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NEET PG · India

NEET PG pathology questions

50 original NEET PG-style questions on pathology, each with the correct answer, an explanation, why the other options are wrong and the educational objective. Solve them in the quiz or read them below, grouped by section.

Questions50 with answers
Sections8
Options per question4
Time≈ 75 min
NEET PG Pathology: practice quizPick an answer to see the correct option and the explanation at once. Your best score is saved in this browser.

Cell Injury & Adaptation

10 questions
1

A 55-year-old man undergoes a 40-minute episode of myocardial ischemia followed by reperfusion. Electron microscopy of the affected myocytes performed within minutes of onset of ischemia is most likely to show which of the following earliest ultrastructural abnormalities?

  1. ASwelling of mitochondria with loss of cristae
  2. BNuclear chromatin clumping
  3. CRupture of lysosomal membranes with release of hydrolytic enzymes
  4. DBlebbing of the plasma membrane with loss of microvilli
Show answer
Answer: A. Swelling of mitochondria with loss of cristae

The earliest ultrastructural change in reversible cell injury is mitochondrial swelling due to loss of ion homeostasis, which impairs oxidative phosphorylation and ATP generation. This precedes plasma membrane blebbing, nuclear changes, and lysosomal rupture, all of which are later and often irreversible changes. ATP depletion from mitochondrial dysfunction triggers failure of the Na+/K+ ATPase pump, leading to cellular swelling. This sequence is classically demonstrated in ischemic injury models.

  • B. Chromatin clumping is a nuclear change seen after significant injury, later than mitochondrial swelling.
  • C. Lysosomal membrane rupture is a late, often irreversible event associated with cell death.
  • D. Membrane blebbing occurs later and is a feature closer to irreversible injury.
Educational objective: Mitochondrial swelling is the earliest ultrastructural change in reversible cell injury, preceding membrane and nuclear alterations.
2

A 30-year-old man ingests carbon tetrachloride (CCl4) accidentally at an industrial site and develops acute hepatocellular injury. The hepatotoxic effect of this compound is primarily mediated by which of the following mechanisms?

  1. ADirect inhibition of mitochondrial electron transport chain complex I
  2. BCytochrome P450-mediated conversion to a trichloromethyl free radical
  3. CDepletion of glutathione leading to osmotic cell swelling
  4. DCovalent binding to DNA causing mutagenesis
Show answer
Answer: B. Cytochrome P450-mediated conversion to a trichloromethyl free radical

CCl4 is metabolized by cytochrome P450 in the hepatocyte smooth endoplasmic reticulum to a highly reactive trichloromethyl free radical (CCl3•). This free radical causes lipid peroxidation of the endoplasmic reticulum and plasma membrane, leading to loss of membrane integrity, decreased protein synthesis (seen as fatty change), and ultimately cell death. This is the classic teaching example of free radical-mediated cell injury. Antioxidants like vitamin E can mitigate this damage by scavenging free radicals.

  • A. CCl4 toxicity is not primarily due to direct electron transport chain inhibition.
  • C. Glutathione depletion is central to paracetamol toxicity, not the primary mechanism of CCl4 injury.
  • D. CCl4 does not act through direct DNA covalent binding as its principal mechanism.
Educational objective: CCl4 hepatotoxicity results from cytochrome P450-mediated generation of free radicals causing lipid peroxidation of cell membranes.
3

A 60-year-old man dies of a massive myocardial infarction. On autopsy, the infarcted myocardium shows preserved tissue architecture with cell outlines visible despite loss of nuclei. In contrast, if a similar infarct had occurred in the brain, the expected histological pattern would most likely show which of the following?

  1. ALiquefactive necrosis with cystic cavity formation
  2. BCaseous necrosis with granulomatous rim
  3. CFibrinoid necrosis with vessel wall deposition
  4. DCoagulative necrosis with preserved cell outlines
Show answer
Answer: A. Liquefactive necrosis with cystic cavity formation

Cerebral infarcts characteristically undergo liquefactive necrosis because the brain has high lipid content and lacks sufficient structural (collagen) framework, and because microglia/macrophages digest the dead tissue, resulting in a soft, liquefied area that often evolves into a cystic cavity. In contrast, the heart and kidney undergo coagulative necrosis after infarction, where protein denaturation predominates over enzymatic digestion, preserving the basic tissue architecture and cell outlines for several days. This distinction (coagulative in solid organs, liquefactive in brain) is a classic morphological correlation.

  • B. Caseous necrosis is typical of tuberculosis, not ischemic infarction of the brain.
  • C. Fibrinoid necrosis is seen in immune-mediated vascular injury, not in simple ischemic infarcts.
  • D. This pattern describes myocardial/renal infarcts, not the expected pattern of a cerebral infarct.
Educational objective: Coagulative necrosis typically occurs in solid organs like heart and kidney, whereas liquefactive necrosis is characteristic of cerebral infarcts.
4

A 45-year-old woman with acute pancreatitis has serum lipase of 2200 U/L (normal less than 160 U/L) and chalky white deposits noted in the peripancreatic fat at surgery. These deposits form through which of the following biochemical processes?

  1. ACombination of free fatty acids released by lipase action with calcium to form soaps
  2. BPrecipitation of unconjugated bilirubin with calcium
  3. CBinding of calcium to denatured collagen fibers in fibrous tissue
  4. DDeposition of calcium phosphate in areas of prior tissue damage
Show answer
Answer: A. Combination of free fatty acids released by lipase action with calcium to form soaps

In acute pancreatitis, activated pancreatic lipase hydrolyzes triglycerides in peripancreatic and intra-abdominal fat into free fatty acids and glycerol. These free fatty acids then chelate with calcium ions to form chalky white deposits of calcium soaps, a reaction analogous to soap formation (saponification). This process also contributes to hypocalcemia seen in severe acute pancreatitis as serum calcium is consumed. Grossly, these appear as white, chalky foci in the fat.

  • B. Bilirubin-calcium precipitation describes pigment gallstone formation, not fat necrosis.
  • C. Calcium-collagen binding is not the mechanism of fat necrosis in pancreatitis.
  • D. Calcium phosphate deposition in damaged tissue describes dystrophic calcification in general, not the specific saponification mechanism of fat necrosis.
Educational objective: Fat necrosis in acute pancreatitis results from lipase-mediated release of fatty acids that combine with calcium to form soaps (saponification).
5

A lymph node biopsy from a 28-year-old woman with chronic cervical lymphadenopathy shows granulomas with central areas of structureless, granular, cheese-like eosinophilic material, surrounded by epithelioid cells, lymphocytes, and Langhans giant cells, with complete loss of tissue architecture. This necrosis pattern is most characteristically associated with which of the following conditions?

  1. ATuberculosis
  2. BThrombosed hemorrhoid
  3. CRheumatoid nodule
  4. DAcute appendicitis
Show answer
Answer: A. Tuberculosis

Caseous necrosis is the classic necrosis pattern seen in tuberculosis, where the combination of coagulative and liquefactive necrosis produces a soft, cheese-like, friable material with complete loss of cellular and tissue architecture. Histologically, it appears as amorphous, structureless eosinophilic debris surrounded by a granulomatous rim of epithelioid macrophages, lymphocytes, and multinucleated Langhans giant cells. It is also seen in other granulomatous infections but is most strongly associated with mycobacterial disease in standard teaching. This pattern differs from both coagulative and liquefactive necrosis in that neither cell outlines nor architecture persist.

  • B. Thrombosed hemorrhoids show coagulative necrosis/thrombosis, not caseation.
  • C. Rheumatoid nodules show fibrinoid necrosis, not caseous necrosis.
  • D. Acute appendicitis shows suppurative/liquefactive inflammation, not caseation.
Educational objective: Caseous necrosis, characterized by cheese-like structureless debris within a granuloma, is the hallmark of tuberculosis.
6

A 65-year-old man with diabetes mellitus and peripheral arterial disease develops progressive discoloration of his right foot over three weeks. The affected toes are shrunken, dry, and black, with a clear line of demarcation from viable tissue, and there is no foul odor or significant swelling. These features are most consistent with which of the following?

