Pathology Review for USMLE

Answer: E (X-linked agammaglobulinemia)

Explanation: X-linked agammaglobulinemia occurs in males and presents after the age of 6 months, once passively transferred maternal antibodies are no longer available. Patients present with recurrent infections with encapsulated bacteria, and can also develop some viral infections. Laboratory testing will reveal very low concentrations of immunoglobulin and absence of B cells, since the defect in X-linked agammaglobulinemia prevents the maturation of B cells from pre-B cells. Chronic granulomatous disease affects the ability of white blood cells to make activated oxygen compounds for the killing of bacteria; in many cases it is also X-linked and would be present in males, but the diagnosis is usually made in early childhood. Selective IgA deficiency involves only a deficiency of IgA and presents later in life. Leukocyte adhesion deficiency involves a protein that enables neutrophils to adhere to the endothelium. Symptoms begin in infancy, and at birth, delayed detachment of the umbilical cord can occur, but the disease would not cause low concentrations of immunoglobulin, or absence of B cells. In X-linked severe combined immunodeficiency, most patients present at less than 6 months, with failure to thrive due to disseminated infections such as candidiasis and Pneumocystis. Laboratory testing would reveal decreased T cells. The number of B cells can be normal or increased.

Answer: C (Giardia lamblia induced diarrhea)

Explanation: given that the patient is male and greater than 6 months of age, and presenting with recurrent bacterial infections of the lung and ear, including with an encapsulated organism (Haemophilus influenzae), and with the laboratory testing revealing an absence of B cells, the clinical presentation would be consistent with a diagnosis of X-linked agammaglobulinemia. Patients with X-linked agammaglobulinemia are most susceptible to infections with encapsulated bacterial organisms (Streptococcus pneumoniae, Haemophilus influenzae, Pseudomonas aeruginosa, Neisseria meningitidis), GI parasites (Giardia and Cryptosporidium) due to lack of IgA, and enteroviruses (which enter the body through the gastrointestinal tract). Infection with Epstein-Barr virus requires the presence of B cells.

Answer: D (Selective IgA deficiency)

Explanation: given the past history of sinusitis and the age of the patient, selective IgA deficiency is the best choice. In patients with selective IgA deficiency, transfusion with blood can lead to the development of anti-IgA antibodies, which, when transfused again, can lead to an anaphylactic reaction. X-linked agammaglobulinemia as well as Wiskott-Aldrich are X-linked and would not normally present in a female, and Wiskott-Aldrich normally presents in infants. Severe combined immunodeficiency would present in infancy and, while ataxia-telangiectasia is autosomal recessive and can present in females and at a later age, her past medical history does not include mention of ataxia or telangiectasias.

Answer: A (normal IgG; normal IgM)

Explanation: given his age and the fact that he developed two Giardia lamblia infections, which is an intestinal disease process, a selective IgA deficiency is possible, which is confirmed by the laboratory testing. Patients with selective IgA deficiency also are at risk for the development of an autoimmune disease. Given his age and the presence of the autoimmune disorder, a common variable immunodeficiency would also be possible, and with this condition, the IgG concentration could be decreased; however, as the patient mounted an appropriate response to his vaccinations, common variable immunodeficiency is much less likely as these patients usually have a poor response to vaccination. In selective IgA deficiency, the IgG and IgM concentrations will usually be normal.

Answer: B (hyperplastic follicles)

Explanation: given the age of the patient and recurrent infections of the sinopulmonary and GI tract, combined with the laboratory testing, the most likely diagnosis is a common variable immunodeficiency, which can present in females and normally presents later in life. In common variable immunodeficiency, the number of B cells is increased; however, no differentiation to plasma cells occurs. B cells can proliferate in response to an infection, so areas that contain B cells (e.g., lymphoid follicles) can be hyperplastic. These patients do not overproduce a protein, so amyloidosis would not be expected.

Answer: B (autosomal recessive)

Explanation: the child has ataxia-telangiectasia. This disease process often presents after the age of 5 years. His difficulty walking is a manifestation of the ataxia and the small nodules on his eyelid and ear are telangiectasias. Patients often have decreased IgA. Ataxia-telangiectasia is inherited in an autosomal recessive pattern.

Answer: C (Alpha-fetoprotein)

Explanation: the red nodules on her eyelids and skin are telangiectasias, and her difficulty walking and slurred speech are manifestations of ataxia. Patients with ataxia-telangiectasia are at risk for development of neoplasms, especially lymphoid neoplasms. Individuals with ataxia-telangiectasia can have decreased concentrations of IgA and IgG2. An increased concentration of alpha-fetoprotein is used in the diagnosis of ataxia-telangiectasia.

