NBME FORM 30 EXAM QUESTIONS WITH VERIFIED ANSWERS || A VERIFIED A+ PASS.

NBME FORM 30 EXAM QUESTIONS WITH VERIFIED ANSWERS || A VERIFIED A+ PASS. Exam Section 1: Item 1 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment "1 Exam Section 1: Item 1 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 41. A 14-year-old boy is brought to the physician by his parents because of increasingly severe left knee pain during the past week. The pain is exacerbated by running or kneeling. He plays multiple sports and is currently in the middle of soccer season. There has been no recent trauma to the area or related sports injuries. Physical examination shows full range of motion of both lower extremities. There is a discrete area of swelling just below the left patella over the proximal portion of the tibia. Palpation of the area produces pain. Pain is also reproduced when he does a full squat. Examination of the right knee shows no abnormalities. Which of the following is the most likely cause of this patient's condition? A) Chondromalacia B) Ligamental tear C) Osgood-Schlatter disease D) Osteochondritis E) St - CORRECT ANSWER C. Osgood-Schlatter disease refers to osteochondrosis or traction apophysitis of the tibial tubercle that typically occurs in adolescent, athletic children. Repetitive tension via the patellar tendon transmits to the tibial tubercle, presenting as pain reported during activities that increase the stress on the tubercle, such as kneeling, squatting, kicking, or similar activities that increase the extensor force transmitted by the quadriceps. Patients localize pain to the anterior aspect of the proximal tibia and knee. Physical examination typically discloses tenderness over an enlarged tibial tubercle. The condition is diagnosed clinically; x-rays, if obtained, may show increased lucency in the area of the tibial tubercle. Treatment is through rest, cryotherapy, and non-steroidal anti-inflammatory medications, as the condition resolves with time and unloading. Incorrect Answers: A, B, D, and E. Chondromalacia (Choice A) as related to the knee refers to deterioration of cartilage along the posterior aspect of the patella. It is common in young athletes, especially runners, and presents with knee pain that is worse with bending. It can be differentiated from Osgood-Schlatter disease by the location of pain, whereas chondromalacia will not present with pain at the tibial tuberosity. Ligamental tear (Choice B) is a broad term that in the context of the knee could describe damage to any of the collateral ligaments, such as the anterior or posterior cruciate or the medial or lateral collateral ligaments. Tear or rupture of these ligaments presents with pain which is worse when placing the affected ligament under strain, tenderness in the area of the injury, and laxity when evoking the motion restricted by the ligament itself. Osteochondritis (Choice D) describes the inflammation of bone or cartilage within a joint; it may al" "2 Exam Section 1: Item 2 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 42. A 40-year-old man is evaluated because his skin is extremely sensitive to sunlight. Exposure to the sun causes the formation of vesicles and blisters on the skin, which frequently take weeks to heal. He is diagnosed with a disorder caused by the increased synthesis of compounds in the skin that are subject to excitation by visible light. Which of the following biochemical pathways is most likely defective in this patient? A) Bile acid synthesis B) Bilirubin degradation C) Heme synthesis D) Melanin synthesis E) Riboflavin metabolism - CORRECT ANSWER C. Porphyria cutanea tarda is characterized by severe cutaneous photosensitivity with blistering and hyperpigmentation after exposure to sunlight and is the most common of the porphyrias. It is caused by decreased activity of uroporphyrinogen decarboxylase, an enzyme used to in the production of heme. The initial substrates for heme are glycine and succinyl-CoA. Their conversion to heme begins in the mitochondria with a rate limiting step catalyzed by aminolevulinate synthase. A series of additional steps then occurs which take place in the cellular cytoplasm. In one of these intermediate steps, uroporphyrinogen decarboxylase catalyzes the conversion of uroporphyrinogen III to coproporphyrinogen III. Not only will a deficiency in uroporphyrinogen decarboxylase prevent correct heme synthesis, it will also cause uroporphyrinogen III to accumulate. Accumulated uroporphyrinogen IIl is then deposited in the skin. Upon exposure to light of wavelength 400nm, the molecule enters an excited state and releases photons which in turn create reactive oxygen species within the skin. These reactive oxygen species damage the basement membrane, lipids, and proteins nearby resulting in dermoepidermal separation and blister formation. Incorrect Answers: A, B, D, and E. Bile acid synthesis (Choice A) uses cholesterol as the initial substrate. Cholesterol 7a-hydroxylase is the rate-limiting step in the creation of bile acids. Porphyrins are not intermediate byproducts of this pathway. Impaired bile acid production will lead to decreased intestinal absorption of lipids and fat-soluble vitamins, but not photosensitivity and blister formation. Bilirubin forms as a result of the degradation of heme. Bilirubin degradation (Choice B) is the process by which bilirubin is first conjugated in the liver to become water soluble and then excreted into" "3 Exam Section 1: Item 3 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 43. A 70-year-old man with metastatic prostate cancer has been taking leuprolide for the past 3 months. Which of the following best describes this patient's current serum luteinizing hormone (LH) and testosterone concentrations compared with concentrations before treatment? LH Testosterone A) ↑ ↑ B) ↑ C) No change ↑ D) no change E) - CORRECT ANSWER E. Leuprolide is a gonadotropin-releasing hormone (GNRH) analog. If given in a pulsatile fashion, mimicking the physiologic secretion of GNRH, it will act as a GNRH receptor agonist and increase follicle-stimulating hormone (FSH) and luteinizing hormone (LH). When leuprolide is initially started in a patient, the hypothalamus reacts as if it were an agonist and there is a transient rise in LH and FSH for the first week of treatment. However, with continued use in a non-pulsatile fashion, as in this case, it will act as a GNRH receptor antagonist and subsequently decrease FSH and LH. By decreasing the production of LH and the stimulation of Leydig cells, leuprolide indirectly lowers testosterone. Prostate cancer is a hormonally sensitive cancer; androgens play a critical role in its growth. Thus, by inhibiting testosterone production, this driver of cancer growth is removed, and sensitive tumors begin to shrink. This approach to treating prostate cancer is termed medical castration. Some tumors may become castration-resistant, in which mutations develop that allow the cancer to continue growing without hormonal stimulation. Incorrect Answers: A, B, C, and D. While there is a transient increase in LH and testosterone (Choice A) for the first week of treatment, after three months of treatment leuprolide will be exerting GNRH antagonist effects. This will lead to decreased LH and testosterone concentrations. Because LH stimulates the Leydig cells to produce testosterone, when LH concentration is decreased, testosterone concentrations will also decrease (Choice B). A medication which continuously increases testosterone would not be appropriate for use in prostate cancer as it is an androgen-dependent malignancy and would continue to grow under androgen stimulation. LH decreases rather than showing no change (Choice C) when" "4 Exam Section 1: Item 4 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 44. A 34-year-old woman comes to the physician because of a 10-year history of constipation. She usually has