QUESTION 10
A researcher wants to develop a drug that prolongs the life of neutrophils, knowing that the survival time of a neutrophil after release from the bone marrow is days, and thinking that a prolonged survival time may assist in combating infection. Which of the following drug mechanisms would be most likely to accomplish this goal?
A. Inhibition of bcl-2 function
B. Enhancement of bcl-2 function
C. CD31 agonist
D. CD31 antagonist
E. Competitive inhibitor of phosphatidylserine binding
Answer for Question 10
Answer: B (Enhancement of bcl-2 function)
Explanation: at the end of the acute inflammatory process, the neutrophils die. This is accomplished through apoptosis. A trigger for apoptosis is release of cytochrome c from the mitochondria, which activates caspases. Bcl-2 inhibits this release, therefore, a drug that enhanced the function of bcl-2 would inhibit apoptosis and provide for a longer life span of the neutrophil. (A) being the opposite is therefore, incorrect. CD31 (PECAM) plays a role in neutrophil migration through the wall of the vessel to reach the site of the inflammatory trigger, but an agonist or antagonist of CD31 would not prolong the life of the neutrophil. Phosphatidylserine plays a role in apoptosis being the marker that identifies what debris is to be consumed by phagocytes. A competitive inhibitor would inhibit this uptake and affect the process of apoptosis (i.e., impairing its orderly progression), but it would not affect the life span of a neutrophil.
QUESTION 11
A 17-year-old football player fractures his left femur during a game. The fracture is open and requires a surgical procedure to repair. Following the surgery, he is unable to move his left lower extremity for 4 weeks, but physical therapy is performed on his right lower extremity so as not to lose muscle tone. When he is first able to be mobile, measurements are taken of his left and right thigh, and the left thigh is noted to be 5 cm smaller in circumference than the right thigh. Of the following, which most likely played the major role in the development of this size discrepancy?
A. Coagulative necrosis
B. Liquefactive necrosis
C. Compensatory hyperplasia
D. Autophagy
E. Apoptosis
Answer for Question 11
Answer: D (autophagy)
Explanation: with immobility, the use of the muscles in the left thigh will have decreased physiologic stimulus and thus will undergo atrophy (i.e., decrease in size), which accounts for the decrease in size of the left thigh. There is no increase in number of skeletal muscle cells possible, since they do not divide; thus compensatory hyperplasia will not occur, and, if it did occur, this would result in an increase in the size of the muscle, not decrease. Atrophy is an adaptation–the cell is adapting to its new environment by decreasing in size. Cell death would not occur, so A, B, and E would not play a major role in the development of the size discrepancy. One mechanism by which atrophy occurs is autophagy.
QUESTION 12
A 52-year-old male presents to the emergency room with chest pain, which began 15 minutes ago. Cardiac catheterization is performed and reveals an occlusive thrombus in the left anterior descending coronary artery. In the myocardium downstream of the change in the coronary artery, the activity of which of the following enzymes would have the most significant decrease in activity?
A. Cytochrome c oxidase
B. Lactate dehydrogenase
C. Glucokinase
D. Pyruvate kinase
E. HMG-CoA reductase
Answer for Question 12
Answer: A (cytochrome C oxidase)
Explanation: a thrombus in the coronary artery would block blood flow through the coronary artery leading to ischemia of the downstream myocardium. Without oxygen, oxidative phosphorylation to produce ATP would not occur. Cytochrome c oxidase is a component of the electron transport chain, which performs oxidative phosphorylation; therefore, the activity of this enzyme would be reduced in this scenario. Lactate dehydrogenase reduces pyruvate to lactate, regenerating NAD+. It’s activity would increase in anaerobic conditions. Glucokinase converts glucose to Glu-6-P, the first step in glycolysis. Pyruvate kinase produces ATP in the final step of anaerobic glycolysis. Neither would have a decrease in activity more significant than the impairment of the electron transport chain. HMG-CoA reductase is the rate-limiting step in the production of cholesterol.