Answer: A (22q11)

Explanation: hypocalcemia and heart defects are characteristic of Tetralogy of Fallot. The disease can be either complete or partial, depending upon the involvement of the immune system. In partial DiGeorge syndrome, patient’s have preserved T cell function, and their hypoparathyroidism can be treated with calcium and vitamin D supplementation. Congenital heart defects can include Tetralogy of Fallot, truncus arteriosus, and atrial or ventricular septal defects. The deletion occurs at 22q11.

Answer: B (thymic aplasia)

Explanation: hypocalcemic tetany (i.e., the muscle spasms) occurring shortly after birth, in association with hypocalcemia and in the background of Tetralogy of Fallot is consistent with DiGeorge syndrome, with the hypocalcemia being due to parathyroid gland hypoplasia. Patients also have thymic abnormalities, manifested as thymic aplasia and resultant defects in T lymphocyte function, predisposing to infections with viruses, fungi, parasites, and intracellular bacteria. The B cell function is un-disturbed, and thus, patients will have normal numbers of B lymphocytes. Scoliosis is not a characteristic feature of this disease process, and they are most at risk for fungal, parasite, and viral infections. While the aorta can be abnormal (i.e., an interrupted aortic arch), absence of the aorta would be incompatible with life.

Answer: A (acute congestion)

Explanation: thrombosis of major hepatic veins should present acutely as there is greatly reduced ability for blood to exit the liver, effectively blocking its flow. With obstruction of venous outflow from the liver, blood will back up into the central veins, which is a passive process, and hence congestion. Given that it happened suddenly, it is an acute process. So, the appropriate process would be acute congestion. Hyperemia is an active process, such as occurs with vasodilation during acute inflammation. Chronic hyperemia would essentially never occur, because it would require an active process to constantly stimulate dilation of the vessels.

Answer: B (Oxygen)

Explanation: in hyperemia, patients have an active process that causes dilation of blood vessels (e.g., acute inflammation). Because of the dilation of the blood vessels, the vessels become filled with oxygenated blood, which imparts a lighter red-pink coloration to the tissue. In congestion, patients have a passive process that causes dilation of blood vessels (e.g., failure of the heart leading to back-up in the venous system). The blood that accumulates is deoxygenated blood, which will impart a dark red blue discoloration to the tissue.

Answer: A (increased hydrostatic pressure)

Explanation: with congestive heart failure, failure of the ventricles to be able to effectively pump blood leads to congestion within the venous systems (i.e., systemic venous system with failure of the right side of the heart and pulmonary venous system with failure of the left side of the heart). An increased amount of blood in the veins will distend the vessel, thereby increasing the hydrostatic pressure, which will promote the leakage of fluid from the vessel into the surrounding tissue and hence, edema.

Answer: A (increased hydrostatic pressure)

Explanation: a pulmonary thromboembolus is a complication of a venous thrombosis, most commonly of the deep veins in the lower extremities. Because of the thrombosis of the vein, blood will pool in the venous system distal to the thrombosis as it cannot go around the site of blockage. An increased amount of blood in the veins will distend the vessel, thereby increasing the hydrostatic pressure, which will promote the leakage of fluid from the vessel into the surrounding tissue and hence, edema of the lower extremities, which would appear on physical examination as an increased girth of the affected extremity.

Answer: B (increased hydrostatic pressure; decreased plasma osmotic pressure)

Explanation: a certain proportion of alcoholics develop cirrhosis of the liver and patients with cirrhosis of the liver can develop ascites (accumulation of watery fluid in the abdominal cavity). The scarring of the liver associated with cirrhosis causes some degree of obstruction of venous blood flow through the liver, which contributes to increased hydrostatic pressure, leading to edema. In addition, the liver produces albumin, which is the main protein in the blood and hence, the largest contributor to the plasma osmotic pressure. Damage to the liver leads to a decreased production of albumin and thus, decreased plasma osmotic pressure, leading to edema.