fewer than three bowel movements weekly. She often has bloating and cramping, and she has to strain and apply perineal pressure to defecate. She says that she does not have problems with sleep, appetite, or energy. Palpation of the abdomen produces diffuse discomfort; bowel sounds are present. Which of the following is the most likely cause of this patient's gastrointestinal symptoms? A) Colon polyps B) Congenital megacolon (Hirschsprung disease) C) Factitious disorder D) Gluten enteropathy E) Irritable bowel syndrome F) Major depressive disorder - CORRECT ANSWER E. Irritable bowel syndrome (IBS) is characterized by recurrent, intermittent abdominal pain and alteration of bowel habits as either a change in stool frequency or consistency. The abdominal pain may be either exacerbated or relieved by defecation and can vary widely in location and character. IBS commonly presents with intermittent episodes of diarrhea and/or constipation, often which alternate. When diarrhea does occur, it usually does so during waking hours which serves to differentiate it from other causes of diarrhea. It is most common in middle-aged women and the pathogenesis is unknown. It is not caused by a structural abnormality, and physical examination will show nonspecific abdominal discomfort rather than localized pain. Stress management may improve bowel habits in patients with irritable bowel syndrome, which is often associated with concomitant anxiety. If medication is needed, dicyclomine is an antispasmodic and diphenoxylate slows gut motility. Both are used in the management of irritable bowel syndrome. Incorrect Answers: A, B, C, D, and F. Colon polyps (Choice A) present in a variety of subtypes, from non-neoplastic polyps (eg, hamartomatous, mucosal, inflammatory, hyperplastic) to potentially malignant polyps (adenomatous, serrated). Polyps are typically asymptomatic and are only recognized after screening colonoscopy. They are not a common cause of chronic constipation. Inflammatory polyps may be seen in inflammatory bowel disease, but not irritable bowel syndrome. Congenital megacolon (Hirschsprung disease) (Choice B), or intestinal aganglionosis, is caused by the congenital absence of the distal portion of the myenteric plexus, a part of the enteric nervous system located between the inner and outer layers of the muscularis externa. This often leads to a failure to pass stool within the first fe" "5 Exam Section 1: Item 5 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 45. Human gene fragments in appropriate vectors may be introduced into bacterial cells by transformation. Which of the following is most likely to be used to transfer cloned human genetic material to bacteria? A) DNA complexed with bacterial histones O B) Human chromosomal fragments C) Naked DNA D) Purified euchromatin E) Purified heterochromatin - CORRECT ANSWER C. Transformation describes the ability of bacteria to uptake and incorporate exogenous genetic material (naked DNA) from the environment. Following lysis of a cell DNA may exist extracellularly; a bacterium may uptake such material through the cell membrane. Transformation is one of three methods of horizontal gene transfer by which bacteria can acquire novel genes which may confer a survival benefit. Notably, Streptococcus pneumoniae and Haemophilus influenzae type B have demonstrated acquisition of genetic material by transformation. In order for transformation to occur, a bacterium must demonstrate competence (the ability to uptake such material). Competence occurs in states of stress such as starvation. Beyond transformation, horizontal gene transfer may also occur via conjugation (in which two cells in direct contact, typically involving a sex pilus), exchange material, and transduction (transfer of genetic material via a viral vector such as a bacteriophage). Incorrect Answers: A, B, D, and E. DNA complexed with bacterial histones (Choice A), human chromosomal fragments (Choice B), purified euchromatin (Choice D), and purified heterochromatin (Choice E) all describe fragments of DNA associated with additional protein sizes. The process of transformation requires naked DNA, which describes DNA that is not associated with proteins, lipids, or molecules that may shield it. exes and of variable Educational Objective: Transformation describes the ability of bacteria to uptake and incorporate genetic material (naked DNA) that is not associated with proteins or lipids from the environment. This mechanism may promote genetic diversity and acquisition of survival advantage. %3D Previous Next Score Report Lab Values Calculator Help Pause" "6 Exam Section 1: Item 6 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 46. A 4-year-old girl has acute lymphoblastic leukemia that has not responded to aggressive treatment. The physician recommends palliative care to the parents and suggests that they talk to their daughter about her impending death and answer any questions that she may have. The parents ask the physician to help them respond to her questions. When speaking with the child, the parents should anticipate that she will most likely understand death as which of the following? A) She will blame God for her illness and death B) She will comprehend that all living things die C) She will have no understanding of death D) She will view death as temporary and reversible - CORRECT ANSWER D. Most children between the ages of 3 and 5 years view death as temporary and reversible. Partly related to a lack of experience of death, they also do not understand that all living things inevitably die. Children of this age range who are dying may believe dying is a punishment from their parents, signifying an inability to abstract. However, these gaps in understanding can be filled by adults explaining death to children and by children's experiences of death. It is suggested that the parents of dying children help their child understand the situation. Incorrect Answers: A, B, and C. This patient is unlikely to blame God for her illness and death (Choice A) as it is too abstract a concept for a child this age. The patient is more likely to blame her parents. She may, depending on her parents' religious beliefs, believe that death represents going to Heaven without an abstract understanding of Heaven. Starting at age five, children begin to understand that death is inevitable and irreversible (Choice B). At this age, children may continue to demonstrate an incomplete understanding of death. Media images of death (eg, ghosts) may be part of this incomplete understanding. At age ten, children typically understand that death is universal, irreversible, and renders people inanimate (versus believing in ghosts). Only infants have no understanding of death (Choice C). Toddlers typically begin to become aware of the concept of death over time. Educational Objectives: Most children between the ages of 3 and 5 years view death as temporary and reversible. Further, preschool age children are frequently unaware of the inevitability of death. %3D Previous Next Score Report Lab Values Calculator Help Pause" "7 Exam Section 1: Item 7 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 47. A 12-year-old girl is brought to the emergency department 15 minutes after she accidentally sliced her left palm with a knife. Physical examination shows a 2-cm laceration over the left palm. The wound is cleansed and sutured. One week later, the sutures are removed. At this time, which of the following factors is most instrumental in the migration of fibroblasts into the area of wound healing? A) Collagen B) Fibronectin C) Нераrin D) Immunoglobulin O E) Plasminogen - CORRECT ANSWER B. Wound healing occurs via a staged process. In early wound healing, platelet aggregation and platelet plug formation occur to achieve hemostasis. In the subsequent 1 to 7 days, neutrophils and macrophages infiltrate the area and release growth factors and