  1. AWet gangrene
  2. BFournier gangrene
  3. CDry gangrene
  4. DGas gangrene
Show answer
Answer: C. Dry gangrene

Dry gangrene occurs when a slowly progressive arterial occlusion causes coagulative necrosis of tissue, which then becomes desiccated, shrunken, and black-brown due to hemoglobin degradation products. There is a clear line of demarcation between gangrenous and viable tissue because the process is slow and generally aseptic, without significant superimposed bacterial infection. In contrast, wet gangrene occurs when bacterial infection complicates the necrotic tissue, producing liquefaction, foul odor, swelling, and poorly defined margins, carrying a higher risk of systemic sepsis. The absence of odor, swelling, and infection with a sharp demarcation line favors the dry pattern here.

  • A. Wet gangrene would show swelling, foul odor, and poor demarcation due to superimposed infection, which are absent here.
  • B. Fournier gangrene is a specific necrotizing fasciitis of the perineal/genital region, not the foot.
  • D. Gas gangrene is caused by Clostridium perfringens with gas production and crepitus, not described here.
Educational objective: Dry gangrene shows a sharp demarcation line from aseptic coagulative necrosis, while wet gangrene shows poor demarcation, swelling, and foul odor from bacterial superinfection.
7

During development, interdigital mesenchymal cells undergo programmed cell death to separate the digits. Electron microscopy of these dying cells shows cell shrinkage, chromatin condensation, and formation of membrane-bound fragments containing intact organelles, without associated inflammation. The final common executioner enzyme responsible for this morphological pattern is which of the following?

  1. ACaspase-3
  2. BCaspase-9
  3. CCathepsin D
  4. DCaspase-8
Show answer
Answer: A. Caspase-3

Caspase-3 is the principal executioner caspase in apoptosis, activated by cleavage from either the extrinsic pathway (via caspase-8) or the intrinsic mitochondrial pathway (via caspase-9). Activated caspase-3 cleaves nuclear and cytoskeletal proteins, leading to the characteristic morphological features of apoptosis: cell shrinkage, chromatin condensation, nuclear fragmentation, and formation of apoptotic bodies that are phagocytosed without eliciting inflammation. Caspase-8 and caspase-9 are initiator caspases that activate caspase-3 but are not themselves the final executioners.

  • B. Caspase-9 is an initiator caspase of the intrinsic (mitochondrial) pathway, activating downstream executioner caspases.
  • C. Cathepsin D is a lysosomal protease involved in some apoptotic pathways but is not the principal executioner caspase.
  • D. Caspase-8 is an initiator caspase of the extrinsic (death receptor) pathway, activating downstream executioner caspases.
Educational objective: Caspase-3 is the key executioner caspase responsible for the characteristic morphological features of apoptosis.
8

A 50-year-old woman with chronic kidney disease on hemodialysis has a serum calcium of 11.8 mg/dL and serum phosphate of 6.5 mg/dL. Imaging reveals calcium deposits in previously normal lung parenchyma, gastric mucosa, and renal tubules. This pattern of calcification is best explained by which of the following?

  1. ADystrophic calcification in necrotic tissue with normal serum calcium
  2. BAge-related degenerative calcification of elastic tissue
  3. CMetastatic calcification due to hypercalcemia
  4. DCalcification secondary to localized tissue hypoxia
Show answer
Answer: C. Metastatic calcification due to hypercalcemia

Metastatic calcification occurs in previously normal tissues as a consequence of systemic hypercalcemia or hypercalcemia-hyperphosphatemia, commonly due to chronic kidney disease with secondary hyperparathyroidism, vitamin D toxicity, or multiple myeloma. Calcium salts deposit preferentially in tissues with an alkaline environment, such as lung, kidney, gastric mucosa, and blood vessel walls. This is distinct from dystrophic calcification, which occurs in dead or degenerating tissue despite normal serum calcium levels, as seen in old tuberculous lesions or atherosclerotic plaques. The elevated serum calcium and phosphate here, with deposition in normal tissue, confirms the metastatic pattern.

  • A. Dystrophic calcification occurs with normal serum calcium in damaged tissue, which does not match this patient's hypercalcemia and normal parenchyma.
  • B. This is not a simple age-related process; it is driven by the patient's metabolic derangement from renal failure.
  • D. Localized hypoxia causing calcification without systemic hypercalcemia describes dystrophic, not metastatic, calcification.
Educational objective: Metastatic calcification occurs in normal tissues due to systemic hypercalcemia, whereas dystrophic calcification occurs in damaged tissue with normal serum calcium.
9

A 52-year-old man with a 15-year history of heartburn and acid regurgitation undergoes upper GI endoscopy for persistent symptoms despite proton pump inhibitor therapy. The distal esophageal mucosa shows salmon-pink, velvety patches extending proximally from the gastroesophageal junction, contrasting with the surrounding pale squamous mucosa. Biopsy from this area is most likely to show which histological change?

  1. AReplacement of squamous epithelium by columnar epithelium with goblet cells
  2. BSquamous cell hyperplasia with increased basal cell layer thickness
  3. CSubmucosal fibrosis with chronic lymphoplasmacytic infiltrate
  4. DDysplastic squamous epithelium with nuclear atypia
Show answer
Answer: A. Replacement of squamous epithelium by columnar epithelium with goblet cells

Chronic gastroesophageal reflux causes repeated injury to the squamous epithelium of the distal esophagus, leading to its replacement by metaplastic columnar epithelium containing goblet cells, a condition called Barrett esophagus. This is a classic example of intestinal metaplasia, an adaptive response to chronic chemical injury. Barrett esophagus is clinically important because it carries an increased risk of progression to dysplasia and adenocarcinoma. Diagnosis requires endoscopic recognition of salmon-pink mucosa plus histological confirmation of goblet cells on biopsy.

  • B. Squamous hyperplasia is seen in reflux esophagitis without metaplasia, not the defining change of Barrett esophagus.
  • C. Submucosal fibrosis is a feature of healed esophagitis or stricture, not the characteristic epithelial change of Barrett esophagus.
  • D. Dysplasia may develop later in Barrett esophagus but is not the initial metaplastic change and the epithelium described here is squamous, not the columnar epithelium of Barrett.
Educational objective: Chronic GERD leads to Barrett esophagus, an intestinal (columnar with goblet cells) metaplasia of the distal esophageal squamous epithelium that predisposes to adenocarcinoma.
10

A 60-year-old woman with long-standing rheumatoid arthritis presents with bilateral pedal edema, proteinuria of 4.2 g/24 hours, and hepatosplenomegaly. A renal biopsy is performed to confirm suspected amyloid deposition. Which histochemical finding is most specific for establishing this diagnosis?

  1. AApple-green birefringence under polarized light after Congo red staining
  2. BStrong PAS-positive, diastase-resistant staining of glomerular deposits
  3. CBlack staining of deposits with Masson-Fontana stain
  4. DMetachromatic staining with toluidine blue
Show answer
Answer: A. Apple-green birefringence under polarized light after Congo red staining

Amyloid deposits bind Congo red dye and, because of the beta-pleated sheet configuration of amyloid fibrils, show characteristic apple-green birefringence when viewed under polarized light. This is the gold-standard confirmatory test for amyloidosis across all organs and amyloid types. PAS stains amyloid pink but is not specific, since other glycoprotein-rich deposits also stain PAS-positive. Electron microscopy showing non-branching fibrils of 7.5-10 nm diameter supports the diagnosis but Congo red birefringence remains the standard screening and confirmatory stain.