Answer: F (lymphatic obstruction)

Explanation: with a lymph node dissection, the flow of lymph can be disrupted; as lymph serves to return fluid to the heart that does not make its way back into the venous vasculature, obstruction of lymphatic flow will cause edema, which could produce enlargement of the right upper extremity. While it could be argued that increased hydrostatic pressure due to loss of veins and thereby resultant obstruction of venous flow, which would lead to edema, could result, during the surgery, preservation of the vasculature as best as possible could occur, whereas, a lymph node dissection, by its nature, would be damaging to the lymphatic system. Current techniques very greatly minimize the risk of such occurrences (i.e., lymphatic obstruction post surgery); however, understanding the complication and its mechanism is still useful.

Answer: E (petechiae)

Explanation: various terms are used to describe hemorrhages in the skin and mucous membranes. These terms are based upon the size. Pinpoint hemorrhages (1-2 mm) are referred to as petechiae. Purpura are larger (3-5 mm) and can be raised (i.e., palpable). Ecchymoses are even large (1-2 cm or more). A hemorrhage is a non-specific term that applies to all three. Bruising is another term for ecchymosis.

Answer: C (adhesion).

Explanation: the child has leukocyte adhesion deficiency, the most common form of which is due to an autosomal recessive mutation involving the β sub-unit of CD18. CD18 is a component of integrins (both LFA-1 and MAC-1). LFA-1 is present on neutrophils and plays a role in adhesion. ICAM-1 on the activated endothelium binds to LFA-1 on the neutrophil causing the neutrophil to adhere to the endothelium. Margination occurs because of vasodilation. Rolling is due to the interaction of selectins and sialyl-Lewis X. Migration through the endothelium involves CD31 (PECAM-1) and agents of chemotaxis (i.e., movement of the leukocyte to the site of injury) are C5a, LTB4, and bacterial peptides.

Answer: D (22q11.2)

Explanation: cardiac defects and hypocalcemia along with T cell immunodeficiency and facial abnormalities are consistent with 22q11.2 deletion syndrome. Previous terminology included DiGeorge syndrome, when the main manifestations were the hypocalcemia and the T cell immunodeficiency, and velocardiofacial syndrome, when the main manifestations were facial dysmorphology and congenital cardiac disease. Both were due to deletions at 22q11.2, with the exact location of the deletion determining the subsequent phenotype of the patient.

Answer: A (histamine)

Explanation: because of the fall and striking her left elbow, this girl has initiated an acute inflammatory response. The first manifestation is usually swelling, which is due to vasodilation of the vasculature and increased membrane permeability through separation of endothelial cells, which allows for fluid to leave the blood vessel and enter the surrounding soft tissue, hence, the swelling. Histamine is an important mediator in causing this change. CD11a/C18 plays a role in attachment of neutrophils to the endothelium; leukotriene B4 is a chemoattractant; Toll-like receptor plays a role in the body’s ability to identify microbes, and C3b is an opsonin, which allows phagocytes to ingest microbes and other foreign substances.

Answer: B (a purulent exudate)

Explanation: one complication of pneumonia is pleuritis. In this case, the left pleural cavity was filled with pus (i.e., an empyema). Being cloudy, the fluid is filled with cells. A serous fluid is watery and should be clear. An exudate has an increased amount of protein and most often cells; whereas a transudate is mostly just fluid, with a low protein content and essentially no cells. So, fibrinous and transudate should be mutually exclusive as would purulent and transudate, as a purulent fluid is one filled with neutrophils. A serous exudate is also mutually exclusive.

Answer: C (P-selectin)

Explanation: the researcher hopes to block the rolling stage of acute inflammation. In the rolling stage, Sialyl-Lewis-X molecules on the neutrophils bind to E- or P-selectin molecules on the surface of the endothelium causing loose binding of the white blood cells to the capillary wall. LFA-1 (on neutrophils) binds to ICAM-1 on endothelial cells, leading to adhesion. CD31 (PECAM) plays a role in transmigration and Toll-like receptor plays a role in recognition of microbes.

Answer: E (endothelial necrosis)

Explanation: both individuals have evidence of increased vascular permeability, with fluid leaking from the vessel into the surrounding tissue. Increased vascular permeability can occur by two mechanisms: endothelial retraction or endothelial damage. Endothelial retraction as mediated by histamine rapidly leads to increased vascular permeability but the effect is short lived, such as occurred in the individual with the bruise. Endothelial damage occurs rapidly, but is long lived, requiring replacement of the damaged endothelium during the repair process. This can occur in burns, or due to the effects of a bacterial, or other microbial infection. While a bacterial superinfection can occur in a burn patient, it is not required to have continued loss of fluid.

