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2025 NBME 27 ACTUAL EXAM WITH COMPLETE QUESTIONS AND CORRECT ANSWERS RATED A+, Exams of Medicine

NBME 27 NEWEST COMPREHENSIVE STUDY GUIDE WITH COMPLETE QUESTIONS AND CORRECT ANSWERS RATED A +

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2024/2025

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2025 NBME 27 ACTUAL EXAM WITH

COMPLETE QUESTIONS AND CORRECT

ANSWERS RATED A+

Exam Section 1: Item 1 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment

  1. A 40-year-old woman at 5 months' gestation comes to the physician for amniocentesis. Results show a normal 46,XY karyotype of the fetus. Four months later, the newborn is delivered and examination shows a female phenotype with mild clitoral enlargement. Ultrasonography shows the presence of male but not female genital ducts. This newborn most likely has a mutation of the gene of which of the following factors? A) Anti-paramesonephric (müllerian) hormone B) 5a-Reductase C) SRY protein D) Steroid sulfatase E) Zinc finger transcription factor Wilms tumor 1 Correct Answer B. 5a-Reductase is an enzyme that catalyzes the metabolism of testosterone to dihydrotestosterone (DHT). Testosterone promotes the development of the mesonephric (wolffian) duct that develops

into the seminal vesicles, epididymis, vas deferens, and ejaculatory duct. DHT promotes the development of male external genitalia and the prostate from the genital tubercle and urogenital sinus. Individuals with 5a-reductase deficiency have defective conversion of testosterone to DHT, leading to decreased concentrations of DHT and impaired virilization of the male urogenital tract. Individuals with 5a-reductase deficiency therefore appear phenotypically female externally or may have ambiguous external genitalia. Individuals have normal male internal genitalia as a result of normal concentrations of testosterone. InCorrect Answers: A, C, D, and E. Anti-paramesonephric (müllerian) hormone (Choice A) or müllerian inhibiting factor (MIF) is secreted by Sertoli cells and suppresses the development of the paramesonephric (müllerian) duct that would have developed into female internal genital structures such as the fallopian tubes, uterus, and upper vagina. The SRY gene is located on the Y chromosome and is responsible for producing testis-determining factor, also known as the SRY protein (Choice C), for testes development. Following testes development, hormones secreted by Sertoli cells (MIF) and Leydig cells (testosterone and DHT) promote the development of male internal and external genitalia and suppress the development

of female structures. SRY gene translocation can occur during recombination in which the SRY gene on the Y chromosome becomes part of the X chromosome, leading to an XX embryo developing male characteristics. Steroid sulfatase (Choice D) is an enzyme that catalyzes the conversion of sulfated steroid precursors to the 2 Exam Section 1: Item 2 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment

  1. A 16-year-old girl with cystic fibrosis is brought to the physician because of a 3-week history of generalized weakness, numbness and tingling of her arms and legs, and difficulty walking. She has not adhered to her medication regimen during the past 6 months. She appears alert and oriented. Her vital signs are within normal limits. Physical examination shows bilateral weakness and decreased deep tendon reflexes in the upper and lower extremities. She walks with an ataxic gait. The most likely cause of these findings is a deficiency of which of the following? A) Folic acid B) Vitamin A C) Vitamin B, (pyridoxine) D) Vitamin D E) Vitamin E Correct Answer E.