  • B. PAS positivity is seen with amyloid but is nonspecific, as other glomerular deposits like basement membrane material also stain PAS-positive.
  • C. Masson-Fontana stains melanin and argentaffin granules, not amyloid.
  • D. Toluidine blue metachromasia is used to detect mast cell granules and mucopolysaccharides, not amyloid.
Educational objective: Congo red staining with apple-green birefringence under polarized light is the specific confirmatory test for amyloid deposits regardless of the underlying amyloid protein type.

Inflammation & Repair

9 questions
11

In an experimental model, intradermal injection of histamine produces rapid leakage of plasma proteins from venules within minutes, which resolves within 15-30 minutes. Which mechanism best explains this early transient increase in vascular permeability?

  1. AContraction of endothelial cells widening interendothelial junctions in venules
  2. BLeukocyte-mediated endothelial injury via released proteases
  3. CIncreased transcytosis through endothelial cell vesiculo-vacuolar organelles
  4. DDirect toxic injury to endothelial cells causing necrosis
Show answer
Answer: A. Contraction of endothelial cells widening interendothelial junctions in venules

Histamine, along with bradykinin and leukotrienes, acts on venular endothelium causing endothelial cell contraction that widens the interendothelial junctions, producing the immediate transient response of increased vascular permeability. This response predominantly affects venules of 20-60 micrometers diameter and is rapid in onset and short-lived, typically resolving within 15-30 minutes. It differs from the delayed prolonged leakage seen with direct endothelial injury, which affects arterioles, capillaries, and venules and can last for hours to days. Histamine release occurs from mast cells, basophils, and platelets in response to physical injury, immune reactions, and various mediators.

  • B. Leukocyte-mediated endothelial injury occurs later in inflammation as leukocytes adhere and degranulate, not in the immediate histamine-induced phase.
  • C. Increased transcytosis via vesiculo-vacuolar organelles is induced by VEGF and contributes to chronic leakage, not the immediate histamine response.
  • D. Direct endothelial necrosis causes immediate sustained leakage from direct injury such as burns, not the transient histamine-mediated response.
Educational objective: Histamine produces the immediate transient phase of increased vascular permeability through endothelial cell contraction that widens venular interendothelial gaps.
12

During acute inflammation, circulating neutrophils begin to tether loosely to the venular endothelium and tumble along the vessel wall before firm adhesion occurs. Which molecular interaction is primarily responsible for this initial rolling step?

  1. APlatelet endothelial cell adhesion molecule-1 homotypic binding
  2. BSelectins on endothelium binding to sialylated oligosaccharide ligands on leukocytes
  3. CChemokine binding to G-protein coupled receptors on leukocytes
  4. DIntegrin binding to intercellular adhesion molecule-1 on endothelium
Show answer
Answer: B. Selectins on endothelium binding to sialylated oligosaccharide ligands on leukocytes

Margination and rolling of leukocytes along the endothelium are mediated by selectins, specifically P-selectin and E-selectin on activated endothelial cells, which bind to sialylated Lewis-X modified glycoprotein ligands on the surface of leukocytes. P-selectin is rapidly redistributed to the endothelial surface from Weibel-Palade bodies upon stimulation by histamine or thrombin, while E-selectin expression requires cytokine-induced transcription and takes longer to appear. These low-affinity interactions allow the leukocyte to roll along the vessel wall, slowing it down before stronger integrin-mediated firm adhesion occurs. This sequential cascade of rolling, activation, and adhesion is essential for leukocyte recruitment to the site of injury.

  • A. PECAM-1 homotypic interactions mediate transmigration of leukocytes through the endothelium, occurring after firm adhesion.
  • C. Chemokine receptor signaling activates integrins to increase their affinity for firm adhesion but does not mediate the initial rolling interaction.
  • D. Integrin-ICAM-1 binding mediates firm adhesion and arrest of the leukocyte, a step that follows rolling, not the rolling step itself.
Educational objective: Selectin-ligand interactions mediate the initial rolling of leukocytes along inflamed endothelium, preceding integrin-mediated firm adhesion.
13

A 6-month-old infant presents with recurrent bacterial skin and soft tissue infections without pus formation, delayed separation of the umbilical cord, and a peripheral blood neutrophil count of 35,000/microliter. Wound sites show absent neutrophil infiltration on biopsy despite the marked neutrophilia. Which molecular defect best explains this presentation?

  1. ADeficiency of myeloperoxidase in neutrophil granules
  2. BDeficiency of CD18 beta-2 integrin subunit on leukocytes
  3. CDeficiency of C3 complement component
  4. DDeficiency of NADPH oxidase components in phagocytes
Show answer
Answer: B. Deficiency of CD18 beta-2 integrin subunit on leukocytes

Leukocyte adhesion deficiency type 1 results from a mutation in the gene encoding the CD18 beta-2 integrin subunit, which is shared by LFA-1, Mac-1, and other beta-2 integrins required for firm adhesion of leukocytes to endothelium. Without functional integrins, neutrophils cannot adhere firmly to endothelium or transmigrate into tissues, leading to recurrent bacterial infections without pus formation, delayed umbilical cord separation, and impaired wound healing despite a markedly elevated peripheral neutrophil count. The hallmark clinical clue is neutrophilia with absence of neutrophils at the site of infection. Diagnosis is confirmed by flow cytometric demonstration of absent or reduced CD18 expression on leukocytes.

  • A. Myeloperoxidase deficiency typically causes a mild or asymptomatic phagocytic defect, not severe recurrent infections with impaired neutrophil migration.
  • C. C3 deficiency impairs opsonization and chemotaxis via complement but does not cause the integrin-mediated adhesion defect seen here.
  • D. NADPH oxidase deficiency causes chronic granulomatous disease, characterized by recurrent infections with catalase-positive organisms and granuloma formation, not an adhesion defect.
Educational objective: Leukocyte adhesion deficiency type 1 is caused by a defect in the CD18 beta-2 integrin subunit, causing recurrent infections without pus and neutrophilia with impaired neutrophil migration into tissues.
14

A 2-year-old boy has recurrent pneumonia and lymphadenitis caused by Staphylococcus aureus and Aspergillus species, with biopsy of an affected lymph node showing granulomas containing pigment-laden macrophages. His phagocytes are found to have a defect in generating microbicidal reactive oxygen species. Which test is used to screen for this underlying enzyme defect?

  1. ANitroblue tetrazolium reduction test
  2. BTzanck smear
  3. CLeishman-Donovan body staining
  4. DDirect Coombs test
Show answer
Answer: A. Nitroblue tetrazolium reduction test

Chronic granulomatous disease results from a defect in NADPH oxidase, the enzyme complex responsible for generating superoxide and subsequent reactive oxygen species needed for intracellular killing of catalase-positive organisms such as Staphylococcus aureus and Aspergillus. The nitroblue tetrazolium test screens for this defect: normal neutrophils reduce the yellow dye to a blue-black precipitate during the respiratory burst, whereas neutrophils from patients with chronic granulomatous disease fail to reduce the dye. The flow cytometric dihydrorhodamine assay has largely replaced nitroblue tetrazolium in modern practice because it is more sensitive and quantitative, but nitroblue tetrazolium remains the classic screening test taught for this defect. Failure of catalase-positive organisms to be killed despite normal phagocytosis is the hallmark of this disease.

  • B. Tzanck smear detects multinucleated giant cells in herpes virus infections, not phagocyte oxidative burst defects.
  • C. Leishman-Donovan bodies are the intracellular amastigote form of Leishmania seen in visceral leishmaniasis, unrelated to this enzyme defect.
  • D. The direct Coombs test detects antibody or complement bound to red cell surfaces and is used in autoimmune hemolytic anemia, unrelated to phagocyte oxidative function.
Educational objective: The nitroblue tetrazolium test screens for NADPH oxidase deficiency in chronic granulomatous disease by demonstrating failure of the respiratory burst in neutrophils.
15

A patient with bronchial asthma experiences acute bronchospasm during an exacerbation triggered by mast cell degranulation. Mast cell activation leads to lipoxygenase pathway activity on membrane arachidonic acid, generating mediators responsible for sustained smooth muscle contraction and increased vascular permeability in the airway. Which product of this pathway is primarily responsible for this bronchoconstriction?

