USMLE USMLE Certifications USMLE-STEP-1 Questions & Answers

• Question 641:

  Which of the following persons will have a negative free-water clearance?

  A. one who begins excreting large volumes of urine with an osmolality of 100 mosm/kg H2O after severe head injury

  B. one who drinks 2 L of distilled water in 30 minutes

  C. one who is receiving lithium treatment for depression, and who develops polyuria that is unresponsive to administration ADH

  D. one with an oat cell carcinoma of the lung, who excretes urine with an osmolality of 1000 mosm/kg H2O

  E. one with neurogenic diabetes insipidus

  Correct Answer: D

  Section: Physiology A negative free-water clearance occurs when the kidney is reabsorbing more water than it is excreting into the urine and is regulated by ADH. Choice D is correct because lung cancer can cause exogenous production of ADH, which will stimulate increased renal water reabsorption and thus a negative freewater clearance. Choice A is incorrect because the decreased ADH secretion that may occur following a severe head injury will result in a positive free-water clearance. Choice B is incorrect because rapid intake of 2 L of water will inhibit ADH secretion and thus trigger a positive free-water clearance. Choice C is incorrect because the insensitivity to ADH that occurs during lithium treatment will increase water excretion and thus cause a positive free-water clearance. Choice E is incorrect because a person with neurogenic diabetes insipidus has a reduced secretion of ADH, which will cause reduced renal water reabsorption and thus a positive free-water clearance.

• Question 642:

  A healthy 27-year-old male is given a treadmill stress test. His cardiovascular values at rest and during

  exercise were:

  When compared to rest, what was the change in his estimated mean arterial pressure?

  A. decreased by 10 mm Hg

  B. increased by 5 mm Hg

  C. increased by 25 mm Hg

  D. increased by 35 mm Hg

  E. increased by 140 mm Hg

  Correct Answer: B

  Section: Physiology True mean arterial pressure cannot be determined without a pressure transducer, which averages the systolic and diastolic pressures throughout the cardiac cycle. However, mean arterial pressure can be estimated by adding one-third of pulse pressure to the minimum diastolic pressure. At rest his mean pressure is 30/3 + 75 = 85; during exercise mean pressure is 75/3 + 65 = 90. Hence, estimated mean pressure increased by 5 mm Hg.

• Question 643:

  Normal metabolism by the body generates large quantities of acid. In spite of this, normal blood pH is a slightly alkaline 7.4. This extracellular fluid alkalinity is maintained primarily by the body's removal of which of these?

  A. ammonia

  B. carbon dioxide

  C. keto acids such a acetoacetic acid

  D. lactic acid

  E. titratable acids such as phosphoric acid

  Correct Answer: B

  Section: Physiology The vast majority of metabolic acid excretion is in the form of the volatile acid carbon dioxide which is removed via the lungs. Much smaller quantities of nonvolatile acids must be excreted in the urine. Titratable acids, such as phosphoric acid (choice E), make up a large fraction of the acids normally excreted in the urine. When stored fats are metabolized in large quantities, such as in diabetes mellitus, keto acids such as acetoacetic acid (choice C) make up a larger portion of the renal excreted titratable acids. Likewise, when the ischemic tissues pursue anaerobic metabolism, lactic acid (choice D) makes up a larger portion of the renal excreted titratable acids. Ammonia (choice A) is not a titratable cid, but it provides a mechanism for the kidney to excrete large amounts of acid during chronic metabolic acidosis.

• Question 644:

  Exhibit: *missing*

  A healthy 27-year-old male is given a treadmill stress test. His cardiovascular values at rest and during

  exercise were:

  

  When compared to rest, what was the change in his arterial pulse pressure?

  A. decreased from 140 to 65

  B. increased from 30 to 75

  C. increased from 68 to 140

  D. increased from 115 to 140

  E. remained unchanged since heart rate and systolic pressure both increased

  Correct Answer: B

  Section: Physiology By definition, arterial pulse pressure is the difference between peak systolic and minimum diastolic pressure. His resting pulse pressure was 105 - 75 = 30 and his exercise pulse pressure increased to 140

  -65 = 75.

• Question 645:

  The parents bring a 5-month-old baby to the emergency room. It is their first child and they are insecure. The boy vomits frequently, seems to be constantly constipated, and has difficulties in defecation. A barium enema study reveals a region in the bowel that is collapsed and an enlarged colon above this area. Abiopsy from the part of the bowel 1 in above the anus is sent to the laboratory and histological analysis reveals the absence of ganglia in this tissue. What is the most likely diagnosis?

