Cisco
CompTIA
ISC
EC-Council
EXIN
Fortinet
ITIL
Juniper
Microsoft
VMware
Avaya
SAP
Google
NVIDIA
Salesforce
Amazon
Palo Alto Networks
ISC2
Splunk
ServiceNow
All Vendors
  2. Home
  3. Exams
  4. USMLE
  5. STEP1

USMLE STEP1 Exam Questions
Step1 (Page 33 )

Updated On: 31-Mar-2026
Page 33 of 171
PDF with 845 Questions

An advertisement promotes energy bars containing fructose as an ideal food to take on extreme mountain- climbing expeditions. Which of the following statements concerning fructose absorption is true?

  1. absorption of fructose into an intestinal epithelial cell is by facilitated transport and thus does not require energy
  2. metabolism of fructose generates more energy than glucose
  3. some fructose is already absorbed in the mouth and hence is the fastest way to get energy
  4. the presence of fructose aids in absorption of vitamin A, C, and D
  5. the presence of fructose inhibits reabsorption of glucose, which is then more readily available for muscle activity

Answer(s): A

Explanation:

Carbohydrate absorption occurs at enterocytes of the upper region of small intestinal villi. Fructose absorption is via the facilitated transporters GLUT5 across the apical enterocyte membrane and GLUT2 across the basolateral enterocyte. Glucose and galactose on the other hand are transported into enterocytes on carriers in combination with a sodium ion. The energy for this secondary active transport is provided by the electrochemical sodium gradient that is created by Na/K-ATPases. Experimental conditions that collapse the sodium electrochemical gradient, hypoxia, or poisoning of the Na-KATPase by ouabain inhibit glucose, but not fructose absorption. Nevertheless, the physiological importance of "saving energy" under extreme conditions such as mountain climbing through the use of fructose as energy source is questionable. For instance, fructose absorption is much slower than absorption of glucose and galactose.
The statements in choices B, C, D, and E are incorrect.



Patients with functional dyspepsia (disturbed indigestion) and prominent nausea frequently experience spurts of excessive acid exposure to the upper duodenum. This results in pancreatic secretion, mainly through the action of which of the following substances?

  1. cholecystokinin
  2. gastrin
  3. glucagon
  4. secretin
  5. vasoactive intestinal polypeptide (VIP)

Answer(s): D

Explanation:

The strongest stimulator for the release of secretin from cells in the upper small-intestinal mucosa is the contact with acidic chyme. Increased serum secretin levels stimulate water and alkali secretions from the pancreas and the hepatic ducts and inhibit gastrin release. The release of pancreatic enzymes is stimulated by cholecystokinin (choice A). The most potent stimulators for the release of cholecystokinin are not acid, but digestion products of fat and protein. Strong stimulators for gastrin secretion (choice B) are vagus nerve excitation, distention of the stomach, and protein digestion products. Gastrin then stimulates acid secretion and mucosal growth. The major effect of glucagon (choice C) is to increase blood glucose levels. Hence, it is secreted in response to hypoglycemia and protein digestion products, which are then used for gluconeogenesis. VIP (choice E) indeed stimulates intestinal and pancreatic secretion. However, it acts as neurotransmitter in the enteric nervous system and is mainly released by mechanical and neuronal stimulation.



A 41-year-old male suffering from unstable angina is undergoing cardiac testing. At rest, his heart rate is 70 beats per minute. His left ventricular end-diastolic volume is estimated to be 150 mL, while his end systolic volume is 80 mL. What is his estimated cardiac output?

  1. 3900 mL/min
  2. 4400 mL/min
  3. 4900 mL/min
  4. 5400 mL/min
  5. 5900 mL/min

Answer(s): C

Explanation:

Cardiac output is calculated by multiplying heart rate by stroke volume. Stroke volume is the difference between end-diastolic ventricular volume and end-systolic volume. Hence 150 - 80 = 70 mL stroke volume.
70 mL × 70 beats/min = 4900 mL/min.



A dislocated shoulder of a middle-aged man was corrected under conscious sedation with etomidate. Since etomidate has been reported to affect adrenocortical function as a side effect, the patient's ACTH and cortisol were measured in blood samples taken at the indicated times on the following day. What do the ACTH results indicate?

  1. inverted circadian pulsatile release
  2. normal circadian pulsatile release
  3. normal circadian non-pulsatile release
  4. primary adrenal insufficiency
  5. secondary adrenal insufficiency

Answer(s): A

Explanation:

The differences in the hormone concentrations of blood samples taken half-an-hour apart indicate pulsatile release. Hence, non-pulsatile release (choice C) is excluded. However, the normal circadian rhythm for ACTH (choice B) is likely to be highest early in the morning, with cortisol production following ACTH by a short delay, opposite to the patient's values. Such a reversed pattern could, for instance, be observed for night workers and is not likely to be caused by etomidate. Etomidate has been associated with suppression of glucocorticoid and mineralcorticoid synthesis in the adrenal cortex. Substantial damage to the adrenal gland might indeed lead to adrenocortical hypofunction, called Addison disease. However, primary adrenal insufficiency (disorder of the gland, choice D), and secondary adrenal insufficiency (inadequate ACTH secretion, choice E) both present with low levels of cortisol.



A biopsy from a 55-year-old woman with breast cancer was submitted to an estrogen receptor assay. The binding of estrogen to the cancer cells below figure, solid line was plotted in the form of a Scatchard plot and compared to control cells below figure, dotted line). Which of the following would you conclude for the patient's estrogen receptors?

  1. they are equal in number and have a higher affinity
  2. they are equal in number and have a lower affinity
  3. they are fewer in number and have a higher affinity
  4. they are fewer in number and have a lower affinity
  5. they are higher in number and have an equal affinity

Answer(s): B

Explanation:

The slope of the patient's line is decreased compared to normal. This indicates a low affinity of estrogen for the estrogen receptor, which makes choice B correct and choice Aincorrect. The x- intercept is called Bmax and represents the total number of estrogen receptors. This is unchanged in the patient compared to normal so that choices C, D, and E are excluded. The information about the numbers and status of estrogen and progesterone receptors on cancer cells is of high clinical value to determine the efficiency of consequent hormonal therapy.



Viewing page 33 of 171
Viewing questions 161 - 165 out of 845 questions



Post your Comments and Discuss USMLE STEP1 exam dumps with other Community members:

STEP1 Exam Discussions & Posts

AI Tutor 👋 I’m here to help!