  1. AProstacyclin
  2. BThromboxane A2
  3. CLeukotrienes C4, D4, and E4
  4. DProstaglandin E2
Show answer
Answer: C. Leukotrienes C4, D4, and E4

The 5-lipoxygenase pathway converts arachidonic acid to leukotrienes, and the cysteinyl leukotrienes LTC4, LTD4, and LTE4 are the most potent bronchoconstrictors and also markedly increase vascular permeability, together constituting what was classically called slow-reacting substance of anaphylaxis. These mediators act on airway smooth muscle to produce sustained bronchospasm, making them central to the pathophysiology of bronchial asthma exacerbations. Leukotriene receptor antagonists such as montelukast are used clinically to block this pathway in asthma management. In contrast, thromboxane A2 is a cyclooxygenase pathway product that promotes platelet aggregation and vasoconstriction rather than bronchoconstriction via this mechanism.

  • A. Prostacyclin is a cyclooxygenase product from endothelium causing vasodilation and inhibiting platelet aggregation, opposite to bronchoconstriction.
  • B. Thromboxane A2 is a cyclooxygenase pathway product from platelets causing vasoconstriction and platelet aggregation, not the principal bronchoconstrictor of asthma.
  • D. Prostaglandin E2 is a cyclooxygenase pathway product mainly involved in vasodilation and pain sensitization, not the primary mediator of asthmatic bronchoconstriction.
Educational objective: Cysteinyl leukotrienes LTC4, LTD4, and LTE4 generated via the 5-lipoxygenase pathway are the principal mediators of bronchoconstriction and vascular permeability in bronchial asthma.
16

A 35-year-old man with pulmonary tuberculosis has a lung biopsy showing well-formed granulomas with central caseous necrosis surrounded by epithelioid macrophages, Langhans giant cells, and a peripheral lymphocytic rim. Formation and maintenance of these granulomas depend critically on a cytokine secreted by activated T-helper 1 lymphocytes that activates macrophages to differentiate into epithelioid cells. Which cytokine is this?

  1. AInterferon-gamma
  2. BTransforming growth factor-beta
  3. CInterleukin-4
  4. DInterleukin-10
Show answer
Answer: A. Interferon-gamma

Granuloma formation in tuberculosis is driven by the Th1 immune response, in which activated CD4+ T lymphocytes secrete interferon-gamma, the key cytokine that activates macrophages and induces their transformation into epithelioid cells capable of forming organized granulomas. Macrophage-derived interleukin-12 drives differentiation of naive T cells toward the Th1 phenotype, and in turn the macrophages release tumor necrosis factor-alpha, which helps recruit and organize cells to form and maintain the granuloma. This interferon-gamma and tumor necrosis factor-alpha axis is central to granulomatous inflammation, and its disruption, for example with anti-TNF biologic therapy, can lead to reactivation of latent tuberculosis due to granuloma breakdown. Epithelioid cells and multinucleate Langhans giant cells are the morphological result of this sustained macrophage activation.

  • B. Transforming growth factor-beta is primarily involved in fibrosis and immune regulation, not the central macrophage-activating step of granuloma formation.
  • C. Interleukin-4 drives the Th2 response associated with allergic and parasitic immunity, not granuloma formation in tuberculosis.
  • D. Interleukin-10 is an immunosuppressive cytokine that downregulates macrophage activation, opposing granuloma formation.
Educational objective: Interferon-gamma secreted by Th1 lymphocytes is the key cytokine activating macrophages to form epithelioid cells and granulomas in tuberculosis.
17

A 54-year-old man with diabetes mellitus has a diabetic foot ulcer that was left open and allowed to heal without surgical closure. Six weeks later, the base of the ulcer shows a pink, granular, friable surface composed of proliferating capillaries and fibroblasts, and the wound is contracting in size. Which process best describes the type of healing occurring in this wound?

  1. AHealing by first intention with minimal scarring
  2. BHealing by second intention with wound contraction
  3. CHealing by third intention after delayed primary closure
  4. DHealing by regeneration without any fibrosis
Show answer
Answer: B. Healing by second intention with wound contraction

Wounds left open heal by secondary intention, characterized by a larger tissue defect, abundant granulation tissue formation, and wound contraction mediated by myofibroblasts, followed by re-epithelialization from the margins. This differs from first intention healing, which occurs in clean, surgically approximated wounds with minimal granulation tissue and a thin linear scar. The pink, granular, friable tissue described is classic granulation tissue, rich in new capillaries, fibroblasts, and inflammatory cells. Secondary intention healing results in a larger, more noticeable scar due to the greater volume of connective tissue deposited.

  • A. First intention applies to clean, approximated surgical wounds with minimal granulation tissue, not an open ulcer.
  • C. Third intention (delayed primary closure) is used when a contaminated wound is initially left open and later surgically closed, which did not happen here.
  • D. All cutaneous wound healing involves some fibrosis and scar formation; true regeneration without fibrosis does not occur in skin.
Educational objective: Open wounds healing without surgical closure undergo secondary intention healing, marked by abundant granulation tissue and wound contraction.
18

A 68-year-old man who lives alone and eats mostly canned and processed food presents with gum bleeding, perifollicular hemorrhages on his legs, and a poorly healing surgical wound from 3 weeks ago that has partially dehisced. Laboratory studies show normal platelet count and normal prothrombin time. Deficiency of which vitamin best explains this presentation?

  1. AVitamin K
  2. BVitamin A
  3. CVitamin C
  4. DVitamin B12
Show answer
Answer: C. Vitamin C

Vitamin C (ascorbic acid) is a cofactor for prolyl and lysyl hydroxylase, enzymes required for hydroxylation of proline and lysine residues on procollagen, which is essential for stable collagen triple helix formation and cross-linking. Deficiency results in weak, poorly cross-linked collagen, causing capillary fragility (perifollicular hemorrhages, bleeding gums) and impaired wound healing with dehiscence. Platelet count and coagulation studies remain normal because scurvy is not a clotting factor or platelet disorder but a connective tissue defect. This classic presentation in an elderly person with poor dietary intake is typical of scurvy.

  • A. Vitamin K deficiency causes a coagulopathy with prolonged prothrombin time, which is normal in this patient.
  • B. Vitamin A deficiency causes night blindness and squamous metaplasia of epithelium, not impaired collagen hydroxylation.
  • D. Vitamin B12 deficiency causes megaloblastic anemia and subacute combined degeneration, not defective collagen synthesis.
Educational objective: Vitamin C is essential for hydroxylation of proline and lysine in collagen synthesis, and its deficiency (scurvy) causes fragile vessels and poor wound healing.
19

A 45-year-old woman is being monitored after an elective cholecystectomy performed 2 days ago. Her temperature is 37.2°C, pulse 82/min, and blood pressure 118/76 mmHg. Serum CRP, which was 2 mg/L preoperatively, rises to 96 mg/L on postoperative day 2 and then falls to 18 mg/L by day 5 as recovery proceeds uneventfully. Which statement best explains the clinical utility of this marker?

  1. ACRP is an acute phase protein produced by the liver that rises rapidly and tracks the course of inflammation or tissue injury
  2. BCRP is a cell-surface receptor that identifies the pathogen causing infection
  3. CCRP level reflects chronic antibody-mediated autoimmune activity rather than acute inflammation
  4. DCRP is a specific tumor marker used to screen for postoperative malignancy recurrence
Show answer
Answer: A. CRP is an acute phase protein produced by the liver that rises rapidly and tracks the course of inflammation or tissue injury

C-reactive protein is a hepatocyte-derived acute phase reactant induced by interleukin-6, rising within 6 to 12 hours of tissue injury, infection, or surgery, and falling as the inflammatory stimulus resolves. Its short half-life makes it a useful, nonspecific marker for tracking the trajectory of inflammation, as illustrated by its sharp postoperative rise followed by an expected decline during uneventful recovery. A persistently elevated or re-rising CRP after surgery raises suspicion for ongoing infection or complications such as an abscess. CRP is nonspecific and does not identify a causative organism or indicate malignancy by itself.