Vitamin E is an antioxidant that protects erythrocytes and cells from free radical damage. Deficiency may present with hemolytic anemia and generalized muscle weakness. It can have a similar presentation to vitamin B12 (cobalamin) deficiency with posterior column and spinocerebellar tract demyelination. In contrast to vitamin B12 deficiency, patients do not have megaloblastic anemia, hypersegmented neutrophils, or increased serum methylmalonic acid concentrations. On peripheral smear, patients with vitamin E deficiency may have acanthocytosis. InCorrect Answers: A, B, C, and D. Folic acid (Choice A) is converted to tetrahydrofolic acid and used as a coenzyme in the synthesis of nucleotides and nucleosides. Folate is contained in leafy vegetables and absorbed in the jejunum. Folate deficiency is often seen in patients with malnutrition, alcoholism, and patients taking anti-folate medications (eg, phenytoin, methotrexate). Megaloblastic anemia occurs in the setting of impaired DNA synthesis. Vitamin A (Choice B) is an antioxidant necessary for differentiation of epithelial cells into specialized tissue. Deficiency is characterized by ocular manifestations including night blindness, corneal degeneration, Bitot spots on the conjunctiva, dry skin, and immunosuppression. Vitamin B6 (pyridoxine) (Choice C) deficiency limits synthesis of histamine, hemoglobin, and neurotransmitters including epinephrine, norepinephrine, dopamine, serotonin, and GABA. Deficiency

commonly presents with peripheral neuropathy, dermatitis, sideroblastic anemia, glossitis, and seizures (especially in the setting of isoniazid use). Vitamin D (Choice D) deficiency commonly comes from low exposure to UV radiation and low dietary vitamin D intake. Fat malabsorption syndromes such as celiac disease or cystic fibrosis can impair absorption of vitamin D 3 Exam Section 1: Item 3 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment

  1. A 55-year-old woman comes to the physician because of a 6- month history of progressive shortness of breath while climbing stairs and constant fatigue. She is 163 cm (5 ft 4 in) tall and weighs 107 kg (235 lb); BMI is 40 kg/m2. Her blood pressure is 170/110 mm Hg on three separate readings. Physical examination shows no other abnormalities. Echocardiography shows an ejection fraction of 45%. There are no regional wall motion or valvular abnormalities. An ECG shows that the sum of the magnitudes of the S wave in lead V, and the R wave in lead Vs is 45 mm (scale=10 mm/mV). Which of the following cellular changes is most likely in this patient's left ventricle compared with normal? A) Decreased sarcomeres and increased fibroblasts B) Increased sarcomeres in an array in parallel with each other

C) Increased s Correct Answer B. Left ventricular hypertrophy (LVH) is a common disorder with numerous causes, including chronic systemic hypertension, aortic stenosis, coarctation of the aorta, chronic aortic or mitral insufficiency, ventricular septal defect, infiltrative cardiomyopathy (eg, amyloidosis, sarcoidosis, Fabry disease, hereditary hemochromatosis), or hypertrophic cardiomyopathy. Hypertension and aortic stenosis are the most common causes. In these conditions, hypertrophy develops secondary to chronic contraction against increased afterload. This stimulates growth and remodeling of cardiomyocytes, resulting in increased sarcomeres in an array in parallel with each other in order to generate greater contractile force to overcome the afterload. The condition can progress to left-sided heart failure if untreated. Diagnostic testing includes ECG; one of the criteria suggestive of LVH is if the sum of the magnitude of the S wave in lead V2 and the R wave in lead V5 is greater than 35 mm as in this patient. Echocardiography is more sensitive than ECG for the presence of LVH. Cardiac MRI is the gold standard for diagnosis, though availability, cost, and logistics limit its use as a first-line screening tool. InCorrect Answers: A, C, D, and E. Decreased sarcomeres and increased fibroblasts (Choice A) occurs in myocardial scar formation, which may result after a myocardial infarction or acute inflammatory injury. It is also seen in Fabry disease as a result of dysfunctional metabolism of sphingolipids.

Increased sarcomeres in an array in series with each other (Choice C) occurs in dilated cardiomyopathy as the cardiac myofibrils lengthen. Dilated cardiomyopathy may be caused by an underlying genetic predisposition, toxin exposure (eg, chronic alcohol use, chronic cocaine use, doxorubicin chemotherapy), infections (eg, Chagas disease, viral myo 4 Exam Section 1: Item 4 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment

  1. A 55-year-old man with long-standing type 2 diabetes mellitus comes to the physician because of a 2-day history of fever, chills, nausea, and swelling of his left leg. His temperature is 38.9°C (102°F), and pulse is 110/min. Examination of the left lower extremity shows a warm, tender, erythematous, blanching rash. A photograph of the left lower extremity is shown. A photomicrograph of a Gram stain of a blood culture is shown. Which of the following is the most likely causal organism? A) Clostridium perfringens B) Enterococcus faecalis C) Haemophilus influenzae D) Pasteurella multocida E) Pseudomonas aeruginosa F) Staphylococcus aureus G) Streptococcus pyogenes (group A) Correct Answer G.