cytokines that stimulate fibroblast proliferation. Fibronectin is essential for fibroblast migration by providing a pathway for migration during wound healing. Fibroblasts bind to peptide sequences within fibronectin, which guide them to the site of healing. Granulation tissue forms as collagen is deposited into the area by fibroblasts and neovascularization begins to occur. During this time, wound edges contract from the action of myofibroblasts. Epidermal cells migrate across the newly deposited collagen matrix to reconstitute normal skin appearance. In the following weeks and months, scar formation and remodeling occur via metalloproteinase-mediated collagen breakdown. Incorrect Answers: A, C, D, and E. Collagen (Choice A) is synthesized by fibroblasts and contributes to scar formation in the process of wound healing. Dermal collagen fibers increase as a result of scarring. While the majority of collagen fibers in healthy skin are type I collagen, scarring is initially created by type III collagen. Heparin (Choice C) is a common anticoagulant that potentiates the action of antithrombin III to inhibit multiple coagulation factors. It does not relate to fibroblast migration in wound healing. Immunoglobulins (Choice D) are found on B lymphocyte membranes or are secreted into the serum by plasma cells to recognize antigens and activate the immune system in response to a pathogen. Plasminogen (Choice E) is converted to plasmin, which subsequently degrades fibrin clots leading to clot dissolution. Plasminogen is made in the liver and its role is to degrade and prevent" "8 Exam Section 1: Item 8 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 48. The local health department is investigating an outbreak of acute hepatitis A in two city districts by reviewing the medical records at a health center in each district. The investigators plan to analyze the clinical and epidemiologic characteristics of patients who tested positive for hepatitis A serum IgM antibody compared with characteristics of those who tested negative for the antibody. Which of the following best describes this study design? A) Case series B) Case-control study C) Prospective study D) Randomized trial E) Retrospective cohort - CORRECT ANSWER B. This study compares one group of patients with an outcome under study (seropositive for Hepatitis A IgM antibody) (cases) against a second matched group without that outcome (controls) and identifies the associated exposure within each group. This study design is known as a case-control study. Case-control studies can be conducted in a prospective or retrospective manner but are always observational studies. By grouping patients by outcome and comparing differences in the odds of exposure, case-control studies can detect associations between exposure and outcome, such as exposure to certain risk factors and outcome of contracting hepatitis A as in this study. This is described statistically as an odds ratio (OR). Two unrelated variables will have an OR of 1.0, whereas positive association between an exposure and an outcome will have an OR greater than 1.0 and negatively associated variables will have an OR less than 1.0. Case-control studies are therefore capable of establishing association between exposure and outcome, but they do not establish causality. Incorrect Answers: A, C, D, and E. A case series study (Choice A) is a descriptive study that describes the history, possible exposures, and clinical findings of a group of patients with a similar diagnosis. Case series are non-analytic studies. They do not test a hypothesis and do not generally contain a control group. A prospective study (Choice C) is one in which one group of patients experiences an intervention or exposure and an associated control group does not. The two groups are followed and the desired outcome tracked. A randomized trial (Choice D) is a stringent type of prospective study wherein patients are randomly assigned to receive a particular intervention. The intervention may be compared against placebo therapy or against standard therapy, depend" "9 Exam Section 1: Item 9 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 49. A 1-day-old female newborn with aniridia undergoes genetic testing. Her father and several other paternal family members also have aniridia. Chromosomal microarray analysis of the patient shows no gain or loss of gene dosage throughout the genome, including the chromosome 11p15 region that contains the PAX6 commonly mutated in patients with aniridia. Routine chromosomal analysis shows a balanced translocation between chromosomes 1 and 11, with the chromosomal breakpoint on chromosome 11 being 11p15. The same balanced translocation is also found in all of the affected family members but not in the unaffected family members. However, DNA sequencing of the PAX6 located at 11p15 shows no mutation in any of the exons of PAX6 in the patient. Which of the following is the most likely cause of this patient's - CORRECT ANSWER C. Balanced translocations occur when a segment of a chromosome has switched places with another broken chromosomal segment. In this patient scenario, a segment of chromosome 1 has been relocated and joined onto chromosome 15 just as a segment of chromosome 15 has been joined onto chromosome 1 in place of the translocated segment. Frequently, chromosomal translocations carry no significant consequence unless the break point affects a regulatory sequence or section of an exon that results in reduced or absent translation of the encoded protein. In this case, chromosomal analysis showed that the exons of PAX6, a highly conserved gene that encodes the iris, demonstrated no mutations. Therefore, the translocation must affect a promoter, enhancer, silencer, or similar regulatory sequence that prevents appropriate expression of the gene itself. Aniridia is an autosomal dominant mutation; homozygosity at this gene locus is often severe and fatal. This patient inherited one copy of the abnormal chromosome from her father and a normal copy from the mother, leading to the observed phenotype. Regulatory sequences in DNA include promoters, the sites bound by RNA polymerase and transcription factors, enhancers, sites bound by transcription factors that generally upregulate transcription, and silencers, regions bound by repressor proteins. Incorrect Answers: A, B, D, and E. Creation of a new stop codon (Choice A) would result in cessation of ribosomal translation of the protein prior to completion. A stop codon mutation would likely be detected in the chromosomal analysis of the patient's PAX6 gene, which was reportedly found to be without mutation. Deletion of an intron (Choice B) would generally have no effect on the phenotype, as introns are non-translated intervening regions of DNA between exons that are spliced out during pre-m" "10 Exam Section 1: Item 10 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 50. The graph shows the mean steady state plasma concentration of Drug X as a function of time in a subject who has normal weight (solid curve) and a subject who is obese (dashed curve). Both subjects take 10 mg of Drug X daily. Which of the following best describes the disposition of Drug X in the obese patient compared with the patient who has normal weight? A) Greater volume of distribution B) Higher clearance C) Lower bioavailability D) Slower absorption rate E) Shorter half-life 1 3 4 6 7 Time (days) - CORRECT ANSWER A. Volume of distribution is the theoretical volume in which a drug exists given its plasma concentration following a dose of medication. There are three main compartments into which a medication can distribute: the intravascular compartment, the interstitial compartment, and the intracellular compartment (including fat). With an increased volume of distribution, the same dose of medication will lead to a smaller plasma concentration (ie, it will distribute more widely into tissue), and