Answers: A, B, C, and D. Adrenal insufficiency (Choice A) presents with hypotension, hypothermia, hypoglycemia, weakness, and variable metabolic and acid-base derangements on the basis of the level of the causative lesion (primary versus secondary). It does not present with oligomenorrhea and hirsutism. Congenital adrenal hyperplasia (Choice B) is a disorder of cortisol synthesis, with variable effects on mineralocorticoids and androgens. Symptoms include virilization, ambiguous genitalia, salt- wasting, hirsutism, precocious puberty, and abnormal uterine bleeding, depending on the subtype. It would be a less likely cause of oligomenorrhea in this patient with otherwise normal development. Hyperaldosteronism (Choice C) presents with hypertension, hypokalemia, and metabolic alkalosis, most commonly secondary to an adrenal adenoma or bilateral adrenal hyperplasia. It does not cause hirsutism or irregular menstrual periods. Hypothyroidism (Choice D) presents with weight gain, fatigue, constipation, and cold intolerance. It may also demonstrate heavy or irregular menses. Physical examination typically shows bradycardia, dry, edematous skin, and delayed relaxation of deep tendon reflexes. This patient does not demonstrate any additional signs to suggest hypothyroidism. Educational Objective: PCOS presents with menstrual irregularities, signs of androgen excess, obesity, insulin resistance, and polycystic ovaries on ultrasonography. Treatment includes weight loss and combined oral contraceptives for menstrual regulation, spironolactone or ketoconazole for hirsutism, and antihyperglycemic medications or insulin for the management of diabetes mellitus, if diagnosed. Previous Next Score Report Lab Values Calculator Help Pause

126 Exam Section 3: Item 26 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 26. A 34-year-old woman has a 3-year history of recurring episodes of tenesmus, mucoid bloody diarrhea, and cramping abdominal pain. A section of bowel removed following a severe episode is shown. Which of the following is the most likely diagnosis? A) Crohn disease B) Diverticulitis C) Pseudomembranous enterocolitis D) Superior mesenteric artery occlusion E) Ulcerative colitis 1cm

E. Subacute or chronic bloody diarrhea, abdominal pain, abdominal tenderness on examination, and a portion of bowel showing diffuse ulceration and inflammation is consistent with a diagnosis of ulcerative colitis. Ulcerative colitis is a chronic inflammatory condition within the spectrum of inflammatory bowel disease that results in inflammation and ulcers of the colonic mucosa, typically beginning with the rectum and advancing proximally. Symptoms include chronic abdominal pain, weight loss, bloody diarrhea, abdominal bloating, and tenesmus. Extraintestinal manifestations (eg, uveitis) also occur. In addition, because of the chronic inflammation of the colon, patients are at increased risk for developing colorectal carcinoma. Severe presentations can involve high fever, tachycardia, hypotension, leukocytosis, enlarged and dilated bowel, and impaired colonic motility, a constellation known as toxic megacolon. Diagnosis of ulcerative colitis is by colonoscopy and biopsy, and treatment includes immunomodulators, biologics, and steroids. In severe cases, a partial or total colectomy may be required, especially in cases of toxic megacolon. Incorrect Answers: A, B, C, and D. Crohn disease (Choice A) is characterized by transmural inflammation of the intestine that can involve any portion of the gastrointestinal tract, manifesting with cobblestone mucosa and skip lesions. It particularly involves the terminal ileum and typically spares the rectum. The section of bowel shown shows no skip lesions or cobblestone mucosa. Diverticulitis (Choice B) presents because of infection or inflammation of colonic diverticula, which are small outpouchings of the bowel wall commonly located in the sigmoid colon. Diverticulitis presents with acute left lower quadrant abdominal pain, sometimes accompanied by diarrhea, fever, and tenderness to palpation. Clostridium difficile infections can present with pseudomembranous enterocolitis (Choice C). On colonoscopy, yellow pseudomembranes are visualized amidst inflamed, ulcerated colonic mucosa. It typically presents with acute, nonbloody, profuse watery diarrhea. Superior mesenteric artery occlusion (Choice D) may describe the pathophysiology of acute mesenteric ischemia, which presents with acute, severe abdominal pain, nausea, vomiting, and bloody diarrhea, often in a patient with atrial fibrillation, splanchnic atherosclerosis, or a hypercoagulable state. Acute mesenteric ischemia is life-threatening because of complications of bowel necrosis including sepsis, peritonitis, and perforation. Educational Objective: Ulcerative colitis is a chronic inflammatory condition within the spectrum of inflammatory bowel disease that results in inflammation and ulcers of the colonic mucosa, typically beginning with the rectum and advancing proximally. Symptoms include chronic abdominal pain, weight loss, bloody diarrhea, abdominal bloating, and tenesmus. Previous Next Score Report Lab Values Calculator Help Pause

46 Exam Section 1: Item 46 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 46. A 28-year-old woman, gravida 1, para 1, comes to the physician because of a 1-week history of pain with sexual intercourse. She had a spontaneous vaginal delivery 5 weeks ago and is currently breast-feeding. Pelvic examination shows vaginal dryness and atrophy. The most likely cause is a decreased concentration of which of the following hormones? A) Estrogen B) Follicle-stimulating hormone C) Luteinizing hormone D) Progesterone E) Prolactin

A. Breastfeeding women are at an increased risk for atrophic vaginitis caused by the antagonistic effects of prolactin on estrogen production through the inhibition of gonadotropin-releasing hormone (GNRH). In non-breastfeeding women, atrophic vaginitis commonly occurs after the onset of menopause. Hypoestrogenemia leads to atrophic vulvovaginal changes and can present with symptoms of irritation, itching, dryness, and dyspareunia. Examination shows atrophy of the labia, decreased secretions, and mucosal erythema and friability. Treatment includes topical estrogen and the use of lubricants during sexual intercourse. Women who also require contraception can be treated with an estrogen-containing oral contraceptive. Incorrect Answers: B, C, D, and E. Both follicle-stimulating hormone (FSH) (Choice B) and luteinizing hormone (LH) (Choice C) deficiencies would present with infertility and amenorrhea caused by anovulation. They are decreased in hypogonadotropic hypogonadism and in the early postpartum state, especially if the patient is consistently breastfeeding. Prolactin inhibits GNRH release from the hypothalamus, which in turn results in the decreased production of FSH and LH from the pituitary. Decreased FSH release does play a role in the patient's decreased estrogen concentrations, but deficiencies of FSH and LH do not directly cause atrophic vaginitis themselves. Progesterone (Choice D) deficiency presents with infertility and irregular menstrual bleeding. While progesterone is increased in pregnancy, its concentration decreases after delivery with the loss of the placenta. It is not associated with vaginal atrophy. Prolactin (Choice E) deficiency would prevent successful breastfeeding as the production of milk in the postpartum state is stimulated by prolactin. Its deficiency does not cause vaginal dryness or atrophy. Educational Objective: Hypoestrogenemia in breastfeeding women is caused by the negative feedback of prolactin on GNRH. This results in an increased risk for the development of atrophic vaginitis. Treatment includes topical estrogen and the use of lubricants during sexual intercourse. %3D Previous Next Score Report Lab Values Calculator Help Pause

92 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment Exam Section 2: Item 42 of 50 42. An 18-year-old man is brought to the emergency department 30 minutes after a motorcycle collision. He has lost a significant amount of blood. His pulse is 120/min, and blood pressure is 90/70 mm Hg. Physical examination shows a severe laceration of the left calf. Rapid infusion of 0.9% saline is started to restore his vascular volume deficit. The solid curves in the graph show the relationship between right atrial pressure and cardiac output (cardiac function curve). The dashed lines show the relationship between cardiac output and right atrial pressure (vascular function curve). Point X is the preinfusion equilibrium point. Which of the following is the most likely new cardiac and vascular equilibrium point following the infusion? E wwww. ww. Central venous pressure Right atrial pressure and cardiac output (Cardiac function curve) -- Cardiac output and right atrial pressure (Vascular function curve) A) B) C) D) E)

A. Hypovolemic shock occurs secondary to a significant decrease in intravascular volume, commonly related to hemorrhage, dehydration, or gastrointestinal volume losses, which results in decreased cardiac preload and therefore, decreased stroke volume and cardiac output. In turn, there is arteriolar vasoconstriction that increases the systemic vascular resistance (SVR) in an attempt to maintain mean arterial pressure; an increase in pulse also occurs to attempt to maintain cardiac output. If progressive and severe hypovolemia ensues, compensation fails, which leads to the development of hypotension and inadequate organ perfusion. The initial step in management is fluid resuscitation to prevent cardiovascular collapse. The cardiac function curve demonstrates the resultant increased cardiac output secondary to increased blood return, as represented by central venous/right atrial pressure (eg, preload) via Frank-Starling mechanics. An infusion of crystalloid fluid will increase preload. In this young patient with an otherwise healthy heart, cardiac output will increase and remain on the same cardiac function curve as the origin point X. The vascular function curve represents an opposing phenomenon, which refers to the decrease in venous return from the body that occurs as central venous/right atrial pressure increases. This is because the return of venous blood is largely determined by the pressure gradient between the systemic capillary beds and the right atrium. As pressure in the right atrium increases, the gradient decreases and less venous blood returns to the right atrium. The intercept of the cardiac function curve and the vascular function curve represents the mean circulatory pressure (ie, the pressure that would exist throughout the whole cardiovascular system if cardiac output dropped to zero). The vascular function curve shifts position on the basis of the blood volume, venous compliance, and systemic arterial pressure. An increase in blood volume shifts the curve to the right because the mean circulatory pressure increases with the increased volume. The overall effect of the crystalloid infusion in this case is best represented by a shift to point A on the graph. Incorrect Answers: B, C, D, and E. Choice B represents a shift of the cardiac function curve to a lower contractility state with a shift of the vascular function curve to a greater mean circulatory pressure. This may occur if a negative inotrope is inappropriately administered at the same time as the crystalloid infusion or if systolic heart failure occurs because of a mismatch of myocardial oxygen supply and demand. Similarly, Choice C represents a shift of the cardiac function curve to a state of lower contractility with no change in the vascular function curve, which may occur if cardiac function is depressed with no change in blood volume or venous compliance. Choices D and E represent a shift of the vascular function curve to a lower mean circulatory pressure, which may occur if blood volume suddenly decreases or a venous vasodilator is administered, rapidly decreasing the pressure gradient between the venous system and the right atrium. Choice D also shows a shift of the cardiac function to a lower contractility state, while Choice E shows no change in cardiac contractility. Educational Objective: Central venous pressure is used as an estimation of right atrial pressure and preload. Increased central venous pressure causes an increase in cardiac output along a cardiac function curve because of greater end-diastolic volume and cardiomyocyte fiber stretch, which generates increased contractility during systole. Conversely, increased central venous pressure leads to decreased venous return of blood caused by a drop in the pressure gradient between the postcapillary venules and the right atrium. This is represented by the vascular function curve. The intersection of the cardiac and vascular function curves represents the mean circulatory pressure. Calculator Help Pause Lab Values Previous Next Score Report --ww Cardiac output

96 Exam Section 2: Item 46 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 46. A 10-year-old boy is brought to the emergency department 30 minutes after sustaining injuries in a skateboard accident. His vital signs are within normal limits. Physical examination shows tenderness to palpation over the right wrist. An x-ray shows a fracture of the right radius and ulna. He is treated with cast immobilization extending over the right elbow. He is no longer able to move his forearm muscles. Which of the following is most likely to occur in the immobilized muscles after 3 weeks in the cast? A) Decreased capillarity B) Decreased intracellular Po, C) Decreased myoglobin oxygen saturation D) Increased metabolic rate E) Increased production of vascular endothelial growth factor F) Increased tissue adenosine concentration G) Increased total blood flow

A. Muscle mass and strength are lost during cast immobilization, prolonged bed rest, and unloading, and as a natural component of aging. The effect of immobilization on muscle is broad and debilitating. Lower limb atrophy in states of immobilization tends to be worse than that experienced in upper limbs; however, skeletal muscle, regardless of location, requires stretch, load bearing, and depolarization to maintain a balance between growth and atrophy. Immobilization places muscle in a passive, metabolically hypoactive state, characterized by loss of all load bearing, stretch, and reflex activity. The result is loss of concentric and eccentric sarcomeric bulk and devascularization, as the inert muscle experiences disuse atrophy. Hypoactive muscle does not maintain the same need for blood supply as active muscle. Loading, reflex, and depolarization result in transcription of MRNA coding for actin and myosin filaments, with sarcomeres added in series and parallel to muscle under frequent and recurrent load. Ās the myocytes enlarge, bear loads, or experience hypoxia, growth factors (eg, vascular endothelial growth factor [VEGF]) promote neovascularization and capillarity. An absence of such signals during states of disuse results in decreased capillarity because atrophic muscle has lower demand for glucose and oxygen compared with metabolically active muscle tissue. Incorrect Answers: B, C, D, E, F, and G. Decreased intracellular Po2 (Choice B) and decreased myoglobin oxygen saturation (Choice C) are not as affected by immobilization since these properties relate to blood supply and the affinity of myoglobin for oxygen. While capillarity may be decreased, myofiber bulk decreases in parallel, pairing decreased supply with decreased demand. Myoglobin oxygen saturation would decrease in states of imbalance between myocyte oxygen demand and blood supply. For example, exercise beyond the anaerobic threshold would result in consumption of myoglobin-bound oxygen, decreased saturation, and conversion from aerobic oxidative phosphorylation to anaerobic glycolysis. Increased metabolic rate (Choice D) and increased production of VEGF (Choice E) are incorrect as these are properties of metabolically active muscle under frequent load, stretch, and depolarization. Immobilized muscle would show a decreased metabolic rate and decreased synthesis of VEGF. Increased tissue adenosine concentration (Choice F) is a property of metabolically active muscle. As adenosine triphosphate molecules are hydrolyzed during the excitation-contraction coupling cycle, adenosine diphosphate, monophosphate, and dephosphorylated adenosine nucleosides result. Adenosine triggers local vasodilation to decrease vascular resistance within local blood supply, a means of autoregulation of blood flow to metabolically active tissues. Its concentration would be decreased with immobilization and lack of muscle use. Increased total blood flow (Choice G) is incorrect as decreased demand for blood results in decreased capillarity and total vascular arbor, while limited production of local mediators of blood flow such as adenosine, lactate, carbon dioxide, and hydrogen ions decrease overall local blood flow by favoring vasoconstriction in the remaining vessels. Educational Objective: Metabolically inactive muscle experiences disuse atrophy through the loss of trophic signals from stretch, reflex, and load bearing. Sarcomeres are catabolized, and vascular arbor diminishes, reflected histologically as decreased capillarity. %3D Previous Next Score Report Lab Values Calculator Help Pause

4 Exam Section 1: Item 4 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 4. Cholestyramine prevents the reabsorption of bile acids from the lumen of the intestine. This decreases serum cholesterol concentrations by which of the following mechanisms? A) Activating lecithin-cholesterol acyltransferase (LCAT) B) Decreasing VLDL production OC) Inhibiting hepatic cholesterol synthesis D) Stimulating HDL production E) Upregulating hepatic LDL receptors

E. Cholestyramine, along with colestipol and colesevelam, are bile acid resins that inhibit the reabsorption of bile acids through the intestinal lumen. Bile acids are synthesized from cholesterol and, when conjugated with glycine or taurine to become water soluble, are the principle component of bile. Bile is made by hepatocytes, collected through the biliary tract, and excreted into the lumen of the duodenum. It aids in the digestion and absorption of lipids and fat-soluble vitamins and provides some antimicrobial activity. Bile is also the body's primary source of cholesterol excretion, though all but 10% of bile acids are typically reabsorbed through the intestinal lumen and recycled by the liver. Bile acid resins, such as cholestyramine, decrease the percentage of bile acids reabsorbed by the intestinal lumen. This lowers the pool of intrahepatic cholesterol and, thus, stimulates the liver to increase expression of LDL receptors in order to capture LDL from the plasma and restore intrahepatic cholesterol concentrations. This decreased blood concentrations of LDL. Through this mechanism, bile acid resins may be used to treat hypercholesterolemia. Incorrect Answers: A, B, C, and D. Activating lecithin-cholesterol acyltransferase (LOCAT) (Choice A) increases the activity of this enzyme, which converts cholesterol to cholesterol esters that can be packaged into lipoproteins. Activation of this enzyme does not lead to decreased serum cholesterol concentrations and is not the mechanism of action of bile acid resins. Decreasing VLDL production (Choice B) is one of the mechanisms of action of niacin (vitamin B3). VLDL is produced in the liver from cholesterol and niacin inhibits this catalytic conversion. Niacin also inhibits hormone-sensitive lipase and thus decreases lipolysis, the process by which triglycerides stored in adipose tissue are degraded. Inhibiting hepatic cholesterol synthesis (Choice C) is the mechanism of action of HMG-COA reductase inhibitors such as simvastatin, atorvastatin, and rosuvastatin. Statins inhibit the enzyme HMG-COA reductase, which catalyzes the rate limiting step in cholesterol biosynthesis, thereby resulting in decreased intrahepatic concentrations of cholesterol. As is the case when bile acid resins are used, decreased intrahepatic cholesterol concentrations lead to increased expression of LDL receptors on the surface of hepatocytes with a consequent decrease of circulating concentrations of LDL. Bile acid resins stimulate HDL production (Choice D) to a small degree. However, this is not the mechanism by which bile acid resins lower the total serum cholesterol. Educational Objective: Bile acid resins decrease serum cholesterol by inhibiting bile acid recycling through the intestinal lumen. The decreased reabsorption of cholesterol results in decreased intrahepatic cholesterol concentrations; this causes the liver to increase expression of the LDL receptors that bind plasma LDL and ultimately lower serum cholesterol. Previous Next Score Report Lab Values Calculator Help Pause

180 Exam Section 4: Item 30 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 30. A 39-year-old woman, gravida 5, para 5, with type 2 diabetes mellitus and a history of gallstones comes to the physician because of a 4-day history of nausea, vomiting, and midabdominal pain radiating to her back. She is 165 cm (5 ft 5 in) tall and weighs 82 kg (180 lb); BMI is 30 kg/m2. Physical examination shows bluish discoloration of both flanks and epigastric tenderness. Which of the following is the most likely diagnosis? OA) Acute cholangitis B) Acute pancreatitis C) Ruptured abdominal aortic aneurysm D) Ruptured ectopic pregnancy E) Ruptured splenic artery aneurysm

B. Acute pancreatitis typically presents with epigastric abdominal pain that radiates to the back, along with nausea and emesis, often in a patient with a history of gallstones, alcoholism, trauma, hypertriglyceridemia, or hypercalcemia. Less common clinical manifestations are Grey-Turner sign (bruising of the flanks) and Cullen sign (bruising around the umbilicus). Acute pancreatitis can be complicated by necrosis, hemorrhage, abscess, or the formation of pseudocysts. Laboratory studies often show increased amylase and lipase activity. Recurrent episodes of acute pancreatitis may lead to chronic pancreatitis and insufficiency of both the exocrine and endocrine pancreas. The exocrine pancreas produces digestive enzymes, the absence of which leads to malnutrition (eg, cachexia, muscle wasting), weight loss, vitamin deficiencies, and steatorrhea (frequent, loose, foul-smelling, sometimes oily stools). The endocrine pancreas produces insulin, glucagon, and somatostatin. The loss of endogenous insulin leads to the development of diabetes mellitus, with symptoms including polyuria, polydipsia, and polyphagia. The diagnosis of chronic pancreatitis and pancreatic insufficiency is through clinical history, imaging, and biopsy. Exocrine pancreatic insufficiency is treated with oral pancreatic enzyme and fat-soluble vitamin replacement, and the development of diabetes mellitus caused by the loss of pancreatic insulin requires insulin replacement. Any modifiable inciting causes of pancreatitis (eg, gallstones, alcohol abuse, hypertriglyceridemia, hypercalcemia) should be mitigated to delay disease progression and preserve remaining native pancreatic function. Incorrect Answers: A, C, D, and E. Acute cholangitis (Choice A) occurs secondary to acute bacterial infection of the biliary tree, typically in biliary instrumentation or choledocholithiasis. It commonly presents with fever, right upper quadrant pain, and jaundice (Charcot triad) with possible shock and altered mental status (Reynold pentad). Ruptured abdominal aortic aneurysm (AAA) (Choice C) may present with a variety of symptoms including abdominal pain, flank pain, and chest pain, often in a patient with a degree of hemodynamic instability. Risk factors for AAA include smoking, being male, atherosclerosis, increasing age, family history, and a connective tissue disorder. This patient demonstrates no risk factors for AAA, and pancreatitis is a more likely diagnosis. Ruptured ectopic pregnancy (Choice D) is an abnormal extrauterine pregnancy (eg, in the fallopian tube, cervix, ovary, or peritoneum), which then ruptures, presenting with vaginal bleeding, abdominopelvic pain, and hemodynamic instability. This patient's presentation is more consistent with acute pancreatitis. Splenic artery aneurysms develop more commonly in women during periods of increased blood flow such as pregnancy. Rupture of a splenic artery aneurysm (Choice E) presents with nausea, vague epigastric or left upper quadrant abdominal pain, splenomegaly, and potentially hemoperitoneum or hemodynamic instability. Educational Objective: Acute pancreatitis typically presents with epigastric abdominal pain that radiates to the back, along with nausea and emesis, often in a patient with a history of gallstones, alcoholism, trauma, hypertriglyceridemia, or hypercalcemia. Grey-Turner sign (bruising of the flanks) and Cullen sign (bruising around the umbilicus) are less common clinical manifestations. Previous Next Score Report Lab Values Calculator Help Pause

135 Exam Section 3: Item 35 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 35. A 67-year-old man comes to the physician because of a 2-month history of pain in his feet. The discomfort is more severe in bed at night and is relieved by taking a hot bath. He has type 2 diabetes mellitus treated with glipizide. His pulse is 60/min, respirations are 12/min, and blood pressure is 130/88 mm Hg. Strength is normal and symmetric in the distal and proximal upper and lower extremities. The Achilles deep tendon reflexes are decreased, and quadriceps deep tendon reflexes are normal. Sensation to pinprick and vibration is decreased from just above the ankles distally. This patient is most likely to describe his pain as which of the following? A) Aching B) Burning C) Colicky D) Cramping E) Sharp F) Squeezing

B. This patient's presenting findings of distal extremity pain, decreased deep tendon reflexes, and diminished sensation to pinprick and vibration are suggestive of diabetic neuropathy. In diabetic patients, chronic hyperglycemia leads to the deposition of advanced glycation end products in capillary endothelium. This leads to microvascular damage to capillaries throughout the body, including the vasa nervorum in the distal extremities. Patients experience neuropathic pain, which is commonly described as burning or similar to a sensation of electricity, or pins and needles. Diabetic neuropathy both impairs the healing response of the distal extremities and reduces the patient's awareness of injuries to the extremities, thereby creating the conditions that underlie the increased incidence of foot ulcerations in diabetic patients. Neuropathic pain is commonly responsive to tricyclic antidepressants or serotonin-norepinephrine reuptake inhibitors. In most cases, it is not appropriate to treat neuropathic pain with opioids. Incorrect Answers: A, C, D, E, and F. All of the other descriptors (Choices A, C, D, E, and F) may be used to describe other forms of chronic pain but are not the terms most commonly used by patients to describe neuropathic pain. A description of pain as burning, electric, or similar to pins and needles, should alert the clinician to the possibility of neuropathic pain. Other causes of neuropathic pain include shingles, nutritional or toxic neuropathy, stroke, multiple sclerosis, malignancy, and trauma. Educational Objective: Patients with diabetes mellitus experience microvascular damage to the vasa nervorum, resulting in neuropathic pain. Neuropathic pain is classically described as burning or similar to a sensation of electricity, or pins and needles. Neuropathic pain is commonly responsive to tricyclic antidepressants or serotonin-norepinephrine reuptake inhibitors. Previous Next Score Report Lab Values Calculator Help Pause

158 Exam Section 4: Item 8 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 8. A 25-year-old man is admitted to the hospital after he sustained second-degree burns over 70% of his body while fighting a fire. Administration of fluid and antibiotics is started. Two days later, he develops dyspnea, tachypnea, and hypoxemia. Diffuse crackles are heard on auscultation of the chest. A chest x-ray shows fluffy alveolar infiltrates. Despite appropriate intervention, he dies. Autopsy shows diffusely firm and hyperemic lungs. Microscopic examination of the lung tissue shows widening of the alveolar septa and eosinophilic hyaline membranes lining the alveolar spaces. Which of the following pathologic mechanisms is the most likely cause of the pulmonary disorder in this patient? A) Bacterial invasion of the distal pulmonary parenchyma B) Diffuse pulmonary atelectasis C) Injury to the alveolar epithelium and endothelium D) Proteolytic destruction of the alveolar walls E) Type I (immediate) hypersensitivity reaction

C. Acute respiratory distress syndrome (ARDS) may occur in response to severe trauma, including burns. ARDS results from cytokine-mediated injury to the pulmonary parenchyma, especially the alveolar epithelium and endothelium, resulting in diffuse alveolar damage with consequent exudative fluid filling the alveolar spaces. On chest x-ray, this is characterized by diffuse, fluffy airspace and reticular interstitial opacities. Biopsy demonstrates widening of the alveolar septae and the deposition of hyaline material. Patients present with severe hypoxia and respiratory distress, and they frequently require mechanical ventilation. The severity of ARDS may be stratified using the Berlin criteria. Incorrect Answers: A, B, D, and E. Bacterial invasion of the distal pulmonary parenchyma (Choice A) occurs in pneumonia. While pneumonia is a common cause of pulmonary opacification, it is more commonly focal or lobar and is less likely than ARDS to result in rapid pulmonary injury and death following a burn injury. Diffuse pulmonary atelectasis (Choice B) may occur because of airway obstruction. Atelectasis refers to distal collapse of alveoli, often caused by hypoventilation or a lack of surfactant. While atelectasis is characterized by alveolar collapse, ARDS is characterized by the filling of alveoli with exudative fluid. Proteolytic destruction of the alveolar walls (Choice D) is observed in emphysema. Young patients with emphysema may possess a deficiency of a-antitrypsin, an inhibitor of elastase. Emphysema does not manifest acutely and would not result from a burn injury or other trauma. Type I (immediate) hypersensitivity reaction (Choice E) is the underlying cause of anaphylaxis, which may result in airway spasticity and diffuse endothelial dysfunction leading to anaphylactic shock. Type I hypersensitivity reactions are, by definition, mediated by the cross-linking of IgE, a mechanism that is not shared by ÅRDS. Educational Objective: Acute respiratory distress syndrome may occur in response to severe trauma, including burns. ARDS results from cytokine-mediated injury to the pulmonary parenchyma, especially the alveolar epithelium and endothelium, causing diffuse alveolar damage with consequent exudative fluid filling the alveolar spaces. II Previous Next Score Report Lab Values Calculator Help Pause

62 Exam Section 2: Item 12 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 12. A previously healthy 72-year-old man is brought to the physician because of weakness and fatigue. His hemoglobin concentration is 9.2 g/dL, leukocyte count is 5400/mm3, and platelet count is 350,000/mm3. A peripheral blood smear is shown. Which of the following is the most likely cause of these findings? A) Aplastic anemia B) Chronic myelogenous leukemia C) Gastrointestinal blood loss D) B-Thalassemia major 88 E) Vitamin B12 (cobalamin) deficiency 00

C. Gastrointestinal blood loss resulting in chronic iron deficiency most likely explains this patient's weakness, fatigue, and blood smear demonstrating hypochromic, microcytic erythrocytes. Iron is required for the synthesis of heme, which is a necessary component of the hemoglobin molecule, and thus, of erythrocytes. It functions to shuttle oxygen to and from peripheral tissues. Erythrocytes on the peripheral blood smear are hypochromic as a result of decreased hemoglobin concentration. It is postulated that erythrocytes are microcytic as a result of continuing erythrocyte division in order to reach an adequate mean corpuscular hemoglobin concentration; because hemoglobin stores are inadequate, cell division continues beyond what would normally occur and causes the cells to be smaller than normal. In patients with unexplained iron deficiency anemia (IDA), colonoscopy is required to rule out colorectal carcinoma. Incorrect Answers: A, B, 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 decreased reticulocyte count but not with hypochromia and microcytosis. Chronic myelogenous leukemia (Choice B) is defined by the presence of the Philadelphia chromosome, a translocation between chromosomes 9 and 22, which causes constitutive activation of the ABL1 kinase. Patients have significant leukocytosis, with concentrations often exceeding 100,000 cells/mm3 Basophilia and eosinophilia are often present. These features are not seen in this blood smear. B-Thalassemia major (Choice D) is caused by homozygous mutations in the B-globin gene. This leads to an absence of B-globin and the development of severe, transfusion-dependent, microcytic anemia. Signs of extramedullary hematopoiesis such as frontal bossing are also present. This condition is usually detected at a young age. Vitamin B 12 (cobalamin) deficiency (Choice E) causes a macrocytic anemia and often presents with hypersegmented neutrophils. The erythrocytes in this peripheral smear are microcytic, not macrocytic. Educational Objective: IDA, which commonly is the result of gastrointestinal blood loss in older individuals, causes a microcytic, hypochromic anemia. In patients with unexplained IDA, colonoscopy is required to rule out colorectal carcinoma. Previous Next Score Report Lab Values Calculator Help Pause 00

170 Exam Section 4: Item 20 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 20. An 18-month-old girl is brought to the physician for a follow-up examination because of macrosomia. She was 53 cm (21 in) in length and weighed 4200 g (9 lb 4 oz) at birth. She is now at the 95th percentile for length and weight. Physical examination shows a nevus flammeus on the forehead, a large tongue, and an abdominal mass. Abdominal ultrasonography shows a tumor in the left kidney. A biopsy specimen of the resected mass shows nephroblastoma (Wilms tumor). Genetic testing of the patient shows abnormal methylation at KCNQ10T1 (DMR2) of maternal 11p15. Which of the following genetic mechanisms is the most likely cause of these findings? A) Anticipation B) Heteroplasmy C) Imprinting D) Non-paternity E) Pleiotropy

C. Imprinting refers to suppression of either the maternal or paternal allele for a particular gene leading to monoallelic expression of that gene. Imprinting may occur by methylation of DNA or modification of histone proteins, and results in under-expression or complete repression of the affected allele. There are over one hundred genes normally imprinted in the human genome, and Beckwith-Wiedemann syndrome (BWS) occurs as a result of inappropriate imprinting of one or more of these genes. There are several imprinted genes in the 11p15 region that, when abnormally methylated, can result in BWS. KCNQ10T1 is a gene that encodes potassium channel voltage-gated KQT-like subfamily member 1. Regulation of this gene is under the control of the IC2 imprinting center that lies within the KCNQ10T1 intron 10. Normally, a single allele is repressed by methylation. The loss of methylation results in expression of the KCNQ10T1 gene from both parents (biallelic) with subsequent development of BWS, which is an overgrowth syndrome that predisposes to the development of cancer. The most commonly seen tumors in BWS are Wilms tumors (as in this patient), neuroblastomas, and rhabdomyosarcomas. Incorrect Answers: A, B, D, and E. Anticipation (Choice A) is a genetic phenomenon whereby each affected generation manifests the symptoms of the disease at an earlier age. It is most common in diseases caused by abnormal trinucleotide repeats such as Huntington disease, and it occurs as a result of expansion of these repeats through successive generations. Symptoms may not only present earlier but also may be more severe. Heteroplasmy (Choice B) refers to the presence of more than one type of mitochondrial DNA (mtDNA) in a cell. Each cell contains many mitochondria, but the differential presence of mitochondria with mutated versus normal mtDNA can determine the severity and clinical manifestations of mitochondrial diseases. Non-paternity (Choice D) refers to a situation in which the patient's presumed father is not their biological father. In this case, the patient will not share any genetic material with the presumed father, so evaluation for genetically transmitted diseases will be inaccurate. Evaluation should include the biological father if known. Pleiotropy (Choice E) refers to genes that influence one or more different and seemingly unrelated traits. One example is phenylketonuria which is caused by mutations that affect phenylalanine hydroxylase, resulting in failure to convert phenylalanine to tyrosine. Neurologic manifestations are a result of accumulated phenylalanine, but tyrosine is also a precursor to melanin, so patients commonly have fair skin and hair. Educational Objective: BWS is caused by abnormal imprinting of one or more different genes in the 11p15 region, resulting in biallelic expression of genes that are normally expressed in a monoallelic fashion. This results in an overgrowth syndrome with hemihyperplasia and predisposition to malignancies, including Wilms tumors. Previous Next Score Report Lab Values Calculator Help Pause

153 Exam Section 4: Item 3 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 3. A 45-year-old man develops multiple punctate, erythematous, nonblanching, 1-mm lesions on the upper and lower extremities and the belt line. A defect in which of the following is the most likely cause of the patient's symptoms? A) Metabolism of protein catabolic byproducts B) Oxidative metabolism of endogenous toxins C) Production of platelets D) Synthesis of coagulation factors E) Synthesis of protein C

C. Production of platelets is most likely impaired in this patient, leading to thrombocytopenia manifesting as diffuse petechiae. Platelets are small cellular fragments that are derived from bone marrow megakaryocytes. They function in primary hemostasis by binding to injured endothelium directly and via interactions with von Willebrand factor. Following activation, they recruit additional platelets through the release of a- and õ-granules. The binding of fibrinogen to Gpllb/llla receptors on the surface of activated platelets functions to recruit and tether other platelets to form a mature platelet plug. This process occurs in conjunction with the coagulation cascade that ends with the formation of a fibrin superstructure. Patients with deficient platelets have a tendency to develop spontaneous mucocutaneous bleeding, which can manifest as bleeding gums, petechiae, or purpura. A severely decreased platelet count can predispose to intracranial bleeding, but this is less common. Thrombocytopenia can develop as a result of myriad pathologic processes including drug reactions, viral infections, conditions associated with primary bone marrow failure, and autoimmune processes. Incorrect Answers: A, B, D, and E. Metabolism of protein catabolic byproducts (Choice A) describes a ubiquitous process of protein recycling. Disorders such as lysosomal storage disorders prevent breakdown of proteins in lysosomes. Disorders such as pancreatic insufficiency prevent degradation of proteins in the gastrointestinal lumen because of the absence of digestive enzymes. Oxidative metabolism of endogenous toxins (Choice B) describes a process that primarily occurs in the liver via the action of cytochrome P450 enzymes. Deficiency would not result in petechiae. Synthesis of coagulation factors (Choice D) and protein C (Choice E) occur in the liver. Deficiency of coagulation factors occurs in conditions such as disseminated intravascular coagulopathy, hemophilia, and acute liver failure, and would increase the risk for bleeding. Protein C is an endogenous anticoagulant and deficiency results in a heightened risk for venous and arterial thrombosis. Neither deficiency would result in petechiae from thrombocytopenia. Educational Objective: Platelets are cellular fragments derived from megakaryocytes and are involved in primary hemostasis. Thrombocytopenia can manifest with mucocutaneous bleeding and petechiae. Previous Next Score Report Lab Values Calculator Help Pause

100 Exam Section 2: Item 50 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 50. A healthy 30-year-old woman is traveling on a commercial airliner where the cabin's barometric pressure is measured at 565 mm Hg. The inspiratory partial pressure of alveolar oxygen in this woman is most likely which of the following (in mm Hg)? A) 27 B) 60 C) 109 D) 515 E) 713

C. The alveolar gas equation can be used to calculate the inspiratory partial pressure of oxygen at a given atmospheric pressure. The equation is P;02 = F;02 x (PATM - PH20) where P;02 is the partial pressure of inspired oxygen, F;O2 is the fraction of inspired oxygen, PATM is the atmospheric pressure, and PH2O is the partial pressure of water vapor. The fraction of inspired oxygen is 21%, or 0.21, and is unchanged at modest altitude, as is the water vapor pressure, which is estimated at 47 mm Hg. Inserted into the equation, P;O2 = 0.21 x (565 mm Hg - 47 mm Hg) = 109 %3D mm Hg. This represents the partial pressure of oxygen that is inspired by the young woman within the pressurized cabin. Incorrect Answers: A, B, D, and E. 27 (Choice A) is not the correct answer and represents the use of erroneous numbers or miscalculations. 60 (Choice C) represents the alveolar partial pressure of oxygen, not the inspiratory partial pressure, which is calculated by adjusting the inspiratory partial pressure of alveolar oxygen to accommodate the partial pressure of alveolar carbon dioxide: PAO2 = F;02 x (PATM - PH20) - (PACO2/respiratory exchange ratio), where PAO2 is the alveolar partial pressure of oxygen, and PACO2 is the arterial pressure of carbon dioxide). The patient is a young, healthy adult, and the expected PACO2 would be 40 mm Hg. 515 (Choice D) represents the difference between the atmospheric pressure in the cabin and the ratio of the arterial partial pressure of carbon dioxide to respiratory exchange ratio. This calculation, 565 mm Hg - (40 mm Hg / 0.8) = 565 mm Hg - 50 mm Hg = 515 mm Hg, neglects the fraction of inspired oxygen and the water vapor pressure. 713 (Choice E) represents the difference between the atmospheric pressure and water vapor pressure at sea level. These values are typically approximated as 760 mm Hg - 47 mm Hg = 713 mm Hg. %3D Educational Objective: The inspiratory partial pressure of oxygen can be calculated for a given atmospheric pressure. This can then be utilized the calculate the alveolar partial pressure of oxygen, which is useful for the assessment of an alveolar-arterial gradient in the assessment of hypoxia. %3D Previous Next Score Report Lab Values Calculator Help Pause

93 Exam Section 2: Item 43 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 43. A 28-year-old man comes to the physician because of a 2-day history of an abscess at the site of a splinter on his right index finger. The leukocytes in this abscess initially localized to the site of inflammation because of an endothelial adhesion molecule that binds ligands on the surface of the leukocyte. Which of the following molecules is the most likely cause of the adhesion to the endothelial surface at the site of inflammation? A) Cadherins B) Extracellular matrix proteins C) G protein D) Proteoglycans E) Selectins

E. Selectin is the most likely molecule implicated in cellular adhesion to the endothelial surface. Following introduction of a pathogen into the skin, as might occur with a foreign body such as a splinter, endothelial cells in the surrounding area upregulate expression of P and E selectins. Additionally, release of cytokines such as tumor necrosis factor-a and interleukin-1 stimulate migration of inflammatory cells to the site of infection. Leukocytes that are attracted to the site of infection express cell surface receptors, such as Sialyl-LewisX that bind to P and E selectins. This is the first step in migration of inflammatory cells across the endothelium and into the area of infection. Incorrect Answers: A, B, C, and D. Cadherins (Choice A) are glycoproteins that extend through cellular membranes and anchor neighboring cells together. They are intimately associated with the cytoplasmic actin cytoskeleton through interaction with a- and B-catenin. They do not play a role attracting and binding leukocytes. Extracellular matrix proteins (Choice B) include collagen, fibronectin, and laminin among others. Together, they provide a superstructure that acts as a scaffolding for cells. They are not implicated in the adhesion of leukocytes to the endothelial surface. G proteins (Choice C) are often associated with extracellular receptors that, upon binding of a ligand, initiate a signaling cascade. They are a highly conserved group of proteins in numerous signaling cascades. Proteoglycans (Choice D) are a component of the extracellular matrix and can interact with integrins on the surface of inflammatory cells. They are important for cellular migration. Educational Objective: E and P selectins are expressed on the surface of endothelial cells in response to local infection. They function to bind specific receptors on the surface of leukocytes, thereby halting their travel through the vasculature. This initiates the cascade that results in migration of inflammatory cells across the endothelium to the site of infection or injury. I3D Previous Next Score Report Lab Values Calculator Help Pause

23 Exam Section 1: Item 23 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment ITIT 23. The table shows survival information for patients who had an operation for a particular form of cancer: Patients at the Beginning of the Interval 1000 800 700 630 Patients Surviving This Interval (%) 80 87.5 Patients Who Died During the Interval 200 100 Interval 0-1 year 1-2 years 2-3 years 3-4 years 70 62 90 90.1 Based on this information, if a patient survives for 3 years, which of the following best represents the probability that he or she will survive for 4 years? A) 0.80 B) 0.8 x 0.875 x 0.9 x 0.901 C) (0.8 + 0.875 + 0.9 + 0.901)/4 D) (0.8 x 0.875 + 0.9 + 0.901)/4 E) 0.901 F) 0.91 - (0.9 + 0.875 + 0.80)/3

E. A survival analysis is used to convey the fraction of a group of patients who are alive at a given time point after an intervention or in the natural history of disease. Graphically, this is often done with a Kaplan-Meier survival curve, the data for which is shown in the table. For this particular cancer, the time interval of 0 to 1 year is accompanied by a 20% death rate, or 80% survival rate. This means that if 100 patients were given the same diagnosis and treated with the same operation, 80 of them would be living at one year. After surviving to 1 year, there is then a likelihood of 87.5% that they would survive to 2 years, and so on. This analysis allows prognostication for a patient newly diagnosed with this cancer when that patient's outcome is yet unknown. However, if a patient is known to have survived for three years, then the survival rate of the whole group for that interval is irrelevant because that patient's survival is known. Therefore, the likelihood that a patient who has survived for 3 years will survive through the 4th year is simply the percentage of patients surviving the 3-4 years interval: 90.1%, or 0.901. Incorrect Answers: A, B, C, D, and F. 0.80 (Choice A) is the likelihood that a patient will survive for 1 year after diagnosis and operation. The patient in this case has been known to survive to three years, so the predicted survival at one year is no longer relevant. 0.8 x 0.875 x 0.9 x 0.901 (Choice B) would be the probability that a patient with a new diagnosis who undergoes treatment would survive until 4 years. It is the combined probability of surviving until 1 year, then 2 years, then 3 years, and finally 4 years. (0.8 + 0.875 + 0.9 + 0.901)/4 (Choice C) is the average of the survival rates of each individual interval. This does not accurately represent the overall four-year survival rate. (0.8 x 0.875 + 0.9 + 0.901)/4 (Choice D) does not generate the survival rate at four years. Neither does it generate the survival rate for the interval from year 3 to year 4. 0.91 - (0.9 + 0.875 + 0.80)/3 (Choice F) does not reflect the probably of survival at four years in either a newly diagnosed patient or in an individual known to survive to three years. Educational Objective: Kaplan-Meier estimates are used to provide prognostic data to patients with new diagnoses. The survival rate is the percentage of patients who survive a given time interval. In the case of a patient who is known to have survived to a certain time point, the prior survival likelihoods are irrelevant. Previous Next Score Report Lab Values Calculator Help Pause

53 Exam Section 2: Item 3 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 3. In a series of 1000 patients with colonic polyposis, three developed colonic malignancies over a 2-year follow-up period. In 1000 control patients without colonic polyposis, one developed a colonic malignancy over the 2-year follow-up period. Which of the following factors most limits the ability to conclude that polyposis is associated with an increased risk for subsequent malignancies? A) The follow-up period is too short B) The investigators are not blinded C) The subjects are not blinded D) There is no control group

A. A cohort study identifies a group of patients with an exposure, compares them with a matched group of patients without the exposure, and follows them over time, seeking to identify whether the exposure is associated with an outcome of interest. In this case, patients with or without polyposis were followed for a 2-year period to assess the incidence of developing colonic malignancy. For any study to be valid, regardless of design, internal and external validity must be considered. Internal validity describes the degree to which a study establishes or supports a faithful conclusion about cause and effect-in other words, whether the design of the study was sufficient to support the conclusion drawn. Sufficient internal validity should limit the degree to which alternative explanations of cause and effect can be reasonably drawn. Internal validity thus refers to whether the exposure, treatment, or variable being tested makes a difference, and whether the experiment was designed in a way to actually support the claim. External validity determines whether an experiment can be generalized and its conclusions applied to groups beyond those considered in the study population. In this example, colonic malignancy is known to have a long latency period, often of 10 or more years. This study observed the cohort for a 2-year period of time, which raises the question of whether the timeframe was long enough to determine whether a difference between those patients with and without polyposis and the risk for developing colonic malignancy actually exists. This insufficient time period risks a Type Il error, stating that no difference exists, when in fact a difference does exist. Incorrect Answers: B, C, and D. The investigators are not blinded (Choice B) and the subjects are not blinded (Choice C) cannot be concluded from the description of the study; however, both statements can be assumed to be true given that the study involves persons with known polyposis, malignancy, or otherwise. Blinding may limit the internal validity of the study; however, this study's conclusions would be easily replicated by external analysis and reexamination of the imaging and histopathology involved. Blinding, therefore, would not likely significantly alter the results. Additionally, the ethics of blinding in this study would be questionable, since known malignancy should be disclosed to the patients at a minimum. There is no control group (Choice D) is incorrect, since the group of patients without polyposis are serving as the control group. Educational Objective: Colonic malignancy has a long latency period, which suggests that any study intended to detect the difference in risk factors and exposures should also have a similarly long evaluation period. An absence of sufficient duration may limit internal validity and the ability to conclude the presence or absence of an association between the exposure and outcome in question. %3D Previous Next Score Report Lab Values Calculator Help Pause

108 Exam Section 3: Item 8 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 8. A 21-year-old man comes to the physician because of intermittent episodes of light brown urine during the past year associated with upper respiratory tract infections. His blood pressure is 145/85 mm Hg. Physical examination shows no other abnormalities. Urinalysis shows microscopic blood, 3+ protein, and several RBC casts. Immunohistochemical staining of a renal biopsy specimen is most likely to show granular deposits in the glomerular mesangium predominantly consisting of which of the following immunoglobulins? A) IgA B) IgD C) IgE D) IgM

A. Episodic hematuria that occurs concurrently with upper respiratory tract infections is consistent with IgA nephropathy or Berger disease. Immunohistochemical staining of a renal biopsy would show immune complex deposits containing IgA in the glomerular mesangium. Hematuria can occur during respiratory or gastrointestinal tract infections since both are mucosal linings that secrete IgA during periods of acute inflammation or infection. When IgA deposition occurs in the renal mesangium, glomerulonephritis may ensue, causing hypertension, microscopic hematuria, RBC casts, and proteinuria. Incorrect Answers: B, C, and D. IgD (Choice B) is an immunoglobulin found on the surface of B lymphocytes and in serum. It is not associated with nephropathy. IgE (Choice C) is an immunoglobulin associated with hypersensitivity reactions. It binds mast cells and basophils and mediates the release of histamine and inflammatory mediators during type I (immediate) hypersensitivity reactions. IgM (Choice D) is an immunoglobulin produced during the initial B lymphocyte response to a new antigen. It can be found on the surface of B lymphocytes as a monomer or as a pentamer when secreted into the serum, allowing for binding of the antigen during the primary response. Educational Objective: IgA nephropathy is characterized by episodic hematuria that occurs concurrently with upper respiratory or gastrointestinal tract infections. Immunohistochemical staining of a renal biopsy typically shows immune complex deposits containing IgA in the glomerular mesangium. %D Previous Next Score Report Lab Values Calculator Help Pause

139 Exam Section 3: Item 39 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 39. An investigator is conducting a randomized, double-blind, placebo-controlled clinical trial of a new medication for the treatment of mild insomnia in adults. A total of 2000 participants are enrolled in the study and randomized to one of two groups: 1000 participants receive the new medication (Group 1), and 1000 participants receive a placebo that appears identical to the medication (Group 2). Participants in both groups are instructed to take one pill 30 minutes before bedtime each day and to keep a sleep diary. After 1 month, each participant is interviewed, and the daily sleep diaries are reviewed. At follow-up, it is determined that 200 participants in Group 1 and 50 participants in Group 2 did not take the pill as directed. In accordance with intention-to-treat analysis, how should the data pertaining to all individuals who did not adhere to the instructions be treated? A) Analyze all nonadherent participants according to the group of the study to which each was randomized B) Exclude all nonadherent participants from analysis C) Exclude only nonadherent participants in Group 1 from analysis D) Exclude only nonadherent participants in Group 2 from analysis E) Perform separate analyses of the 250 nonadherent participants and the 1750 adherent participants

A. Intention-to-treat analysis is used to preserve randomization when attrition or crossover is introduced into the study. According to intention-to-treat analysis, all patients should be analyzed as part of the group to which they were initially randomized. Although randomization is preserved, intention-to-treat analysis leads to problems as it ignores protocol deviations, withdrawal from the study, and noncompliance. In this study, the intention-to-treat analysis ignores the fact that a proportion of each of the placebo and control groups did not adhere to the protocol as prescribed. Because of this, the true effectiveness of the new medication may be diluted. In this example, intention-to-treat analysis would dictate that the 200 participants in Group 1 and the 50 participants in Group 2 should be analyzed in the manner of their original allocation, which preserves initial randomization. Incorrect Answers: B, C, D, and E. Excluding all nonadherent participants from analysis (Choice B) may create significant differences between the populations analyzed in Groups 1 and 2 because of potentially asymmetric exclusion of certain participants. For example, if the 200 patients in Group 1 were meaningfully different in some way from the remainder of the group, and from their respective counterparts in Group 2, bias could be introduced into the study. Exclude only nonadherent participants in Group 1 from analysis (Choice C) and exclude only nonadherent participants in Group 2 from analysis (Choice D) may create significant differences between the analyzed populations. If, for example, the 200 participants in Group 1 were all over the age of 80 years and simply forgot to take the new medication, the remaining participants in Group 1 may have a collectively different age and therefore different set of comorbidities as compared with the control group. Perform separate analyses of the 250 nonadherent participants and the 1750 adherent participants (Choice E) may again introduce bias and significant differences in participant selection between the two groups, limiting potential conclusions about the effectiveness of the new medication. Educational Objective: Intention-to-treat analysis groups patients as initially randomized regardless of the actual occurrence of treatment within the study. This analysis strategy preserves randomization but may dilute the true effects of the intervention. II Previous Next Score Report Lab Values Calculator Help Pause

119 Exam Section 3: Item 19 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 19. A male newborn develops vomiting and convulsions 4 hours after delivery. There is slight bulging of the fontanelles. Which of the following organisms is most likely responsible for these findings? A) Escherichia coli B) Haemophilus influenzae C) Neisseria meningitidis D) Staphylococcus aureus E) Streptococcus pneumoniae

A. Neonatal meningitis is commonly caused by bacteria and presents with fever or hypothermia, lethargy, irritability, a bulging fontanelle, poor feeding, and seizures, as in this patient. Leukocytosis is common on laboratory examination and evaluation of the cerebrospinal fluid will show an increased white blood cell count with predominant neutrophils, increased protein, and decreased glucose. Thorough evaluation also includes a complete blood count, blood culture, and urine culture to evaluate for other causes of sepsis, as well as culture of the cerebrospinal fluid to determine the causative organism. Common causes of neonatal meningitis include Escherichia coli, Streptococcus agalactiae, and Listeria monocytogenes. Risk factors for the development of neonatal meningitis include preterm birth, low birth weight, premature rupture of membranes, and maternal infection. Meningitis occurring shortly after delivery is most commonly caused by Group B streptococci (GBS) or E. coli, the frequency of GBS meningitis has decreased, however, through the routine use of intrapartum antibiotic therapy for mothers that test positive for GBS in the antepartum period. Neonatal meningitis is commonly treated with ampicillin and gentamicin, and complications include lifelong neurologic deficits, such as developmental delay and seizures. Incorrect Answers: B, C, D, and E. Haemophilus influenzae (Choice B) more commonly causes meningitis in children aged six months to six years. The incidence of H. influenzae meningitis has decreased with the widespread use of vaccination in early childhood. Neisseria meningitidis (Choice C) is a common cause of meningitis in children and adults that is spread via respiratory droplets and oral secretions. It is more common in areas of shared housing, such as college dormitories and military barracks. Vaccination against the bacteria is typically administered to patients 11 to 12 years old with a booster dose at age 16 years. Staphylococcus aureus (Choice D) is an uncommon cause of neonatal meningitis. It is more commonly associated with nosocomial infection following neurosurgical intervention, but its risk is also increased in prematurity and immunocompromise. It typically presents more than 72 hours after birth. Streptococcus pneumoniae (Choice E) does not frequently cause meningitis in neonates. It is a common cause of meningitis in children and adults. Educational Objective: Escherichia coli is a common cause of neonatal meningitis. It typically presents shortly after delivery with temperature instability, lethargy or irritability, poor feeding, a bulging fontanelle, and potential seizures. %3D Previous Next Score Report Lab Values Calculator Help Pause

188 Exam Section 4: Item 39 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 39. A 32-year-old woman comes to the physician because of a 1-year history of palpitations and intermittent headaches. She has a 10-year history of poorly controlled hypertension treated with hydrochlorothiazide and metoprolol. Her blood pressure is 180/105 mm Hg. Physical examination shows no other abnormalities. CT scans of the abdomen are shown; the arrows indicate an abnormality. An abnormality of which of the following structures is the most likely cause of these findings? A) Adrenal gland B) Caudate lobe of the liver C) Gallbladder D) Inferior vena cava E) Portal vein F) Right kidney

A. Pheochromocytoma presents with episodic headaches, sweating, palpitations, tremors, agitation, tachycardia, tachypnea, and treatment-refractory hypertension. It is a common primary tumor of the adrenal medulla, and is characterized by the functional secretion of epinephrine, norepinephrine, and dopamine, which cause episodic hyperadrenergic symptoms. Signs and symptoms include hyperthermia, hypertension, tachycardia, tachypnea, diaphoresis, tremor, and headache. Laboratory studies may show increased urine and plasma catecholamines and metanephrines. Imaging studies may show a tumor within the adrenal gland. Treatment includes alpha antagonists such as phenoxybenzamine, an irreversible inhibitor, followed by beta blockade before resection. Failure to administer appropriate antiadrenergic therapy may result in a hypertensive crisis or tachydysrhythmia at the time of resection. Incorrect Answers: B, C, D, E, and F. The caudate lobe of the liver (Choice B) may be the site of hepatocellular carcinoma or adenoma, focal nodular hyperplasia, or metastasis. This patient's symptoms are suggestive of an inappropriate production of catecholamines, which would not occur secondary to a hepatic neoplasm. The gallbladder (Choice C) may present with gallbladder (cholangio-) carcinoma. Like the adrenal glands, it is located beneath the liver. The gallbladder is located anteriorly in comparison to the adrenal glands, which are retroperitoneal. Additionally, gallbladder carcinoma would be unlikely to produce this patient's symptoms of palpitations and headache, which are more consistent with a diagnosis of pheochromocytoma. The inferior vena cava (Choice D) is a retroperitoneal structure that is found to the right of the aorta and vertebrae. The inferior vena cava, however, is an unlikely site for pathology that could produce this patient's symptoms of palpitations, headache, and uncontrolled hypertension. The portal vein (Choice E) is a potential site of thrombosis. Portal vein thrombosis presents with nausea, abdominal pain, and ascites. The right kidney (Choice F) is the potential site of renal cell carcinoma or oncocytoma, which may present with hematuria, flank pain, or a palpable mass. The right kidney may be visualized immediately posterior and inferior to the labeled structure. Educational Objective: Pheochromocytoma is a common primary tumor of the adrenal medulla. Excessive production of catecholamines is characteristic and results in episodic hyperadrenergic signs and symptoms, including hyperthermia, hypertension, tachycardia, tachypnea, diaphoresis, tremor, and headache. %3D Previous Next Score Report Lab Values Calculator Help Pause

163 Exam Section 4: Item 13 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 13. A 17-year-old boy has septic shock that is unresponsive to intravenous ADH (vasopressin) therapy. This treatment is discontinued, and high-dose dopamine therapy is initiated. Stimulation of which of the following receptors is most likely to be of benefit to this patient? OA) a-Adrenoreceptors O B) a-Adrenoreceptors OC) BrAdrenoreceptors D) BzAdrenoreceptors O E) D,Receptors

A. Septic shock is a form of distributive shock, which is characterized by a primary reduction in systemic vascular resistance (SVR). This reduction in SVR is secondary to cytokine-mediated endothelial dysfunction. Cardiac output increases to compensate for decreased SVR in the early stages of distributive shock. If the inciting state is not appropriately treated, further decompensation can occur, resulting in progressive hypotension and insufficient perfusion with associated end-organ damage. The a-adrenergic receptor is a G protein-coupled adrenoreceptor. It is located on vascular smooth muscle, intestinal and bladder sphincter muscles, and the pupillary dilator muscle. It is coupled with a Gq protein, which results in the activation of phospholipase C and leads to increased intracellular calcium and protein kinase C activity with smooth muscle contraction and vasoconstriction. The resultant vasoconstriction results in increased SVR and mean arterial pressure. Therefore, vasopressors that act on the a-adrenergic receptor increase SVR and would be beneficial to patients experiencing septic shock. Incorrect Answers: B, C, D, and E. a-Adrenoreceptors (Choice B) are G protein-coupled receptors localized to the central nervous system that decrease sympathetic outflow when activated. Examples of a-agonists include clonidine and guanfacine, which can treat ADHD and hypertensive urgency. BrAdrenoreceptors (Choice C) are G protein-coupled receptors located primarily on cardiac myocytes that increase cardiac output, heart rate, atrial and ventricular contractility, and atrioventricular node conduction. Septic shock is not characterized by decreased cardiac output; vasopressors acting on Bradrenoreceptors would be most beneficial in cardiogenic shock. B,Adrenoreceptors (Choice D) are G protein-coupled receptors involved in smooth muscle relaxation in blood vessels, muscle, the gastrointestinal tract, and the bronchi. Albuterol, a bronchodilator used in asthma, is one example of a Bzagonist. Bzagonists do not act on skeletal muscle, though tremors can be seen as a consequence of increased sympathetic tone. Intravenous administration of dopamine can be used in cases of shock or refractory hypotension. It may theoretically prevent splanchnic vasoconstriction caused by D, and D Receptor agonism (Choice E); however, this has not been shown in clinical practice. Dopamine does bind a-adrenoreceptors at higher concentration, but its effects are variable and less predictable than a direct a- agonist. Educational Objective: a-Adrenergic receptor activation leads to vascular smooth muscle contraction and vasoconstriction, resulting in increased systemic vascular resistance to maintain mean arterial pressure in septic shock. Previous Next Score Report Lab Values Calculator Help Pause

79 Exam Section 2: Item 29 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 29. A male newborn is found to have an increased serum phenylalanine concentration on routine screening. The increased serum phenylalanine concentration is best explained by a defect in which of the following processes? A) Metabolism of tetrahydrobiopterin B) Metabolism of tetrahydrofolate C) Transport of branched-chain amino acids across the enterocytes D) Transport of phenylalanine across the enterocytes E) Transport of tyrosine across the enterocytes

A. Metabolism of tetrahydrobiopterin (BH) is deficient in a small subset of patients with phenylketonuria (PKU). The majority of patients with PKU have a deficiency in the enzyme phenylalanine hydroxylase (PAH) that converts phenylalanine to tyrosine. Deficiency results in accumulation of phenylalanine and its metabolites phenylacetate and phenyllactate. However, defects in enzymes that either produce or recycle BH, can have similar clinical manifestations since BH, is a required cofactor for the function of PAH. It is also used as a cofactor for the enzymes tyrosine hydroxylase and tryptophan hydroxylase. BH, is synthesized from guanosine triphosphate via the action of three separate enzymes, and deficiency in any of these enzymes can cause symptoms in the spectrum of PKU. However, BH, is also recycled via the action of pterin-4a-carbinolamine dehydratase (PCD) and dihydropteridine reductase (DHPR). Autosomal recessive mutations in these enzymes result in impaired metabolism and recycling of BH, Clinical manifestations include neurotransmitter deficiency in addition to symptoms normally associated with PKU such as microcephaly, skin disease, seizures, and intellectual disability. Treatment is with oral replacement of BH4 neurotransmitter precursors, and folinic acid. Incorrect Answers: B, C, D, and E. Metabolism of tetrahydrofolate (THF) (Choice B) is not the correct answer. THF is synthesized from dihydrofolate (DHF) by the enzyme dihydrofolate reductase. DHF is, in turn, derived from dietary folate. THF is used in pathways that end in the synthesis of nucleic acids. It is not involved in the metabolism of phenylalanine. Transport of branched-chain amino acids (Choice C), phenylalanine (Choice D), and tyrosine (Choice E) across the enterocytes is mediated by sodium-dependent amino acid transporters, with different subtypes used for acidic, basic, and neutral amino acids. These transporter proteins must bind sodium first, which induces a conformational change and allows them to bind specific amino acids. They then transport amino acids into the cytoplasm where they diffuse throughout the cell and are transported across the basolateral membrane of the enterocyte via an alternate set of transporter proteins that do not require sodium. Defects in transport of amino acids do not cause PKU. Educational Objective: Deficiencies in the enzymes required to synthesize or recycle BH4 result in a clinical syndrome that is largely indistinguishable from PKU, as BH, is a necessary cofactor in the reactions catalyzed by PAH. These disorders are inherited in an autosomal recessive fashion and treatment is with oral replacement of BH4 %3D Previous Next Score Report Lab Values Calculator Help Pause

132 Exam Section 3: Item 32 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 32. A 17-year-old boy comes to the physician because of a 3-month history of worsening acne. Treatment over the past year with tetracycline had been effective. His symptoms began when he started taking the medication in the morning with a glass of milk. Which of the following best explains the ineffectiveness of tetracycline in this patient? A) Milk calcium chelates with tetracycline B) Milk carbonate inactivates tetracycline C) Milk lipids bind tetracycline D) Milk peptides induce drug-metabolizing enzymes E) Milk proteins bind tetracycline

A. Tetracycline antibiotics are broad-spectrum protein synthesis inhibitors that act by binding to the 30S ribosomal subunit and inhibiting translation. These medications include tetracycline, doxycycline, and minocycline. They can be used in a variety of infections, such as Lyme disease, Rocky Mountain spotted fever, and infections from methicillin-resistant Staphylococcus aureus. They can also be used to treat acne, as in this patient. Common adverse effects include photosensitivity, esophagitis, and gastrointestinal upset. They should not be used in pregnant patients or in patients under the age of eight years because they have a propensity for causing discoloration of teeth and inhibiting long bone growth. They should not be taken with iron, calcium, or magnesium because they chelate with divalent cations, which prevents proper absorption. Incorrect Answers: B, C, D, and E. Milk carbonate inactivates tetracycline (Choice B) is incorrect, as milk does not contain calcium carbonate. Calcium carbonate is commonly used as an antacid and would diminish the absorption of tetracycline, not inactivate it. Milk lipids bind tetracycline (Choice C) is incorrect, as tetracycline is water soluble and is not bound by lipids. Milk peptides induce drug-metabolizing enzymes (Choice D) is incorrect, as tetracycline is not metabolized but rather excreted in its biologically active form through the bile, urine, and feces. Milk proteins bind tetracycline (Choice E) is incorrect, as tetracycline is water soluble and not bound by proteins prior to absorption. It is primarily bound in the serum. Educational Objective: Tetracycline antibiotics should not be taken with iron, calcium, or magnesium; they chelate with divalent cations, which prevents proper absorption. %3D Previous Next Score Report Lab Values Calculator Help Pause

147 Exam Section 3: Item 47 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 47. A72-year-old woman is admitted to the hospital because of an acute myocardial infarction. She undergoes cardiac catheterization. Angiography shows a left dominant circulation, and a 90% narrowing of the artery supplying the diaphragmatic surface and atrioventricular node of the heart. A balloon angioplasty is scheduled during which a stent will be inserted in the narrowed vessel. The catheter and the balloon must be passed through which of the following vessels (stated in order) to reach the narrowed vessel? A) Left coronary, circumflex, anterior interventricular (left anterior descending) B) Left coronary, circumflex, posterior interventricular (posterior descending) C) Left coronary, posterior interventricular (posterior descending), circumflex D) Right coronary, circumflex, anterior interventricular (left anterior descending) E) Right coronary, circumflex, posterior interventricular (posterior descending) F) Right coronary, posterior interventricular (posterior descending), circumflex

