Month: October 2019

Answer: C (Tyrosine kinase)

Explanation: patients with X-linked agammaglobulinemia will present after the age of 6 months with recurrent infections, including with Streptococcus pneumoniae, Haemophilus influenzae, and enteroviruses, which are acquired through the gastrointestinal tract. Laboratory testing will reveal very low concentrations of the immunoglobulins and very low levels or absence of B cells in the blood. X-linked agammaglobulinemia is due to lack of Bruton’s tyrosine kinase, the loss of which inhibits the production of light chains, and subsequently the production of intact immunoglobulins. Heavy chains are still produced.

Answer: E (rheumatoid arthritis)

Explanation: the history of recurrent sinus and gastrointestinal infections, and the laboratory test results, are consistent with a diagnosis of selective IgA deficiency, which presents after the age of 4 years, but can be asymptomatic also. These patients are at increased risk for allergies and the development of autoimmune disorders, which would include rheumatoid arthritis.

Answer: B

Explanation: recurrent episodes of diarrhea and sinopulmonary infections first diagnosed in an older child are consistent with a selective IgA deficiency. Laboratory testing would reveal normal concentrations of IgG and IgM and normal number of T cells.

Answer: B (T lymphocytes)

Explanation: Hypocalcemic tetany (i.e., the muscular spasms) presenting shortly after birth is one feature of DiGeorge syndrome. Another feature is congenital heart disease, which can include several conditions, of which a ventricular septal defect is one. DiGeorge syndrome will produce a T cell deficiency, but does not affect the other cell types of the immune system.

Answer: E (plasma cells)

Explanation: given the history of recurrent otitis media and pulmonary infections occurring in an adolescent female, common variable immunodeficiency is a likely diagnosis. In this disease process, B cells are produced; however, they do not differentiate into plasma cells, and patients are hypogammaglobulinemic. In the normal lymph node, all of the above cell types are present; however, as patients with common variable immunodeficiency do not produce plasma cells, they would not be identified in a biopsy from this patient.

Answer: C (very small tonsils)

Explanation: the patient has X-linked (Bruton’s) agammaglobulinemia, which presents in males after the age of 6 months. Patients develop recurrent lung and sinus infections, most commonly with encapsulated organisms. The laboratory diagnosis is made by confirming very low levels of immunoglobulins and absence of B cells in the blood. In addition to an inability to develop lymph nodes, patients will often have very small tonsils. The other conditions listed are not characteristically associated with X-linked agammaglobulinemia.

Answer: C (X-linked)

Explanation: the development of recurrent pulmonary infections with encapsulated organisms in a male greater than 6 months of age is consistent with Bruton’s agammaglobulinemia. The essential absence of two or more types of Ig and the absence of B cells by flow cytometry, as confirmed by the absence of CD19+ cells (CD19 is a B cell marker), confirms the diagnosis. Bruton’s agammaglobulinemia is X-linked, and, thus, present in males. As the presence of mother’s antibodies protects the infant for a time period after birth, this disease usually presents after the age of 6 months.

Answer: C (phagocyte oxidase)

Explanation: the patient has chronic granulomatous disease, which is most commonly X-linked and thus presents in males, usually in early childhood. Patients develop granulomas, because the defective phagocyte oxidase does not allow them to produce reactive oxygen species, such as superoxide radical, and thus their ability to kill bacteria is diminished. Instead, the body’s macrophages respond and wall off the infections, creating granulomas. Integrin defects (Leukocyte adhesion deficiency) and LYST defects (Chediak-Higashi syndrome) are not commonly associated with the formation of granulomas.

Answer: D (abnormal fusion of phagosomes and lysosomes)

Explanation: the albinism and recurrent bacterial infections are consistent with Chediak-Higashi syndrome. Patients with Chediak-Higashi syndrome can enter an accelerated phase, which can include jaundice, neurological changes, fever, hepatosplenomegaly, and lymphoadenopathy. Chediak-Higashi is autosomal recessive involving the LYST gene, which produces a protein that plays a role in lysosomal trafficking, resulting in abnormal fusion of phagosomes and lysosomes. Patients can have enlarged granules in the cytoplasm of neutrophils, which represent the abnormal fusion of the phagosome and the lysosome.

Answer: C (Caseous necrosis)

Explanation: given the fact that he is a homeless chronic alcoholic with a persistent cough, tuberculosis is a distinct possibility. When the disease process erodes into a vessel, patients will often develop hemoptysis. Of the above, caseous necrosis is associated with tuberculosis. Liquefactive necrosis could occur in a person with severe necrotizing pneumonia, which chronic alcoholics also can get; however, given the more chronic nature of his symptoms, this is less likely. Fat necrosis essentially does not occur in the lungs and would not be associated with hemoptysis. An infarct could be associated with hemoptysis and could occur associated with a pulmonary thromboembolus; however, he has no other symptoms, no past medical history, and no other risk factors. Fibrinoid necrosis is associated with some forms of vasculitis and is possible; however, much less common than tuberculosis in this patient population.

Answer: B (Gangrenous necrosis)

Explanation: Gangrenous necrosis is a term used most commonly for coagulative necrosis of the lower extremities (if there is no bacterial infection, the term dry gangrene can be used and if there is superimposed bacterial infection, the term wet gangrene can be used. Wet gangrene would essentially be coagulative necrosis with superimposed liquefactive necrosis). Gangrenous necrosis can also be used for the pelvic region (i.e., Fournier’s gangrene) and can also be used for some lesions of the gastrointestinal tract (e.g., gas gangrene). Gangrene most often occurs secondary to ischemic injury, such as could occur in diabetics, who often have vascular damage.

Answer: E (swollen hepatocytes)

Explanation: the well-defined circular clear spaces are vacuoles, containing triglyceride (i.e., steatosis). Fat accumulation is a sign of reversible injury. For example, microscopic examination of a person who binge drinks one weekend would reveal steatosis in the hepatocytes; however, days later, the steatosis would clear, unless they drank heavily again. Cytoplasmic swelling is also a feature of reversible cell injury. A pale nucleus (karyolysis) and a fragmented nucleus (karyorrhexis) are signs of irreversible injury. Amyloid deposition in the liver would be rare, and would not necessarily accompany steatosis. Mallory’s hyaline is best associated with alcoholic liver disease; both Mallory’s hyaline and steatosis can be seen in the same liver; however, most livers with steatosis will not have Mallory’s hyaline.

Answer: E (a kidney).

Explanation: The description fits that of an infarct (i.e., an area of coagulative necrosis). In organs with a single blood supply, such as a kidney (but also the spleen and the heart), the infarct will be yellow-white, yellow-tan, or otherwise pale. As infarcts are due to ischemia most commonly, no blood is passing to the dead tissue and therefore, it will be more pale, and not hemorrhagic, such as when an infarct occurs in an organ with dual blood supply (e.g., the liver, the lungs, or the colon). An infarct in the brain will always have markedly soft tissue, as, on a gross level, the brain undergoes liquefactive necrosis, with dissolution of architecture, and not coagulative necrosis, in which tissue architecture is preserved for some time after the infarct occurs.

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.