it will take longer for the medication to reach a steady state within the plasma. Large molecules or protein-bound molecules generally remain within the vasculature and quickly reach a steady state concentration given the relatively low volume of the intravascular compartment. Smaller hydrophilic molecules tend to move into the interstitial space, which increases their volume of distribution and increases the amount of time it takes for the medication to reach a steady state concentration. Small lipophilic molecules, however, exhibit the largest volume of distribution as they are commonly taken up by adipocytes, which greatly increases the volume in which they are distributed. Therefore, it takes these molecules longer to reach a steady state concentration, especially in obese patients in which the volume of distribution is even greater given the presence of increased fat stores, as seen in this graph. Incorrect Answers: B, C, D, and E. Higher clearance (Choice B) of a medication indicates that more medication is removed from the body in a given period of time. Differences in clearance are typically because of changes in hepatic or renal function and are commonly displayed in a graph that shows the decline in drug concentration. An increase in clearance with the same volume of distribution would require a higher maintenance dose to reac" "11 Exam Section 1: Item 11 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 11. A 26-year-old man comes to the physician because of fever, cough, chest pain, and malaise for 2 weeks. He moved to central California 6 months ago. A complete blood count shows mild eosinophilia. A chest x-ray shows patchy bronchopneumonia. Culture of the sputum grows a mold with the morphology shown in the photomicrograph, with the arrow indicating the infectious particle. Which of the following is the most likely causal organism? A) Actinomyces israelii B) Coccidioides immitis C) Histoplasma capsulatum D) Legionella pneumophila E) Mycobacterium tuberculosis OF) Nocardia brasiliensis G) Staphylococcus aureus H) Streptococcus pneumoniae - CORRECT ANSWER B. Coccidioides immitis is an endemic fungus of the Southwestern United States and central valley of California that typically causes coccidioidomycosis, a self-limited respiratory illness. Signs and symptoms include fever, fatigue, cough, arthralgia, and myalgia. Patients may also present with erythema nodosum. Coccidioidomycosis can potentially present with disseminated disease, especially in immunocompromised patients, and cause infections of the skin, bone, and central nervous system. Silver stain of sputum or tissue biopsy demonstrates large spherules containing endospores. Diagnosis can be facilitated with enzyme-linked immunosorbent assay (ELISA) testing and be confirmed with polymerase chain reaction. Treatment is with oral or intravenous antifungals, including agents from the azole or polyene classes. Incorrect Answers: A, C, D, E, F, G, and H. Actinomyces israelii (Choice A) is a Gram-positive, anaerobic bacillus that forms branching filaments and yellow granules. It typically causes abscesses of the head and neck, often with draining fistulous tracts. Histoplasma capsulatum (Choice C) is a fungus native to the Ohio river and Mississippi river valleys that can cause pneumonia and is spread by the droppings of birds or bats. It is identifiable as oval yeasts within macrophages. Legionella pneumophila (Choice D) is a Gram-negative bacillus that is transmitted primarily through aerosols from water sources. It causes Legionnaire disease, characterized by fever, pneumonia, and gastrointestinal symptoms. It can be identified by positive staining with silver stains and its fastidious nature, requiring culture on charcoal yeast extract with supplemental iron and cysteine. Mycobacterium tuberculosis (Choice E) is an acid-fast bacillus that causes tuberculosis. It is identifiable by its thick, waxy capsule, positive ac" "12 Exam Section 1: Item 12 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 12. An investigator is conducting a study of the effects of a newly discovered spider venom on neuromuscular transmission. An isolated nerve muscle preparation is bathed in a solution containing the venom; there is a decrease in the end-plate potential amplitude following stimulation of the nerve. The presence of the venom does not change the amplitude of the nerve action potential or the muscle potential change in response to the exogenous application of acetylcholine to the neuromuscular junction. Blockade of which of the following by the venom best explains the decreased end-plate potential amplitude in this study? A) Chloride conductance in the nerve terminal B) Function of acetylcholinesterase C) Inactivation of voltage-gated sodium channels in the muscle D) Insertion of acetylcholine receptors into - CORRECT ANSWER E. The blockade of presynaptic, voltage-gated calcium channels would lead to decreased end-plate potential amplitude. Normally, action potentials lead to depolarization in the terminal bouton of the axon. Presynaptic, voltage-gated calcium channels consequently open and allow calcium influx, which triggers the exocytosis of acetylcholine-filled vesicles into the synaptic cleft. Acetylcholine diffuses across the synaptic cleft and binds nicotinic acetylcholine receptors (NÁCHRS) on the skeletal muscle cell membrane. The bound NACHRS allow sodium influx (and a lesser degree of potassium efflux), leading to depolarization of the postsynaptic membrane potential. Once the threshold membrane potential is reached, voltage-gated sodium channels open and the depolarization propagates down the postsynaptic membrane, representing the muscle action potential. Each NACHR is associated with its own end-plate potential, or voltage change, that results from the binding of acetylcholine. The summation of the end-plate potentials represents the membrane potential. Decreased synaptic acetylcholine concentrations result in decreased end-plate potential amplitudes. Incorrect Answers: A, B, C, and D. Blockade of chloride conductance in the nerve terminal (Choice A) would lead to decreased chloride influx and a more positive presynaptic membrane potential. Consequently, more voltage-gated calcium channels would open, leading to calcium influx and increased release of acetylcholine vesicles. The amplitude of end-plate potentials would increase. Blockade of the function of acetylcholinesterase (Choice B) leads to decreased degradation of synaptic acetylcholine and is the mechanism of medications (eg, pyridostigmine) utilized for myasthenia gravis. An increased synaptic acetylcholine concentration would increase the amplitude of end-plate pote" "13 Exam Section 1: Item 13 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 13. A 38-year-old woman with asthma comes to the physician for advice concerning contraception. She is sexually active with one partner, and they use condoms inconsistently. Current medications include inhalational fluticasone and albuterol. She has smoked one-half pack of cigarettes daily for 20 years and drinks two alcoholic beverages daily. She does not use illicit drugs. Her maternal grandmother developed breast cancer at the age of 58 years. She is 157 cm (5 ft 2 in) tall and weighs 82 kg (180 lb); BMI is 33 kg/m2 Her vital signs are within normal limits. Physical examination shows no other abnormalities. This patient should be advised to avoid the use of an oral contraceptive because of which of the following historical factors? A) Age B) Alcohol use C) Asthma D) Family history E) Obesity F) Tobacc - CORRECT ANSWER F. Combined oral contraceptive pills (OCPS) contain a combination of estrogen and progesterone. They are primarily used for contraceptive purposes but can also be used in a variety of gynecologic disorders, such as polycystic ovarian syndrome (PCOS), menorrhagia, and endometriosis. A sustained release of