Answer: D (1 day)

Explanation: although an acute inflammatory infiltrate begins immediately after the time of injury, the gross and microscopic changes take some time to develop. As vasodilation and fluid exudation is rapid, red discoloration and swelling can be seen within minutes; however, development of the neutrophilic response described would take a little time. Following the acute inflammatory response, macrophages would move in to clean up the mess. After 5 seconds and 15 minutes, no neutrophils would be expected. While neutrophils may be present at 3 hours, a significant infiltrate on microscopic examination would not be identified; at 1 week, the neutrophilic response should have ended (usually by a few days), the macrophage response should be complete or nearly completed (with macrophages beginning to enter the picture around 2-3 days).

Answer: D (new blood vessel formation).

Explanation: the first three answers (neutrophilic infiltration, edema of the tissue, and dilated blood vessels) are found in acute inflammation. New blood vessel formation along with fibrosis, infiltrates of mononuclear cells (e.g., macrophages and lymphocytes), and tissue destruction are features of chronic inflammation.

Answer: B (IFN-γ)

Explanation: the macrophages present at the infarct would be classically activated macrophages, responsible for removing the dead cells. IFN-γ is a major mediator for activating macrophages through this pathway. IL-13 activates macrophages through the alternative pathway, and macrophages activated via this pathway (M2 macrophages) promote wound repair, including fibrosis and angiogenesis. Eotaxin is chemoattractive for eosinophils and C5a stimulates release of histamine, and is also chemoattractive for neutrophils, monocytes, and basophils. TGF-β is secreted by activated macrophages and promotes fibrosis during wound repair.

Answer: E (epithelioid histiocytes)

Explanation: a granuloma is composed of epithelioid histiocytes, which are large eosinophilic macrophages [histiocyte is another term for macrophage]. While granulomas can individually possess any of the other features listed (giant cells, a fibrous rim, a rim of lymphocytes, or caseous necrosis), the only feature listed that all granulomas possess is the epithelioid histiocytes.

Answer: A (IL-1)

Explanation: IL-1 (also TNF and IL-6) are responsible for producing the acute phase response in patients with acute infections. The acute phase responses include fever, elevated acute phase proteins (e.g., C-reactive protein and fibrinogen), leukocytosis, and also increased heart rate and blood pressure.

Answer: E (basement membrane).

Explanation: With the squamous epithelium of the skin having such good regenerative capabilities, injury to the epidermis will result in complete resolution, however, once the injury perforates the basement membrane, complete resolution is no longer possible and a scar will result.

Answer: B (physiologic hypertrophy)

Explanation: as a member of the track and field team, hypertrophy of the muscle would be expected. As skeletal muscle essentially cannot divide, hyperplasia would not be present. Because athleticism is considered normal stimulation of the musculature, the changes would be considered physiologic and not pathologic.

Answer: B (cardiac hypertrophy)

Explanation: given his past medical history of hypertension for 15 years, he is most likely to have some degree of cardiac hypertrophy. The cardiac myocytes essentially do not divide, so cardiac hyperplasia would not occur. While in some cases, cells within the lung can divide, pulmonary hyperplasia and pulmonary hypertrophy are not often described; also, with only a past medical history of hypertension, involvement of the lungs would not be likely.

Answer: B (β-myosin heavy chain)

Explanation: β-myosin heavy chain is the fetal form of the myosin heavy chain, and is more effective in its use of energy and slower in its contractions, and hence, in a hypertrophied heart, as would occur in a patient with hypertension, and as a 540 gram heart represents, would be in increased amount. The switch from the adult α-myosin heavy chain to the fetal β-myosin heavy chain represents the body’s adaptation to the increased stress placed on the heart by the systemic hypertension.

Answer: C (atrophy)

Explanation: Stenosis of the left femoral artery would cause ischemia of the down-stream musculature. Because the stenosis has developed slowly, the skeletal muscle cells would have a chance to adapt to the change; therefore, necrosis would not be expected. Skeletal muscle cells essentially do not divide, so hyperplasia could not occur–also, ischemia would decrease the amount of oxygen and nutrients available to the cells, so there would be no stimulus for hyperplasia or hypertrophy. Metaplasia does not occur in this context (i.e., there is not a better cell type that the skeletal muscle cells can switch to to better handle the decreased blood flow). The skeletal muscle cells would undergo atrophy.

