USMLE USMLE-STEP-1 Online Practice Questions and Exam Preparation

USMLE USMLE-STEP-1 Online Questions & Answers

• Question 321:

  An 18-year-old woman presented with 1 week of history of fever and malaise. She had mild jaundice and elevated temperature. Hemoglobin was 13.8 g/dL, leukocyte count 13 ?109 per liter. Serum bilirubin was elevated (42 mmol/L) and contained 95% unconjugated bilirubin. Liver enzyme tests were normal. Which of the following is the most likely cause of these signs and symptoms?

  A. alcohol poisoning
  B. decreased glucuronyl transferase
  C. increased lactate dehydrogenase (LDH)
  D. excessive hemolysis
  E. obstruction of bile flow
  B. decreased glucuronyl transferase

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  Explanation

  Section: Physiology Glucuronyl transferase is the enzyme that conjugates bilirubin in the liver, after which it is excreted in bile or urine. Ahereditary defect in glucuronyl transferase concentration, or activity, is called Gilbert syndrome. It may lead to mild jaundice and general discomfort with typical onset in childhood or early adulthood. Alcohol poisoning (choice A) leads to liver damage, and an elevation of conjugated bilirubin. Abnormalities of liver enzyme tests would be expected. Lactate dehydrogenase (choice C) catalyzes the conversion of lactate to pyruvate as part of cellular energy production. Since many cells including red blood cells are rich in LDH, increased serum LDH levels could point toward excessive hemolysis, but would not be a cause for it. Although hemolysis that exceeds the capacity of the liver to clear bilirubin from serum (choice D) would lead to increased unconjugated bilirubin, it is not the best choice due to the woman's normal hemoglobin (1216 g/dL for females). Obstruction of bile flow (choice E) leads to backup of largely conjugated bilirubin in the blood stream.

• Question 322:

  All patients who have trisomy of autosome 21 who survive to adulthood develop which one of the following conditions?

  A. Alzheimer's disease
  B. lewy body dementia
  C. multi-infarct dementia
  D. Pick disease
  E. secondary dementia
  A. Alzheimer's disease

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  Explanation

  Section: Behavioral Science and Biostatics Down syndrome is a congenital mental retardation associated with trisomy of autosome 21. All patients with Down syndrome who survive into adulthood develop the brain pathologies of Alzheimer's disease, rendering support to the notion that at least one form of Alzheimer's disease may be associated with an autosome 21 abnormality. Choices B, C, D, and E are not associated with trisomy of autosome 21.

• Question 323:

  An infant admitted to the emergency room has been found to be suffering from ammonia intoxication, which was verified by measurement of an elevation of N + in the serum. Treatment of this infant

  

  with arginine results in a reduction serum N + and a lessening of the effects of the ammonia

  

  toxicity. The ability of arginine to render this effect stems from its role in the synthesis of an allosteric activator of the the urea cycle enzyme, carbamoylphosphate synthetase-I (CPS-I). Which of the following represents this potent allosteric effector of CPS-I?

  A. argininosuccinate
  B. bicarbonate ion
  C. fumarate
  D. N-acetylcysteine
  E. N-acetylglutamate
  E. N-acetylglutamate

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  Explanation

  Section: Biochemistry CPS-I is absolutely dependent on the allosteric effector N-acetylglutamate for its activity. This allosteric effector is synthsized by the enzymeN-acetylglutamate synthetase, which is activated by the urea cycle amino acid arginine. None of the other compunds (choices A, B, C, and D) have any effect on CPS-I activity.

• Question 324:

  Under conditions of anaerobic glycolysis, the NAD+ required by glyceraldehyde-3-phosphate dehydrogenase is supplied by a reaction catalyzed by which of the following enzymes?

  A. glycerol-3-phosphate dehydrogenase
  B. alpha-ketoglutarate dehydrogenase
  C. lactate dehydrogenase
  D. malate dehydrogenase
  E. PDH
  C. lactate dehydrogenase

• Question 325:

  Carbidopa is often used in patients with Parkinson's disease. Which of the following best describes the mechanism of action of carbidopa?