B. Acute myocardial infarction is a medical emergency, and rapid coronary artery catheterization and revascularization of the occluded vessel is essential for preserving myocardial tissue. The inferior walls of the ventricles form the diaphragmatic surface of the heart and are supplied by the posterior descending artery. The atrioventricular (AV) node is supplied by the AV nodal artery, which is most commonly a branch of the right coronary artery but may arise from other coronary arteries as well. The circumflex artery originates from the left main coronary artery and primarily provides perfusion to the lateral and posterior walls of the left ventricle, the anterolateral papillary muscle, and some blood flow to the AV node. In a left dominant circulation (-8% of patients), the circumflex artery gives rise to the posterior descending artery. Thus, to reach the posterior descending artery in a left dominant circulation, the catheter and balloon must first be passed through the left main coronary artery, into the left circumflex artery, and then into the posterior interventricular (posterior descending) artery. Incorrect Answers: A, C, D, E, and F. Choice A is incorrect as the anterior interventricular (left anterior descending) artery and the circumflex artery branch from the left coronary artery at the same juncture. Choice C is incorrect as the posterior interventricular (posterior descending) artery is distal to the left circumflex artery in a left dominant circulation. Choices D, E, and F are incorrect as the posterior descending artery arises from the left circumflex artery in a left dominant circulation. In a right dominant circulation, which is most common (-85%), the correct order of passage would be right coronary to posterior interventricular (posterior descending) artery. A small percentage of patients have a codominant circulation, in which case both the circumflex and the right coronary arteries contribute to the posterior descending artery. Educational Objective: The circumflex artery originates from the left main coronary artery and primarily provides perfusion to the lateral and posterior walls of the left ventricle, the anterolateral papillary muscle, and some blood flow to the AV node. In a left dominant circulation (-8% of patients), the circumflex artery gives rise to the posterior descending artery. II Previous Next Score Report Lab Values Calculator Help Pause

166 Exam Section 4: Item 16 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 16. A67-year-old woman is brought to the emergency department 30 minutes after she fainted while grocery shopping. On arrival, she is conscious. Her blood pressure is 115/85 mm Hg. Physical examination shows carotid pulses with small amplitude and a delayed upstroke. A grade 4/6, crescendo systolic murmur that is heard throughout the precordium is loudest at the upper right sternal border and radiates to the carotid arteries. There is a loud S. A chest x-ray shows no cardiomegaly. This patient most likely has which of the following cardiac valve abnormalities? A) Aortic regurgitation B) Aortic stenosis C) Mitral regurgitation D) Mitral stenosis E) Pulmonic regurgitation F) Pulmonic stenosis G) Tricuspid regurgitation H) Tricuspid stenosis

B. Aortic stenosis (AS) is a common disorder of the aortic valve that results from fibrosis and calcification. These changes of the aortic valve are a pathologic consequence of mechanical stresses on the heart valves, and result from repetitive microtrauma from the opening and closing of valve leaflets with associated chronic inflammation. Many people will develop some degree of AS over time, but structural abnormalities of the valve, such as a bicuspid aortic valve, alter the biomechanics of valve opening and closing and increase the likelihood of earlier fibrosis and calcification with consequent stenosis. While many patients may be asymptomatic, those with severe AS may complain of fatigue, shortness of breath, cough, diminished exercise tolerance, angina, or syncope with exertion. Examination findings include a crescendo-decrescendo systolic murmur best heard at the upper right sternal border with radiation to the carotids, and pulsus parvus et tardus (weak and delayed) may be noted on examination of peripheral pulses. Because of the chronic increased afterload from a fixed obstruction by the valve, left ventricular hypertrophy and resultant diastolic dysfunction can occur. Patients may exhibit an S, heart sound caused by a noncompliant, hypertrophied left ventricle. Incorrect Answers: A, C, D, E, F, G, and H. Aortic regurgitation (Choice A) presents with an early diastolic decrescendo murmur best heard in the right second intercostal space. It is most commonly associated with endocarditis, acute rheumatic fever, and aortic root dilation. Mitral regurgitation (Choice C) presents with a holosystolic murmur best heard in the left fourth or fifth intercostal space along the midclavicular line and radiates to the left axilla. It is commonly associated with mitral valve prolapse, acute rheumatic fever, endocarditis, and prior myocardial infarction. Mitral stenosis (Choice D) is classically heard as an opening snap followed by a diastolic rumble that is loudest over the cardiac apex and radiates to the axilla. If severe enough, it can result in left atrial enlargement, cardiogenic pulmonary edema, and arrhythmias such as atrial fibrillation and flutter. Pulmonic regurgitation (Choice E) presents with a diastolic murmur best heard in the left second intercostal space. This is an uncommon valvular disorder. Pulmonic stenosis (Choice F) presents with a systolic murmur best heard in the second left intercostal region. This murmur is also often crescendo-decrescendo, though it is quieter and radiates less to the lower neck because of the lower pressure in the pulmonary circulation. This is typically seen in tetralogy of Fallot. Otherwise, symptomatic aortic stenosis is more common. Tricuspid regurgitation (Choice G) demonstrates a holosystolic murmur best heard in the left lower sternal border; it may result from severe pulmonary hypertension or infective endocarditis in intravenous drug use. Tricuspid stenosis (Choice H) is an uncommon valvular disorder that presents with a diastolic murmur with an opening snap that is best heard at the lower left sternal border. Educational Objective: Calcification and fibrosis of the aortic valve occur in most people as they age as a result of chronic mechanical stress on the valve leaflets. This can sometimes result in AS, which presents with a systolic crescendo-decrescendo murmur heard best at the right upper sternal border with radiation to the carotid arteries. Symptoms depend on the severity of AS and are graded by echocardiography. Previous Next Score Report Lab Values Calculator Help Pause

194 Exam Section 4: Item 45 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 45. A 65-year-old woman with rheumatoid arthritis comes to the physician for a follow-up examination. In addition to methotrexate therapy, a course of rituximab is initiated. Approximately 45 minutes after the first infusion of rituximab, the patient's temperature is 38.5°C (101.3° F), and she develops diffuse muscle pain. She does not have any difficulty breathing or swallowing, and there are no signs of urticaria or angioedema. Which of the following is the most likely cause of this patient's symptoms? A) Basophil degranulation B) B-lymphocyte release of cytokines C) Natural killer cell activation D) Neutrophil activation E) Platelet aggregation

B. B-lymphocyte release of cytokines most likely accounts for this patient's myalgias and fever, which is consistent with an infusion reaction. Rituximab is a monoclonal antibody directed against CD20, which is present on the surface of B lymphocytes and makes this medication ideally suited for the treatment of B-cell lymphomas and diseases in which B-lymphocyte activity plays a role in pathology (eg. rheumatoid arthritis). Infusion reactions occur in nearly fifty percent of patients who receive rituximab for the first time and are graded from one to four on the basis of the severity of symptoms; higher numbers represent more severe reactions. These reactions are thought to be caused by an initial antibody-antigen interaction between rituximab and CD20. The reaction causes B lymphocytes to release cytokines, contributing to symptoms of myalgias, fevers, and occasionally anaphylactoid symptoms. Severe reactions are more common in patients with high numbers of circulating B lymphocytes, as seen in diseases such as chronic lymphocytic leukemia. Infusion reactions with subsequent infusions are less common, but patients can be premedicated with antihistamines and acetaminophen and the infusion rate of rituximab can be slowed. Incorrect Answers: A, C, D, and E. Basophil degranulation (Choice A) is a component of anaphylaxis and occurs simultaneously with mast cell degranulation. Antigen-bound IgE can bind Fc receptors on the surface of basophils and trigger degranulation with release of substances such as histamine. Infusion reactions are not allergic reactions. Natural killer (NK) cell activation (Choice C) occurs when NK cells recognize cells lacking major histocompatibility complex I molecules on their surfaces. This triggers NK cell activation leading to the release of granzyme and perforin. NK cell activation is not a feature of rituximab infusion reactions. Neutrophil activation (Choice D) occurs in bacterial infections and results in the production of hypochlorite in the respiratory burst. It is not a feature of rituximab infusion reactions. Platelet aggregation (Choice E) is a part of primary hemostasis whereby platelets are trapped by developing fibrin strands deposited at the site of endothelial injury. Clotting is not a feature of rituximab infusion reactions. Educational Objective: Rituximab is a monoclonal antibody against B-lymphocyte CD20. Infusion reactions are common and occur as a result of rituximab interaction with its antigen on the surface of B lymphocytes, leading to cytokine release. Symptoms are variable but are graded on a scale of one to four, with higher numbers indicating more severe reactions. Previous Next Score Report Lab Values Calculator Help Pause

120 Exam Section 3: Item 20 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 20. A 23-year-old woman comes to the physician because of a 2-week history of increasingly severe itching and yellow-tinged skin. Three months ago, she started taking a combination oral contraceptive. She does not smoke. She consumes two to three alcoholic beverages only on the weekends. Physical examination shows mild jaundice and scleral icterus. The liver span is 11 cm with no tenderness. Serum studies show: Bilirubin, total Direct Alkaline phosphatase AST ALT 7.3 mg/dL 5.8 mg/dL 440 U/L 80 U/L 70 U/L 234 U/L (N=0-30) y-Glutamyltransferase (GGT) Serologic studies for acute viral hepatitis are negative. Which of the following is the most likely cause of these findings? A) Autoimmune hemolytic anemia B) Dilated canaliculi with bile plugs C) Hepatic vein thrombosis D) Hepatocellular necrosis E) Portal vein thrombosis

B. Drug-induced liver injury (DILI) can be divided into hepatocellular and cholestatic subtypes. Cholestatic injury is identified by a disproportionate rise in bilirubin and alkaline phosphatase compared to AST and ALT. In this case, the alkaline phosphatase is approximately four times the upper limit of normal compared to the AST and ALT, which are roughly two times the upper limit of normal, suggesting the diagnosis of drug-induced cholestasis. In addition, an increased y-glutamyltransferase (GGT) concentration indicates damage to the bile duct walls. Cholestasis is accompanied by jaundice, visible at a bilirubin concentration of 2 mg/dL or greater, and pruritus. In this patient with acute cholestatic injury secondary to combination oral contraceptives, the primary histologic finding is bile plugging with dilation of the canaliculi proximal to the plugs. In addition to oral contraceptives, antibiotics, particularly amoxicillin-clavulanate, are a common cause of DILI. In order to attribute liver disease to a drug exposure, the drug must have preceded the onset of liver injury, underlying liver disease must be excluded, and stopping the drug must lead to improvement in the liver injury. Incorrect Answers: A, C, D, and E. Autoimmune hemolytic anemia (Choice A) is a type II (complement-mediated cytotoxic) hypersensitivity reaction in which autoantibodies form to antigens on the surface of erythrocytes. Antibody binding to these antigens causes the erythrocytes to be targeted for destruction by complement, leading to the breakdown of erythrocytes, or hemolysis. Hemolysis leads to the release and subsequent degradation of hemoglobin, a byproduct of which is bilirubin. While autoimmune hemolytic anemia would explain the hyperbilirubinemia seen in this case, it would not explain the increased alkaline phosphatase or GGT. Hepatic vein thrombosis (Choice C), or Budd-Chiari syndrome, occurs when there is obstruction of the venous outflow from the liver. Hepatic vein thrombosis typically presents with fever, right upper quadrant pain, distention, and jaundice. This vascular congestion would lead to hepatocellular injury, marked by an increase in AST and ALT rather than bilirubin, alkaline phosphatase, and GGT. Hepatocellular necrosis (Choice D) is seen in the hepatocellular subtype of DILI. In contrast to the cholestatic subtype, the hepatocellular subtype demonstrates a disproportionate increase of AST and ALT compared with that of alkaline phosphatase. Portal vein thrombosis (Choice E) is a potential cause of portal hypertension in the liver. The increased pressure in the portal system causes portosystemic shunts to develop, including esophageal varices, gastric varices, caput medusae, and rectal varices, along with splenomegaly and ascites. While combined oral contraceptives increase the risk for venous thromboembolism, there are no signs of portal hypertension in this case. Educational Objective: Drug-induced liver injury (DILI) may be suspected when a drug preceded the onset of liver injury, underlying liver disease has been excluded, and stopping the drug improves the liver injury. Antibiotics are the most common cause of DILI, though estrogens also are a common cause of this condition. It can be divided into hepatocellular and cholestatic subtypes. Previous Next Score Report Lab Values Calculator Help Pause

191 Exam Section 4: Item 42 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 42. In order to investigate possible treatment options for acute respiratory distress syndrome (ARDS), experimental animals are anesthetized and subjected to smoke inhalation (cotton smoke) according to a well-established experimental protocol. The experimental animals are then given compounds designed to upregulate specific mediators involved in the inflammatory cascade. Upregulation of which of the following mediators would most likely prevent or ameliorate the lung injury induced in this experimental model of ARDS? A) Interleukin-1 (IL-1) B) IL-10 C) Nuclear factor-kappa B D) Toll-like receptors E) Tumor necrosis factor

B. IL-10 upregulation would be most likely to ameliorate the lung injury induced in the described experimental model of ARDS. ARDS is characterized by acute onset, diffuse inflammatory lung injury, and bilateral pulmonary infiltrates on chest imaging leading to respiratory failure. One potential cause of ARDS is inhalational lung injury. Alveolar damage and inflammation lead to increased permeability of the alveolar-capillary interface with subsequent development of a protein-rich exudate. Hypothetically, upregulation of anti-inflammatory cytokines such as IL-10 could attenuate this response. IL-10 expression occurs via the action of nuclear factor-KB (NF-KB) and anti-inflammatory genes within Th, helper T lymphocytes. IL-10 attenuates the immune response by inhibiting activated macrophages, decreasing major histocompatibility complex (MHC) class II antigen expression, and decreasing the expression of Th-associated cytokines. Incorrect Answers: A, C, D, and E. Interleukin-1 (IL-1) (Choice A) is also known as osteoclast activating factor. IL-1 leads to an increase in RANK ligand signaling and subsequent osteoclast-mediated bone resorption along with its role as a pyrogen. It would not likely be implicated in an anti-inflammatory response to ARDS as it is a proinflammatory cytokine. Nuclear factor-kappa B (NF-KB) (Choice C) comprises a family of transcription factors that induce the expression of several proinflammatory genes, promoting the production of cytokines, chemokines, and inflammatory mediators in response to infection or tissue injury. While it is involved in anti-inflammatory IL-10 signaling, its other downstream pathways are primarily proinflammatory. Toll-like receptors (Choice D) 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 inflammatory mediators. Activation results in further inflammation. Tumor necrosis factor (TNF) (Choice E) is a potent activator of macrophages and neutrophils, enhancing their cytotoxic effects and expression of endothelial adhesion molecules to promote migration into peripheral tissues. Upregulation would likely worsen ARDS. Educational Objective: IL-10 is an anti-inflammatory cytokine that is expressed by helper T lymphocytes and acts to attenuate the immune response by inhibiting activated macrophages, decreasing MHC class Il antigen expression, and decreasing the expression of Th-associated cytokines. %3D Previous Next Score Report Lab Values Calculator Help Pause

161 Exam Section 4: Item 11 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 11. A 40-year-old man with alcoholism is admitted to the hospital because of a 2-day history of confusion. Serum studies show a sodium concentration of 99 mEq/L. He is treated with 0.9% saline. Four days later, he develops slurred speech. Physical examination shows mild-to-moderate muscle weakness of all extremities and dysarthria. Sensation is intact. Babinski sign is present bilaterally. These findings are most likely caused by a lesion in which of the following locations? A) Bilateral cerebral hemispheres B) Brain stem C) Medial diencephalon D) Muscle E) Neuromuscular junction OF) Peripheral nerve

B. Overly rapid correction of sodium can lead to central pontine myelinolysis. When serum sodium concentrations decrease, water flows into neurons as osmolar equilibrium occurs between cells and the interstitial space. When hyponatremia is corrected too rapidly, extracellular sodium exerts osmotic pressure that leads to rapid efflux of water from neurons; this cancan lead to demyelination. The pons is particularly vulnerable to demyelinating injury. Demyelination of the pontine corticospinal and corticobulbar tracts can lead to acute, irreversible quadriparesis or quadriplegia, paraparesis or paraplegia, dysarthria, dysphagia, and confusion. Severe damage can result in alertness with the inability to move or communicate (known as locked-in syndrome), coma, or death. Prevention of central pontine myelinolysis is essential and centers around adhering to a slow rate of sodium correction in patients with severe hyponatremia. If too rapid a correction occurs, treatment involves decreasing the serum sodium acutely along with long-term supportive measures such as physical therapy to regain neurologic function. A safe rate of sodium correction is generally no more than 6-12 mEq/day. Incorrect Answers: A, C, D, E, and F. Though lesions of the bilateral cerebral hemispheres (Choice A) could lead to quadriparesis and bulbar muscle weakness, the pons is more vulnerable than the cerebral hemispheres to demyelinating injury. Damage to the medial diencephalon (Choice C) refers to the medial thalamus and hypothalamus. These regions control homeostatic mechanisms such as hunger, thermoregulation, and sexual activity. Lesions of the medial diencephalon would not result in quadriparesis or bulbar dysfunction. Damage to the muscles (Choice D) rarely causes acute quadriparesis. Most degenerative muscle disease (eg, muscular dystrophy) occurs over a chronic time course and is unrelated to the rapid correction of hyponatremia. Neuromuscular junction disease (Choice E) includes myasthenia gravis and Lambert-Eaton syndrome and typically presents with slowly progressive muscle weakness. Further, the rapid correction of hyponatremia does not lead to neuromuscular junction dysfunction. Peripheral nerve diseases (Choice F) such as Guillain-Barre syndrome can present with the acute onset of bulbar and lower extremity motor dysfunction that ascends over the course of days. Rapid correction of hyponatremia is not a precipitating factor for Guillain-Barre syndrome. Educational Objective: Rapid correction of hyponatremia can result in central pontine myelinolysis, which typically involves demyelination of the corticospinal and corticobulbar tracts. This can result in acute, irreversible quadriparesis or quadriplegia, dysarthria, dysphagia, and/or confusion. In severe cases, patients may die or remain alert but without the ability to move or communicate. Previous Next Score Report Lab Values Calculator Help Pause

33 Exam Section 1: Item 33 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 33. The cartilage surrounding the knee supports considerable pressure because of the presence of chondroitin sulfate, which creates a gel-like medium resilient to shock. Which of the following properties of chondroitin sulfate is responsible for this substance occupying a larger volume in solution than in the dehydrated solid? A) It adheres to positively charged molecules of collagen in the extracellular matrix B) It forms a covalent bond with the core protein of the proteoglycan complex C) It has many surface negative charges D) It is a high-molecular-weight polymer of N-acetylgalactosamine-sulfate monomers E) It is a highly branched N-linked oligosaccharide attached to cell-surface proteins

C. Chondroitin sulfate is a highly polar molecule with many negative charges. Because of its polar nature, it has a strong affinity with water. In solution, it is heavily hydrated which increases the overall volume of the complex, in contrast to its closer-packed form when it is a dehydrated solid. The negative charges self-repel, which results in increased space in molecular environments that permit separation of the molecules such as in aqueous solution. Chondroitin sulfate organizes into an extended, linear, rod-like structure, and the absence of rigidity between its bonds when in solution also accounts for its function at the molecular level as a macroscopic shock absorber. It maintains great ability to compress and conform when in solution because of its ability to reorganize as hydrogen bonds, and van der Waals interactions form and deform. Incorrect Answers: A, B, D, and E. It adheres to positively charged molecules of collagen in the extracellular matrix (Choice A) may be a factual statement, but it does not explain the reorientation of chondroitin sulfate molecules when hydrated as compared to when dehydrated. It forms a covalent bond with the core protein of the proteoglycan complex (Choice B) describes the formation of chondroitin sulfate proteoglycans (CSPGS), which are proteoglycans with a chondroitin side chain. They are present in many tissues, including those of the musculoskeletal and central nervous system. In pathology, they are known to inhibit neuronal regeneration following axonal injury. Their existence does not explain the volume-occupying difference of solution-to-solid chondroitin sulfate. It is a high-molecular-weight polymer of N-acetylgalactosamine-sulfate monomers (Choice D), while a true statement, does not explain its affinity with water, bonding characteristics, and molecular organization; it does, however, account for the presence of negatively charged groups that make up the molecular basis for such interactions. It is a highly branched N-linked oligosaccharide attached to cell-surface proteins (Choice E) does not explain the structural characteristics when hydrated in solution. N-glycosylation is an important modification to proteins and functions in immune recognition, cell signaling, and protein trafficking. Educational Objective: Chondroitin sulfate is a highly polar molecule with many negative charges. Because of its polar nature, it has a strong affinity with water. In solution, it is heavily hydrated, which increases the overall volume of the complex, in contrast to its closer-packed form when it is a dehydrated solid. %3D Previous Next Score Report Lab Values Calculator Help Pause

151 Exam Section 4: Item 1 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 1. Cytogenetic analysis of 20 metaphases from the peripheral lymphocytes of a 15-year-old girl with gonadal dysgenesis (Turner syndrome) shows the karyotype 45,X in 11 metaphases and 46,XX in 9 metaphases. Which of the following is the most likely explanation for this result? A) Genetic recombination B) Loss of heterozygosity C) Postfertilization nondisjunction D) Translocation E) X-inactivation

C. Postfertilization nondisjunction explains the presence of cells with both a 45,X and 46,XX karyotype, and occurs as a result of errors during mitosis. Nondisjunction, or the inappropriate separation of chromosomes into daughter cells during mitosis, can result in two separate somatic cell lines: one that possesses the karyotype 46,XX and another that possesses the karyotype 45,X. As the embryo develops, these disparate cell lines will give rise to daughter cells that are distributed widely throughout the body. This is referred to as mosaicism, which can account for the various phenotypic features of Turner syndrome. Patients with Turner syndrome without mosaicism in which the karyotype of all cells is 45,X most commonly have experienced errors during meiosis, which results in daughter cells with an identical genotype distributed throughout the body. Characteristic physical features of this syndrome include short stature, a wide webbed neck, and a broad chest with widely spaced nipples. Patients with Turner syndrome may also present with a bicuspid aortic valve, aortic coarctation, or a fused (horseshoe) kidney. Incorrect Answers: A, B, D, and E. Genetic recombination (Choice A) describes the process in which homologous chromosomes line up during meiosis and exchange alleles of various genes to form a unique chromosome that is then passed on to the fetus. This process accounts for the inheritance of certain dominant and recessive mutations but does not cause Turner syndrome, as the error in Turner syndrome occurs during chromosomal separation, referred to as nondisjunction. Loss of heterozygosity (Choice B) refers to the complete loss of one copy of a gene. This term is most commonly applied to the "two-hit" hypothesis of cancer, whereby the loss of one tumor suppressor gene predisposes to the development of malignancy when a second chromosomal mutation occurs. Translocation (Choice D) refers to the movement of one part of a chromosome onto a separate chromosome. Translocations result in an abnormal chromosome and are often implicated in cancer because of the placement of certain genes next to promoter regions. This results in constitutive activation of oncogenes, but it is not the means by which Turner syndrome develops. X-inactivation (Choice E) is a normal process in female development whereby one copy of the X chromosome is effectively suppressed by dense folding into heterochromatin. While this chromosome may become transcriptionally inactive, it would still be apparent on karyotype analysis. Educational Objective: Turner syndrome can result from nondisjunction during mitosis or meiosis whereby one copy of the X-chromosome is lost. If nondisjunction occurs during postfertilization mitosis, then there will be two distinct populations of daughter cells, one with the karyotype 45,X and one with 46,XX, which is referred to as mosaicism. If nondisjunction occurs during meiosis, all cells will possess the 45,X karyotype. Previous Next Score Report Lab Values Calculator Help Pause

178 Exam Section 4: Item 28 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 28. A 24-year-old woman, gravida 1, para 1, comes to the physician because of nervousness and tremor since delivering a healthy female newborn 6 weeks ago. She also has had a 5-kg (11- Ib) weight loss during this period. She has a history of panic disorder and carpal tunnel syndrome. Her temperature is 37°C (98.6°F), pulse is 100/min, and blood pressure is 140/80 mm Hg. Physical examination shows a firm thyroid gland that is twice the normal size. Serum studies show a free thyroxine (FT) concentration of 2.4 ng/dL (N=0.8-2.4), thyroid-stimulating hormone concentration of less than 0.03 uU/mL, and 24-hour thyroid radioactive iodine uptake of 1% (N=8-25%). Which of the following is the most likely explanation for this patient's symptoms? A) Anaplastic thyroid carcinoma B) Hypothyroidism C) Release of stored thyroid hormone from a thyroid gland infiltrated by lymphocytes D) Release of thyroid hormone from a lymphomatous thyroid gland E) Release of thyroid hormone from a thyroid gland stimulated by antibodies

C. Postpartum thyroiditis most commonly presents within one year of delivery with abnormalities in thyroid function. It can present with hyperthyroidism, hypothyroidism, or transient hyperthyroidism followed by hypothyroidism. It is associated with an increased serum concentration of antithyroid peroxidase antibodies, but the syndrome is caused by lymphocytic infiltration of the thyroid gland that leads to the release of preformed thyroxine (T and triiodothyronine (T3). This release presents clinically as hyperthyroidism with weight loss, heat intolerance, tachycardia, tremor, and palpitations. When the preformed hormone stores are exhausted, the patient typically enters a hypothyroid phase caused by the continued inhibition of thyroid-stimulating hormone (TSH) by the previously released thyroid hormone and underlying thyroid follicle damage and dysfunction. This presents with fatigue, constipation, cold intolerance, and weight gain. Physical examination will show a mildly enlarged but nontender thyroid gland. Laboratory studies will show increased concentrations of T and T3 with a decreased TSH in the hyperthyroid phase. Radioiodine uptake will be minimal in the hyperthyroid phase, as hormone is not being actively synthesized, which drastically decreases the need for uptake of iodine into the gland. Most women recover and become euthyroid. Incorrect Answers: A, B, D, and E. Anaplastic thyroid carcinoma (Choice A) is a type of thyroid cancer that typically occurs in older patients. It exhibits a poor prognosis and has the ability to invade local structures, such as the esophagus and trachea, leading to symptoms of dysphagia and hoarseness. It does not typically present with symptoms of hyperthyroidism. Hypothyroidism (Choice B) presents with cold intolerance, weight gain, dry skin, and fatigue. Primary hypothyroidism is associated with increased concentrations of TSH and decreased concentrations of free T3 while secondary hypothyroidism, related to hypopituitarism, exhibits decreased concentrations of both TSH and T4 Release of thyroid hormone from a lymphomatous thyroid gland (Choice D) is not a likely cause of the hyperthyroidism seen in this patient. A lymphomatous thyroid gland typically presents with a thyroid nodule and occasionally thyroiditis. It is a less common cause than postpartum thyroiditis in a patient who recently delivered a child. Release of thyroid hormone from a thyroid gland stimulated by antibodies (Choice E) is the pathophysiology of Graves disease, in which thyroid-stimulating immunoglobulin binds to TSH receptors in the thyroid, causing it to release thyroxine. This presents with symptoms of hyperthyroidism, such as tachycardia, weight loss, palpitations, and tremor. However, it is associated with markedly increased radioiodine uptake in the thyroid, as compared with the minimal uptake in this patient. Educational Objective: Postpartum thyroiditis presents within one year of delivery with hyperthyroidism, hypothyroidism, or transient hyperthyroidism followed by hypothyroidism. It is characterized histologically by the infiltration of lymphocytes into the thyroid gland. The hyperthyroid phase is secondary to the release of preformed thyroid hormones, which is associated with decreased radioiodine uptake. The hypothyroid phase is caused by persistent negative feedback on thyroid-stimulating hormone along with damage and dysfunction of the thyroid follicles. Most patients recover to a euthyroid state. Previous Next Score Report Lab Values Calculator Help Pause

155 Exam Section 4: Item 5 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 5. A 20-year-old woman comes to the emergency department because of pain in the right lower quadrant of the abdomen and vaginal bleeding for 2 days. Her last menstrual period was 8 weeks ago. Which of the following historic findings further increases the likelihood of an ectopic pregnancy in this patient? A) Cigarette smoking B) Habitual abortion C) HIV seropositivity D) Previous gonorrhea infection E) Previous vaginal delivery

D. Pregnancy is suspected when there is a missed or delayed menstrual period. An ectopic pregnancy often presents with vaginal bleeding and abdominopelvic pain, as in this patient. Ectopic pregnancy is an abnormal pregnancy in which the fertilized ovum implants in the fallopian tube (most common), on the ovary, within the peritoneal cavity, or in any non-endometrial location. Risk factors for the development of ectopic pregnancy include a previous sexually transmitted infection (eg, gonorrhea or chlamydia), especially when it led to pelvic inflammatory disease, along with prior ectopic pregnancy, prior tubal surgery, and the presence of an intrauterine device. Ultrasound may be used to assess for an ectopic pregnancy. Ectopic pregnancy is nonviable and can be managed medically, using methotrexate for small, early pregnancies. Salpingectomy or evacuation via laparoscopy may be required for larger pregnancies, in cases of medical treatment failure, or in any case of complication. Complications of ectopic pregnancy include rupture, bleeding, and hemoperitoneum. Severe cases can result in hemorrhagic shock. Incorrect Answers: A, B, C, and E. Cigarette smoking (Choice A) increases the risk for premature labor and delivery, premature rupture of membranes, uteroplacental insufficiency, abruptio placentae, and intrauterine growth restriction. It does not increase the risk for ectopic pregnancy. Habitual abortion (Choice B), now more commonly called recurrent abortion or miscarriage, is defined as three or more consecutive pregnancy losses before 20 weeks' gestation. It does not increase the risk of ectopic pregnancy. Elective abortion can cause hemorrhage, uterine perforation, retained products of conception with sepsis, or in rare circumstances, maternal death, but it does not increase the likelihood of ectopic pregnancy with subsequent ovum fertilization. HIV seropositivity (Choice C) increases the risk for opportunistic infections such as Toxoplasma gondii, which may cause congenital malformations, and herpes simplex, which can result in neonatal infection. It can also cause congenital HIV infection, but it does not increase the risk for ectopic pregnancy. Previous vaginal delivery (Choice E) does not increase the risk for ectopic pregnancy. Educational Objective: Ectopic pregnancy presents with amenorrhea, abdominopelvic pain, and vaginal spotting or bleeding. Risk factors for its development include previous ectopic pregnancy, prior tubal surgery, and a history of sexually transmitted infections, especially if these infections led to pelvic inflammatory disease. %3D Previous Next Score Report Lab Values Calculator Help Pause

177 Exam Section 4: Item 27 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 27. Amildly obese 42-year-old woman comes to the physician 2 days after the sudden onset of chest pain and intermittent cough while she was doing laundry. Medications include oral contraceptives. She has smoked 12 packs of cigarettes daily for 24 years. She appears somewhat anxious. An x-ray of the chest shows normal findings, and an ECG shows sinus tachycardia. Ventilation-perfusion lung scans show a defect in the anterior inferior segment of the right upper lobe. Collateral arterial flow is provided to this area by which of the following arteries? A) Brachiocephalic B) Internal thoracic C) Left pulmonary D) Right bronchial E) Right coronary

D. Pulmonary thromboembolism is a common disorder caused by occlusion of a pulmonary artery and impaired blood flow to a segment of lung. Risk factors include immobility, obesity, recent surgery, pregnancy, use of oral contraceptives, trauma, fracture of long bones, and inherited thrombophilia disorders. Pulmonary embolism classically presents with acute chest pain, shortness of breath, and hypoxemia. Small pulmonary emboli are often undetectable on chest x-ray, and the ECG may be normal or show sinus tachycardia. Blood flow to the lungs is accomplished via two systems: the pulmonary and the bronchial circulations. The pulmonary circulation is a low-pressure system originating from the pulmonary trunk. The pulmonary capillary bed interacts with the alveoli for gas exchange, and the venous network returns oxygenated blood to the left atrium. The bronchial circulation is a branch of the high-pressure systemic circulation that arises from an intercostal artery on the right side and from the aorta on the left. The bronchial circulation delivers oxygenated blood to the bronchi, the lung parenchyma, and the pleura. The right bronchial artery provides collateral arterial flow to the right upper lobe. Incorrect Answers: A, B, C, and E. The brachiocephalic artery (Choice A) is the first major branch of the aortic arch and provides blood supply to the head, neck, and right upper extremity. It branches into the right common carotid and right subclavian arteries. The bronchial artery on the right side typically arises from an intercostal artery. The internal thoracic artery (Choice B) is a paired artery that courses along the anterior chest wall after branching from the subclavian artery. It provides blood flow to structures of the mediastinum and the sternum. It is commonly utilized for coronary artery bypass grafting. The pulmonary artery (Choice C) is one of the principal branches of the pulmonary trunk that carries deoxygenated blood from the right ventricle to the pulmonary capillary bed of the left lung for gas exchange. It does not supply collateral circulation to the right lung. The right coronary artery (Choice E) originates at the right aortic sinus and provides blood flow to the right ventricle, right atrium, sinoatrial node, and atrioventricular node. Educational Objective: The lungs contain a dual circulatory supply from the bronchial and pulmonary systems. The pulmonary circulation is a low-pressure system involved in gas exchange with the alveoli and oxygenation of blood returning from the systemic venous circulation. The bronchial circulation branches from the systemic arterial system and perfuses the bronchi, lung parenchyma, and visceral pleura. Previous Next Score Report Lab Values Calculator Help Pause

142 Exam Section 3: Item 42 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 42. A61-year-old man is prescribed fluoxetine for major depressive disorder. This drug has its initial effects on neurons arising in which of the following structures? A) Amygdala B) Insular cortex C) Locus caeruleus D) Raphe nuclei E) Substantia nigra

D. The raphe nuclei synthesize and release serotonin, which is modulated by selective serotonin reuptake inhibitors (SSRIS) such as fluoxetine. The raphe nuclei are located within the reticular formation of the brainstem. Neurons in the raphe nuclei convert L-tryptophan to serotonin, which is released onto postsynaptic serotonin receptors throughout the brain and spinal cord. Presynaptic serotonin transporters (SERTS) reuptake serotonin from the synaptic cleft. SSRIS inhibit the presynaptic reuptake of serotonin and thereby increase serotonin concentration in the synaptic cleft, which improves mood. Symptoms of major depressive disorder (MDD) typically include two or more weeks of depressed mood, anhedonia, guilt or worthlessness, difficulty concentrating, suicidal thoughts, and/or neurovegetative symptoms (eg, decreased energy, sleep disturbance, appetite disturbance). SSRIS are first-line medications for major depressive disorder and work best when combined with psychotherapy such as cognitive-behavioral therapy. Incorrect Answers: A, B, C, and E. The amygdala (Choice A), a region in the mesial temporal lobe that mediates fear responses, is modulated by serotonin. However, the neurons in the amygdala with serotonin receptors are postsynaptic, since the amygdala does not synthesize or release serotonin. SSRIS would therefore not affect the amygdala initially. The insular cortex (Choice B) resides within the lateral fissure and integrates sensory information with emotions. Serotonin binding to postsynaptic neurons in the insular cortex may mediate anxiety. However, the inhibition of presynaptic serotonin reuptake would not affect the insular cortex. The locus caeruleus (Choice C) is the primary brain area that synthesizes norepinephrine and is located in the dorsal wall of the pons. The noradrenergic locus caeruleus neurons project widely to the cortex, brainstem, and spinal cord to modulate attention and motivation as part of the reticular activating system. The substantia nigra (Choice E), located in the anterior midbrain, is a primary brain area that synthesizes dopamine. The dopaminergic neurons of the substantia nigra project to the basal ganglia as part of the nigrostriatal pathway and modulate motor function. These neurons degenerate in Parkinson disease. Educational Objective: Selective serotonin reuptake inhibitors prevent the reuptake of serotonin into presynaptic neurons located in the raphe nuclei. Neurons within the raphe nuclei synthesize serotonin from L-tryptophan. Previous Next Score Report Lab Values Calculator Help Pause

45 Exam Section 1: Item 45 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 45. A 27-year-old man comes to the emergency department because of a 3-day history of numbness and a 2-day history of increasing weakness in his arms and legs. He had a "bad cold" 2 weeks ago. He tells the physician that his oral examinations for graduate school are in 1 month and that he broke up with his girlfriend 1 week ago. His temperature is 37.1°C (98.8°F). Neurologic examination shows symmetrical weakness that is greater in his legs than in his arms. Muscle stretch reflexes are absent in all four extremities, and vibration sense and proprioception are diminished in his lower extremities. Which of the following is most likely the greatest contributor to his current symptoms? A) Anterior horn cell degeneration B) Axonal polyneuropathy C) Conversion disorder D) Demyelinating polyneuropathy E) Muscle fiber degeneration F) Neuromuscular junction disorder

D. This patient most likely has the acute inflammatory demyelinating polyradiculopathy (AIDP) subtype of Guillain-Barré syndrome, leading to polyneuropathy. In AIDP, a preceding infection (eg, Campylobacter jejuni) leads to autoimmune cross-reactivity with the myelin of the nerve roots exiting the spinal cord. The ventral nerve roots, containing motor and sympathetic efferents, are most commonly involved, but the dorsal nerve root (controlling sensation) may also be affected. Patients typically present with acute symmetric muscle weakness with depressed or absent deep tendon reflexes reflecting a lower motor neuron pattern of dysfunction; his begins in the lower extremities and may rapidly ascend to involve the upper extremities, bulbar muscles, and/or respiratory muscles. Autonomic dysfunction (eg, blood pressure fluctuations, cardiac irregularities) is common. Sensory deficits and pain can also occur, though these symptoms are typically mild compared with motor dysfunction. The diagnosis is confirmed by increased cerebrospinal fluid protein with normal cell counts on lumbar puncture (also called albuminocytologic dissociation), and these findings are thought to reflect generalized inflammation. Management includes respiratory support and, for severe cases, plasma exchange or intravenous immunoglobulin therapy. Incorrect Answers: A, B, C, E, and F. Anterior horn cell degeneration (Choice A) can be caused by poliovirus infection or spinal muscular atrophy (a congenital disease presenting with progressive anterior horn disease). Patients present with a lower motor neuron pattern of weakness with decreased reflexes, but the weakness would not rapidly ascend, and dysautonomia and sensory symptoms would be atypical. Axonal polyneuropathy (Choice B) is a rare form of Guillain-Barré syndrome that involves the autoimmune attack of the spinal nerve root axons themselves. Patients may present with motor symptoms only or motor and sensory symptoms that are similar to those of AIDP. However, reflexes are more likely to be preserved in axonal polyneuropathy, and autonomic dysfunction is less likely. AIDP and axonal polyneuropathies are distinguished using nerve conduction studies. Conversion disorder (Choice C), also called functional neurologic disorder, involves acute neurologic symptoms such as sensory or motor dysfunction that are not fully explained by objective findings on physical examination or imaging. Decreased reflexes and symmetrical weakness are uncommon physical examination findings in conversion disorder, and this patient's presentation is more suggestive of AIDP. Most disorders of muscle fiber degeneration (Choice E), such as muscular dystrophy, begin in early childhood and affect the proximal before the distal muscles. Acute ascending weakness, autonomic dysfunction, and sensory symptoms would be atypical. Neuromuscular junction disorders (Choice F) include myasthenia gravis and Lambert-Eaton syndrome and typically present with slowly progressive muscle weakness. Acute ascending weakness, autonomic dysfunction, and sensory symptoms would be atypical. Educational Objective: Acute inflammatory demyelinating polyradiculopathy (AIDP) arises from the autoimmune attack of the spinal nerve roots and typically presents with acute, symmetric, ascending muscle weakness with hyporeflexia. Many patients also have autonomic dysfunction and/or sensory symptoms. The diagnosis is confirmed by increased cerebrospinal fluid protein with normal cell counts on lumbar puncture. Previous Next Score Report Lab Values Calculator Help Pause

174 Exam Section 4: Item 24 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 24. A2-month-old male infant has jaundice and pale stools caused by congenital biliary atresia. Deficiency of which of the following is most likely in this patient? A) Folic acid B) Vitamin B, (pyridoxine) C) Vitamin B 12 (cobalamin) D) Vitamin C E) Vitamin E

E. Bile acids are synthesized from cholesterol and, when conjugated with glycine or taurine to become water-soluble, are the principle component of bile. Bile is made by hepatocytes, collected through the biliary tract, and excreted into the lumen of the duodenum. It aids in the digestion and absorption of lipids and fat-soluble vitamins and provides some antimicrobial activity. In congenital biliary atresia, the common bile duct is occluded, and bile is unable to pass into the duodenum. This precludes the use of bile in the digestion of dietary fats and causes fat malabsorption. Vitamins A, D, E, and K are fat-soluble and absorbed along with fat in the small intestine. Without bile to facilitate fat absorption, these vitamins become deficient. Congenital biliary atresia also presents with jaundice caused by hyperbilirubinemia and pale stools caused by the absence of stercobilin, a byproduct of bilirubin in the stool which gives it a brown color. This condition must be managed surgically and may be complicated by liver failure requiring transplantation. Incorrect Answers: A, B, C, and D. The B-complex vitamins, including folic acid (Choice A), vitamin B6 (pyridoxine) (Choice B), and vitamin B12 (cobalamin) (Choice C), as well as vitamin C (Choice D) are water-soluble. They do not require the absorption of fat for their own absorption and thus will not be deficient in the absence of bile. Educational Objective: The absence of bile because of a congenital narrowing or complete obstruction of the common bile duct causes decreased intestinal absorption of fat. Without fat absorption, the absorption of vitamins A, D, E, and K is also inhibited. Deficiency of these vitamins is a possible complication of congenital biliary atresia. %D Previous Next Score Report Lab Values Calculator Help Pause

159 Exam Section 4: Item 9 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 9. A 65-year-old woman with type 1 diabetes mellitus suddenly develops continuous involuntary ballistic movements of her right arm and leg. She is alert, and neurologic examination shows no other abnormalities. The most likely cause of this patient's movement disorder is infarction involving which of the following areas of the brain on the contralateral side? A) Caudate nucleus B) Globus pallidus C) Putamen D) Substantia nigra E) Subthalamus

E. Lesions to the subthalamus lead to contralateral hemiballismus, an involuntary flailing movement. The subthalamus, located near the junction of the diencephalon and midbrain, modulates basal ganglia output and thus mediates the control of voluntary movements. Lesions therefore result in uncontrollable movements. In Parkinson disease, the glutamatergic neurons of the subthalamus may fire in abnormal oscillatory patterns, resulting in tremor and motor symptoms such as rigidity and bradykinesia. Deep brain stimulation inactivates the subthalamic nucleus and alleviates parkinsonian symptoms. Incorrect Answers: A, B, C, and D. The caudate nucleus (Choice A) is a basal ganglia structure located immediately lateral to the lateral ventricles that modulates motor control along with higher order functions such as learning, memory, motivation, and emotion. Lesions of the caudate nucleus lead to disinhibition and agitation. Caudate atrophy is associated with Huntington disease, and decreased caudate activity is associated with memory loss in Parkinson disease. The globus pallidus (Choice B) and putamen (Choice C) together form the lentiform nucleus of the basal ganglia and are located deep within the cerebral hemispheres. The inhibitory neurons contained within both structures coordinate with the subthalamic nucleus to control purposeful movement. Lesions in both locations typically lead to parkinsonism (eg, dystonia and tremor). The substantia nigra (Choice D), located in the anterior midbrain, is the primary brain area that synthesizes dopamine. The dopaminergic neurons of the substantia nigra project to the basal ganglia as part of the nigrostriatal pathway and modulate motor function. Lesions lead to parkinsonism. Educational Objective: The subthalamus modulates basal ganglia output and thus mediates the control of voluntary movements. Lesions to the subthalamus typically lead to hemiballismus. Previous Next Score Report Lab Values Calculator Help Pause

192 Exam Section 4: Item 43 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment Hematoxylin and eosin Trichrome 43. A 58-year-old woman comes to the physician because of a 6-month history of shortness of breath and chronic nonproductive cough. She has a 2-year history of difficulty swallowing, joint stiffness, and diffuse tightening of the skin of the face, neck, shoulders, arms, and fingers. She has had significant sensitivity to cold weather for 20 years, and she says that her hands turn white when they are exposed to the cold. Current medications include a histamine (H,) blocking agent for a long-standing history of esophageal reflux. Biopsy specimens of the skin from 1 year ago are shown in the photomicrographs. Examination of the hands shows cutaneous ulceration, clawlike flexion deformity, and decreased joint mobility. An autoimmune disorder is suspected. Which of the following pulmonary disorders is most likely in this patient? A) Adenocarcinoma B) Bronchiectasis C) Emphysema D) Granulomatous inflammation E) Interstitial fibrosis F) Pneumonia G) Pulmonary embolism H) Squamous cell carcinoma

E. Systemic sclerosis (scleroderma) is an autoimmune disorder characterized by collagen deposition and progressive fibrosis of the skin, soft tissues, and internal organs as well as noninflammatory vasculopathy. There are multiple phenotypes ranging from localized scleroderma to diffuse disease. Multiple organ systems may be affected, with involvement of the renal, pulmonary, gastrointestinal, and cardiovascular systems being common. Interstitial fibrosis of the lungs classically presents with dyspnea on exertion and a nonproductive cough. Patients often present with an increased mean pulmonary artery pressure secondary to chronic hypoxic vasoconstriction and increased pulmonary vascular resistance along with left ventricular dysfunction. Gastrointestinal manifestations include esophageal dysmotility with symptoms of dysphagia and acid reflux. Vascular manifestations include Raynaud phenomenon caused by cutaneous vasospasm. The skin may be taut without wrinkles on physical examination with flexion deformities of the digits. Women are typically affected more than men. Skin biopsy may assist with diagnosis; histologic features include an increased deposition of collagen in the dermis, along with perivascular inflammatory cell infiltration and fibrosis. Incorrect Answers: A, B, C, D, F, G, and H. Adenocarcinoma (Choice A) of the lung is the most common primary lung cancer and the most common among nonsmokers. It typically presents as a chronic consolidation in the periphery of the lung. Squamous cell carcinoma (Choice H) is the second most common type of primary lung cancer and typically presents as a central lesion. Scleroderma is not associated with an increased risk for primary lung cancer. Bronchiectasis (Choice B) is abnormal dilatation of the bronchi that may result from recurrent necrotizing infections and chronic inflammation. Risk factors for bronchiectasis include cystic fibrosis, Kartagener syndrome, and allergic bronchopulmonary aspergillosis. Emphysema (Choice C) refers to the destruction of alveolar walls resulting in abnormal and enlarged air spaces, which manifests as obstructive lung disease. There is noted absence of fibrotic changes. Emphysema is associated with smoking and a-1 antitrypsin deficiency. Granulomatous inflammation (Choice D) in the lung is not associated with scleroderma, but it is associated with other autoimmune disorders such as eosinophilic granulomatosis with polyangiitis (Churg-Strauss syndrome), sarcoidosis, granulomatosis with polyangiitis, microscopic polyangiitis, and rheumatoid arthritis. Granulomas may also form in infection, such as tuberculosis and the endemic mycoses, as well as with environmental exposures to beryllium or talc. Pneumonia (Choice F), as well as other infections, occurs more frequently in patients treated with immunosuppressants for autoimmune disorders; however, a six-month history of shortness of breath and cough is atypical for infectious pneumonia which generally presents acutely. Pulmonary embolism (Choice G) risk is increased in autoimmune disorders such as antiphospholipid antibody syndrome and systemic lupus erythematosus. It is not directly associated with scleroderma and generally presents acutely. Educational Objective: Systemic sclerosis (scleroderma) is an autoimmune disorder characterized by collagen deposition and progressive fibrosis of the skin, soft tissues, and internal organs. It is associated with pulmonary interstitial fibrosis and consequent pulmonary arterial hypertension. II Previous Next Score Report Lab Values Calculator Help Pause

121 Exam Section 3: Item 21 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 21. A 17-year-old boy is brought to the emergency department because of a 2-day history of fever and shortness of breath. He underwent bone marrow transplantation for acute myelogenous leukemia 2 months ago. His temperature is 39°C (102.2°F), and respirations are 32/min. Diffuse crackles are heard over the lung fields on auscultation. A chest x-ray shows interstitial pneumonia. A photomicrograph of a biopsy specimen of the lung tissue is shown. Decreased function of which of the following most likely predisposed this patient to infection? A) Dendritic cells B) Eosinophils C) Mast cells D) Neutrophils E) T lymphocytes

E. T-lymphocyte deficiency as a result of immunosuppressive medications following bone marrow transplantation has predisposed this patient to the development of cytomegalovirus (CMV) pneumonia. Cytomegalovirus (CMV), also known as human herpesvirus-5 (HHV-5), is an opportunistic infection that commonly occurs in immunocompromised patients with solid-organ or allogeneic stem cell transplantation, severe ulcerative colitis, or HIVIAIDS infection. CMV pneumonitis with or without diffuse alveolar damage is one of the many presentations of CMV that can occur in patients taking immunosuppressants following transplant and typically occurs in the first 3 to 6 months. Radiologic imaging findings are nonspecific but may show an interstitial pneumonitis pattern with the potential for infiltrative consolidations representing diffuse alveolar hemorrhage. The classic histologic findings of CMV pneumonia are mononuclear cell infiltrates with infected cells showing prominent basophilic nuclear inclusions with a nuclear halo, 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, B, C, and D. Dendritic cells (Choice A) express antigens via class II major histocompatibility complex molecules that are recognized by CD4+ T lymphocytes. Persistence of these interactions following transplant can lead to graft versus host disease when donor lymphocytes recognize host antigens as foreign. Immunosuppressive regimens following transplant are designed to suppress T lymphocyte function, not dendritic cells. Eosinophils (Choice B) offer a primary defense against infection with helminths and parasites, although patients with abnormal eosinophils do not appear to have clinical symptoms. Overactivation of eosinophils, however, can result in acute and chronic eosinophilic pneumonia and hypereosinophilic syndrome. Mast cell (Choice C) deficiency does not cause any symptoms, although overactivation of mast cells results in mast cell activation syndrome, which is characterized by frequent bouts of urticaria and anaphylaxis. Neutrophil (Choice D) deficiency (ie, neutropenia) occurs frequently after chemotherapy, including ablative therapies used before stem cell transplant. Prolonged neutropenia predisposes primarily to bacterial infections although it increases the risk for fungal and viral infections as well. Educational Objective: CMV pneumonia or pneumonitis is occasionally encountered following stem cell transplant as a result of the prolonged use of immunosuppressant medications that preferentially target T lymphocytes. Characteristic features on lung biopsy include infected cells with prominent basophilic nuclear inclusions. Previous Next Score Report Lab Values Calculator Help Pause

185 Exam Section 4: Item 36 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 1000 800 600 400 200 200 400 600 800 1000 IgM 36. A 2-year-old boy has had recurrent sinopulmonary infections since 8 months of age. Flow cytometric analysis of his peripheral blood lymphoid cells with anti-CD3 and anti-IgM (with blockade of Fc receptor-mediated binding of IgG) is shown. Which of the following is the most likely diagnosis? A) Adenosine deaminase deficiency B) Hyper-IgM syndrome C) IgA deficiency D) Myeloperoxidase deficiency E) X-linked agammaglobulinemia

E. X-linked agammaglobulinemia is the most likely diagnosis suggested by flow cytometry, which shows a large population of CD3+ T lymphocytes and relative absence of IgM, a marker for B lymphocytes that express IgM on their cell surface. X-linked agammaglobulinemia is typically defined by defects in the Bruton tyrosine kinase (BTK) gene. BTK is necessary for all stages of B lymphocyte development and proliferation, and a deficiency of this enzyme results in failed B lymphocyte development. As B lymphocytes are responsible for secreting immunoglobulins once they differentiate into plasma cells, a defect in B lymphocyte maturation leads to a dearth of plasma cells and severely decreased immunoglobulin concentrations of every class, including IgM, IgG, IgE, and IgA. Patients present with recurrent infections caused by impaired humoral immunity and treatment is with intravenous immunoglobulin concentrates. Incorrect Answers: A, B, C, and D. Adenosine deaminase (ADA) deficiency (Choice A) results in the accumulation of a purine derivative called deoxyadenosine, which prevents DNA synthesis in B and T lymphocytes. Patients often exhibit severe combined immunodeficiency and present with failure to thrive and opportunistic infections. These patients would show severely decreased concentrations of CD3+ cells (T lymphocytes) on flow cytometry, and some B lymphocytes would be present. Hyper-IgM syndrome (Choice B) patients have a normal amount of IgM and a paucity or absence of immunoglobulins from different classes, which predisposes to infections. They lack CD40, a T-lymphocyte ligand that interacts with B lymphocytes to induce class switching of immunoglobulins. Patients have normal populations of T and B lymphocytes. IgA deficiency (Choice C) describes a condition in which patients lack IgA but have normal concentrations of other immunoglobulins. IgA deficiency can result in anaphylaxis following transfusion of plasma and blood products containing IgA as patients with this disease possess antibodies against IgA. It may also predispose to infection of mucosal surfaces. Flow cytometry would show normal populations of T and B lymphocytes. Myeloperoxidase deficiency (Choice D) is an inherited immunodeficiency syndrome characterized by the inability to produce hydroxy-halide free radicals within phagolysosomes. Patients often presents with recurrent fungal infections. Patients have normal populations of T and B lymphocytes. Educational Objective: Patients with X-linked agammaglobulinemia have mutations in the BTK gene, which encodes a tyrosine kinase that is essential for all stages of B lymphocyte development and proliferation. Deficiency on flow cytometry is represented by an absence of IgM-expressing B lymphocytes and a normal population of CD3+ T lymphocytes. Previous Next Score Report Lab Values Calculator Help Pause CD3

127 Exam Section 3: Item 27 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 27. A 22-year-old woman is admitted to the hospital because of a 10-day history of polydipsia and polyuria. She says that the urge to urinate often awakens her at night. She has been taking lithium carbonate for 2 years for bipolar disorder; her dosage was increased 6 months ago because of recurrent severe manic episodes. Her vital signs are within normal limits. Physical examination shows no abnormalities. Over the next 24 hours, urine excretion totals 6.5 L. Laboratory studies at this time show a serum sodium concentration of 148 mEq/L, serum osmolality of 315 mOsmol/kg, and urine osmolality of 75 mOsmol/kg. After administration of desmopressin, urine output and osmolality do not change. Which of the following findings in the nephron best describes the tubular osmolality, compared with serum, in this patient? Proximal Tubule Juxtaglomerular Apparatus Medullary Collecting Duct A) Hypertonic hypertonic hypertonic B) Нуpertonic hypertonic hypotonic C) Hypertonic hypotonic hypotonic OD) Isotonic isotonic isotonic E) Isotonic hypotonic hypertonic F) Isotonic hypotonic hypotonic G) Нуpotonic hypertonic hypertonic H) Hypotonic hypotonic hypertonic Нуpotonic hypotonic hypotonic

F. Lithium concentrates within the cells of the renal collecting duct and interferes with the signaling pathway related to ADH (vasopressin), leading to limited aquaporin insertion and resultant nephrogenic diabetes insipidus (NDI). Normally, ADH triggers insertion of aquaporin channels into the collecting duct membrane via a CAMP-mediated pathway. When this process fails (as in NDI) the ability of the kidney to reclaim free water is compromised. Dehydration results with symptoms of polydipsia, a normal compensatory response to decreased serum volume or increased serum osmolarity, and polyuria or incontinence, caused by the large volume of dilute urine produced. Laboratory studies show increased serum osmolarity, hypernatremia, and inappropriately dilute urine. Normally, urine should be concentrated in states of hyperosmolar serum, as the main function of ADH is to reclaim water, thereby maintaining serum osmolarity and sodium balance in addition to plasma volume. Desmopressin, an ADH analog, is used as a means of differentiating nephrogenic from central DI. Central DI occurs because of an absence of ADH production, not a failure of the renal collecting tubule to respond to it. If urine output and serum osmolarity decrease following desmopressin, a central cause of DI (eg, head trauma, hypothalamic damage, pituitary tumor) should be investigated. No change in response to desmopressin indicates that the cause of DI is nephrogenic. These derangements result in isotonic filtrate in the proximal tubule, since this is upstream of the lesion in DI, and hypotonic filtrate in the juxtaglomerular apparatus and medullary collecting duct. Incorrect Answers: A, B, C, D, E, G, H, and I. The proximal tubule is the site for reabsorption of the majority of the electrolytes (HCO;, Na+, Cl; PO, K*), water, uric acid, glucose, and amino acids in the initial filtrate. It is generally isotonic with respect to the serum osmolarity (Choices A, B, C, G, H, and I). The juxtaglomerular apparatus consists of juxtaglomerular cells that secrete renin, the macula densa, and extraglomerular mesangial cells. The macula densa contains specialized cells as part of the distal convoluted tubule (DCT) that are sensitive to the concentration of sodium chloride. The DCT contains hypotonic filtrate, as solute has been reabsorbed by the thick ascending loop of Henle just prior to this segment. In addition, the DCT reabsorbs sodium and chloride, further making the filtrate hypotonic (Choices A, B, D, and G). Typically, the collecting duct is the site of ADH action with concentration of filtrate; however, in NDI this process fails, resulting in hypotonic filtrate in the medullary collecting duct (Choices A, D, E, G, and H). Educational Objective: Nephrogenic diabetes insipidus is a known adverse effect of lithium. Lithium impairs the ability of the collecting tubule to reclaim free water by interfering with the pathway of ADH in the cells. Patients present with polydipsia, polyuria, concentrated serum, dilute urine, and failure to respond to desmopressin. II Previous Next Score Report Lab Values Calculator Help Pause

186 Exam Section 4: Item 37 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 37. A 16-year-old girl comes to the physician because of worsening acne over her forehead for the past month. She began a vegetarian diet 6 months ago; she also has been craving and eating large amounts of chocolate. She currently works outdoors as a flag person for a local road construction company and is required to wear a helmet. Her brother received a chinchilla for a pet 2 months ago. Physical examination shows erythematous papules and pustules on the forehead. Which of the following is the most likely cause of the exacerbation of this patient's condition? A) Allergic reaction to materials used in road construction B) Allergy to the family pet C) Chocolate consumption D) Excessive sun exposure E) Vegetarian diet F) Wearing a helmet

F. The initial lesion of acne vulgaris is a comedone, a hair follicle which has been blocked by keratin debris. The inciting event in comedone formation is hyperproliferation of the epidermis and abnormal keratinization. This is compounded by androgenic stimulation of sebaceous glands associated with the hair follicle, which are together called a pilosebaceous unit, leading to increased sebum production. This increased sebum provides a substrate for bacterial overgrowth of normal skin flora including Staphylococcus epidermidis and Propionibacterium acnes. With accumulation of keratin debris, increased sebum production, and bacterial overgrowth, the comedone becomes inflamed and ruptures, causing a papule or cyst to form. Clinically, areas with increased sebaceous activity, including the face, upper back, and chest, are prone to acne. It is a common disorder that affects the vast majority of adolescents. Acne mechanica encompasses acne vulgaris that is caused or exacerbated by mechanical friction, which can be caused by a helmet, jaw strap, or face mask. Friction caused by these items results in increased abnormal keratinization, which initiates comedone formation. Treatment of acne mechanica includes decreasing exposure to mechanical friction in addition to the mainstays of acne vulgaris treatment, which are aimed at decreasing comedone formation (retinoids), decreasing sebum production (antiandrogens), and mitigating bacterial overgrowth (antibiotics). Incorrect Answers: A, B, C, D, and E. Allergic reaction to materials used in road construction (Choice A) may be a cause of allergic contact dermatitis, not acne. Allergic contact dermatitis presents as an eczematous rash in the distribution of the exposure. Papules and pustules are not typically seen. Allergy to the family pet (Choice B) may cause features of Type I hypersensitivity such as allergic rhinitis, conjunctivitis, or urticaria. It does not contribute to acne and does not present with papules or pustules. Neither chocolate consumption (Choice C) nor a vegetarian diet (Choice E) have been definitively shown to worsen acne. A diet rich in high glycemic index foods and fat free dairy milk are the only proven dietary risk factors for acne vulgaris. Excessive sun exposure (Choice D) may worsen the post-inflammatory hyperpigmentation often left by acne lesions, but it is not a risk factor for the development of acne vulgaris itself. Educational Objective: The primary lesion of acne vulgaris, the comedone, is formed when the epithelial cells of the follicle accumulate and produce excess keratin debris. This clogs the follicle, allowing for the proliferation of bacteria and, eventually, follicle rupture. This process is exacerbated by mechanical irritation and obstruction of the follicles. Previous Next Score Report Lab Values Calculator Help Pause

118 Exam Section 3: Item 18 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 18. Following a stroke, a 68-year-old man has a language problem. His speech is fluent but contains many grammatical errors, word substitutions, and neologisms. He is unable to repeat words after the examiner and is apparently unable to comprehend other verbal requests. Which of the following labeled sites on the photograph of the left hemisphere is most likely to be damaged? E F B. A- A) OB) C) D) E) F) G) H) 1) J)

H. Choice H identifies the left posterior, superior temporal gyrus, an area in the dominant hemisphere that makes up one portion of the Wernicke area, a brain region involved in the understanding of language. In a cerebrovascular accident (CVA), lesions involving this area can result in Wernicke aphasia. Wernicke aphasia is an aphasia of comprehension that features fluent speech which does not make logical sense and that may contain paraphasic errors (ie, errors of word or phrase selection) and neologisms (ie, made-up words). Patients with Wernicke aphasia demonstrate an impaired understanding of written and spoken language and are typically unaware of their deficits. Lesions of the Wernicke area also result in impaired repetition. Classically, CVAS manifest as a neurologic deficit related to loss of function of the affected part of the brain. Incorrect Answers: A, B, C, D, E, F, G, I, and J. Choice A identifies the left posterior, inferior frontal lobe. In the dominant hemisphere (typically the left hemisphere), this brain region represents the Broca area. Lesions can result in Broca aphasia, an aphasia of expression that involves nonfluent, hesitant speech with intact comprehension. Lesions of the Broca area also result in impaired repetition. Choice B identifies the prefrontal cortex. This area is associated with functions including learning, reasoning, problem solving, emotion, behavioral control, memory, self-regulation, and personality. Lesions affecting this area may result in behavioral dysregulation and the development of psychiatric symptoms (eg, poststroke depression). Choice C identifies the left posterolateral frontal lobe, representing the primary motor cortex in the precentral gyrus. This brain region mediates motor function of the right-sided (contralateral) face. Lesions of this area result in weakness, facial droop, and an upper motor neuron pattern of dysfunction (eg, hyperreflexia) in the lower two-thirds of the face. Choice D identifies the left posteromedial frontal lobe, representing the primary motor cortex in the precentral gyrus. This brain area controls the motor function of the right (contralateral) lower extremity. Lesions lead to an upper motor neuron pattern of weakness of the right leg. Choice E identifies the left anteromedial parietal lobe, representing the primary sensory cortex in the postcentral gyrus. This brain area controls sensation of the right (contralateral) leg. Lesions lead to sensory deficits of the right leg. Choice F identifies the left anterolateral parietal lobe, representing the primary sensory cortex in the postcentral gyrus. This brain region mediates sensation in the right (contralateral) face. Choice G identifies the left angular gyrus in the inferior parietal lobe. This brain area mediates multimodal sensory integration and assists in mental spatial orientation, paying attention, and solving problems. Patients with left angular gyrus damage may demonstrate agraphia, acalculia, finger agnosia, and left-right disorientation (known as Gerstmann Syndrome). Choice I identifies the middle temporal gyrus, which assists in semantic memory processing, visual perception, and sensory integration. Lesions in this area have been associated with deficits in visual perception and semantic memory processing. Choice J identifies the superior temporal gyrus, which represents the auditory association area. Lesions of this area may disrupt spoken word recognition. Educational Objective: The posterior, superior temporal gyrus in the dominant hemisphere represents the Wernicke area, which mediates language comprehension. Lesions of the Wernicke area lead to fluent speech that does not make logical sense and includes errors of word and phrase selection. Previous Next Score Report Lab Values Calculator Help Pause