Answer: D (pathologic endometrial hyperplasia)

Explanation: the elevated concentration of estrogen would promote division and proliferation of the endometrial glandular epithelium (hyperplasia). Because the stimulus for the proliferation is a tumor, the hyperplasia would be considered pathologic. Physiologic hyperplasia of the endometrium occurs during the proliferative phase of the menstrual cycle.

Answer: E (vascular stenosis)

Explanation: some of the skeletal myocytes are smaller than normal, hence the decreased size on cross section with the resultant more closely packed nuclei. This is atrophy. While in some way, each of the choices could potentially be associated with skeletal muscle atrophy (e.g., due to wasting, or nerve involvement), the only condition listed that could be consistently associated with atrophy of the skeletal muscle would be vascular stenosis, which would result in ischemia of the downstream tissue.

Answer: B (squamous metaplasia of the bronchi)

Explanation: due to prolonged exposure to the toxins in cigarette smoke, the lining of the trachea and bronchi can switch from respiratory type epithelium to squamous metaplasia, as the squamous metaplasia is better designed to handle such exposures. With no history of a condition producing elevated serum concentrations of calcium (e.g., parathyroid gland adenoma) metastatic calcification would not occur. While dystrophic calcification could be present as aged tissue is damaged to some extent and can calcify, it should not be extensive in distribution. Patients who smoke can develop chronic bronchitis, which would be associated with hyperplasia of submucosal glands. With no history of a neuromuscular condition predisposing to aspiration, chronic aspiration is very unlikely, also, if gastric metaplasia did occur in patients, it would be very rare to find.

Answer: A (abundant anthracotic pigment and focal squamous metaplasia)

Explanation: a patient with a long history of heavy cigarette use would be expected to have prominent amount of anthracotic pigment (i.e., carbonaceous pigment). In addition, with long term exposure to cigarette smoke, the respiratory epithelium can undergo squamous metaplasia. This process can lead to the development of a squamous cell carcinoma in the lung (i.e., one form of non-small cell carcinoma). Respiratory metaplasia does not occur, as the respiratory epithelium is the expected normal type of epithelium. Honey comb change can have respiratory epithelium, but this would be expected in a person with a history of interstitial lung disease. Steatosis is most commonly encountered in the liver, but also the heart and skeletal muscle, but essentially not in the lung. Russell bodies would be found in plasma cells, and plasma cells could be found in the lung in this case; however, the number of Russell bodies would not be prominent and scant anthracotic pigment would not be what was expected.

Answer: C (menstrual phase endometrium)

Explanation: apoptosis is an orderly destruction of a cell. During the menstrual cycle, the endometrium thickens in preparation for implantation; however, if implantation does not occur, the proliferated tissue must be removed. Apoptosis would allow for an orderly removal of tissue without promoting a significant inflammatory infiltrate. As a part of apoptosis, the nucleus is fragmented, and the DNA when run on a gel will then appear as numerous discrete bands whereas in necrosis, which is uncoordinated destruction of the nucleus, the DNA run on a gel would appear more as a smear.

Answer: C (healed bruise)

Explanation: the pigment described is hemosiderin, which accumulates due to the breakdown of red blood cells. A collection of hemosiderin would be associated with a healed bruise (due to the breakdown of red blood cells that would have accumulated at the site). While tattoo pigment can occur in the dermis, it would be more of a finely stippled black coloration, as would melanin, which would be finely stippled and brown to black in color. Lipofuscin, which is a finely stippled yellow-brown pigment, is associated with increasing age; however, the dermis is not a location where it is identified, instead it is commonly found in the heart, liver, and neurons. A parathyroid adenoma would be associated with metastatic calcification; however, deposits of calcium are basophilic.

Answer: E (caspases)

Explanation: given that the inflammatory process has been going on for around three days, some of the neutrophils should be at the end of their life expectancy, and will be undergoing apoptosis. Nuclear fragmentation is characteristic of apoptosis. Histamine, CD31, and LFA-1 play a role in the inflammatory process, vasodilation, transmigration (diapedesis), and leukocyte adhesion respectively, but do not play a direct role in apoptosis. Bcl-2 inhibits apoptosis. Caspases are activated in apoptosis, by either the release of cytochrome c or through adaptor proteins activated by Fas-FasLigand binding. Caspases activate endonucleases (among other enzymes), resulting in nuclear fragmentation.

Answer: A (phosphatidylserine).

Explanation: the description of the cells fits that of apoptosis (i.e, shrunken cells with fragmented nuclei, and with no associated adjacent neutrophilic infiltrate). Apoptotic cells express phosphatidylserine on their surface as a signal to phagocytic cells that they should be engulfed and removed, and thus, do not trigger an acute inflammatory reaction. The other answers are incorrect.