progestin and estrogen prevents ovulation and causes thickening of the cervical mucus along with thinning of the endometrial layer by inhibiting the release of gonadotropin-releasing hormone, follicle- stimulating hormone, and luteinizing hormone. OCPS are typically administered as a daily pill for three weeks, followed by a daily placebo pill for one week, during which time withdrawal bleeding occurs. They are associated with increased hypercoagulability and are therefore contraindicated in patients with a history of deep venous thrombosis, pulmonary embolism, stroke, or myocardial infarction. They are also contraindicated in patients who are 35 years of age or older and concomitantly smoke 15 or more cigarettes per day, along with patients diagnosed with hypertension, migraine with aura, severe cirrhosis, and breast cancer. Common adverse effects include nausea, breast tenderness, irregular menstrual periods, and a mildly delayed return to fertility following cessation. OCPS are contraindicated in this patient caused by her tobacco use. Incorrect Answers: A, B, C, D, and E. Age (Choice A) can increase the risks associated with OCPS, as there is an increased baseline risk for cardiovascular disease. However, this patient is only 38 years old, and age alone is not a contraindication to OCP use without smoking history. Alcohol use (Choice B) can preclude the use of OCPS if severe and associated with consequent cirrhosis, as they are generally contraindicated in severe decompensated cirrhosis. However, alcohol use alone" "14 Exam Section 1: Item 14 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 14. A 27-year-old man with AIDS comes to the physician because of a 6-week history of progressive memory loss, imbalance, clumsiness, and difficulty finding words. The symptoms began with an inability to concentrate 3 months ago. He is not adherent to combination antiretroviral therapy and recently recovered from Pneumocystis jirovecii (formerly P. carinii) pneumonia. An MRI shows diffuse, poorly defined, bilateral hyperintensities within cerebral white matter. A stereotactic biopsy specimen shown in the photomicrograph depicts cells scattered predominantly along the perivascular spaces. Which of the following is the most likely diagnosis? A) Cryptosporidiosis B) HIV encephalopathy C) HIV protease inhibitor toxicity O D) Lymphoma E) P. jirovecii infection - CORRECT ANSWER B. HIV encephalopathy is the most likely diagnosis in this patient with AIDS, progressive memory loss, and motor deficits who does not adhere to antiretroviral therapy. HIV encephalopathy is a diagnosis of exclusion in patients with HIV, and other causes of encephalopathy such as toxoplasmosis, meningitis, encephalitis, bacterial abscess, progressive multifocal leukoencephalopathy (PML), or primary central nervous system (CNS) lymphoma should be ruled out. The classic triad of HIV encephalopathy involves movement disorders, psychomotor impairment, and memory deficits. While it can occasionally be confused with PML, patients with PML tend to have a more rapid disease progression, focal deficits, and different findings on MRI. In HIV encephalopathy, the MRI demonstrates multiple, symmetric, and poorly demarcated T2 hyperintense lesions scattered in the subcortical white matter. Brain biopsy characteristically demonstrates microglial nodules with multinucleated giant cells as in this case. Incorrect Answers: A, C, D, and E. Cryptosporidiosis (Choice A) is caused by Cryptosporidium parvum that presents with severe diarrhea in patients with AIDS. Symptoms include fever, weight loss, symptoms of dehydration and orthostasis, severe, watery diarrhea, cramping abdominal pain, nausea, and vomiting. Disseminated infection can involve the lungs and liver but encephalopathy or encephalitis does not occur. HIV protease inhibitor toxicity (Choice C) can present acutely with nausea and vomiting, but long-term complications commonly include lipodystrophy and increased cardiovascular risk. They do not cause encephalopathy. Lymphoma (Choice D) in patients with AIDS can take on many forms, but primary CNS lymphoma is common. It often presents with seizures, lethargy, subacute memory loss, and headache. Physical examination may show neurol" "15 Exam Section 1: Item 15 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 15. A 4-year-old boy has had a viral upper respiratory tract infection for the past 3 days. Clearance of the virus is most dependent on which of the following? A) Activation of macrophages B) Class I MHC-viral peptide complex presentation C) Class II MHC-viral peptide complex presentation D) Phagocytosis of viral particles by CD8+ T lymphocytes E) Production of memory B lymphocytes F) Proliferation of memory T lymphocytes G) Proliferation of plasma cells - CORRECT ANSWER B. Class I MHC-viral peptide complex presentation with subsequent activation of CD8+ T lymphocytes will most likely contribute to clearance of the virus. Viruses are obligate intracellular pathogens and must enter a cell to proliferate. Viral particles that are released from infected cells are taken up by professional antigen presenting cells (APCS) such as dendritic cells. These large proteins are shuttled to the proteasome, which generates smaller fragments of protein that are in turn shuttled to the endoplasmic reticulum (ER). In the ER, these peptides are attached to class I MHC molecules before being transferred to the Golgi apparatus and then to the cytoplasmic membrane where they are expressed. Viral particles may also be presented on class I MHC molecules on the surface of infected cells, although these cells often lack the costimulatory capabilities of APCS. CD8+ T lymphocytes possess unique T-cell receptors (TCRS) that recognize particular viral antigens, and binding of the TCR to the class I MHC-antigen complex activates the CD8+ T lymphocytes. In turn, they release cytokines such as interferon-y and tumor necrosis factor; they also interact with cell surface Fas and FasL to induce apoptosis of infected cells. They can release granzyme and perforin, which cause membrane pore formation and lead to apoptosis of infected cells. Incorrect Answers: A, C, D, E, F, and G. Activation of macrophages (Choice A) occurs primarily through IL-12/IFN-y signaling. Activated macrophages can become histiocytes and may form granulomas in the case of some infections (eg, Mycobacterium tuberculosis). Class II MHC-viral peptide complex presentation (Choice C) activates CD4+ T lymphocytes, which assist B lymphocytes in making antibodies, and recruit macrophages, CD8+ T lymphocytes, and leukocytes to the site of activation. While t" "16 Exam Section 1: Item 16 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 16. The graph shows diastolic and end-systolic relationships of the left ventricle. The solid line shows the control pressure-volume loop in a single cardiac cycle. If a healthy person is given nitroprusside and reflex responses are blocked, which of the following labeled end-systolic pressure points is most likely? 