171 Exam Section 4: Item 21 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 21. A randomized, double-blind, clinical trial of a new medication for acute stroke was just completed. The results are summarized as follows: Relative Risk Relative Risk (95% Confidence Interval) Treatment Group Placebo (n=90) Active drug (n=89) Disability 40% (95% Confidence Interval) Mortality 17% 30% 0.67 (0.44-0.89) 13% 0.77 (0.43-1.11) Compared with the placebo group, which of the following sets best summarizes the relative risks in the active drug group? Relative Risks of Disability Relative Risks of Mortality A) Significantly higher significantly higher B) Significantly higher no significant difference C) Significantly higher significantly lower D) No significant difference significantly higher E) No significant difference no significant difference F) No significant difference significantly lower G) Significantly lower significantly higher H) Significantly lower no significant difference 1) Significantly lower significantly lower

H. Relative risk (RR) describes the difference in likelihood of the occurrence of a particular disease outcome between two groups of patients with or without a particular exposure. In this case, the outcomes of disability and mortality are considered between patients with acute stroke receiving or not receiving a new medication. Calculations of relative risk are commonly performed in cohort studies. RR is calculated by dividing the fraction of patients with a positive exposure and who developed disease (a) amongst all patients who were exposed which includes those exposed who did not develop disease (b), (a + b), by the fraction of patients with a negative exposure and who developed disease (c) amongst all patients who were not exposed which includes those who did not develop disease (d), (c + d). RR thus equals [a / (a + b)] / [c / (c + d)]. RR values greater than 1.0 indicate an increased risk for developing disease in association with the exposure, whereas values less than 1.0 indicate a decreased risk for developing disease; an RR equal to 1.0 indicates that the disease outcome and the exposure are not related. The confidence interval (CI) must be considered in the interpretation of the results. The Cl defines the range of values in which repeated measurements are expected to fall with a specified confidence. In this case, the 95% confidence interval defines the range in which repeated analyses of RR would be expected to fall 95% of the time. In the case of application to RR, a confidence interval that does not encompass the value 1.0 maintains statistical significance. The RR of the new medication decreasing the risk for disability is 0.67, with the 95% Cl suggesting that repeated analyses would fall between 0.44 and 0.89 (always less than 1.0); this permits the conclusion that the treatment decreases the risk for disability. By contrast, the RR of the new medication decreasing mortality is 0.77, which is less than 1.0, but statistically, it cannot be concluded that this value would be less than 1.0 more than 95% of the time on repeated measurements because the 95% Cl ranges from 0.43 (less than 1.0) to 1.11 (greater than 1.0). Incorrect Answers: A, B, C, D, E, F, G, and I. Choices A, B, C, D, E, and F are incorrect, since the RR of disability is statistically significantly lower with the new treatment as compared with the placebo. This is shown by an RR less than 1.0 with the Cl range also less than 1.0. Choices A, C, D, F, G, and I are incorrect, since the RR of mortality is not statistically significant, and therefore, no significant difference can be concluded on the basis of the new treatment's contribution to any mortality risk or benefit.v Educational Objective: The Cl defines the range of values in which repeated measurements are expected to fall with a specified confidence. RR computations for which the Cl includes 1.0 are not statistically significant. Previous Next Score Report Lab Values Calculator Help Pause

167 Exam Section 4: Item 17 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 17. A previously healthy 73-year-old man with benign prostatic hyperplasia comes to the physician because of a 2-day history of pain with urination and a 12-hour history of fever and chills. His temperature is 38.3°C (101°F). Physical examination shows moderate tenderness of the left flank. Urinalysis shows 50 to 100 WBC/hpf. Culture of urine grows 100,000 colonies/mL of a gram-positive organism. Which of the following is the most likely causal organism? A) Enterococcus faecalis B) Escherichia coli C) Shigella dysenteriae D) Staphylococcus aureus E) Streptococcus pyogenes (group A)

A. Acute pyelonephritis refers to a urinary tract infection that has spread to the kidneys, which is characterized by flank pain, fever, chills, nausea, vomiting, and costovertebral tenderness on physical examination. Urinary tract infections are much more common in women because of the shorter urethra and favorable regional environment for bacterial growth. In men, benign prostatic hyperplasia can contribute to urinary retention and incomplete bladder emptying, increasing the risk for urinary tract infections. If untreated, bacteria can travel up the ureter to infect the kidney, resulting in pyelonephritis. Urinalysis typically shows leukocyte esterase and nitrites along with many WBCS and WBC casts on microscopy. Treatment includes antibiotics covering common pathogens. Urinary tract infections are most commonly caused by Escherichia coli and gram-negative enteric flora, but some gram-positive organisms may also be the culprit. Enterococcus faecalis is a gram-positive organism found in normal colonic flora and may cause urinary or biliary tract infections. Incorrect Answers: B, C, D, and E. Escherichia coli (Choice B) is a common cause of urinary tract infections such as cystitis and pyelonephritis. However, it is a gram-negative rod, whereas this patient has a urine culture growing a gram-positive organism. Shigella dysenteriae (Choice C) is a gram-negative rod that can cause bloody diarrhea with systemic symptoms such as fever and myalgias. Staphylococcus aureus (Choice D) is a gram-positive organism that is a common cause of cutaneous infections including impetigo, cellulitis, folliculitis, and furunculosis. It can also produce an exotoxin leading to toxic shock syndrome. Its enterotoxins commonly cause nausea, vomiting, diarrhea, and intestinal cramping after ingestion of food contaminated by S. aureus. Streptococcus pyogenes (group A) (Choice E) is a gram-positive bacterium that can result in various diseases, most commonly cellulitis and pharyngitis. Additional conditions related to S. pyogenes include scarlet fever, necrotizing fasciitis, glomerulonephritis, and rheumatic fever. It is not associated with urinary tract infections. Educational Objective: Acute pyelonephritis refers to a urinary tract infection that has spread to the kidneys, which is characterized by flank pain, fever, chills, nausea, vomiting, and costovertebral tenderness on physical examination. Enterococcus faecalis is a gram-positive organism found in normal colonic flora and may cause urinary tract infections. Previous Next Score Report Lab Values Calculator Help Pause

66 Exam Section 2: Item 16 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 16. A 3-month-old girl is brought to the physician by her father because of developmental delay and delayed growth. She is below the 3rd percentile for length and weight. Physical examination shows coarse facial features, generalized hypotonia, bilateral hip dislocation, and inguinal hernias. Laboratory studies show an increased serum activity of lysosomal enzymes. Urine studies show no mucopolysaccharides. This patient most likely has a defect in which of the following biochemical pathways? A) Addition of mannose phosphate to lysosomal enzymes B) Degradation of dermatan sulfate C) Glycosaminoglycan degradation D) Lysosomal acid lipase E) Lysosomal trafficking of exogenous cholesterol

A. Addition of mannose phosphate to lysosomal enzymes is impaired in patients with mucolipidosis II, also known as l-cell (inclusion cell) disease. It results from accumulation of oligosaccharides, lipids, and glycosaminoglycans such as heparan sulfate and dermatan sulfate within cells throughout the body. I-cell disease is a type of lysosomal storage disease. It is similar in phenotype and pathophysiology to Hurler syndrome. I-cell disease results from a failure of Golgi-based post-translational modification of proteolytic enzymes that would typically be targeted to the lysosome. Proteases targeted to the lysosome are tagged with phosphate at the sixth carbon on their mannose residues, forming mannose 6-phosphate. Defective Golgi N-acetylglucosaminyl-1- phosphotransferase enzymes are unable to join phosphate into mannose residues. This causes the synthesized enzymes to be abnormally routed into vesicles for exocytosis instead of the İysosome. With the absence of these proteins in the lysosome, normal cellular debris that requires lysosomal degradation accumulates within the cells, causing inclusions seen on light microscopy. The resulting accumulation of such products leads to widespread cellular and organ dysfunction. Signs and symptoms of l-cell disease include failure to thrive, developmental delay, coarse facial features, restricted skeletal development, hepatosplenomegaly, cardiac structural defects, corneal clouding, and dwarfism. I-cell disease follows an autosomal recessive inheritance pattern; there is no treatment for the condition other than supportive and symptomatic care. Incorrect Answers: B, C, D, and E. Degradation of dermatan sulfate (Choice B) is dysfunctional in mucopolysaccharidosis |I (Hunter syndrome). Dermatan sulfate is a type of glycosaminoglycan (mucopolysaccharide) that must be degraded in the lysosome by the enzyme iduronate sulfatase. Mutations encoding for this enzyme result in accumulation of dermatan sulfate. Within the first year of life, patients develop coarse facial features, hepatosplenomegaly, hernias, and dysostosis multiplex. These patients would have dermatan sulfate present in the urine, which this patient lacks. Glycosaminoglycan degradation (Choice C) defects occur in numerous diseases. Glycosaminoglycans are ubiquitous molecules and must be constantly created and recycled. The degradation of glycosaminoglycans requires many enzymes, and mutations in any of them can cause disease. Collectively, these diseases are referred to as mucopolysaccharidoses, otherwise known as lysosomal storage disorders. Lysosomal acid lipase (Choice D) deficiency causes accumulation of lipids throughout the body but primarily in the liver. Early symptoms include hepatosplenomegaly, failure to thrive, steatorrhea, and malabsorption in addition to fatty liver followed by the early development of cirrhosis. In its most severe form, patients do not live beyond their first year. Lysosomal trafficking of exogenous cholesterol (Choice E) requires the presence of several enzymes including Niemann-Pick C1 and 2 (NPC1/2) and lysosomal-associated membrane protein (LAMP-2). Deficiencies in NPC1 or 2 cause Niemann-Pick disease, which has variable manifestations depending upon the severity. These may include hepatomegaly, vertical supranuclear gaze palsy, epilepsy, and psychiatric conditions. Educational Objective: I-cell disease is an autosomal recessive inherited lysosomal storage disease that results from a defect in N-acetylglucosaminyl-1-phosphotransferase enzymes. This results in the failure of mannose residue phosphorylation of lysosomal hydrolases, which subsequently leads to their abnormal expulsion from the cell instead of to their normal site of action within the lysosome. Previous Next Score Report Lab Values Calculator Help Pause

99 Exam Section 2: Item 49 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 100.0 0.0 LA 100.0 2 0.0 100.0 3 0.0 100.0 4 0.0 100.0 5 0.0 100.0 6 0.0 100.0 7 0.0 100.0 0.0 98.17 97.21 111.29 118.33 Time 49. A healthy 22-year-old man participates in a study that analyzes the effects of various therapies on gastrointestinal motility. A manometric tracing of this man is shown. This manometric pattern is most likely due to binding of which of the following neurotransmitters with its respective receptor in the stomach? A) Acetylcholine B) Motilin C) Norepinephrine D) Serotonin E) Vasoactive intestinal polypeptide

B. Gastrointestinal manometry involves a pressure transducer and is used to measure pressure and contractility within different parts of the gastrointestinal tract. The manometry tracing of this patient shows increased contractility in the stomach followed by mild and sustained increases in contractility in the small intestine over time. Motilin is a neurotransmitter that acts on G protein- coupled receptors and increases gastrointestinal motility by triggering the migratory motor complex. It stimulates contraction of the gastric antrum and fundus to accelerate gastric emptying and stimulates peristalsis in the small bowel. Its effects are more pronounced in the upper gastrointestinal tract smooth muscle, marked by the increased pressure in the stomach and proximal small bowel. Incorrect Answers: A, C, D, and E. Acetylcholine (Choice A) is released by parasympathetic fibers in the vagus nerve and acts on muscarinic receptors that stimulate parietal cell secretion of gastric acid in the stomach. Norepinephrine (Choice C) typically acts via the sympathetic nervous system in times of stress, and generally causes a decrease in digestive activity, including decreases in gastrointestinal motility and secretion of digestive enzymes. It acts on alpha receptors on splanchnic blood vessels decreasing blood flow to the gastrointestinal tract. It would lead to decreased peristalsis and contractility on manometry. Serotonin (Choice D) can have a variety of complex actions in the gastrointestinal tract including stimulation of motility, secretion from the epithelial lining, and vasodilation. In contrast, motilin has a predictable effect in increasing gastric contraction and motility, especially in the upper gastrointestinal tract, as demonstrated by its effect in this manometry tracing. Vasoactive intestinal polypeptide (VIP) (Choice E) is released by parasympathetic ganglia and leads to increased secretion of water and electrolytes by the intestines as well as relaxation of smooth muscle fibers in the gastrointestinal tract. Patients with excess VIP, such as in a VIPoma, can present with chronic and profound watery diarrhea, electrolyte disturbances (eg, hypokalemia, hypercalcemia), achlorhydria, alkalosis, flushing, and vasodilation. Educational Objective: Motilin is a neurotransmitter that acts on G protein-coupled receptors and increases gastrointestinal motility. It stimulates contraction of the gastric antrum and fundus to accelerate gastric emptying as well as peristalsis in the small bowel. Previous Next Score Report Lab Values Calculator Help Pause Location of electrode Stomach Small bowel Pressure

1 Exam Section 1: Item 1 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 1. Patients with prolonged starvation or untreated type 1 diabetes mellitus overproduce ketone bodies. Which of the following is a common factor that is responsible for ketosis in patients with these two conditions? A) Depletion of pentose phosphate pathway intermediates B) Increased availability of acetyl COA C) Inhibition of fatty acid oxidation D) Inhibition of gluconeogenesis E) Inhibition of glycogenolysis

B. Increased availability of acetyl CoA is common to states of starvation and untreated type 1 diabetes mellitus. In starvation, glycogen stores are gradually depleted. The body then relies on the breakdown of fat through oxidation of fatty acids to provide energy. Fatty acid oxidation occurs in the mitochondria of the cell. In each cycle of oxidation, two-carbon fragments are cleaved to form acetyl CoA, 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 produced in each cycle. Ketogenesis occurs in starvation via two-carbon fragments, with formation of ketoacids such as acetoacetyl CoA and B-hydroxybutyrate. The ability to generate energy from adipose during a period of starvation is critical to maintaining the function of vital organs, such as the brain. In untreated type 1 diabetes mellitus, there is an absolute deficiency of insulin, which renders many cells incapable of shuttling glucose into the cell for use in cellular respiration. While there are several glucose transporters that do not require insulin for glucose entry, GLUT4 is commonly expressed in many cell types and requires insulin to function. Patients with untreated type 1 diabetes mellitus will have serum glucose concentrations in the high-normal to high range as a result of ongoing gluconeogenesis and glycogenolysis, but insulin deficiency prevents entry of circulating glucose into cells through glucose transport proteins, effectively resulting in a state of starvation. This results in ketogenesis and accounts for the presence of ketone bodies in the blood and urine of patients with starvation or untreated type 1 diabetes mellitus. Incorrect Answers: A, C, D, and E. Depletion of pentose phosphate pathway intermediates (Choice A) is not correct. This pathway is active in many tissues including the liver, adrenal cortex, and in erythrocytes. It produces NADPH, ribose-5-phosphate, and erythrose-4-phosphate from glucose. Decreased activity of this pathway would not lead to generation of ketones. Inhibition of fatty acid oxidation (Choice C) is not correct. Fatty acid oxidation is increased during states of starvation or untreated type 1 diabetes mellitus. Inhibition of gluconeogenesis (Choice D) and inhibition of glycogenolysis (Choice E) are not features of starvation or untreated type 1 diabetes mellitus. In fact, these processes are accelerated in both states, although in starvation, this is caused by a deficiency of glucose. In untreated type 1 diabetes mellitus, these pathways are upregulated as a result of insulin deficiency, perceived by the cell as a hypoglycemic state. Educational Objective: Both starvation and untreated type 1 diabetes mellitus result in an increased rate of fatty acid oxidation, which generates acetyl CoA to be used in the citric acid cycle and NADH. Concomitant ketogenesis results in the formation of ketone bodies that can be detected in serum and urine. %3D Next Score Report Lab Values Calculator Help Pause

37 Exam Section 1: Item 37 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 37. A 66-year-old woman comes to the physician because of a 3-week history of difficulty getting up from a seated position. Her temperature is 37°C (98.6°F), pulse is 70/min, and blood pressure is 145/95 mm Hg. Physical examination shows global weakness of the proximal muscles of the extremities. Sensation and reflexes are intact. Her fasting serum glucose concentration is 150 mg/dL. A chest x-ray shows a large mass in the hilum of the left lung. Examination of an endobronchial biopsy specimen shows small cell carcinoma of the lung. Increased secretion of which of the following hormones from the lung carcinoma is the most likely cause of the findings in this patient? A) ADH (vasopressin) B) Adrenocorticotropic hormone C) Calcitonin D) Parathyroid hormone E) Serotonin

B. Small cell carcinoma of the lung is a neoplasm of neuroendocrine cells and is associated with numerous paraneoplastic syndromes, including Cushing syndrome caused by adrenocorticotropic hormone (ACTH) production, syndrome of inappropriate antidiuretic hormone (SIADH), Lambert-Eaton myasthenic syndrome caused by presynaptic calcium channel antibody production, paraneoplastic myelitis, encephalitis, and subacute cerebellar degeneration. ACTH produced by small cell carcinoma stimulates the adrenal gland zona fasciculata to increase the production of cortisol. Hypercortisolism, or Cushing syndrome, ensues with its many manifestations, including proximal muscle weakness, osteoporosis, hypertension, hyperglycemia, truncal obesity, moon facies, striae, amenorrhea, and/or immunosuppression. Increased cortisol with a concomitant increase in ACTH suggests either ectopic tumor production of ACTH or Cushing disease (the formation of an ACTH-secreting pituitary adenoma) as the source of hypercortisolism. The two can be distinguished from each other using a high-dose dexamethasone suppression test. If the ACTH is being produced by a structure within the typical endocrine pathway, then it will likely be inhibited by negative feedback, just at a high dose. If ACTH production is coming from an ectopic site, such as a small cell carcinoma, no amount of corticosteroid will cause negative feedback on ACTH production. Thus, cortisol continues to increase under the constant stimulation of ACTH production. Incorrect Answers: A, C, D, and E. ADH (vasopressin) (Choice A) may be produced inappropriately by small cell carcinoma of the lung in paraneoplastic SIADH. SIADH does not cause signs or symptoms of hypercortisolism. Rather, hyponatremia may develop along with its manifestations, such as altered mental status or even coma if severe enough. Calcitonin (Choice C) is secreted by medullary thyroid carcinoma. This tumor is derived from the parafollicular C cells of the thyroid and opposes parathyroid hormone. It functions by mitigating the effects of PTH on the resorption of calcium from bone. Medullary thyroid carcinoma and/or increased calcitonin concentrations would not cause hypercortisolism or proximal muscle weakness. Parathyroid hormone (Choice D) related peptide is produced as a paraneoplastic syndrome by squamous cell carcinoma of the lung, not small cell carcinoma. This causes hypercalcemia, which would present with diffuse, mild muscle weakness, bone pain, nephrolithiasis, fatigue, and confusion. Serotonin (Choice E) is secreted by carcinoid tumors. If the tumor is contained within the intestines, the serotonin is removed by first-pass metabolism within the liver. If no metastasis occurs, then symptoms may not be noticed. In contrast, when it has metastasized to the liver, the serotonin bypasses first-pass metabolism and enters the systemic circulation. This classically presents with diarrhea, cutaneous flushing, and wheezing. Proximal muscle weakness is not a typical finding of carcinoid tumors. Educational Objective: Small cell carcinoma of the lung is a neoplasm of neuroendocrine cells and is associated with numerous paraneoplastic syndromes, including Cushing syndrome caused by ACTH production. Cushing syndrome has many manifestations, including proximal muscle weakness, osteoporosis, hypertension, hyperglycemia, truncal obesity, moon facies, striae, amenorrhea, and/or immunosuppression. Previous Next Score Report Lab Values Calculator Help Pause

5 Exam Section 1: Item 5 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 5. A70-year-old woman comes to the physician for a follow-up examination. She had breast cancer 5 years ago and underwent partial mastectomy of the left breast at that time. She has received treatment with tamoxifen since then. Physical examination and mammography show no evidence of recurrence. Which of the following mechanisms most likely explains the beneficial effect of tamoxifen in treating this patient? A) Competitive inhibition of estradiol activation of cyclin D1 and E2 proteins B) Competitive inhibition of estradiol binding to its receptor C) Competitive inhibition of estradiol synthase in ovaries and adrenal cortex D) Downregulation of bcl-2 E) Downregulation of estrogen receptors F) Increased estradiol catabolism by CYP3A G) Increased renal excretion of synthesized estradiol

B. Tamoxifen is a selective estrogen receptor modulator that is used in the treatment of breast cancer. It acts as an estrogen receptor antagonist in the breast but as an estrogen receptor agonist in the endometrium and bone. Because of its unopposed estrogen stimulation of the endometrium, it increases the risk for endometrial hyperplasia with the potential for atypia and consequent endometrial cancer. Other risks of tamoxifen use include deep venous thrombosis and pulmonary embolism as a result of the hypercoagulability caused by alterations to estrogen receptors. It decreases the risk for recurrent invasive breast cancer in bilateral breasts by approximately 30% to 50%, although it results in little difference in overall mortality. Incorrect Answers: A, C, D, E, F, and G. Competitive inhibition of estradiol activation of cyclin D1 and E2 proteins (Choice A) would decrease cell proliferation and the risk for recurrence of invasive breast carcinoma. Both cyclins are overexpressed or deregulated in tamoxifen-resistant breast cells. This is not the mechanism of tamoxifen. Competitive inhibition of estradiol synthase in ovaries and adrenal cortex (Choice C) is the mechanism of action of aromatase inhibitors, such as anastrozole, which are also used in the management of estrogen receptor-positive breast cancer. Downregulation of bcl-2 (Choice D) would decrease the risk for invasive carcinoma, since bcl-2 inhibits pro-apoptotic proteins, promoting cellular survival. However, this is not the mechanism of action of tamoxifen therapy. Downregulation of estrogen receptors (Choice E) would decrease the responsiveness of breast cancer cells to estradiol in the serum. This is not a mechanism currently employed in the hormonal treatment of breast cancer. Increased estradiol catabolism by CYP3A (Choice F) and increased renal excretion of synthesized estradiol (Choice G) would decrease the available estradiol to bind to estrogen receptor- positive breast cancer cells and would theoretically decrease the risk for recurrent breast carcinoma. They are not, however, mechanisms currently utilized by medications used to treat breast carcinoma. Educational Objective: Tamoxifen is a selective estrogen receptor modulator that acts as an antagonist at breast tissue and an agonist at the endometrium and bone. Risks of its use include the development of endometrial hyperplasia, endometrial cancer, and venous thromboembolism. Previous Next Score Report Lab Values Calculator Help Pause

68 Exam Section 2: Item 18 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 18. A 32-year-old man comes to the physician for a health maintenance examination. His maternal uncles and grandfather had hypertension, renal calculi, enlargement of both kidneys, and decreased renal function. His maternal grandfather died of a ruptured intracranial berry aneurysm. Physical examination shows no abnormalities. Ultrasonography shows five to seven cysts in both kidneys. This patient most likely has a mutation in the gene that encodes which of the following proteins? A) Fibrocystin B) Nephrocystin C) Polycystin D) Sodium chloride cotransporter E) Sodium-potassium-2 chloride transporter F) Voltage-gated chloride channel G) Wilms tumor 1 (WT1)

C. Autosomal dominant polycystic kidney disease is caused by genetic mutations in polycystin (PKD1 or PKD2) that result in the development of multiple cysts in the kidney because of structural abnormalities of the renal tubules. These large cysts generally form in adulthood in the autosomal dominant variant, and eventually, the cysts compress adjacent normal renal parenchyma. Patients can present with flank pain, hematuria, hypertension, and progressive chronic kidney disease. The disease is associated with aneurysmal vascular disease (eg, berry aneurysm), mitral valve prolapse, and hepatic cysts. Patients may experience complications of chronic hypertension or chronic kidney disease. Incorrect Answers: A, B, D, E, F, and G. A mutation in fibrocystin (Choice A) results in autosomal recessive polycystic kidney disease. Cysts generally form in infancy or childhood in the autosomal recessive variant. Mutations in nephrocystin (Choice B) result in familial juvenile nephronophthisis. Patients with this disease demonstrate early-onset end-stage kidney disease and can manifest with symptoms including polyuria, polydipsia, anemia, and growth retardation. In contrast to polycystic kidney disease, familial juvenile nephronophthisis is characterized by growth retardation, the absence of hypertension, and manifestations of the disease typically before the age of 20 years. Defects in the sodium chloride cotransporter (Choice D) result in Gitelman syndrome, a renal tubular defect characterized by decreased serum potassium and magnesium, metabolic alkalosis, and hypocalciuria. Defects in the sodium-potassium-2 chloride transporter (Choice E) result in Bartter syndrome and abnormal reabsorption in the thick ascending limb of the loop of Henle. Barter syndrome results in decreased serum potassium, metabolic alkalosis, and hypercalciuria. Voltage-gated chloride channels (Choice F) are predominantly found on skeletal muscle, and defects may result in congenital myotonia. Zinc finger transcription factor Wilms tumor 1 (WT1) (Choice G) is a tumor suppressor gene on chromosome 11. Nephroblastoma is the most common renal malignancy in childhood caused by mutations in tumor suppressor genes WT1 or WT2. It is characterized by a large, often palpable, unilateral flank mass and hematuria. Educational Objective: Polycystic kidney disease is caused by genetic mutations in polycystin that result in the development of multiple cysts in the kidney because of structural abnormalities of the renal tubules. Cysts may form in infancy, childhood, or adulthood and eventually compress adjacent normal renal parenchyma. %3D Previous Next Score Report Lab Values Calculator Help Pause

72 Exam Section 2: Item 22 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 22. During a clinical study, an investigator examines the effects of monoamine oxidase inhibitors to treat certain types of major depressive and phobic anxiety disorders. The concentration of which of the following compounds is most likely to be increased as a result of this treatment? O A) y-Aminobutyric acid (GABA) O B) Dihydroxyphenylalanine C) Epinephrine D) Tryptophan E) Tyrosine

C. Monoamine oxidase inhibitors (MAOIS) inhibit the breakdown of epinephrine in the brain. MAOIS, which are structurally similar to biogenic amines, bind to monoamine oxidase and inhibit its enzymatic degradation of biogenic amines. This increases synaptic concentrations of serotonin, dopamine, norepinephrine, and epinephrine. MAÓIS are effective for depression and anxiety but are typically used only after other agents have failed because of the risk for serious adverse effects. MÁOIS decrease peripheral degradation of tyramine (a sympathomimetic biogenic amine that does not cross the blood-brain barrier), which may lead to hypertensive crisis if combined with dietary tyramine (eg, cheese, wine). Therefore, patients on MAOIS must adhere to a low-tyramine diet. Combining MAOls with other serotonergic agents also increases the risk for serotonin syndrome. Incorrect Answers: A, B, D, and E. V-Aminobutyric acid (GABA) (Choice A) is the primary inhibitory neurotransmitter. It is an amino acid rather than a biogenic amine and is not metabolized by monoamine oxidase. Dihydroxyphenylalanine (Choice B) is the precursor to the catecholamines (eg, dopamine, norepinephrine, epinephrine). Tryptophan (Choice D) is the precursor to serotonin, and tyrosine (Choice E) is another dopamine precursor. MAOIS directly increase synaptic concentrations of fully formed biogenic amine neurotransmitters, not neurotransmitter precursors. Educational Objective: Monoamine oxidase degrades biogenic amine neurotransmitters (eg, serotonin, dopamine, norepinephrine, epinephrine). Monoamine oxidase inhibitors (MAOIS) inhibit the enzymatic activity of monoamine oxidase and thereby increase the synaptic concentrations of these neurotransmitters, which target depressive and anxiety symptoms. MAOIS increase the risk for serotonin syndrome and hypertensive crisis from dietary tyramine. %D Previous Next Score Report Lab Values Calculator Help Pause

94 Exam Section 2: Item 44 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 44. A 25-year-old man is lost in the desert for 1 week with an ample supply of water but no food. Which of the following changes in enzyme activities and regulatory molecule concentrations in liver is most likely in this patient? Fructose Glucose Phosphoenolpyruvate Pyruvate Carboxykinase 2, 6-bisphosphate 6-phosphatase Kinase A) ↑ ↑ B) ↑ C) ↑ ↑ D) ↑ OE) F) ↑ ↑ G) ↑ H)

D. In the fed state, sufficient dietary carbohydrates, fats, and amino acids promote anabolism through the action of insulin. By contrast, in the fasting or starvation state, an absence of sufficient dietary precursors leads to upregulation of glucagon and epinephrine with downregulation of insulin, promoting catabolism. Catabolism occurs through intracellular signaling that results in activation or deactivation of enzymes involved in cellular metabolic pathways. For example, in the fed state, glycolysis, glycogen synthesis, fatty acid synthesis, and protein synthesis are upregulated by virtue of increased activity of the enzymes involved in these pathways. In the same state, gluconeogenesis, glycogen degradation, fatty acid degradation, protein degradation, and ketogenesis are downregulated. In order to maintain blood glucose concentrations and supply critical organs such as the brain, heart, and kidneys with nutrients, gluconeogenesis, glycogen degradation, fatty acid degradation, protein degradation, and ketogenesis will be active in this patient. Once glycogen stores have depleted, which usually occurs within the first day, hepatic gluconeogenesis will be the primary means of maintaining serum glucose. In this case, upregulation of hepatic glucose 6-phosphatase and phosphoenolpyruvate carboxykinase (PEPCK) will occur. Similarly, there will be a lower concentration of fructose 2,6-bisphosphate caused by decreased production in glycolysis, and pyruvate kinase, a key regulatory enzyme in glycolysis, will be downregulated. Incorrect Answers: A, B, C, E, F, G, and H. Fructose 2,6-bisphosphate (F 2,6-BP) is produced in the fed state from fructose 6-phosphate by phosphofructokinase-2. In turn, F 2,6-BP activates phosphofructokinase-1, a key regulatory step in glycolysis, promoting ongoing degradation of glucose monomers. In the fasting state, F 2,6-BP concentration will be decreased (Choices A, B, and C) because of the lack of available upstream glucose as a substrate for its synthesis. This will downregulate glycolysis. Glucose 6-phosphatase (G6Pase) is upregulated in hepatic gluconeogenesis and glycogen degradation, as it serves the final critical step in cleaving phosphate from glucose and permitting release into the bloodstream. It will be upregulated, not downregulated (Choices C, F, G, and H) in the fasting state. PEPCK is the first committed step in gluconeogenesis, converting oxaloacetate to phosphoenolpyruvate. In the fasting state, hepatic gluconeogenesis is the principal method of maintenance of serum glucose concentration and PEPCK will, in turn, be upregulated, not downregulated (Choices B, C, E, G, and H). Pyruvate kinase is the final regulatory step in glycolysis, producing pyruvate. It is bypassed in gluconeogenesis and is downregulated, not upregulated, in the fasting state (Choices C, F, and G). Educational Objective: In the fasting state, enzymes and regulatory molecules that promote gluconeogenesis, glycogen degradation, fatty acid degradation, protein degradation, and ketogenesis are upregulated or increased in concentration. Those related to gluconeogenesis include the upregulation of glucose 6-phosphatase and phosphoenolpyruvate carboxykinase. %3D Previous Next Score Report Lab Values Calculator Help Pause

44 Exam Section 1: Item 44 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 44. A 35-year-old African American man comes to the physician for a routine examination. He recently read a pamphlet at a health fair on the importance of screening for skin cancer. He enjoys sailing and usually goes out on his boat every weekend when the weather is nice. He does not use sunscreen, but he states that he does not "burn." Physical examination shows no abnormalities. This patient is at increased risk for melanoma at which of the following locations? A) Back B) Chest C) Forehead D) Palms E) Scalp

D. Malignant melanoma is likely to be present when a lesion has asymmetry, irregular appearing borders, variable coloration, a diameter greater than 6 mm, and rapid evolution in characteristics. Malignant melanoma has the ability to rapidly invade and metastasize, which carries a poor prognosis when diagnosed late. Subtypes include superficial spreading, nodular, lentigo maligna, and acral lentiginous. While most melanoma subtypes are associated with ultraviolet (UV) radiation, the acral lentiginous subtype is not. In patients with darker skin types, increased melanin density in the keratinocytes serves to shield the DNA from the harmful effects of UV radiation. Thus the subtypes of melanoma whose pathophysiology are dependent on UV radiation are seen less commonly in this patient cohort. Likewise, those subtypes that occur regardless of UV radiation, such as the acral lentiginous subtype, have a higher rate of diagnosis in patients with darker skin. The acral lentiginous subtype of melanoma has a predilection for the distal extremities, often referred to as acral sites. Acral lentiginous melanoma may present as a new band of pigment on a nail or as an enlarging dark brown macule or patch on the fingers, palms, soles, or toes. These melanomas commonly go unnoticed, particularly if on the feet, and often are diagnosed at more advanced stages. Any lesion with features suggestive of malignant melanoma should be surgically excised with negative margins and pathologically examined for the depth of dermal invasion. Incorrect Answers: A, B, C, and E. The back (Choice A), chest (Choice B), forehead (Choice C), and scalp (Choice E) are all sun-exposed sites likely to receive UV radiation on a regular basis. UV radiation penetrates clothing, so even areas covered with clothing may be affected. Superficial spreading, nodular, and lentigo maligna types of melanoma have a predilection for these sun-exposed areas. Involvement of sun- exposed anatomical sites by these subtypes are not typically seen in individuals with darker skin types because of the protective effect of increased melanin pigment. Acral lentiginous melanoma involving the distal extremities (ie, acral sites) is more common in this patient cohort. Educational Objective: Acral lentiginous melanoma is a subtype of melanoma that occurs on the distal extremities (ie, acral sites). Its development is not associated with UV radiation, and it is the most common subtype of melanoma in individuals with darker skin. %3D Previous Next Score Report Lab Values Calculator Help Pause

50 Exam Section 1: Item 50 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 50. In a recent study of patients with congestive heart failure, the investigators found that these patients showed a diminished capacity to restore low basal calcium concentration during diastole. Function of which of the following is most likely to be altered in these patients? A) Fast sodium channel B) Inward rectifier C) L-type calcium channel D) Nat-Ca2+ exchanger E) Ryanodine receptor

D. The myocardial action potential (AP) is the process by which a neuronal action potential causes a myocyte to depolarize, leading to contraction, and then repolarize. The myocardial AP has five phases. During phase 0, voltage-gated Na+ channels ("fast" channels) open and allow Na+ to rush into the cell. This causes the intracellular charge to rapidly become positive or depolarize. Once a threshold of depolarization occurs, Phase 1 begins, in which the voltage-gated Na+ channels are quickly inactivated, preventing further Na+ from entering the cell. Voltage-gated K+channels simultaneously open, allowing positive charge to leave the cell. The combination of these two changes leads to an initial repolarization. A plateau in charge then follows during Phase 2, during which Ca2+ influx through voltage-gated Ca+ channels is equivalent to K* efflux through K+channels. When these voltage-gated Ca2+ channels increase Ca2+ myocyte influx, it leads to Ca2+ release from the sarcoplasmic reticulum and consequent myocyte contraction. During Phase 3, rapid repolarization occurs through a large K+efflux out of the cell through voltage-gated K+ channels. The voltage-gated Ca2+ channels also close at this time and contraction stops. This must be accompanied by activity of a Na-Ca2+ exchanger, which brings Na*into the cell in exchange for Ca2+ leaving the cell. The ability of Ca2+ to be removed from the cytoplasm is critical to cessation of contraction and preparation to be depolarized again. The Na-Ca2+ exchanger helps to re-establish the baseline resting potential, which is the key step in the final phase, Phase 4, of the myocardial AP. At this point, the myocyte has now reached its baseline resting potential again and awaits the next AP. Incorrect Answers: A, B, C, and E. Fast sodium channel (Choice A) is another name for the voltage-gated Na channels that allow Na* to rush quickly into the myocyte and create the initial rapid depolarization. These channels do not directly affect the calcium concentration in the cell. Inward rectifier (Choice B) is a type of K+ channel. It is referred to as inward rectifying because it favors influx of K rather than efflux. The K+channel that is active in Phase 4 of the action potential, when the efflux of K+ brings the cell back to its initial resting membrane potential, is an outward rectifier, not an inward rectifier. Neither of these types of Kt channels restore decreased intracellular calcium concentrations. L-type calcium channel (Choice C), also known as the dihydropyridine channel, is a voltage-gated Ca2+ channel. This channel opens during phase 2 of the myocardial action potential. It is coupled with the ryanodine receptor so that when the L-type calcium channel opens during phase 2, it also triggers release of Ca2+ from the sarcoplasmic reticulum and causes contraction. It is not used in restoring decreased basal calcium concentration during the resting phase. The ryanodine receptor (Choice E) is located on the sarcoplasmic reticulum membrane and allows the movement of Ca2+ from inside the sarcoplasmic reticulum into the cytoplasm. This efflux of Ca2+ into the cytoplasm initiates excitation-contraction coupling. However, this receptor is not responsible for the reuptake of Ca2+ into the sarcoplasmic reticulum or re-establishment of resting potential. Educational Objective: The myocardial action potential has five phases, each of which is characterized by activity of a different ion channel. Calcium influx into the cytoplasm from the sarcoplasmic reticulum causes contraction of the myocyte. Removal of calcium from the cytoplasm when contraction is no longer desired, such as during myocardial diastole, is critical. Previous Next Score Report Lab Values Calculator Help Pause

61 Exam Section 2: Item 11 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 11. A 32-year-old man sustains a head injury in a motorcycle accident and is admitted to the intensive care unit. Six days later he becomes confused and is very thirsty. His serum sodium concentration is 158 mEq/L, and his urine output is 6.1 L/day. These findings most likely indicate that he has a disorder that is affecting which of the following labeled parts of the nephron shown? E A F A) B) C) D) E) F)

F. Central diabetes insipidus may result from head trauma and is characterized by inadequate secretion of antidiuretic hormone (ADH, also known as vasopressin) from the posterior pituitary. ADH binds V, receptors of the late distal convoluted tubule and the collecting tubule cells, which results in the insertion of aquaporin channels in the luminal surface. This results in increased free water absorption and maintains tight regulation of the serum osmolality. Inadequate production of ADH by the posterior pituitary gland may occur in various pathologies, such as pituitary tumors, trauma, tuberculosis, or intracranial surgery. This results in a decreased concentration of aquaporin channels inserted within the principal cells of the collecting tubule, leading to an inability to absorb adequate amounts of free water; this, in turn, causes dilute urine and increased serum osmolality. The hypothalamic osmoreceptors detect this increased serum osmolality, which causes the patient to report thirst. Patients typically present with the production of large volumes of dilute urine. Treatment is with desmopressin, an analog of vasopressin (ADH), which results in recovery of free water via aquaporin insertion in the collecting tubule. Incorrect Answers: A, B, C, D, and E. The proximal tubule (Choice A) is the closest to the glomerulus. It is the site for reabsorption of the majority of the electrolytes (HCO3, Na+, Cl; PO,3, K*), water, uric acid, glucose, and amino acids in the initial filtrate. It does not rely on the action of ADH. The thin descending loop of Henle (Choice B) passively reabsorbs water and is impermeable to sodium. It does not rely on the action of ADH. The thin ascending loop of Henle (Choice C) is impermeable to water and reabsorbs small amounts of sodium. It does not rely on the action of ADH. The thick ascending loop of Henle (Choice D) reabsorbs Na+, Cl, and K+through the Na-K-CI cotransporter (NKCC). It is impermeable to water. It does not rely on the action of ADH. The distal convoluted tubule (Choice E) is the main site for sodium and chloride reabsorption in the nephron through the Na-Cl symporter, making dilute and hypotonic filtrate. It does not rely on the action of ADH. Educational Objective: Central diabetes insipidus may result from head trauma and is characterized by inadequate secretion of ADH. Lack of ADH leads to decreased free water reabsorption from collecting tubules, resulting in an abnormally increased serum osmolality. %3D Previous Next Score Report Lab Values Calculator Help Pause

7 Exam Section 1: Item 7 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 7. A 60-year-old woman comes to the physician because of intermittent joint pain during the past 5 years. The pain is abrupt in onset and involves principally her hands, knees, and toes. She has a history of pyelonephritis; there is no history of trauma. Examination of the left knee shows tenderness, swelling, and warmth with dusky, erythematous overlying skin. A photograph of the left hand is shown, along with a photomicrograph of joint fluid aspirate. Which of the following is most likely in excess concentration in the joint fluid of this patient? A) Calcium apatite B) Calcium oxalate C) Calcium pyrophosphate D) Cholesterol E) Sodium chloride F) Sodium monourate

F. Gout is an acute, monoarticular arthropathy resulting from an intra-articular inflammatory reaction to the precipitation of monosodium urate crystals (MSU) within the joint space. This can result from the overproduction of uric acid, but it more commonly occurs following the underexcretion of uric acid. This patient's pyelonephritis may have predisposed her to decreased clearance of MSU. Gout typically presents with acute, atraumatic single-joint pain, erythema, and swelling, and it can recur in patients that are under- or untreated; polyarticular presentations are also seen. The most common joint involved is the first metatarsophalangeal, but other common joints include the knees and elbows. Excess serum MSU may also precipitate in tissues as tophi, seen as yellow subdermal inclusions and masses in the photograph of this patient's left hand. Arthrocentesis generally shows 10,000 to 50,000 leukocytes/mm3 and birefringent, needle-shaped crystals, as seen in this photomicrograph of synovial fluid. Treatment consists of nonsteroidal anti-inflammatory drugs (NSAIDS), colchicine, and/or corticosteroids for acute exacerbations. Chronic preventive management involves the use of xanthine oxidase inhibitors such as allopurinol. Additionally, patients should avoid triggering risk factors such as purine-rich meals and heavy alcohol consumption. Incorrect Answers: A, B, C, D, and E. Calcium apatite (Choice A) crystals deposit within joints and tendons in hydroxyapatite deposition disease. The shoulder is often involved and presents as calcific tendonitis with decreased range of motion and tenderness to palpation. Calcium oxalate (Choice B) is the most common crystal that precipitates in urine, presenting as nephro- or ureterolithiasis. It does not commonly precipitate in joints. Calcium pyrophosphate (Choice C) deposits in joints, leading to acute monoarticular arthritis that commonly involves larger joints such as the knee. Calcium pyrophosphate deposition disease occurs more frequently in patients who have systemic inflammatory or endocrine diseases (eg, hyperparathyroidism). Arthrocentesis findings are similar to those seen in gout; however, rhomboid crystals are seen on microscopy. Chondrocalcinosis may be seen on x-ray. Cholesterol (Choice D) crystals are seen on biopsy of atherosclerotic plaques within blood vessels and would be less likely to occur within a joint. Similar to MSU crystals, they have a needle-like shape. Sodium chloride (Choice E) is ionized in solution at the common osmolality and pH of the human body. It would be unlikely to deposit as a crystal. Educational Objective: Gouty arthritis occurs because of the precipitation of monosodium urate crystals in a joint and may be accompanied by deposits in soft tissue called tophi. On microscopy, MSU crystals are needle-shaped and birefringent. %3D Previous Next Score Report Lab Values Calculator Help Pause

25 Exam Section 1: Item 25 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment In Airflow Out Phrenic nerve activity Abdominal pressure Time 25. A study is conducted to determine the effects of drug X on respiratory function. The respiratory tracing of an experimental animal is shown. Drug X was administered at the arrow. Drug X is most likely which of the following? A) Lidocaine B) Morphine sulfate C) Pentobarbital D) Potassium chloride E) Tetrodotoxin F) Tubocurarine

F. Tubocurarine is a historical nondepolarizing neuromuscular blocker that is no longer in clinical use. Other nondepolarizing neuromuscular blockers include rocuronium, vecuronium, atracurium, and cis-atracurium. These medications cause muscular paralysis via competitive antagonism at the postsynaptic acetylcholine receptor sites in the neuromuscular junction. They have no effect on the presynaptic nerve which will continue to be active, as shown by the continued phrenic nerve activity on the graph. Paralysis of the diaphragm leads to respiratory failure if the patient is not mechanically ventilated, which can also be seen on the graph with the absence of airflow. It can be reversed with neostigmine, which acts to increase the acetylcholine available at the neuromuscular junction by inhibiting acetylcholinesterase. Neostigmine is commonly paired with glycopyrrolate or atropine to minimize muscarinic effects, such as profound bradycardia. Incorrect Answers: A, B, C, D, and E. Lidocaine (Choice A) is a class Ib antiarrhythmic and local anesthetic that inhibits sodium channels. It can be used intravenously to treat ventricular tachyarrhythmias or subcutaneously to provide local anesthetic for minor surgical procedures. It does not inhibit diaphragmatic movement. Morphine sulfate (Choice B) is an opioid analgesic used to manage moderate to severe pain. It acts on opioid receptors to provide pain relief and is available in oral, intravenous, intramuscular, and suppository formulations. With high doses, it can cause central respiratory depression, which would present with features similar to this graph, although phrenic nerve activity would be decreased. Pentobarbital (Choice C) is a barbiturate used for sedation or treatment of seizures. Its mechanism of action is potentiation of GABA at its receptor, leading to an increase in the duration of chloride channel opening. With rapid administration, it can cause apnea as a result of central depression of the medullary ventilatory center. This would cause decreased activity of the phrenic nerve. Potassium chloride (Choice D) is used to supplement depleted potassium concentrations, which occurs frequently in patients on diuretics or with poor oral intake. When given in excess, it can cause cardiac dysrhythmias and cardiac arrest. It does not affect the innervation of the diaphragm. Tetrodotoxin (Choice E) is a toxin secreted by pufferfish that inhibits cardiac and neuronal sodium channels and leads to weakness, paresthesia, loss of reflexes, hypotension, and, when severe, respiratory failure caused by paralysis. However, this would cause decreased activity of the phrenic nerve. Educational Objective: Tubocurarine is a nondepolarizing neuromuscular blocker that causes muscular paralysis at the postsynaptic neuromuscular junction by competitively antagonizing the acetylcholine receptors. %3D Previous Next Score Report Lab Values Calculator Help Pause

21 Exam Section 1: Item 21 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 21. A6-year-old boy is brought to the physician because of a 2-year history of progressive leg stiffness and difficulty walking. Physical examination shows dystonia of the lower extremities. An MRI of the brain shows no abnormalities. A lumbar puncture is done. Cerebrospinal fluid analysis shows a markedly decreased homovanillic acid concentration. A deficiency of which of the following enzymes is the most likely cause of this patient's symptoms? A) CoQ reductase B) Cytochrome c reductase C) Glutamic acid decarboxylase D) Pyruvate decarboxylase E) Tryptophan hydroxylase F) Tyrosine hydroxylase

F. Tyrosine hydroxylase deficiency is an autosomal recessive disorder that results from mutations of the tyrosine hydroxylase gene on chromosome 11. Tyrosine hydroxylase, the rate-limiting enzyme of catecholamine synthesis, normally converts tyrosine to L-DOPA, the precursor to dopamine. Dopamine is a precursor to norepinephrine and epinephrine and is degraded to homovanillic acid. Consequently, analysis of cerebrospinal fluid shows decreased catecholamines and homovanillic acid in patients with tyrosine hydroxylase deficiency. Tyrosine hydroxylase deficiency results in progressive dystonia (from dopamine deficiency) and possible cognitive impairment (from catecholamine deficiency). Treatment centers around the replacement of L-DOPA. Incorrect Answers: A, B, C, D, and E. CoQ reductase (Choice A) and cytochrome c reductase (Choice B) both participate in the mitochondrial electron transport chain that synthesizes adenosine triphosphate (ATP). Deficiencies in electron transport enzymes commonly manifest as exercise intolerance or as mitochondrially-inherited neuromuscular syndromes (eg, myoclonic epilepsy with ragged red fibers) rather than dystonia. Glutamic acid decarboxylase (Choice C) catalyzes the rate-limiting step of gamma-aminobutyric acid (GABA) synthesis. Glutamic acid decarboxylase deficiency may lead to syndromes of central nervous system hyperexcitability, such as stiff-person syndrome. However, homovanillic acid concentrations would be normal. Pyruvate decarboxylase (Choice D) in an enzyme involved in the conversion of pyruvate to acetaldehyde. Pyruvate decarboxylase deficiency is rare and may present with lactic acidosis, failure to thrive, seizures, developmental delay, and ataxia rather than leg stiffness. Tryptophan hydroxylase (Choice E) catalyzes the rate-limiting step of serotonin synthesis. Tryptophan hydroxylase deficiency may underlie major depressive disorder. Educational Objective: Tyrosine hydroxylase normally converts tyrosine to L-DOPA, the precursor to dopamine, norepinephrine, and epinephrine. Tyrosine hydroxylase deficiency leads to decreased dopamine and consequent progressive dystonia. A decrease in dopamine leads to a decrease in dopamine degradation products, such as homovanillic acid. II Previous Next Score Report Lab Values Calculator Help Pause

30 Exam Section 1: Item 30 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 30. An overweight 50-year-old man comes to the physician because of a 3-day history of intermittent severe abdominal and interscapular pain associated with nausea and vomiting. He has had an 18-kg (40-lb) weight loss during the past 6 months by caloric restriction and exercise. He has not used appetite suppressants. He is in mild distress. Abdominal examination shows tenderness of the right upper quadrant. Laboratory studies show: Hemoglobin Leukocyte count Serum Total bilirubin 13.8 g/dL 14,500/mm3 Alkaline phosphatase AST ALT 4 mg/dL 200 U/L 70 U/L 68 U/L Test of the stool for occult blood is negative. Which of the following is the most likely diagnosis? O A) Acute pancreatitis B) Cholecystitis C) Esophageal reflux D) Fatty liver disease E) Peptic ulcer disease

B. Acute cholecystitis occurs secondary to obstruction of the cystic duct from a gallstone, which results in subsequent inflammation of the gallbladder wall. Risk factors for developing gallstones include being female, multiparity, increasing age, and obesity. High rates of gallstone formation are also seen after periods of rapid weight loss. Acute cholecystitis typically presents with fever, abdominal pain, and tenderness to palpation in the right upper quadrant, often in a patient with a history of biliary colic and cholelithiasis. The pain may also radiate to the right shoulder or interscapular region. It may be distinguished from biliary colic by its constant rather than intermittent nature and its associated findings of fever and leukocytosis. Jaundice is a less common finding, and suggests an alternative, or complicated, diagnosis such as choledocholithiasis or ascending cholangitis. Abdominal ultrasonography is the first-line diagnostic examination for the evaluation of potential acute cholecystitis. Characteristic ultrasound findings of acute cholecystitis include gallbladder wall thickening and hyperemia, pericholecystic fluid, presence of gallstones, and a sonographic Murphy sign (pain with compression directly over the gallbladder). If ultrasound findings are equivocal, hepatobiliary scintigraphy (HIDA) scan can be performed, which is a nuclear medicine scan that evaluates for patency of the cystic duct and biliary tree. Treatment includes supportive therapy, intravenous fluids, intravenous antibiotics, and cholecystectomy. Incorrect Answers: A, C, D, and E. Acute pancreatitis (Choice A) presents with epigastric abdominal pain, nausea, and emesis, often in patients with a history of gallstones, alcoholism, trauma, hypertriglyceridemia, or hypercalcemia. It can be complicated by necrosis, hemorrhage, abscess, or the formation of pseudocysts. Laboratory studies often show increased lipase. Esophageal reflux (Choice C) typically presents with burning epigastric and lower chest pain, often in association with the consumption of a large meal or trigger food such as an acidic beverage, coffee, chocolate, or tomatoes. It is often worse with supine positioning. Fatty liver disease (Choice D) may present with insidious right upper quadrant pain and tenderness, nausea, vomiting, anorexia, and jaundice, or it may be asymptomatic and indolent. In severe cases, liver function testing may show transaminitis, indirect hyperbilirubinemia, hypoalbuminemia, and coagulopathy. However, it would not typically cause the leukocytosis demonstrated by this patient. Peptic ulcer disease (Choice E) describes the presence of ulcers in the stomach or duodenum, which classically present with worsening epigastric pain related to the consumption (gastric) or lack of consumption (duodenal) of food. Peptic ulcer disease is strongly associated with infection from the bacterium Helicobacter pylori. It would not explain the interscapular or right upper quadrant pain seen in this case. Educational Objective: Acute cholecystitis is caused by obstruction of the cystic duct from a gallstone with subsequent gallbladder wall inflammation. Obesity, being female, multiparity, increasing age, and rapid weight loss are risk factors for gallstone formation. Previous Next Score Report Lab Values Calculator Help Pause

195 Exam Section 4: Item 46 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 46. An investigator is conducting a study of natural killer cells and their role in destroying malignant tumors in experimental animals. This study is most likely to show that natural killer cells are particularly effective in destroying malignant tumors with which of the following characteristics? A) Decreased expression of class I MHC molecules B) Expression of adherent forms of cellular proteins C) Expression of altered cell-surface glycolipids D) Expression of viral antigens E) Increased expression of oncofetal antigens

A. Decreased expression of class I MHC molecules on the surface of malignant tumors most likely accounts for targeting by natural killer (NK) cells. Nonmalignant cells express class I MHC molecules that interact with inhibitory receptors on the surface of NK cells, including killer cell immunoglobulin-like receptors and leukocyte immunoglobulin-like receptors, which prevents NK cell activation. When class I MHC molecules are absent or appear in decreased numbers, NK cells are activated to synthesize perforin and granzyme, which are both proapoptotic. This is a common first-line defense against malignant cells, although it is not uncommon for cancer cells to escape destruction by NK cells as they continue to accumulate mutations and tumor growth outpaces the rate of NK cell activity. Incorrect Answers: B, C, D, and E. Expression of adherent forms of cellular proteins (Choice B) does not account for NK cell activity against malignant cells. Cellular proteins are constantly expressed on the surface of cells via class I and II MHC molecules to allow for activation of CD8+ and CD4+ T lymphocytes if these proteins are recognized as foreign; however, NK cells are not activated by particular antigens, but by the absence of class I MHC molecules. Expression of altered cell-surface glycolipids (Choice C) occurs frequently in malignant cells. Increased expression of glycolipids may actually help malignant cells evade the immune system and augment cellular growth and differentiation. Recognition of glycolipids on the surface of malignant cells is not a feature of NK cell activity. Expression of viral antigens (Choice D) occurs via MHC class I and Il molecules that are expressed on most cells, but virally infected cells often downregulate the expression of MHC molecules in an attempt to evade detection by preventing presentation of viral antigens. The downregulation of MHC molecules allows NK cells to recognize these virally infected cells as abnormal and trigger apoptosis. Increased expression of oncofetal antigens (Choice E) is a common feature of many cancers, but expression of these antigens does not activate NK cells. Oncofetal proteins are those that are normally expressed only during fetal development. Because of this, detection of these in adult subjects indicates the presence of abnormal cells. Common examples of oncofetal proteins include a-fetoprotein and carcinogenic embryonic antigen. Educational Objective: NK cells are activated by an absence of class I MHC molecules on the surface of tumor cells. Normally, NK cell activation is inhibited by the binding of inhibitory receptors to class I MHC molecules, but an absence of class I MHC shifts the balance toward activation, with the subsequent release of granzyme and perforin leading to the apoptosis of tumor cells. Previous Next Score Report Lab Values Calculator Help Pause

35 Exam Section 1: Item 35 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 35. A 58-year-old man is brought to the emergency department 2 hours after having a generalized tonic-clonic seizure. He has been treated numerous times in the past for seizures caused by alcohol withdrawal. On arrival, he is awake but confused and tremulous. His pulse is 110/min, and blood pressure is 160/90 mm Hg. Physical examination shows bilateral asterixis. Laboratory studies show: Hemoglobin Hematocrit Mean corpuscular volume Leukocyte count Reticulocyte count Platelet count 11 g/dL 32% 110 pm3 3800/mm3 1% 160,000/mm3 Which of the following is the most likely cause of this patient's anemia? A) Folic acid deficiency B) Iron deficiency C) Lead toxicity D) B-Thalassemia E) Vitamin B, (thiamine) deficiency

A. Folic acid, or vitamin Bg, is converted in the body to tetrahydrofolic acid and used as a coenzyme in the synthesis of nucleotides and nucleosides. Folate is contained in leafy vegetables and primarily absorbed in the jejunum. Folate deficiency is often seen in patients with malnutrition, alcoholism, and patients taking anti-folate medications (eg, phenytoin, methotrexate). Megaloblastic anemia occurs in impaired DNA synthesis, commonly related to folate or vitamin B 12 deficiency, and is classically characterized by erythrocyte macrocytosis and hypersegmented neutrophils. Laboratory studies show anemia, increased mean corpuscular volume, and normal white cell and platelet indices. In this patient's case, the megaloblastic anemia is likely caused by folic acid deficiency, as these stores in the body can be depleted quickly in cases of malnourishment, while B12 reserves are substantial and take longer to deplete. Incorrect Answers: B, C, D, and E. Iron deficiency (Choice B) anemia commonly results from occult blood loss through the gastrointestinal tract or via menstruation. It can also result from severe nutritional iron deficiency. Iron deficiency causes erythrocytes to be normocytic (early) or microcytic (late), but not macrocytic as in this case. Lead toxicity (Choice C) can occur secondary to workplace exposure, lead paint in a residence, contaminated drinking water, or other contaminated sources. Symptoms of lead exposure can be variable and nonspecific, including abdominal pain, constipation, anorexia, myalgias and arthralgias, fatigue, headache, difficulty concentrating, and sideroblastic (microcytic) anemia. B-Thalassemia (Choice D) is caused by one or more mutations of the two paired B-globin chains. Mutation of one gene results in B-thalassemia minor. Mutation of both genes causes B- thalassemia major, which is characterized by severe, transfusion-dependent microcytic anemia and signs of extramedullary hematopoiesis such as frontal bossing. Vitamin B, (thiamine) deficiency (Choice E) caused by poor nutritional intake and thiamine malabsorption is seen in patients who chronically abuse alcohol. A characteristic clinical presentation is Wernicke encephalopathy, a triad of altered mental status, ophthalmoplegia, and ataxia. Thiamine deficiency would not explain the macrocytic anemia seen in this patient. Educational Objective: Megaloblastic anemia occurs in impaired DNA synthesis, most commonly related to folate or vitamin B 12 deficiency. Folic acid deficiency leading to anemia is faster to develop than B,12 deficiency in malnutrition, given that the body's reserve of folate is less extensive than that of vitamin B12 Previous Next Score Report Lab Values Calculator Help Pause

73 Exam Section 2: Item 23 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment Affected female Affected male O Unaffected female O Unaffected male 23. A 28-year-old woman and her 40-year-old husband both have albinism; none of their three children is affected. A pedigree is shown. Which of the following genetic mechanisms best explains the absence of albinism in their children? A) Genetic heterogeneity B) Germinal mosaicism C) Pleiotropy D) Variable expressivity

A. Genetic heterogeneity explains why both patients have albinism but none of their three children have albinism. There are numerous mutations that can cause oculocutaneous albinism, including mutations in OCA1, TYRP1, MATP, and SLC24A5 genes among others. While the phenotype remains the same, with lack of retinal pigmentation and absent production of melanin, the phenotype can be produced by numerous different gene mutations. Albinism is an autosomal recessive disease, so it is possible that two individuals with albinism who have mutations in different genes can have offspring who do not exhibit any disease. If albinism were a monogenic autosomal recessive disorder, all offspring of the parents would be expected to also have albinism. In this case, the father's transmission of one autosomal recessive allele would be paired with a normal allele from the mother, and vice versa, eliminating the phenotype in the heterozygous children. Incorrect Answers: B, C, and D. Germinal mosaicism (Choice B) refers to mosaicism that is found only within gametes but not within somatic cells. In this way, deleterious mutations can be harbored in the gametes of unaffected individuals and passed on to offspring. This would be a reasonable explanation if the pedigree showed unaffected parents with multiple affected children. Pleiotropy (Choice C) refers to genes that influence one or more different and seemingly unrelated traits. One example is phenylketonuria which is caused by mutations that affect phenylalanine hydroxylase, resulting in failure to convert phenylalanine to tyrosine. Neurologic manifestations are a result of accumulated phenylalanine, but tyrosine is also a precursor to melanin, so patients commonly have fair skin and hair and may resemble patients with albinism. Variable expressivity (Choice D) refers to genetically inherited conditions that have different phenotypes despite identical underlying mutations. Common examples include neurofibromatosis and Marfan syndrome, each of which can manifest with a spectrum of disease. Educational Objective: The inheritance pattern of albinism is autosomal recessive, but its genetic heterogeneity can lead to inheritance patterns that do not fit classic autosomal recessive patterns. As there are multiple mutations that can cause disease, it is possible that two affected individuals can exhibit the same symptoms of albinism but have entirely disparate genetic mutations. Previous Next Score Report Lab Values Calculator Help Pause

54 Exam Section 2: Item 4 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 4. During a clinical study, nine patients are given a new antihypertensive drug. One week later, their serum creatinine concentrations are measured, and the results are shown: Serum Creatinine (mg/dL) 0.9 0.9 0.9 1.0 1.1 1.2 1.2 1.3 1.4 Patient 1 5 If one of the 0.9 values were now mistakenly entered as 1.2, which of the following would the most likely effect be on the mean, median, and mode? Mean Median Mode A) ↑ ↑ ↑ no change no change B) ↑ ↑ C) ↑ no change D) No change ↑ E) No change ↑ no change F) No change no change no change

A. In statistical analysis, measures of central tendency are useful in describing a data set. Mode, median, and mean are all measures of central tendency. The mode is defined as the value that occurs most frequently within the data. In this data set, the serum creatinine value that occurs most frequently is 0.9 mg/dL, making this number the mode. The median is defined as the middle value of the data set when ordered, meaning that there are the same number of values to the left and right (lesser than and greater than) of the median. If the median falls between two numbers, these numbers are summed and divided by two. In this data set, the median is 1.1 mg/dL. The mean is defined simply as the average of all the values included in the data set; in this set, it is equal to 1.1 mg/dL. Inadvertent entry of a value of 0.9 mg/dL as 1.2 mg/dL would alter all three measures of central tendency in this data set. The mode, which was 0.9 mg/dL, would change to 1.2 mg/dL, which would now be the most commonly occurring value. The median value would become 1.2 mg/dL, since this would be the central point of the data set with an equivalent number of greater and lesser values. Finally, the mean would increase to 1.13 mg/dL. Thus, all three measures of central tendency would increase. Incorrect Answers: B, C, D, E, and F. Choices B, C, E, and F are incorrect since the mode would increase. One fewer value of 0.9 mg/dL would occur, for a count of two data points of this value, whereas the count of data points equal to 1.2 mg/dL would increase to three. Choices C and F are incorrect since the median would increase by the skew of data toward higher values. Choices D, E, and F are incorrect since the mean would be increased by the inadvertent substitution of a lower data point for a higher data point within the set. Educational Objective: Median, mode, and mean are measures of the central tendency of data. Mode is the most common value, median is the central value of the data, and mean is the average of all values in the data set. Inadvertent substitution of a higher data point would potentially shift all values away from the correct value and to falsely increased values. Previous Next Score Report Lab Values Calculator Help Pause 234 6789

156 Exam Section 4: Item 6 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 6. An otherwise healthy 15-year-old boy is brought to the physician 2 days after the onset of a headache. His vital signs are within normal limits. Physical examination shows mild tenderness over the right occipital area. An x-ray of the skull shows a 2-cm lytic lesion in the right occipital bone. A photomicrograph of a biopsy specimen of the lesion stained with hematoxylin and eosin is shown. Which of the following is the most likely diagnosis? A) Langerhans cell histiocytosis B) Meningioma C) Metastatic carcinoma D) Osteosarcoma E) Solitary bone cyst