Cellulitis presents with cutaneous erythema, warmth, and induration, often following inoculation from an injury such as an abrasion. Systemic symptoms, including fever, tachycardia, and leukocytosis are often present. Most commonly, it is acute and results from a bacterial infection of the skin. The two most common pathogens to cause cellulitis are Staphylococcus aureus and Streptococcus pyogenes (group A). S. pyogenes is a gram- positive bacterium that forms chains. It is part of the B-hemolytic group of Streptococcus species and is bacitracin sensitive. Clinically, the cellulitis caused by S. aureus is typically purulent while that caused by S. pyogenes is not. In addition to cellulitis, infection with S. pyogenes can result in various diseases such as pharyngitis, scarlet fever, necrotizing fasciitis, glomerulonephritis, and rheumatic fever. InCorrect Answers: A, B, C, D, E, and F. Clostridium perfringens (Choice A) is an anaerobic gram-positive bacillus. It often infects an open wound and causes necrotizing fasciitis. Clostridium perfringens secretes alpha toxin, a phospholipase that degrades cell membranes to cause gas gangrene. Enterococcus faecalis (Choice B) belongs to Group D Streptococcus, which are gamma hemolytic. Gamma-hemolytic species are identified by their ability to grow in bile and 6.5% sodium chloride solution. Clinically, Enterococcus faecalis is a common cause of nosocomial urinary tract infections.

Haemophilus influenzae (Choice C) is an encapsulated gram- negative bacterium that can result in various mucosal infections such as conjunctivitis, otitis media, bronchitis, and pneumonia. It was the leading cause of epiglottitis in children prior to widespread immunization. It is an uncommon cause of cutaneous infections. Pasteurella multocida (Choice D) is a gram-negative coccobacillus, which is fre 5 Exam Section 1: Item 5 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment

  1. A 26-year-old woman is brought to the emergency department by her husband after she ingested an overdose of narcotics. She is arousable only with deep painful stimuli. Her pulse is 110/min, respirations are 10/min, and blood pressure is 110/ mm Hg. Cerebral blood flow is increased. Which of the following is the most likely cause of the increase in cerebral blood flow? A) Decreased blood pressure B) Decreased hydrogen ion concentration in the blood C) Decreased oxygen concentration in the blood D) Increased carbon dioxide concentration in the blood E) Increased heart rate Correct Answer D.

The intracranial compartment is fixed, surrounded by a rigid skull containing brain parenchyma, cerebrospinal fluid, and intravascular blood. Serum carbon dioxide concentration directly increases or decreases cerebral blood flow (CBF). Hypocapnia induces cerebral vasoconstriction, whereas hypercapnia induces cerebral vasodilation, increasing the amount of CBF and the volume of intracranial blood. In a patient with decreased respirations secondary to narcotic-induced respiratory depression, decreased air exchange and ventilation will lead to an increased amount of carbon dioxide in the blood. The resulting hypercapnia leads to vasodilation and increased CBF. InCorrect Answers: A, B, C, and E. Decreased blood pressure (Choice A) would not increase CBF. Mean arterial pressure (MAP) can have an effect on CBF, but as a result of cerebral autoregulation, CBF is maintained at a constant state over a wide range of MAP. Small changes or decreases in blood pressure have no effect on CBF; only blood pressure below or above the autoregulatory range will affect CBF. Decreased hydrogen ion concentration in the blood (Choice B) is the opposite of what would occur with opioid-induced respiratory depression, which leads to hypercapnia and increased hydrogen ion concentration. Decreased oxygen concentration in the blood (Choice C) results from hypoventilation, but CBF is more directly affected by concentrations of carbon dioxide. Increased heart rate (Choice E), just like blood pressure and MAP, has little effect on CBF provided that MAP remains within

the autoregulatory range. Extreme bradycardia or tachycardia leading to severe hypotension could decrease CBF. Educational Objective: As a result of cerebral autoregulation, changes in MAP have minimal impact on CBF. Serum carbon dioxide concentration elicits a reversible effect on CBF 6 Exam Section 1: Item 6 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment

  1. The incidence of Haemophilus influenzae type b meningitis in children below age 2 years has decreased since the introduction of the current vaccination for this infection. This vaccine is composed of which of the following? A) Anti-idiotype B) Attenuated bacterial strain OC) Conjugated capsular polysaccharide D) Killed whole cell E) Outer membrane complex F) Peptide G) Recombinant cell protein H) Toxoid Correct Answer C. The Haemophilus influenzae type b vaccination uses conjugated capsular polysaccharide to provoke an adaptive immune

response. Vaccination is a form of active immunity designed to prevent disease by stimulating the immune system to develop and maintain the ability to respond to a foreign antigen. The maintenance of immunity is referred to as immune memory. Vaccinations may contain the whole infectious organism in a killed or a live- attenuated state that is unable to produce virulent disease. They may also be fractional, containing a specific component of a pathogen that the immune system can recognize. Fractional vaccines may be developed from a particular protein or polysaccharide associated with the organism. Bacterial vaccinations commonly employ a capsular or cell-wall polysaccharide, or a denatured toxin that is produced by the bacterium (referred to as a toxoid). Examples of capsular polysaccharide-based vaccines include vaccinations against Haemophilus influenzae type b, Neisseria meningitidis, and Streptococcus pneumoniae. Examples of toxoid-based vaccines include those against Corynebacterium diphtheriae, Clostridium tetani, and Bordetella pertussis. The presence of the polysaccharide alone is usually insufficient to provoke a lasting immune response as the polysaccharide is not usually well- presented to T cells on major histocompatibility complexes (MHC). Conjugation of the pathogen-specific polysaccharide with another antigen that is known to be highly immunogenic (presented on MHC) leads to a robust B- and T-cell response, with lasting immune memory. Toxoids are commonly used as conjugates.

InCorrect Answers: A, B, D, E, F, G, and H. Anti-idiotype (Choice A) antibodies are those that bind to the variable region of another antibody. Attenuated bacterial strain (Choice B) is a form of vaccine in which the 7 Exam Section 1: Item 7 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment

  1. A 24-year-old man with AIDS comes to the physician because of a 1-month history of abdominal cramps and watery diarrhea. These symptoms have become increasingly severe during the past 5 days. His pulse is 95/min. Physical examination shows dry mucous membranes and decreased skin turgor. A biopsy specimen of the colon is shown in the photomicrograph. Which of the following is the most likely causal organism? A) Candida albicans B) Cryptococcus neoformans C) Cryptosporidium parvum OD) Cytomegalovirus E) Giardia lamblia OF) Mycobacterium avium complex Correct Answer C.

The photomicrograph demonstrates small, round oocysts on the colonic epithelial surface, consistent with cryptosporidiosis, an AIDS-defining illness. Cryptosporidium is a genus of protozoal pathogens that cause cryptosporidiosis, a diarrheal illness that is mild and self-limited in immunocompetent hosts but which is severe and often life-threatening in immunocompromised hosts. Signs and symptoms include fever, weight loss, symptoms of dehydration and orthostasis, severe watery diarrhea, cramping abdominal pain, nausea, and vomiting. Immunocompromised patients may develop disseminated infection with involvement of the liver and lungs. Diagnosis is made with stool antigen testing or with detection of luminal oocysts by acid-fast staining. Treatment is supportive, with repletion of fluids and electrolytes, as well as antiparasitic therapy. Nitazoxanide is the treatment of choice in immunocompetent patients, whereas immunocompromised patients require antiretroviral therapy to improve their immune status and CD4+ cell counts, as well as combination regimens that involve nitazoxanide and other agents, such as azithromycin or paromomycin. InCorrect Answers: A, B, D, E, and F. Candida albicans (Choice A) can produce oral thrush and esophagitis in immunocompromised patients. Candida esophagitis is an AlIDS-defining illness. Staining with Gomori methenamine silver or periodic acid-Schiff shows pseudohyphae and budding yeasts.