Answer: A (lysis of BH3 proteins)

Explanation: cells exposed to radiation develop DNA damage and can undergo apoptosis. The mitochondrial pathway of apoptosis is one of two pathways by which apoptosis can occur, but is the pathway most commonly utilized, including by radiation induced damage. Cell injury, including DNA damage, activates BH3 sensors, which promotes Bax and Bak, which lead to release of cytochrome C from the mitochondria and subsequent activation of caspases. Lysis of BH3 proteins would directly inhibit apoptosis. Bcl-2 inhibits apoptosis, so antagonists would block this negative influence, and better allow for apoptosis to occur. While FAS ligand binding to FAS promotes apoptosis through the death receptor pathway, and an antagonist should inhibit this pathway, the death receptor pathway is not as commonly used as the mitochondrial pathway. Binding to phosphatidylserine, depending upon its effects, could inhibit or promote the cleaning up of cells that have undergone apoptosis, but it would not affect whether or not a cell became apoptotic. Apoptosis is an important mechanism by which the body can remove dead or damaged cells. In the case of radiation induced DNA damage, hindering apoptosis to allow for the survival of the cells would not necessarily be a good idea.

Answer: C (production of hypoxia-inducible factor 1 (HIF-1))

Explanation: stenosis of the femoral artery would lead to ischemia of the lower extremity. To adapt to ischemia, the skeletal muscle cells would produce hypoxia-inducible factor 1, which would then induce several changes to lead to increased blood flow, including the up-regulation of VEGF, to promote develop of new vasculature. HIF-1 would also inhibit oxidative phosphorylation and promote aerobic glycolysis.

Answer: D (increased production of reactive oxygen species)

Explanation: the patient has sustained a myocardial infarct. Following lysis of the clot, blood will flow to the damaged tissue; however, because the tissue is damaged, an increased number of reactive oxygen species may be generated, potentially from dysfunction of the electron transport chain, or through increased infiltration with inflammatory cells. This process is termed reperfusion injury. The release of troponin I indicates necrosis has occurred, with release of enzymes from the cell. Inhibition of bcl-2 and release of FAS ligand would increase apoptosis, which may play a role, but would be less significant, if they occurred, than the increased production of reactive oxygen species.

Answer: C (superoxide)

Explanation: the description is consistent with lipofuscin. Lipofuscin, a finely stippled yellow-brown pigment, also called wear and tear pigment, is commonly found in the heart of older individuals and in a perinuclear distribution. Lipofuscin derives from lipid peroxidation of membranes, which is caused by oxygen-derived free radicals, including superoxide.

Answer: D (centrilobular hepatocytes)

Explanation: toxins affecting the body can be either direct acting or latent. Direct acting toxins are those which themselves exert the deleterious effect on the body (e.g., mercuric chloride). Latent toxins are those which first have to be modified by the body (often the cytochrome p-450 system) before they exert effects. Carbon tetrachloride is an example of a latent toxin. Carbon tetrachloride is converted to a free radical, which is toxic to the liver (as the liver is the organ the alters the carbon tetrachloride, that it is the organ most affected first is understandable). When blood enters the liver, it flows from the portal tract through the hepatic sinusoids to the central veins. Thus, when carbon tetrachloride enters the liver, it is non-toxic, but, as it is metabolized, it becomes toxic, so, the centrilobular hepatocytes would be expected to sustain the majority of the injury. Acetaminophen functions in the same way, becoming toxic when metabolized by the liver.

Answer: A (decreased production of chaperones)

Explanation: chaperones guide proteins through the endoplasmic reticulum. A decreased number of chaperones could contribute to an increase in the amount of misfolded protein, and therefore, the accumulation of tau protein tangles in neurons. The brain is very sensitive to hypoxia and not capable of significant adaptation to hypoxia. The production of hypoxia-inducible factor 1 would play little role in the development of neurofibrillary tangles.

Answer: C (misfolded protein)

Explanation: the patient has family hypercholesterolemia. Familial hypercholesterolemia occurs due to a loss of the LDL receptor (which impairs LDL from being removed from the blood). This loss is due to misfolding of the LDL receptor protein.

Answer: B (cysteine)

Explanation: free radicals cause protein cross-linking through sulfhydryl groups. Of the amino acids listed, only cysteine has a sulfhydryl group.

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.