  • B. CRP is a soluble plasma protein, not a cell-surface receptor, and it does not identify specific pathogens.
  • C. CRP reflects acute inflammatory activity driven by IL-6, not specifically chronic autoimmune antibody activity.
  • D. CRP is a nonspecific marker of inflammation and is not used as a screening tool for cancer recurrence.
Educational objective: CRP is a hepatic acute phase protein induced by IL-6 that rises and falls with the course of inflammation, useful for monitoring disease activity and complications.
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Hemodynamic Disorders

3 questions
20

A 26-year-old man sustains a closed fracture of the femoral shaft in a road traffic accident. He is stable initially, but 36 hours later he develops sudden breathlessness with an oxygen saturation of 88% on room air, confusion, and a petechial rash over the chest, neck, and conjunctivae. Chest X-ray shows diffuse bilateral infiltrates. Which clinical triad is characteristic of this condition?

  1. AJaundice, hemolytic anemia, and splenomegaly
  2. BRespiratory distress, neurologic abnormalities, and petechial rash
  3. CHypertension, bradycardia, and irregular respiration
  4. DFever, hemoptysis, and pleuritic chest pain
Show answer
Answer: B. Respiratory distress, neurologic abnormalities, and petechial rash

Fat embolism syndrome classically presents 24 to 72 hours after a long bone or pelvic fracture with a triad of respiratory distress (hypoxemia from pulmonary fat microemboli causing capillary damage and inflammation), neurologic symptoms ranging from confusion to coma (from cerebral fat emboli or hypoxia), and a petechial rash typically over the upper body, axillae, and conjunctivae (from occlusion of dermal capillaries and thrombocytopenia). The latent period between the injury and symptom onset is a key diagnostic clue, differentiating it from an immediate pulmonary embolism of thrombotic origin. Management is largely supportive, with early fracture fixation and oxygen support reducing the risk of its occurrence.

  • A. Jaundice, hemolytic anemia, and splenomegaly describe a hemolytic process, unrelated to fat embolism syndrome.
  • C. Cushing triad of hypertension, bradycardia, and irregular respiration reflects raised intracranial pressure, not fat embolism syndrome.
  • D. Fever and hemoptysis with pleuritic pain are more suggestive of pulmonary infarction or infection, not the classic fat embolism triad.
Educational objective: Fat embolism syndrome presents 24-72 hours after long bone fracture with the triad of respiratory distress, neurologic impairment, and petechial rash.
21

A 58-year-old man is brought to the emergency department after sustaining extensive blood loss from a road traffic accident. On arrival, his pulse is 128/min, blood pressure is 78/50 mmHg, and skin is cold and clammy. Invasive monitoring shows central venous pressure of 2 mmHg, cardiac output of 2.8 L/min, and systemic vascular resistance of 1800 dyn.s/cm5. Which of the following best describes this hemodynamic pattern?

  1. ALow cardiac output, low filling pressure, low systemic vascular resistance
  2. BLow cardiac output, low filling pressure, high systemic vascular resistance
  3. CLow cardiac output, high filling pressure, high systemic vascular resistance
  4. DHigh cardiac output, low filling pressure, low systemic vascular resistance
Show answer
Answer: B. Low cardiac output, low filling pressure, high systemic vascular resistance

Hypovolemic shock results from reduced intravascular volume, leading to decreased venous return and low central venous pressure (low preload), low cardiac output, and compensatory vasoconstriction causing high systemic vascular resistance. This contrasts with cardiogenic shock, where pump failure causes low cardiac output but high filling pressures due to backward failure. Septic shock, in its early warm phase, shows high cardiac output with low systemic vascular resistance due to vasodilation. Recognizing these patterns is essential for guiding fluid resuscitation and vasopressor therapy.

  • A. Systemic vascular resistance rises, not falls, as a compensatory response to volume loss.
  • C. High filling pressure is seen in cardiogenic shock where the heart fails to pump effectively, not in volume loss.
  • D. High cardiac output with low resistance describes early hyperdynamic septic shock.
Educational objective: Hypovolemic shock is characterized by low cardiac output, low central venous pressure, and high systemic vascular resistance due to compensatory vasoconstriction.
22

A 34-year-old woman develops sudden breathlessness and oozing from venipuncture sites 2 hours after an amniotic fluid embolism during labor. Laboratory evaluation shows platelet count of 42,000/mm3, prothrombin time of 24 seconds (control 12 seconds), activated partial thromboplastin time of 58 seconds (control 30 seconds), fibrinogen of 80 mg/dL, and markedly elevated D-dimer. Which of the following best explains this laboratory profile?

  1. AImmune thrombocytopenic purpura
  2. BDisseminated intravascular coagulation
  3. CHemophilia A
  4. DVitamin K deficiency
Show answer
Answer: B. Disseminated intravascular coagulation

Disseminated intravascular coagulation is characterized by widespread activation of the coagulation cascade, leading to consumption of platelets and clotting factors (prolonging PT and aPTT), depletion of fibrinogen, and secondary activation of fibrinolysis producing elevated D-dimer. Amniotic fluid embolism is a classic obstetric trigger that releases tissue factor into the circulation. The combination of thrombocytopenia, prolonged PT and aPTT, low fibrinogen, and raised D-dimer together is diagnostic of consumptive coagulopathy rather than an isolated factor or platelet defect. Treatment addresses the underlying cause along with blood product support.

  • A. Immune thrombocytopenic purpura causes isolated thrombocytopenia with normal PT, aPTT, and fibrinogen.
  • C. Hemophilia A causes isolated prolongation of aPTT with normal platelet count, PT, and fibrinogen.
  • D. Vitamin K deficiency prolongs PT and aPTT but does not cause thrombocytopenia or elevated D-dimer.
Educational objective: DIC is diagnosed by the combination of thrombocytopenia, prolonged PT and aPTT, low fibrinogen, and elevated D-dimer reflecting simultaneous consumption and fibrinolysis.

Neoplasia

11 questions
23

A pathology resident is asked to name tumors arising from mesenchymal tissue based on standard nomenclature rules. Which of the following suffix conventions is correctly applied for a malignant tumor of adipose tissue origin?

  1. ALipomatous carcinoma
  2. BLipoma
  3. CLipoadenoma
  4. DLiposarcoma
Show answer
Answer: D. Liposarcoma

Malignant mesenchymal tumors are named by attaching the suffix "-sarcoma" to the cell type of origin, so a malignant tumor of adipocytes is called liposarcoma. In contrast, benign mesenchymal tumors take the suffix "-oma" added to the cell of origin, as in lipoma for a benign fatty tumor. The suffix "-carcinoma" is reserved for malignant tumors of epithelial origin, not mesenchymal tissue. This nomenclature distinction between mesenchymal and epithelial malignancies is fundamental to tumor classification.

  • A. Carcinoma is the suffix for malignant epithelial tumors, not mesenchymal tumors like those of fat.
  • B. Lipoma denotes a benign adipose tumor, not a malignant one.
  • C. Lipoadenoma is not a standard term; "adenoma" suffix applies to benign glandular epithelial tumors.
Educational objective: Malignant mesenchymal tumors are named using the suffix "-sarcoma", while benign mesenchymal tumors use "-oma", distinct from epithelial tumor nomenclature using "-carcinoma".
24

A 52-year-old woman is diagnosed with invasive ductal carcinoma of the breast. Histopathology reports it as a moderately differentiated tumor, while clinical and imaging evaluation shows a 3.5 cm primary tumor with metastasis to two axillary lymph nodes and no distant spread. Which statement best distinguishes the two types of information provided in this report?