Cryptococcus neoformans (Choice B) is a potential cause of meningitis in AIDS patients. Staining with India ink or mucicarmine shows an encapsulated yeast with narrow-based budding. Cytomegalovirus (CMV) (Choice D) is a herpesvirus (human herpesvirus-5) that causes esophagitis, pneumonia, and retinitis in AIDS patients. Light microscopy may show intracellular inclusion bodies known as Cowdry bodies. Diag 8 Exam Section 1: Item 8 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment

  1. A 60-year-old woman comes to the physician because of vaginal bleeding for 6 weeks. Menarche occurred at the age of 12 years, and menses had occurred at regular 30-day intervals. She had three term pregnancies by the age of 25 years. She has been receiving conjugated equine estrogen therapy since menopause 10 years ago. She has smoked 1 pack of cigarettes daily for 40 years. She drinks alcoholic beverages five times weekly. Physical examination and laboratory studies show no abnormalities. An endometrial biopsy specimen shows endometrial adenocarcinoma. Which of the following is the strongest predisposing risk factor for the biopsy findings? A) Alcohol use B) Cigarette smoking

C) Early menarche D) Estrogen therapy E) Multiparity Correct Answer D. Vaginal bleeding in a postmenopausal woman should raise suspicion for malignancy, including uterine or cervical carcinoma. Estrogen excess without adequate opposition by progestin is a strong risk factor for atypical endometrial hyperplasia and endometrial carcinoma, as this malignancy is estrogen-sensitive. This is one reason that estrogen therapy without progestin is no longer generally recommended for long-term treatment of menopausal symptoms or prevention of osteoporosis. Additionally, estrogen therapy also increases the risk for thromboembolic events in postmenopausal women. Increased exposure to estrogen can come from an exogenous supplement as in this case, or an endogenous source, such as increased peripheral production of estrogen by adipose tissue or from an increased number of lifetime menstrual cycles. Thus, risk factors for endometrial carcinoma include estrogen supplementation, obesity, early menarche or late menopause, nulliparity, advanced age, and family history. Endometrial biopsy is used to confirm the diagnosis. With biopsy and imaging, the stage of endometrial carcinoma is determined and follows TNM (tumor, nodes, metastases) staging. If the disease is confined to the uterus, surgical excision is the first-line therapy. This is generally achieved through hysterectomy with bilateral salpingo-oophorectomy. InCorrect Answers: A, B, C, and E.

While alcohol use (Choice A) and cigarette smoking (Choice B) are risk factors for cancers such as hepatocellular carcinoma and squamous cell carcinoma, respectively, they have not been shown to increase the risk for endometrial carcinoma. Early menarche (Choice C) is a risk factor for endometrial carcinoma as it increases the total number of ovulatory cycles and thus endometrial estrogen exposure over a lifetime. However, this patient's use of estrogen supple 9 Exam Section 1: Item 9 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment

  1. A 3-year-old girl is brought to the emergency department by her mother 1 hour after she was found with a half-empty bottle of her grandmother's diabetes medication. The mother tells the physician that the child consumed approximately 25 metformin tablets. Physical examination shows no abnormalities. This patient is at greatest risk for which of the following serum abnormalities? A) Decreased calcium concentration B) Decreased glucose concentration C) Decreased sodium concentration D) Increased AST and ALT activities E) Increased creatinine concentration