200 A •B 100- E• C D 100 300 400 200 Left ventricular volume (mL) A) B) C) D) E) - CORRECT ANSWER D. Nitroprusside is an intravenous, titratable vasodilator that can be used for the treatment of hypertensive emergencies. It breaks down in circulation to release nitric oxide, which in turn activates guanylate cyclase in vascular smooth muscle to result in vascular smooth muscle relaxation and vasodilation via a cyclic GMP pathway. It typically induces a reflex tachycardia. Sodium nitroprusside preferentially dilates arterial vessels over venous vessels, resulting in a decrease in afterload without significant change in preload. An isolated decrease in afterload results in a shortened pressure-volume loop (less left ventricular pressure is needed to open the aortic valve) and a decreased end-systolic left ventricular volume (because of an increased stroke volume and ejection fraction). The associated end-systolic pressure point on this graph is represented by point D. Incorrect Answers: A, B, C, and E. Choice A represents a point that may be observed if afterload is increased and cardiac contractility is increased. The left ventricular pressure needed to open the aortic valve is increased, and the increased contractility maintains the stroke volume to result in the same end-systolic left ventricular volume as the control. Choice B represents the end-systolic pressure point associated with an isolated increase in afterload. The left ventricular pressure needed to open the aortic valve is increased and the stroke volume is reduced, resulting in an increased end-systolic left ventricular volume. Choice C represents an end-systolic pressure point that may be seen in the case of decreased afterload and decreased cardiac contractility, with reduced end-systolic left ventricular pressure and a reduced stroke volume resulting in increased end-systolic left ventricular volume. Choice E represents a point that may occur with a" "17 Exam Section 1: Item 17 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 17. The patient whose cardiac function is illustrated most likely has which of the following? Normal A) Arteriovenous malformation B) Cardiac tamponade C) Congestive heart failure Patient D) Cor pulmonale E) Restrictive cardiomyopathy Left ventricular end-diastolic volume - CORRECT ANSWER C. The Frank-Starling mechanism describes the phenomena by which cardiac output is dependent on the amount of cardiomyocyte fiber stretch prior to contraction, as represented by the left ventricular end-diastolic volume. A greater pre-contraction stretch results in a greater force of contraction (to a point), and the relationship is demonstrated by Frank-Starling curves. A given Frank-Starling curve applies for constant afterload and inotropy. Changes in afterload and/or inotropy shift the curve up or down. This patient has a Frank-Starling curve that is shifted down, indicating that for a given preload, there is reduced cardiac output relative to normal. This may occur in decreased inotropic states such as congestive heart failure, with the administration of negative inotropes, or in the setting of increased afterload. The curve shifts up in positive inotropic states and/or with decreased afterload. Incorrect Answers: A, B, D, and E. Arteriovenous malformation (Choice A) results in low-resistance, high-volume flow of blood from the arterial to the venous system with greatly increased venous return. The increase in preload causes a greater distension in the cardiomyocyte fibers at the end of diastole, which results in increased cardiac output per the Frank-Starling relationship. Cardiac tamponade (Choice B) result in decreased ventricular filling because of compression of the heart by fluid in the pericardium. In the absence of other factors affecting afterload or cardiac contractility, the Frank-Starling curve would not be depressed. Cor pulmonale (Choice D) describes right ventricular failure resulting from chronic pulmonary hypertension. Left ventricular contractility and afterload are not affected, and the Frank-Starling curve for the left ventricle would not shift. Restrictive cardiomyopathy (Choice E) results in" "18 Exam Section 1: Item 18 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 18. A 28-year-old woman has had fatigue and heavy menstrual periods during the past year. She is a vegetarian who eats eggs and dairy products. Her diet contains little fruit. Which of the following erythrocyte types is most likely to be present on a peripheral blood smear from this patient? OA) Macrocytes B) Microcytes C) Ovalocytes D) Spherocytes E) Stippled cells - CORRECT ANSWER B. Microcytes are most likely to be present in this patient's peripheral blood smear, which is suggestive of iron deficiency anemia (IDA) in the setting of heavy menstrual periods and low dietary iron intake (vegetarianism). Erythrocytes in the setting of IDA are pale (hypochromic) and small in size (microcytic) but normal in shape. Iron is required for the synthesis of heme, which is a necessary component of the hemoglobin molecule, and thus, of erythrocytes. It functions to shuttle oxygen to and from peripheral tissues. In individuals who do not have an adequate intake of dietary iron in the form of heme obtained from animal meat, especially when other causes of ongoing blood loss such as heavy menstruation are present, deficiency can develop. IDA may also develop as the result of chronic blood loss from colorectal bleeding, as a result of malabsorption syndromes (eg, celiac disease), and in patients who have undergone gastric bypass surgery. Erythrocytes on the peripheral blood smear are hypochromic and microcytic as a result of deficient hemoglobin concentration. It is hypothesized that erythrocytes are microcytic as a result of continuing erythrocyte division in order to reach an adequate hemoglobin concentration; because hemoglobin stores are inadequate, cell division continues beyond what would normally occur and causes the cells to be smaller than normal. Treatment for this patient would include management of her menstrual bleeding and oral iron supplementation. Incorrect Answers: A, C, D, and E. Macrocytes (Choice A) refer to larger than normal erythrocytes and are seen in the setting of folate or vitamin B 12 deficiency. This patient has chronic blood loss most likely leading to iron deficiency and is likely to have adequate stores of folate and vitamin B 12 which are consumed in high amounts in most vegetari" "19 Exam Section 1: Item 19 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 19. A 13-year-old girl is brought to the physician by her mother because of a mass in her armpit for 3 days. She has a history of several similar masses since early childhood that have resolved with treatment. She also has had several episodes of pneumonia, rhinitis, and perianal abscesses since birth. Her temperature is 37°C (98.6°F). Physical examination shows a 2-cm, raised, red, tender, fluctuant mass in the right axilla. There is mild hepatosplenomegaly. A complete blood count shows a mildly increased leukocyte count. A chest x-ray shows bilateral interstitial infiltrates. A photomicrograph of lung tissue obtained on biopsy is shown. A nitroblue tetrazolium test shows deficient reduction by granulocytes. The axillary mass is incised and drained, and culture of the fluid grows Staphylococcus aureus - CORRECT ANSWER B. Chronic granulomatous disease (CGD) is the most likely diagnosis in this patient with a Staphylococcus aureus abscess, history of prior similar infections, and an abnormal nitroblue tetrazolium test. CGD is defined by deficiency of the NADPH oxidase complex, which is essential for normal neutrophil intracellular killing of pathogens. NADPH oxidase uses oxygen as a substrate for the generation of free radicals (superoxide anions). Free radical oxygen species are subsequently used for the creation of hydrogen peroxide and hypochlorous acid. Activation of this pathway leads to the respiratory burst which results in bacterial death. Deficiency of NADPH oxidase renders phagocytes incapable of neutralizing catalase-positive bacteria, which are capable of neutralizing their own hydrogen peroxide, thus leaving the host cells without the substrate necessary to complete the respiratory burst. Diagnosis is made by an abnormal dihydrorhodamine test or a nitroblue tetrazolium reduction test. In this latter test, normal phagocytes use the action of NADPH to reduce nitroblue, which leads to a color change from yellow to blue. Patients with CGD will not demonstrate color change. Recurrent pneumonia is the most common presenting infection in patients with CGD, and the most common infecting bacteria include Staphylococcus species, Aspergillus species, Burkholderia cepacia, and Nocardia species. Patients with CGD are also at risk for