  1. AHistologic differentiation reflects stage, while tumor size and nodal/metastatic spread reflect grade
  2. BBoth histologic differentiation and tumor size/spread together constitute the tumor stage
  3. CBoth histologic differentiation and tumor size/spread together constitute the tumor grade
  4. DHistologic differentiation reflects grade, while tumor size and nodal/metastatic spread reflect stage
Show answer
Answer: D. Histologic differentiation reflects grade, while tumor size and nodal/metastatic spread reflect stage

Tumor grade refers to the degree of histologic differentiation and cytologic resemblance to the tissue of origin, reflecting how aggressive the tumor appears microscopically, as in "moderately differentiated." Tumor stage, assessed by the TNM system, describes the anatomic extent of disease based on primary tumor size (T), regional lymph node involvement (N), and distant metastasis (M). Staging has greater prognostic value than grading in most solid tumors and directly guides treatment decisions such as the need for adjuvant therapy. Grade and stage are assessed independently and reported separately in pathology and oncology practice.

  • A. This reverses the definitions; differentiation describes grade, not stage, and anatomic spread describes stage, not grade.
  • B. Histologic differentiation is a component of grading, not staging.
  • C. Tumor size and nodal/metastatic spread are components of staging, not grading.
Educational objective: Tumor grade reflects histologic differentiation while tumor stage, determined by the TNM system, reflects anatomic extent of disease and carries greater prognostic significance.
25

In a colorectal adenocarcinoma specimen, molecular testing reveals a point mutation in codon 12 of a proto-oncogene encoding a membrane-associated GTP-binding protein. This mutation impairs the protein's intrinsic GTPase activity. Which of the following best describes the functional consequence of this mutation?

  1. AThe protein remains locked in an active, GTP-bound state, causing constitutive downstream signaling
  2. BThe protein translocates to the nucleus and directly binds DNA as a transcription factor
  3. CThe protein is permanently degraded by the proteasome, eliminating downstream signaling
  4. DThe protein loses its ability to bind GTP, remaining permanently inactive
Show answer
Answer: A. The protein remains locked in an active, GTP-bound state, causing constitutive downstream signaling

RAS is a proto-oncogene encoding a membrane-bound GTP-binding protein that normally cycles between an inactive GDP-bound state and an active GTP-bound state, with intrinsic GTPase activity hydrolyzing GTP to switch itself off. Point mutations at codons 12, 13, or 61 impair this GTPase activity, trapping RAS in its active GTP-bound conformation. This leads to constitutive activation of downstream mitogen-activated protein kinase signaling, driving uncontrolled cell proliferation independent of upstream growth factor signals. RAS mutations are among the most common oncogenic alterations found in colorectal, pancreatic, and lung cancers.

  • B. RAS functions in the cytoplasm as a membrane-associated signaling protein, not as a nuclear transcription factor.
  • C. The mutation enhances rather than eliminates RAS activity; it is not targeted for degradation.
  • D. The mutation does not impair GTP binding; it impairs the hydrolysis of bound GTP.
Educational objective: Point mutations in RAS impair its intrinsic GTPase activity, locking it in the active GTP-bound state and causing constitutive mitogenic signaling.
26

A 2-year-old child presents with leukocoria in the left eye. Family history reveals that the child's father had bilateral retinoblastoma treated in infancy. Genetic analysis shows the child inherited one mutant allele of the RB1 gene from the father. Which of the following best explains why tumor development still requires an additional somatic event in the retinal cell?

  1. AA single mutant allele of a tumor suppressor gene is sufficient to produce a dominant oncogenic signal
  2. BBoth alleles of the tumor suppressor gene must be inactivated for loss of its growth-regulatory function
  3. CThe mutant allele must translocate to an active chromosomal region to be expressed
  4. DThe mutant allele must undergo gene amplification to increase its copy number
Show answer
Answer: B. Both alleles of the tumor suppressor gene must be inactivated for loss of its growth-regulatory function

The retinoblastoma gene RB1 is a tumor suppressor gene that follows the two-hit hypothesis, requiring inactivation of both alleles for loss of function and subsequent tumorigenesis. In familial retinoblastoma, the first hit is inherited in the germline, present in every cell, so only a single additional somatic mutation in the retina is needed to inactivate the second allele, explaining the early onset and bilateral presentation. In sporadic retinoblastoma, both hits must occur as independent somatic events within the same retinal cell, which is a rarer event and explains the later onset and typically unilateral presentation. This recessive behavior at the cellular level distinguishes tumor suppressor genes from dominantly acting oncogenes.

  • A. Tumor suppressor genes act recessively at the cellular level; a single functional allele is usually sufficient to maintain normal growth control.
  • C. Tumor suppressor inactivation occurs through mutation, deletion, or loss of heterozygosity, not through activating translocation.
  • D. Gene amplification is a mechanism of oncogene activation, not tumor suppressor gene inactivation.
Educational objective: Retinoblastoma exemplifies the two-hit hypothesis, in which both alleles of the RB1 tumor suppressor gene must be inactivated for tumor development.
27

A 28-year-old woman is diagnosed with breast cancer. Her family history includes a sibling who had osteosarcoma at age 14, her mother who had adrenocortical carcinoma at age 32, and a maternal uncle who developed a soft tissue sarcoma at age 25. Genetic testing in the patient reveals a germline mutation in a tumor suppressor gene that normally arrests the cell cycle at the G1-S checkpoint and promotes apoptosis in response to DNA damage. Which of the following hereditary cancer syndromes best fits this presentation?

  1. APeutz-Jeghers syndrome
  2. BLi-Fraumeni syndrome
  3. CLynch syndrome
  4. DVon Hippel-Lindau syndrome
Show answer
Answer: B. Li-Fraumeni syndrome

Li-Fraumeni syndrome results from a germline mutation in the TP53 tumor suppressor gene, which encodes p53, a protein that senses DNA damage and induces cell cycle arrest at the G1-S checkpoint or triggers apoptosis to prevent propagation of genetically damaged cells. Affected families show a characteristic pattern of early-onset, multiple primary cancers including sarcomas, breast carcinoma, brain tumors, and adrenocortical carcinoma across generations, matching the family history described. Loss of functional p53 removes this critical surveillance mechanism, allowing cells with damaged DNA to proliferate and accumulate further mutations. This syndrome follows autosomal dominant inheritance with high penetrance.

  • A. Peutz-Jeghers syndrome involves STK11 mutations causing mucocutaneous pigmentation and gastrointestinal hamartomatous polyps with increased cancer risk, not this tumor spectrum.
  • C. Lynch syndrome results from mismatch repair gene mutations and predisposes mainly to colorectal and endometrial cancer, not this pattern of sarcomas and adrenal carcinoma.
  • D. Von Hippel-Lindau syndrome involves VHL gene mutations causing hemangioblastomas, renal cell carcinoma, and pheochromocytoma.
Educational objective: Germline TP53 mutation causes Li-Fraumeni syndrome, characterized by early-onset sarcomas, breast cancer, brain tumors, and adrenocortical carcinoma across generations.
28

A 48-year-old farmer from eastern India presents with right upper quadrant pain, weight loss and ascites. Imaging reveals a large mass in the right lobe of the liver with underlying cirrhosis. He reports consuming home-stored groundnuts and maize that were visibly moldy during the monsoon season. Which of the following best explains the molecular basis of the hepatic malignancy in this patient?

  1. AFormation of a DNA adduct that causes a G to T transversion in codon 249 of the TP53 gene
  2. BActivation of the Wnt/beta-catenin pathway through APC gene mutation
  3. CIntegration of viral DNA into the host genome activating the MYC oncogene
  4. DMicrosatellite instability due to mismatch repair gene silencing
Show answer
Answer: A. Formation of a DNA adduct that causes a G to T transversion in codon 249 of the TP53 gene

Aflatoxin B1, produced by Aspergillus flavus contaminating improperly stored grains and nuts, is metabolized by hepatic cytochrome P450 to a reactive epoxide that forms a DNA adduct with guanine. This adduct characteristically produces a G to T transversion at codon 249 of TP53, inactivating this tumor suppressor and strongly predisposing to hepatocellular carcinoma, especially in regions with high aflatoxin exposure and chronic hepatitis B carriage.