  A. activates D2 receptors in the basal ganglia
  B. inhibits the metabolism of dopamine in the blood and peripheral tissues
  C. inhibits the metabolism of dopamine in the brain
  D. inhibits the metabolism of levodopa in peripheral tissues
  E. inhibits the metabolism of levodopa in the brain
  D. inhibits the metabolism of levodopa in peripheral tissues

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  Explanation

  Section: Pharmacology Carbidopa is a peripheral inhibitor of dopa decarboxylase. Dopa decarboxylase is an enzyme present in large amounts in the GI tract and peripheral tissues, and in smaller amounts in the nerve terminals of dopaminergic neurons in the basal ganglia. It is required for the conversion of DOPA to dopamine in the biosynthesis of dopamine and norepinephrine. However, when the prodrug levodopa is used for parkinsonism, over 90% is metabolized in the periphery by the enzyme to dopamine and inactive products and only about 3% of the administered dose enters the brain. Because dopamine does not cross the blood- brain barrier, dopamine formed outside the brain is of no value in treating parkinsonism. When carbidopa is given with levodopa, less levodopa is metabolized in the gut and other peripheral tissues, so more (about 10%) is available to enter the CNS. Carbidopa does not cross the bloodbrain barrier, so it does not prevent conversion of levodopa to active dopamine in the basal ganglia.

  Combination therapy thus reduces the peripheral effects of levodopadopamine and allows patients to receive more benefit and less toxicity from a given dose of levodopa. Choices AC and E are incorrect.

• Question 326:

  A2-month-old infant suffering from increased vomiting and diarrhea is seen in the hospital and observed to have significant abdominal distention due to hepatosplenomegaly. Unfortunately, the infant does not survive. Autopsy reveals calcification of the adrenals and massive accumulation of cholesteryl esters and triglycerides in most tissues. Analysis of enzyme activity in fibroblasts and lymphocytes demonstrates a significant acid lipase (cholesteryl ester hydrolase) deficiency. These clinical findings are indicative of which of the following disorders?

  A. hyperlipoproteinemia, type I (familial liproprotein lipase deficiency)
  B. I-cell disease (mucolipidosis type II)
  C. Maroteaux-Lamy syndrome
  D. Sanfilippo syndrome
  E. Wolman disease
  E. Wolman disease

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  Explanation

  Section: Biochemistry Wolman disease results from a deficiency in the lysosomal acid lipase enzyme (also called cholesteryl ester hydrolase) and is very nearly always a fatal disease of infancy. Symptoms of the disorder arise from massive accumulation of cholesteryl esters and triglycerides in most tissues. Clinical manifestations include hepatosplenomegaly leading to abdominal distention, GI abnormalities, steatorrhea, and adrenal calcification. Type I hyperlipoproteinemia (choice A) results from a deficiency in LPL activity. LPL is the enzyme found on the surface of vessel endothelial cells and is responsible for hydrolyzing fatty acids from the triglycerides found in circulating liporpotein particles. This action allows cells to get fatty acids for energy production from lipoproteins in the plasma. This disoder leads to massive accumulation of chylomicrons and triglycerides in the plasma. Symptoms associated with type I hyperlipoproteinemia are usually detected in childhood and include eruptive cutaneous xanthomatosis, hepatosplenomegaly, and repeated episodes of abdominal pain. I-cell disease (choice B) results from a deficiency in the targeting of lysosomal enzymes to the lysosomes. Affected cells have dense inclusion bodies filled with storage material, hence the derivation of the disease name. There are elevated levels of lysosomal enzymes in the plasma and body fluids of I-cell patients. Symptoms include severe psychomotor retardation, coarse facial features, and severe skeletal abnormalities. The rapidly progressing disease can lead to death between 5 and 8 years of age. Maroteaux- Lamy syndrome (choice C) results from a defect in the enzyme N- acetylgalactosamine-4-sulfatase (arylsulfatase B) responsible for the catabolism of complex glycosaminoglycans. There are three distinct forms of Maroteax-Lamy syndrome that range in severity. Symptoms include coarse facial features, skeletal abnormalities corneal clouding, and aortic valve disruption. Sanfillipo syndrome (choice D) comprises at least four distinct genetic defects that result from deficiencies in enzymes required for the catabolism of complex glycosaminoglycans of the heparan sulfate class. In all cases, the clinical symptoms are similar and include severe mental deterioration, hyperactivity, and disorders in the skin, lungs, heart, and skeletal muscle.

• Question 327:

  A 19-year-old man was in a barroom brawl and was punched squarely in the right eye. He comes to the emergency room the next day and complains of diplopia. An X-ray reveals fracture of the orbital floor. Neurological examination shows loss of sensation of the skin of the right face below the right eye and the upper gums. Which of the following nerves may be injured?

  A. frontal nerve
  B. infraorbital nerve
  C. nasociliary nerve
  D. supraorbital nerve
  E. trochlear nerve
  B. infraorbital nerve

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  Explanation

  Section: Anatomy The infraorbital nerve, a branch of the maxillary (V2) division of the trigeminal (fifth cranial) nerve, courses below the orbital floor to reach the area of skin below the eye. It provides superior alveolar branches to supply the upper gums and is vulnerable in fractures involving the floor of the orbit and face area. All the nerves mentioned in the other choices will be spared by this type of injury. The frontal nerve (choice A) and nasociliary nerve (choice C) are branches from the ophthalmic division (V1) of the trigeminal (fifth cranial) nerve and course within the orbit. The supraorbital (choice D) nerve is a continuation branch of the frontal nerve onto the forehead, providing sensory innervation for this area. The trochlear (fourth cranial) nerve is also located within the orbit.