fungal infections, especially Aspergillus species. Incorrect Answers: A, C, D, and E. presentingChédiak-Higashi syndrome (Choice A) is a rare, autosomal recessive disorder of the immune system caused by mutations in the lysosomal trafficking regulator gene (LYST) that encodes a protein essential for normal formation and transportation of lysosomes within the cell. The clinical manifestations include freq" "20 Exam Section 1: Item 20 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 20. A 67-year-old white woman undergoes punch biopsy of an urticarial lesion on her back. When the biopsy specimen container is received in the laboratory, it is determined that it contains two specimens instead of one. Review of the additional specimens received from the physician's office indicates that specimens from two separate patients were inadvertently shipped in the same container. The other specimen is from the back of a dark-skinned, 58-year-old Nigerian woman. An increase in which of the following in the specimen from the Nigerian patient is most likely to differentiate these two specimens? A) Number of melanocytes B) Number of melanophages C) Number of melanosomes D) Size of melanocytes - CORRECT ANSWER C. Melanin is a pigment made within a specialized cellular organelle called a melanosome found in melanocytes, which are dendritic cell derivatives that reside in the basal layer of the epidermis. Melanocytes are derived from neural crest cells, which originate at the dorsal neural tube. During embryologic development, they migrate from dorsal to ventral, and then to the epidermis. The initial substrate in the creation of melanin is the amino acid tyrosine. After producing melanin from tyrosine, the melanosomes are transferred to the surrounding keratinocytes. The variation in skin tones seen in different ethnic groups is based on the number of the melanosomes and the distribution of melanin within them in the keratinocytes, not the number of melanocytes themselves. In patients with darker skin, the melanosomes are more numerous and more densely packed with melanin. In patients with lighter skin, the melanin is distributed with less density, and there are fewer melanosomes. Melanin production is regulated in part by melanocyte-stimulating hormone (MSH). MSH is a byproduct of proopiomelanocortin (POMC) which also produces adrenocorticotropic hormone (ACTH). In adrenal insufficiency, stimulation of ACTH production by POMC simultaneously produces MSH, leading to diffuse hyperpigmentation in affected patients. Incorrect Answers: A, B, and D. Neither the number of melanocytes (Choice A) nor their size (Choice D) differs between patients with different skin types. Increased number, size, and nuclear/cytoplasmic ratio of melanocytes would be seen in melanoma, a neoplasm of melanocytes. Lesions concerning for melanoma are characterized clinically by asymmetry, irregular-appearing borders, variable coloration, a diameter greater than 6 mm, and rapid evolution in characteristics. Melanophages are macrophages which have taken up" "21 Exam Section 1: Item 21 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 21. An investigator is studying the adverse effects of a proteasome inhibitor, bortezomib, on immune responses. Bortezomib is administered to a group of patients with relapsed multiple myeloma, and immune responses are observed. Which of the following immunologic processes is primarily affected by this drug? A) Activation of the complement cascade B) Activation of perforin C) Presentation of antigens to CD8+ T lymphocytes D) Secretion of histamine by mast cells E) Secretion of interleukin-1 (IL-1) by macrophages - CORRECT ANSWER C. Presentation of antigens to CD8+ T lymphocytes would be affected by the use of the proteasome inhibitor bortezomib. Proteasomes are large proteases with numerous domains that are present in both the cytoplasm and the nucleus. They are made of a and B subunits and have an extensive role in cell cycle regulation, but also in immune function. The caspase, chymotrypsin, and trypsin-like protease capabilities of this enzyme complex are also critical to their role as antigen processing centers; intracellular peptides are processed in the proteasome for presentation on class I major histocompatibility complex (MHC). Antigen-bound class I MHC molecules are recognized by cytotoxic CD8+ T lymphocytes, so proteasome inhibitors blunt the response of CD8+ T lymphocytes by altering antigen presentation on the surface of MHC I molecules. Incorrect Answers: A, B, D, and E. Activation of the complement cascade (Choice A) is not a feature of proteasome inhibitors. The complement cascade is activated by the classical, alternative, or lectin pathways. Activation leads to opsonization of invading microbes, direct microbial killing via formation of the membrane attack complex (MAC), and proinflammatory signaling (eg, C5a mediation of neutrophil chemotaxis). Activation of perforin (Choice B) occurs as a result of activated natural killer cells, which are immune cells that primarily respond to an absence of MHC I molecules on the surface of cells. Perforins, along with granzyme, are proapoptotic agents and induce cellular death. Proteasome inhibitors interfere with antigen presentation on MHC I molecules but do not affect the presence of MHČ I molecules. Secretion of histamine by mast cells (Choice D) occurs as a result of mast cell degranulation, a process that primarily occurs during allergic reactions but can also occur in the setting" "22 Exam Section 1: Item 22 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 22. A 3-month-old girl is brought to the physician 2 days after the mother felt masses in the infant's groin while bathing her. Physical examination shows bilateral inguinal hernias and a shallow vagina that ends blindly. Serum studies show increased concentrations of luteinizing hormone, follicle-stimulating hormone, testosterone, and dihydrotestosterone. During repair of the hernias, gonads are found. A biopsy specimen of the gonads shows bilateral testes. The karyotype is 46,XY. Ultrasonography shows no uterus or fallopian tubes. Which of the following is the most likely cause of the findings in this patient? A) Decreased 21-hydroxylase activity B) Decreased 5a-reductase activity C) Defective androgen receptor D) Failure of testis to secrete anti-müllerian hormone E) Mutation of the sex-determining r - CORRECT ANSWER C. Androgen insensitivity syndrome is caused by a defect in the androgen receptor complex resulting in a genotypic XY male to develop external female or ambiguous genitalia and female secondary sexual characteristics. Testes are present and produce testosterone normally, but absence of a functioning androgen receptor prevents hormone binding and thereby prevents the development of male sexual characteristics. Patients present with female external genitalia, scant pubic and axillary hair, absent uterus and fallopian tubes, and a rudimentary vagina. Results of laboratory studies show increased concentrations of testosterone, estrogen, and luteinizing hormone. Menses will not occur because of the lack of cycled progesterone and estrogen, and the lack of a functional uterus with endometrial lining. Incorrect Answers: A, B, D, and E. Decreased 21-hydroxylase activity (Choice A) occurs in the setting of congenital adrenal hyperplasia, of which the most common form is 21-hydroxylase deficiency. Lack of this enzyme prevents production of aldosterone and cortisol, and results in excessive androgen production. Genetically female patients present with hypoaldosteronism as well as virilization during infancy, and genetically male patients present with precocious puberty in childhood. Decreased 5a-reductase activity (Choice B) results in the insufficient conversion of testosterone to dihydrotestosterone (DHT), leading to decreased concentrations of DHT and impaired virilization of the male