A. Langerhans cell histiocytosis is a rare proliferation of Langerhans cells and histiocytes (activated dendritic cells and macrophages) and includes multiple syndromes and subtypes such as Hand- Schüller-Christian disease and histiocytosis X. The presentation can include isolated bone lesions, as in this patient, or diffuse, multisystemic, invasive disease. The condition results from a clonal proliferation of dendritic cells and macrophages. Lymphocytes and eosinophils are also commonly present on biopsy. Symptoms, when present, may include nonspecific fever, fatigue, bone pain, cough, and weight loss, but they may also be specific to the affected organ system. Bone is the most commonly affected, with bone pain reported by patients and swelling noted on examination. The skull is often affected, with lytic lesions demonstrated on imaging, as in this patient's case. Perilesional skin eruptions are common on the skull and include rash, erythematous scale, and papules. Bone marrow, lungs, lymph nodes, and endocrine glands are commonly also involved. Diagnosis is through tissue biopsy, with Langerhans cells and eosinophils seen on hematoxylin and eosin staining, as seen in this patient's photomicrograph. Birbeck granules may be present on electron microscopy, and cytologic markers include CD1, CD1A, and S-100. Treatment includes excision for local disease, radiation, or systemic chemotherapy. Incorrect Answers: B, C, D, and E. Meningioma (Choice B) is a common, benign primary brain tumor arising from the meninges that occurs along the surface of the brain. It may present with focal neurologic deficits from compression of the underlying brain parenchyma or with seizures. It would not present with a lytic skull lesion or involve the skull tables unless invasive, which is rare. Metastatic carcinoma (Choice C) is unlikely, as this patient is young and healthy, with a solitary lesion on imaging and histologic features consistent with Langerhans cell histiocytosis. When metastatic disease is present involving the bones or brain, lesions are often multiple. Osteosarcoma (Choice D), the most common type of primary bone cancer, often occurs in persons under age 30 years. Osteosarcoma typically occurs in long bones, frequently above the knee in the distal femur or below the knee in the proximal tibia. The metaphysis is often affected. The initial presentation often includes pain and swelling, although sometimes a fragility fracture is the declaring presentation. Causes include familial cases, especially involving the retinoblastoma or p53 tumor-suppressor genes, bony dysplasia such as in

8 Exam Section 1: Item 8 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 8. A 12-year-old boy is brought to the office by his mother for an annual well-child examination. He has asthma well controlled with inhaled budesonide. The mother says that the boy recently has voiced concern about his height and is bothered by the fact that he is considerably shorter than the other boys on his basketball team. The patient's father is 182 cm (6 ft) tall and his mother is 168 cm (5 ft 6 in) tall. The patient is 135 cm (4 ft 5 in; 3rd percentile) tall and weighs 32 kg (71 lb; 10th percentile); BMI is 17 kg/m2 (35th percentile). The patient's growth has remained steady at the 3rd percentile during the past 8 years. Vital signs are within normal limits. Sexual maturity rating is 1 for pubic hair and genital development. The remainder of the physical examination shows no abnormalities. Which of the following is the most likely diagnosis? A) Constitutional growth delay B) Familial short stature C) Growth hormone deficiency D) Hypopituitarism E) Hypothyroidism F) Steroid-induced growth failure

A. Short stature can be hereditary, or a result of a variety of causes, such as constitutional growth delay, chronic steroid use, endocrine abnormalities, cardiopulmonary disease, and genetic disorders. Familial (hereditary) short stature and constitutional growth delay are the most common causes of short stature. Familial short stature is associated with short stature in both parents, a normal gain of height velocity, and a bone age that corresponds with chronological age, whereas a constitutional growth delay is independent of parental stature, has a decreased growth velocity, and a bone age younger than the patient's chronological age. Those with constitutional growth delay typically reach a normal adult height, with more rapid increases in height gain velocity after they reach puberty. Other causes of short stature include growth hormone deficiency, hypothyroidism, cystic fibrosis, severe congenital heart disease, Turner syndrome, and Down syndrome. This patient has parents with normal stature and has had a decreased growth velocity but does not have other signs or symptoms to suggest another cause of short stature, making constitutional growth delay the most likely diagnosis. Incorrect Answers: B, C, D, E, and F. Familial short stature (Choice B) typically presents in a patient whose parents are both of short stature with a normal growth velocity and bone age consistent with chronological age. They will remain of short stature into adulthood. This patient has a diminished growth velocity, making this answer less likely. Growth hormone deficiency (Choice C) can present in infancy or childhood, depending on the severity of deficiency. Acquired growth hormone deficiency is typically caused by a nonfunctional pituitary tumor and presents with severe growth delay and a bone age younger than chronological age. It is associated with a rapid decline in the height-for-age curve, crossing percentiles. This patient's growth velocity has been slow but consistent, making growth hormone deficiency a less likely cause of his short stature. Hypopituitarism (Choice D) presents with short stature, fatigue, hypotension, cold intolerance, constipation, amenorrhea or oligomenorrhea in women, and decreased libido in men, as well as decreased development of secondary sexual characteristics. This patient does not have signs or symptoms to suggest hypopituitarism. Hypothyroidism (Choice E) in adolescents can present with short stature and growth delay, but it would be expected to present with other signs of hypothyroidism, such as fatigue, weight gain, cold intolerance, and constipation. Steroid-induced growth failure (Choice F) is seen in patients on chronic systemic steroid therapy as a result of the negative feedback on growth hormone secretion and inhibition of collagen and bone formation. While this patient takes daily inhaled steroids, this local delivery of steroids to the respiratory tract would be unlikely to cause steroid-induced growth failure. Educational Objective: Constitutional growth delay presents in a patient who has parents of normal stature and demonstrates short stature, delayed pubertal development, slow but consistent growth velocity, and a bone age less than chronological age. It is associated with normal development and stature once the patient reaches puberty. Previous Next Score Report Lab Values Calculator Help Pause

69 Exam Section 2: Item 19 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 19. An 18-year-old man develops a painless 1x1-cm ulcer on his penis, followed 4 weeks later by an erythematous maculopapular rash involving the palms and soles. The drug of choice for treatment of this disease inhibits which of the following microbial processes? A) Cell wall synthesis B) DNA replication C) Glycosylation D) Pore formation E) Translation

A. This patient's clinical findings of a painless genital ulcer followed several weeks later by a diffuse maculopapular rash involving the palms and soles is consistent with a diagnosis of syphilis. Syphilis is caused by an infection with the spirochete Treponema pallidum. It demonstrates multiple stages with varying symptoms, including primary with a painless chancre, secondary with a fever, a diffuse maculopapular rash that involves that palms and soles, and condylomata lata, and tertiary with tabes dorsalis, aortitis, and gummas. First-line treatment for syphilis is with penicillin. Penicillin binds to bacterial transpeptidase and interferes with bacterial cell wall synthesis through the inhibition of peptidoglycan cross-linking. Cephalosporins and carbapenems exert their antimicrobial effects through a similar mechanism of action. Incorrect Answers: B, C, D, and E. Inhibition of DNA replication (Choice B) is the mechanism of action of quinolone antibiotics. Quinolone antibiotics inhibit bacterial DNA topoisomerase II, also known as DNA gyrase, thereby preventing DNA unwinding and replication. Inhibition of glycosylation (Choice C) is another mechanism of interfering with the synthesis of peptidoglycan, which is rich in glycosylated proteins. This is the mechanism of vancomycin; by binding to D-Ala-D-Ala motifs, it prevents the transglycosylation of n-acetylglucosamine and n-acetylmuramic acid necessary for bacterial cell wall integrity. Vancomycin is not used for the treatment of syphilis. Inhibition of pore formation (Choice D) is demonstrated by some antibiotics. For example, by inhibiting bacterial protein synthesis, clindamycin inhibits the production of bacterial toxins. In particular, clindamycin decreases group A streptococcal production of streptolysin O and streptolysin S, which form pores in red blood cells. Clindamycin is not used for treating syphilis. Inhibition of translation (Choice E) is the mechanism of action of several classes of antibiotics, including macrolides, aminoglycosides, and tetracyclines. Macrolides inhibit the bacterial ribosomal 50s subunit, while aminoglycosides and tetracyclines inhibit the 30s subunit. Educational Objective: Penicillin is the first-line treatment for syphilis. Penicillin binds to bacterial transpeptidase and interferes with the ability of the bacterial cell wall to form cross-links in peptidoglycan. %3D Previous Next Score Report Lab Values Calculator Help Pause

83 Exam Section 2: Item 33 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 33. A 69-year-old man is admitted to the hospital after having a myocardial infarction. He cannot take aspirin because of a history of bronchospasm. At discharge, a substitute drug that inhibits platelet function is prescribed. This drug most likely has which of the following mechanisms of action? A) Activates adenosine deaminase B) Blocks the ADP receptors C) Blocks serotonin release D) Inhibits gamma-carboxylation of coagulation factors O E) Inhibits prostaglandin synthesis

B. Acute coronary syndrome and myocardial infarction most commonly occur secondary to atherosclerotic plaque rupture and thrombus formation in the coronary arteries, which occlude the vessels. Deposition of cholesterol in the endothelial walls promotes inflammatory cell migration and formation of an atherosclerotic plaque characterized by fibroblasts, smooth muscle cells, and lipid-laden macrophages that transform into foam cells. A fibrous cap forms over the plaque, which may rupture because of stress depending on its thickness and stability. This exposes subendothelial collagen and initiates thrombus formation. Antiplatelet agents are indicated in the treatment of acute coronary syndrome to decrease the risk for thrombus progression. ADP receptor (also called P2Y 12 receptor) antagonists are commonly utilized antiplatelet agents for this indication. Clopidogrel and ticagrelor are examples of agents in this class. These drugs block ADP receptors, impairing platelet aggregation by decreasing the expression of glycoprotein Ilb/Illa. Incorrect Answers: A, C, D, and E. Activates adenosine deaminase (Choice A) is incorrect as this is not a known mechanism of antiplatelet agents. Adenosine is an important physiologic inhibitor of platelet activity through G protein-coupled receptor signaling. Adenosine deaminase is the main enzyme responsible for adenosine degradation, and deficiency of it is associated with lymphocyte dysfunction in severe combined immunodeficiency. Blocks serotonin release (Choice C) is incorrect as this is not a known mechanism of antiplatelet agents. Serotonin is contained within platelets in dense granules and is released upon activation, enhancing platelet aggregation. Inhibits gamma-carboxylation of coagulation factors (Choice D) is the mechanism of vitamin K antagonists such as warfarin. These medications are used in the treatment of other thromboembolic diseases, such as deep venous thrombosis and pulmonary embolism, but they are not indicated for acute coronary syndrome. Inhibits prostaglandin synthesis (Choice E) is the mechanism of aspirin and cyclooxygenase (COX) inhibitors. COX is responsible for producing arachidonic acid derivatives, and dysregulation of this process can exacerbate symptoms of asthma. A drug with this activity would be contraindicated in this patient with a history of bronchospasm. Educational Objective: Antiplatelet therapy is indicated in acute coronary syndrome to decrease thrombus progression. Multiple agents are available. ADP receptor blockers (also called P2Y 12 receptor blockers) inhibit platelet aggregation by decreasing the expression of glycoprotein Ilb/Illa on the platelet surface. Previous Next Score Report Lab Values Calculator Help Pause

15 Exam Section 1: Item 15 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 15. A 17-year-old boy has had fever, malaise, and a sore throat for 1 week. Laboratory studies show: Hemoglobin Hematocrit 10 g/dL 30% Leukocyte count Erythrocyte count Mean corpuscular volume 85 um3 Platelet count 70,000/mm3 3.52 million/mm3 90,000/mm3 A representative leukocyte from a peripheral blood smear is shown. Which of the following is the most likely diagnosis? A) Acute lymphocytic leukemia B) Acute myelocytic leukemia C) Chronic lymphocytic leukemia D) Infectious mononucleosis E) Leukemoid reaction

B. Acute myelocytic leukemia (AML) is the most likely diagnosis in this patient with anemia, thrombocytopenia, profound leukocytosis, and malignant blasts present on the peripheral smear. White blood cells are divided into two distinct lineages: lymphoid and myeloid, each with their own unique morphological features. Myeloid stem cells give rise to erythrocytes, neutrophils, basophils, monocytes, megakaryocytes, and eosinophils, while lymphoid stem cells give rise to B and T lymphocytes. AML is the malignant proliferation of a monoclonal population of myeloblasts. Itis most common in adults but may occasionally occur in children. Common presenting symptoms include malaise, weight loss, and easy bruising, and laboratory studies often show anemia and thrombocytopenia with leukocytosis. Peripheral smear can distinguish between AML and acute lymphoblastic leukemia as the cells have different morphologic characteristics. Myeloblasts tend to have a lower nuclear to cytoplasm ratio and are larger as compared with lymphoblasts. The presence of Auer rods is strongly suggestive of acute promyelocytic leukemia (APML), but it can occur in subtypes of AML as well. Incorrect Answers: A, C, D, and E. Acute lymphocytic leukemia (Choice A) is more common in pediatric patients than AML and presents with similar symptoms, including weight loss, fevers, and malaise. Anemia and thrombocytopenia occur. Circulating blasts appear different, and tend to be smaller and darker, with a prominent nucleus and an increased nuclear to cytoplasm ratio. Chronic lymphocytic leukemia (Choice C) most commonly occurs in adults over 60 years old. Patients are often asymptomatic and show slow progression, with characteristic ruptured lymphocytes on peripheral blood smear. Diffuse nontender lymphadenopathy is a characteristic feature. Infectious mononucleosis (Choice D) is caused either by Epstein-Barr virus (EBV) or cytomegalovirus (CMV). Clinically, the syndrome commonly presents with fever, lymphadenopathy (typically involving the posterior cervical lymph nodes), and hepatosplenomegaly along with pharyngitis. Peripheral smear will show atypical lymphocytes, not myeloblasts. Leukemoid reaction (Choice E) is a response to stress or infection that results in a significant rise in peripheral circulating leukocytes. It can sometimes be confused with acute leukemia but is differentiated by the fact that, in a leukemoid reaction, leukocytes are mature and functional, and they show increased leukocyte alkaline phosphatase activity. Educational Objective: AML commonly presents with malaise, fever, weight loss, thrombocytopenia, anemia, and leukocytosis. Peripheral blood smear shows circulating myeloblasts, which are characterized by a lower nuclear to cytoplasmic ratio than lymphoblasts. Auer rods are occasionally also seen. %3D Previous Next Score Report Lab Values Calculator Help Pause

88 Exam Section 2: Item 38 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 38. An investigator is studying the efficacy of distinct vaccine formulations directed against the capsular polysaccharides of Neisseria meningitidis. The carbohydrates are chemically conjugated to various compounds and injected into laboratory mice. Titers of anticapsular IgG antibodies are then measured. Which of the following compounds is most likely to induce increased titers of these antibodies when conjugated to the polysaccharides? A) Dinitrophenol B) Flagellin C) Lactose D) Lipopolysaccharide E) Palmitic acid OF) Polyinosinic acid-polycytidylic acid

B. Flagellin is a highly conserved protein present on the surface of many bacterial cells that aids in motility and adhesion. While Neisseria meningitidis does not possess flagellin, conjugating polysaccharide components unique to N. meningitidis to flagellar components can induce a robust and lasting immune response to these specific polysaccharides. Many flagellin proteins comprise a flagellum, which includes both intracellular and extracellular components that are recognized by the immune system. Flagellin proteins bind to toll-like receptor 5 (TLR5) and initiate signaling cascades that stimulate antigen-presenting cells and activate the inflammasome. This results in a mixed Th, and Th2 immune response which allows for lasting and robust immunity. Incorrect Answers: A, C, D, E, and F. Dinitrophenol (Choice A) uncouples oxidative phosphorylation in mitochondria. It was classically used as a weight loss supplement but can result in severe toxicity. It is not used in vaccine formulations. Lactose (Choice C) is a component present in several vaccines including those for adenovirus and cholera, but it is used as a stabilizer. It does not have immunogenic properties. Lipopolysaccharide (LPS) (Choice D) is a highly immunogenic component of the bacterial cell wall in gram-negative bacteria. Despite its immunogenic properties, several components of the LPS are similar to moieties found on the surface of normal human cells, including erythrocytes. This property aids N. meningitidis in escaping the immune system. Palmitic acid (Choice E) is formed during fatty acid synthesis. It is a precursor to long fatty acids. It would not be an ideal component in a vaccine formulation as it is found ubiquitously in the human body. Polyinosinic acid-polycytidylic acid (Choice F) stimulates the immune system and is commonly used in research. It is similar in structure to dsRNA present in some viruses, and it binds to toll-like receptor 3 expressed on antigen-presenting cells and B lymphocytes. Educational Objective: Flagellin is a conserved bacterial component that possesses unique immunogenic properties by stimulating TLR5 and inducing a Th1- and Th2-mediated immune response. Conjugating unique polysaccharide components from N. meningitidis to flagellin would be most likely to induce a robust immune response with the formation of lasting IgG antibodies. Previous Next Score Report Lab Values Calculator Help Pause

26 Exam Section 1: Item 26 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 26. A 95-year-old man who is a resident of a skilled nursing care facility is brought to the emergency department because of a 1-day history of temperatures to 39.4°C (102.9°F), headache, cough, and muscle aches. Several other residents have similar symptoms. Physical examination shows no other abnormalities. Treatment with a neuraminidase inhibitor is begun. This drug most likely will inhibit which of the following processes, therefore decreasing the duration of this patient's symptoms? A) Nucleocapsid-matrix protein interactions B) Release of virus from infected epithelial cells C) Replication of genomic RNA D) Synthesis of surface glycoproteins E) Transcription of MRNA

B. Influenza virus is a contagious orthomyxovirus that is transmitted via respiratory droplets and contact with contaminated surfaces. Infection classically presents with respiratory, gastrointestinal, and systemic symptoms, including cough, nausea, vomiting, diarrhea, fever, chills, headache, myalgias, and arthralgias; it can be complicated by bacterial superinfection causing severe bacterial pneumonia. The virus initially binds to sialic acid residues on respiratory epithelial cells using the viral glycoprotein hemagglutinin, which allows for membrane fusion and viral entry into the cell. The virus is transported to the host cell nucleus where its genome is replicated. The newly synthesized viral particles are transported back to the cell membrane, and a membrane bud forms. Viral neuraminidase cleaves the sialic acid residues, allowing the release of virus from the infected epithelial cells. Neuraminidase inhibitors such as oseltamivir inhibit this process and may decrease the duration of symptomatic illness. Influenza virus is responsible for seasonal pandemics, and vaccination is recommended for all individuals aged 6 months or older on an annual basis. Incorrect Answers: A, C, D, and E. Nucleocapsid-matrix protein interactions (Choice A) are important in the lifecycle of retroviruses such as HIV. The gag gene encodes a polyprotein that is cleaved into nucleocapsid and matrix proteins (p24 and p17, respectively). These interactions are not affected by neuraminidase inhibitors. Replication of genomic RNA (Choice C) is performed by viral RNA-dependent RNA polymerases. Neuraminidase inhibitors do not inhibit this process. Synthesis of surface glycoproteins (Choice D) is impaired by protease inhibitors, which are a common component of highly active antiretroviral therapy (HAART) for HIV. The HIV-1 protease cleaves synthesized proteins into their functional parts. Transcription of MRNA (Choice E) is an essential step in the replication of DNA viruses and retroviruses. Orthomyxoviruses, such as influenza, are RNA viruses, which are directly translated in the cytoplasm. Educational Objective: Neuraminidase and hemagglutinin are the two major virulence factors of influenza viruses. Neuraminidase is a surface glycoprotein that allows for the release of virus from infected cells. Neuraminidase inhibitors are used in the treatment of influenza infection. Previous Next Score Report Lab Values Calculator Help Pause

78 Exam Section 2: Item 28 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 28. A7-year-old boy is brought to the physician by his parents because of a 1-month history of progressive double vision and a 2-week history of imbalance. Physical examination shows weakness of abduction of the right eye, mild weakness of the left extremities, increased muscle stretch reflexes of the left extremities, and gait ataxia. Which of the following is the most likely diagnosis? A) Arnold-Chiari malformation B) Brain stem glioma C) Guillain-Barré syndrome D) Multiple sclerosis E) Myasthenia gravis

B. This patient most likely has a brain stem glioma. Gliomas are the most common pediatric brain tumor and are typically low-grade pilocytic astrocytomas. Pediatric gliomas commonly grow in the cerebellum but can occur in the brain stem. Gliomas typically present with focal neurologic deficits caused by mass effect. These deficits may progress over the course of months as the tumor grows larger and affects additional brain regions. This patient's brainstem glioma is likely affecting several structures within or adjacent to the right-sided anterior pons: the right abducens nerve as it exits the pons (weakness of right eye abduction), the right-sided pontocerebellar tract (gait ataxia), and the right-sided corticospinal tract (left-sided weakness and hyperreflexia). Treatment centers around surgical resection. Incorrect Answers: A, C, D, and E. Arnold-Chiari malformations (Choice A) are rare congenital malformations that typically present in infancy with feeding and breathing problems, myelomeningocele, and progressive hydrocephalus. Imaging shows downward displacement of the cerebellar vermis. Some Arnold-Chiari malformations present in adolescence with mostly bilateral cranial nerve palsies, weakness, and cerebellar dysfunction caused by the downward displacement of the cerebellar tonsils and consequent brainstem compression. However, this 7-year-old patient's unilateral deficits are more consistent with a brain stem glioma, and gliomas are more common than Arnold-Chiari malformations. Guillain-Barré syndrome (Choice C) arises from the autoimmune attack of the spinal nerve roots and typically presents with acute-onset, ascending, bilateral muscle weakness with hyporeflexia. Many patients also show autonomic dysfunction and/or sensory symptoms. This patient's insidious onset of eye movement abnormalities, upper motor neuron-pattern of dysfunction, and ataxia are more consistent with a brain stem glioma. Multiple sclerosis (Choice D) is an autoimmune demyelinating disease that presents with focal neurologic deficits separated in space and time. Multiple sclerosis would affect separate brain regions rather than multiple areas in the anterior pons and typically occurs in patients who are older than 20 years of age. Myasthenia gravis (Choice E) is an autoimmune disorder of neuromuscular transmission that results in progressive muscle weakness and fatigability. The extraocular muscles may be affected. However, an age of onset before the teenage years, gait ataxia, and an upper motor neuron pattern of dysfunction would be atypical of myasthenia gravis. Educational Objective: Gliomas are the most common pediatric brain tumor and are typically low-grade pilocytic astrocytomas that may occur in the cerebellum or brain stem. Gliomas commonly present with focal neurologic deficits caused by mass effect; these may progress over the course of months as the tumor grows larger and affects additional brain regions. II Previous Next Score Report Lab Values Calculator Help Pause

16 Exam Section 1: Item 16 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 16. A 50-year-old man with a temperature of 38.9°C (102°F) has produced 120 mL of rust-colored sputum over the past 2 days. Which of the following physical findings most clearly indicates a diagnosis of pneumonia in the right lower lobe? A) Hyperresonance on percussion over the right lower lung field B) Increased tactile fremitus over the right lower lung field C) Muffled whispered sounds over the right lower lung field D) A shift of the trachea to the right E) Vesicular breath sounds over the right lower lung field

B. This patient's findings of fever and productive cough are concerning for pneumonia. In addition to typical radiographic findings, pneumonia can be identified by a variety of bedside examination techniques. Tactile fremitus relies on the increased transmission of sound waves and vibrations through consolidated lung tissue as compared with normal, aerated lung tissue. The examiner will note increased vibrations against the hand when the patient vocalizes diphthong phrases. Increased vibration is indicative of the presence of consolidation and pneumonia underlying the region of the examiner's hands. Incorrect Answers: A, C, D, and E. Hyperresonance on percussion (Choice A) may be noted in situations when there is increased air beneath the percussed area, such as in patients with emphysema or pneumothorax. By contrast, consolidation is characterized by dullness to percussion because the increased density of consolidated lung tissue produces less resonance (eg, in pneumonia). Muffled whispered sounds over the right lower lung field (Choice C) is a normal finding. Voice should typically be muffled during lung auscultation. If the whispered voice sound is clear and easily audible it may indicate the presence of consolidation and pneumonia, a finding known as whispered pectoriloquy. A shift of the trachea to the right (Choice D) is found in a left tension pneumothorax or a large left hemothorax or pleural effusion. It can also occur in a right-sided complete lung collapse. Pneumonia does not typically lead to positional shifts of mediastinal structures, and if this were to occur, the finding would be nonspecific. Vesicular breath sounds (Choice E) are a normal finding. Pneumonia is characterized by rales and rhonchi on pulmonary auscultation. Educational Objective: Tactile fremitus relies on the increased transmission of sound waves and vibrations through consolidated lung tissue as compared with normal, aerated lung tissue. The examiner will note increased vibration against the hand when the patient vocalizes diphthong phrases, indicative of the presence of consolidation and pneumonia. %3D Previous Next Score Report Lab Values Calculator Help Pause

38 Exam Section 1: Item 38 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 38. A 15-year-old boy has nasal polyps and develops bronchospasms after taking a drug. The drug is most likely to inhibit which of the following enzymes in the biochemical scheme shown? Phospholipid Enzyme A Arachidonate Enzyme B Enzyme C Cyclic endoperoxides 5-Hydroperoxyeicosatetraenoic acid Enzyme D Enzyme E Prostacyclin Thromboxane A2 A) B) C) D) E)

B. This patient's presenting findings, including asthma, exacerbation of respiratory symptoms by a medication (likely aspirin), and nasal polyps, are consistent with a diagnosis of aspirin- exacerbated respiratory disease (AERD). Aspirin is a nonsteroidal anti-inflammatory drug that nonselectively and irreversibly inhibits cyclooxygenase 1 and 2, thereby leading to decreased conversion of arachidonate to cyclic endoperoxides and, subsequently, to prostacyclins and thromboxanes. Some patients develop AERD as a hypersensitivity reaction to aspirin. In affected patients, the blockade of cyclooxygenases leads to increased concentrations of arachidonates, which are then converted to leukotrienes by 5-lipoxygenase. Several leukotrienes, including LTC LTD4 and LTE4, increase bronchial tone and produce bronchospasm; they also lead to the production of proinflammatory cytokines. Patients are treated by desensitization to aspirin and may be placed on a chronic maintenance dose of aspirin to maintain the desensitization. Incorrect Answers: A, C, D, and E. Enzyme A (Choice A) corresponds to phospholipase A, which converts phospholipids to arachidonate. Phospholipase A, is inhibited by glucocorticoids. Enzyme C (Choice C) corresponds to 5-lipoxygenase, which converts arachidonate to 5-hydroperoxyeicosatetraenoic acid and, subsequently, to leukotrienes. This reaction is inhibited by the drug zileuton. Enzyme D (Choice D) corresponds to prostacyclin synthetase. This enzyme is not a common pharmacologic target within the arachidonate pathway. Enzyme E (Choice E) corresponds to thromboxane synthetase. This enzyme is inhibited by the antiplatelet agent picotamide. Educational Objective: AERD is defined by the presence of asthma, exacerbation of respiratory symptoms by aspirin, and nasal polyps. Aspirin is a nonsteroidal anti-inflammatory drug that nonselectively and irreversibly inhibits cyclooxygenase 1 and 2, thereby leading to the decreased conversion of arachidonate to cyclic endoperoxides and, subsequently, to prostacyclins and thromboxanes. Previous Next Score Report Lab Values Calculator Help Pause

141 Exam Section 3: Item 41 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 41. A 42-year-old man is brought to the emergency department 1 hour after police found him wandering the street. He has a 5-year history of gout. There is no odor of alcohol, but he is drowsy and ataxic. Physical examination shows severe jaundice, prominent periumbilical veins, and several cutaneous spider angiomata. Abdominal examination shows splenomegaly and ascites. There is a flapping, up-and-down motion of the hands when his arms are outstretched horizontally. His hemoglobin concentration is 10.8 g/dL. Test of the stool for occult blood is positive. The arm motions and drowsiness are best explained by which of the following? A) Alcohol intoxication B) Anemia C) Hyperammonemia OD) Hyperbilirubinemia E) Hyperuricemia

C. Cirrhosis can present with edema, ascites, increased bilirubin, jaundice, spider angiomata, and sequelae of portal hypertension such as esophageal varices, splenomegaly, caput medusae, rectal varices, and hepatic encephalopathy (HE). It typically occurs in patients with preceding conditions such as alcohol abuse or chronic hepatitis. In portal hypertension, there is the potential for retrograde blood flow between tributaries of the portal system and systemic veins, forming portosystemic anastomoses. This results in the formation of esophageal varices, prominent periumbilical veins (caput medusae), and rectal varices. Another consequence of portosystemic shunting in cirrhosis is HE. HE occurs as a result of ammonia (NH) build up in the body because of decreased NH, metabolism. In cirrhosis, the blood carrying NH3 does not entirely pass through the liver; rather, it enters directly into the systemic circulation because of portosystemic shunting of blood. When NH3 accumulates, it causes neuropsychiatric disturbances ranging from subtle cognitive defects like impaired attention span and memory deficits to altered mental status and coma. Asterixis, a flapping tremor in which the hands move up and down when the arms are outstretched, is another clinical manifestation of hyperammonemia. While the measurement of serum ammonia is commonly performed, venous levels do not reliably correlate with the degree of encephalopathy. Lactulose is used to prevent further uptake of NH3 by trapping ammonia in the colon. Degradation of lactulose by colonic bacteria generates hydrogen ions that create an acidic environment within the gut. The abundance of extra hydrogen ions converts ammonia (NH3) to ammonium (NH4), which cannot freely be absorbed by the bowel and diffuses into the bloodstream, causing it to be readily excreted in the feces. Lactulose is given prophylactically to many patients with cirrhosis to prevent the development of HE and is titrated to a target number of bowel movements daily. Incorrect Answers: A, B, D, and E. Alcohol intoxication (Choice A) may cause emotional lability, slurred speech, ataxia, and even coma. While it may be a potential cause of the patient's drowsiness and ataxia, it does not explain his asterixis. Anemia (Choice B) caused by a vitamin B 12 deficiency may present with neurologic symptoms. This condition is called subacute combined degeneration, and it occurs because of demyelination of the spinocerebellar tracts, lateral corticospinal tracts, and dorsal columns. While ataxia may be seen in this type of anemia, asterixis is not. Hyperbilirubinemia (Choice D) causes scleral icterus and jaundice, which are seen in this patient, but it is not responsible for causing asterixis or HE. Cirrhosis of the liver prevents bile, which contains bilirubin, from moving down the bile canaliculi into the biliary tract. It instead spills into the bloodstream and deposits in the tissues, causing yellowing of the skin and eyes. Hyperuricemia (Choice E) may be seen in gout but is not a diagnostic criterion because it is often normal. Despite this patient's history of gout, his acute presentation is much more related to his cirrhosis and the accumulation of NH3 Educational Objective: One consequence of portosystemic shunts created by liver cirrhosis is hepatic encephalopathy caused by decreased metabolism of NH, When NH, accumulates, it causes neuropsychiatric disturbances ranging from subtle cognitive defects, like impaired attention span and memory deficits, to altered mental status and coma. Asterixis is a physical manifestation of hyperammonemia. Previous Next Score Report Lab Values Calculator Help Pause

176 Exam Section 4: Item 26 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 26. A 57-year-old man undergoes coronary artery bypass grafting. The stress of the operation results in increased secretion of cortisol from the adrenal gland postoperatively. Which of the following cells is the most likely origin of the stimulus of this cortisol secretion? A) Acidophil of the anterior pituitary gland B) Alpha cell of the islet of Langerhans C) Basophil of the anterior pituitary gland D) Beta cell of the islet of Langerhans E) Delta cell of the islet of Langerhans F) Pituicytes of the posterior pituitary gland

C. Cortisol is produced by the adrenal zona fasciculata under the control of the hypothalamic-pituitary-adrenal axis. The two important upstream hormones within this axis include hypothalamic corticotropin-releasing hormone, which is produced by neurosecretory cells within the paraventricular nucleus, and adrenocorticotropic hormone, which is produced under the direction of corticotropin-releasing hormone by basophil cells of the anterior pituitary gland. Basophil cells, some of which are corticotropic cells, produce adrenocorticotropic hormone through the cleavage of pro-opiomelanocortin. Basophil cells of the anterior pituitary gland are also responsible for the production of follicle-stimulating hormone, luteinizing hormone, and thyroid-stimulating hormone. Incorrect Answers: A, B, D, E, and F. Acidophils of the anterior pituitary gland (Choice A) include both somatotropic and lactrotropic cells, which are responsible for producing growth hormone and prolactin, respectively. Alpha cells of the islet of Langerhans (Choice B) produce glucagon, a catabolic hormone that is responsible for increasing serum glucose concentration. Beta cells of the islet of Langerhans (Choice D) produce insulin, an anabolic hormone that is responsible for decreasing serum glucose concentration. Delta cells of the islet of Langerhans (Choice E) produce somatostatin, a hormone that is responsible for decreasing the gastric and pancreatic secretion of digestive hormones and for decreasing gastrointestinal motility. Pituicytes of the posterior pituitary gland (Choice F) are glial cells that regulate the release of oxytocin and vasopressin from the posterior pituitary. Educational Objective: Adrenocorticotropic hormone is produced by basophil cells of the anterior pituitary gland. Basophil cells produce adrenocorticotropic hormone through the cleavage of pro- opiomelanocortin. Adrenocorticotropic hormone promotes the release of cortisol from the adrenal gland. Previous Next Score Report Lab Values Calculator Help Pause

184 Exam Section 4: Item 34 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 34. A 44-year-old man with a 1-year history of angina pectoris comes to the emergency department because of increasingly severe chest pain during the past 2 days. He has had five previous similar episodes, which required treatment with increasing doses of nitroglycerin to resolve. His temperature is 37°C (98.6°F), pulse is 105/min, respirations are 16/min, and blood pressure is 150/90 mm Hg. Cardiac examination shows an S, An ECG shows ST-segment depression in the precordial leads. In addition to aspirin, heparin, and nitroglycerin, he is given a dose of a monoclonal antibody against the platelet Ilb/Illa receptor. This antibody will most likely prevent binding of which of the following substances to platelets? A) Adenosine B) ADP C) Fibrinogen D) Serotonin E) Thrombin F) Thromboxane A2

C. Fibrinogen binding to platelets will be prevented by administration of abciximab, a monoclonal antibody against the platelet Ilb/Illa receptor. Following rupture of unstable plaque within a coronary artery, exposure of the denuded endothelial surface initiates the process of clot formation. Platelets bind and adhere to the endothelial surface through cell surface receptor interactions with the endothelium and von Willebrand factor (vWF). However, further aggregation of platelets is required to form a stable clot. Platelet activation after binding initiates signaling cascades, which result in conformational changes to the cytoplasmic component of the Gpllb/illa receptor. Such changes allow for the extracellular component of this receptor to bind fibrinogen, which in turn allows platelets to link together. Inhibition of the extracellular component of the Gpllb/illa receptor by a monoclonal antibody effectively prevents platelets from binding to one another using fibrinogen as a bridge, and thus, it prevents further clot propagation. Incorrect Answers: A, B, D, E, and F. Adenosine (Choice A) is an inhibitor of platelet aggregation. It binds to adenosine receptors and increases the intracellular concentration of adenosine monophosphate, which inhibits platelet activation. It does not interact directly with the Gpllib/lla receptor. ADP (Choice B) acts as a stimulator of platelet aggregation and clot formation. Binding of ADP to platelet receptors induces conformational changes in platelets leading to release of platelet granules. Additionally, ADP induces platelets to form thromboxane A, (Choice F) via the action of cyclooxygenase and thromboxane A, synthase. Thromboxane A, is a potent stimulator of further platelet activation and recruitment. Neither of these molecules is the target of monoclonal antibodies against the Gpllb//lla receptor. Serotonin (Choice D) is contained within platelets and is released upon activation. It causes vasoconstriction and enhances platelet aggregation. It is not the target of abciximab. Thrombin (Choice E) is formed from prothrombin in a reaction requiring factor Xa and Va at the terminal part of the coagulation cascade. Thrombin, in turn, is required for the formation of fibrin from fibrinogen, which stabilizes the newly formed clot. Thrombin is inhibited by direct thrombin inhibitors such as argatroban. Educational Objective: Damaged and exposed endothelium binds and causes the activation of platelets. Subsequently, conformational changes to the Gpllb/Illa receptor uncover the extracellular domain, allowing fibrinogen to bind. Fibrinogen acts as a crosslink between multiple activated platelets and is critical in clot formation. Monoclonal antibodies against the extracellular domain of the Gpllb/llla receptor prevent the binding of fibrinogen. Previous Next Score Report Lab Values Calculator Help Pause

60 Exam Section 2: Item 10 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 10. An 82-year-old woman comes to the physician because of constant severe lower abdominal pain and fever for 24 hours. Laparoscopic examination shows severe diverticulosis and perforated diverticulitis. In spite of appropriate therapy, she dies 2 days later. Her liver at autopsy is shown. Which of the following is the primary component of the material shown on the hepatic surface? A) Collagen, type I B) Collagen, type III C) Fibrin D) Fibronectin O E) Proteoglycans

C. Fibrin deposition as a result of secondary bacterial peritonitis from a perforated viscus is the most likely material observed on the surface of the liver. In cases of perforated diverticulitis, intestinal contents including a variety of pathogenic microbes are able to enter the normally sterile peritoneal cavity. This induces a robust immune response against these organisms, and the production of fibrinous exudate is one component of this response. Fibrin acts to trap and sequester bacteria within the peritoneal cavity and prevents their spread to other vital organs. However, this also walls off bacteria from access by immune cells and may contribute to ongoing infection with the potential to develop intra-abdominal abscesses. Incorrect Answers: A, B, D, and E. Collagen, type I (Choice A) is the most common variety of collagen found in the human body and is a structural component of scar tissue, ligaments, tendons, and skin. Collagen, type I (Choice B) is found in the extracellular matrix (ECM) along with type I collagen and is secreted by fibroblasts. It is implicated in pathologic fibrosis of various organs in response to infection or autoimmune disease and is deficient in the Ehlers-Danlos syndrome vascular subtype. It is also a component of granulation tissue. Fibronectin (Choice D) is a component of the ECM that is critical in cellular adhesion and migration through interaction with integrins. It also interfaces with collagen and fibrin. Proteoglycans (Choice E) are a component of the ECM and are complexes of large sugar moieties and proteins. They form large complexes and interact with collagen to form cartilage. Educational Objective: The introduction of pathogenic bacteria into the normally sterile environment of the peritoneum induces a robust inflammatory response with the secretion of fibrin that subsequently traps and encases bacteria. While important in limiting the spread of infection, it can also predispose to the formation of abscesses. Previous Next Score Report Lab Values Calculator Help Pause

67 Exam Section 2: Item 17 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 17. During an investigational study, antibodies are induced that bind to and inactivate inhibin in female experimental animals. On examination of these animals, which of the following is the direct result of the inactivation of inhibin? A) Decreased serum follicle-stimulating hormone concentration B) Decreased serum gonadotropin-releasing hormone concentration C) Increased serum follicle-stimulating hormone concentration D) Increased serum gonadotropin-releasing hormone concentration E) Theca cell atrophy F) Theca cell hypertrophy

C. Inhibin is secreted by both granulosa cells of the ovary in women and the Sertoli cells of the testis in men. It acts to inhibit follicle-stimulating hormone (FSH) at the level of the pituitary. Inhibin is also secreted by the placenta during pregnancy and is measured in the maternal quad screen to evaluate for chromosomal disorders. Thus, an antibody that inactivates inhibin would result in increased serum FSH concentration, as the negative feedback of inhibin on FSH release would be removed. Incorrect Answers: A, B, D, E, and F. Decreased serum follicle-stimulating hormone concentration (Choice A) would be caused by an increased inhibin concentration or secondary to a decrease in gonadotropin-releasing hormone (GNRH) concentration, such as in hypogonadotropic hypogonadism or hyperprolactinemia. These animals received an antibody that inactivates inhibin, rendering it ineffective, which would result in an increased serum FSH concentration. Decreased serum gonadotropin-releasing hormone concentration (Choice B) would be present in hypogonadotropic hypogonadism and hyperprolactinemia. Increased serum gonadotropin- releasing hormone concentration (Choice D) is present in central precocious puberty. Inhibin does not directly affect the release or concentration of GNRH. Theca cell atrophy (Choice E) and hypertrophy (Choice F) would not be caused by changes in serum inhibin concentration or activity. While inhibin does increase androgen production in the theca cells, it does not affect theca cell growth. Educational Objective: Inhibin is released by the granulosa cells of the ovary and Sertoli cells of the testis. It primarily acts to inhibit the production and release of FSH. Previous Next Score Report Lab Values Calculator Help Pause

77 Exam Section 2: Item 27 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment LR HEL 27. A 24-year-old man comes to the emergency department because of a 2-hour history of severe abdominal pain. His pulse is 110/min. Abdominal examination shows distention. An x-ray of the abdomen while the patient is sitting upright and an image from upper gastrointestinal series are shown. Based on the findings shown, this patient most likely has which of the following developmental anomalies? A) Gastroschisis B) Intestinal aganglionosis C) Malrotation D) Persistent vitelline duct E) Situs inversus viscerum

C. Malrotation is a congenital anomaly in which the midgut incompletely rotates during development. This leaves the midgut attached to a thin stalk of mesentery at the base of the superior mesenteric artery (SMA) instead of the normally broad-based mesenteric stalk. The cecum also remains in an abnormal position within the mid-upper abdomen instead of the right lower quadrant and is attached to the lateral abdominal wall by fibrous bands that cross over the top of the duodenum and can also cause constriction. This leaves the small intestine at risk for volvulus originating at the base of the SMA. While most patients present in the first year of life, some individuals do not present until later childhood or adulthood. In this case, the patient's upright abdominal x-ray demonstrates air-fluid levels throughout the small intestine, suggesting a small bowel obstruction. Additionally, the supine upper gastrointestinal barium series demonstrates the presence of the small intestine only within the right abdomen. In its usual anatomic position, the duodenum crosses the midline toward the ligament of Treitz and the remaining small intestine courses throughout the abdominal quadrants. Thus, these imaging findings suggest malrotation of the small intestine, complicated by midgut volvulus and a small bowel obstruction. Incorrect Answers: A, B, D, and E. Gastroschisis (Choice A) is a ventral wall defect in which the abdominal contents herniate through the abdominal folds, typically to the right of the umbilicus. It is typically identified on prenatal ultrasound. The viscera outside the abdominal cavity are not covered by peritoneum and urgent surgical intervention is required following birth. Intestinal aganglionosis (Choice B), also known as Hirschsprung disease, is caused by the congenital absence of the distal portion of the myenteric plexus, a part of the enteric nervous system located between the inner and outer layers of the muscularis. This often leads to an inability to pass meconium within the first few days of life. An explosive expulsion of feces on rectal examination is a classic clinical finding. Persistent vitelline duct (Choice D) is the cause of a Meckel diverticulum. This is the most common congenital anomaly of the gastrointestinal tract and is characterized by a small diverticulum projecting from the terminal ileum that is approximately two inches in length and two feet proximal to the ileocecal valve. The diverticulum may contain ectopic gastric mucosa, which secretes gastric acid, leading to erosion of the intestinal mucosa and subsequent painless hematochezia or melena. The diverticulum may also act as a lead point for intussusception, obstruction, or volvulus. Situs inversus viscerum (Choice E) is a finding of Kartagener syndrome, or primary ciliary dyskinesia. The viscera are completely reversed from the standard anatomical laterality. In this case, the laterality of the abdominal viscera is anatomically appropriate, reflected by the liver shadow in the right upper quadrant of the abdomen on x-ray. Educational Objective: Malrotation is a congenital anomaly in which the midgut incompletely rotates during development, leaving the midgut attached to a thin stalk of mesentery at the base of the SMA instead of the normally broad-based mesenteric stalk. It can be recognized on imaging by the presence of small intestine in the right abdomen only. Malrotation increases the risk for volvulus of the small bowel at the base of the SMA (midgut volvulus) causing small bowel obstruction. Previous Next Score Report Lab Values Calculator Help Pause

64 Exam Section 2: Item 14 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 14. A 52-year-old woman has a 10-year history of recurrent urinary tract infections. Urinalysis shows: pH Specific gravity Blood 7.6 1.019 1+ Glucose Protein Ketones none none none Microscopic analysis of urine sediment shows numerous erythrocytes, leukocytes, and crystals. Her symptoms do not respond to therapy, and she undergoes a left nephrectomy. A photograph of the resected specimen is shown. Which of the following is the most likely causal agent? A) Escherichia coli B) Mycobacterium tuberculosis C) Proteus vulgaris D) Pseudomonas aeruginosa E) Streptococcus pyogenes (group A) 1 сm

C. Proteus vulgaris is a nonlactose fermenting, oxidase-negative, gram-negative rod, which frequently causes disease of the urinary tract. Proteus vulgaris and Proteus species produce urease. Urease catalyzes the conversion of urinary urea to ammonia and carbon dioxide, which results in urine alkalinization, demonstrated in this patient by increased urinary pH. The solubility of magnesium ammonium phosphate, or struvite, is decreased under alkaline conditions and precipitates as a crystal. Struvite crystals have an orthorhombic configuration under light microscopy. Struvite calculi are radiopaque and can be potentially identified by an x-ray or CT scan. Struvite calculi may adopt a branching morphology that molds itself to the shape of the collecting system. Given their large size and often ramified structure, it is rare for struvite calculi to pass through the urinary tract, and surgical removal is often necessary along with treatment of the underlying infection. Incorrect Answers: A, B, D, and E. Escherichia coli (Choice A) is a common cause of urinary tract infections such as cystitis and pyelonephritis. However, it is not associated with staghorn calculi like Proteus species. Mycobacterium tuberculosis (Choice B) is a slow-growing organism transmitted via respiratory secretions. It presents with primary, latent, and reactivation patterns. In primary disease, patients present with subacute fevers, weight loss, night sweats, cough, and malaise. It is not associated with urinary tract infections. Pseudomonas aeruginosa (Choice D) is a gram-negative rod that can result in numerous clinical infectious conditions, including urinary tract infections, osteomyelitis, otitis externa, folliculitis, hospital- or ventilator-associated pneumonia, and sepsis. Streptococcus pyogenes (group A) (Choice E) is a gram-positive bacterium that can result in various diseases, most commonly cellulitis and pharyngitis. Additional conditions related to S. pyogenes include scarlet fever, necrotizing fasciitis, glomerulonephritis, and rheumatic fever. It is not associated with urinary tract infections. Educational Objective: Struvite precipitates as renal calculi in alkaline urine, which can be caused by urinary tract infections with urease-producing organisms such as Proteus. Urease catalyzes the conversion of urinary urea to ammonia and carbon dioxide, which results in alkalinization. Struvite calculi may be large and branching, often requiring surgical removal. Previous Next Score Report Lab Values Calculator Help Pause

160 Exam Section 4: Item 10 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 10. A healthy 8-year-old boy is brought to the physician by his parents because of a long history of disruptive behavior at school and poor academic performance. During class, he talks out of turn, wanders around the room, and pesters classmates. He is easily distracted and cannot keep his attention on a topic very long. His homework is sloppy and usually incomplete; he often explains that he leaves segments of assignments undone because he "just didn't notice them." He has had similar problems since kindergarten. He has normal developmental milestones. He is bright and cheerful and has a caring family. A drug with which of the following actions is most likely to be useful for this patient? A) Stimulation of the release of excitatory amino acid neurotransmitters B) Blockade of postsynaptic excitatory amino acid neurotransmitter receptors C) Stimulation of the release of biogenic amine neurotransmitters D) Blockade of postsynaptic biogenic amine neurotransmitter receptors E) Stimulation of the release of inhibitory amino acid neurotransmitters F) Blockade of postsynaptic inhibitory amino acid neurotransmitter receptors

C. 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, and/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 dopamine reward pathway), whereas children with predominant inattentive symptoms tend to daydream, process information slowly, and have difficulty in completing tasks (related to norepinephrine dysregulation). This patient is experiencing symptoms of both inattention and hyperactivity. Combination treatment with medications and behavioral therapy may be the most effective option, though medication monotherapy is also appropriate for children older than six 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, D, E, and F. Stimulation of the release of excitatory amino acid neurotransmitters (Choice A) happens during glutamate excitotoxicity, which may occur in response to neuronal injury (eg, traumatic brain injury or stroke) and result in neuronal apoptosis. No medications utilize this mechanism of action. Blockade of postsynaptic excitatory amino acid neurotransmitter receptors (Choice B) refers to the blockade of the glutamatergic N-methyl-D-aspartic acid (NMDA) receptor. Some NMDA antagonists lead to dissociation (eg, ketamine, phencyclidine) while others may protect the brain from glutamate excitotoxicity in dementia (eg, memantine). Dopamine and norepinephrine signaling play larger roles in ADHD than glutamate signaling. Blockade of postsynaptic biogenic amine neurotransmitter receptors (Choice D) includes norepinephrine receptor antagonists (eg, B-adrenergic blockers), dopamine receptor antagonists (eg, antipsychotics), and serotonin receptor antagonists (eg, trazodone, which additionally blocks serotonin reuptake). Since patients with ADHD have deficient dopamine and norepinephrine signaling, dopamine and norepinephrine antagonists would not be helpful. Stimulation of the release of inhibitory amino acid neurotransmitters (Choice E) refers to the increased release of y-aminobutyric acid (GABA) or glycine. Though many substances modulate postsynaptic GABA receptors (eg, alcohol, benzodiazepines), no commonly used medications are known to increase presynaptic GABA release. Blockade of postsynaptic inhibitory amino acid neurotransmitter receptors (Choice F) refers to medications that antagonize GABA receptors. Flumazenil, which is utilized to reverse benzodiazepine overdose, is one example. GABA does not play a large role in the pathogenesis of ADHD. Educational Objective: The first-line treatment for ADHD is stimulant medication such as amphetamine salts and methylphenidate. These medications increase presynaptic release of dopamine and norepinephrine, which improve hyperactivity and inattention, respectively. Previous Next Score Report Lab Values Calculator Help Pause

145 Exam Section 3: Item 45 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 45. A 2-year-old boy is brought to the physician by his mother because of a 3-week history of increased irritability. He is at the 10th percentile for height, 5th percentile for weight, and 80th percentile for head circumference. Physical examination shows evidence of hydrocephalus. A CT scan of the head shows a cyst-like dilation of the fourth ventricle, hypoplastic cerebellar hemispheres, and an enlarged posterior fossa with high tentorium cerebelli. Which of the following subdivisions of the neural tube was most likely affected in this patient during prenatal development? A) Diencephalon B) Mesencephalon C) Metencephalon D) Prosencephalon E) Spinal cord O F) Telencephalon

C. The metencephalon is the embryologic origin of the pons, cerebellum, and fourth ventricle. During week four of embryonic development, the neural tube forms three primary brain vesicles: the prosencephalon (forebrain), mesencephalon (midbrain), and rhombencephalon (hindbrain). The rhombencephalon further differentiates into the metencephalon and myelencephalon during week five. Within the metencephalon, the rhombic lips fuse to form the cerebellar vermis, and over several weeks, neuroblasts from the ventricle proliferate to form the cerebellar hemispheres. In Dandy-Walker syndrome (DWS), the cerebellar vermis and fourth ventricle are malformed, resulting in cystic enlargement of the fourth ventricle and posterior fossa. The malformed fourth ventricle lacks the foramina of Magendie or Luschka, which prevents normal cerebrospinal fluid drainage and can result in hydrocephalus. Patients typically present in infancy to early childhood with developmental delay, progressive skull enlargement, noncommunicating hydrocephalus, and potentially cerebellar and cranial nerve dysfunction. DWś may be associated with other congenital defects such as spina bifida. Management is symptomatic and includes ventriculoperitoneal shunting. The mortality rate is approximately 25%. Incorrect Answers: A, B, D, E, and F. The diencephalon (Choice A) and telencephalon (Choice F) are formed from the prosencephalon (Choice D) during week five of gestation. The diencephalon ultimately forms the thalamus, hypothalamus, and third ventricle, while the telencephalon forms the cerebral hemispheres and lateral ventricles. DWS does not lead to prosencephalon malformations. The mesencephalon (Choice B) forms the midbrain and cerebral aqueduct. DWS does not lead to mesencephalon malformations. The spinal cord (Choice E) is not directly affected in DWS. However, spina bifida does occur with DWS in some patients. Educational Objective: The metencephalon is the embryologic origin of the pons, cerebellum, and fourth ventricle. In Dandy-Walker syndrome, the cerebellar vermis and fourth ventricle are malformed, resulting in cystic enlargement of the fourth ventricle and posterior fossa with consequent hydrocephalus. II Previous Next Score Report Lab Values Calculator Help Pause

2 Exam Section 1: Item 2 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 2. A 65-year-old man comes to the emergency department because of a 1-week history of blood in his sputum. A mass is found on a radiograph. Bronchoscopy is planned. In order to pass the bronchoscope through the oropharynx to the lungs without eliciting a gag reflex, the pharynx is anesthetized. The afferent limb of the reflex is most likely to be blocked by anesthesia to which of the following cranial nerves? A) Trigeminal B) Facial C) Glossopharyngeal D) Vagus E) Hypoglossal

C. The glossopharyngeal nerve (cranial nerve IX) contains somatic afferent fibers that control the sensation of the palate and upper pharynx, providing afferent information for the gag reflex. The efferent nerve for the gag reflex is the vagus nerve, which innervates the soft palate and pharyngeal muscles to control gagging and swallowing. In addition to its role in the gag reflex, the glossopharyngeal nerve contains other afferent and efferent nerve fibers. Components of the glossopharyngeal nerve include general visceral afferent fibers (carotid body and sinus chemo- and baroreceptors), special sensory fibers (taste from posterior third of the tongue), general visceral efferent fibers (parasympathetic innervation of the parotid gland), and special visceral efferent fibers (stylopharyngeus muscle). Lesions of the glossopharyngeal nerve lead to impaired sensation and taste in the posterior third of the tongue, impaired sensation of the palate and pharynx, parotid gland dysfunction, difficulty swallowing, and an absent gag reflex. Incorrect Answers: A, B, D, and E. The trigeminal nerve (Choice A), or cranial nerve V, provides somatic sensory innervation to the face and scalp. The mandibular nerve (the V3 branch of the trigeminal nerve) additionally controls the muscles of mastication and mediates pain and temperature sensation in the anterior two-thirds of the tongue. The trigeminal nerve does not innervate the pharynx. The facial nerve (Choice B), or cranial nerve VII, contains afferent and efferent fibers that carry somatic and visceral information. Motor axons innervate the muscles of facial expression and the stapedius muscle, while sensory axons innervate the outer ear and mediate taste sensation from the anterior two-thirds of the tongue. The facial nerve does not innervate the pharynx. The vagus nerve (Choice D), or cranial nerve X, contains afferent and efferent fibers that carry somatic and visceral information to the palate, pharynx, larynx, heart, gastrointestinal system, pancreas, spleen, kidneys, and adrenal glands. The vagus nerve serves as the efferent nerve for the gag reflex. The hypoglossal nerve (Choice E) is primarily a somatic efferent nerve to the tongue muscles, mediating tongue protrusion, depression, and retraction. The hypoglossal nerve does not innervate the pharynx. Educational Objective: The glossopharyngeal nerve (cranial nerve IX) contains somatic afferent fibers that control the sensation of the palate and upper pharynx, providing afferent information for the gag reflex. The efferent nerve for the gag reflex is the vagus nerve, which innervates the soft palate and pharyngeal muscles to control gagging and swallowing. Previous Next Score Report Lab Values Calculator Help Pause

110 Exam Section 3: Item 10 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 10. A 25-year-old man comes to the emergency department 5 hours after developing shortness of breath and chest pain during exercise; he has had no cough or bloody mucus. He has asthma and major depressive disorder. Current medications include fluticasone inhaler and albuterol. His temperature is 37.1°C (98.8°F), pulse is 110/min, respirations are 30/min, and blood pressure is 90/60 mm Hg. Pulse oximetry on room air shows an oxygen saturation of 93%. Cardiac examination shows a normal S, and S, with no murmurs and no increase in jugular venous pressure. Laboratory studies show: Hemoglobin Hematocrit 13 g/dL 39% Arterial blood gas analysis on room air: pH Pco2 Po2 7.46 26 mm Hg 60 mm Hg A chest x-ray is shown. Which of the following pulmonary findings are most likely in this patient? A) Crackles on the left lung base and apex B) Crackles on the right lung base C) Decreased breath sounds on the left D) Increased wheezes on the left E) Rhonchi on the right

C. The patient's chest x-ray shows a left-sided pneumothorax, and physical examination would likely disclose decreased breath sounds on the left with hyperresonance to percussion. A pneumothorax results from air entering the intrapleural space from either an external (eg, stab wound) or internal (eg, ruptured bleb or bronchus, bronchopleural fistula) source. A spontaneous primary pneumothorax can occur in a tall, thin, young person because of the rupture of a pulmonary bleb, classically in a young athlete and often during physical exertion. As the air accumulates within the space, the lung parenchyma becomes compressed, resulting in respiratory distress, hyperventilation, decreased breath sounds, hyperresonance to percussion, and, in severe cases, tension on the mediastinal structures that impedes pulmonary venous return and results in hypotension. Signs of tension pneumothorax include tachycardia, hypotension, hypoxemia, jugular venous distention, and contralateral tracheal deviation. In cases of acute tension pneumothorax, evacuating the accumulated air via needle decompression and aspiration permits expansion of the lung and eliminates tension on the mediastinum. Needle decompression is followed by tube thoracostomy, which prevents reaccumulation and promotes repair of the pleural defect. Incorrect Answers: A, B, D, and E. Crackles on the left lung base and apex (Choice A) or crackles on the right lung base (Choice B) may be appreciated in pulmonary disorders such as multifocal pneumonia, interstitial lung disease, atelectasis, or pulmonary edema. These disorders would appear as opacities on a chest x-ray. Pneumothorax presents with decreased breath sounds in the affected region. Increased wheezes on the left (Choice D) or rhonchi on the right (Choice E) may occur during an asthma exacerbation, chronic obstructive pulmonary disease exacerbation, pulmonary edema, or with foreign body aspiration and partial airway obstruction. The chest x-ray shows a pneumothorax as the cause of his symptoms. 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 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 hyperresonance to percussion on physical examination. Previous Next Score Report Lab Values Calculator Help Pause

101 Exam Section 3: Item 1 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 1. A 20-year-old man is brought to the physician by his mother because of bizarre behavior for 6 months. Although he had always been a good student, he failed all his classes at college last term. He no longer showers or shaves, and his appearance has become disheveled and unkempt. He appears distracted at the beginning of the interview and then begins talking nonstop to unseen people in the corner of the examining room. His mother looks upset and says, "It seems as if he's hearing voices all the time now. He's really scaring our family." Which of the following initial responses by the physician is most appropriate? A) "He must be schizophrenic. Does schizophrenia run in your husband's family or yours?" B) "He needs to take a shower. Have you tried sitting down with him and talking to him about taking better care of himself?" C) "How frightening for you to see your son like this. Do you have any idea about what might be causing his behavior?" D) "I am concerned he might be on drugs. Do you know anything about the friends he is hanging out with now?" E) "Was he under any unusual stress at home or at school or did he have any problems with a girlfriend when all of this started?"