  A. cholecystitis

  B. gastroesophageal reflux disease

  C. hirschsprung disease

  D. polymyositis

  E. temporary problem with no treatment required

  Correct Answer: C

  Section: Physiology Hirschsprung disease is a genetic disorder caused by the absence of enteric nerve cells in the wall of the sigmoid colon and/or rectum. The portion of the bowel wall without nerve ganglia (aganglionic) cannot relax in response to bowel content so that the stool builds up behind the obstruction. In some children the problems begin shortly after birth, other infants are not acutely ill, but develop chronic symptoms such as constipation or anemia. Cholecystitis (choice A), caused by inflammation of the gallbladder, gastroesophageal reflux disease GERD (choice B), and polymyositis (choice D), a disorder affecting esophageal skeletal muscle, do not affect the neuronal regulation of the large intestine. Hirschsprung disease is almost always treated by surgical removal of the affected bowel segment and then joining the healthy bowel segments (choice E). A GI motility disorder might improve on its own due to the ability of the enteric nervous system in healthy GI tract portions to learn new motility patterns. However, it takes a very long time and the success is not certain.

• Question 646:

  A motor vehicle accident caused complete pituitary stalk transection. Secretion of all pituitary hormones is lost except for one, the blood level of which actually increases. Which one of the following pituitary hormones is distinctive in that its primary control is by inhibition rather than stimulation by the hypothalamus?

  A. gonadotropin-releasing hormone

  B. growth hormone

  C. prolactin

  D. proopiomelanocortin

  E. thyroid-stimulating hormone

  Correct Answer: C

  Section: Physiology The primary control over prolactin secretion is inhibition by hypothalamic dopamine; all other anterior pituitary hormones are primarily controlled by hypothalamic hormone stimulation. Hence, with stalk transection, loss of connection of the hypothalamus to the pituitary is associated with decreased secretion of all pituitary hormones except prolactin, the secretion of which increases in the absence of dopamine. Secretion of growth hormone (choice B), proopiomelanocortin (choice D), and thyroid- stimulating hormone (choice E) all decrease. Gonadotropinreleasing hormone (choice A) is a hypothalamic hormone, and is not made in the pituitary gland.

• Question 647:

  A 25-year-old male athlete has just completed a long distance bicycle race during a hot, humid day. At the conclusion of the race, he provides a urine sample for testing. Assuming that his fluid intake during the race was zero, in what portion of the nephron shown in below figure is the tubular fluid osmolality the lowest?

  

  A. A

  B. B

  C. C

  D. D

  E. E

  Correct Answer: D

  Section: Physiology The bicycle racer is likely dehydrated from his exercise and ADH would be secreted by the posterior pituitary in response to increased extracellular osmolality. The area of the nephron with the lowest osmolality will be the early distal tubule (choice D), a nephrondiluting segment. The fluid in Bowman's capsule (choice A) will be at the same osmolality as the plasma entering the glomerular capillaries. The fluid at the end of the proximal tubule (choice B) is isoosmotic with the Bowman's capsule fluid while the fluid in the thin descending loop of Henle (choice C) is greatly concentrated due to water reabsorption into the hyperosmotic medullary interstitium. The fluid at the end of the collecting duct (choice E) can be very concentrated or very dilute, but since ADH greatly increases water reabsorption in this segment, the fluid becomes very concentrated.

• Question 648:

  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

  Correct Answer: B

  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 649:

  Which of the following is likely to cause a negative free-water clearance by the kidney?

  A. central diabetes insipidus

  B. demeclocycline, an inhibitor of the renal tubular actions of ADH

  C. nephrogenic diabetes insipidus

  D. water deprivation

  E. water drinking

  Correct Answer: D

  Section: Physiology Water deprivation will eventually increase extracellular fluid osmolality, which triggers release of ADH. ADH acts on the collecting duct to increase water reabsorption, thus making the excreted urine more concentrated via a negative free-water clearance. Choices B and C will reduce the sensitivity of the collecting duct to normal circulating ADH. Choices A and E will reduce the amount of ADH secreted by the posterior pituitary.

• Question 650:

  below figure illustrates uptake of two gases (nitrous oxide and carbon monoxide) from alveolar air to

  pulmonary capillary blood. Based on this information what can we conclude about carbon monoxide?

  

  A. it does not dissolve in blood

  B. it does not interact with hemoglobin

  C. it has equilibrated with pulmonary capillary blood

  D. it is a diffusion-limited gas

  E. it is a perfusion-limited gas

  Correct Answer: D

  Section: Physiology The plot shows that carbon monoxide fails to equilibrate with pulmonary capillary blood during the time that blood is exposed to alveolar air. Since pulmonary partial pressure rises to only a few percent of alveolar partial pressure, carbon monoxide is a diffusion-limited gas. In contrast, nitrous oxide, which does equilibrate, is a perfusion-limited gas (choice E). Carbon monoxide does cross the respiratory membrane (choice C), dissolve in pulmonary capillary blood (choice A), and binds with hemoglobin (choice B). In fact, its affinity for hemoglobin is about 250-fold greater than that of oxygen.