Pathology Review for USMLE

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.

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.

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.

Answer: D (Necrosis).

Explanation: Given his symptoms (pressure sensation in his chest with radiation to the upper extremities and jaw) and his past medical history of hypertension and hyperlipidemia, he is most likely having an acute myocardial infarct. During an acute myocardial infarct, ischemia of the cardiac myocytes leads to necrosis. While his heart would have likely been enlarged (i.e., cardiac hypertrophy) given his history of hypertension, it would not be the cause of his chest pain. The heart essentially does not undergo hyperplasia as the cardiac myocytes are not capable of division. Ischemia represents an uncontrolled mechanism for causing cell death, and apoptosis, while some could be present, would not be a prominent response compared to necrosis. Atrophy of the heart would be more common in an elderly individual and would not necessarily lead to chest pain; also, given the history of hypertension, hypertrophy and not atrophy would be present.

Answer: B (an occlusive thrombus of the coronary artery with successful thrombolysis 15 minutes after its occurrence)

Explanation: the features listed (some swelling of the cell, some swelling of the rough endoplasmic reticulum and mitochondria, and a few myelin figures) are all consistent with a reversible injury. In a reversible injury the damage done is either relatively minor or quickly reversed or short term (i.e., the cell can recover). Although an occlusive thrombus of the coronary artery would cause the downstream myocytes to become ischemic, and thus, injured, if it is resolved quickly, blood flow will return and no permanent damage is done. Answers C-E would all cause permanent damage (i.e., cell death), and A. would not cause significant blockage of blood flow, and thus, most likely, no injury at all.

Answer: D (endometrial biopsies with menstrual phase endometrium)

Explanation: necrosis is a disorganized breakdown of dead cells whereas apoptosis is an organized breakdown of dead cells. While apoptosis does occur as a result of pathologic processes (e.g., some viral infections, some accumulations), it usually only occurs when the cellular death is on a cellular level; whereas, when the cellular death is on more of an organ level (e.g., a relatively large segment of an organ), necrosis is the predominant form. Apoptosis also occurs as a result of physiologic processes. During growth of an organism, death of cells must take place at times, but, apoptosis does so without inducing inflammation, which is a damaging process. After endometrium thickens in the proliferative phase, if no implantation occurs, the cells must die, and the best way is via apoptosis. A, B, C, and E are organ-level insults, and would trigger predominantly necrosis.

Answer: A (hypoxia)

Explanation: given her heavy menstrual bleeding, she has developed iron deficiency anemia. Being anemic, she has decreased oxygen carrying capacity in her blood, and with exercise, is unable to meet the oxygen requirements of her active tissue, and is thus hypoxic. Her blood flow would be normal; thus, she is not ischemic. And, given the scenario presented, there is no indication of thrombosis, an immunologic reaction, or sepsis.

Answer: B (ischemia).

Explanation: the patient is having an acute myocardial infarct. The left anterior descending coronary artery has a thrombus that is obstructing blood flow to the myocardium, causing ischemia. While hypoxia is a component of ischemia, as, when blood cannot flow to the tissue, oxygen cannot be delivered; however, the better answer is ischemia, which is reduced blood flow. While trauma, immunologic processes, and infectious agents can affect the heart and potentially cause a myocardial infarct, given the obstruction to the blood flow described and that the obstruction is most commonly due to atherosclerosis, C, D, and E are not the best answer.

Answer: B (Na/K pump).

Explanation: The lack of blood flow to the cardiac myocytes (i.e., ischemia) would restrict production of ATP by the mitochondria. Without ATP, the Na/K pump would not be able to effectively maintain the normal sodium gradient across the plasma membrane; sodium would enter the cell, followed by water, leading to swelling of the cell. The primary role of the Na/K pump is to maintain this gradient, and it is the primary system responsible for maintaining the gradient. The Na/Ca counter-transporter would transport sodium into the cell (and calcium out). The Na/H counter-transporter and Na/glucose co-transporter would transport sodium into the cell and are more commonly found on renal and intestinal cells and not cardiac myocytes. Cyclin D plays a regulatory role in the cell cycle and is not responsible for maintaining the sodium gradient, and thus, would not contribute significantly to cellular swelling.

Answer: E (muscle).

Explanation: Desmin is a type of intermediate filament that is found in the cytoskeleton of muscle cells. Desmin forms scaffolding for the contraction of actin and myosin. The other tissues listed (hyaline cartilage, squamous epithelium, mucous producing glandular epithelium, and brain) should not stain positive for desmin as this form of intermediate filament would not be abundantly present, such as it is in muscle.

Answer: C (intermediate filaments).

Explanation: intermediate filaments are components of the cytoskeleton. Common intermediate filaments include vimentin, desmin, neurofilaments, glial fibrillary acidic protein, and cytokeratins. The presence of a specific intermediate filament within a tumor type can help a pathologist identify the cell of origin, and thus, often, the tumor type.

Answer: A (hypoxia)

Explanation: carbon monoxide binds to hemoglobin, restricting the normal binding of oxygen; thus, although blood is flowing through the vessels, an insufficient amount of oxygen can be delivered to the tissues. The man was hypoxic. There would be no arterial or venous obstruction, and thus, ischemia would not occur. Also, carbon monoxide poisoning in the short term does not cause thrombosis, an immunologic reaction, or sepsis.