• Question 328:

  Which of the following statements is true concerning infant mortality in the United States?

  A. Infant mortality in the United States is high compared to other developed nations.
  B. Prenatal care, disappointingly, has not been shown to reduce infant mortality.
  C. The leading cause of infant mortality in the United States is tetanus.
  D. The mortality rate for white infants is greater than that of black infants.
  E. Very few Asian Americans receive prenatal care as compared to African Americans or Hispanic Americans.
  A. Infant mortality in the United States is high compared to other developed nations.

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  Explanation

  Section: Behavioral Science and Biostatics The infant mortality rate in the United States is 8.9 deaths per 1000 live births as of 1989, which ranks the United States behind 11 other developed countries. The leading causes of infant death are congenital anomalies, respiratory distress syndrome, and sudden infant death syndrome; therefore, choice C is incorrect. Choices B, D, and E are incorrect because prenatal care does reduce infant mortality, but only 80% of Caucasian women receive prenatal care, followed by 75% of Asian American women, 65% of African American women, and 60% of Native American women. The infant mortality rate for black infants is greater than that of white infants.

• Question 329:

  The rate-limiting step in glycolysis occurs at the step catalyzed by which of the following enzymes?

  A. glyceraldehyde-3-phosphate dehydrogenase
  B. 6-phosphofructo-1 kinase, PFK-1
  C. 6-PFK-2
  D. phosphoglycerate kinase
  E. pyruvate kinase
  B. 6-phosphofructo-1 kinase, PFK-1

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  Explanation

  Section: Biochemistry There are three reactions of glycolysis that are thermodynamically irreversible. These are the hexokinase (glucokinase), PFK-1, and PKcatalyzed reactions. Reactions that are essentially irreversible in most metabolic pathways are subject to complex regulatory controls and represent rate-limiting steps in the pathway. The primary site of regulation of glycolysis occurs at the level of the PFK-1-catalyzed step. Hence, this reaction is the rate-limiting step in glycolysis. PFK-1 is subject to allosteric control by numerous compounds. Citrate and ATP inhibit the activity of PFK-1, while AMP and fructose 2,6- bisphosphate (F2,6- BP) activate the enzyme. The principal control of PFK-1 activity is exerted by alterations in the level of F2,6- BP. This compound is synthesized from fructose-6-phosphate by the bifunctional enzyme, PFK-2/fructose- 2,6-bisphosphatase (PFK-2/F2,6-BPase). PFK-2 (choice C) contains two catalytic domains, one a kinase and the other a phosphatase, the activities of which are affected by the state of phosphorylation. The phosphatase domain is active when the enzyme is phosphorylated and converts F2,6-BP back to F6-P, thereby reducing the levels of this powerful activator of PFK-1. Thus, although the activity of PFK-2 will determine the rate of activity of PFK-1, it is itself not the rate-limiting enzyme in glycolysis. Glyceraldehyde- 3-phosphate dehydrogenase (choice A) and PGK (choice D) are not regulated enzymes of glycolysis. PK (choice E) is regulated during glycolysis, but does not constitute a rate-limiting step.

• Question 330:

  The striped area in below figure indicates a red light on the receptive fields of a red on-center, green off- center surround retina ganglion cell. How will the light spot be perceived?

  

  A. brown (red + green)
  B. green
  C. no color
  D. red
  E. reddish green
  D. red

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  Explanation

  Section: Physiology A light spot exclusively hitting the center would be perceived as deep red. Some of the light hits the inhibitory surround area. In this area, light is processed coming from cones that are best excited by green light (middle wavelength) and inhibited by red light (long wavelength). In the below figure the on-response dominates the off-response so that the perceived light is different from deep red, but not green (choice B). At the retina ganglion level, the trichromate signals from the cones are transferred into the neuronal stage of color processing, the opponent color stage. The cone photoreceptors are linked together to form opposing color pairs, in this case red versus green. Activation of one member of the pair inhibits the activity in the other so that a color pair cannot be seen at the same location. We cannot experience reddish green (choice E). Physiological color image formation differs from the physical process of coloring a piece of paper with a red and green pencil (choice A), one on top of the other, called subtractive color mixture. It starts with the presence of all colors of light and then dyes subtract some of the reflected light so that adding more dyes results in a darker image such as brown. For physiological color mixing, the presence of all colors is perceived as white, the absence as black (choice C).