  • B. APC mutation and Wnt pathway activation underlie familial adenomatous polyposis and sporadic colorectal carcinoma.
  • C. Viral integration with MYC activation is the mechanism of EBV-associated Burkitt lymphoma, not aflatoxin-related HCC.
  • D. Microsatellite instability is characteristic of Lynch syndrome-associated colorectal and endometrial cancers.
Educational objective: Aflatoxin B1 causes hepatocellular carcinoma through a characteristic G to T transversion at codon 249 of the TP53 gene.
29

An 8-year-old boy from Uganda presents with a rapidly enlarging, painless jaw swelling of 3 weeks' duration. Biopsy of the mass shows a diffuse infiltrate of medium-sized lymphoid cells with a high mitotic rate, interspersed with pale-staining macrophages containing ingested apoptotic debris, giving a characteristic microscopic appearance. Which of the following investigations would most specifically confirm the presumed underlying viral association in this patient?

  1. APolymerase chain reaction for EBV DNA in peripheral blood
  2. BImmunohistochemistry for latent membrane protein-1 (LMP-1)
  3. CIn situ hybridization for EBV-encoded RNA (EBER) in tumor cells
  4. DSerum heterophile antibody test (Monospot test)
Show answer
Answer: C. In situ hybridization for EBV-encoded RNA (EBER) in tumor cells

The clinical picture and the classic 'starry sky' histology described point to Burkitt lymphoma, which in its endemic African form is strongly associated with EBV. EBER in situ hybridization detects the abundant, non-polyadenylated EBV-encoded RNA transcripts present in virtually all latently infected tumor cells and is the gold-standard, most sensitive and specific method to demonstrate EBV within the tumor tissue itself. Peripheral blood tests reflect systemic viral load, not tumor-cell infection, and are far less specific for confirming tumor association.

  • A. PCR for EBV DNA in blood reflects systemic viral load and is used for monitoring, not for confirming EBV presence within the tumor cells.
  • B. LMP-1 is expressed in EBV latency type II/III programs such as nasopharyngeal carcinoma, but endemic Burkitt lymphoma typically shows latency type I with minimal LMP-1 expression, making EBER a more reliable marker.
  • D. The Monospot test detects heterophile antibodies in acute infectious mononucleosis, not latent EBV within tumor cells, and is often negative in Burkitt lymphoma.
Educational objective: EBER in situ hybridization is the most specific test to demonstrate EBV infection within tumor cells in EBV-associated malignancies like endemic Burkitt lymphoma and nasopharyngeal carcinoma.
30

A 62-year-old male smoker presents with a central lung mass on chest radiograph, along with generalized weakness, polyuria and constipation. Serum calcium is 12.8 mg/dL, serum phosphate is low, and parathyroid hormone (PTH) level is suppressed. Biopsy of the lung mass confirms squamous cell carcinoma. Which of the following is the most likely mechanism underlying this patient's hypercalcemia?

  1. ASecretion of parathyroid hormone-related protein by tumor cells acting on PTH receptors in bone and kidney
  2. BDirect osteolytic destruction of bone by metastatic tumor deposits
  3. CEctopic secretion of authentic parathyroid hormone by tumor cells
  4. DTumor secretion of 1,25-dihydroxyvitamin D causing increased intestinal calcium absorption
Show answer
Answer: A. Secretion of parathyroid hormone-related protein by tumor cells acting on PTH receptors in bone and kidney

Squamous cell carcinoma of the lung is the classic cause of humoral hypercalcemia of malignancy, mediated by tumor secretion of PTH-related protein (PTHrP). PTHrP shares homology with the amino-terminal region of PTH and binds the same PTH/PTHrP receptor (PTH1R), increasing osteoclastic bone resorption and renal calcium reabsorption while suppressing renal phosphate reabsorption, mimicking the biochemical profile of hyperparathyroidism but with a suppressed native PTH level. This distinguishes it from local osteolytic hypercalcemia, which occurs with direct bony metastases such as from breast carcinoma or multiple myeloma.

  • B. Local osteolytic hypercalcemia is typically seen with extensive bone metastases, as in breast carcinoma or myeloma, not usually in squamous cell lung cancer without bone involvement.
  • C. Ectopic secretion of authentic PTH by non-parathyroid tumors is exceedingly rare and would not explain the suppressed PTH level seen here.
  • D. Excess 1,25-dihydroxyvitamin D production causing hypercalcemia is characteristic of certain lymphomas, not squamous cell lung carcinoma.
Educational objective: Squamous cell carcinoma of the lung causes paraneoplastic hypercalcemia through tumor secretion of PTHrP, which mimics PTH action while native PTH remains suppressed.
31

A 55-year-old woman presents with abdominal distension and a pelvic mass on ultrasound suggestive of an ovarian tumor. Which of the following serum tumor markers would be most useful for monitoring treatment response and detecting recurrence in this patient?

  1. ACA-125
  2. BAlpha-fetoprotein
  3. CCarcinoembryonic antigen
  4. DCA 19-9
Show answer
Answer: A. CA-125

CA-125 is the tumor marker most closely associated with epithelial ovarian carcinoma, particularly serous subtype, and is widely used to monitor treatment response and detect recurrence, though it lacks sensitivity and specificity for primary screening in the general population. CA 19-9 is more specific for pancreatic and biliary tract carcinomas, AFP for hepatocellular carcinoma and yolk sac tumors, and CEA for colorectal carcinoma. Correct correlation of marker to tumor type is essential since no marker listed is diagnostic in isolation but each has established clinical utility for monitoring specific malignancies.

  • B. AFP is elevated in hepatocellular carcinoma and germ cell tumors such as yolk sac tumor, not epithelial ovarian carcinoma.
  • C. CEA is used mainly for colorectal carcinoma monitoring, not as the primary marker for ovarian cancer.
  • D. CA 19-9 is primarily elevated in pancreatic and biliary tract carcinomas, not ovarian cancer.
Educational objective: CA-125 is the tumor marker of choice for monitoring treatment response and recurrence in epithelial ovarian carcinoma.
32

A 16-year-old boy presents with pain and swelling around the right knee for 2 months. Imaging shows a destructive lesion in the distal femur with a periosteal reaction and soft tissue extension. Biopsy confirms osteosarcoma. Staging CT of the chest reveals multiple well-defined nodules in both lung fields. Which of the following best explains the mechanism of spread to the lungs in this patient?

  1. ATumor cells invade regional lymphatics and travel through the thoracic duct into the systemic circulation
  2. BTumor cells enter thin-walled veins and venules draining the tumor and are carried via the systemic venous circulation to the pulmonary capillary bed
  3. CTumor cells disseminate along perineural sheaths to reach mediastinal structures
  4. DTumor cells spread by direct transcoelomic seeding across the pleural cavity
Show answer
Answer: B. Tumor cells enter thin-walled veins and venules draining the tumor and are carried via the systemic venous circulation to the pulmonary capillary bed

Sarcomas, being mesenchymal in origin, characteristically spread hematogenously because they arise in tissues rich in thin-walled veins and lack well-developed lymphatic drainage. Tumor cells invade these vessels, enter the systemic venous circulation, and lodge in the pulmonary capillary bed, making the lung the most common site of metastasis for sarcomas such as osteosarcoma. This contrasts with carcinomas, which more commonly metastasize initially via lymphatic channels to regional lymph nodes.