C. This physician should validate the mother's emotion and then ask an open-ended question. When patients or family members show or describe an emotion while discussing a medical concern, the physician should validate the emotion. There are several methods to validate emotion: being present (actively listening and indicating understanding via nonverbal cues such as nodding), accurate reflection (restating a patient's words, as this physician did), mind-reading (inferring the emotional content behind the patient's words), and normalizing (recognizing the patient's reaction as a response any other person would have). The next step in investigating any medical concern is to ask an open-ended question so that the patient or family member can elaborate on their concern. This patient may have an underlying primary psychiatric (eg, schizophrenia) or substance-induced psychotic disorder, and the physician should allow the mother to provide more details and identify her concerns and priorities. Patients with schizophrenia show two clusters of symptoms: positive symptoms (addition of mental dysfunction including delusions, hallucinations, and disorganized thinking and behavior) and negative symptoms (absence of certain mental functions such as affect, volition, and speech). Negative symptoms may manifest as poor personal hygiene. Incorrect Answers: A, B, D, and E. Immediately mentioning the suspicion that the patient is "schizophrenic" (Choice A) would most likely startle this patient's mother unnecessarily given that the diagnosis of schizophrenia has not been confirmed and because of the abruptness of the statement. Asking closed-ended questions about family history, drugs, and stress to narrow the differential diagnosis (Choices A, D, and E) may be appropriate after first validating this family member's distress and asking an open-ended question. This patient's hygiene is not itself a significant medical concern (Choice B). Instead, it potentially represents a symptom of a more serious problem such as schizophrenia. As well, the physician should first validate the mother's distress and ask an open-ended question. Educational Objective: When patients or family members show or describe an emotion while discussing a medical concern, the physician should validate the emotion. Next, the physician should ask an open-ended question to elucidate the concern. Previous Next Score Report Lab Values Calculator Help Pause

137 Exam Section 3: Item 37 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 37. A 35-year-old man comes to the physician because of a 2-week history of daily left-sided headaches that last 30 to 60 minutes. The headaches usually occur at night and often awaken him. They are accompanied by redness and tearing of the left eye and nasal congestion on the left. Alcohol seems to trigger their occurrence. Vital signs are normal. Neurologic examination shows no abnormalities. Which of the following is the most likely diagnosis? A) Alcohol-induced vasodilation B) Allergic rhinitis C) Central nervous system neoplasia D) Cluster headache E) Temporal arteritis

D. Cluster headaches are primary headaches that typically present with short-lived, unilateral, severe pain in the orbital, supraorbital, or temporal region of the head and are accompanied by ipsilateral autonomic symptoms. The pathogenesis is incompletely understood but likely involves activation sensory fibers of the trigeminal nerve and parasympathetic nerves. Autonomic symptoms commonly include ptosis, miosis, lacrimation, conjunctival injection, rhinorrhea, and/or nasal congestion. Patients typically present between the ages of 20 and 50 years, and smoking and alcohol use may trigger the headaches in patients with known cluster headaches. The headaches often recur over a series of weeks to months at the same time each year. Oxygen therapy using 100% oxygen is the first-line abortive treatment, and verapamil is the first-line prophylactic treatment. Incorrect Answers: A, B, C, and E. Alcohol-induced vasodilation (Choice A) is an unlikely trigger of cluster headaches. Alcohol use can be associated with headaches (including cluster and migraine headaches), but the pathogenesis likely involves cortical or subcortical activation rather than vasodilation. Allergic rhinitis (Choice B) may present with conjunctival injection, lacrimation, nasal congestion, and headache. However, these symptoms would be relatively mild and insidious rather than abrupt in onset. Further, allergic rhinitis is not associated with alcohol use. Central nervous system neoplasia (Choice C) may be associated with focal neurologic deficits and postural headaches, including headaches that are worse at night. However, the onset of these headaches and neurologic symptoms would be insidious rather than acute, and the focal neurologic deficits would be constant rather than episodic. Further, headaches related to neoplasms are not associated with alcohol use. Temporal arteritis (Choice E), also known as giant cell arteritis, is a systemic vasculitis that affects the temporal artery and typically presents with the acute onset of headache, transient monocular vision loss, and/or jaw claudication in patients older than 50 years. Autonomic symptoms as in this patient would be uncommon. Further, headaches related to temporal arteritis are not associated with alcohol use. Educational Objective: Cluster headaches typically present with short-lived, unilateral, severe pain in the orbital, supraorbital, or temporal region of the head and are accompanied by ipsilateral autonomic symptoms such as ptosis, miosis, conjunctival injection, lacrimation, rhinorrhea, and nasal congestion. Previous Next Score Report Lab Values Calculator Help Pause

84 Exam Section 2: Item 34 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 34. A 28-year-old woman comes to the physician for a health maintenance examination. During the interview, she tearfully tells the physician that she and her new husband are having problems. She says she wants to go out with friends, but he does not enjoy being with people, preferring individual activities such as hiking. She says he seems indifferent to sexual intimacy and neither shows much emotion nor understands her feelings at all. Which of the following personality styles best explains the husband's behavior? A) Avoidant B) Narcissistic C) Paranoid D) Schizoid E) Schizotypal

D. Schizoid personality disorder is a cluster A personality disorder that manifests with extreme social detachment and a cold, restricted affect. These patients typically experience a limited range of emotions and therefore have difficulty understanding others' larger range of emotions. They are indifferent to praise and criticism. Patients with schizoid personality disorder do not desire to form social relationships, do not enjoy relationships, and avoid interpersonal activities. Patients may benefit from psychotherapy to learn social skills. Incorrect Answers: A, B, C, and E. Avoidant personality disorder (Choice A) is a cluster C personality disorder, the anxious or fearful cluster, that features the avoidance of interpersonal contact because of feelings of social inadequacy and hypersensitivity to rejection. This patient instead has social apathy and a decreased emotional range. Narcissistic personality disorder (Choice B) is a cluster B personality disorder, the emotional or dramatic cluster, that is characterized by fragile self-esteem and compensatory arrogant, self- aggrandizing behavior to gain admiration, sometimes at others' expense. Patients with schizoid personality disorder are instead apathetic to praise. Paranoid personality disorder (Choice C) is a cluster A personality disorder characterized by pervasive mistrust of others, viewing others as possessing malicious intent. Patients with schizoid personality disorder are apathetic, not mistrustful. Schizotypal personality disorder (Choice E) is a cluster A personality disorder that features odd or magical thinking, perceptual disturbances, social isolation (because of mistrust of others), and a constricted affect. Though patients with schizoid and schizotypal personality disorders both present with constricted affects and social isolation, patients with schizoid personality do not illustrate magical thinking, perceptual disturbances, or a pervasive mistrust of others. Educational Objective: Schizoid personality disorder is a cluster A personality disorder that manifests with social indifference, social isolation, and a decreased range of emotions. Previous Next Score Report Lab Values Calculator Help Pause

152 Exam Section 4: Item 2 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 2. A healthy 20-year-old man comes to the physician with his wife for genetic counseling prior to conception. His sister died of cystinosis, an autosomal recessive disorder affecting cystine transport across lysosomal membranes. The incidence of this disorder in the general population is approximately 1/40,000. The wife's history is noncontributory. The wife's risk for being a carrier of this disorder is closest to which of the following? A) 1/2 B) 2/3 C) 1/50 D) 1/100 E) 1/200

D. The likelihood that the patient's wife is a carrier is 1/100. The Hardy-Weinberg equation can be used to predict the frequency of heterozygous carriers in a population for diseases that are caused by recessive alleles. The equation is typically written as p2 + 2pq + q= 1, where p is the frequency of the dominant (normal) allele and q is the frequency of the recessive allele. In this instance, the incidence of cystinosis, an autosomal recessive disorder requiring two alleles for manifestation is 1/40,000, which represents q2 The square root of 1/40,000 is 1/200, which is equal to q. Since the sum of the allele frequencies, p+ q = 1, p therefore equals 199/200, or 0.995. The frequency of heterozygous carriers is 2pq, which in this instance is 2 x 0.995 x (1/200) = 1/100. This means that she has a one in one hundred chance of being a heterozygous carrier. Incorrect Answers: A, B, C, and E. 1/2 (Choice A) represents the likelihood that a phenotypically normal patient is a carrier of an autosomal recessive disease assuming that both parents were carriers and that the condition is not homozygous-lethal. 2/3 (Choice B) represents the likelihood that an unaffected child of two heterozygous parents for a homozygous-lethal condition is a carrier of the allele. 1/50 (Choice C) and 1/200 (Choice E) are incorrect and do not represent the risk that this patient's wife will carry the recessive allele for cystinosis. They may reflect mathematical errors involving computations within the Hardy-Weinberg allele frequency. Educational Objective: The Hardy-Weinberg equation can be used to estimate the frequency of heterozygosity of recessive alleles in a population using the incidence of the disease in a given population. Previous Next Score Report Lab Values Calculator Help Pause

91 Exam Section 2: Item 41 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 41. A 55-year-old man is admitted to the hospital because of a 2-day history of vomiting and severe abdominal pain in the right upper quadrant. He drinks six 12-ounce beers daily. Abdominal examination shows spider angiomata. The liver is hard and nodular on palpation. A CT scan of the abdomen is shown; the arrow indicates a mass. Hypertension of which of the following veins is most likely in this patient? A) Hepatic B) Inferior phrenic C) Renal D) Short gastric O E) Suprarenal

D. The portal venous system describes a series of interconnected veins that drain blood from the colon, small intestines, spleen, stomach, and inferior esophagus. While this blood eventually makes its way to the systemic circulation via the inferior vena cava, it must first pass through the liver. In cirrhosis, obliteration of the hepatic sinusoids through progressive fibrosis increases the resistance to blood flow through the liver. Blood is then transmitted to the portal venous system, causing portal hypertension (PH). Increased venous pressure is transmitted to all veins that drain into the portal vein. Cirrhosis is also a risk factor for hepatocellular carcinoma, which can potentially narrow or obstruct the portal vein further as a result of mass effect. Anatomically, the portal vein begins at the confluence of the left gastric, splenic, and superior mesenteric veins. The short gastric veins drain into the splenic vein, which then contributes to the formation of the portal vein. Manifestations of PH include esophageal varices, gastric varices, caput medusae, hemorrhoids, splenomegaly, and ascites. In patients with cirrhosis who have ascites that is refractory to diuretic therapy or life-limiting complications of PH (eg, bleeding varices), the placement of a transjugular intrahepatic portosystemic shunt (TIPS) is an option for relieving portal hypertension. In this procedure, a stent is placed between the portal vein and the hepatic vein, effectively allowing blood to bypass the liver. Incorrect Answers: A, B, C, and E. The hepatic veins (Choice A) are not part of the portal venous system and act to drain deoxygenated blood from the liver into the inferior vena cava. Portal hypertension is not transmitted to the hepatic veins because the hepatic veins originate distal to the fibrotic liver parenchyma. Budd-Chiari syndrome, caused by an obstruction of the hepatic veins, is a cause of hepatic vein hypertension. The inferior phrenic veins (Choice B) either drain directly into the inferior vena cava (right inferior phrenic vein) or into the left suprarenal vein (left inferior phrenic vein), which then drains directly into the inferior vena cava. Neither the inferior phrenic veins nor the suprarenal veins (Choice E) drain into the portal vein, and thus, neither develop hypertension secondary to cirrhosis. The renal veins drain directly into the inferior vena cava (Choice C). Portal hypertension is not transmitted to these veins. Educational Objective: Portal vein hypertension may be seen in cirrhosis secondary to obliteration of the hepatic sinusoids through progressive fibrosis, which increases the resistance to blood flow through the liver. The increased pressure is transmitted to the entire portal venous system which includes the left gastric, splenic, and superior mesenteric veins and their tributaries. Previous Next Score Report Lab Values Calculator Help Pause

169 Exam Section 4: Item 19 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 19. A full-term newborn has respiratory distress. Imaging studies show herniation of abdominal contents into the left pleural cavity. Maldevelopment of which of the following structures is most likely to have caused the defect of the diaphragm? A) Esophageal mesoderm B) Left diaphragmatic crus C) Left pleuropericardial fold D) Left pleuroperitoneal membrane E) Septum transversum

D. This newborn presents with a congenital diaphragmatic hernia, which most commonly occurs because of impaired development of the pleuroperitoneal membrane. The diaphragm develops from the pleuroperitoneal membrane, the septum transversum, and somites. Without the appropriate formation of the diaphragm, the abdominal contents herniate into the ipsilateral hemithorax, most commonly on the left because the liver limits passage of contents into the right hemithorax. A congenital diaphragmatic hernia places the newborn at risk for pulmonary hypoplasia, in which herniation of abdominal organs into the thorax compresses the lungs while they are developing in utero. Neonates with pulmonary hypoplasia commonly develop persistent pulmonary hypertension caused by increased pulmonary vascular resistance; this can result in potentially severe heart failure. The associated respiratory and circulatory complications may be rapidly life- threatening to the newborn. Incorrect Answers: A, B, C, and E. Esophageal mesoderm (Choice A) forms the muscular and vascular layers of the esophagus; it also contributes to the formation of the diaphragmatic crura. The diaphragmatic septum transversum is of mesodermal origin. However, defects of these structures are not typically implicated in congenital diaphragmatic hernia. The left diaphragmatic crus (Choice B) is a derivative of esophageal mesoderm. The crura contribute structural support to the diaphragm by connecting to the vertebral bodies. The left pleuropericardial fold (Choice C) contributes to the separation of the pericardial space from the pleural space. This structure does not contribute to the formation of the diaphragm. The septum transversum (Choice E) forms much of the central tendon of the diaphragm and is of mesodermal origin. It is not commonly implicated in congenital diaphragmatic hernia. Educational Objective: Congenital diaphragmatic hernia commonly occurs because of impaired development of the pleuroperitoneal membrane. The diaphragm develops from the pleuroperitoneal membrane, the septum transversum, and somites. Congenital diaphragmatic hernia places the newborn at risk for pulmonary hypoplasia. %3D Previous Next Score Report Lab Values Calculator Help Pause

129 Exam Section 3: Item 29 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 29. A newborn born with ambiguous external genitalia becomes severely dehydrated shortly after birth. Serum sodium and chloride concentrations are decreased, and serum potassium concentration is increased. The newborn's karyotype is 46,XX. The most likely cause of these findings is a genetic defect in which of the following labeled enzymes? Cholesterol Al Side-chain cleavage enzyme C CH3 CH3 HO HO HO Pregnenolone 17-Hydroxypregnenolone Dehydroepiandrosterone DI 38-Hydroxysteroid dehydrogenase: A, isomerase CH3 CH3 HO- Progesterone 17-Hydroxyprogesterone A4 Androstene-3.17-dione E 21-Hydroxylase 17-Ketoreductase | CH2OH CH2OH C=0 C%3O он HO OH 11-Deoxycorticosterone 11-Deoxycortisol Testosterone FC 116-Hydroxylase CH2OH CH2OH C=0 но, 0-? Corticosterone Cortisol G[ 18-Hydroxylase H18-Hydroxydehydrogenase CH2OH H-C HO Aldosterone A) B) C) D) E) F) G) H) I)

E. 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. Lack of this enzyme prevents the conversion of progesterone and 17-hydroxyprogesterone to precursors that ultimately produce aldosterone and cortisol in the adrenal gland. In deficiency, precursors are shunted away from the production of mineralocorticoids and glucocorticoids and toward the production of androgens. Consequently, genetically female patients undergo virilization because of androgen excess and may also demonstrate salt wasting, alkalosis, hyperkalemia, and hypovolemia caused by aldosterone deficiency. In the classic form of 21-hydroxylase deficiency, genetically female patients present with hypoaldosteronism as well as virilization during infancy, and genetically male patients present with precocious puberty in childhood. Treatment consists of exogenous glucocorticoid and mineralocorticoid supplementation. Incorrect Answers: A, B, C, D, F, G, and H. Cholesterol side-chain cleavage enzyme (Choice A) results in impaired production of pregnenolone and deficient synthesis of mineralocorticoids, glucocorticoids, and androgens. Deficiency of this enzyme is a rare cause of lipoid congenital adrenal hyperplasia. While genetically female patients with this enzyme deficiency may also present with salt wasting, they have normal external genitalia. Deficiency in 17a-hydroxylase (Choice B) causes increased concentrations of mineralocorticoids (eg, aldosterone) and decreased concentrations of cortisol and androgens, resulting in ambiguous genitalia and undescended testes in XY chromosome infants. Infants with an XX karyotype lack secondary sexual development later in life, observed as delayed puberty, amenorrhea, and failure to develop breasts and pubic hair. 17,20-Lyase (Choice C) is important for androgen synthesis and is a rare cause of CAH. Genetically male patients may present with a feminine external appearance caused by lack of androgens. Genetically female patients typically present with amenorrhea. Salt wasting is not a typical feature of this disorder. 3B-Hydroxysteroid dehydrogenase (Choice D) deficiency leads to decreased synthesis of mineralocorticoids, glucocorticoids, and androgens. However, an intermediate in the androgen synthetic pathway, dehydroepiandrosterone (DHEA), is increased (because of decreased metabolism of DHEA by 3B-hydroxysteroid dehydrogenase), which can lead to virilization of genetically female infants, children, or young adults (depending on the severity of the deficiency). 11B-Hydroxylase (Choice F) deficiency is a rare form of CAH in which the production of cortisol and aldosterone are interrupted later in the pathway in comparison with 21-hydroxylase deficiency. The result is decreased cortisol and aldosterone concentrations with an increased concentration of the intermediate mineralocorticoid, 11-deoxycorticosterone. 11-Deoxycorticosterone acts at the kidney in a manner similar to aldosterone, with resultant sodium and water retention and excretion of potassium and protons. 18-Hydroxylase (Choice G), also known as aldosterone synthase, and 18-hydroxydehydrogenase (Choice H) are important for the synthesis of aldosterone. Deficiency of these enzymes may present with salt wasting, hyperkalemia, and signs of cortisol excess. Deficiency of either of these enzymes is rare and much less common than 21-hydroxylase deficiency. 17-Ketoreductase (Choice I) deficiency is a rare autosomal recessive disorder that results in the inability to produce testosterone. In genetic males, this results in the development of ambiguous or female external genitalia with the presence of male internal genitalia (testes). 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 and increased concentrations of 17-hydroxyprogesterone. Treatment includes the exogenous administration of mineralocorticoids and glucocorticoids. I3D Previous Next Score Report Lab Values Calculator Help Pause 17a-Hydroxylase Ol asek70-"

52 Exam Section 2: Item 2 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 2. An 8-year-old boy is brought to the physician by his parents for a well-child examination. His sister died of leukemia 4 months ago. The parents state that he is sad at times, but he has had only minor performance difficulty at school since his sister died. During the interview, the patient talks about his sister's death and cries. He then says that he has a headache. Physical examination shows no abnormalities. Which of the following best describes this patient's condition? A) Abnormal grief reaction B) Conversion disorder C) Major depressive disorder D) Somatoform disorder E) Normal emotional response

E. Acute, uncomplicated grief presents in diverse ways and can include sadness, shock, anguish, guilt, regret, difficulty concentrating, decreased interest in usual activities, preoccupation with thoughts and memories of the deceased, and neurovegetative symptoms such as sleep, appetite, and energy disturbances. Uncomplicated grief also commonly presents with somatic symptoms that are not explained by objective findings such as headache, dizziness, chest pain, and constipation. In some patients, these symptoms can endure for months. However, in uncomplicated grief reactions, the full criteria for major depressive disorder are not satisfied. This patient's sadness is intermittent, which makes uncomplicated grief more likely than major depressive disorder. Management includes frequent reassessment, encouraging reconnection with family, and, in severe cases, referral for grief counseling. Acute grief ultimately becomes integrated grief, in which thoughts of the deceased continue to engender sadness but acute symptoms resolve. Incorrect Answers: A, B, C, and D. An abnormal grief reaction (Choice A) refers to prolonged (lasting between 6 and 12 months), intense, and disabling grief symptoms including yearning for the deceased and feeling upset by memories of the deceased. These symptoms may convince the patient that life is empty and result in dysfunction in daily life. This patient's sadness and dysfunction are relatively mild. Conversion disorder (Choice B), more commonly called functional neurologic disorder, involves neurologic symptoms such as sensory or motor dysfunction that are not fully explained by objective findings on physical examination or imaging. A headache alone would not be classified as a conversion disorder and is instead attributable to uncomplicated grief in this patient. Major depressive disorder (Choice C) features depressed mood, decreased energy, psychomotor retardation, anhedonia, sleep disturbances, appetite disturbances, guilt, poor concentration, and/or suicidal ideation. Patients must demonstrate five of these symptoms for two or more weeks. This patient with intermittent sadness and without other depressive symptoms does not meet major depressive disorder criteria. In somatoform disorder (Choice D), also called somatic symptom disorder, the patient is preoccupied with one or more somatic symptoms such that these symptoms disrupt the patient's daily life. This patient is not clearly preoccupied with his headache. Educational Objective: Uncomplicated grief can include sadness, guilt, difficulty concentrating, preoccupation with thoughts and memories of the deceased, and disturbed neurovegetative symptoms such as sleep, appetite, and energy changes without meeting the criteria for major depressive disorder. Uncomplicated grief also commonly presents with somatic symptoms that are not explained by objective findings. Previous Next Score Report Lab Values Calculator Help Pause

11 Exam Section 1: Item 11 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 11. A 47-year-old man comes to the emergency department because of the sudden onset of severe pain of his lower chest and upper abdomen, and dry heaves. He also has a 4-year history of intermittent heartburn. His pulse is 118/min, and blood pressure is 80/50 mm Hg. Physical examination shows epigastric tenderness. A nasogastric tube cannot be passed into the stomach. The diagnosis of gastric volvulus is made, and appropriate treatment is begun. This patient most likely also has which of the following underlying conditions? A) Cholecystitis B) Duodenal ulcer C) Gastric cancer D) Pancreatitis E) Paraesophageal hernia

E. Gastric volvulus occurs when the stomach rotates along its short or long axis causing obstruction. It may be caused by anomalies of the gastric ligaments (primary gastric volvulus) or anatomic abnormalities (eg, paraesophageal hernia). An acute gastric volvulus presents with severe abdominal pain and nonproductive vomiting or heaving. These symptoms with the inability to pass a nasogastric tube into the abdomen are called the Borchardt triad and are present in -70% of patients with gastric volvulus. The most common cause of secondary gastric volvulus is a paraesophageal hernia, followed by a diaphragmatic hernia or phrenic nerve paralysis. In children, a congenital diaphragmatic hernia is the most commonly associated abnormality. In the case of a paraesophageal hernia, the stomach rotates on its long axis, which can be imagined as a line from the gastroesophageal junction to the pyloric sphincter. This causes the greater curvature of the stomach to be superior to the inferior curvature of the stomach and obstructs the gastric outlet. With prolonged volvulus, strangulation of the stomach can occur, which is followed by ischemia, necrosis, perforation, and finally peritonitis. The first step in treatment is to attempt decompression via a nasogastric tube to decrease pressure on the gastric mucosa and prevent ischemia. However, if unsuccessful, surgical intervention is required. Incorrect Answers: A, B, C, and D. Cholecystitis (Choice A) occurs secondary to obstruction of the cystic duct from a gallstone, which results in subsequent inflammation of the gallbladder wall. It typically presents with fever, abdominal pain, and tenderness to palpation in the right upper quadrant, often in a patient with a history of biliary colic and cholelithiasis. The pain may also radiate to the right shoulder or interscapular region. It would not contribute to the development of gastric volvulus. Duodenal ulcer (Choice B) classically presents with worsening abdominal pain related to a lack of consumption of food. The most common cause of duodenal ulcers is infection with the bacterium Helicobacter pylori. Persistent inflammation related to a duodenal ulcer can result in complications including fibrosis, stricture, and hemorrhage. Perforation is another potential complication, which commonly presents with acute abdominal pain, peritonitis, fever, and free air on x-rays and CT. While a stricture or fibrosis from a duodenal ulcer could cause a small bowel obstruction or be a lead point for intussusception, it would not cause gastric volvulus. Gastric cancer (Choice C) would be more likely to manifest with weight loss, early satiety, gastrointestinal bleeding, and systemic symptoms such as fatigue and malaise. Gastric cancer is less likely to cause gastric volvulus than a paraesophageal hernia since it is usually confined to the lumen of the stomach and does not cause redundancy in the gastric wall or an anatomic change that allows the stomach to rotate on its axis. Pancreatitis (Choice D) presents with epigastric abdominal pain that radiates to the back, along with nausea and emesis, often in a patient with a history of gallstones, alcoholism, trauma, hypertriglyceridemia, or hypercalcemia. Acute pancreatitis can be complicated by necrosis, hemorrhage, abscess, or the formation of pseudocysts, but not by gastric volvulus. Educational Objective: Gastric volvulus occurs when the stomach rotates along either its short or long axis to cause gastric outlet obstruction. It may be caused by anatomic abnormalities (secondary gastric volvulus), the most common of which is a paraesophageal hernia. Gastric volvulus classically presents with a combination of severe abdominal pain, dry heaving, and inability to pass a nasogastric tube (referred to as the Borchardt triad). Previous Next Score Report Lab Values Calculator Help Pause

36 Exam Section 1: Item 36 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 36. A 20-year-old woman comes to the physician because of easy bruising for 2 weeks. She had been in good health except for a viral respiratory illness 1 month ago. Laboratory studies show thrombocytopenia. A bone marrow smear is shown. Which of the following is the most likely cause of her thrombocytopenia? A) Acute myeloblastic leukemia B) Aplastic anemia C) Metastatic adenocarcinoma D) Parvovirus infection E) Peripheral destruction of platelets

E. Peripheral destruction of platelets as a result of circulating antiplatelet antibodies accounts for this patient's presentation, which is consistent with immune thrombocytopenic purpura (ITP). Anti- Gpllb/lla antibodies bind to platelets, which results in their opsonization. Macrophages within the spleen then phagocytose the platelet-antibody complex, resulting in thrombocytopenia. ITP is diagnosed with a decreased platelet count in the absence of other explanatory causes for thrombocytopenia. Bone marrow biopsy will classically show an increased number of megakaryocytes indicating adequate platelet production but increased peripheral destruction of platelets, as shown in this patient's photomicrograph. ITP presents with petechiae, purpura, and prolonged bleeding time on laboratory analysis, but normal prothrombin time and partial thromboplastin time. Severe ITP may lead to uncontrolled hemorrhage. Acute ITP often follows an infection, and generally is self-limited. Chronic ITP is first treated with steroids, which may be required long-term, intravenous immunoglobulin, or immunomodulators. Refractory ITP that is nonresponsive to therapies is treated with splenectomy, which results in remission of thrombocytopenia in the majority of cases because of the spleen's primary role in the underlying pathophysiology. Incorrect Answers: A, B, C, and D. Acute myeloblastic leukemia (AML) (Choice A) is a malignancy of myeloid precursors and commonly presents with pancytopenia and circulating blast cells. Bone marrow biopsy would show replacement of the normal bone marrow with myeloid blasts, not increased numbers of megakaryocytes. Aplastic anemia (Choice B) 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 decreased reticulocyte count, not thrombocytopenia. Bone marrow biopsy would show a paucity of erythroid precursors. Metastatic adenocarcinoma (Choice C) with involvement of the bone marrow can occur in a variety of cancers including lung, breast, and colorectal adenocarcinoma. Bone marrow biopsy would show the presence of malignant cells. Parvovirus infection (Choice D) is the cause of erythema infectiosum and is most common in children, although it may affect adults and cause arthralgias of the small joints of the fingers. It can also cause an aplastic crisis in patients with sickle cell disease that would present with anemia. Educational Objective: ITP is a diagnosis of exclusion, and it presents with mucocutaneous bleeding and thrombocytopenia often following a viral illness. It is caused by the antibody-mediated destruction of circulating platelets, and bone marrow biopsy will disclose an increased number of megakaryocytes. II Previous Next Score Report Lab Values Calculator Help Pause

85 Exam Section 2: Item 35 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 35. A 55-year-old woman comes to the physician because of a 2-day history of fever and the recent onset of cough productive of blood-tinged sputum. She also has had progressive shortness of breath during physical activity. She has smoked 2 packs of cigarettes daily for 41 years. Her temperature is 38.6°C (101.5°F), pulse is 95/min, respirations are 22/min, and blood pressure is 135/84 mm Hg. Breath sounds are decreased, and crackles are heard in the right middle lung field. A chest x-ray shows consolidation of the right middle lobe and a possible mass in the right hilar region. A CT scan of the lungs shows a 4-cm mass involving and obstructing the proximal right middle lobe bronchus. Which of the following is the most likely type of lung neoplasm? A) Adenocarcinoma B) Bronchioalveolar carcinoma C) Large cell carcinoma D) Lymphoma E) Squamous cell carcinoma

E. Squamous cell carcinoma of the lung, which typically presents as a central lesion, is the second most common type of primary lung cancer after adenocarcinoma, which more often presents as a peripheral lesion. Risk factors for all major types of lung cancer include tobacco use, secondhand smoke, asbestos, radon exposure, and family history of lung cancer. Features associated with squamous cell carcinoma of the lung include pulmonary cavitations, central location, and hypercalcemia caused by paraneoplastic parathyroid hormone-related peptide (PTHPP) production. Histologic characteristics include polygonal cells with intercellular bridges, eosinophilic cytoplasm, keratin pearls, and extensive necrosis. Lung cancer, in general, typically presents with cough, unintentional weight loss, hemoptysis, chest pain, dyspnea, and hoarseness; occasionaly, wheezing, focal rhonchi, or hypertrophic osteoarthropathy may be noted on examination. Obstruction of the airways can lead to postobstructive pneumonia, as seen in this patient. Diagnosis is made by chest imaging and examination of a biopsy specimen. Prognosis is a function of the cancer type along with grading and staging of the disease. It is often detected once metastatic, at which point the prognosis is poor. Incorrect Answers: A, B, C, and D. Adenocarcinoma (Choice A) of the lung is the most common overall primary lung cancer and the most common among nonsmokers. It typically presents as a chronic consolidation in the periphery of the lung rather than centrally. A glandular pattern is classically seen on histology with mucin-positive staining. Bronchioalveolar carcinoma (Choice B) refers to a subtype of adenocarcinoma that arises from the terminal bronchioles and spreads along intact alveolar septae. Large cell carcinoma (Choice C) is a highly anaplastic, undifferentiated neoplasm of epithelial cells that lacks glandular, squamous, or neuroendocrine characteristics. It most often presents as a peripheral lesion. Lymphoma (Choice D) may arise from B or T lymphocytes and often presents with constitutional symptoms such as fatigue, unintentional weight loss, fevers, and night sweats. Physical examination findings may include diffuse lymphadenopathy and hepatosplenomegaly. Involvement of the hilar lymph nodes may result in a hilar mass and postobstructive pneumonia caused by extrinsic compression of the bronchus. A primary lung malignancy is more likely in this patient with a history of smoking and tumor involvement of the bronchus. Educational Objective: Centrally located primary lung cancers include squamous cell carcinoma and small cell carcinoma of the lung. Squamous cell carcinoma is the more common subtype. These can result in obstruction of bronchi, which increases the risk for postobstructive pneumonia. Previous Next Score Report Lab Values Calculator Help Pause

71 Exam Section 2: Item 21 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 21. A 3-year-old girl is found to have a grade 4/6, loud, harsh, high-pitched holosystolic murmur that radiates over the precordium and a palpable thrill at the left sternal border. Which of the following defects is most likely in this patient? A) Aortic regurgitation B) Aortic stenosis C) Atrial septal defect D) Coarctation of the aorta E) Mitral regurgitation F) Mitral stenosis G) Patent ductus arteriosus H) Pulmonic stenosis OI) Tricuspid regurgitation J) Ventricular septal defect

J. Ventricular septal defects (VSDS) are amongst the most common congenital heart malformations, and they may occur in isolation or in association with complex structural abnormalities. The genetic cause of VSDS is heterogeneous and includes known associations with well-defined conditions such as Down, DiGeorge, and Holt-Oram syndromes. It may also occur in isolation and via nonspecific, random mutations or individual point mutations that are de novo. VSDS are characterized by a holosystolic murmur best heard in the left lower sternal border and may have an associated palpable thrill. VSDS result in a left-to-right shunt, which increases pulmonary blood flow. If uncorrected, the increased pulmonary blood flow may result in pulmonary artery remodeling and consequent pulmonary arterial hypertension, which may cause a reversal of the shunt to right-to-left. This process is referred to as Eisenmenger syndrome. Incorrect Answers: A, B, C, D, E, F, G, H, and I. Aortic regurgitation (Choice A) presents with an early diastolic decrescendo murmur best heard at the right second intercostal space, and it is most commonly associated with endocarditis, acute rheumatic fever, and aortic root dilation. Aortic stenosis (Choice B) presents with a crescendo-decrescendo systolic murmur best heard at the upper right sternal border that radiates to the carotid arteries. It classically occurs secondary to age-related fibrotic and calcific changes of the valve but can occur earlier in life in cases of bicuspid aortic valve or chronic rheumatic heart disease. Atrial septal defect (Choice C) results in a left-to-right shunt with abnormal flow of blood from the left atrium to the right atrium. This causes relative volume overload of the right atrium and ventricle, which presents as a fixed, split S, and low-grade physiologic ejection murmur on cardiac auscultation. Coarctation of the aorta (Choice D) refers to a narrowing of the aorta. It is associated with a bicuspid aortic valve and Turner syndrome. It typically presents with a systolic murmur along with differential pulses and blood pressures between extremities. Mitral regurgitation (Choice E) presents with a holosystolic murmur best heard at the left fourth or fifth intercostal space along the midclavicular line and radiates to the left axilla. It is commonly associated with mitral valve prolapse, endocarditis, acute rheumatic fever, and prior myocardial infarction. Mitral stenosis (Choice F) is classically heard as an opening snap followed by a diastolic rumble that is loudest over the cardiac apex and radiates to the axilla. If severe enough, it can result in left atrial enlargement, cardiogenic pulmonary edema, and arrhythmias such as atrial fibrillation and flutter. Patent ductus arteriosus (Choice G) results in a continuous, machine-like murmur best heard at the left second intercostal space radiating to the clavicle. Pulmonic stenosis (Choice H) presents with a systolic murmur best heard at the second left intercostal region; it is also often crescendo-decrescendo, though it is quieter and radiates less to the lower neck than aortic stenosis because of the lower pressure in the pulmonary circulation. Pulmonic stenosis is typically seen in tetralogy of Fallot. Tricuspid regurgitation (Choice I) discloses a holosystolic murmur best heard at the left lower sternal border; it may result from severe pulmonary hypertension or infective endocarditis in intravenous drug use. Educational Objective: Ventricular septal defects are common congenital heart malformations and may occur in isolation or with other defects. They are characterized by a harsh, holosystolic murmur best heard at the left lower sternal border. Previous Next Score Report Lab Values Calculator Help Pause

81 Exam Section 2: Item 31 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 31. A73-year-old woman is admitted to the hospital 2 days after she fell while walking down her front steps. She has severe back pain. She has a 7-year history of polymyalgia rheumatica and a 20-year history of hypertension. Current medications include lisinopril, prednisone, vitamin D, and a multivitamin. She appears frail. Physical examination shows thoracic kyphosis and multiple ecchymoses over the dorsa of the hands. There is a tender point on percussion of the mid spine. A spinal x-ray shows multiple vertebral compression fractures. If a bone biopsy specimen were obtained from this patient, it would most likely show which of the following findings? A) Defective mineralization B) Increased collagen staining C) Increased mineralization D) Thickening at the cortex E) Thinned trabeculae

E. Thinned trabeculae are characteristic of osteoporosis, a metabolic bone disease with decreased bone density. Osteoporosis occurs because of the loss of bone mineral density, most commonly appearing in elderly women of European descent, and exacerbated in those with chronic steroid use and limited physical activity, as in this patient. The condition results from the interplay of three possible mechanisms: failure to achieve peak bone density, increased bone resorption, and decreased new bone formation. It is often asymptomatic until a minimally traumatic injury results in a fracture, most commonly involving the hip, distal radius, or vertebrae, again seen in this patient. Classically, serum calcium, PTH, vitamin D, and phosphate are within normal limits; however, calcium handling in the bloodstream is potentially key in the pathogenesis of osteoporosis. Hypocalcemia and hypovitaminosis D both lead to increased concentrations of parathyroid hormone (PTH), which results in increased osteoclast activity and bone resorption. Decreased concentrations of vitamin D are also associated with increased concentrations of PTH, even when serum calcium is within normal limits. The diagnosis of osteoporosis is made on the basis of dual-energy x-ray absorptiometry (DEXA) once the bone density falls below 2.5 standard deviations of the mean for a young adult. In cases where the diagnosis is unclear or confounded, a biopsy may be pursued, though this is seldom necessary. Biopsy may exclude confounding diagnoses such as osteomalacia, multiple myeloma, and

173 Exam Section 4: Item 23 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 23. A 45-year-old man with type 1 diabetes mellitus comes to the physician because of a 6-week history of swelling of his face, legs, and left testicle. He says that his blood pressure has been higher than normal (2170/100 mm Hg) during the past 2 months. At his last visit 1 year ago, his serum creatinine concentration was 2.1 mg/dL. He is 180 cm (5 ft 11 in) tall and weighs 100 kg (220 lb); BMI is 31 kg/m2. His blood pressure is 172/93 mm Hg. Physical examination shows marked edema around the eyes and of the lower extremities to the level of the thighs. The left side of the scrotum is 1½ times the size of the right side and enlarges further when the patient bears down. Palpation shows a lumpy mass just superior to the normal-sized left testis. Which of the following is the most likely cause of these scrotal findings? O A) Epididymal cyst B) Epididymitis C) Hydrocele D) Testicular cancer E) Testicular torsion F) Varicocele

F. A varicocele is a dilation of the pampiniform plexus of spermatic veins that are typically located just superior to the testis. It more commonly presents on the left side, and symptoms include scrotal fullness and an irregular, multi-tubular scrotal mass that is more prominent in the standing position or with a Valsalva maneuver. The left spermatic vein enters the renal vein, whereas the right spermatic vein enters the inferior vena cava directly. The left renal vein, in turn, courses between the aorta and the superior mesenteric artery, which causes it to have an increased intravascular pressure, especially in fluid overload, as in this patient. This pressure is transmitted to the left spermatic vein and is the most common cause of varicocele. Varicoceles do not transilluminate and can be further evaluated with ultrasound if the diagnosis is unclear. They are associated with infertility, and treatment options include conservative management with scrotal support, surgical ligation, or percutaneous venous embolization. Incorrect Answers: A, B, C, D, and E. Epididymal cysts (Choice A) are commonly asymptomatic but can present with a soft, palpable mass commonly within the epididymal head. They do not increase in size with a Valsalva maneuver. Epididymitis (Choice B) presents with scrotal pain, erythema, swelling, and tenderness to palpation, commonly secondary to a sexually transmitted infection, although it can also be caused by other bacterial infections, such as Escherichia coli. It does not cause a lumpy mass in the scrotum. Hydrocele (Choice C) refers to abnormal fluid between the visceral and parietal layers of the tunica vaginalis. It typically presents with scrotal swelling that may or may not be painful. They also increase in size with Valsalva maneuver and would transilluminate on examination. Testicular cancer (Choice D) typically presents with a painless, firm nodule in the testis that is commonly asymptomatic. The size of the mass would not increase with Valsalva maneuver. Testicular torsion (Choice E) classically presents with acute testicular pain, tenderness to palpation, scrotal swelling, and an absent cremasteric reflex. Physical examination will show a high- riding testis that is oriented transversely with surrounding erythema. It would not increase in size with a Valsalva maneuver. Emergent detorsion and orchipexy is required for the treatment of testicular torsion. Educational Objective: A varicocele refers to the dilation of the pampiniform plexus of spermatic veins. It typically presents with scrotal fullness and an irregular scrotal mass, which increases in size with a Valsalva maneuver and standing. It more commonly occurs on the left. %3D Previous Next Score Report Lab Values Calculator Help Pause

90 Exam Section 2: Item 40 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 40. A 12-year-old boy with severe deafness is brought to the physician because of a 3-month history of episodes of loss of consciousness. Physical examination shows no abnormalities except for severe bilateral sensorineural hearing loss. An ECG shows a long QT interval. This patient most likely has a mutation in the gene that encodes which of the following proteins? A) a-Adrenergic receptor B) B-Adrenergic receptor C) Gap junction channel D) Muscarinic cholinergic receptor E) Neurotransmitter-gated calcium channel F) Voltage-gated potassium channel

F. Long QT syndrome (LQTS) is a disorder of prolonged ventricular depolarization and repolarization. It can be congenital or acquired. There are several mutations that may result in LQTS, some of which involve loss of function mutations of voltage-gated potassium channels (LQTS type 1). Romano-Ward syndrome is a subtype of LQTS type 1 that is inherited in an autosomal dominant pattern and purely affects cardiomyocytes. Jervell and Lange-Nielsen syndrome is another subtype that is inherited in an autosomal recessive pattern; it affects cardiomyocytes as well as cells of the inner ear. It is associated with congenital sensorineural hearing loss. LQTS, in general, carries an increased risk for ventricular arrhythmias, syncope, and sudden cardiac death. Patients are often asymptomatic. ECG will show a prolonged QT interval, and genetic testing is used to confirm the diagnosis. Treatment consists of B-adrenergic blockers and implantable cardioverter defibrillators for high-risk patients. Incorrect Answers: A, B, C, D, and E. a-Adrenergic receptors (Choice A) and B-adrenergic receptors (Choice B) are the target receptors for postganglionic fibers of the sympathetic nervous system and are ubiquitous throughout the body. Mutations in adrenergic receptors are not associated with long QT syndrome. Gap junction channels (Choice C) are specialized membrane structures in cardiomyocytes that connect adjacent cells and allow for rapid chemical and electrical signaling. The presence of gap junctions allows for coordinated action potential propagation. Alteration in gap junction expression and distribution occurs in many cardiac disease states and increases the risk for arhythmias, but the mutations in long QT syndrome involve voltage-gated potassium channels. Muscarinic cholinergic receptors (Choice D) are ubiquitous throughout the body and are the target receptors for the parasympathetic nervous system (with the exception of sweat glands, which are innervated by the sympathetic nervous system). There are five primary subtypes in humans, found in the central nervous system, heart, smooth muscle, exocrine glands, and sweat glands. In the heart, muscarinic receptors act to decrease the rate of cardiomyocyte depolarization, slowing the heart rate. Muscarinic receptors are not involved in prolongation of the QT interval. Neurotransmitter-gated calcium channels (Choice E) may be mutated in some types of Brugada syndrome, which may present with arrhythmias or sudden cardiac death. Other forms of Brugada syndrome involve disruption of normal cardiomyocyte sodium channel activity. Educational Objective: LQTS is characterized by delayed ventricular depolarization and repolarization with an increased risk for ventricular arrhythmias. It may be congenital or acquired. The most common subtype of congenital long QT syndrome involves mutations in voltage-gated potassium channels. Previous Next Score Report Lab Values Calculator Help Pause

157 Exam Section 4: Item 7 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 7. Patients with a history of deep venous thrombosis are recruited as part of a research study on chronic venous insufficiency. The patients are evaluated at the beginning of the study and reevaluated 2 years later. At the second visit, they are asked if they wore compression stockings since the study began. The prevalence of venous insufficiency in those who wore the stockings is compared to those who did not. Which of the following best describes this type of study? A) Case-control study B) Case series study C) Clinical trial D) Cohort study

D. A cohort study identifies a group of patients with or without a given exposure or intervention and follows them over time to identify whether that exposure or intervention 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. This case provides an example of a prospective cohort study design, as the investigator identifies a group of individuals with a history of deep venous thrombosis, and then evaluates them at the beginning of the study and 2 years later to evaluate their use of the compression stockings and associated development of venous insufficiency. This is different than a retrospective cohort study, which would only evaluate the cohort of patients in the past, not sequentially in real-time at the beginning of the study and 2 years later. Incorrect Answers: A, B, and C. A case-control study (Choice A) investigates an association between an exposure and an outcome. In this study design, a group of patients with the disease (cases) are identified. A group of patients without the disease (controls) are matched on baseline characteristics to the cases. Exposure data for the two groups is collected, and these data are compared to determine association with the outcome (disease) in question. An odds ratio may be calculated to compare exposures between groups. A case-control study design in this case would have identified patients with chronic venous insufficiency, compared them with a group of age-matched controls, and then surveyed both groups for use of compression stockings in the past. 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, since the small population size may not permit conduction of a larger cohort or randomized trials with sufficient power. A clinical trial (Choice C) is an experimental study design. Patients are randomly allocated to two or more interventional arms or control arms, and these patients are followed over time to evaluate an outcome of interest. Randomized design minimizes opportunity for bias; thus, a randomized interventional study can be used to imply causation. Common examples of randomized trials include therapeutic comparisons between a new drug and the previous standard of care. Educational Objective: A cohort study may be retrospective or prospective. The exposure is chosen, and patients both with and without that exposure are followed, either in real-time or retrospectively, to evaluate for the development of a given outcome. %3D Previous Next Score Report Lab Values Calculator Help Pause

181 Exam Section 4: Item 31 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 31. A 35-year-old woman who is being examined because of infertility receives an injection of contrast material into her cervix to visualize the reproductive tract before undergoing radiographic tests. On the hysterosalpingogram shown, the contrast material (as indicated by the arrows) is also seen in the peritoneal cavity. Which of the following best explains this finding? A) Rupture of the fallopian tube B) Rupture of the uterine body C) Spillage of contrast, which is an artifact D) Spillage of contrast, which is normal

D. A hysterosalpingogram is a radiographic technique used to visualize the structure of the uterine cavity and fallopian tubes, including tube patency. It is commonly used in the evaluation of infertility to determine whether there are any anatomic abnormalities of the uterus or fallopian tubes, such as polyps, synechiae, leiomyomas, septations, or peritubal adhesions. Contrast media is injected into the cervical canal, then fluoroscopy is used to image the pelvis. The fallopian tubes empty into the peritoneal cavity and are not directly connected to the ovaries; therefore, spillage of the contrast into the peritoneal cavity confirms normal tube patency. Incorrect Answers: A, B, and C. Rupture of the fallopian tube (Choice A) would appear with incomplete fallopian tube filling on one side of the uterus, accompanied by spillage of contrast into the peritoneal cavity. The narrow lines of contrast extending to the right and left of the uterus in this image are patent fallopian tubes. Rupture of the uterine body (Choice B) would appear with an irregular outline of the uterine cavity with excessive spillage of contrast into the peritoneal cavity, rather than through the fallopian tubes. The uterus in this image has a regular contour. Spillage of contrast is not an artifact (Choice C) since the movement of contrast through the fallopian tubes and into the peritoneal cavity confirms fallopian tube patency, which is a normal finding. Educational Objective: A hysterosalpingogram is used to evaluate the structure of the uterine cavity and the patency of the fallopian tubes with fluoroscopy. A normal finding includes a regular contour of the uterine cavity, narrow linear contrast stripes extending from the uterus (fallopian tubes), and spillage of the contrast into the peritoneal cavity. This confirms normal fallopian tube patency. Previous Next Score Report Lab Values Calculator Help Pause

138 Exam Section 3: Item 38 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 38. A 4-year-old girl with chronic otitis media is brought to the physician by her mother because of a 5-day history of recurrence of symptoms. The mother says that her daughter had a mild cough that began 8 days ago and progressed to a sore throat and ear pain during the next 3 days. The patient appears irritable and is pulling at her right ear. Otoscopic examination shows fluid behind the tympanic membrane. Which of the following best describes the anatomic pathway of the spread of infection from this patient's throat? A) Laryngopharynx - auditory tube - inner ear B) Larynx oropharynx sinuses inner and middle ear | C) Lungs - paranasal sinuses inner ear D) Nasopharynx auditory tube middle ear E) Nose sinuses outer ear

D. Acute otitis media is an acute infection of the middle ear and usually spreads to this space via the eustachian (auditory) tubes from the nasopharynx. The middle ear is bounded externally by the tympanic membrane and internally by the oval window. The eustachian tube connects the middle ear to the posterior aspect of the nasopharynx near the torus tubarius. Acute otitis media classically occurs in children, often from viral or bacterial pathogens, commonly Streptococcus pneumonia or Haemophilus influenza. It presents with ear pain, often with fever, and with an erythematous tympanic membrane with retrotympanic pus or a middle ear effusion. Recurrent episodes may lead to inflammation of the adjacent mastoid bone, a condition known as mastoiditis. First-line therapy includes antibiotic treatment with amoxicillin with or without clavulanate, as well as symptomatic treatment. Incorrect Answers: A, B, C, and E. Choices A, B, C, and E describe incorrect anatomical pathways that either do not exist or are less plausible as a mechanism of otitis media. There are, for example, no direct anatomical pathways between the laryngopharynx and the eustachian tube, or between the outer, middle, or inner ear and the paranasal sinuses. Educational Objective: Acute otitis media is an acute infection of the middle ear and usually spreads to this space via the eustachian tubes from the nasopharynx. Commonly implicated pathogens include Streptococcus pneumonia and Haemophilus influenza. Treatment is with antibiotics, typically amoxicillin. %D Previous Next Score Report Lab Values Calculator Help Pause

58 Exam Section 2: Item 8 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 8. A 78-year-old man comes to the physician because of a 1-week history of blisters over several areas of his body. A photograph of the lesions is shown. Direct immunofluorescence of the affected areas shows linear deposition of IgG and C3b at the dermal-epidermal junction. Which of the following best describes the function of the target structure to which IgG is binding in this patient? A) Attaches keratinocytes to the extracellular matrix B) Provides tensile strength to the skin C) Secretes the chitinous cuticle D) Serves as a chemokine receptor on keratinocytes E) Serves as a ligand for cutaneous lymphocyte antigen

A. Bullous pemphigoid is caused by antibodies against the hemidesmosome. The hemidesmosome is an intricate complex of proteins whose function is to anchor the basal keratinocytes of the epidermis to the dermis and its extracellular matrix 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 that do not easily rupture with lateral shear forces. Bullous pemphigoid often presents in elderly men with severe pruritus, as in this case. The mucosa is usually spared. Early in the disease course, bullae may be absent and only urticarial plaques and severe itching will be present. On histopathology, the bullae will show eosinophils. Direct immunofluorescence shows a linear pattern of IgG deposition along the basement membrane. Treatment of bullous pemphigoid includes topical corticosteroids, systemic corticosteroids, and steroid sparing immunosuppressants such as azathioprine or mycophenolate mofetil. Incorrect Answers: B, C, D, and E. Collagen provides tensile strength to the skin (Choice B), which confers resistance to tearing, a vital function. Conditions caused by autoantibody formation to various types of collagen include mucous membrane pemphigoid (collagen, type IV), epidermolysis bullosa (collagen, type VII), and bullous lupus erythematosus (collagen, type XVII). Mutations in collagen cause disorders including Alport syndrome and osteogenesis imperfecta. A chitinous cuticle (Choice C) is not a feature of human keratinocytes or skin cells. Chitinous cuticles are seen in ectoparasites (eg, scabies) or arthropods (eg, spiders). Antibodies to a chitinous cuticle may be seen in scabies infestation. The pruritus that accompanies scabies is secondary to this allergic reaction rather than a feature of the ectoparasite itself. Scabies is characterized by pruritic papules and burrows in the web spaces of the hands, groin, and umbilicus. It does not present with tense bullae. The hemidesmosome, the target of bullous pemphigoid, serves as a structural link between the epidermis and dermis. It does not serve as a chemokine receptor on keratinocytes (Choice D) and does not contribute to the function of cells of either the innate immune system or adaptive immune system. Cutaneous lymphocyte antigen (Choice E) is present on the surface of memory T lymphocytes. It is a ligand for E-selectin and plays an integral role in the ability of memory T lymphocytes to target inflamed skin. It is not the target of autoantibody formation in bullous pemphigoid. Educational Objective: Bullous pemphigoid is caused by antibodies against the hemidesmosome, an intricate complex of proteins that anchors the basal keratinocytes of the epidermis to the extracellular matrix of the dermis at the dermal-epidermal junction. Previous Next Score Report Lab Values Calculator Help Pause

122 Exam Section 3: Item 22 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 22. Which of the following curves shows the highest cooperativity between a hormone and its receptor? (A B. Hormone A) B) C) D) E)

A. Cooperative binding describes an interaction between a ligand and receptor in which binding of the ligand to one receptor site increases the receptor's affinity to bind more of the same ligand. The most common example of this is hemoglobin's increased affinity for oxygen molecules after sequentially binding oxygen at one or more of its four binding sites. This is best demonstrated by Curve A in the graph, as once the receptor begins binding the ligand A, the amount of receptor binding rapidly increases to a level much above other receptor/ligand pairs. Incorrect Answers: B, C, D, and E. While the amount of receptor binding increases with an increased hormone concentration in Curves B, C, and D (Choices B, C, and D), the relative increase in binding for the hormone by the receptor does not increase from the decreased concentration baseline as much as Curve A. This indicates that Curves B, C, and D do not demonstrate as high a degree of cooperativity as Curve А. The amount of receptor binding in Curve E (Choice E) decreases with increased ligand binding, indicating that this does not show characteristics of cooperativity. Educational Objective: Cooperativity is the mechanism by which a ligand binding to a receptor increases the receptor's affinity to bind additional ligand. Previous Next Score Report Lab Values Calculator Help Pause Receptor binding

107 Exam Section 3: Item 7 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 7. A 68-year-old man comes to the physician because of difficulty swallowing solids for 2 months. He has a history of dilated cardiomyopathy. X-rays of the esophagus with barium contrast show indentation and posterior displacement of the esophagus. Enlargement of which of the following structures is the most likely cause of the dysphagia? A) Left atrium B) Left ventricle C) Right atrium D) Right ventricle E) Superior vena cava

A. In the typical anatomical orientation, the left atrium is the most posterior chamber of the heart and lies anterior to the esophagus in the mediastinum. Enlargement of the left atrium in dilated cardiomyopathy can result in esophageal distortion and displacement. Dilated cardiomyopathy can have numerous causes. It is often familial or idiopathic. Other causes include toxins (eg, alcohol, cocaine), adverse effects of medications, infections (eg, Chagas disease), chronic ischemia, thyrotoxicosis, thiamine deficiency, sarcoidosis, and hemochromatosis. Patients may present with signs and symptoms of heart failure, including dyspnea on exertion, orthopnea, peripheral edema, an S, heart sound, and increased jugular venous pressure. The heart will appear dilated with a thin myocardium on echocardiography. Incorrect Answers: B, C, D, and E. The left ventricle (Choice B) is located inferior and lateral to the left atrium. It comprises the left pulmonary surface and the majority of the left border of the heart. The right atrium (Choice C) makes up the right pulmonary surface of the heart and the right heart border. The right ventricle (Choice D) is the most anterior part of the heart. It forms the inferior (or diaphragmatic) surface of the heart along with the left ventricle. It lies just posterior to the sternum. The superior vena cava (Choice E) may become congested in dilated cardiomyopathy secondary to heart failure. It is not located directly adjacent to the esophagus. Educational Objective: The left atrium is the most posterior chamber of the heart and lies directly anterior to the esophagus in the mediastinum. In left atrial enlargement, the esophagus can be distorted and displaced posteriorly. The proximity of these structures allows for the evaluation of the heart via transesophageal echocardiography. %3D Previous Next Score Report Lab Values Calculator Help Pause

104 Exam Section 3: Item 4 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 4. A 35-year-old woman is brought to the emergency department because of a 3-hour history of sharp cramping pain in the right flank. Urinalysis shows blood. A noncontrast CT scan is obtained, and a renal calculus is found to be lodged in the ureter just above the sacroiliac joint. At this point, the ureter narrows as it passes over which of the following structures? A) External iliac vessels B) Infundibulopelvic ligament C) Internal pudendal vessels D) Left colic vessels E) Ovary

A. Renal calculi can cause symptoms of acute flank or back pain and hematuria. They especially cause pain or complications when lodged along the path of the ureter. The ureter descends in the abdomen anteriorly and medially to the psoas muscle and enters the pelvis anteriorly to the sacroiliac joint at the bifurcation of the common iliac vessels. In the pelvis, the ureter enters the posterolateral wall of the bladder at the internal ureteric orifice. There are three main sites of narrowing along the ureter that may obstruct a calculus: at the ureteropelvic junction of the renal pelvis and the ureter, the ureterovesical junction as the ureter enters the posterior bladder, and the pelvic inlet when the ureter crosses the external iliac vessels. The external iliac vessels cause narrowing of the ureter at the pelvic brim. Calculi smaller than 5 mm are likely to pass spontaneously, whereas those of larger size may require lithotripsy or operative removal to prevent hydronephrosis and kidney injury. Incorrect Answers: B, C, D, and E. The infundibulopelvic ligament (Choice B) passes laterally to the ureter. They do not course in close anatomic proximity. The internal pudendal vessels (Choice C) are a branch of the internal iliac vessels and supply the perineum and external genitalia. They run along the lateral pelvic wall and would not cause ureteral obstruction. The left colic vessels (Choice D) pass anteriorly to the left ureter. The left colic artery is a branch of the inferior mesenteric artery. It does not cause narrowing of the ureter. The ovary (Choice E) is typically anterior to the ureter and would be unlikely to cause obstruction. Educational Objective: There are three main sites of narrowing along the course of the ureter that may obstruct a calculus: at the ureteropelvic junction of the renal pelvis and the ureter, the ureterovesical junction as the ureter enters the posterior bladder, and the pelvic inlet when the ureter crosses the external iliac vessels. %3D Previous Next Score Report Lab Values Calculator Help Pause

103 Exam Section 3: Item 3 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment Peak A Peak B 0 1 2 3 4 6. 24 Hours 3. In an experimental model of mild thermal injury to skin (a single application of 54°C for 20 seconds), the following pattern of extravasation of intravenously administered dye is observed. Administration of an antihistamine prior to thermal injury would have significantly blunted the extent of dye extravasation observed at which of the following labeled sites? A) Peak A only B) Peak B only C) Both Peak A and B D) Neither Peak A nor B

A. The initial vasodilation that occurs in thermal injury is mediated by histamine, which promotes vasodilation of arterioles and leakage from capillaries leading to initial extravasation and tissue edema. Mast cell degranulation results in the release of preformed histamine which binds its receptors on the vascular endothelium and smooth muscle. An antihistamine would blunt this effect and limit initial vasodilation and extravasation, which would be marked in this model by a decrease in Peak A. Incorrect Answers: B, C, and D. Peak B (Choice B), delayed extravasation, occurs because of the cytokine milieu produced by damaged skin cells and by responding macrophages and inflammatory cells. Tumor necrosis factor, interleukins, and prostaglandins have all been implicated in the inflammatory response to a burn, which is marked by increased vascular permeability, tissue edema, and local vasodilation. Choices C and D are incorrect, as the initial peak is mediated by histamine with the delayed peak mediated by cytokine release. Educational Objective: The initial vasodilation that occurs in thermal injury is mediated by histamine, which promotes vasodilation of arterioles and leakage from capillaries leading to extravasation and tissue edema. Antihistamines can attenuate this initial response. Previous Next Score Report Lab Values Calculator Help Pause

123 Exam Section 3: Item 23 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 23. A 17-year-old girl with type 1 diabetes mellitus comes to the physician for a follow-up examination. She has had multiple admissions to the hospital during the past 5 months for diabetic ketoacidosis. She appears well. Physical examination shows no abnormalities. Laboratory studies show a hemoglobin A, of 11.5%. She tells the physician that she does not take her insulin regularly because she either forgets to or is too busy. Which of the following is the most appropriate initial step by the physician? A) Acknowledge the patient's reasons for missing insulin injections B) Discuss the patient's noncompliance with her parents C) Educate the patient about the long-term complications of untreated diabetes D) Tell the patient's parents to ensure that their daughter is compliant E) Refer the patient to a psychologist

A. The physician should validate this patient's barriers to adherence. Regularly administering insulin can be challenging, especially for teenagers who possess underdeveloped executive functioning abilities. Acknowledging this struggle can improve therapeutic alliance and help the physician and patient problem-solve around these barriers to adherence. Barriers to adherence may be individual (eg, age, stigma, problems with routine, concern about adverse effects), interpersonal (eg, family belief systems), or systemic (eg, cost, transportation, sociodemographic factors). Physicians should target dynamic (eg, changeable) risk factors and recognize static risk factors. Incorrect Answers: B, C, D, and E. Involving the patient's parents (Choices B and D) or a psychologist (Choice E) constitutes a paternalistic approach that may trigger the patient's defensiveness and preclude an open and effective discussion. Physicians should attempt to engage adolescent patients in open-ended discussions about barriers to adherence, as behavior change is more likely to happen when patients are engaged and intrinsically motivated. Further, difficulty adhering to medical recommendations is typical for teenagers, so a psychologist referral would not be indicated (Choice E). Educating the patient about the long-term complications of untreated diabetes (Choice C) may be appropriate later in the discussion. The physician should first acknowledge the challenge of adhering to insulin to engage the patient and promote open discussion about the barriers to adherence. Educational Objective: When patients experience barriers to adhering to medical recommendations, physicians should first acknowledge that adherence is challenging and then collaboratively problem-solve around these barriers. %D Previous Next Score Report Lab Values Calculator Help Pause

183 Exam Section 4: Item 33 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 33. An investigator is studying the effects of several different drugs on human brain activity. During the study, positron emission tomography is used to measure real-time brain activity in human participants after administration of various drugs with abuse potential. Results show a common pattern of neurologic effects for drugs such as cocaine, amphetamines, opioids, sedative- hypnotics, alcohol, and nicotine. Activation of which of the following neurotransmitter pathways best explains the rewarding or positive reinforcing effects of using these drugs? A) A dopaminergic from the nucleus accumbens to the prefrontal cortex B) An endocannabinoid from the anterior thalamic nuclei to the somatosensory cortex C) An endorphin (opioid) from the substantia gelatinosa to the ventral tegmental area D) A GABAergic from the amygdala to the limbic cortex E) A noradrenergic from the locus caeruleus to the frontal cortex

A. The release of dopamine by the nucleus accumbens to the prefrontal cortex mediates the conscious experience of the pleasurable effects of drugs, drug craving, and compulsive drug administration. This mesocortical circuit acts in concert with the mesolimbic pathway (dopaminergic projections from the ventral tegmental area to the nucleus accumbens), which mediates anticipatory reward. All drugs of abuse increase mesocortical and mesolimbic dopamine pathway activity. These dopaminergic pathways are therefore thought to underlie addiction, as the activation of these pathways can lead to drug-seeking behavior at the expense of a patient's health or social functioning. Incorrect Answers: B, C, D, and E. An endocannabinoid from the anterior thalamic nuclei to the somatosensory cortex (Choice B) may induce synaptic plasticity of the cortex. Dopaminergic pathways more directly mediate the reinforcing effects of drugs than the endocannabinoid system. An endorphin (opioid) from the substantia gelatinosa to the ventral tegmental area (Choice C) is not a known neurotransmitter pathway. The ventral tegmental area is involved in dopamine signaling in the mesolimbic pathway. A GABAergic from the amygdala to the limbic cortex (Choice D) may lead to decreased fear and anxiety. However, GABAergic mechanisms in the limbic system are not directly related to the reinforcing effects of drugs. A noradrenergic from the locus caeruleus to the frontal cortex (Choice E) may increase motivation and attention. However, norepinephrine does not directly mediate the rewarding effects of drugs of abuse. Educational Objective: The mesocortical pathway refers to the projection of dopaminergic neurons from the nucleus accumbens to the prefrontal cortex. This dopaminergic pathway mediates the salience of drugs' pleasurable effects, drug cravings, and compulsive drug use. %3D Previous Next Score Report Lab Values Calculator Help Pause

140 Exam Section 3: Item 40 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 40. A 64-year-old man undergoes surgical repair of an abdominal aortic aneurysm. During the repair, the left testicular artery is ligated. Anastomotic supply from which of the following arteries will maintain adequate arterial supply to the left testis in this patient? A) Artery of the ductus deferens B) Inferior vesical artery C) Obturator artery D) Posterior scrotal artery E) Superficial circumflex iliac artery

A. The testicular arteries originate directly from the abdominal aorta bilaterally. They travel through the internal and external inguinal rings to reach the scrotum, where they supply blood to the epididymis and the testis. In the scrotum, they anastomose with the artery of the ductus deferens, which is a branch of the anterior division of the internal iliac artery. Therefore, in this patient with a ligated testicular artery, blood supply to the testis and epididymis will be maintained through the artery of the ductus deferens and its anastomosis with the testicular artery. Incorrect Answers: B, C, D, and E. The inferior vesical artery (Choice B) is a branch of the anterior division of the internal iliac artery that provides blood to the bladder. It is responsible for supply of the prostate and seminal vesicles but does not anastomose with the testicular artery. The obturator artery (Choice C) is a branch of the anterior division of the internal iliac artery that supplies blood to the iliac bones and pelvic muscles prior to exiting the pelvis through the obturator foramen. It anastomoses with the inferior epigastric artery but does not anastomose with the testicular artery. The posterior scrotal artery (Choice D) is a branch of the internal pudendal artery and supplies blood to the scrotum. However, it does not anastomose with the testicular artery. The superficial circumflex iliac artery (Choice E) is a branch of the femoral artery that supplies blood to the skin of the groin. It anastomoses with the deep iliac circumflex, superior gluteal, and lateral femoral circumflex arteries but not the testicular artery. Educational Objective: The testicular arteries arise directly from the aorta bilaterally and travel to the scrotum through the internal and external inguinal rings, where they supply blood to the testis and epididymis. There, they anastomose with the artery of the ductus deferens, which is a branch of the anterior division of the internal iliac artery. Previous Next Score Report Lab Values Calculator Help Pause

128 Exam Section 3: Item 28 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 28. A 58-year-old woman with type 2 diabetes mellitus is brought to the emergency department because of an 18-hour history of fever, confusion, and pain of her left leg. Her temperature is 39.6°C (103.3°F), pulse is 124/min, respirations are 22/min, and blood pressure is 94/60 mm Hg. Physical examination shows a diffuse, erythematous rash and conjunctival injection. Erythema and edema extend from the dorsum of the left foot to the knee. She is oriented to person but not to place or time. Neurologic examination shows no other abnormalities. Serum studies show: Urea nitrogen Creatinine Total bilirubin 36 mg/dL 2.8 mg/dL 2.2 mg/dL 470 U/L 385 U/L AST ALT One day later, blood cultures grow gram-positive cocci in chains. These bacteria most likely produced a toxin that caused this patient's condition via which of the following mechanisms? A) Binding to the T-lymphocyte receptor and class II MHC molecules B) Blockade of protein synthesis by depurination of ribosomal RNA C) Catalyzing of the formation of CAMP from ATP D) Cleavage of kinases that activate mitogen-activated protein kinase to induce macrophage apoptosis E) Disruption of cytoskeleton architecture by glycosylating Rho1

A. This patient presents with signs and symptoms of toxic shock syndrome secondary to cellulitis with Streptococcus pyogenes. S. pyogenes is capable of producing exotoxin A. Similar to the toxic shock syndrome toxin (TSST-1) produced by Staphylococcus aureus, exotoxin A functions as a superantigen by activating both class II major histocompatibility complex (MCH) molecules and T- lymphocyte receptors, leading to excessive release of proinflammatory cytokines. The ensuing cytokine storm results in dysfunction of multiple organs and can lead to shock and death. Patients present with fever, hypotension, diffuse erythematous rash, desquamation, and clinical and laboratory evidence of organ dysfunction, such as altered mental status, myalgia, thrombocytopenia, azotemia, and transaminitis. Toxic shock secondary to streptococcal infection may be fatal, even with appropriate treatment. Treatment includes antibiotics, aggressive hemodynamic support, and support of organ function. Incorrect Answers: B, C, D, and E. Blockade of protein synthesis by depurination of ribosomal RNA (Choice B) is the mechanism of action of Shiga toxin produced by Shigella species and of Shiga-like toxin produced by Escherichia coli. These infections present with bloody diarrhea and potentially hemolytic uremic syndrome (HŪS). While HUS may present with altered mental status and renal dysfunction, this diagnosis is less likely given this patient's transaminitis and lower extremity rash, which is more suggestive of streptococcal cellulitis and toxic shock. Catalyzing of the formation of CAMP from ATP (Choice C) is the mechanism of action of several bacterial toxins, including heat labile and heat stable toxins produced by E. coli and cholera toxin produced by Vibrio cholerae. By increasing the production of CAMP, these toxins cause disordered electrolyte handling by gastrointestinal enterocytes, leading to watery diarrhea and solute loss. Cleavage of kinases that activate mitogen-activated protein kinase to induce macrophage apoptosis (Choice D) is the mechanism of lethal toxin, a subunit of the anthrax toxin produced by Bacillus anthracis. Anthrax manifests with pulmonary, cutaneous, or gastrointestinal syndromes. Cutaneous anthrax presents with painless, necrotic eschars. Disruption of cytoskeleton architecture by glycosylating Rho1 (Choice E) is the mechanism of action of C3 toxin produced by Clostridium botulinum. C3 toxin alters cytoskeletal function through glycosylation of Rho1 GTPase. Several other bacterial species also produce C3-like toxins. Educational Objective: Toxic shock syndrome may result from infection with Streptococcus pyogenes and presents with fever, hypotension, diffuse erythematous rash, desquamation, and clinical and laboratory evidence of organ dysfunction. Streptococcal exotoxin A functions as a superantigen by activating both class II MCH molecules and T-lymphocyte receptors, leading to a cytokine storm. %3D Previous Next Score Report Lab Values Calculator Help Pause