F) Increased lactic acid concentration Correct Answer F. Metformin is an oral biguanide agent used in the management of type 2 diabetes mellitus. It decreases gluconeogenesis, increases peripheral tissue glucose uptake, and decreases serum free fatty acid concentration. Because of its efficacy, tolerability, safety, and low cost, it serves as the first-line treatment for type 2 diabetes mellitus. Although it is generally safe and well-tolerated, there are a few contraindications to metformin, including impaired renal function (caused by its excretion in the urine) and active liver disease. Metformin may cause several adverse effects, including nausea, abdominal discomfort, diarrhea, and (rarely) increased lactic acid concentration resulting in lactic acidosis. Although the incidence is rare, metformin-associated lactic acidosis has a high mortality rate. Treatment is supportive, and patients with severe lactic acidosis may require hemodialysis. Metformin does not generally cause hypoglycemia. InCorrect Answers: A, B, C, D, and E. Decreased calcium concentration (Choice A) is not a common side effect of metformin. Since metformin does not stimulate insulin release, hypoglycemia and decreased glucose concentration (Choice B) are not common. Hypoglycemia can be an adverse effect in the setting of an insulin or sulfonylurea overdose. Decreased sodium concentration (Choice C) can occur in syndrome of inappropriate antidiuretic hormone secretion (SIADH) resulting in excessive free water retention and euvolemic

hyponatremia. SIADH is a common side effect of many medications (eg, cyclophosphamide) but not metformin. Increased AST and ALT activities (Choice D), seen in medication- induced hepatotoxicity, can be caused by oral hypoglycemic drugs such as glitazones (eg, pioglitazone, rosiglitazone). They increase peripheral tissue insulin sensitivity and can be hepatotoxic. Increased 10 Exam Section 1: Item 10 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment

  1. A 28-year-old man comes to the physician for a routine health maintenance examination. He is from Jamaica. He has practiced Rastafarianism all his life and follows a diet consisting only of natural juices, fruits, and grains. He also regularly smokes marijuana as part of his religion. Physical examination shows no abnormalities. His hematocrit is 32%, and serum vitamin B (cobalamin) concentration is within the reference range. Which of the following best describes the erythrocytes in this patient? Color Size (in Diameter) Shape A) Normal

small normal B) Normal small 10% sickle, 90% normal C) Normal large normal D) Normal large 10% sickle, 90% normal E) Pale small normal F) Pale small 10% sickle, 90% normal G)

Pale large normal H) Pale large 10% sickle, 90% normal Correct Answer E. Choice E corresponds with a diagnosis of iron deficiency anemia (IDA). Erythrocytes in the setting of IDA are pale (hypochromic) and small in size (microcytic) but are normal in shape. Iron is required for the synthesis of heme, which is a necessary component of the hemoglobin molecule and, thus, of erythrocytes. It functions to shuttle oxygen to and from peripheral tissues. In individuals who do not have an adequate intake of dietary iron in the form of heme obtained from animal meat, deficiency can develop. IDA may also develop as the result of chronic blood loss from menstruation, colorectal bleeding, or can occur as a result of malabsorption syndromes such as celiac disease and in patients who have undergone gastric bypass surgery. Erythrocytes on the peripheral blood smear are hypochromic as a result of deficient hemoglobin concentration. It is thought that erythrocytes are microcytic as a result of continuing erythrocyte division in order to reach an adequate hemoglobin concentration; because hemoglobin stores are inadequate, cell division continues beyond what would normally occur and causes the cells to be smaller than

normal. Treatment in the case of poor dietary intake is with oral iron supplementation. If severe, intravenous iron can be given. InCorrect Answers: A, B, C, D, F, G, and H. Normal color, small size, and normal shape (Choice A) describes several other causes of microcytosis without hypochromia, including thalassemia and anemia of chronic disease. While thalassemia may present with target cells on peripheral smear, the majority of erythrocytes will appear as microcytic. The degree of hypochromia is dependent on the severity of decreased hemoglobin concentration and mean corpuscular volume. Normal color and small size with 10% sickle and 90% normal shape (Choice B) can be seen in patients with s 11 Exam Section 1: Item 11 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment

  1. A 22-year-old man is brought to the emergency department 30 minutes after collapsing at home. He has a 3-year history of intravenous drug use. He appears confused. His temperature is 39.5°C (103.1°F). Physical examination shows subungual hemorrhages of the fingers and jugular venous distention. A holosystolic murmur is heard at the lower left sternal border. Blood cultures are positive for Staphylococcus aureus. A chest x-ray shows multifocal airspace opacities. The most likely cause of these findings is infection of which of the following labeled

cardiac valves in the CT scans of the chest?

  • B C A A) B) C) D) Correct Answer C. Intravenous drug use is a risk factor for infective endocarditis and valvular dysfunction as a result of the nonsterile injection of material into the venous system. Though left-sided infective endocarditis is more common in general, the tricuspid valve is most commonly involved in the setting of intravenous drug use. Tricuspid valve endocarditis is typically associated with Staphylococcus aureus, Pseudomonas aeruginosa, and Candida organisms. The formation of vegetations on the valve and local inflammatory damage can lead to severe tricuspid regurgitation, which presents as a holosystolic murmur best heard in the left lower sternal border. Severe tricuspid regurgitation is a risk factor for the development of right-sided heart failure, which is characterized by increased jugular venous pressure, hepatomegaly, ascites, and peripheral extremity edema. Other complications include septic pulmonary emboli, which present as multifocal peripheral cavitary lung lesions. The tricuspid valve can be identified on CT imaging by its location between the right ventricle

and right atrium. The right ventricle is the most anterior and inferior part of the heart, as seen in the axial slices of the patient's CT scan. InCorrect Answers: A, B, and D. Choice A represents the aortic valve, which separates the left ventricle from the ascending thoracic aorta. It is located superior to the left ventricle, anterior to the left atrium, and posterior to the pulmonic valve. It may be, along with the mitral valve, more commonly affected in cases of endocarditis that are unrelated to intravenous drug abuse. Systemic septic emboli are common. Choice B represents the pulmonic valve, which separates the right ventricle from the pulmonary trunk. It is anterior to the aorta. It is the least common valve affected by endocarditis. Choice D represents t 12 Exam Section 1: Item 12 of 50 National Board of Medical Examiners' Comprehensive Basic Science Self-Assessment

  1. A73-year-old woman has a 3-month history of increasing fatigue, weakness, and a 9.0-kg (20-lb) weight loss. Examination shows hepatosplenomegaly. An M-protein spike is observed on serum electrophoresis. Which of the following is the most likely diagnosis? A) Acute myelogenous leukemia

B) B-lymphocyte neoplasia C) Chronic myelogenous leukemia D) Leukemoid reaction OE) T-lymphocyte neoplasia Correct Answer B. B-lymphocyte neoplasia likely accounts for this patient's presentation with an M-protein spike (M-spike). While an M-spike may occur in a variety of B cell malignancies including Waldenstrom macroglobulinemia, plasma cell leukemia, smoldering myeloma, amyloidosis, or plasmacytoma, it is most commonly recognized in the setting of multiple myeloma (MM). Other signs and symptoms of MM include bone pain, anemia, hypercalcemia, renal dysfunction, and lytic bone lesions. Hepatosplenomegaly is an occasional finding. B cells are precursors to plasma cells, which secrete specific immunoglobulins of different classes. MM is a malignancy caused by the neoplastic proliferation of a single plasma cell clone, which overproduces monoclonal immunoglobulin and light or heavy chains. These clonal immunoglobulins are secreted in high numbers and appear as a monoclonal spike on protein electrophoresis in the gamma region. Diagnosis of multiple myeloma is suspected based on the presence of an M-protein spike in the presence of concerning symptoms, but is confirmed by bone marrow biopsy, which must demonstrate at least 10% clonal plasma cells. As a result of the hypersecretion of a single immunoglobulin, these patients often have a relative