urogenital tract. Individuals with 5a-reductase deficiency exhibit either female or ambiguous external genitalia, although demonstrate normal male internal genitalia because of normal concentrations of testosterone. Increased concentrations of DHT, as in this infant, would not be found. Failure of testis to secrete anti-müllerian hormone (Choice" "23 Exam Section 1: Item 23 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 23. A 35-year-old man comes to the physician because of a 3-year history of an enlarging nose, coarsening of his facial features, muscle weakness, and increased hand and foot size. Physical examination shows a large fleshy nose and prognathism. Fasting serum studies show increased insulin-like growth factor-l, glucose, and triglyceride concentrations. Increased serum growth hormone concentrations do not decrease after the administration of oral glucose. Compared with a healthy patient, which of the following best describes the metabolic changes in this patient? Muscle Glucose Adipose Lipolysis Неpatic Gluconeogenesis Uptake O A) ↑ ↑ ↑ B) ↑ ↑ C) ↑ D) ↑ ↑ E) ↑ F) - CORRECT ANSWER D. Acromegaly is caused by a growth hormone (GH) secreting pituitary adenoma that stimulates the excessive production of insulin-like growth factor-1 (IGF-1) by the liver. IGF-1 interacts with its receptor, a tyrosine kinase-based receptor, that stimulates cell growth, proliferation, and growth of the axial and appendicular skeleton. Acromegaly occurs after the closure of growth plates, and in this context excessive IGF-1 leads to expansion of flat bones and tissues. Excessive GH and IGF-1 leads to deranged glucose homeostasis by increasing peripheral insulin resistance, impairing muscle and adipose uptake of glucose, increasing adipose lipolysis, and increasing hepatic gluconeogenesis. Consequently, up to half of patients with acromegaly develop diabetes mellitus. Patients with acromegaly also exhibit increased rates of hypertriglyceridemia. Incorrect Answers: A, B, C, E, and F. Choices A, B, and C reflect states of increased muscle glucose uptake, which would occur in the setting of increased peripheral insulin concentrations or receptor sensitivity. Excessive concentrations of GH and IGF-1 lead to reduced peripheral insulin sensitivity. Choices B, C, and F reflect states of decreased lipolysis, whereas deficient insulin signaling will lead to increased lipolysis. Choices C, E, and F reflect states of decreased hepatic gluconeogenesis. GH directly stimulates gluconeogenesis, which is increased in acromegaly. Educational Objective: Acromegaly is associated with excessive GH and IGF-1 concentrations that lead to deranged glucose homeostasis by increasing peripheral insulin resistance, impairing muscle and adipose uptake of glucose, and increasing lipolysis and hepatic gluconeogenesis. Patients with acromegaly experience increased rates of diabetes mellitus and hypertriglyceridemia. %3D Previous Next Score Report Lab Va" "24 Exam Section 1: Item 24 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 24. A 45-year-old woman is brought to the emergency department 30 minutes after the sudden onset of left-sided facial drooping and left arm and leg weakness. One week ago, she returned from a business trip that involved a 6-hour airplane flight. She has no history of major medical illness. She does not smoke. Her only medications are an oral contraceptive and occasional acetaminophen for headache. Her pulse is 88/min and regular, and blood pressure is 160/90 mm Hg. Physical examination shows left- sided facial paralysis. Examination of the left lower extremity shows swelling, tenderness, and localized erythema over the calf. Muscle strength is 3/5 in the left upper and lower extremities. Cardiac examination shows no abnormalities. Echocardiography is most likely to show which of the following findings in - CORRECT ANSWER E. This patient with a deep venous thrombosis (DVT) most likely possesses a patent foramen ovale, allowing a thromboembolism to bypass the pulmonary circulation and cause a thromboembolic cerebrovascular accident (CVA, or stroke). DVT typically presents with lower extremity edema, erythema, warmth, and pain at rest along with calf pain on dorsiflexion of the foot (Homan sign). Blood clots typically occur in patients who possess some or all of the risk factors of the Virchow triad: stasis (long flight, postoperative immobility), hypercoagulability (oral contraceptive use, hypercoagulable disorders), and endothelial damage (smoking, hypertension, atherosclerosis, injury). DVT may lead to pulmonary embolism if the clot embolizes to the right heart and subsequently to the pulmonary circulation. In patients with a patent foramen ovale (a congenital defect between the right and left atria), the clot may travel from the right atrium to the left atrium and then to the systemic circulation. This patient's clot likely embolized to the right internal carotid artery and then the right middle cerebral artery (MCA), occluding blood flow to the right primary motor cortex resulting in left-sided weakness. CVAS occur because of ischemic or hemorrhagic loss of blood supply to the brain. Approximately 80 to 85% of CVAS are ischemic, commonly arising from thromboembolic disease, whereas 15 to 20% of CVAS are hemorrhagic as a result of blood vessel rupture (eg, hypertension-related intraparenchymal hemorrhage from a perforating artery). Classically, CVAS manifest as a neurologic deficit related to the affected part of the brain. Incorrect Answers: A, B, C, and D. Aortic stenosis (Choice A), hypertrophy of the interventricular septum (Choice C), and mitral valve prolapse (Choice D) are structural abnormalities of the heart that lead to abno" "25 Exam Section 1: Item 25 of 50 National Board of Medical Examiners® Comprehensive Basic Science Self-Assessment 25. A 57-year-old man dies 5 days after a stroke. He had progressively severe hypertension over the past 2 years but refused treatment. Blood pressure just prior to the stroke was 220/110 mm Hg. Which of the following is the most likely histologic finding in his kidneys at autopsy? A) Amyloidosis B) Chronic pyelonephritis C) Hyperplastic arteriolitis D) Nodular glomerulosclerosis E) Renal papillary necrosis - CORRECT ANSWER C. Chronic hypertension is associated with vascular changes in the small arteries and arterioles called arteriosclerosis, marked by thickening of the vessel walls and loss of elasticity. There are two primary types, which are hyaline arteriosclerosis and hyperplastic arteriosclerosis (also called hyperplastic arteriolitis). Hyaline arteriosclerosis is characterized by protein deposition in the vessel walls. Hyperplastic arteriosclerosis is characterized by concentric thickening of the vessel wall with a laminar appearance caused by the proliferation of smooth muscle cells and is associated with severe, chronic hypertension, as in this patient. Incorrect Answers: A, B, D, and E. Amyloidosis (Choice A) is an infiltrative disorder caused by the deposition of abnormal proteins in tissue. The kidneys are commonly involved, with abnormal protein deposits in the mesangium that display an apple-green birefringence with Congo red dye on histologic analysis. Renal impairment and nephrotic syndrome may develop. It is not caused by chronic hypertension. Chronic pyelonephritis (Choice B) may develop from recurrent infections of the genitourinary tract with reflux into the renal pelvis, which may occur in the setting of obstructive uropathy, nephrolithiasis, and vesicoureteral reflux. The kidneys display atrophy, calyceal deformities, and fibrosis of the renal parenchyma. Nodular glomerulosclerosis (Choice D) is associated with diabetic nephropathy and amyloidosis. It may progress to nephrotic syndrome. Renal papillary necrosis (Choice E) may result from severe ischemic injury to the kidney. Risk factors include sickle cell disease, obstructive nephropathy, nonsteroi