143 Exam Section 3: Item 43 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 43. A 29-year-old woman with major depressive disorder develops constipation and urinary retention 1 week after beginning antidepressant therapy. The most likely cause of these adverse effects is blockade of which of the following neurotransmitters by the antidepressant? OA) Acetylcholine B) Dopamine C) Histamine D) Norepinephrine E) Serotonin

A. Treatment for major depressive disorder involves influencing the action of neurotransmitters. Some medications used include selective-serotonin reuptake inhibitors (SSRIS), whereas others block both serotonin and norepinephrine reuptake (eg, serotonin-norepinephrine reuptake inhibitors and tricyclic antidepressants). Less selective antidepressants have a higher risk for adverse effects secondary to their action on other neurotransmitters, such as acetylcholine. Tricyclic antidepressants (TCAS) have a high risk for anticholinergic effects by antagonizing muscarinic acetylcholine receptors. Muscarinic receptors stimulate the contraction of smooth muscle in the urinary bladder wall. The anticholinergic adverse effects can antagonize this action, resulting in impaired urinary bladder contractility which leads to urinary retention. Other anticholinergic effects can include dry mouth, sedation, constipation, hallucinations, delirium, ataxia, flushed skin, and visual disturbances (cycloplegia). Incorrect Answers: B, C, D, and E. Blockade of dopamine (Choice B) is the mechanism of action for most antipsychotic drugs. Antidepressants do not typically block dopamine, and dopamine antagonism would not lead to urinary retention. Antidepressants can also antagonize histamine (Choice C) H1 and H2 receptors. Histaminergic blockade may present with dry mouth and sedation. It would not cause urinary retention. Some antidepressants such as serotonin-norepinephrine reuptake inhibitors (SNRIS) and TCAS inhibit both serotonin and norepinephrine (Choice D) reuptake. The action of norepinephrine is not blocked; rather, it is promoted. Treatment for major depressive disorder involves the inhibition of serotonin reuptake at nerve terminals, leading to increased concentrations of serotonin in the synapse, not the blockade of serotonin activity (Choice E). Educational Objective: Less selective antidepressants have a higher risk for adverse effects secondary to action on other neurotransmitter pathways such as acetylcholine. Muscarinic receptors stimulate the contraction of smooth muscle in the urinary bladder wall. The anticholinergic adverse effects of nonspecific antidepressants can antagonize this action, resulting in impaired bladder contractility and urinary retention. Previous Next Score Report Lab Values Calculator Help Pause

193 Exam Section 4: Item 44 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 44. An 80-year-old woman undergoes a brain biopsy of a cerebral mass. Immunofluorescence studies of a biopsy specimen show that the tumor cells contain intermediate filaments of the cytokeratin type. This patient's cancer cells most likely arose from which of the following cell types? A) Astrocytes B) Epithelial cells C) Mesenchymal cells D) Muscle cells E) Neurons

B. Solitary brain lesions may be primary central nervous system malignancies, metastatic disease, infections, or abscesses. Clinical features and analysis of a biopsy specimen can narrow the diagnosis. Intermediate filaments are one of the three major structural polymers of the cellular cytoskeleton, the other two being microtubules and microfilaments. Microfilaments have the smallest diameter (-7 nm), intermediate filaments the next smallest (~10 nm), and microtubules the largest diameter (~25 nm). Intermediate filaments are able to stretch and contract but do not undergo constant dynamic remodeling like the other polymer types. They provide strength to the cytoskeleton. There are five types of intermediate filaments categorized by their protein composition. Types I and Il are composed of keratin and stain positive with cytokeratin immunohistochemical stains. Keratin-based intermediate filaments are found in epithelial cells. Incorrect Answers: A, C, D, and E. Astrocytes (Choice A) are glial neuronal support cells. They may form malignant astrocytomas which are the most common type of primary brain tumor; glioblastoma multiforme is an example of a high-grade, aggressive glial malignancy. Intermediate filaments of astrocytes are composed of glial fibrillary acidic protein. Mesenchymal cells (Choice C) are pluripotent stem cells located in the bone marrow that differentiate into chondrocytes and osteocytes. Intermediate filaments in mesenchymal cells are formed by the protein vimentin. Muscle cells (Choice D) have intermediate filaments formed by a similar protein called desmin. Neurons (Choice E) form nerve tissue in the central and peripheral nervous systems. Intermediate filaments in neurons of the central nervous system are composed of neurofilament, while peripheral neurons contain peripherin. Educational Objective: Intermediate filaments are essential components of the cell cytoskeleton and are composed of different proteins dependent on the tissue of origin. Keratin is the primary component of intermediate filaments in epithelial tissue. Previous Next Score Report Lab Values Calculator Help Pause

182 Exam Section 4: Item 32 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 32. Fatty acids are infused into the proximal small intestine of an experimental animal. Which of the following is the most likely change in peptide secretion in this animal? A) Increased amylase B) Increased cholecystokinin C) Increased pancreatic polypeptide OD) Increased secretin E) Suppressed amylase O F) Suppressed cholecystokinin G) Suppressed motilin H) Suppressed pancreatic polypeptide 1) Suppressed secretin

B. The presence of increased fatty acids in the small intestine would stimulate the secretion of cholecystokinin. Cholecystokinin is released from cells in the duodenum and jejunum and stimulates the cholecystokinin receptor. Stimulation of this receptor leads to increased gastric acid secretion, pancreatic enzyme secretion, gallbladder contraction and bile secretion, and sphincter of Oddi relaxation. Pancreatic digestive enzymes include amylase which aids in carbohydrate digestion, lipase which aids in the digestion of fatty acids, and proteases which help digest proteins. Trypsinogen is also secreted by the pancreas; its active form, trypsin, activates proenzymes for digestion. Incorrect Answers: A, C, D, E, F, G, H, and I. Amylase is secreted by the pancreas; however, its release is not directly stimulated by fatty acids in the duodenum. Its secretion is a downstream response to cholecystokinin release (Choice A). Fatty acids directly regulate and stimulate cholecystokinin release. Suppressed amylase (Choice E) is the opposite of what should be seen in fatty acid stimulation of cholecystokinin release. Pancreatic polypeptide is secreted by pancreatic cells and functions to self-regulate secretion of both exocrine and endocrine pancreatic products. Increased pancreatic polypeptide (Choice C) is seen in states of fasting, exercise, hypoglycemia, and after a protein-rich meal. It is not directly stimulated by fatty acids. Suppressed pancreatic polypeptide (Choice H) is seen in states of hyperglycemia and its release is inhibited by somatostatin. Secretin is released by S cells in the duodenum. It increases pancreatic bicarbonate secretion and decreases gastric acid secretion. Increased secretin (Choice D) is stimulated by acid in the lumen of the duodenum, and the bicarbonate released by the pancreas neutralizes gastric acid, allowing pancreatic enzymes to function. Suppressed secretin (Choice I) would lead to decreased pancreatic bicarbonate secretion (an acidic milieu in the duodenum) and decreased activity of pancreatic enzymes. Suppressed cholecystokinin (Choice F) would be the opposite of what should be expected with the presence of fatty acids in the proximal small intestine. Fatty acids in the small intestine stimulate cholecystokinin release. Motilin (Choice G) stimulates contraction of the gastric antrum and fundus to promote gastric emptying. It also stimulates peristalsis in the small bowel. It would not be suppressed by the infusion of fatty acids into the small intestine. Educational Objective: The presence of fatty acids in the small intestine stimulates the secretion of cholecystokinin, which leads to increased gastric acid secretion, pancreatic enzyme secretion, gallbladder contraction with subsequent bile secretion, and sphincter of Oddi relaxation. Previous Next Score Report Lab Values Calculator Help Pause

130 Exam Section 3: Item 30 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 30. A 25-year-old man comes to the physician because of a 2-month history of recurrent episodes of chest pain and shortness of breath. He has a 2-year history of nodules on his hands and feet. Physical examination shows numerous 1- to 2-cm yellow nodules that are palpable on both Achilles tendons and on the extensor tendons of the hands. Ophthalmologic examination shows corneal arcus. Serum studies show: 35 U/L 350 mg/dL 50 mg/dL 300 mg/dL 100 mg/dL 0.1 ng/mL (N<0.4) Creatine kinase Cholesterol, total HDL-cholesterol LDL-cholesterol Triglycerides Troponin I An ECG shows no abnormalities. This patient most likely has an impairment in function involving which of the following structures? A) Caveolae B) Clathrin-coated pit C) COPII-coated pit D) Phagosome E) Snap receptors (SNARES)

B. The transport of substances from the extracellular to the intracellular compartment involves multiple possible pathways, such as diffusion through the semipermeable membrane, active or passive passage through membrane channels, or endocytosis with protein-coated vesicle formation. Endocytosis involves binding of the target substance, invagination and cleavage of the cell membrane bud, and finally, transport to the correct area of the cell. Clathrin-coated vesicles are responsible for the majority of receptor-mediated endocytosis and transport from the cell membrane, including the uptake of LDL by hepatocytes. Familial hypercholesterolemia (also known as type Il familial dyslipidemia) is a relatively common, autosomal dominant disorder of cholesterol metabolism. The disorder is caused by mutations in the LDLR, PCSK9, or APOB genes. Defects in these genes lead to the decreased clearance of cholesterol by the liver and increased concentrations of cholesterol in the serum. LDLR encodes the LDL receptor (LDLR) expressed on hepatocytes, and loss of function mutations decrease LDL uptake by the liver through endocytosis involving clathrin-coated pits. Gain-of-function mutations in PCSK9 lead to decreased LDLR expression through formation of the PCSK9-LDLR complex, which is internalized by the cell and unable to be recycled to the cell surface. Apolipoprotein B-100 (ApoB-100) defects decrease the ability of the LDL receptor to bind LDL. Incorrect Answers: A, C, D, and E. Caveolae (Choice A) are cell membrane structures composed of glycolipids, cholesterol, and the protein caveolin. They are found on many but not all cell types and participate in clathrin- independent endocytosis. Hepatocytes primarily utilize clathrin-mediated endocytosis. COPII-coated pits (Choice C) are involved in the molecular transport of substances from the endoplasmic reticulum to the Golgi apparatus. Phagosome (Choice D) formation occurs in phagocytosis, where specialized cells engulf large particles such as bacteria, protozoa, or cellular debris. It involves actin-mediated extension of pseudopodia and membrane fusion around the target. In contrast, LDL uptake is a form of pinocytosis, or the uptake of small macromolecules and substances. Snap receptors (SNARES) (Choice E) are primarily involved in vesicle fusion with a target membrane. For example, SNARE complexes mediate the release of neurotransmitters from vesicles into the synaptic cleft. Educational Objective: Clathrin-coated pits are the main structures on the cell membrane involved in receptor-mediated endocytosis. Binding of a ligand to a receptor, such as LDL to the LDL- receptor on hepatocytes, results in vesicle formation and transport of the ligand-receptor complex into the cell. Previous Next Score Report Lab Values Calculator Help Pause

102 Exam Section 3: Item 2 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 2. A 79-year-old woman comes to the physician because of swollen, painful joints in her hands and knees for 6 months. Her hematocrit is 39%. Serum protein electrophoresis shows a monoclonal IgG spike of 1.8 g; serum IgA and IgM concentrations are within the reference range. Her urine does not contain Bence Jones protein. A bone marrow biopsy specimen shows no abnormalities. Which of the following is the most likely explanation for the increased serum IgG concentration in this patient? A) Chronic hepatic disease B) Chronic inflammation C) Monoclonal gammopathy of uncertain significance D) Multiple myeloma E) Waldenström macroglobulinemia

C. Monoclonal gammopathy of uncertain significance (MGUS) is the most likely explanation of this patient's monoclonal IgG spike of 1.8 g/dL in the absence of other signs or symptoms of multiple myeloma. MGUS, smoldering myeloma, and multiple myeloma are all diseases existing on a spectrum defined by the presence of a monoclonal gammopathy, but they have specific distinguishing characteristics, criteria, and prognoses. MGUS is diagnosed when patients have a monoclonal spike of either IgG or IgM that is less than 3 g/dL and do not have any signs or symptoms of multiple myeloma (ie. Iytic lesions, renal insufficiency, anemia, or hypercalcemia). Smoldering myeloma diagnosis requires a monoclonal spike (M-spike) greater than 3 g/dL but no signs or symptoms. Multiple myeloma diagnosis requires the patient to have signs and symptoms along with either an M-spike of greater than 3 g/dL or a bone marrow biopsy with greater than 10% plasma cells. MGUS is generally a benign condition with a decreased rate of transformation to smoldering myeloma or multiple myeloma. Those with an M-spike of less than 1.5 g/dL are monitored with history-taking and physical examination alone. All patients with greater M-spikes require annual laboratory screening to evaluate for progression. Incorrect Answers: A, B, D, and E. Chronic hepatic disease (Choice A) can cause increased concentrations of all immunoglobulins. A monoclonal spike of IgG or IgM would not be expected. Additionally, this patient has no sequelae of chronic liver disease such as palmar erythema, ascites, or encephalopathy. Chronic inflammation (Choice B) can increase the concentrations of all immunoglobulins but would not be associated with a monoclonal protein spike, which suggests clonal proliferation of plasma cells. Multiple myeloma (Choice D) is a disorder of clonal plasma cell proliferation. It presents with signs and symptoms of multiple myeloma, including renal insufficiency, hypercalcemia, anemia, and lytic lesions. Additionally, laboratory studies show an M-spike greater than 3 g/dL or examination of a bone marrow biopsy specimen shows greater than 10% plasma cells. This patient does not meet these criteria. Waldenström macroglobulinemia (Choice E) is a B-lymphocyte malignancy that presents with a monoclonal IgM spike, not IgG. Additionally, common presenting features include neuropathy, cryoglobulinemia, and hyperviscosity syndrome. Educational Objective: MGUS is defined by an M-spike of less than 3 g/dL in the absence of signs or symptoms of multiple myeloma, including renal insufficiency, anemia, hypercalcemia, and lytic bone lesions. Progression to multiple myeloma does occur but at a decreased rate, and patients are followed clinically without the need for treatment. Previous Next Score Report Lab Values Calculator Help Pause

190 Exam Section 4: Item 41 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 41. The diagram shows the major factors that determine blood pressure. Which of the following labeled factors is most responsible for the antihypertensive effect of a dihydropyridine calcium- channel blocking agent? Central nervous system A Blood pressure = Cardiac output x Peripheral resistance (arteriolar) Stroke volume Heart rate B Contractility Venous return Capacitance vessel tone (venular) Blood volume D) OA) B) C) D) E)

C. Dihydropyridine calcium-channel blockers (CCBS), such as amlodipine and nifedipine, are potent arteriolar vasodilators that decrease blood pressure by decreasing vascular peripheral resistance. CCBS, in general, inhibit L-type calcium channels on smooth muscle cells and cardiomyocytes, decreasing the transmembrane calcium current and resulting in vasodilation and decreased myocyte contractility. The dihydropyridine CCBS have minimal cardiomyocyte activity and primarily act on vascular smooth muscle cells. Adverse effects of ČCBS include peripheral edema, headache, light-headedness, and facial flushing. They are also associated with gingival hyperplasia. Incorrect Answers: A, B, D, and E. The central nervous system (Choice A) is the site of action for a-adrenergic agonists such as clonidine. a-Adrenergic agonists decrease sympathetic outflow from the central nervous system by inhibiting the presynaptic release of norepinephrine. Heart rate (Choice B) is affected by B-adrenergic antagonists, such as metoprolol and propranolol, and by the nondihydropyridine CCBS, such as verapamil and diltiazem. Adverse effects include bradycardia, heart block, and/or impaired cardiac output. Blood volume (Choice D) may be decreased in the treatment of hypertension through the use of diuretics. CCBS do not have a significant effect on blood volume. Venular capacitance vessel tone (Choice E) is affected by nitrates, which act preferentially on veins more than arteries. Nitrates result in vascular smooth muscle relaxation and venous pooling of blood, which decreases cardiac preload. Educational Objective: Dihydropyridine calcium-channel blockers (CCBS) function as effective antihypertensive medications by inhibiting calcium influx into vascular smooth muscle cells, resulting in vasodilation and a decrease in peripheral arteriolar resistance. In contrast with nondihydropyridine CCBS, they have minimal activity on the cardiomyocytes. I3D Previous Next Score Report Lab Values Calculator Help Pause

168 Exam Section 4: Item 18 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 18. A 53-year-old woman is diagnosed with gastric carcinoma and receives a chemotherapeutic regimen including doxorubicin. Six months later, she develops cardiomyopathy. Dose-limiting toxicity of doxorubicin, resulting in cardiomyopathy, is produced by which of the following mechanisms? A) Downregulation of epidermal growth factor receptors B) Formation of abnormal myoglobin C) Generation of free radicals D) Impairment of actin synthesis E) Reduction of formation of nitric oxide in cardiac cells

C. Doxorubicin (and other anthracyclines) have a well-established cardiotoxicity. Myocyte damage occurs because the increased generation of free radicals and consequent reactive oxygen species increase oxidative stress, leading to lipid peroxidation of myocyte membranes and myocyte apoptosis. Patients may present with decompensated heart failure from dilated cardiomyopathy, characterized by dyspnea on exertion, pulmonary crackles (rales), tachycardia, tachypnea, and peripheral edema on physical examination. Diagnostic evaluation with chest imaging and echocardiography may show cardiomegaly, pulmonary edema, and a decreased ejection fraction indicating abnormal systolic function. Anthracycline-induced dilated cardiomyopathy confers a poor prognosis. Patients managed with doxorubicin should be monitored for cardiotoxicity with routine echocardiography. Incorrect Answers: A, B, D, and E. Downregulation of epidermal growth factor receptors (EGFRS) (Choice A) is the mechanism of EGFR inhibitors used in the treatment of lung and colon cancer. The most common adverse effect of EGFR inhibitors is a papulopustular rash. Formation of abnormal myoglobin (Choice B) is not associated with doxorubicin. Myoglobin is a sensitive marker for muscle tissue injury and rhabdomyolysis, which may occur with several chemotherapeutic agents including doxorubicin. The abnormal release of normal myoglobin can cause nephrotoxicity, leading to an acute kidney injury from acute tubular necrosis. Impairment of actin synthesis (Choice D) is not associated with doxorubicin. Other chemotherapeutic agents such as paclitaxel, vincristine, and vinblastine target microtubules (composed of tubulin). The major adverse effects of paclitaxel are bone marrow suppression and neuropathy, while vincristine is associated with neurotoxicity and vinblastine is associated with bone marrow suppression. Reduction of formation of nitric oxide in cardiac cells (Choice E) is associated with numerous biological effects. Nitric oxide is synthesized by nitric oxide synthase, which is regulated by complex signaling pathways. Doxorubicin is not associated with changes in these pathways. Educational Objective: Cardiotoxicity leading to dilated cardiomyopathy is a well-known adverse effect of doxorubicin chemotherapy. This occurs secondary to the increased production of free radicals and consequent reactive oxygen species resulting in lipid peroxidation of myocyte membranes and myocyte apoptosis. I3D Previous Next Score Report Lab Values Calculator Help Pause

164 Exam Section 4: Item 14 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 14. A 5-year-old boy is stung on his foot by a bee. Within 30 minutes the local area is edematous. The extravascular accumulation of fluid is most directly related to which of the following? A) Demargination of leukocytes B) Fibrin thrombi C) Gap formation between endothelial cells D) Vasoconstriction E) Vasodilation

C. Gap formation between endothelial cells following a bee sting allows for the accumulation of extravascular fluid. Bee venom contains melittin, which hydrolyzes cellular membranes and causes local damage at the site of envenomation. It also increases cellular permeability and induces synthesis of arachidonic acid through the action of phospholipase-A, which damages the vascular endothelium. Additionally, mast cell degranulation with release of histamine causes local vasodilation. Collectively, these mechanisms weaken junctions between endothelial cells and allow for the extravasation of fluid into the interstitial space, which causes local swelling. Incorrect Answers: A, B, D, and E. Demargination of leukocytes (Choice A) is not a feature of bee stings. Certain medications such as steroids can cause massive demargination of leukocytes from the vascular endothelium, as do local bacterial infections. Fibrin thrombi (Choice B) are present in blood clots. Damage to vascular endothelium activates the clotting cascade, leading to deposition of fibrin strands in conjunction with trapping and activation of platelets to form a clot. This does not occur in bee stings. Vasoconstriction (Choice D) does not occur in bee stings, although vasodilation (Choice E) does. While vasodilation contributes to increased blood flow to the area of the envenomation, it does not cause increased vascular permeability with extracellular accumulation of fluid. Educational Objective: Bee venom contains several active peptides that cause direct cellular membrane damage to endothelial cells, resulting in the formation of gaps between cells with subsequent extravasation of fluid into the extravascular space. %D Previous Next Score Report Lab Values Calculator Help Pause

112 Exam Section 3: Item 12 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 12. A 6-year-old girl of Mediterranean ancestry has severe anemia. Analysis shows normal production of a-globin gene MRNA. Production of B-globin gene MRNA is decreased 75% because of a splice-site mutation within both of the patient's B-globin genes. Which of the following is the most likely explanation for the anemia? A) Decreased production of a-4 tetramers compared with normal individuals B) Decreased production of B-4 tetramers compared with normal individuals C) Imbalance in the synthesis of the a- and B-globin gene products D) Imbalance in the synthesis of the four a-globin genes

C. Imbalance in the synthesis of the a- and B-globin gene products most likely accounts for this patient's anemia, which is consistent with a diagnosis of B-thalassemia major. The normal adult hemoglobin consists of two a-globin subunits and two B-globin subunits, which together form aß2 (HbA), a unique conformation that has a high affinity for oxygen in the lungs and can offload oxygen effectively in the peripheral tissues. In patients who have a deficient B-globin gene, the normal HbA molecule cannot be effectively assembled, and the excess a-globin chains combine with one another to form tetramers that are insoluble. They precipitate in erythrocytes and contribute to ineffective erythropoiesis. Additionally, excess a-globin molecules can associate with other various globin molecules including y- and o- to form other varieties of hemoglobin such as fetal hemoglobin (HbF) and HbA, respectively. Patients with B-thalassemia major can be identified by hemoglobin electrophoresis. Incorrect Answers: A, B, and D. Decreased production of a-4 tetramers compared with normal individuals (Choice A) would not occur in this patient. In fact, patients with B-thalassemia major will show increased concentrations of a-4 tetramers. These tetramers are not soluble, precipitate in the erythrocyte, and contribute to ineffective erythropoiesis. Decreased production of B-4 tetramers compared with normal individuals (Choice B) does occur in B-thalassemia major, although B-4 tetramers form an abnormal hemoglobin molecule known as hemoglobin H (HbH) that is not typically found in healthy individuals. Increased concentrations of HbH are found in individuals with a-thalassemia. Imbalance in the synthesis of the four a-globin genes (Choice D) describes the pathophysiology of a-thalassemia. This patient has a normal concentration of a-globin, which essentially rules out this diagnosis. Educational Objective: Normal human hemoglobin A consists of two a- and two B-globin subunits. Mutations in both B-globin alleles lead to B-thalassemia and a relative excess of a-globin molecules, which pair with other a-globin molecules to form a-globin tetramers. These are insoluble in erythrocytes, do not carry oxygen effectively, and interfere with erythropoiesis, thereby contributing to anemia. Previous Next Score Report Lab Values Calculator Help Pause

172 Exam Section 4: Item 22 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 22. A 65-year-old man with hypertension volunteers to participate in a clinical trial of a newly developed loop diuretic. As part of the study, his acid-base/volume status is monitored. After 3 days of treatment, which of the following sets of findings is most likely in this patient? Acid-base Volume Contraction A) Metabolic acidosis yes B) Metabolic acidosis no C) Metabolic alkalosis yes D) Metabolic alkalosis no

C. Loop diuretics (eg, furosemide, torsemide, bumetanide, ethacrynic acid) act at the thick ascending limb of the loop of Henle to decrease solute and fluid reabsorption. They inhibit the sodium- potassium-chloride cotransport channel. As a result of decreased solute reabsorption, less water is reabsorbed by osmotic pull. This leads to a total decrease in body fluid volume or volume contraction. Metabolic alkalosis occurs secondary to dehydration and subsequent volume contraction; there is also a relative excess of bicarbonate. Volume depletion leads to activation of the renin-angiotensin-aldosterone system, which in turn stimulates renal tubule bicarbonate reabsorption and new bicarbonate synthesis. Metabolic alkalosis, volume depletion, and increased concentrations of aldosterone also lead to potassium excretion and hypokalemia. Volume contraction is thus marked by a hypokalemic, metabolic alkalosis on laboratory studies. Incorrect Answers: A, B, and D. The resulting bicarbonate reabsorption secondary to volume depletion and activation of the renin-angiotensin-system leads to metabolic alkalosis, not metabolic acidosis (Choices A and B). Loop diuretics lead to a decrease in body fluid volume and cause volume contraction, excluding options without volume contraction (Choices B and D). Educational Objective: Loop diuretics act at the thick ascending limb of the loop of Henle to decrease solute and fluid reabsorption. This leads to volume contraction, which activates the renin- angiotensin-aldosterone system, which in turn, stimulates bicarbonate reabsorption and new bicarbonate synthesis. This results in metabolic alkalosis. %D Previous Next Score Report Lab Values Calculator Help Pause

179 Exam Section 4: Item 29 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 29. A 32-year-old woman comes to the physician because of a 1-week history of a progressive, itchy rash. The rash initially appeared as an oval patch on her abdomen and has since spread over her abdomen and back. She has not had fever, chills, or sweating. She has no history of major medical illness. She has been sexually active with two male partners during the past year, and she uses an oral contraceptive. Her vital signs are within normal limits. Physical examination shows the findings in the photograph. Which of the following is the most likely diagnosis? A) Erythema infectiosum (fifth disease) B) Molluscum contagiosum C) Pityriasis rosea D) Roseola (exanthema subitum) E) Scabies F) Secondary syphilis

C. Pityriasis rosea is a cutaneous eruption characterized by the acute onset of a pruritic patch, the herald patch, which is followed by the development of numerous smaller macules and patches. It most commonly involves the trunk and proximal extremities. The individual patches of pityriasis rosea are oval in shape, and their longitudinal axis is often lined up with the relaxed skin tension lines bilaterally on the trunk. They are typically scaly and have a scaly rim around the leading edge of any new lesion. In patients of darker skin types, pityriasis rosea may take on a papular appearance or may show hyperpigmentation. It is usually induced by a viral infection including human herpesviruses 6 and 7, though often the exact infection is not determined. Pityriasis rosea self-resolves within 6 to 8 weeks without treatment, though topical corticosteroids may be used for symptomatic management of pruritus. Incorrect Answers: A, B, D, E, and F. Erythema infectiosum (Choice A), or fifth disease, is caused by parvovirus B19. This presents with pink patches on the bilateral cheeks of children, giving an appearance of slapped cheeks. This facial rash may be followed by a lacy, macular eruption on the trunk and extremities. In adults, cutaneous manifestations are uncommon and are replaced by mild joint pain and myalgias. Molluscum contagiosum (Choice B) is a viral infection of the skin caused by poxvirus. Lesions are characteristically round and smooth, and they exhibit a central umbilication. It is transmitted through close contact with infected individuals and thus is commonly seen in multiple children within a family or in children who play contact sports. Molluscum contagiosum will eventually resolve on its own as a result of immune system activation, but resolution can be hastened by using destructive techniques such as liquid nitrogen therapy or curettage. Roseola (exanthem subitum) (Choice D) is a viral syndrome characterized by several days of high fevers, sometimes accompanied by seizures; as the fevers subside, a characteristic, blanching, macular eruption forms on the neck and trunk, which spreads to the face and extremities. It is most frequently caused by human herpesvirus 6 (HHV-6) infection. In a minority of cases, this viral syndrome may also be caused by human herpesvirus 7, enteroviruses, and adenoviruses. Scabies (Choice E) is an ectoparasite infection that causes severe itching through a type IV (delayed) hypersensitivity response to components of the body or saliva of the scabies mite. Classic anatomic sites involved include the digital web spaces, the umbilicus, and the groin. Syphilis is caused by an infection with the spirochete Treponema pallidum and demonstrates multiple stages with varying symptoms. In the secondary stage (Choice F), a scaly rash may develop over the entire body with accentuation on the palms and soles. The rash is characterized by scaly, light brown macules and patches. Pityriasis rosea and secondary syphilis are both included in the differential diagnosis of scaly patches on the trunk and extremities. While this patient has had multiple sexual encounters this year, the presentation of a herald patch makes pityriasis rosea more likely. Educational Objective: Pityriasis rosea is characterized by the acute onset of a pruritic patch followed by the development of numerous smaller macules and patches in a recent viral illness. The individual patches of pityriasis rosea are oval in shape and scaly, and they have a scaly rim around the leading edge. This patient should be reassured that the condition will self-resolve in 6 to 8 weeks and can be treated symptomatically with topical corticosteroids in the meantime. Previous Next Score Report Lab Values Calculator Help Pause

111 Exam Section 3: Item 11 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 11. A 42-year-old woman with type 2 diabetes mellitus comes to the physician because of a 1-day history of increasingly severe right knee pain. She now has difficulty walking because of the pain. She recalls no trauma to the knee and has never had similar symptoms. She is 152 cm (5 ft) tall and weighs 100 kg (220 lb); BMI is 43 kg/m2. Her temperature is 39.3°C (102.7°F), pulse is 110/min, respirations are 12/min, and blood pressure is 160/90 mm Hg. Physical examination shows a 10-cm area of erythema surrounding the right knee, which is warm to the touch and exquisitely tender with movement. There is a moderate effusion. Arthrocentesis yields cloudy yellow synovial fluid. Analysis of the synovial fluid shows a leukocyte count of 90,000/mm3 (95% neutrophils) and no crystals. A Gram stain of the fluid shows numerous segmented neutrophils but no organisms. Which of the following is the most likely cause of the findings in this patient? A) Deep venous thrombosis of the right common femoral vein B) Gout C) Infectious arthritis D) Pseudogout E) Ruptured Baker cyst

C. Septic (infectious) arthritis classically presents with joint pain (worse with loading, ranging, or use), swelling, and overlying erythema. Children, persons with immunocompromising conditions, those with a history of traumatic injury to the joint, and those with existing abnormal joint anatomy (eg, rheumatoid arthritis, gout) may be more commonly affected. The mechanism of infection may be direct inoculation, hematogenous spread, or extension from a neighboring local infection (eg, osteomyelitis). Vital signs often show fever and tachycardia. Physical examination classically shows an erythematous, edematous joint with a palpable effusion, and tenderness elicited on range of motion or palpation. If direct trauma preceded the infection, a punctum or laceration may be present in association with the joint findings. Laboratory studies typically show leukocytosis, along with increased erythrocyte sedimentation rate and c-reactive protein concentrations. Arthrocentesis is essential to diagnose septic arthritis and to differentiate the condition from osteoarthritis and inflammatory arthropathies, such as gout and rheumatoid arthritis. Synovial fluid can also be sent for culture and Gram stain plus sensitivity testing of offending organisms, permitting targeted treatment. The leukocyte count in synovial fluid is generally greater than 50,000/mm3 and predominantly neutrophilic when the infectious agent is bacterial, as in this patient. Treatment includes broad-spectrum antibiotics, and often arthrocentesis with operative irrigation and debridement is necessary to minimize damage to the joint. Incorrect Answers: A, B, D, and E. Deep venous thrombosis of the right common femoral vein (Choice A) would present with unilateral lower extremity edema, tenderness along deep venous tracts, dilated superficial or collateral veins, and pain with passive stretch of the limb, often in an immobilized, hypercoagulable patient. This patient's localized erythema and swelling is more associated with the joint than with a draining vein. Gout (Choice B) is an acute, monoarticular arthropathy resulting from an intra-articular inflammatory reaction to precipitation of monosodium urate crystals. It typically presents with atraumatic joint pain, erythema, and swelling, which can recur in patients who are under- or untreated. The most common joint involved is the first metatarsophalangeal, but other joints include the knees and elbows. Arthrocentesis generally shows 10,000 to 50,000 leukocytes/mm3 with a mixture of neutrophils and lymphocytes, and needle-like, birefringent crystals on microscopy. Treatment consists of nonsteroidal anti-inflammatory drugs (NSAIDS), colchicine, and/or corticosteroids for acute exacerbations. Pseudogout (Choice D), also known as calcium pyrophosphate deposition disease, often results in acute monoarticular arthritis. It commonly involves larger joints in patients who have systemic inflammatory or endocrine diseases (eg, hyperparathyroidism). Arthrocentesis findings are similar to those seen in gout; however, rhomboid crystals are seen on microscopy, and chondrocalcinosis may be seen on x-ray. Ruptured Baker cyst (Choice E) typically presents with acute knee pain and swelling, often in the popliteal fossa. Baker cysts are synovial outpouchings that occur commonly in patients with rheumatoid arthritis. Diagnosis is through ultrasound which will show the cyst. Educational Objective: Septic (infectious) arthritis classically presents with joint pain (worse with loading, ranging, or use), swelling, and overlying erythema. Arthrocentesis generally shows greater than 50,000 leukocytes/mm3 with neutrophilic predominance. In contrast, inflammatory arthropathies generally show fewer leukocytes and often a mixture of lymphocytes and neutrophils. Previous Next Score Report Lab Values Calculator Help Pause

198 Exam Section 4: Item 49 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 49. A 19-year-old woman comes to the physician because of a 2-day history of pain in her left index finger. She injured it during a softball game when catching a ball. Physical examination shows erythema of the left index finger. The patient is unable to flex the distal phalanx when the proximal interphalangeal joint and metacarpophalangeal joints are restrained. Laboratory studies and x-rays of the left hand show no abnormalities. Which of the following structures is most likely injured in this patient? A) Extensor digitorum indicis B) Extensor digitorum tendon C) Flexor digitorum profundus tendon D) Flexor digitorum superficialis tendon E) Median nerve F) Ulnar nerve

C. The abrupt loading of a tendon can result in rupture. Tendons have high tensile strength, but when exceeded, the collagen fibers are unable to maintain connective integrity of the structure, resulting in tears or complete rupture if the loading force is not eliminated. Rupture of the Achilles tendon is a common example, often occurring as a consequence of athletic injury. Tendons are continuations of the connective tissue investing the muscle fibers (epimysium, perimysium, endomysium), and they act to transmit force generated by a muscle. They originate from within the muscle and insert onto the structure that the muscle acts to move, thereby transmitting the force. In this example, the patient is unable to flex the distal phalanx of her index finger, a segment inserted upon by the tendon of the flexor digitorum profundus (FDP). The FDP muscle originates in the forearm, and its tendons traverse the carpal tunnel, palm, and proximal and middle segments of the fingers to insert on the anterior distal phalanges of digits two through five. When the FDP contracts, tension is passed along its tendons to flex the distal phalanx about the distal interphalangeal (DIP) joint. When the tendon is partially or completely ruptured, force cannot be transmitted, and the phalanx will move weakly or will not move at all, respectively. Injuries involving the FDP and extensor digitorum tendons are common in athletes and among those who work with their hands when an inadvertent load is applied to an already tense tendon. The injury can be diagnosed through physical examination, which will disclose tenderness, erythema, or swelling in the area of injury and weakness with attempted motion of the affected tendon. Treatment involves splinting and operative tendon repair. Incorrect Answers: A, B, D, E, and F. Extensor digitorum indicis (Choice A) is involved in extension of the index finger, not flexion. It originates from the ulna within the posterior compartment of the forearm and inserts on the extensor hood of the index finger. Injury to its tendon would result in weakness with extension of the index finger. Similarly, injury to the tendons of the extensor digitorum (Choice B), which originates on the lateral epicondyle and inserts on the extensor expansion of digits two through five, would result in failure to extend the affected finger or fingers. The flexor digitorum superficialis (Choice D) originates in the anterior forearm from the medial humeral epicondyle, and its tendons course through the carpal tunnel and palm to insert on the bases of the anterior middle phalanges of digits two through five. Contraction of this muscle results in flexion of the proximal interphalangeal joints, not the DIP. The median nerve (Choice E) supplies motor fibers to the majority of the muscles of the anterior forearm, along with muscles in the thenar eminence and the first two lumbrical muscles. It transmits sensory input from the anterior palm, the anterior and distal aspects of digits one through three, and the radial aspect of digit four. It supplies motor fibers to the radial half of FDP. Injury to this nerve would result in motor and sensory loss over a larger territory. The ulnar nerve (Choice F) supplies motor fibers to the intrinsic muscles of the hand (with the exception of the thenar eminence and first two lumbrical muscles), to the ulnar aspect of FDP, and to the flexor carpi ulnaris muscle. It transmits sensory input from the anterior medial palm, from digit five, and from the ulnar aspect of digit four. Injury to this nerve would not affect motor or sensory function of the second digit. Educational Objective: Tendon injuries present with functional loss related to the area of injury and affected muscle. In the case of an injury to the flexor digitorum profundus, the patient will not be able to transmit force to flex the distal phalanx of the affected finger. Sensory loss or paresthesia would not be expected with an isolated tendon injury and would indicate that a nerve was also affected. Previous Next Score Report Lab Values Calculator Help Pause

150 Exam Section 3: Item 50 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 50. A hospital laboratory is comparing two methods for measuring serum sodium concentrations. Using method I, the coefficient of variation for the analysis of a series of known samples is 3.5%. Using method II, the coefficient of variation is 2.5% using the same samples. Which of the following best describes the precision of these methods? A) Both are equally precise B) Method I is more precise than method II C) Method Il is more precise than method I D) Precision cannot be determined from information given

C. The coefficient of variation is a standardized measure of dispersion of data. It is generally applied to probability distributions following the characteristics of a normal, Gaussian distribution. It is calculated by dividing the standard deviation by the mean of the data and is expressed as a percentage. In a data set with a constant mean, as in this set of sodium samples, a smaller standard deviation will yield a smaller coefficient of variation, which indicates more precise measurements through reduced dispersion of data. Precision describes the consistency or reproducibility of a test and indicates the absence or minimization of random variation. Increased precision of a test decreases the standard deviation and therefore the coefficient of variation, while also increasing statistical power. Method II shows a smaller coefficient of variation as compared with method I, which indicates that it is more precise. Incorrect Answers: A, B, and D. Both are equally precise (Choice A) could be potentially true if the coefficient of variation was equal between the two tests. Method I is more precise than method II (Choice B) is not true; the higher coefficient of variation for method I indicates a wider dispersion of data and therefore less precision of measurement. Precision cannot be determined from information given (Choice D) is incorrect, as the coefficient of variation is a measure of data dispersion and therefore precision. Educational Objective: The coefficient of variation is a standardized measure of dispersion of data that can be used as a means of assessing the precision of a test. Precision describes the consistency or reproducibility of a test and indicates the absence or minimization of random variation. %D Previous Next Score Report Lab Values Calculator Help Pause

131 Exam Section 3: Item 31 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 31. Patients who have recurrent infections involving Neisseria species are most likely to have a genetic deficiency in which of the following serum factors? O A) C1 esterase inhibitor B) Factors B and H O C) Immunoglobulins D) Late component of complement OE) BzMicroglobulin

D. Late component complement deficiency (C5 to C9) is a common finding in patients who have recurrent infections with Neisseria species. Complement activation can occur through the classical, lectin, or alternative pathways, with each pathway converging at the point where C3 is cleaved by C3 convertase, releasing C3a and C3b. Following this, C3b is incorporated into the C3 convertase complex, which subsequently acts on C5 convertase to lower the Km (increase the affinity) of C5 convertase for its substrate, C5. Cleavage of C5 results in formation of C5a, a potent chemoattractant, and C5b. C5b complexes with C6, C7, C8, and C9 (terminal complement factors) to form the membrane attack complex (MAC). This allows for creation of a transmembrane pore that interferes with the integrity of the cellular membrane, leading to cell lysis. Deficiencies of terminal complement factors impair the formation of the MAC and preferentially predispose patients to infection with Neisseria species, although infections tend to occur with uncommon serotypes. Incorrect Answers: A, B, C, and E. C1 esterase inhibitor (Choice A) deficiency causes hereditary angioedema. Activated factor XII and kallikrein cleave kininogen to form bradykinin, a potent vasodilator that causes angioedema. C1 esterase inhibitor normally inhibits the production of bradykinin, so an absence predisposes to random and recurrent episodes of angioedema. It does not predispose to infection with Neisseria species. Factors B and H (Choice B) are involved in the alternative complement pathway. Factor B binds to C3b to cleave factor B and create C3 convertase. Factor H functions as an inhibitory molecule to prevent overactivation of the alternative pathway. Immunoglobulin (Choice C) deficiency syndromes are myriad. They include selective deficiencies such as IgA deficiency, which may have minimal symptoms, and combined variable immunodeficiency syndrome (CVID) in which patients have decreased concentrations of all immunoglobulins and are susceptible to recurrent sinopulmonary and gastrointestinal infections. Immunoglobulin deficiency does not specifically increase the risk for infection with Neisseria species. B2-Microglobulin (Choice E) deficiency causes a rare immunodeficiency syndrome because B2-microglobulin is a component of class I major histocompatibility complex (MHC) molecules. MHC class I molecules are present on the surface of most cell types and function to present antigens to CD8+ T lymphocytes. Deficiency results in ineffective activation of T lymphocytes. Educational Objective: Deficiency of the terminal complement proteins prevents the formation of the MAC, which is a critical component in the complement cascade leading to the formation of a pore in the membrane of pathogenic organisms. Deficiency predisposes to recurrent infections with atypical serotypes of Neisseria species. %3D Previous Next Score Report Lab Values Calculator Help Pause

189 Exam Section 4: Item 40 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 40. A 41-year-old woman comes to the physician for a follow-up examination 6 weeks after the diagnosis of hypertension was made. Her blood pressure today is 162/104 mm Hg. A right abdominal bruit is heard. Physical examination shows no other abnormalities. Serum studies show: 135 mEq/L 3.2 mEq/L 105 mEq/L 28 mEq/L 9 mg/dL 0.9 mg/dL Na+ K+ CI- HCO,- Urea nitrogen Creatinine On renal arteriography, sampling shows a left renal vein renin activity of 5 µU/mL (N=5-97) and right renal vein renin activity of 176 µU/mL (N=5-97). The systemic hypertension in this patient is directly mediated by a vasoconstrictor that emerges from which of the following? A) Adrenal medullary chromaffin cells B) Glomerular afferent arteriole C) Glomerular efferent arteriole D) Pulmonary vasculature E) Renal juxtaglomerular cells

D. Renal artery stenosis is a cause of secondary hypertension related to activation of the renin-angiotensin-aldosterone system. Decreased afferent blood flow to the affected kidney leads to abnormal stimulation of the juxtaglomerular apparatus and excessive production of renin. Renin converts angiotensinogen to angiotensin I, which is then converted to angiotensin II by angiotensin-converting enzyme (ACE) in the pulmonary vasculature. Angiotensin Il is a potent vasoconstrictor that directly mediates systemic hypertension. Angiotensin II also leads to increased Na+ reabsorption in the renal tubules, aldosterone release by the adrenal cortex with subsequent increased Na+ and fluid retention and hypokalemia, 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. In younger patients, fibromuscular dysplasia is the most common cause of renal artery stenosis, while in older patients it is typically caused by atherosclerosis. An abdominal bruit may be appreciated in the region of the stenosis. Renal vein sampling may detect increased renin concentrations on the affected side. Diagnosis is established with renal artery Doppler ultrasonography or magnetic resonance angiography, either of which can quantify the degree of renal artery stenosis and associated renal atrophy. 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 bilateral renal artery stenosis. Incorrect Answers: A, B, C, and E. Adrenal medullary chromaffin cells (Choice A) are neuroendocrine cells that secrete norepinephrine and epinephrine in response to sympathetic stimulation. They are not involved in the renin- angiotensin-aldosterone system. Aldosterone is secreted by cells of the adrenal cortex, not the medulla. The glomerular afferent arteriole (Choice B) experiences decreased blood flow in renal artery stenosis with subsequent stimulation of the juxtaglomerular apparatus. It is not the site of angiotensin I to angiotensin II conversion. The glomerular efferent arteriole (Choice C) constricts in response to angiotensin II, increasing the filtration pressure across the glomerulus. It is not the site of angiotensin I to angiotensin II conversion. Renal juxtaglomerular cells (Choice E) produce increased concentrations of renin in response to decreased afferent blood flow. Renin indirectly leads to hypertension by converting angiotensinogen to angiotensin I, which is then converted to angiotensin Il in the pulmonary vasculature. Educational Objective: Renal artery stenosis causes secondary hypertension through abnormal activation of the renin-angiotensin-aldosterone system. Angiotensin II is a potent vasoconstrictor that is produced from angiotensin I by ACE in the pulmonary vasculature. Previous Next Score Report Lab Values Calculator Help Pause

117 Exam Section 3: Item 17 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 17. Diisopropyl fluorophosphate, a potent inhibitor of acetylcholinesterase, reacts with which of the following specific amino acids at the active site of the enzyme? A) Cysteine B) Glutamic acid C) Histidine D) Serine E) Tyrosine

D. Serine is the amino acid in acetylcholinesterase (ACHE) targeted by diisopropyl fluorophosphate. ACHE is a ubiquitous enzyme found primarily at neuromuscular junctions and acts to hydrolyze acetylcholine, thereby terminating nerve conduction and allowing free choline to be taken up by the presynaptic neuron to generate more acetylcholine. It is a serine protease with an active site consisting of three amino acids: serine, histidine, and glutamate. This combination of amino acids is conserved throughout most serine proteases. Diisopropyl fluorophosphate is a potent and irreversible inhibitor of ACHE and acts by binding to the serine residue at the active site of the ACHE enzyme. Incorrect Answers: A, B, C, and E. Cysteine (Choice A) is found at the active site of cysteine proteases, also known as thiol proteases. They are ubiquitous enzymes used for cellular apoptosis but are also present in a variety of fruits. They have been used for diverse applications including treatment of helminth infections and as meat tenderizers. Glutamic acid (Choice B) is found at the active site of glutamic proteases, which are found commonly in pathogenic fungi including Aspergillus spp. Histidine (Choice C) contains an imidazole side chain that confers amphoteric properties at the normal pH of the human body. It is present at the active site of many enzymes but is not affected by diisopropyl fluorophosphate. Tyrosine (Choice E) is not found in the catalytic sites of enzymes as often as histidine, cysteine, serine, and glutamate are. It is a non-essential amino acid that is a precursor to neurotransmitters and melanin. Educational Objective: Serine proteases such as ACHE contain a serine residue at their active site. Diisopropyl fluorophosphate is a potent and irreversible inhibitor of ACHE and serine proteases because it binds directly to the serine residue in the active site. Previous Next Score Report Lab Values Calculator Help Pause

165 Exam Section 4: Item 15 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 15. A 43-year-old woman comes to the physician for a routine health maintenance examination. The physician is 30 minutes late for the appointment because of an emergency, and when he enters the examining room, the patient checks her watch. Which of the following initial statements by the physician is most appropriate? A) "I apologize for being behind schedule, but this won't take long." B) "I apologize for being late. Why don't we reschedule for another time?" C) "I apologize for running behind. They made my schedule really tight today." D) "I'm sorry I got delayed. I hope I haven't made you late somewhere else." E) "I'm sorry I got tied up. The emergency room always seems to call me for their most difficult cases." |

D. This physician should apologize and use an empathetic statement. If a patient has experienced hardship while seeking the physician's care (even if the hardship is not the physician's fault), apologizing may help the patient feel that their struggle has been acknowledged. The most effective apologies include an empathic statement that recognizes the inconvenience or suffering the patient may have experienced. Before apologizing, physicians should take a moment to consider how the patient may be feeling. When patients' feelings are acknowledged, more open and effective discussion about the patient's medical concerns may result. Incorrect Answers: A, B, C, and E. Reassuring the patient that the appointment won't take long (Choice A) implies that the short duration of the appointment would make up for the delay. This statement avoids acknowlledging the inconvenience the physician may have caused the patient. As well, the patient may have medical concerns to discuss and may not want a short appointment. Rescheduling the appointment (Choice B) would eliminate this patient's opportunity to undergo the health maintenance examination expeditiously, and the patient may feel that their care is a low priority for the physician. The physician should instead address this patient's medical needs while acknowledging the impact of the delay on the patient. Providing explanations for the delay (Choices C and E) addresses the physician's guilt about the delay rather than the patient's frustration. The patient may feel that their situation is less important to the physician than the physician's other priorities. The physician should instead empathize with the patient's feelings. Educational Objective: If a patient has experienced hardship while seeking the physician's care, physicians should apologize. The most effective apologies include an empathetic statement that recognizes the inconvenience or suffering the patient may have experienced and avoids providing explanations or excuses. %D Previous Next Score Report Lab Values Calculator Help Pause

136 Exam Section 3: Item 36 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 36. A 35-year-old woman comes to the physician for advice on smoking cessation. She has smoked 2 packs of cigarettes daily for 15 years. She has tried to quit smoking numerous times in the past without success. The physician plans to prescribe a medication that decreases the craving for cigarettes and also blocks the central nervous system stimulation caused by smoking. This drug most likely has which of the following nicotinic effects? A) Agonist B) Antagonist C) Inverse agonist D) Partial agonist

D. Varenicline is a medication used to help treat tobacco use disorder and to assist in tobacco use cessation. It is a partial agonist at the nicotinic acetylcholine receptor, which causes decreased cravings for cigarettes. The partial agonism also prevents nicotine from binding to the reward center in the central nervous system. This allows it to be used in patients who are still smoking to decrease their tobacco use. Typical dosing regimens include a loading dose of varenicline so that a steady state can be reached prior to the patient's smoking cessation. This is commonly followed by a 12-week course of gradually increasing dosages of varenicline, which can be continued for longer if the patient is successful in ceasing tobacco use. Adverse effects include nausea and the risk for developing behavioral changes, depression, and suicidal ideation. Other pharmacologic agents used for tobacco cessation include nicotine replacement therapy and bupropion. Incorrect Answers: A, B, and C. Agonism (Choice A) at the nicotinic receptor site is the mechanism of action of nicotine replacement therapy. While in use, it prevents cigarette cravings. However, nicotine replacement therapy would not block the central nervous system stimulation caused by smoking, as full agonism would still occur at nicotinic receptor sites. Antagonism (Choice B) at the nicotinic receptor site would not prevent cravings. However, it would act to prevent the nervous system stimulation caused by smoking because it would prevent nicotine from binding to the nicotinic receptor sites. Bupropion acts as a nicotinic receptor antagonist. An inverse agonist (Choice C) binds to the nicotinic receptor, blocks nicotine from binding, and induces action opposite that of nicotine. This would likely increase cravings for nicotine; thus, it is not a method currently utilized for tobacco cessation therapy. Educational Objective: Varenicline is a medication used for tobacco cessation that acts as a partial agonist at the nicotinic acetylcholine receptor to prevent nicotine cravings and decrease the central nervous stimulation caused by smoking. %3D Previous Next Score Report Lab Values Calculator Help Pause

146 Exam Section 3: Item 46 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 46. A 1-year-old girl is brought to the physician by her mother because of a 1-day history of blisters on her skin. The mother says that she has had several previous similar episodes earlier in the summer. Physical examination shows areas of dense freckling over sun-exposed areas, with hypo- and hyperpigmented macules. The skin appears dry and atrophic. This patient's disorder is most likely caused by a defect in which of the following DNA repair mechanisms? A) Base excision repair B) Mismatch repair C) Nonhomologous end joining D) Nucleotide excision repair E) Recombination repair

D. Xeroderma pigmentosum is caused by aberrant DNA nucleotide excision repair, a form of single-strand DNA repair, which prevents mutations induced by ultraviolet radiation from being corrected. It is a genetic disorder with 100% penetrance that is inherited in an autosomal recessive pattern. Clinically, xeroderma pigmentosum is characterized by a prolonged, severe sunburn reaction, freckling in sun-exposed areas, and signs of premature photoaging such as dryness, lentigos, and mixed hypopigmentation and hyperpigmentation. These findings appear as early as 1 year of age. Patients are also at a significantly higher risk for developing both nonmelanoma and melanoma skin cancers. Nucleotide excision repair is the pathway by which thymidine dimers induced by ultraviolet radiation are repaired. The site of a pyrimidine dimer is marked for repair, and the DNA in that region is unwound. Small fragments of DNA that are affected are then excised by endonucleases. These segments are repaired through DNA synthesis pathways using DNA polymerase and ligase enzymes. This process takes place in the G, phase of the cell cycle. Incorrect Answers: A, B, C, and E. Base excision repair (Choice A) is a second form of single-strand DNA repair, which removes altered bases rather than the entire nucleotide. This is done by a base-specific glycosylase, which leaves an apurinic/apyrimidinic (AP) site instead. The AP-endonuclease then cleaves the 5' end and lyase cleaves the 3' end. The remaining hole is filled by DNA polymerase and sealed by DNA ligase. Unlike nucleotide excision repair, this form of mutation repair occurs throughout the cell cycle. Mismatch repair (Choice B) is a third form of single-strand DNA repair in which mismatched nucleotides are removed from a newly generated strand of DNA. Dysfunctional mismatch repair proteins increase the risk for spontaneous mutations that may lead to cancer. Lynch syndrome is an example of a condition with defective mismatch repair, which leads to the instability of microsatellite regions of the genome. Lynch syndrome is associated with an increased risk for endometrial, ovarian, colon, and skin cancers, but it does not present with the same severe reaction to ultraviolet radiation as xeroderma pigmentosum. Nonhomologous end joining (Choice C) is a repair mechanism for double-stranded DNA breaks. The term nonhomologous indicates that the DNA strands are ligated together without requiring a homologous strand to use as a template. Because of this, a small portion of DNA where the strands are brought together may be lost. Nonhomologous end joining is mutated in ataxia telangiectasia syndrome and in breast cancers with a BRCA1 mutation, not xeroderma pigmentosum. Recombination repair (Choice E) is a second mechanism used to repair double-stranded DNA breaks. Unlike nonhomologous end joining, however, a homologous strand is required to be used as a template in this repair process. However, the initial defect in xeroderma pigmentosum is the generation of pyrimidine dimers by ultraviolet radiation, and homologous recombination repair is not used to correct this type of mutation. Educational Objective: Nucleotide excision repair, base excision repair, and mismatch repair are three modalities used to repair single strand DNA mutations. Nucleotide excision repair is the pathway by which thymidine dimers induced by ultraviolet radiation are repaired. It is aberrant in xeroderma pigmentosum, resulting in a prolonged, severe sunburn reaction, freckling in sun- exposed areas, signs of premature photoaging, and an increased risk for skin cancers. Previous Next Score Report Lab Values Calculator Help Pause

105 Exam Section 3: Item 5 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 5. An inexpensive screening test for a disease is available through analysis of venous blood. The receiver operating characteristic (ROC) curve for five different cut points is shown. The disease is irreversible and fatal if not discovered and treated early. Which of the following is the most appropriate cut point between normal and abnormal venous blood analysis results for this screening test? 100 DE C A 0 1-specificity 100 A) B) C) D) E)

E. Receiver operating characteristic (ROC) curves are used in biostatistics to illustrate the diagnostic ability of a test on the basis of chosen cut points to define binary positive and negative results. ROC curves are plots of sensitivity (the true positive rate) against 1-specificity (the true negative rate) and therefore, the false positive rate. Sensitivity equals the true positive test results divided by the sum of true positive and false negative test results. Therefore, as sensitivity approaches 1.0, all patients who have the disease will be detected by the test (there will be no false negatives). In this example, high sensitivity is critical, since the disease is fatal if not detected. Therefore, a priority should be placed on sensitivity when deciding the cut point threshold for positive and negative results. Specificity is calculated by the true negative test results divided by the sum of true negative and false positive results. Therefore, as specificity approaches 1.0, there will be no false positive tests. Applying these principles to the ROC curve, as values increase on the y-axis, sensitivity approaches 1.0 (or 100%) with few, if any, false negatives. Increasing sensitivity comes at the expense of specificity; by setting the cut point to include all persons with the disease, many false positives will be introduced since persons who have the venous blood biomarker without having the disease will test falsely positive. This is reflected on the x-axis; as 1-specificity increases, many false positives will be detected. However, in this example, the irreversible and fatal condition should be detected and treated early; therefore, a high-sensitivity test should be prioritized. Patients testing negative on a sensitive test can be safely considered disease-free, whereas patients who test positive should receive additional diagnostic evaluation to determine whether their results were false positive or they actually have the underlying disease. Incorrect Answers: A, B, C, and D. Choices A, B, C, and D have lower sensitivity than point E, and therefore, would result in a higher number of false negative test results. These patients falsely testing negative would have their disease persist undetected, causing morbidity and mortality. By contrast, these cut points all offer a greater degree of specificity, that is, the false positive rate is lower, indicating that those testing positive can be considered to have the disease. Educational Objective: ROC curves plot the true positive rate against the false positive rate. With a sensitive test, increased sensitivity comes at the expense of increased false positive tests; however, for fatal conditions for which early detection is key, sensitivity should be prioritized. Previous Next Score Report Lab Values Calculator Help Pause Sensitivity םם

115 Exam Section 3: Item 15 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 15. A 72-year-old man with severe congestive heart failure is brought to the emergency department because of a 5-day history of malaise and vomiting. He takes no medications. His pulse is 104/min, respirations are 35/min, and blood pressure is 90/64 mm Hg. Physical examination shows dusky-colored skin, peripheral cyanosis, and 10-cm jugular venous distention. Crackles are heard over the lung bases bilaterally. Cardiac examination shows an S, There is 2+ pitting edema to the knees bilaterally. Laboratory studies show: Serum Na+ K+ CI- HCO3- Urea nitrogen Glucose Creatinine Urine Glucose Ketones 127 mEq/L 5.2 mEq/L 79 mEq/L 17 mEq/L 100 mg/dL 171 mg/dL 8.4 mg/dL 1+ absent Arterial blood gas analysis on 28% oxygen pH Pco2 Po2 HCO,- 7.35 32 mm Hg 154 mm Hg 17 mEq/L Which of the following is the most likely cause of the anion gap in this patient? A) Alcoholic ketoacidosis B) Diabetic ketoacidosis O C) Hyperperfusion D) Hypoalbuminemia E) Metabolic alkalosis F) Milk-alkali syndrome G) Renal failure H) Respiratory alkalosis

G. Metabolic acidosis occurs when there is an accumulation of an organic acid in the serum that exceeds the capacity of the primary buffer, bicarbonate (HCO,), or in abnormal loss of bicarbonate. The normal plasma pH is approximately 7.4; a pH of less than 7.35 indicates acidemia while a pH of more than 7.45 indicates alkalemia. In metabolic acidosis, there is typically compensatory hyperventilation to exhale carbon dioxide (an acid) in an attempt to neutralize the abnormally decreased pH. A primary metabolic acidosis is thus identified when the HCO3 is decreased and the PCO, is also decreased. Metabolic acidosis can occur in multiple causes, either with or without an associated anion gap. The anion gap is calculated as Na+- (Cl-+ HCO3). In a healthy patient, the typical amount of anion in the blood that is not accounted for by either CI - or HCO, is 12 mEq/L or less. An increased anion gap indicates that there is accumulation of an anion that is not accounted for in the equation, such as lactic acid, ketoacid, or salicylic acid. An anion gap of more than 12 mEq/L is considered increased. In this case, the anion gap is calculated as 127 - (79 + 17) = 31. Causes of a metabolic acidosis with an increased anion gap include methanol ingestion, uremia, diabetic or alcoholic ketoacidosis, ethylene or propylene glycol ingestion, iron or isoniazid administration, lactic acidosis, ethylene glycol ingestion, and late salicylate overdose. In this case, the patient has developed an acute kidney injury secondary to decreased renal perfusion from congestive heart failure and vomiting. The increased creatinine and blood urea nitrogen (uremia) support this diagnosis. Acidosis secondary to uremia in renal failure is an indication for emergent dialysis. Incorrect Answers: A, B, C, D, E, F, and H. Alcoholic ketoacidosis (Choice A) is a starvation state induced by excessive alcohol intake that results in increased ketone production and an anion gap metabolic acidosis without hyperglycemia. Diabetic ketoacidosis (Choice B) typically occurs in patients with type 1 diabetes mellitus as a result of increased insulin requirements during times of physiologic stress or insulin nonadherence. Diabetic ketoacidosis results in a metabolic acidosis with an increased anion gap, ketonuria, and ketonemia. Hypoperfusion, not hyperperfusion (Choice C), may cause lactic acidosis, an anion gap metabolic acidosis. Lactate accumulates when tissues perform anaerobic metabolism in the absence of охудen. Hypoalbuminemia (Choice D) is seen in cirrhosis. Albumin is synthesized in the liver, which is impaired in patients with cirrhosis. Hypoalbuminemia does not result in a metabolic acidosis with an anion gap. Metabolic alkalosis (Choice E) may result from loss of protons (eg, vomiting) or from volume contraction. Respiratory compensation for metabolic alkalosis leads to the retention of carbon dioxide. In metabolic alkalosis, the pH is greater than 7.45, HCO3 is increased, and PCO 2 is increased, none of which are present in this patient. Milk-alkali syndrome (Choice F) is caused by ingestion of calcium and absorbable alkali. Antacids contain calcium and alkali and are the most common culprit of milk-alkali syndrome. Respiratory alkalosis (Choice H) is a form of alkalosis (pH > 7.45) in which the PCO, is abnormally removed by hyperventilation. Increased conversion of HCO3 to CO2 occurs, leading the serum HCO, concentration to decrease. This patient's serum is acidotic, not alkalotic. Educational Objective: The accumulation of organic acid that exceeds buffering capacity is the cause of an anion gap metabolic acidosis. The differential diagnosis for metabolic acidosis with an increased anion gap includes methanol ingestion, uremia, ketoacidosis, propylene or ethylene glycol ingestion, iron or isoniazid administration, lactic acidosis, and late salicylate overdose. %3D Previous Next Score Report Lab Values Calculator Help Pause

97 Exam Section 2: Item 47 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 47. A previously healthy 12-year-old girl is brought to the physician because of a 1-week history of brief generalized tonic-clonic seizures. Her family recently emigrated from Mexico. Physical examination shows no abnormalities.. A CT scan of the head shows multiple calcified nodules and some cysts. The most likely cause of her seizures is infection with which of the following causal organisms? A) Cryptococcus neoformans B) Herpes simplex virus C) HIV D) Listeria monocytogenes E) Mycobacterium tuberculosis F) Streptococcus pneumoniae G) Taenia solium H) Toxoplasma gondii

G. Neurocysticercosis is a central nervous system (CNS) infection caused by the pork tapeworm Taenia solium that occurs most commonly in endemic rural regions in Latin America, Asia, and Africa. Šymptoms include headache, seizures, signs of increased intracranial pressure, and, rarely, altered vision. Certain patients are asymptomatic, and other patients present with muscle or subcutaneous nodules. Typical manifestations on CT or MRI include cystic, ring-enhancing, calcified lesions in the brain parenchyma, meninges, ventricles, spinal cord, and/or orbits.. A scolex (larval head) may also be seen on imaging. Treatment may include symptomatic management of increased intracranial pressure and seizures, antiparasitic therapy, and/or steroids. Incorrect Answers: A, B, C, D, E, F, and H. Cryptococcus neoformans (Choice A) and Listeria monocytogenes (Choice D), typically cause meningoencephalitis rather than intraparenchymal cystic lesions. Cryptococcal meningitis is an invasive fungal infection commonly seen in immunocompromised patients. Listeria monocytogenes also invades the CNS in immunocompromised patients, the elderly, and infants. Both infections may present with mass lesions, but cystic, ring-enhancing lesions would be atypical. Herpes simplex virus (Choice B) causes viral encephalitis in patients of all ages and may present with fever, headache, seizures, focal neurologic signs, and confusion. Imaging may show inflammation in the temporal lobe. HIV (Choice C) does not directly infect the parenchyma of the brain. Instead, HIV infection may lead to opportunistic infections such as Cryptococcus neoformans, which would not typically cause cystic lesions. Mycobacterium tuberculosis (Choice E) may cause meningitis or tubercles in patients visiting endemic areas such as southeast Asia or Africa. Tubercles may appear similar on neuroimaging to neurocysticercosis lesions with ring enhancement and calcifications. However, tuberculosis is less prevalent in Mexico, tubercles do not show a scolex, and patients with tuberculosis commonly show an extraneural site of disease such as the lungs. Streptococcus pneumoniae (Choice F) is a common cause of bacterial meningitis in children and adults. Focal lesions of the parenchyma would be atypical. Toxoplasma gondii (Choice H) is a common parasitic infection in patients with HIV infection that may show ring-enhancing lesions on imaging. Immunocompetent patients are typically asymptomatic. Further, Toxoplasma gondii is common in the United States; travel to Mexico does not constitute a risk factor. Educational Objective: Neurocysticercosis is a central nervous system infection caused by the pork tapeworm Taenia solium that occurs most commonly in endemic rural regions in Latin America, Asia, and Africa, and may infect immunocompetent patients. Typical manifestations on CT or MRI include cystic, ring-enhancing, calcified lesions in the brain parenchyma, meninges, ventricles, spinal cord, and/or orbits. Previous Next Score Report Lab Values Calculator Help Pause

12 Exam Section 1: Item 12 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 12. A 45-year-old man comes to the physician for a follow-up examination 2 weeks after beginning treatment with hydrochlorothiazide for hypertension. He says that he feels faint, light-headed, and dizzy when he gets out of bed and when he rises from a seated position too quickly. His pulse is 75/min, respirations are 12/min, and blood pressure is 130/85 mm Hg while supine. Physical examination shows no abnormalities. Which of the following sets of changes best characterizes changes in the cardiovascular system as this man goes from the supine to the standing position? Carotid Sinus Cerebral Venous Return Baroreceptor Activity Blood Flow A) ↑ ↑ ↑ B) ↑ ↑ C) ↑ ↑ OD) E) ↑ ↑ F) ↑ G) ↑ H)

H. Hydrochlorothiazide is a thiazide diuretic commonly used in the treatment of hypertension. Thiazide diuretics act in the distal convoluted tubule to decrease NaCl reabsorption by inhibiting the sodium-chloride cotransporter channel. These drugs result in urinary losses of sodium and potassium with associated water loss. A common complication of diuretics is volume depletion with orthostatic hypotension. The vasculature must respond to postural changes in order to maintain perfusion to the brain and organs of the upper body in response to gravity. Rising from supine to standing results in decreased venous return, decreased carotid sinus baroreceptor activity, and decreased cerebral blood flow. The body normally responds to these changes with increases in mean arterial and diastolic pressures, stroke volume, pulse, and cardiac output. In patients with volume depletion, these changes are not adequate to maintain the blood pressure and cerebral perfusion. Patients may experience symptoms such as light-headedness and dizziness on standing; there is also a greater than expected increase in pulse and a decrease in blood pressure as measured at the upper extremity. Incorrect Answers: A, B, C, D, E, F, and G. Choices A, B, C, and D are incorrect since venous return decreases on standing because of the effect of gravity. Increased venous return can be accomplished with fluid resuscitation. It also occurs during inspiration as a result of chest wall expansion and the generation of negative intrathoracic pressure. Choices A, B, E, and F are incorrect since carotid sinus baroreceptors are stimulated by increases in pressure or stretch on the vascular wall. Increased pressure triggers an action potential that travels along afferent pathways to the brain, stimulating increased parasympathetic outflow and decreased sympathetic outflow. A drop in blood pressure results in decreased baroreceptor activity, completing the negative feedback loop. Choices A, C, E, and G are incorrect since cerebral blood flow decreases in orthostatic hypotension because of the effect of gravity in the absence of adequate blood volume to maintain a compensatory response. Educational Objective: The cardiovascular system must adjust to postural changes to maintain adequate perfusion pressure to the head and upper body. Baroreceptors in the carotid sinus and aortic arch detect changes in pressure and stimulate the body's response. Inadequate intravascular volume results in symptomatic orthostatic hypotension. Previous Next Score Report Lab Values Calculator Help Pause

80 Exam Section 2: Item 30 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 30. The table summarizes the result of a study of the relationship between smoking and death from lung cancer. Annual Death Rate (Per 1000 Persons) Nonsmokers 0.07 Heavy Smokers 2.27 Lung cancer Which of the following best describes the attributable (excess) risk per 1000 persons from heavy smoking in those with lung cancer? A) 2.27 - 0.07 | B) 2.27 - 0.07 C) 0.07 2.27 D) (2.27 - 0.07) 2.27 E) Cannot be determined from the data provided

A. Attributable (excess) risk (AR) describes the risk for developing the disease or outcome under study (in this case, death from lung cancer) that can be attributed to an exposure as compared with the risk that exists without exposure. AR is calculated as the difference between disease or outcome incidence among those with the exposure and those without the exposure. In biostatistics, the lowercase letters a, b, c, and d are frequently used to represent numbers of patients exposed to a risk, receiving an intervention, or having a disease. The number of persons having the exposure or receiving the intervention who have the disease or condition in question is represented by 'a'. The number of persons having the exposure or receiving the intervention who do not have the disease or condition in question is represented by 'b'. The number of persons who do not have the exposure or who do not receive the intervention in question and who do have the disease or condition is represented by 'c'. The number of persons who do not have the exposure or who do not receive the intervention in question and who do not have the disease is represented by 'd'. By this convention, the incidence of those with the exposure is computed by determining the number of patients with the exposure and who developed disease (a) and dividing it by the total number of patients who were exposed (a + b). The incidence of those without the exposure who developed disease is computed by determining the number of patients without the exposure who developed disease (c) and dividing it by the total number of patients who were not exposed (c + d). AR is defined as AR = [a / (a + b)] - [c/ (c + d)]. In this scenario, the incidence rates of death from lung cancer are already computed, thus AR is equal to the difference in the death rate among smokers compared to that among nonsmokers, or AR = 2.27 - 0.07 deaths per 1000 persons. Incorrect Answers: B, C, D, and E. 2.27/0.07 (Choice B) calculates the relative risk (RR) of death from lung cancer among smokers as compared to nonsmokers. RR = [a / (a + b)]/ [c/ (c + d)]. 0.07/2.27 (Choice C) is the inverse of the RR. (2.27-0.07)/2.27 (Choice D) reflects incorrect mathematical assumptions and divides the AR by the death rate from lung cancer among smokers. The statement that AR cannot be determined from the data provided (Choice E) is incorrect since the data provides the risk for death in each group and, from this, the difference reflects the AR. Educational Objective: Attributable risk (AR) describes the risk for developing the disease or outcome under study that can be attributed to the exposure as compared with the risk that exists without exposure. AR is calculated as the difference between disease or outcome incidence among those with the exposure and those without the exposure. Previous Next Score Report Lab Values Calculator Help Pause

55 Exam Section 2: Item 5 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 5. A 28-year-old woman comes to the physician because of heavy menstrual bleeding since menarche at the age of 13 years. She also has a lifelong history of easy bleeding with minor cuts. She says several family members of both genders have prolonged bleeding with minor surgical procedures. Physical examination shows no abnormalities. Laboratory studies show a deficiency of von Willebrand factor. Which of the following mechanisms is the most likely cause of this patient's bleeding disorder? A) Decreased platelet adhesion B) Decreased platelet glycoprotein Ilb/Illa C) Decreased secretion of platelet granular contents D) Increased antiplatelet antibodies E) Increased number of giant platelets F) Increased platelet sequestration in the spleen

A. Decreased platelet adhesion occurs in patients with deficiencies in von Willebrand factor (vWF). Von Willebrand disease is one of the most common hereditary bleeding disorders and is caused by quantitative or qualitative abnormalities of vWF, which binds platelets and subendothelial collagen in primary hemostasis. It interacts with the Gplb receptor complex and Gpllb/Illa receptors on the surface of activated platelets and tethers them to the endothelium. Impaired platelet adherence to the subendothelial lining leads to a prolonged bleeding time. vWF also transports factor VIII in plasma, which degrades rapidly when unbound. Factor VIII is a critical component of the intrinsic coagulation pathway and decreased concentrations can occasionally lead to a prolonged partial thromboplastin time (PTT), although PTT can also be normal (as in this patient). Von Willebrand disease is inherited in an autosomal dominant pattern. It can present with epistaxis, gingival bleeding, petechiae, easy bruising, and menorrhagia. Additional diagnostic testing can be performed with the ristocetin cofactor assay, which requires functional vWF for piatelet aggregation to occur. Treatment depends on the site and severity of bleeding and may include desmopressin (promotes release of additional vWF from endothelial cells) or administration of vWF concentrates. Incorrect Answers: B, C, D, E, and F. Decreased platelet glycoprotein Ilb/lla (Choice B) is characteristic of Glanzmann thrombasthenia. Gpllb/llla receptors play a primary role in binding to vWF, thereby tethering platelets to the endothelium. These receptors are also involved in binding fibrinogen and joining platelets together. While a deficiency does cause mucocutaneous bleeding, it is not the pathologic mechanism of bleeding in von Willebrand disease. Decreased secretion of platelet granular contents (Choice C) can result from deficiencies of specific platelet granules, including a- and õ-granules. Inherited conditions such as Hermansky- Pudlak syndrome (deficient õ-granules) and gray platelet syndrome (deficient a-granules) result in mild mucocutaneous bleeding, but they also have additional findings such as oculocutaneous albinism and early onset myelofibrosis, respectively. Increased antiplatelet antibodies (Choice D) are seen in immune thrombocytopenic purpura (ITP). Patients present with severe thrombocytopenia from autoimmune destruction of circulating platelets but do not have a deficiency in vWF. Increased number of giant platelets (Choice E) is a characteristic feature of myelodysplasia and Bernard-Soulier syndrome. They do not frequently cause severe bleeding problems. Patients may have thrombocytopenia but increased mean platelet volume. Increased platelet sequestration in the spleen (Choice F) can cause thrombocytopenia and is frequently seen in liver cirrhosis with portal hypertension. This does not cause a deficiency in vW. Educational Objective: Von Willebrand disease is a common inherited cause of mucocutaneous bleeding and results from a deficiency in vWF concentration or function, impairing platelet adhesion to the endothelium. It is reflected by increased bleeding time, decreased factor VIII activity, and suppressed ristocetin cofactor activity. Treatment depends on the site and severity of bleeding and may include desmopressin or vWF concentrates. Previous Next Score Report Lab Values Calculator Help Pause

40 Exam Section 1: Item 40 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 40. A 30-year-old woman, gravida 1, para 0, at 32 weeks' gestation comes to the emergency department because of a 2-day history of fever, nausea, and headache. Physical examination shows a uterus consistent in size with a 30-week gestation. Serologic testing prior to pregnancy showed a positive lIgG antibody titer to Toxoplasma gondii. Blood cultures grow a small gram-positive rod, which grows as pinpoint B-hemolytic colonies on sheep blood agar. The organism is catalase positive and exhibits tumbling motility. This infection could have been prevented by avoiding which of the following? A) Consumption of delicatessen meats B) Consumption of grapefruit juice C) Contact with cats D) Contact with dogs E) Contact with rabbits

A. Listeria monocytogenes is an aerobic gram-positive rod that is capable of living inside or outside of a cell. It is catalase-positive, B-hemolytic, and translocates between cells via actin polymerization to move through cell membranes. It shows characteristic tumbling motility when viewed under light microscopy. Risk factors for infection include extremes of age (eg, neonate or elderly), pregnancy, malignancy, immunosuppression (eg, AIDS or medication-induced), diabetes, and end-stage renal disease. In nonpregnant patients, it typically causes a febrile gastroenteritis, but can also cause bacteremia, meningitis, and encephalitis. In pregnancy, however, it commonly presents in the third trimester with gastroenteritis or a flu-like illness caused by bacteremia, consisting of fever, chills, and back pain. L. monocytogenes can spread across the placenta and lead to chorioamnionitis, premature birth, fetal demise, or neonatal meningitis or sepsis. The most severe form of neonatal infection that occurs in utero is granulomatosis infantiseptica, which causes multiple granulomas in various internal organs, as well as papular skin lesions. Mortality for the infant is high with maternal infection. L. monocytogenes is typically contracted from unpasteurized cheeses or cold delicatessen meats. It is treated with ampicillin. Incorrect Answers: B, C, D, and E. Consumption of grapefruit juice (Choice B) will inhibit the cytochrome P450 enzymes responsible for the degradation of many medications, which will lead to increased concentrations of these medications. However, consumption of grapefruit juice does not increase the risk for the development of L. monocytogenes infection. Contact with cats (Choice C) increases the risk for Pasteurella multocida infection (following a cat bite), Bartonella infections, and Toxoplasma gondii. Of these, T. gondii infection is most dangerous during pregnancy, as it can cause chorioretinitis, hydrocephalus, and intracranial calcifications of the newborn. It is typically asymptomatic in the mother. Contact with dogs (Choice D) increases the risk for Pasteurella multocida infection (following a dog bite), Echinococcus granulosa, Campylobacter jejuni, and rabies. It does not increase the risk for the development of L. monocytogenes infection. Contact with rabbits (Choice E) increases the risk for tularemia, which presents with fever, chills, loss of appetite, fatigue, nausea, emesis, and diarrhea, as well as various other symptoms depending on the form of tularemia. Francisella species, which are responsible for causing tularemia, are gram-negative coccobacilli. Educational Objective: L. monocytogenes is an aerobic, gram-positive rod that is catalase-positive, B-hemolytic, and shows characteristic tumbling motility under light microscopy. It is a common infection in pregnancy as well as in neonates and elderly patients, and it can cause febrile gastroenteritis, bacteremia, meningitis, or encephalitis. Infection in pregnancy can cause severe consequences to the developing fetus, including in utero infection and fetal demise. Previous Next Score Report Lab Values Calculator Help Pause

86 Exam Section 2: Item 36 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 36. An investigator conducts a study of cranial nerve regeneration among two groups of experimental animals. During the study, the facial nerves are severed in one group (Group X) and the optic nerves are severed in the second group (Group Y). Three months later, it is found that the severed facial nerves among animals in Group X have regenerated, but the severed optic nerves among animals in Group Y have not regenerated. Which of the following best explains the absence of nerve regeneration among the animals in Group Y? A) Axons of the optic nerve are myelinated by oligodendrocytes B) The blood vessels that supply the retina cannot grow to supply the optic nerve C) Regenerating optic nerve fibers cannot cross the optic chiasm D) Retinal ganglion neurons are formed by neural crest cells E) Retinal photoreceptors are nonpermissive for optic nerve regeneration

A. Oligodendrocytes and central nervous system (CNS) myelin release inhibitory factors that prevent neural regeneration. Consequently, CNS axons do not typically regenerate after injury. The optic nerve (cranial nerve II) forms as an outpouching of the embryonic diencephalon, begins in the retina, travels to the optic chiasm, and continues posteriorly as the optic tract. The optic track synapses in the lateral geniculate nuclei of the thalamus, pretectal nuclei of the midbrain, and superior colliculi. Therefore, unlike most cranial nerves, the optic nerve is part of the CNS rather than the peripheral nervous system (PNS), meaning that axons of the optic nerve are myelinated by oligodendrocytes. The olfactory nerve (cranial nerve I) is the only other cranial nerve that is part of the CNS. The facial nerve, a peripheral nerve, can regenerate because the Schwann cells present can secrete growth factors and guide axonal regeneration. Incorrect Answers: B, C, D, and E. Retinal blood vessels (Choice B) can form collaterals as a result of vascular endothelial growth factor (VEGF). VEGF can improve nerve revascularization after injury. Optic nerve fibers are not derived from neural crest cells (Choice D). The optic nerve is part of the CNS, derived from the neural plate, rather than the peripheral nervous system, derived from neural crest cells. Consequently, optic nerve fibers cannot regenerate (Choice C). Retinal photoreceptors (Choice E) may regenerate after bright light injury. These photoreceptors are not involved in the signaling that prevents optic nerve regeneration. Educational Objective: Oligodendrocytes and CNS myelin release inhibitory factors that prevent neural regeneration. Consequently, CNS axons do not regenerate after injury. The optic and olfactory nerves are the only cranial nerves that are part of the CNS and therefore cannot regenerate. %D Previous Next Score Report Lab Values Calculator Help Pause

57 Exam Section 2: Item 7 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 7. A 20-year-old woman comes to the physician because of a 10-day history of a vaginal discharge and vaginal itching and soreness. Current medications include a multivitamin and an antibiotic for acne. Pelvic examination shows vulvar erythema and edema and a thick, white vaginal discharge. Her vaginal fluid pH is 4.2. A photomicrograph of the vaginal discharge is shown. The most appropriate pharmacotherapy for this patient is an oral agent with which of the following mechanisms of action? A) Inhibition of the cytochrome P450-dependent demethylation reaction B) Inhibition of DNA and RNA synthesis OC) Pore formation in the fungal cell membrane D) Prevention of cross-linking of B-glucans in the cell wall O E) Prevention of tetrahydrofolic acid synthesis

A. The patient's presenting signs and symptoms of vaginal pruritus and thick, white vaginal discharge are suggestive of vulvovaginitis from Candida albicans. While Candida species compose an element of normal vaginal flora, overgrowth can occur and lead to vulvovaginitis. C. albicans is the most frequently associated species. Candida can be distinguished from other potential causes of vulvovaginitis by abnormal, often white, vaginal discharge with normal vaginal pH, negative KOH testing, and evidence of pseudohyphae on microscopy. Risk factors for Candida vulvovaginitis include the use of medications (especially antibiotics and oral contraceptives), use of intravaginal or intrauterine contraceptive devices, and an immunocompromised state. Candida vulvovaginitis is treated with intravaginal antifungal agents such as miconazole or clotrimazole, or with oral fluconazole, which prevents the formation of ergosterol (part of the fungal cell membrane) by inhibiting the cytochrome P450-dependent demethylation reaction that synthesizes ergosterol from lanosterol. Incorrect Answers: B, C, D, and E. Inhibition of DNA and RNA synthesis (Choice B) is the mechanism of action of flucytosine. Flucytosine is used to treat systemic fungal infections, especially cryptococcal meningitis in combination with amphotericin B, but is not used to treat vaginal candidiasis. Pore formation in the fungal cell membrane (Choice C) is the mechanism of action of amphotericin B and nystatin. Amphotericin B is used to treat systemic fungal infections, such as meningitis and endemic mycoses. Nystatin can be used for the treatment of vaginal candidiasis when applied topically. Prevention of cross-linking of B-glucans in the cell wall (Choice D) is the mechanism of action of echinocandins (eg, micafungin). They are used for invasive fungal infections, including invasive Candida infections. However, they play no role in the management of localized vulvovaginitis. Prevention of tetrahydrofolic acid synthesis (Choice E) is the mechanism of action of sulfonamides and trimethoprim. The combination of the two medications is commonly used to treat urinary tract infections or to provide prophylaxis against Pneumocystis jiroveci and toxoplasmosis in immunosuppressed patients. It does not play a role in the treatment of a Candida infection. Educational Objective: Vulvovaginal candidiasis results from the overgrowth of Candida and presents with vulvar erythema, vaginal pruritus, and thick, white discharge. It can be treated with oral fluconazole, which inhibits the production of ergosterol (part of the fungal cell membrane) by inhibiting the cytochrome P450-dependent demethylation reaction that synthesizes ergosterol from lanosterol. %3D Previous Next Score Report Lab Values Calculator Help Pause

47 Exam Section 1: Item 47 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 47. A73-year-old man comes to the physician with his wife because of a 3-year history of daytime sleepiness. His wife says that he snores loudly. He is 183 cm (6 ft) tall and weighs 113 kg (250 lb); BMI is 34 kg/m2. His pulse is 84/min, respirations are 18/min, and blood pressure is 175/105 mm Hg. Physical examination shows congested conjunctivae and centripetal obesity. Serum studies show a glucose concentration of 250 mg/dL, creatinine concentration of 2 mg/dL, and uric acid concentration of 10.1 mg/dL. Polysomnography confirms the diagnosis, and treatment with nasal continuous positive airway pressure at night is started. This therapy will most likely cause a decrease in which of the following? A) Blood pressure B) Pulse rate C) Serum creatinine concentration D) Serum glucose concentration E) Serum uric acid concentration

A. There is a complex, multifactorial relationship between obstructive sleep apnea (OSA) and hypertension. First, obstructed breathing while sleeping leads to strong respiratory effort against the obstruction. This respiratory effort produces large oscillations in nocturnal blood pressure that correspond to apneic events. Additionally, during the apneic events of OSA, there is a precipitous drop in blood oxygen concentration, which results in varying degrees of ischemia to end-organs such as the brain. This likely results in an autonomic-mediated increase in blood pressure in an attempt to maintain adequate perfusion. Furthermore, many patients with OSA also have hypertension during the day, especially during the morning. Treatment with continuous positive airway pressure decreases the number of apneic events during the night, which in turn, decreases the degree of blood pressure variability and hypertension. Untreated OSA patients are at increased risk for adverse cardiovascular outcomes, including stroke. Incorrect Answers: B, C, D, and E. Pulse rate (Choice B) may be affected by surges of sympathetic activity during apneic episodes, however, this relationship has been less clearly demonstrated than that between apnea and hypertension. Serum creatinine concentration (Choice C) is a useful surrogate measure of renal function. While there is some evidence associating OSA with chronic kidney disease, this relationship is possibly mediated by chronic hypertension, and it is less consistently demonstrated than the relationship between OSA and hypertension. Serum glucose concentration (Choice D) is useful for monitoring hyperglycemia and diabetes. While OSA is associated with diabetes, treatment with continuous positive airway pressure does not directly impact serum glucose concentration. Serum uric acid concentration (Choice E) is useful for monitoring gout. While there may be an association between OSA and gout, treatment with continuous positive airway pressure does not directly impact serum uric acid concentration or affect purine catabolism. Educational Objective: Obstructive sleep apnea is associated with nocturnal and morning hypertension, likely caused by high amplitude oscillations in blood pressure and decreased blood oxygen concentrations during apneic episodes. Treatment with continuous positive airway pressure decreases the frequency of apneic episodes and decreases blood pressure in patients with obstructive sleep apnea. Untreated hypertension in these patients is associated with a variety of adverse cardiovascular outcomes, including stroke. Previous Next Score Report Lab Values Calculator Help Pause

19 Exam Section 1: Item 19 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 1 cm 19. A67-year-old man is brought to the physician by his wife because of a 1-year history of progressive difficulty writing and walking. The patient is stooped and talks slowly. Physical examination shows a bland facial expression. There is a fine resting tremor in both hands, but not in the lower extremities. The tremor is not apparent when the patient is moving. When asked to walk, the patient has some difficulty starting and stopping. There is cogwheel rigidity in all four extremities. Testing of postural reflexes shows moderate retropulsion and mild propulsion. Gross and microscopic appearances of brain tissue from a patient with a similar condition are shown. The midbrain on the left is a normal control. The inclusion indicated by the arrow is predominantly composed of which of the following substances? A) Alpha-synuclein B) Amyloid OC) Ataxin D) Prion protein O E) Tau

A. This patient likely has alpha-synuclein inclusions typical of Parkinson disease. Parkinson disease is a neurologic syndrome that features bradykinesia, resting tremor, rigidity (frequently associated with cogwheeling), and postural instability. This patient's writing difficulty likely represents micrographia and is commonly related to bradykinesia in patients with Parkinson disease. Parkinson disease arises from accumulation of alpha-synuclein protein in the substantia nigra, which disrupts neuronal homeostasis and leads to neuronal death. These alpha-synuclein inclusions, also called Lewy bodies, are circular, eosinophilic cytoplasmic inclusions that lead to the destruction of dopaminergic neurons in the substantia nigra. Consequently, the basal ganglia, which rely on dopaminergic input from the substantia nigra, cannot properly modulate voluntary movements, resulting in Parkinson disease symptoms. Treatment centers around dopaminergic agents such as levodopa-carbidopa. Incorrect Answers: B, C, D, and E. Amyloid plaques (Choice B) and abnormal tau accumulation (Choice E) are pathologic markers of dementia, Alzheimer type. Amyloid plaques are extracellular protein accumulations, while tau forms intracytoplasmic and extracellular protein tangles. Though patients with Alzheimer dementia may have difficulty executing complex motor commands because of cortical dysfunction (apraxia), Alzheimer dementia is not typically associated with bradykinesia, tremor, rigidity, or postural instability. Ataxin (Choice C), an intracellular protein, is normally involved in protein production and degradation and is mutated in spinocerebellar ataxias. Spinocerebellar ataxias typically present in middle age with broad-based ataxia and may include an intention tremor. Alternatively, Parkinson disease typically presents with a narrow-based gait and resting tremor. Further, bradykinesia is atypical of early spinocerebellar ataxia. Prion proteins (Choice D) are intra- and extracellular plaques that contribute to the pathogenesis of Creutzfeldt-Jakob disease. Creutzfeldt-Jakob disease shows rapidly progressive dementia, myoclonus, and cerebellar manifestations such as ataxia and nystagmus. Patients with Creutzfeldt-Jakob disease typically die within one year. Educational Objective: Parkinson disease arises from the accumulation of alpha-synuclein protein in the substantia nigra, which leads to the death of dopaminergic neurons that project to the basal ganglia. Čonsequently, patients with Parkinson disease demonstrate dysfunction of voluntary movements in the form of bradykinesia, resting tremor, rigidity, and postural instability. Previous Next Score Report Lab Values Calculator Help Pause

82 Exam Section 2: Item 32 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 32. A full-term male newborn has severe hypotonia. His 20-year-old mother shows little facial expression. When she shakes the physician's hand, she has difficulty releasing her grip. Which of the following genetic phenomena is the most likely cause of these findings? A) Expanded trinucleotide repeats B) Gonadal mosaicism C) Mitochondrial inheritance D) Submicroscopic deletion O E) Uniparental disomy

A. This patient, with a severe phenotype, and his mother, with a milder phenotype, illustrate anticipation, a genetic phenomenon in which disease traits are more severe and/or appear at earlier ages in later generations. Anticipation typically results from trinucleotide repeat expansions, in which later generations possess more trinucleotide repeats, and therefore have more severe phenotypes. Trinucleotide repeat expansion results from dysfunctional DNA mismatch repair, which normally removes these trinucleotides that are mismatches for their paired bases. This patient and his mother likely have myotonic dystrophy, a muscular dystrophy known to demonstrate intergenerational anticipation. Myotonic dystrophy results from an expansion of cytosine-thymine- guanine (CTG) trinucleotide repeats of the dystrophia myotonica protein kinase (DMPK) gene. The end result is progressive skeletal muscle wasting, weakness, and cardiomyopathy with dysrhythmia. Patients with the classic form of myotonic dystrophy demonstrate prolonged muscle contraction with an impaired ability to relax starting between ages 10 and 30 years. Anticipation may result in a congenital form of the disease characterized by hypotonia, poor feeding, and respiratory failure. Treatment is supportive and may include physical therapy and/or ventilatory support. Incorrect Answers: B, C, D, and E. In gonadal mosaicism (Choice B), a postfertilization de novo mutation occurs in gamete precursors only, leading to disease in the offspring of phenotypically normal parents (mimicking autosomal recessive inheritance). In mitochondrial inheritance (Choice C), offspring inherit mutations in the mitochondrial genome from their mother. Mitochondrial inheritance does not typically lead to anticipation. Submicroscopic deletions (Choice D) are chromosomal deletions too small to be detected by light microscopy, requiring specialized testing to identify the deletion. DiGeorge syndrome primarily results from a sporadic submicroscopic deletion of chromosome 22. Though this submicroscopic deletion may be occasionally inherited, the deletion is stable and would not show anticipation. Uniparental disomy (Choice E) refers to the inheritance of both chromosome copies from one parent. Angelman syndrome and Prader-Willi syndrome result from uniparental disomy. In uniparental disomy, the parent who contributes both chromosomes is phenotypically normal, as opposed to this patient's mother with mild disease. Educational Objective: Anticipation is a genetic phenomenon in which disease traits are more severe and/or appear at earlier ages in later generations.. Anticipation typically results from trinucleotide repeat expansions. Myotonic dystrophy, which features intergenerational anticipation, results from a trinucleotide repeat expansion of the DMPK gene in skeletal muscle cells that leads to progressive muscle dysfunction. Previous Next Score Report Lab Values Calculator Help Pause

98 Exam Section 2: Item 48 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 48. A 24-year-old man comes to the physician because of nasal congestion, a watery nasal discharge, and sneezing. He says that these symptoms happen each spring, when grass and tree pollens are abundant. Which of the following types of drugs is most likely to be effective in relieving these symptoms in the short term? A) a-Adrenergic agonist B) a-Adrenergic antagonist C) B-Adrenergic agonist D) B-Adrenergic antagonist E) Nicotinic cholinergic agonist F) Nicotinic cholinergic antagonist

A. a-Adrenergic receptor agonists, such phenylephrine, decrease secretions in the nasal mucosa and are useful as nasal decongestants in the treatment of allergic rhinitis. The aradrenergic receptor is a G protein-coupled receptor. It is located in the vascular smooth muscle, intestinal and bladder sphincter muscles, and pupillary dilator muscle. The a-adrenergic receptor utilizes phospholipase C as a second messenger, whose activity leads to increased intracellular calcium and protein kinase C activity, in turn causing smooth muscle contraction and arteriolar vasoconstriction. Vasoconstriction of the nasal mucosa arterioles decreases nasal secretions, assisting in the treatment of allergic rhinitis and nasal congestion. Incorrect Answers: B, C, D, E, and F. a-Adrenergic antagonists (Choice B), such as prazosin and terazosin, are useful for the treatment of benign prostatic hyperplasia and hypertension. B-Adrenergic agonists (Choice C), such as albuterol and salmeterol, are primarily utilized for the treatment of bronchoconstriction, as in asthma and chronic obstructive pulmonary disease. Bz Adrenergic receptors are primarily located in blood vessels, the gastrointestinal tract, and the bronchi. They are not used in the treatment of allergic rhinitis. B-Adrenergic antagonists (Choice D) are generally used for the treatment of tachyarrhythmias, hypertension, congestive heart failure, and glaucoma, amongst other indications. Blockade of B- adrenergic receptors leads to decreased blood pressure and negative chronotropic and inotropic effects. Nicotinic cholinergic agonists (Choice E), such as nicotine and varenicline, are useful for the treatment of tobacco dependence. Nicotinic cholinergic antagonists (Choice F), such as atracurium, are useful as muscle relaxants and paralytics. Educational Objective: Nasal decongestant sprays work primarily through a-adrenergic receptor activation and rely on signaling through phospholipase C and protein kinase C. The resultant vasoconstriction of nasal mucosal arterioles results in the decreased production of nasal secretions. Previous Next Score Report Lab Values Calculator Help Pause

76 Exam Section 2: Item 26 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 26. Which of the following hormones decreases synthesis of prostaglandins and leukotrienes? A) Aldosterone B) Cortisol OC) Dehydroepiandrosterone D) Desoxycorticosterone E) Pregnenolone

B. Both prostaglandins and leukotrienes are important inflammatory mediators that are eicosanoids, derived from arachidonic acid. Arachidonic acid is converted to prostaglandins by cyclooxygenase enzymes, a reaction that may be inhibited by nonsteroidal anti-inflammatory drugs. In contrast, leukotrienes are synthesized from arachidonic acid by 5-lipoxygenase, a reaction that may be inhibited by the drug zileuton. The binding of cortisol or other steroids to the glucocorticoid receptor results in an increased production of annexin A, which inhibits phospholipase A, Phospholipase A, is responsible for producing arachidonic acid through the cleavage of phospholipids. This mechanism allows cortisol and other glucocorticoids to exert an anti-inflammatory effect by decreasing the synthesis of eicosanoids, such as prostaglandins and leukotrienes, through deprivation of arachidonic acid. Incorrect Answers: A, C, D, and E. Aldosterone (Choice A) is a steroid hormone that acts on mineralocorticoid receptors at the collecting duct of the nephron, leading to increased sodium and water resorption and excretion of potassium and hydrogen ions. Aldosterone does not have a significant impact on the synthesis of prostaglandins and leukotrienes. Dehydroepiandrosterone (Choice C) is an important intermediate in the synthesis of testosterone and estrogen. It is produced in the adrenal zona reticularis by 17a-hydroxylase. It does not play a role in the synthesis of eicosanoids. Desoxycorticosterone (Choice D), also known as 11-deoxycorticosterone, is an important intermediate in the synthesis of both aldosterone and cortisol. Its effects upon synthesis of prostaglandins and leukotrienes are indirect and mediated by cortisol. Pregnenolone (Choice E) is produced from the cleavage of cholesterol by cholesterol desmolase. It is an early intermediate that is necessary for the production of aldosterone, cortisol, testosterone, and estradiol. Its effects upon the synthesis of prostaglandins and leukotrienes are indirect and mediated by cortisol. Educational Objective: Cortisol and glucocorticoids have a significant anti-inflammatory effect that is primarily mediated by decreasing production of prostaglandins and leukotrienes, which are derived from arachidonic acid. Activation of the glucocorticoid receptor increases the production of annexin A, which inhibits phospholipase A, the enzyme responsible for producing arachidonic acid from phospholipid. 1: II Previous Next Score Report Lab Values Calculator Help Pause

87 Exam Section 2: Item 37 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 37. A70-year-old woman comes to the physician because of a 3-month history of frequent constipation and fatigue. She has had a 4.5-kg (10-lb) weight loss during this period. Her vital signs are within normal limits. Abdominal examination shows mild tenderness. Rectal examination shows a 10-cm firm mass. Laboratory studies show: Hemoglobin Hematocrit 7.5 g/dL 21% Mean corpuscular volume 70 µm3 Leukocyte count Platelet count 25,000/mm3 500,000/mm3 This patient is at greatest risk for complications involving which of the following organs? A) Kidney B) Liver C) Lung D) Skin E) Uterus

B. Constipation, fatigue, weight loss, microcytic anemia (decreased hemoglobin, decreased hematocrit, and decreased mean corpuscular volume), and a rectal mass raise suspicion for a colonic malignancy. The most common site of primary malignancy that metastasizes to the liver is the gastrointestinal tract. Hepatic metastases are more common than primary tumors such as hepatocellular carcinoma. Gastrointestinal tract malignancies often metastasize to the liver because of their connection via the portal circulation. The portal venous system includes the superior mesenteric vein, inferior mesenteric vein, splenic vein, short gastric veins, and left gastric vein, which form the hepatic portal vein. Incorrect Answers: A, C, D, and E. Metastases to the kidney (Choice A) can occur from melanoma, solid tumors from the lung or breast, or gynecologic or gastrointestinal primary tumors. However, gastrointestinal cancers more often metastasize to the liver. Osteosarcoma and renal cell carcinoma often metastasize to the lung (Choice C). As in the liver, metastatic cancer to the lung is more common than primary lung cancer. Metastases to the lungs typically present as multiple rounded nodules scattered throughout both lungs. Cutaneous metastasis, or metastasis to the skin (Choice D), can occur secondary to primary skin cancers such as melanoma, or from breast, head, neck, or throat primary tumors. Cutaneous metastases are uncommon and occur late in the course of disease. Endometrial cancer of the uterus (Choice E) is typically a primary malignancy. Metastases to the uterus are rare and can occur from breast cancer, ovarian cancer, or gastrointestinal cancers. Educational Objective: Gastrointestinal tract malignancies often metastasize to the liver because of their connection via the portal circulation. Previous Next Score Report Lab Values Calculator Help Pause

42 Exam Section 1: Item 42 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 42. A 3-month-old boy is brought to the office by his mother because of a 1-week history of a lump on the right side of his groin. The mother says that when the patient cries, the bulge increases in size, and when he is quiet, it disappears. Physical examination shows a soft mass in the right groin region that decreases easily. Which of the following best describes the relationship of the hernia sac? A) Lateral to the inferior epigastric artery and inferior to the inguinal ligament B) Lateral to the inferior epigastric artery and superior to the inguinal ligament C) Medial to the inferior epigastric artery and inferior to the inguinal ligament D) Medial to the inferior epigastric artery and superior to the inguinal ligament

B. Indirect inguinal hernias are common hernias in children, especially in boys. They present with a groin or scrotal mass that increases in size when intra-abdominal pressure increases (eg, crying). Indirect inguinal hernias are caused by failure of the processus vaginalis to close after migration of the testes into the scrotal sac. This allows abdominal contents to exit the abdominal cavity through the internal (deep) inguinal ring and external (superficial) inguinal ring and extend into the scrotum. The internal inguinal ring is located lateral to the inferior epigastric vessels and superior to the inguinal ligament. Complications of inguinal hernias include incarceration and strangulation of the hernia contents, which can result in potential bowel necrosis; thus, most hernias are managed surgically to prevent such complications. Incorrect Answers: A, C, and D. Lateral to the inferior epigastric artery and inferior to the inguinal ligament (Choice A) does not describe a common location for hernia formation. Medial to the inferior epigastric artery and inferior to the inguinal ligament (Choice C) describes the location of a femoral hernia. Femoral hernias are more common in women and are more likely to present with incarceration. They are not common in children. Medial to the inferior epigastric artery and superior to the inguinal ligament (Choice D) describes the location of direct inguinal hernias. They protrude through the Hesselbach triangle, which is bordered by the inferior epigastric vessels, the lateral border of the rectus abdominis, and the inguinal ligament. They usually occur in older patients, rather than in children, because of weakness in the abdominal wall. Educational Objective: Indirect inguinal hernias are common in children and are caused by failure of the processus vaginalis to close after migration of the testes into the scrotal sac. They exit the abdominal cavity through the deep inguinal ring lateral to the inferior epigastric vessels and superior to the inguinal ligament. They allow abdominal contents to extend into the scrotal sac through the superficial inguinal ring. Previous Next Score Report Lab Values Calculator Help Pause

95 Exam Section 2: Item 45 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 45. A 28-year-old African American man comes to the physician because of a 3-month history of mild fatigue and weakness; he has had a 4.5-kg (10-lb) weight loss during this period. He also has a 6-month history of dry cough associated with chest pain and shortness of breath. The patient is married with two children, ages 2 and 3 years. He says that no one else at home has similar symptoms. He is a carpenter. He has several pets, including a dog, two cats, and a bird. He spent a week in Mexico approximately 1 year ago. He has smoked one-half pack of cigarettes daily for 10 years. He is 178 cm (5 ft 10 in) tall and weighs 70 kg (154 lb); BMI is 22 kg/m2. Mild wheezes are heard. Physical examination shows no other abnormalities. A chest X-ray shows bilateral hilar adenopathy and right paratracheal node enlargement. Biopsy specimens obtained via fiberoptic bronchoscopy show noncaseating granulomas. Test results of the biopsy specimen for acid-fast bacilli and fungi are negative. Which of the following is the strongest predisposing risk factor for this patient's condition? A) Employment B) Ethnicity C) Gender D) Living with pets E) Recent travel to Mexico F) Smoking history

B. Sarcoidosis is a noncaseating granulomatous disease that often involves the lung and classically presents with shortness of breath and cough. It tends to affect younger patients, women more than men, and patients with African-American ethnicity. It is characterized by the presence of noncaseating granuloma formation on histology. The clinical presentation ranges from asymptomatic, to mild shortness of breath and cough, to diffuse granulomatous infiltration of multiple organ systems. Manifestations include uveitis, lymphadenopathy, hepatosplenomegaly, hypercalcemia, and erythema nodosum. On diagnostic imaging, patients often have bilateral hilar lymphadenopathy, with or without bilateral upper lobe predominant interstitial infiltrates depending on the stage of the disease. The diagnosis is presumed on the basis of on clinical history and imaging features, although definitive diagnosis of a patient requires tissue sampling, most commonly from involved lymph nodes. Treatment typically requires steroids. Untreated chronic disease can progress to pulmonary fibrosis and a restrictive lung pathophysiology. Incorrect Answers: A, C, D, E, and F. Employment (Choice A) as a carpenter is a risk factor for occupational reactive or restrictive lung disease caused by exposure to wood dust. Work with asbestos-containing materials increases the risk for asbestosis. Asbestosis typically presents 20 to 30 years after the initial exposure with progressive dyspnea on exertion. Chest x-ray may show bilateral reticular opacities with pleural plaques. Gender (Choice C) is a predisposing risk factor for sarcoidosis, though the disease tends to affect women more than men. Living with pets (Choice D), such as birds, is a risk factor for hypersensitivity pneumonitis. It can present in acute, subacute, or chronic forms, with the acute form characterized by shortness of breath, cough, and systemic symptoms such as fever, chills, headache, and rash following exposure to an antigen. The chest x-ray may be normal at this stage or may show interstitial infiltrates. The subacute and chronic forms commonly present with dyspnea, cough, fatigue, and weight loss. Imaging and biopsy show interstitial inflammation and fibrosis of the lung. Recent travel to Mexico (Choice E) may increase the patient's risk for infection with tuberculosis and/or endemic coccidioidomycosis, both of which can form caseating granulomas in the lungs. In this case, the finding of noncaseating granulomas is more suggestive of sarcoidosis, and diagnostic testing for acid-fast bacilli and fungi are negative. Smoking history (Choice F) is a risk factor for primary lung malignancy, emphysema, and interstitial lung disease. Educational Objective: Sarcoidosis is a noncaseating granulomatous disease that often involves the hilar lymph nodes and/or lungs. Risk factors include being female and African-American ethnicity. Previous Next Score Report Lab Values Calculator Help Pause

9 Exam Section 1: Item 9 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 9. A previously healthy 16-year-old girl comes to the physician because of a progressive rash over her cheeks and the bridge of her nose during the past 5 weeks. She also has a 1-week history of stiffness of her fingers in the mornings that gradually resolves during the day. A photograph of the face is shown. Serum studies show an antinuclear antibody titer of 1:1280 and C3 concentration of 71 mg/dL (N=100-200). The pathogenesis of this patient's cutaneous and musculoskeletal symptoms is most similar to that of which of the following conditions? A) Acute rheumatic fever B) Drug-induced serum sickness C) Graft-versus-host disease D) Immune response to PPD skin testing O E) Peanut allergy

B. The malar rash of systemic lupus erythematosus (SLE) is characterized by pink patches involving the malar cheeks and bridge of the nose that spare the nasolabial folds. SLE is an example of type III (immune complex-mediated) hypersensitivity reaction, which is defined by the presence of circulating antigen-antibody complexes that deposit in tissues. These complexes then bind and activate complement, which attracts neutrophils. The neutrophils set into motion an inflammatory cascade and release lysosomal enzymes, causing inflammation and degradation of the surrounding healthy tissue. Another example of type III hypersensitivity is serum sickness. Serum sickness occurs when a drug, such as anti-thymocyte globulin, acts as a hapten to the immune system, triggering antibody formation. The antibodies then bind the drug, deposit in tissue, and attract and activate complement. The result is fever, rash, and arthralgias. Serum sickness reaction usually occurs 1 to 2 weeks after the initial exposure to the drug, reflecting the time required to mount an immune response. Incorrect Answers: A, C, D, and E. Acute rheumatic fever (Choice A) is a late complication of untreated Group A Streptococcus infection characterized by carditis, subcutaneous nodules, erythema marginatum, Sydenham chorea, and migratory polyarthritis. The cause is related to immune cross-reactivity caused by molecular homology between streptococcal M protein and human cardiac myosin proteins. It is a type II (complement-mediated cytotoxic) hypersensitivity reaction. Graft-versus-host disease (Choice C) and immune response to PPD skin testing (Choice D), or tuberculin reaction, are examples of type IV (delayed) hypersensitivity reactions. Type IV hypersensitivity is characterized by a cell-mediated response that involves the maturation of antigen-specific CD4+ or CD8+ T lymphocytes to a specific antigen. When the antigen is encountered, ČD4+ T lymphocytes release cytokines leading to inflammation and macrophage activation, while CD8+ T lymphocytes directly kill cells expressing the antigen. This type of hypersensitivity takes several days to manifest. Allergic contact dermatitis is another example of a type IV hypersensitivity reaction. Peanut allergy (Choice E) is an example of a type I (immediate) hypersensitivity reaction. Type I hypersensitivity occurs when preformed IgE on the surface of mast cells and basophils is crosslinked by an antigen, leading to immediate degranulation and release of histamine. This results in vasodilation, vascular pooling, and increased vascular permeability, which clinically manifests as urticaria, angioedema, or anaphylaxis. Educational Objective: Type III hypersensitivity is defined by the presence of circulating antigen-antibody-complement complexes that deposit in tissues and cause inflammation and destruction. SLE, serum sickness reaction, polyarteritis nodosa, and poststreptococcal glomerulonephritis are all examples of type III hypersensitivity reactions. Previous Next Score Report Lab Values Calculator Help Pause

32 Exam Section 1: Item 32 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 32. A 52-year-old woman with type 2 diabetes mellitus has chronic renal failure. She takes no medications other than glipizide. Creatinine clearance is 20% of normal. Which of the following sets of laboratory findings in serum is most likely in this patient? Ca2+ Phosphate Parathyroid Hormone A) ↑ B) ↑ ↑ C) ↑ ↑ D) E) ↑ ↑ OF) ↑

C. One of the complications of long-term diabetes mellitus is diabetic nephropathy. Diabetic nephropathy is a chronic process that occurs following nonenzymatic glycosylation of the glomerular basement membrane and efferent arterioles, characteristically demonstrating Kimmelstiel-Wilson lesions on light microscopy. It progresses over time in patients with diabetes mellitus, initially beginning as microalbuminuria, which can subsequently lead to macroalbuminuria and then end-stage kidney disease. Secondary hyperparathyroidism is commonly caused by chronic hypocalcemia in end-stage kidney disease. Hypocalcemia in end-stage kidney disease occurs because of decreased native renal production of active vitamin D, which leads to decreased systemic absorption of calcium and increased concentrations of phosphate caused by retention by the kidney. Increased phosphate also complexes with the calcium in serum, decreasing the serum total calcium. In turn, hypocalcemia is sensed by the normally functioning parathyroid gland, which increases synthesis of parathyroid hormone to attempt to normalize serum calcium through osteolysis, conversion of vitamin D, and gastrointestinal absorption. Incorrect Answers: A, B, D, E, and F. End-stage kidney disease results in hypocalcemia caused by decreased vitamin D, which leads to decreased calcium absorption and increased phosphate complexing with serum calcium (Choices A, B, and E). The kidney is essential in phosphate excretion, and in healthy individuals, phosphate is normally filtered through the glomerulus. In patients with end-stage kidney disease, renal elimination of phosphate is impaired, leading to increased concentrations of phosphate in the serum (Choices A, B, D, and F.) The resulting hypocalcemia in end-stage kidney disease stimulates production of parathyroid hormone reflected by increased serum parathyroid hormone concentrations (Choices A and D). Educational Objective: End-stage kidney disease results in hypocalcemia caused by decreased active vitamin D and decreased renal elimination of phosphate. This leads to the decreased absorption of calcium and increased production of calcium-phosphate complexes, respectively. The resulting hypocalcemia in end-stage kidney disease stimulates the synthesis and release of parathyroid hormone. II Previous Next Score Report Lab Values Calculator Help Pause

59 Exam Section 2: Item 9 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 9. A 26-year-old woman is diagnosed with a metastatic anaplastic tumor. Immunohistochemical studies of a biopsy specimen show large quantities of desmin in the tumor cells. Based on these findings, these cells most likely originated in which of the following tissues? A) Bone B) Epithelial C) Fibrous D) Muscle E) Nerve

D. Anaplastic tumors are composed of cells that are poorly differentiated and have lost the majority of the characteristics of the mature cells from which they originated. Anaplastic cells share morphological characteristics including nuclear pleomorphism, an increased nuclear to cytoplasmic ratio, and the presence of nucleoli. It can be challenging to identify the tissue of origin. Immunohistochemical stains for specific proteins are useful when morphological characteristics and cellular organization no longer represent the original cell types. Intermediate filaments are one of the three major structural polymers of the cellular cytoskeleton and are specific to certain cell types. They provide strength to the cytoskeleton. There are five types of intermediate filaments categorized by protein composition. Desmin is the primary protein component of the intermediate filaments of muscle cells. The finding of large quantities of desmin in this patient's biopsy specimen suggests the anaplastic tumor originated from muscle tissue. Incorrect Answers: A, B, C, and E. Bone (Choice A) is derived from mesenchymal cells. Intermediate filaments in mesenchymal cells are formed by the protein vimentin. Epithelial (Choice B) cells contain keratin-based intermediate filaments that stain positive with cytokeratin immunohistochemical stains. Fibrous (Choice C), or connective, tissue is the most abundant tissue in the body and is primarily composed of fibroblasts and extracellular matrix. Intermediate filaments in connective tissue are primarily composed of vimentin, though fibrous connective tissue includes a large portion of acellular extracellular matrix. Nerve (Choice E) tissue includes central neurons, peripheral neurons, and neuronal support (glial) cells. Intermediate filaments are composed of neurofilament in central neurons, peripherin in peripheral neurons, and glial fibrillary acidic protein in glial cells. Educational Objective: Intermediate filaments are essential components of the cell cytoskeleton and are composed of different proteins depending on the tissue of origin. Desmin is the primary component of intermediate filaments in muscle tissue. Previous Next Score Report Lab Values Calculator Help Pause

31 Exam Section 1: Item 31 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 31. A 50-year-old man who has had paraplegia for the past 5 years is admitted to the hospital for treatment of a urinary tract infection. Addressing which of the following factors is most likely to prevent the development of decubitus ulcers in this patient? A) Arterial blood flow B) Bacterial skin flora C) Nutritional status D) Pressure E) Venous stenosis

D. Decubitus ulcers, or pressure sores, are caused by prolonged pressure on a single point such as the buttocks, heels, occipital scalp, or elbows in an individual who has limited mobility. The pressure is confined to a small anatomical site in contact with the bed or chair, and it decreases blood supply to the area causing the skin to become ischemic and break down. Decubitus ulcers are seen in patients who have restricted mobility, such as the paraplegic patient in this case. Decubitus ulcers are diagnosed on the basis of clinical examination and staged on the basis of initial appearance. Stage 1 decubitus ulcers have nonblanchable erythema. There is no granulation tissue or eschar present. Stage 2 ulcers have full epidermal loss with exposed dermis. Stage 3 ulcers have full thickness epidermal and dermal loss with exposed adipose tissue. Stage 4 ulcers have full thickness skin loss with fully exposed fascia, muscle, or bone. Avoiding focal, Tong- standing pressure on a site is the most effective prevention method for decubitus ulcers. This is done by limiting the amount of time that pressure is sustained by any single point on the body. In a patient with some mobility, this may be performed using offloading maneuvers. In a bedbound patient, frequent turning or an offloading mattress may be effective. Incorrect Answers: A, B, C, and E. Arterial blood flow (Choice A) is a component of peripheral vascular disease, which can present with ischemic ulcers on the lower extremities and tips of the digits, particularly the toes. While local ischemia is the first step in formation of decubitus ulcers, this ischemia is caused by pressure, not aberrant blood flow. Bacterial skin flora (Choice B) may contribute to the secondary colonization or infection of decubitus ulcers. However, it does not contribute to the formation of the ulcers. Nutritional status (Choice C) should be optimized in patients with decubitus ulcers so wound healing can occur. However, nutritional status does not significantly affect the initial formation of decubitus ulcers. Venous stenosis (Choice E) does not contribute to decubitus ulcers which arise from pressure, but it may contribute to venous insufficiency ulcers. Ulcers caused by venous insufficiency are caused by decreased venous return and typically involve the lower legs. Educational Objective: Decubitus ulcers are caused by prolonged pressure on a single point such as the buttocks, heels, occipital scalp, or elbows. The primary method of prevention is offloading pressure points through patient maneuvers, turning, or use of an offloading mattress. %3D Previous Next Score Report Lab Values Calculator Help Pause

75 Exam Section 2: Item 25 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 25. A 76-year-old man with a 5-year history of progressive dementia, Alzheimer type, is brought to the emergency department by his son, who found him weak and unable to get out of bed. The patient has had increasing problems caring for himself while living alone and has lost 11 kg (25 lb) over the past year. He takes no medications. He is 173 cm (5 ft 8 in) tall and weighs 42 kg (92 Ib); BMI is 14 kg/m2. He is alert and talkative. Vital signs are normal. Physical examination shows sparse brittle hair and a red smooth tongue. There is general atrophy of interosseous and temporalis muscles, loss of subcutaneous fat, and edema of the lower extremities. He is oriented to person but not to place or time. Laboratory studies show: Total lymphocyte count Serum Albumin Pre-albumin Transferrin Cortisol Liver function tests show no abnormalities. An x-ray of the chest shows a healed granuloma in the right upper lobe of the lung. Which of the following is the most likely explanation for this patient's condition? A) Cushing disease B) Hepatic encephalopathy C) Nephrotic syndrome D) Protein-calorie malnutrition E) Tuberculosis

D. Dementia, Alzheimer type, is characterized by the loss of intellectual function within multiple domains: language, memory, perception, motor skills, attention, orientation, problem solving, and executive function. It is the most common cause of dementia in the elderly. Failure to thrive is characteristic of the final stages of dementia, Alzheimer type, and is often the cause of death. This patient shows weight loss, muscle wasting, loss of fat, and signs of vitamin deficiencies. Glossitis presents with a red, smooth tongue and can be secondary to B-complex vitamin deficiencies. Decreased albumin secondary to inadequate protein intake decreases the oncotic pressure within the blood vessels and fails to pull interstitial fluid back into the vascular space, leading to lower extremity edema. These findings, in combination with his weight loss and muscle wasting, suggest severe protein-calorie malnutrition. Incorrect Answers: A, B, C, and E. Cushing disease (Choice A) refers to the formation of an adrenocorticotropic hormone (ACTH)-secreting pituitary adenoma, which stimulates the adrenal gland to produce excess cortisol. Cushing disease presents with symptoms of hypercortisolism such as proximal muscle weakness, osteoporosis, hypertension, hyperglycemia, truncal obesity, moon facies, striae, and/or immunosuppression. Cortisol has been implicated in memory loss and cognitive disorders such as dementia, Alzheimer type. This is a more likely explanation for increased cortisol in this patient than an ACTH-secreting pituitary adenoma. Hepatic encephalopathy (Choice B) is a consequence of portosystemic shunting in liver cirrhosis. In cirrhosis, the blood carrying ammonia (NH,) does not pass through the liver but rather enters directly into the systemic circulation because of portosystemic shunting of blood. When NH3 accumulates, it causes neuropsychiatric disturbances ranging from subtle cognitive defects, like impaired attention span and memory deficits, to altered mental status and coma. This patient's hypoalbuminemia is caused by malnutrition rather than decreased liver synthetic function. Nephrotic syndrome (Choice C) is characterized by edema secondary to excessive proteinuria, resulting in hypoalbuminemia and diminished intravascular oncotic pressure. While this patient does have hypoalbuminemia, the accompanying weight loss, muscle wasting, and signs of vitamin deficiencies suggest that this is caused by malnutrition and failure to thrive rather than by a loss of protein in the urine as a result of nephrotic syndrome. A past infection with tuberculosis (Choice E) may have caused the scarred granuloma in the right upper lobe of the lung, but this patient does not show signs of active tuberculosis infection. Symptoms of primary tuberculosis infection include subacute fevers, weight loss, night sweats, cough, and malaise. Educational Objective: Protein-calorie malnutrition, or failure to thrive, characterizes the terminal stage of dementia, Alzheimer type. Signs of protein-calorie malnutrition include weight loss, muscle wasting, hypoalbuminemia, and signs of vitamin deficiencies. Previous Next Score Report Lab Values Calculator Help Pause

28 Exam Section 1: Item 28 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 28. A 3-month-old boy is brought to the physician because of frequent loose stools during the past month. His maternal cousin has similar symptoms. He is exclusively breast-fed. He is below the 3rd percentile for length and weight. Physical examination shows mild dehydration. Laboratory studies show acidic stools that are positive for reducing substances. Following the oral administration of glucose, there is no increase in his serum glucose concentration. Following the oral administration of fructose, his stools become normal. A deficiency in which of the following in intestinal mucosal cells is most likely in this patient? A) Glucose transporter-4 (GLUT-4) B) GLUT-5 C) Lactase D) Sodium-glucose cotransporter-1 (SGLT-1) E) Sucrase-isomaltase

D. Galactose, glucose, and fructose are the only monosaccharides absorbed by enterocytes, the cells lining the lumen of the small intestine. Lactose, the primary disaccharide of breastmilk, is broken down to galactose and glucose. Both are taken up into the bloodstream via the sodium-glucose cotransporter-1 (SGLT-1). In this individual, a genetic deficiency in SGLT-1 causes decreased intestinal absorption of galactose and glucose. The symptoms of this deficiency include watery, acidic diarrhea and severe dehydration. Reducing substances, defined as unabsorbed di- or monosaccharides, are present in the stool. The oral administration of glucose does not increase serum glucose as glucose is also absorbed by this cotransporter. However, fructose is absorbed by glucose transporter-5 (GLUT-5) on the enterocytes and is unaffected by an SGLT-1 deficiency. Incorrect Answers: A, B, C, and E. Glucose transporter-4 (GLUT-4) (Choice A) is an insulin-dependent transporter, which is primarily located in adipose tissue and skeletal muscle. It does not play a role in the intestinal absorption of glucose or other monosaccharides. GLUT-5 (Choice B) is an insulin-independent transporter that absorbs fructose from the intestinal lumen. This transporter is unaffected in SGLT-1 deficiency, which explains the improvement of symptoms when fructose is provided as nutrition. Lactase (Choice C) catalyzes the breakdown of lactose, a disaccharide, to galactose and glucose, both monosaccharides, at the brush border of the enterocytes. Deficiency of lactase may be induced by several mechanisms. Primary lactase deficiency occurs after early childhood in children of ethnic groups which evolutionarily received little nutrition from milk (Asian, African, or Native American descent). Gastroenteritis or autoimmune disease may also damage the brush border and lead to effective lactase deficiency. Congenital lactase deficiency is also possible but rare. Lactase deficiency does not typically present this early in childhood and the absorption of galactose and glucose would be intact. Sucrase-isomaltase (Choice E) deficiency leads to an inability to break down the disaccharide sucrose into one glucose and one fructose molecule and maltose into two glucose molecules. Lactose, not sucrose or maltose, is the primary carbohydrate contained in breastmilk. Because of this, sucrase-isomaltase deficiency is not identified until after an infant is weaned and begins eating fruit, juice, and grains. Educational Objective: SGLT-1 is the cotransporter that absorbs galactose and glucose through the intestinal enterocytes. Deficiency of this enzyme manifests early in infancy as watery, acidic diarrhea and severe dehydration. Fructose absorption is unaffected as this utilizes the GLUT-5 transporter instead. Previous Next Score Report Lab Values Calculator Help Pause

34 Exam Section 1: Item 34 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 34. A 60-year-old woman is running her first marathon [42 km (26 mi)]. She does not drink enough liquids during the race and becomes dehydrated. Which of the following segments or channels will most likely be activated in this woman's kidney to help maintain hydration homeostasis? A) ADH (vasopressin) activity in the proximal tubular cells B) K-H*exchange in the distal convoluted tubule C) Proximal tubule carbonic anhydrase activity D) Urea reabsorption in the medullary collecting ducts E) Water reabsorption in the ascending limb of the loop of Henle

D. In this dehydrated patient, increased plasma osmolality will be sensed by hypothalamic osmoreceptors and result in the increased production of ADH (vasopressin) in the hypothalamic paraventricular and supraoptic nuclei. ADH is subsequently released by the posterior pituitary gland. The primary action of ADH is to cause the insertion of aquaporin channels in the renal collecting ducts, leading to increased free water absorption and a subsequent decrease in plasma osmolality. Its secondary action includes causing the insertion of urea transporters into the medullary collecting ducts. Increased reabsorption of urea allows for the maintenance of the urea concentration within the medullary interstitium, contributing to an osmolar gradient that facilitates the reabsorption of water in the thin descending loop of Henle. Incorrect Answers: A, B, C, and E. Activity in the proximal tubular cells (Choice A) is not a characteristic of ADH. ADH primarily exerts its effects on the collecting tubules to maintain osmolar (and volume) homeostasis. K*-H* exchange in the distal convoluted tubule (Choice B) is not the mechanism of maintaining hydration in this patient. K+-H+ exchange occurs in the collecting duct under the regulation of aldosterone. Proximal tubule carbonic anhydrase activity (Choice C) converts CO2 to bicarbonate to maintain acid-base balance in the nephron. Carbonic anhydrase activity is not regulated by ADH and does not play a prominent role in volume retention in the dehydrated state. Water reabsorption in the ascending limb of the loop of Henle (Choice E) does not occur. The ascending limb of the loop of Henle is largely impermeable to water and is an important site of reabsorption of sodium, potassium, and chloride. Educational Objective: In the dehydrated state, increased plasma osmolality triggers the production of hypothalamic ADH (vasopressin). ADH regulates serum osmolality through the reabsorption of free water in the collecting duct, and it also causes reabsorption of urea from the collecting duct into the medullary interstitium, where it contributes to the medullary osmotic gradient. Previous Next Score Report Lab Values Calculator Help Pause

89 Exam Section 2: Item 39 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 39. A79-year-old woman is brought to the emergency department because of pain in her left hip that began after she fell in her home. Physical examination shows a shortened and externally rotated left lower extremity. An x-ray shows an intertrochanteric fracture of the femur. Which of the following muscles best contributes to the lateral rotation of the thigh? A) Gluteus minimus B) Gracilis C) Pectineus D) Piriformis E) Rectus femoris

D. Lateral (external) rotation of the thigh rotates the knee, foot, and ankle away from the midline. The prime movers for this rotation comprise the lateral rotator group, including the piriformis, obturator internus and externus, gemelli, and quadratus femoris. These muscles originate on the bony pelvis and insert on the femur. Many muscles, including those in the gluteal compartment, contribute to lateral rotation; however, these muscles support the motion as opposed to generating it primarily. The piriformis originates from the anterior sacrum lateral to the sacral foramina and inserts on the superior greater trochanter. It is innervated by the nerve to the piriformis as a component of the sacral plexus. An intertrochanteric hip fracture would compromise the bone between the greater and lesser trochanter, which would limit transmission of rotational force by the piriformis to the thigh and leg distal to the fracture site. Hip pain elicited with internal and external rotation about the hip is also a consistent physical examination finding in a hip fracture. Incorrect Answers: A, B, C, and E. Gluteus minimus (Choice A) originates from the gluteal surface of the ilium inferior to gluteus medius and inserts on the femur at the greater trochanter. Its prime movement is hip abduction and medial (internal) rotation of the thigh. Gracilis (Choice B) originates from the ischium and pubis, inserts onto the pes anserinus of the tibia, and functions in hip adduction, medial (internal) rotation, and hip flexion. Pectineus (Choice C) originates from the pectineal line of the pubis and inserts on the femur. Its prime movements include thigh flexion (minor contribution), adduction, and external rotation. Rectus femoris (Choice E) is one of four muscles comprising the quadriceps group in the anterior thigh. It originates from the anterior inferior iliac spine, inserts onto the tibia via the patellar tendon, and is a prime mover of hip flexion and knee extension. Educational Objective: Lateral (external) rotation of the thigh rotates the knee, foot, and ankle away from the midline. The prime movers of this rotation comprise the lateral rotator group, including the piriformis, obturator internus and externus, gemelli, and quadratus femoris. Previous Next Score Report Lab Values Calculator Help Pause

56 Exam Section 2: Item 6 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 6. A 54-year-old woman comes to the physician because of a 5-day history of severe mid-back pain. Physical examination shows point tenderness over the T6 vertebra. Serum studies show a calcium concentration of 13.4 mg/dL; urinalysis shows Bence Jones proteins. Which of the following is the most likely cause of this patient's hypercalcemia? A) Excessive parathyroid hormone production B) Excessive parathyroid hormone-related protein production C) Increased fractional calcium gastrointestinal absorption D) Local interleukin-1 (IL-1) and tumor necrosis factor effects O E) Unregulated 1,25-dihydroxycholecalciferol production

D. Local interleukin-1 (IL-1) and tumor necrosis factor effects are the most likely cause of hypercalcemia in this patient with a probable diagnosis of light-chain multiple myeloma. In multiple myeloma, local and systemic secretion of cytokines including IL-1, tumor necrosis factor a, macrophage inflammatory protein (MIP), and receptor activator of nuclear factor-KB ligand (RANK-L), stimulate osteoclastic activity leading to increased bone turnover and release of calcium and phosphate into circulation. There is also a relative decrease in normal bone turnover and osteoblastic activity which impairs the body's ability to reuse and recycle calcium to create new bone. Patients with multiple myeloma and Bence-Jones proteins, which are free immunoglobulin light-chain proteins that have a tendency to precipitate in the renal tubules, also frequently have renal injury that impairs normal calcium homeostasis and further exacerbates hypercalcemia. Incorrect Answers: A, B, C, and E. Excessive parathyroid hormone production (PTH) (Choice A) is seen in patients with hypercalcemia caused by primary hyperparathyroidism, which is most commonly secondary to a PTH- secreting adenoma. PTH acts to increase calcium resorption from the bone. In patients with hypercalcemia from other causes, PTH is usually suppressed through negative feedback. Excessive parathyroid hormone-related protein production (PTHIP) (Choice B) is the mechanism of hypercalcemia seen in most malignancies other than multiple myeloma. PTHIP is secreted by malignant cells and acts in a similar fashion as endogenous PTH to increase calcium resorption from the bone. PTHPP is usually not detected in patients with multiple myeloma. Increased fractional calcium gastrointestinal absorption (Choice C) occurs in states of increased vitamin D, which directly stimulates absorption of calcium in the gut. This patient's vitamin D concentration is not known but should be measured and replaced with caution if low. Unregulated 1,25-dihydroxycholecalciferol production (Choice E) occurs in conditions where there is increased transformation of 25-hydroxycholecalciferol to its active form of 1,25- dihydroxycholecalciferol, like sarcoidosis, tuberculosis, and lymphoma. These concentrations can be measured directly in the serum, but such increases in the active form of vitamin D are not seen in multiple myeloma. Educational Objective: Malignant plasma cell clones in multiple myeloma locally secrete cytokines such as IL-1, tumor necrosis factor a, and RANK-L to stimulate osteoclastic activity with subsequent bone resorption leading to hypercalcemia. Combined with the relative inactivity of osteoblasts and frequent renal impairment, this derangement primarily accounts for the profound hypercalcemia seen in patients with multiple myeloma. Previous Next Score Report Lab Values Calculator Help Pause

13 Exam Section 1: Item 13 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 13. A 28-year-old woman undergoes PPD skin testing to determine previous infection with Mycobacterium tuberculosis. Results show a 25-mm, firm area of induration at 48 hours. Analysis of this patient's lesion is most likely to show a predominance of which of the following cell types? A) B lymphocytes B) Cytotoxic T lymphocytes OC) Eosinophils D) Macrophages E) Mast cells O F) Neutrophils

D. Macrophage predominance at the site of PPD skin testing is expected in this patient with a positive test. The PPD skin test is intended to identify patients who have been exposed to Mycobacterium tuberculosis (MTB) but does not distinguish between latent infection, treated latent infection, or active infection. The test works by injecting a small amount of mycobacterial antigen, tuberculin, directly under the skin. Prior exposure to MTB results in the formation of memory Th, cells against typical mycobacterial antigens. Injection of tuberculin under the skin results in processing of antigen by dendritic cells and presentation to Th, cells. Recognition by these Th, cells leads to release of cytokines including interferon-Y, which attracts macrophages to the site of injection. This is an example of a delayed hypersensitivity reaction. Erythema and induration will often develop within 24 hours and subside by 72 hours. Incorrect Answers: A, B, C, E, and F. B lymphocytes (Choice A) are not directly involved in the inflammatory reaction against MTB or against the PPD, although they do function as antigen-presenting cells that can activate CD4+ T lymphocytes Cytotoxic T lymphocytes (Choice B), also called CD8+ T lymphocytes, are present at the site of PPD exposure but are not the predominant cell type. In active MTB infection, CD8+ T lymphocytes directly contribute to the control of MTB by releasing substances like granzyme and perforin that kill infected cells . Eosinophils (Choice C) play a minor role in response to MTB. They are not typically found at the site of PPD exposure. Mast cells (Choice E) play several unique roles in response to MTB infection, including recognition of MTB via toll-like receptors, internalization of MTB followed by degranulation and cytokine release, and interaction with other immune cells like macrophages and antigen-presenting cells. They are not the predominant cell type located at the site of PPD exposure. Neutrophils (Choice F) mainly respond to bacterial pathogens and are not the predominant cell type found at the site of PPD exposure. Educational Objective: Macrophages are attracted to the area of PPD exposure via the action of Th, lymphocytes in patients previously exposed to MTB. They are the predominant cell type found at the site of PPD exposure and in the lungs of patients with pulmonary tuberculosis. Previous Next Score Report Lab Values Calculator Help Pause

65 Exam Section 2: Item 15 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 15. A 13-year-old boy is brought to the physician because of a 4-week history of mild left thigh and knee pain that was exacerbated when he jumped off the school bus steps onto the sidewalk this morning. The pain increases with activity. He has no history of serious illness or trauma and takes no medications. He is at the 50th percentile for height, 90th percentile for weight, and 97th percentile for BMI. His temperature is 37°C (98.6°F), pulse is 80/min, respirations are 16/min, and blood pressure is 115/65 mm Hg. Genital development is Tanner stage 3. Physical examination of the left lower extremity shows no muscle tenderness. Muscle tone and strength are normal in both lower extremities. Range of motion of the left knee is full and elicits no pain. Internal rotation of the left hip is limited by pain. Which of the following is the most likely diagnosis? A) Ankylosing spondylitis B) Osgood-Schlatter disease C) Septic arthritis D) Slipped capital femoral epiphysis

D. Slipped capital femoral epiphysis (SCFE) most commonly occurs in overweight children between the ages of 10 and 15 years. SCFE is a displacement of the femoral epiphysis relative to the femoral neck secondary to anterosuperior movement of the metaphysis. It is a Salter-Harris type 1 fracture of the growth plate. SCFE typically presents with a painful limp and, if severe, may result in restricted range of motion of the hip joint (eg, limited abduction and internal rotation of the hip as in this case) and inability to bear weight on the affected lower extremity. Pain often radiates to the knee and thigh. Radiographic evaluation of both hips in frontal and frog-leg views permits side-by-side comparison to diagnose the condition. If untreated, complications include avascular necrosis of the femoral head, limited range of motion, gait impairment, and premature osteoarthritis. To prevent these complications, treatment requires surgical fixation of the femoral epiphysis. Incorrect Answers: A, B, and C. Ankylosing spondylitis (Choice A) is a rheumatologic inflammatory condition involving the vertebrae that presents with subacute to chronic back pain and limited range of motion that is improved with exercise. It often presents in men between the ages of 20 and 30 years and may be associated with uveitis. The resultant arthropathy can lead to frequent falls and fractures; however, without back pain or any systemic inflammatory manifestations, it would be an unlikely diagnosis in this case. Osgood-Schlatter disease (Choice B) refers to osteochondrosis or traction apophysitis of the tibial tubercle that typically occurs in adolescent, athletic children. It presents with pain along the anterior aspect of the proximal tibia and knee that is exacerbated by kneeling or extending the knee. Physical examination typically discloses tenderness over an enlarged tibial tubercle. Septic arthritis (Choice C) results from bacterial infection of the synovial joint space, most commonly by Staphylococcus aureus. Examination of the affected joint discloses tenderness, swelling, and erythema, and patients generally present with fever. Arthrocentesis is indicated for the evaluation of monoarticular erythema and swelling to evaluate for septic arthritis, which characteristically shows purulent synovial fluid with greater than 50,000 leukocytes/mm3. Educational Objective: Slipped capital femoral epiphysis presents with hip, thigh, and knee pain with limited range of motion and an antalgic gait, generally in an overweight child or teenager. It is a Salter-Harris type 1 fracture of the physis, and treatment includes surgical fixation. If untreated, complications include avascular necrosis of the femoral head, limited range of motion, gait impairment, and premature osteoarthritis. %3D Previous Next Score Report Lab Values Calculator Help Pause

3 Exam Section 1: Item 3 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 3. A 62-year-old man with a 4-year history of chronic angina pectoris comes to the emergency department because of severe chest and left shoulder pain. Physical examination shows bradycardia. An ECG shows ST-segment elevation in leads II, III, and aVF. Occlusion of blood flow in which of the following arteries is the most likely cause of the findings in this patient? A) Anterior interventricular (left anterior descending) B) Circumflex C) Left coronary D) Right coronary

D. The right coronary artery originates from the right aortic sinus and primarily provides blood flow to the right ventricle, right atrium, sinoatrial (SA) node, and atrioventricular (AV) node, and in a right-dominant circulation (-85% of patients), it provides a majority of the blood supply to the inferior heart through the posterior descending artery. Occlusion of the right coronary artery can result in infarction of myocardial tissue along the inferior portion of the heart, which leads to ST-segment changes in the inferior ECG leads II, III, and aVF. Impaired perfusion to the SA and AV nodes can also result in bradycardia or heart block. The patient requires admission to the hospital and emergent cardiac catheterization for revascularization or consideration of thrombolytics if not close to a center capable of percutaneous coronary intervention. Incorrect Answers: A, B, and C. The anterior interventricular (left anterior descending) (Choice A) artery branches from the left main coronary artery and provides perfusion to the anterior portion of the interventricular septum, the anterior left ventricle, and the anterolateral papillary muscle. Infarction of this territory may present with ST-segment changes in the precordial ECCG leads. The circumflex (Choice B) artery originates from the left main coronary artery and primarily provides perfusion to the lateral and posterior walls of the left ventricle, the anterolateral papillary muscle, and some blood flow to the AV node. It does not provide perfusion to the inferior heart unless the patient demonstrates a left-dominant circulation (~8% of patients). Infarction of this territory may result in ST-segment changes in the lateral, posterior, and/or inferior ECG leads. The left coronary (Choice C), or left main coronary, artery arises from the left aortic sinus and branches into the left anterior descending and left circumflex arteries. ST-elevation myocardial infarctions involving the left main coronary artery are associated with significant morbidity and mortality, as it would involve a majority of the myocardium. Educational Objective: The right coronary artery provides blood flow to the right ventricle, right atrium, SA node, and AV node, and in a right-dominant circulation (~85% of patients), it provides a majority of the blood supply to the inferior heart through the posterior descending artery. Myocardial infarction in this territory is associated with ST-segment changes in the inferior ECG leads and may also present with bradycardia or heart block. II Previous Next Score Report Lab Values Calculator Help Pause

74 Exam Section 2: Item 24 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 24. A 68-year-old woman with hypertension comes to the physician because of concerns about recent home blood pressure measurements. She says that her systolic blood pressures have ranged from 140 mm Hg to 160 mm Hg during the past 10 days. She has generalized anxiety disorder and major depressive disorder. Medications include amlodipine, citalopram, and hydrochlorothiazide. She appears anxious. Her blood pressure is 145/94 mm Hg. Physical examination shows no other abnormalities. While discussing treatment options, the patient says, "You are not taking good care of me; you must not like me very much." Which of the following is the most appropriate initial response by the physician? A) "Blood pressure often goes up as you get older, so needing additional medication is common." B) "A common reason for poorly controlled blood pressure is not taking medications properly. Can you tell me how you take your medicines?" C) "I am taking the best possible care of you; we just need to make some medication adjustments." D) "I want to help you as much as I can. Why don't we talk about what is going on in your life." E) "Your anxiety seems worse today. Let's talk about some treatment options."

D. When patients bring up concerns about their care, physicians should acknowledge the concerns and ask open-ended questions to explore the concerns. If the physician suspects an outside stressor or psychiatric condition is contributing, the physician should also inquire about how the patient is doing generally. In response to this patient's concern about not being cared for, the physician should reassure the patient that the physician is committed to figuring out how to best help her. Providing this reassurance and commitment will likely improve the therapeutic alliance. Another option would be to validate the patient's distress by making statements such as, "You feel uncared for" or "You are afraid that I will not give you my best." These statements restate the patient's words and infer the underlying emotions, respectively, two proven validation methods. These validating statements would encourage the patient to share more about her concerns. Incorrect Answers: A, B, C, and E. Educating the patient about normal aging (Choice A) or inquiring about medication adherence (Choice B) avoid acknowledging the patient's concerns about her care. Further, implying that the patient may not be properly taking her medications as a response to the patient's concerns about the physician would likely invalidate her concerns and elicit defensiveness. The physician should instead directly respond to the patient's statement and ask open-ended questions to explore the patient's situation. Stating that the patient is receiving the best possible care and providing education about treatment options (Choice C) invalidates this patient's concerns by implying that she should not be dissatisfied with this "best possible care." Before providing education, the physician should explore the patient's concern using an open-ended question. Telling the patient that her anxiety seems worse (Choice E) would imply that her concerns are manifestations of her anxiety and are therefore not valid concerns. The patient may feel defensive and avoid openly discussing her concerns. Educational Objective: When patients bring up concerns about their care, physicians should acknowledge the concerns and ask open-ended questions to explore the concerns. If the physician suspects an outside stressor or psychiatric condition is contributing, the physician should also inquire about how the patient is doing generally. Further, the physician should provide reassurance that they are committed to finding a solution. Previous Next Score Report Lab Values Calculator Help Pause

48 Exam Section 1: Item 48 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 48. Which of the following domains depicted in the schematic diagram of an immunoglobulin molecule shown would be altered by isotype switching? 15 A) 1 B) 1 and 2 26 C) 1 and 5 D) 1, 2, and 6 E) 2, 3, and 4 F) 5 and 6 4

E. Domains 2, 3, and 4 comprise the constant region of the immunoglobulin (Ig) molecule and are exchanged during isotype (class) switching from IgM to another Ig class. The Ig molecule is made of two light chains and two heavy chains that are paired, with both the light and heavy chains containing constant and variable regions. The constant regions are shared among Ig molecules of a certain class. For example, all IgM molecules have identical constant domains while all IgG molecules also share a constant region. The variable sequence represents a unique region for antigen binding such that each individual Ig is specific to an antigen. After exposure to such an antigen, a specific IgM expressed on the surface of a B lymphocyte undergoes class switching from IgM to another class (IgG, IgE, or IgA). During this process, the constant region of the IgM molecule is exchanged for the constant region of the new Ig molecule while leaving the variable regions intact so as not to alter the ability to bind the unique antigen. The B lymphocyte can then differentiate into a plasma cell and begin secreting this new, unique Ig. In the diagram, domains 2, 3, and 4 represent the constant region that would be exchanged during isotype (class) switching. Incorrect Answers: A, B, C, D, and F. Domain 1 (Choice A) is the variable region of the heavy chain and is not exchanged during isotype switching. Similarly, domains 1 and 2 (Choice B) belong to both variable (domain 1) and constant regions (domain 2). Domains 1 and 5 (Choice C) are both variable domains and would remain unchanged. Domains 1, 2, and 6 (Choice D) include those from both constant and variable regions. Similarly, domains 5 and 6 (Choice F) contain both constant and variable regions, although both of these belong to the light chain of the Ig. Educational Objective: Immunoglobulin isotype switching occurs after a particular IgM on the surface of a B lymphocyte recognizes its matching antigen. This induces exchange of the constant region of IgM for the constant region of another Ig subclass that is tissue and target specific. The variable region remains unchanged. Previous Next Score Report Lab Values Calculator Help Pause 3. 88

29 Exam Section 1: Item 29 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 29. A72-year-old woman with hypertension comes to the physician for a follow-up examination. She has had no fever, shortness of breath, weight loss, or symptoms of gastroesophageal reflux disease. She was admitted to the hospital because of tuberculosis at the age of 23 years. Current medications include aspirin, atenolol, hydrochlorothiazide, and lisinopril. Her pulse is 68/min, and blood pressure is 140/74 mm Hg. The lungs are clear. Cardiac examination shows a grade 2/6, systolic ejection murmur that is best heard along the left sternal border. A chest x-ray shows normal lung fields and a calcified aortic valve. Which of the following best explains these cardiac findings? A) Adverse effect of atenolol B) Coronary artery disease C) Idiopathic hypertrophic cardiomyopathy D) Past tuberculosis exposure E) Normal aging

E. Fibrosis and calcification of the aortic valve is a pathologic consequence of mechanical stress. It results from repetitive microtrauma from the opening and closing of valve leaflets with associated chronic inflammation as part of normal aging. Many people will develop some degree of aortic valve stenosis (AS) over time. Underlying structural abnormalities of the valve, such as a bicuspid aortic valve or injury from rheumatic fever, alter the biomechanics of valve opening and closing and increase the likelihood of earlier calcification and resultant stenosis. The severity of AS is graded on echocardiography through the measurement of valve area, peak velocity, and mean pressure gradient across the valve. While many patients may be asymptomatic, those with severe ÁS may complain of fatigue, shortness of breath, cough, diminished exercise tolerance, angina, or syncope with exertion. Examination findings include a crescendo-decrescendo systolic murmur best heard at the upper sternal border, and pulsus parvus et tardus (weak and delayed) may be noted on examination of peripheral pulses. Because of the chronic increased afterload from a fixed obstruction by the valve, left ventricular hypertrophy and resultant diastolic dysfunction can occur. Severe symptomatic aortic stenosis is an indication for valve replacement. Incorrect Answers: A, B, C, and D. Aortic stenosis is not an adverse effect of atenolol (Choice A), a selective Bradrenergic antagonist. The major adverse effects of beta-blockers are bradycardia and fatigue. Coronary artery disease (Choice B) may be assessed by evaluation of calcification of the coronary vessels, but it would not lead to calcification of the aortic valve. Coronary artery disease is a risk factor for acute coronary syndrome, which may be complicated by papillary muscle rupture and mitral valve insufficiency. Idiopathic hypertrophic cardiomyopathy (Choice C) is characterized by concentric hypertrophy of the ventricles and diastolic dysfunction. It is most commonly associated with an underlying genetic predisposition. Patients typically present with an S,gallop and a systolic murmur that increases with Valsalva maneuver because of left ventricular outflow tract obstruction. Diagnosis is made by echocardiography. Past tuberculosis exposure (Choice D) is unlikely to have caused aortic stenosis in this patient. Extrapulmonary tuberculosis infection does not commonly involve the heart. Treatment is combination therapy with isoniazid, rifampin, pyrazinamide, and ethambutol. The major adverse effects of these medications are hepatotoxicity, peripheral neuropathy, and/or optic neuritis. Educational Objective: Aortic stenosis is common with aging because of the calcification and fibrosis of the valve that occurs with repeated opening and closing. Aortic stenosis classically presents with a crescendo-decrescendo systolic murmur best heard at the upper sternal border. Previous Next Score Report Lab Values Calculator Help Pause

27 Exam Section 1: Item 27 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 27. A 14-year-old girl is brought to the physician by her mother because of a 2-month history of hair loss on her head. The mother says that her daughter has been trying to imitate the hairstyles of several pop music stars by using various ties, rubber bands, and curlers, but she is not sure whether any chemical treatments have been used. The mother reports that the patient has been sad since her grandmother died unexpectedly 3 months ago. Vital signs are normal. Physical examination shows decreased hair density over several irregular patches temporally and frontally, but no denuded areas. The underlying skin is normal. The remaining hair shafts in these thinned areas are of varying lengths. There is no frontotemporal recession. A photograph of the affected area is shown. Which of the following is the most likely explanation for this patient's hair loss? A) Alopecia areata B) Androgenetic alopecia C) Telogen effluvium O D) Tinea capitis E) Trichotillomania

E. Focal, irregular patches of hair loss with broken hairs is consistent with trichotillomania. Trichotillomania is a psychiatric impulse control disorder that presents with repetitive hair pulling resulting in noticeable hair loss and several unsuccessful attempts to stop hair pulling. Patients also commonly experience tension prior to hair pulling and relief after hair pulling. Stressors can exacerbate pulling behavior. Physical examination typically shows bizarrely shaped patches of hair loss on the scalp or loss of eyelashes or eyebrows without significant underlying skin changes. Broken hairs are usually different lengths. Trichotillomania is managed with habit reversal therapy, which helps patients identify triggers and develop alternative ways to resolve the tension that precedes hair pulling behavior. Incorrect Answers: A, B, C, and D. Alopecia areata (Choice A) is a chronic, immune-mediated disorder of hair loss that can present with diffuse or focal hair loss. Areas may have smooth or irregular borders. Hairs are typically narrower proximally than distally and are prone to breaking. The broken hairs are usually the same length, while in trichotillomania, the broken hairs are different lengths. In this child with a known stressor, trichotillomania is more likely. Androgenetic alopecia (Choice B) causes male-pattern baldness and uncommonly occurs in women and children. Hair loss progresses over years and results in hairless areas on the temporal, frontal, or vertex areas of the scalp. Follicular miniaturization and a consequent diversity of hair shaft diameters are typical. Telogen effluvium (Choice C) is a disorder of diffuse, nonscarring hair loss that results from inciting factors such as medical illness, childbirth, nutritional deficiencies, or emotional stress. The hair loss is diffuse, as opposed to this patient's focal hair loss, and may be most evident bitemporally, frontally, or on the vertex. The borders of the hair loss typically appear smoother than those of trichotillomania. Tinea capitis (Choice D) is a fungal infection of the scalp common in children. The most common presentation is scaly patches with alopecia, and patches of alopecia with black dots that represent broken hairs. This patient did not have scaly patches or black dots on the scalp. Educational Objective: Trichotillomania is a psychiatric disorder that presents with repetitive hair pulling resulting in noticeable hair loss. Patients typically present with irregularly shaped patches of hair loss with broken hairs of different lengths. %3D Previous Next Score Report Lab Values Calculator Help Pause

10 Exam Section 1: Item 10 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 10. A 33-year-old woman comes to the physician because of fatigue for 2 months; she feels well otherwise. She has a 10-year history of systemic lupus erythematosus treated with prednisone. Physical examination shows no abnormalities. Laboratory studies show: 9.0 g/dL 27% 8500/mm3 0.2% Hemoglobin Hematocrit Leukocyte count Reticulocyte count Platelet count 190,000/mm3 Serum Ferritin Iron Total iron-binding capacity Red cell morphology 250 ng/mL 10 pug/dL 100 ug/dL (N=250-400) normochromic, normocytic Which of the following is the most likely cause of this patient's anemia? A) Accumulation of iron within mitochondria of erythroblasts B) Decreased serum concentrations of hepcidin O C) Decreased serum concentrations of inflammatory cytokines D) Decreased uptake of iron by bone marrow macrophages E) Increased retention of iron within the reticuloendothelial system

E. Increased retention of iron within the reticuloendothelial system (RES) is characteristic of anemia of chronic disease (ACD) and is the most likely cause of anemia in this patient with systemic lupus erythematosus (SLE). ACD is common in patients with rheumatologic disorders such as SLE or rheumatoid arthritis and those with chronic infections, renal disease, or malignancies. Chronic inflammation with release of inflammatory substances such as interleukin-1 and tumor necrosis factor-a lead to an increase in hepcidin production, a decreased responsiveness of the bone marrow to erythropoietin, and retention of iron within the RES. This confluence of factors leads to impaired iron absorption and recycling, with a consequent state of functional iron deficiency despite adequate iron stores. Iron that is retained within the RES cannot effectively be accessed and used for erythropoiesis, so patients present with either normocytic or microcytic anemia and decreased numbers of reticulocytes indicating inadequate bone marrow response to anemia. Other laboratory findings include a normal to slightly increased ferritin concentration, which is the storage form of iron, decreased serum iron concentration, and a decreased total iron-binding capacity (TIBC). If doubt exists in distinguishing between ACD and iron deficiency anemia, measurement of the soluble transferrin receptor can be useful. Incorrect Answers: A, B, C, and D. Accumulation of iron within mitochondria of erythroblasts (Choice A) occurs in patients with porphyria cutanea tarda, sideroblastic anemia, and B-thalassemia major, but not in ACD. Decreased serum concentrations of hepcidin (Choice B) do not occur in ACD. On the contrary, increased concentrations of hepcidin in ACD impair the absorption of iron in the gut and the release of iron from macrophages. This occurs via degradation of ferroportin, which is necessary for iron trafficking. Decreased serum concentrations of inflammatory cytokines (Choice C) do not occur in ACD. In fact, cytokines are known to play an important role in the pathophysiology of ACD, and increased concentrations of cytokines are found in inflammatory states. Decreased uptake of iron by bone marrow macrophages (Choice D) does not occur in ACD. Macrophages in this condition have a tendency to store iron; an increased concentration of hepcidin, which is the primary pathologic change in ACD, inhibits the release of iron from macrophages. Educational Objective: ACD is common in patients with rheumatologic diseases such as SLE; itis the result of increased hepcidin secretion mediated by increased concentrations of inflammatory cytokines. This results in the increased retention of iron within the RES and impaired iron exportation for use in erythropoiesis with consequent anemia. Ferritin concentrations are normal to increased, and serum iron concentrations and TIBC are decreased. Previous Next Score Report Lab Values Calculator Help Pause

18 Exam Section 1: Item 18 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 18. A 13-year-old girl grew 7.5 cm (3 in) over the summer. Which of the following most likely accounts for increased intestinal absorption of calcium during this period? A) Calcitonin-mediated hypercalcemia O B) Calcium-binding proteins in goblet cells C) Cortisol-induced transcription of calcium transporters D) Cyclic AMP generated in the enterocytes in response to parathyroid hormone E) Hormones derived from 7-dehydrocholesterol

E. Intestinal absorption of calcium is regulated by vitamin D. 7-Dehydrocholesterol is converted by ultraviolet radiation in the skin to cholecalciferol, which undergoes two further hydroxylation reactions in the liver and in the kidney via 25-hydroxylase and 1a-hydroxylase, respectively, to form the active hormone 1,25-dihydroxycholecalciferol (vitamin D). Active vitamin D promotes gastrointestinal absorption of calcium and phosphate and also promotes bone mineralization. Thus, because of the patient's increased sun exposure during the summer, an increased amount of ultraviolet radiation-dependent conversion of 7-dehydrocholesterol occurs, which ultimately manifests with increased concentrations of active vitamin D, increased calcium and phosphate intestinal absorption, and increased bone mineralization. Incorrect Answers: A, B, C, and D. Calcitonin-mediated hypercalcemia (Choice A) would only occur from relative deficiency of calcitonin, not overproduction. Calcitonin, produced by thyroid parafollicular C cells, generally opposes parathyroid hormone by decreasing resorption of bone by osteoclasts. Its release is stimulated by increased serum calcium concentration, and it does not function to promote gastrointestinal calcium absorption, which would only further raise serum calcium concentration. Calcitonin may be overproduced in medullary thyroid carcinoma, which would result in hypocalcemia. Calcium-binding proteins in goblet cells (Choice B) may play a role in the regulation of cytoplasmic trafficking of vesicles containing mucin. In intestinal epithelial cells, calcium-binding proteins such as calbindin are produced in response to vitamin D and help facilitate the absorption of calcium. Cortisol-induced transcription of calcium transporters (Choice C) is not a known mechanism related to increased intestinal calcium absorption in puberty. Cyclic AMP generated in the enterocytes in response to parathyroid hormone (Choice D) is not the mechanism of parathyroid hormone. While parathyroid hormone does signal through a G protein-coupled receptor and through cyclic AMP, this takes place primarily within the bone and kidney, rather than the intestinal enterocytes, to promote bone resorption and calcium reabsorption. The gastrointestinal effects of parathyroid hormone are mediated through the increased production of active vitamin D. Educational Objective: 7-Dehydrocholesterol is converted by ultraviolet radiation in the skin to cholecalciferol, which undergoes two further hydroxylation reactions in the liver and in the kidney to form the active hormone 1,25-dihydroxycholecalciferol. Active vitamin D promotes the gastrointestinal absorption of calcium and phosphate via the calcium-binding protein calbindin. Previous Next Score Report Lab Values Calculator Help Pause

43 Exam Section 1: Item 43 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 43. A 40-year-old man with type 2 diabetes mellitus comes to the physician for a follow-up examination. Treatment with glyburide and metformin has been ineffective in controlling his disease. The decision is made to try pioglitazone. This drug produces a beneficial effect through which of the following mechanisms of action? A) Blockade of ATP-sensitive potassium channels in the pancreatic ß cells B) Increased insulin synthesis by promoting the effect of physiologic insulin secretagogues C) Increased insulin uptake by muscle and adipose tissue, thus making the insulin more effective D) Promotion of insulin action interfering with the synthesis of endogenous antagonists, primarily glucagons E) Stimulation of the peroxisome proliferator-activated receptor y

E. Pioglitazone (a thiazolidinedione) is used for the treatment of type 2 diabetes mellitus and achieves its effect by binding to peroxisome proliferator-activated receptors (PPARS), which are expressed in most cell types including adipose cells, pancreatic cells, myocytes, and hepatocytes. PPARS are nuclear receptors that act as transcription factors to regulate metabolism. Activation results in increased cell sensitivity to insulin. Adverse effects of pioglitazone include an increased risk for bone fracture, congestive heart failure, peripheral edema, and weight gain. Incorrect Answers: A, B, C, and D. Blockade of ATP-sensitive potassium channels in the pancreaticß cells (Choice A) is the mechanism of action of sulfonylureas, such as glipizide. Closure of potassium channels leads to depolarization and subsequent opening of voltage-gated calcium channels. Calcium influx promotes the endogenous release of insulin from the pancreatic B cells. Increased insulin synthesis by promoting the effect of physiologic insulin secretagogues (Choice B) is the mechanism of action of glucagon-like peptide 1 (GLP-1) agonists, such as liraglutide and exenatide. Increased insulin uptake by muscle and adipose tissue, thus making the insulin more effective (Choice C) is not a mechanism of action of oral agents for the treatment of diabetes. The insulin receptor is a tyrosine kinase receptor and does not require uptake of insulin by the target cell for its effects to be exerted. Increased muscle and adipose uptake of glucose is one of several mechanisms of action of metformin. Promotion of insulin action interfering with the synthesis of endogenous antagonists, primarily glucagons (Choice D) is the mechanism of action of dipeptidyl peptidase 4 inhibitors, such as sitagliptin and linagliptin. These agents increase the effect of incretins, such as GIP and GLP-1, to inhibit the release of glucagon. Educational Objective: Pioglitazone is an oral thiazolidinedione that is used for the treatment of type 2 diabetes mellitus. It binds to peroxisome proliferator-activated receptors (PPARS), which act as transcription factors to regulate metabolism and promote insulin sensitivity. Previous Next Score Report Lab Values Calculator Help Pause

20 Exam Section 1: Item 20 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 20. During a study of antibiotic treatment for pityriasis rosea, two randomized groups of patients with this condition are established (Groups A and B). Group A receives erythromycin and hydroxyzine, and Group B receives only hydroxyzine. The number of days that lesions are present in each group is then determined, and the average of days is calculated for each group. Which of the following statistics is most likely to establish the difference between the averages of these two groups? A) Chi-square test B) Correlation C) Мean D) Regression E) Student's t-test

E. Statistical tests can be employed to assess whether a difference in values exists. Often, these values are means, though they need not be. Many tests have been derived to assess differences in numerical, quantitative, categorical, and qualitative data sets. Choosing the appropriate test is important to appropriately draw conclusions from data. In this case, comparing the mean number of days to lesion resolution is a comparison of numeric means, which is best assessed by Student's t-test. The t-test determines whether a difference between two means could be reasonably attributed to chance, and if the likelihood of such a conclusion is reasonably low, statistical significance can be inferred. Incorrect Answers: A, B, C, and D. Chi-square test (Choice A) assesses for statistically significant differences between two or more categorical values (eg, proportions, percentages). It is commonly used for larger samples and is also able to compare multiple values (eg, three different experimental groups). It is not used for arithmetic means. Correlation (Choice B) relates the trend of two variables together and is quantified by the Pearson correlation coefficient (r). Correlation represents the increase, decrease, or absence of change in one variable with changes in the other. In cases where variables trend together (both increase), the r value is positive. The closer the value of the correlation coefficient to 1.0, the stronger the trend between the two variables. Negative r values suggest that data trend in opposing fashion; as one variable increases, the other decreases. Uncorrelated data will have a correlation coefficient near zero, which indicates that changes in one variable are not associated with any particular trend in the other variable. Correlation should not be confused with causation. Mean (Choice C) describes a measure of central tendency within a dataset. To determine whether a difference exists between two datasets, the mean can be used as a component of Student's t- test. Regression (Choice D) models the relationship between dependent and independent variables in a linear or nonlinear fashion by deriving the trend from the data. It is not used to compare means. It is used to predict the dependent variable from an independent variable assumed to behave in a similar way to the data used to construct the regression model. Educational Objective: Student's t-test assesses for statistically significant differences between the means of two groups. Previous Next Score Report Lab Values Calculator Help Pause

63 Exam Section 2: Item 13 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 13. A 50-year-old man with septic shock develops diffuse cortical necrosis in the kidneys. He dies 2 weeks later. At autopsy, gross examination of the cortex of the kidneys shows changes indicative of acute ischemic infarction. Histologic examination of the kidney medulla shows no abnormalities. Based on these findings, which of the following structures is most likely to show pathologic changes? A) Collecting ducts B) Interlobular arteries C) Loops of Henle D) Papillary ducts E) Proximal convoluted tubules

E. The kidney cortex is the site of the initial filtration and reabsorption at the glomerulus, Bowman capsule, and proximal convoluted tubule (PCT). The PCT is the site for reabsorption of the majority of electrolytes (HCO3, Na+, Cl; P0,3, K*), water, uric acid, glucose, and amino acids in the initial glomerular filtrate, which occurs by ATP-dependent processes requiring blood flow and oxygen. Therefore, states of acute hypotension or ischemia, such as septic shock, can limit the supply of glucose and oxygen to the highly metabolic cells of the PCT, leading to ischemic cortical necrosis. Incorrect Answers: A, B, C, and D. The kidney cortex is surrounded by the outer capsule of the kidney and is superficial to the kidney medulla. The medulla is the innermost region of the kidney and contains the majority of the nephron, including the collecting ducts (Choice A), loops of Henle (Choice C), and papillary ducts (Choice D). They would not be affected by cortical necrosis. The thin descending loop of Henle passively reabsorbs water and is impermeable to Na - The thick ascending limb of the loop of Henle is impermeable to water and reabsorbs Na+, K+, and Cl: The loops of Henle also create a concentration gradient in the medulla of the kidney via a countercurrent multiplier system. The collecting tubule is the terminal segment of the nephron and is the site of reabsorption of Na+and secretion of K+and H*. It is regulated by aldosterone and antidiuretic hormone. Several collecting ducts converge and drain into a papillary duct, which in turn, empties urine into the renal pelvis for excretion. The interlobular arteries (Choice B) are located in the cortex and supply the cells of the PCT and glomeruli. Their metabolic activity is comparable to that of other arteries, and they require less ATP than the cells of the PCT. Educational Objective: The PCT is the site for reabsorption of the majority of electrolytes (HCO3, Na+, Cl; PO,3, K*), water, uric acid, glucose, and amino acids in the initial glomerular filtrate. It requires a decreased intracellular sodium concentration generated by the NaK-ATPase pump and thus is sensitive to ischemia. II Previous Next Score Report Lab Values Calculator Help Pause

17 Exam Section 1: Item 17 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 17. During an experimental study of oxygen consumption in the kidney, experimental animals are ventilated with 100% nitrogen. Cells from which of the following areas of the kidney are most likely to show the first signs of anoxic injury? A) Bowman capsule B) Distal convoluted tubule C) Efferent arteriole D) Glomerulus E) Proximal tubule

E. The proximal tubule is the closest to the glomerulus. It is the site of reabsorption of the majority of electrolytes (HCO3, Na, Cl; PO,3, K*), water, uric acid, glucose, and amino acids in the initial glomerular filtrate. It contains several sodium-coupled transporters for glucose and amino acids on the apical membrane that depend on the decreased intracellular sodium concentration generated by the Na+/K+ATPase pump on the basolateral membrane. Because of this, the proximal tubule has high metabolic demand and is particularly sensitive to hypoxia. In the ischemic state, cells convert from aerobic to anaerobic metabolism, which results in the production of lactate. Damage to the proximal tubule can result in acute tubular necrosis leading to decreased glomerular filtration rate, glucosuria, aminoaciduria, and impaired reabsorption of bicarbonate (eg, proximal renal tubular acidosis type 2). In severe, prolonged states of hypoxia, terminal damage to the cells of the nephron can occur, preventing regeneration and resulting in irreversible renal failure. Incorrect Answers: A, B, C, and D. The Bowman capsule (Choice A) surrounds the glomerulus (Choice D) and is the site of initial filtration in the nephron. Filtration depends on the relative oncotic and hydrostatic pressures between the glomerulus and the Bowman capsule. It does not depend on an electrolyte gradient or an ATP-dependent pump. The distal convoluted tubule (Choice B) reabsorbs sodium and chloride in an ATP-dependent process; however, it only reabsorbs 5% to 10% of sodium, in contrast to the 65% to 80% that the proximal tubule reabsorbs. It has a resulting lower energy and oxygen demand and is less sensitive to states of ischemia or hypoxia by comparison. The efferent arteriole (Choice C) exits the glomerulus at the Bowman capsule. It is purely an artery with few ATP-dependent cellular processes and less metabolic activity. Educational Objective: The proximal tubule is the site of reabsorption of the majority of electrolytes (HCO3, Nat, Cl; PO,3, K*), water, uric acid, glucose, and amino acids in the initial glomerular filtrate. It requires a decreased intracellular sodium concentration generated by the NaK*-ATPase pump and thus is sensitive to hypoxia. Previous Next Score Report Lab Values Calculator Help Pause

14 Exam Section 1: Item 14 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 14. A 29-year-old woman comes to the physician because of a 4-month history of headaches that occur every few weeks. They consist of unilateral throbbing pain and are accompanied by nausea and photophobia. She has a history of similar episodes, which started with the onset of menarche, but she had been symptom-free for 7 years until now. She says the headaches seem worse now than ever. Physical examination shows no abnormalities. A diagnosis of migraine is made, and sumatriptan is prescribed. The patient then becomes anxious and tells the physician that she expects further evaluation of her symptoms. She says, "How can you know they are nothing dangerous after only examining me?" She refuses the prescription and begins to cry. After empathizing with the patient, which of the following statements by the physician is most appropriate at this time? A) "Have you been getting enough sleep lately or have you had any hormonal changes recently?" B) "I know that migraines are real and painful, but I have all of the information I need to make that diagnosis." %3D C) "I'm concerned that you may be depressed. Would you like a referral to a therapist?" D) "It's all right. There is nothing to worry about since your illness has not changed in character over time." E) "Tell me what you think may be causing the headaches."

E. This physician should first encourage the patient to elaborate on her fears about the headaches. Patients commonly fear that they will contract serious diseases, as most people have relatives with serious diseases or have been exposed to serious illnesses through the media. As such, physicians should empathize with this patient's fears and validate her emotions. Physicians should also initially ask patients open-ended questions about their fears. This patient will feel validated by the physician listening to her worries, and the physician can tailor further discussion to address this patient's specific fears. The patient may reveal important additional details that explain the recurrence of the migraines and can be proactively addressed. After this initial exploration of the patient's fears, the physician may educate the patient about the reasons for opting against a diagnostic workup. Incorrect Answers: A, B, C, and D. Asking about risk factors for mood changes (eg, sleep, hormones) or suggesting that the patient may be depressed (Choices A and C) would imply that her concerns are manifestations of her mental state and are therefore not valid concerns. The patient may feel defensive and avoid openly discussing her fears. Reassuring the patient or providing education about migraines (Choices B and D) may be appropriate after exploring this patient's specific fears. However, when patients express strong emotion, the most effective initial step is to listen. Educational Objective: When patients express fears or other strong emotions, the most effective initial steps are to listen and validate the emotions. Physicians should encourage patients to express their specific fears and then tailor further discussion to these fears. Previous Next Score Report Lab Values Calculator Help Pause

41 Exam Section 1: Item 41 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 41. A 49-year-old man is brought to the emergency department 30 minutes after fainting in the street. He regained consciousness 1 minute after fainting. He says that he has had watery diarrhea during the past 5 days, which has not improved with fasting. He has not changed his diet or travelled overseas recently. His blood pressure is 90/60 mm Hg. Physical examination shows a flushed face and dehydration. Serum studies show a potassium concentration of 2 mEq/L and glucose concentration of 150 mg/dL. He is admitted to the hospital, and intravenous fluid replacement is started. Over the next day, he passes a stool with a volume of 3.5 L. Treatment with which of the following hormones is most likely to control this patient's diarrhea? A) Aldosterone B) Cholecystokinin C) Gastric inhibitory polypeptide D) Gastrin E) Somatostatin

E. Vasoactive intestinal polypeptide (VIP) can be produced by VIPomas. VIPoma syndrome, also called Verner-Morrison syndrome, is caused by a rare tumor often associated with multiple endocrine neoplasia that produces VIP. VIP is released by parasympathetic ganglia and leads to the increased secretion of water and electrolytes by the intestines as well as the increased relaxation of smooth muscle fibers in the gastrointestinal tract. This results in profound fluid and electrolyte secretion into the gastrointestinal tract, severe watery diarrhea, electrolyte disturbances, achlorhydria, alkalosis, flushing, and vasodilation. This syndrome is often termed WDHA syndrome (watery diarrhea, hypokalemia, achlorhydria). Diagnostic evaluation includes CT or MRI imaging of the abdomen, somatostatin receptor scintigraphy, or PET imaging to assist in localization of the tumor; biopsy can confirm histologic diagnosis. Treatment includes correction of the associated fluid, electrolyte, and acid-base derangements, along with surgery and/or chemotherapy. Somatostatin is a hormone that decreases gastric acid and pepsinogen secretion, pancreatic and small intestine fluid secretion, and insulin and glucagon release. Somatostatin analogs, such as octreotide, inhibit the release of VIP and counteract its effects. This can be helpful to control diarrhea in patients with Verner-Morrison syndrome. Incorrect Answers: A, B, C, and D. Aldosterone (Choice A) promotes reabsorption of sodium in the distal convoluted tubule and collecting duct of the nephron, causing indirect reabsorption of water. This generally leads to sodium reabsorption in excess of water reabsorption. Excess production of aldosterone causes hypertension, hypokalemia, and metabolic alkalosis, known as Conn syndrome. Cholecystokinin (Choice B) is released from cells in the duodenum and jejunum and stimulates the cholecystokinin receptor. Stimulation of this receptor leads to increased gastric acid secretion, pancreatic enzyme secretion, gallbladder contraction, and sphincter of Óddi relaxation. Gastric inhibitory polypeptide (Choice C) is released from cells in the duodenum and jejunum and functions to decrease gastric acid secretion and increase insulin release. Gastrin (Choice D) is physiologically produced by G cells in the gastric antrum and functions to stimulate parietal cells within the gastric body to produce hydrochloric acid. It also increases gastric motility. Excess production of gastrin is associated with peptic ulcer disease and Zollinger-Ellison syndrome. Educational Objective: VIP is released by parasympathetic ganglia and leads to the increased secretion of water and electrolytes by the intestines as well as the increased relaxation of smooth muscle fibers in the gastrointestinal tract. Excess production results in profound fluid and electrolyte secretion into the gastrointestinal tract, severe watery diarrhea, electrolyte disturbances, achlorhydria, alkalosis, flushing, and vasodilation. Somatostatin analogs, such as octreotide, inhibit the release of VIP and counteract its effects. Previous Next Score Report Lab Values Calculator Help Pause

49 Exam Section 1: Item 49 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 49. A 16-year-old girl is brought to the emergency department 45 minutes after her mother found her in her room with an empty bottle of acetaminophen. She has a history of major depressive disorder. Physical examination shows no abnormalities. Which of the following events would contribute directly to hepatic toxicity in this patient? A) Activation of protein kinases B) Depletion of glycogen stores C) Increased conversion of triglycerides to fatty acids D) Liberation of lysosomal enzymes E) Peroxidation of lipids in cell membranes F) Swelling of the rough endoplasmic reticulum

E. When taken at therapeutic doses, acetaminophen is safely metabolized through phase IIl conjugation, including glucuronidation and sulfation. In an acetaminophen overdose, saturation of phase Il metabolic pathways leads to excess acetaminophen metabolized by CYP-mediated reactions to N-acetyl-p-benzoquinoneimine (NAPQI), which has strong oxidizing properties and is directly hepatotoxic. Oxidizing free radicals damage hepatocytes through peroxidation of lipids in cell membranes, oxidative damage to intracellular proteins, and strand breaks in DNA. The antioxidant molecule glutathione conjugates NAPQI, allowing it to be safely excreted. The depletion of glutathione is a hallmark of acetaminophen toxicity. Acetaminophen toxicity is treated by repleting hepatic stores of glutathione through intravenous or oral administration of N-acetylcysteine. Incorrect Answers: A, B, C, D, and F. Activation of protein kinases (Choice A) occurs during a variety of cellular responses. However, hepatocellular toxicity caused by acetaminophen overdose is not mediated through cellular protein-mediated stress responses. Rather, it is a direct result of toxicity from reactive oxygen species. Depletion of glycogen stores (Choice B) may occur during a prolonged fasting state. Acetaminophen overdose mediates hepatotoxicity via reactive oxygen species and does not deplete glycogen stores. Increased conversion of triglycerides to fatty acids (Choice C) is observed in hepatic steatosis. Hepatic steatosis is a form of reversible hepatocyte injury that may be observed secondary to toxic injury with ethanol. Liberation of lysosomal enzymes (Choice D) does not occur as a result of acetaminophen toxicity. Lysosomes contain proteolytic enzymes, such as cathepsin, but do not play a major role in either apoptosis or necrosis as a result of acetaminophen toxicity. Swelling of the rough endoplasmic reticulum (Choice F) may occur in cellular ischemia because of depletion of adenosine triphosphate and dysfunction of sodium-potassium pumps, which leads to osmotic shifting within the cell and swelling of the cell and its organelles. Misfolded proteins may also occur, which accumulate in the endoplasmic reticulum. If restoration of blood flow and resolution of hypoxia does not occur, cellular apoptosis or necrosis results. Educational Objective: In acetaminophen overdose, saturation of phase II metabolic pathways leads to excess acetaminophen metabolized by CYP-mediated reactions to N-acetyl-p- benzoquinoneimine (NAPQI), which has strong oxidizing properties and is directly hepatotoxic. Oxidizing free radicals damage hepatocytes through peroxidation of lipids in cell membranes, oxidative damage to intracellular proteins, and strand breaks in DNA. Previous Next Score Report Lab Values Calculator Help Pause

22 Exam Section 1: Item 22 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 22. A 26-year-old man comes to the physician because of palpitations, heat intolerance, and a fine hand tremor for the past month. Examination shows an enlarged thyroid gland. Serum thyroxine (T) concentration is 28 ug/dL, and serum thyroid-stimulating hormone (TSH) concentration is less than 0.03 µU/mL. The most likely cause of his condition is development of antibodies against which of the following? A) T4 B) Trreceptor protein C) Thyroid-releasing hormone (TRH) D) TRH receptor E) TSH F) TSH receptor

F. Graves disease is the most common cause of hyperthyroidism and occurs because of the presence of TSH receptor-stimulating antibodies. This results in a diffusely enlarged thyroid gland on physical examination, along with signs and symptoms of hyperthyroidism including weight loss, insomnia, palpitations, heat intolerance, tremor, and diaphoresis. TSH-receptor stimulating antibodies also stimulate orbital fibroblasts and lead to thyroid ophthalmopathy, which is characterized by the expansion of orbital fat and connective tissue with resultant exophthalmos, lid lag, and restrictive strabismus. Thyroid laboratory studies typically show findings of primary hyperthyroidism, including increased thyroxine (T), increased triiodothyronine (T3), and decreased TSH. Radioactive iodine uptake nuclear scintigraphy typically shows diffuse, abnormally increased uptake in the thyroid gland without a discrete nodule. Incorrect Answers: A, B, C, D, and E. T4(Choice A) is the most abundant form of thyroid hormone in the bloodstream and is converted to T3 by 5'-deiodinase. Antibodies against T4 are not a cause of Graves disease. Tgreceptor protein (Choice B), or thyroid hormone receptor, is expressed widely in peripheral tissues. T3 binds the receptor with greater affinity than T4 Translocation to the nucleus results in transcriptional changes and mediates the metabolic effects of thyroid hormone. Treceptor protein antibodies are not a cause of Graves disease. Thyroid-releasing hormone (TRH) (Choice C) is a hypothalamic hormone that stimulates the release of TSH from anterior pituitary thyrotropic cells. Antibodies against TRH are not a cause of Graves disease. TRH receptor (Choice D) is present on thyrotropic cells of the anterior pituitary and is activated by TRH. Upon activation by TRH, thyrotropic cells release TSH. Antibodies against TRH receptor are not a cause of Graves disease. TSH (Choice E) is produced by pituitary thyrotrophic cells and activates the TSH receptor on thyroid follicular cells, thereby promoting release of T4 and T3 Measurement of serum TSH concentration is useful for distinguishing hypothalamic or pituitary from thyroid-related causes of hyperthyroidism. Educational Objective: Graves disease is caused by TSH receptor-stimulating antibodies. Signs and symptoms include weight loss, insomnia, palpitations, heat intolerance, tremor, diaphoresis, a diffusely enlarged thyroid gland, and thyroid ophthalmopathy. Laboratory studies show increased T4 increased T, decreased TSH, and the presence of TSH receptor-stimulating antibodies. Previous Next Score Report Lab Values Calculator Help Pause

24 Exam Section 1: Item 24 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 24. A 12-year-old boy comes to the physician's office with dependent and periorbital edema, hypoalbuminemia and proteinuria. A biopsy of the kidney is performed. A photomicrograph of a representative glomerulus is shown (periodic acid-Schiff-stained). Which of the following is the most likely diagnosis? A) Diabetic glomerulosclerosis B) Diffuse proliferative nephritis C) Focal glomerulonephritis D) Focal segmental glomerulosclerosis E) Membranous nephropathy F) Minimal change nephrotic syndrome G) Poststreptococcal glomerulonephritis

F. Nephrotic syndrome is characterized by edema secondary to excessive proteinuria resulting in hypoalbuminemia and diminished intravascular oncotic pressure. This occurs because of loss of normal size and charge filtration capability of the glomerular capillary and Bowman capsule interface. Causes of nephrotic syndrome include minimal change disease, membranous nephropathy, and focal segmental glomerulosclerosis. Minimal change disease is the most common cause of nephrotic syndrome in children and can be idiopathic or triggered by a recent infection or immune stimulus. Patients classically present with dependent edema in the buttocks, lower back, and legs; foamy or dark-colored urine, hypoalbuminemia, and proteinuria are also common. Minimal change disease appears normal on light microscopy, as shown in the photomicrograph. Incorrect Answers: A, B, C, D, E, and G. Diabetic glomerulosclerosis (Choice A) is characterized by thickened glomerular basement membranes. It occurs following nonenzymatic glycosylation of the glomerular basement membrane and efferent arterioles, characteristically showing Kimmelstiel-Wilson lesions on light microscopy. It progresses over time in patients with diabetes mellitus, initially beginning as microalbuminuria, which can subsequently lead to macroalbuminuria and, finally, end-stage renal disease. Diffuse proliferative nephritis (Choice B) is a nephritic syndrome characterized by inflammatory processes leading to hematuria, RBC casts in urine, oliguria, and hypertension. It can be associated with proteinuria, but to a lesser extent than nephrotic syndrome. Proliferative nephritis is often caused by systemic lupus erythematosus, and it is marked by granular immunofluorescent deposits of IgG and irregular subepithelial electron-dense deposits. Focal glomerulonephritis (Choice C) describes a nephritic syndrome, which on microscopy shows inflammation involving less than half of the glomeruli with other normal glomeruli present. This photomicrograph shows a normal glomerulus, and the patient does not demonstrate features of nephritic syndrome such as hypertension and hematuria. Focal segmental glomerulosclerosis (FSGS) (Choice D) is a nephrotic syndrome that can result in proteinuria and pitting edema. FSGS is most commonly associated with sickle cell disease, opioid abuse, and HIV, and it is characterized by segmental sclerosis of the glomeruli. Membranous nephropathy (Choice E) is a nephrotic syndrome. However, it usually demonstrates diffuse thickening of the glomerular capillary wall on light microscopy. On immunofluorescence, IgG deposition will occur along the glomerular basement membrane. Neutrophilic infiltration of the glomerular tuft is seen in poststreptococcal glomerulonephritis (Choice G), which is a nephritic syndrome characterized by dark-colored urine, hypertension, and mild proteinuria following an infection with group A Streptococcus. It would be unlikely to cause significant proteinuria and edema. Educational Objective: Minimal change disease is the most common cause of nephrotic syndrome in children and can be idiopathic or triggered by a recent infection or immune stimulus. Patients classically present with dependent edema in the buttocks, lower back, and legs, with foamy or dark-colored urine, hypoalbuminemia, and proteinuria. Minimal change disease appears normal on light microscopy. Previous Next Score Report Lab Values Calculator Help Pause

39 Exam Section 1: Item 39 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 39. A 17-year-old boy is brought to the emergency department 15 minutes after he was found unarousable in the high school's bathroom. Approximately twenty 5-mg oxycodone tablets were found in his pocket. He has a history of using multiple illicit drugs. Which of the following sets of clinical findings is most likely in this patient? Pupil Examination Glasgow Coma Score Pulse Respirations A) Increased normal constricted 12 B) Increased decreased normal 15 C) Normal increased dilated 15 D) Normal normal dilated 7 E) Decreased increased normal 12 F) Decreased decreased constricted 7

F. This patient has likely abused opioids, which are central nervous system (CNS) depressants used as analgesics and recreational drugs. Opioids act throughout the CNS and peripheral nervous system and interact with several neurotransmitter systems. Opioid intoxication causes euphoria (because of interaction with dopamine), altered mental status, sedation (in severe cases manifesting as a decreased Glasgow Coma Score), bradycardia, hypotension, depressed respiratory drive, and constricted pupils. Specifically, the u7opioid receptor, located in the brainstem respiratory centers, controls respiratory depression, which may be severe. Miosis is a distinctive finding that is less common in intoxication with other CNS depressants and is caused by direct opioid receptor activity in brain areas responsible for pupillary control. Opioids also act on receptors within the enteric nervous system, decreasing gut motility and causing constipation. Incorrect Answers: A, B, C, D, and E. An increased pulse (Choices A and B), increased respirations (Choices C and E), and a normal Glasgow Coma Score (Choices B and C) would be atypical in this patient who was unarousable because of opioid intoxication, indicating severe CNS depression. Pupillary dilation is atypical of opioid intoxication (Choices C and D). Such a finding would be more consistent with stimulant abuse. Educational Objective: Opioid intoxication causes symptoms of CNS depression including sedation, respiratory depression, bradycardia, and hypotension. Miosis differentiates opioid intoxication from other CNS depressant toxidromes. %3D Previous Next Score Report Lab Values Calculator Help Pause

70 Exam Section 2: Item 20 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 20. A 22-year-old woman is found to be HIV positive after sexual contact with a partner with HIV infection. Combination therapy with lamivudine (3TC), ritonavir/lopinavir, and zidovudine (AZT) is initiated. Three months later, genomic typing shows that her HIV strain has become resistant to ritonavir/lopinavir. The cause of this resistance is most likely the acquisition of a mutation in a gene that is critical for which of the following viral processes? A) Adsorption and penetration B) Early protein synthesis C) Genome integration D) Late protein synthesis E) Nucleic acid synthesis F) Packaging and assembly G) Protein processing H) Release 1) Uncoating

G. Protein processing is the process in HIV replication that has developed a mutation conferring drug resistance in this patient. Protein processing is targeted by protease inhibitors such as ritonavir/lopinavir. HIV drugs are separated into several classes and work by different mechanisms to prevent viral replication. Drugs of disparate mechanisms are combined to inhibit HIV replication at multiple steps, thereby preventing the development of resistance. HIV infection and replication can be broken into seven primary steps: binding, fusion, reverse transcription, integration, replication, assembly, and budding. After HIV binds to the surface of CD4+ T lymphocytes via interaction of its specific glycoproteins with CD4 receptors, such as CCR5, the virus fuses with the cell membrane and releases its contents into the cytoplasm. HIV RNA is reverse transcribed by reverse transcriptase to create HIV DNA. This DNA is then incorporated into the CD4+ T lymphocyte genome. Once integrated, HIV hijacks the cellular machinery to produce the HIV genome and all the requisite proteins to form a new HIV viral particle. The new virus then buds off of the CD4+ T lymphocyte and is released to infect other lymphocytes. However, the viral particles are immature and contain long proteins that have not yet been cleaved, a task accomplished by HIV protease. Following protein cleavage, the HIV viral particle core has reached maturity. Protease inhibitors such as lopinavir block this step but have no effect on the preceding steps of HIV replication. Ritonavir is a boosting agent given in combination with lopinavir to increase serum drug concentrations. Frequently, several mutations are required for patients to develop resistance to protease inhibitors. Incorrect Answers: A, B, C, D, E, F, H, and I. Adsorption and penetration (Choice A) is blocked by entry inhibitors including CCR5 antagonists and fusion inhibitors. Resistance would be uncommon in patients not taking one of these medications. Early protein synthesis (Choice B) and late protein synthesis (Choice D) describe protein synthesis pre- and post-integration of the HIV genome into T lymphocytes. Early proteins are made before integration of the HIV genome and function to augment HIV replication. Late protein RNA is transcribed early, but it is not translated until the HIV genome has been integrated. These proteins function to create new HIV progeny. Lopinavir does not affect protein synthesis but does affect protein cleaving to form the mature HIV virus. Genome integration (Choice C) is blocked by integrase inhibitors including dolutegravir among others. This patient is not taking any integrase inhibitors. Nucleic acid synthesis (Choice E) is accomplished by the HIV reverse transcriptase enzyme, which converts RNA to DNA prior to integration into the host genome. This step is blocked by nucleoside reverse transcriptase inhibitors (NRTIS) and non-nucleoside reverse transcriptase inhibitors (NNRTIS). Zidovudine and lamivudine are NRTIS, so mutations of HIV reverse transcriptase would result in resistance. Packaging and assembly (Choice F) into the viral capsid can be inhibited by capsid inhibitors, which until recently were only used experimentally. A novel capsid inhibitor called GS-6207 was recently approved for use. Uncoating of viral particles (Choice I) before HIV entry into the cell may also be a mechanism of this drug. Release (Choice H) of the viral particle from T lymphocytes is not inhibited by any drugs that are currently in clinical use. Educational Objective: Protease inhibitors such as lopinavir, which are commonly given with boosters such as ritonavir, are generally effective when used in combination with NRTIS and NNRTIS, but resistance can develop. HIV viral proteins must be cleaved prior to maturity and these medications block this step. II Previous Next Score Report Lab Values Calculator Help Pause

51 Exam Section 2: Item 1 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment 1. A newborn is evaluated for microcephaly, cataracts, and chorioretinitis. The mother developed an illness in the first trimester of pregnancy characterized by low-grade fever, a faint erythematous rash, occipital lymphadenopathy, and joint stiffness. The illness resolved within 1 week without complications. She did not receive any immunizations prior to the pregnancy. Which of the following viruses is the most likely cause of the newborn's illness? A) Cytomegalovirus B) Herpes simplex virus C) HIV D) HTLV-2 E) Measles virus F) Reovirus G) Rubella virus H) Varicella-zoster virus

G. Rubella infection is spread by respiratory droplets and can be asymptomatic or present with an erythematous rash, arthralgias, postauricular lymphadenopathy, and fever. In otherwise healthy patients, the infection is usually self-limited with no long-term sequelae. In pregnant patients, however, especially when the infection occurs in the first trimester, it can cause an infection of the fetus, which manifests with cataracts, sensorineural deafness, and congenital heart disease (eg, patent ductus arteriosus), as well as microcephaly, neurodevelopmental delay, and extramedullary hematopoiesis. Infection with the virus is prevented by administration of the measles-mumps-rubella (MMR) vaccination prior to conception, since the vaccination is a live- attenuated vaccine, making it contraindicated during pregnancy. Ideally, confirmation of rubella immunity should be achieved prior to pregnancy. Incorrect Answers: A, B, C, D, E, F, and H. Cytomegalovirus (CMV) (Choice A) can often be asymptomatic or cause a short febrile, flu-like illness, but it does not commonly present with rash. Fetal infection with CMV is associated with intrauterine growth restriction, hepatosplenomegaly, petechiae, jaundice, thrombocytopenia, microcephaly, and chorioretinitis. Herpes simplex virus (HSV) (Choice B) is characterized by tender vesicles followed by ulceration and crusting, most commonly around the mouth and genitals. A fetal infection with HSV can cause localized lesions of the skin, eyes, and mouth, along with the potential for disseminated disease such as encephalitis and multiorgan dysfunction. HIV (Choice C) presents nonspecifically with fever, lymphadenopathy, gastrointestinal upset, and occasionally rash. Neonatal infection results in an immunocompromised state that puts the newborn at risk for failure to thrive and recurrent infections. It does not cause microcephaly, cataracts, or chorioretinitis. HTLV-2 (Choice D), also called human T-cell leukemia virus 2, is commonly asymptomatic but can result in recurrent respiratory infections and neurologic issues, such as peripheral neuropathy. While it can be spread through the placenta during pregnancy, it does not present in the fetus with microcephaly, cataracts, or chorioretinitis. Measles virus (Choice E) presents with a fever, cough, conjunctivitis, and a maculopapular rash. Complications include subacute sclerosing panencephalitis and giant cell pneumonia. It increases the risk for fetal demise and low birth weight and can cause congenital measles. However, it does not commonly present with postauricular lymphadenopathy or arthralgias. Reovirus (Choice F) is a family of viruses that includes the rotavirus. Rotavirus presents with severe gastrointestinal distress and diarrhea, particularly in children. Maternal infection does not cause congenital malformations. Varicella-zoster virus (Choice H) most commonly presents as chickenpox in children and young adults, with a pruritic vesicular rash spread diffusely over the body and a low-grade fever. It can also present as shingles, with a vesicular rash in a dermatomal distribution caused by reactivation of the virus from a dorsal root ganglion. It can infrequently cause infection of the fetus if maternal infection occurs in the first trimester, which presents with microcephaly, cataracts, chorioretinitis, and limb abnormalities. Educational Objective: Rubella infection in a pregnant patient increases the risk for congenital rubella infection, which characteristically presents with cataracts, sensorineural deafness, and congenital heart disease (eg, patent ductus arteriosus) in the newborn, as well as microcephaly, neurodevelopmental delay, and extramedullary hematopoiesis. Previous Next Score Report Lab Values Calculator Help Pause