  • A. Lymphatic spread to regional nodes is the typical early pattern for carcinomas, not sarcomas, which largely bypass lymphatic spread.
  • C. Perineural spread is a feature of certain carcinomas, such as adenoid cystic carcinoma and prostate carcinoma, not the characteristic route for sarcoma metastasis to lung.
  • D. Transcoelomic spread occurs with tumors involving serosal surfaces such as ovarian carcinoma, not primary bone sarcomas.
Educational objective: Sarcomas characteristically metastasize via the hematogenous route to the lungs due to their origin in vascular mesenchymal tissue with poor lymphatic drainage.
33

A pathologist receives a biopsy of a poorly differentiated malignant tumor in the retroperitoneum, and routine histology alone cannot distinguish between a carcinoma, sarcoma or lymphoma. Immunohistochemical staining is performed. Positivity for which of the following markers would most strongly support an epithelial (carcinomatous) origin of the tumor?

  1. AVimentin
  2. BDesmin
  3. CCytokeratin
  4. DLeukocyte common antigen (CD45)
Show answer
Answer: C. Cytokeratin

Cytokeratins are intermediate filament proteins expressed specifically by epithelial cells and tumors derived from them, making cytokeratin positivity the key immunohistochemical marker supporting a diagnosis of carcinoma in a poorly differentiated malignancy. Vimentin marks mesenchymal cells and tumors such as sarcomas, desmin is specific for cells of myogenic differentiation, and CD45 (leukocyte common antigen) identifies hematolymphoid malignancies. This panel approach of intermediate filament and lineage-specific markers is fundamental to resolving the differential diagnosis of undifferentiated malignant tumors.

  • A. Vimentin is the intermediate filament marker of mesenchymal cells and tumors such as sarcomas, not epithelial tumors.
  • B. Desmin is specific for muscle differentiation, as seen in rhabdomyosarcoma and leiomyosarcoma, not carcinomas.
  • D. CD45 is a pan-leukocyte marker used to identify lymphomas and other hematolymphoid neoplasms, not carcinomas.
Educational objective: Cytokeratin positivity on immunohistochemistry is the key marker confirming epithelial origin in the diagnostic work-up of a poorly differentiated malignant tumor.

RBC Disorders & Hemolytic Anemia

7 questions
34

A 28-year-old woman presents with easy fatigability and pallor for 4 months. She has heavy menstrual bleeding each cycle. Hemoglobin is 8.2 g/dL, MCV is 68 fL, and peripheral smear shows microcytic hypochromic red cells with increased central pallor and mild anisopoikilocytosis. Serum iron is 28 mcg/dL (low), total iron-binding capacity is 480 mcg/dL (high), and serum ferritin is 8 ng/mL (low). Which of the following is the most likely diagnosis?

  1. AIron deficiency anemia
  2. BBeta-thalassemia trait
  3. CAnemia of chronic disease
  4. DSideroblastic anemia
Show answer
Answer: A. Iron deficiency anemia

The combination of microcytic hypochromic anemia with low serum iron, high TIBC and low serum ferritin in a woman with a clear source of chronic blood loss (menorrhagia) is diagnostic of iron deficiency anemia. Ferritin is the most specific and earliest marker to fall in iron deficiency, as it reflects depleted body iron stores, while TIBC rises as transferrin synthesis increases in response to iron deficiency. In anemia of chronic disease, ferritin is typically normal or raised with low TIBC, and in thalassemia trait and sideroblastic anemia, iron stores and ferritin are normal or increased.

  • B. Beta-thalassemia trait presents with microcytosis but normal or increased iron stores and ferritin, not depleted iron studies as seen here.
  • C. Anemia of chronic disease shows low serum iron with low or normal TIBC and normal or raised ferritin, which does not match this iron study pattern.
  • D. Sideroblastic anemia shows increased iron stores with raised ferritin and ring sideroblasts on bone marrow, inconsistent with this low-ferritin picture.
Educational objective: Low serum ferritin with low serum iron and high TIBC is the classic iron study pattern confirming iron deficiency anemia.
35

A 60-year-old strict vegetarian man presents with progressive fatigue, glossitis and tingling in both feet for 6 months. Hemoglobin is 7.5 g/dL, MCV is 122 fL, and peripheral smear shows macro-ovalocytes and neutrophils with 6-7 nuclear lobes. Bone marrow aspirate shows a markedly hypercellular marrow with erythroid hyperplasia, large erythroid precursors with immature-appearing nuclei relative to hemoglobinized cytoplasm, and giant metamyelocytes. Which of the following is the most likely underlying cause?

  1. AAplastic anemia
  2. BVitamin B12 deficiency
  3. CIron deficiency
  4. DChronic lead poisoning
Show answer
Answer: B. Vitamin B12 deficiency

The clinical picture of a vegan diet with neurological symptoms (paresthesias suggesting subacute combined degeneration), macrocytic anemia with hypersegmented neutrophils, and bone marrow findings of megaloblastic change with nuclear-cytoplasmic asynchrony and giant metamyelocytes is classic for vitamin B12 deficiency. B12 is essential for DNA synthesis via the methionine synthase pathway, and its deficiency causes impaired nuclear maturation while cytoplasmic hemoglobinization proceeds normally, producing the characteristic megaloblastic morphology. Strict vegetarians are at risk because B12 is found almost exclusively in animal-derived foods.

  • A. Aplastic anemia presents with pancytopenia and a hypocellular, fatty bone marrow, not erythroid hyperplasia with megaloblastic change.
  • C. Iron deficiency produces microcytic hypochromic anemia, not macrocytosis with hypersegmented neutrophils or megaloblastic marrow changes.
  • D. Lead poisoning causes microcytic anemia with basophilic stippling, not megaloblastic changes.
Educational objective: Vitamin B12 deficiency causes megaloblastic anemia characterized by macro-ovalocytes, hypersegmented neutrophils, and bone marrow nuclear-cytoplasmic asynchrony due to impaired DNA synthesis.
36

A 24-year-old man of Mediterranean descent develops dark-colored urine and yellowish discoloration of eyes two days after starting a sulfonamide antibiotic for a urinary tract infection. Hemoglobin falls from 14 g/dL to 9 g/dL over 48 hours. Peripheral blood smear shows red cells with irregularly distributed, oxidized hemoglobin precipitates adherent to the membrane, along with cells that appear to have a bite taken out of their margin. Which enzyme deficiency best explains this presentation?

  1. AGlucose phosphate isomerase
  2. BPyrimidine 5'-nucleotidase
  3. CPyruvate kinase
  4. DGlucose-6-phosphate dehydrogenase
Show answer
Answer: D. Glucose-6-phosphate dehydrogenase

G6PD deficiency impairs the hexose monophosphate shunt, reducing NADPH generation needed to maintain reduced glutathione, which protects hemoglobin and membrane proteins from oxidative injury. Oxidant stress from drugs such as sulfonamides, dapsone, primaquine, or fava beans causes hemoglobin denaturation, forming Heinz bodies, and splenic macrophages remove these inclusions, producing bite cells. Hemolysis is typically episodic and triggered by exposure rather than chronic. This X-linked enzymopathy is the most common red cell enzyme defect worldwide and classically presents with acute intravascular hemolysis after an oxidant trigger.

  • A. Glucose phosphate isomerase deficiency is a rare glycolytic enzyme defect unrelated to oxidant-induced hemolysis.
  • B. Pyrimidine 5'-nucleotidase deficiency causes basophilic stippling, not Heinz body hemolysis after oxidant drugs.
  • C. Pyruvate kinase deficiency causes chronic hemolysis from birth, not drug-triggered episodic hemolysis with Heinz bodies.
Educational objective: G6PD deficiency causes oxidant drug- or fava bean-induced hemolysis with Heinz bodies and bite cells due to impaired NADPH-dependent glutathione defense.
37

A 6-year-old child from Chhattisgarh presents with recurrent episodes of severe bony pain, pallor, and an enlarged spleen. Peripheral smear shows crescent-shaped red cells and target cells. Family history reveals a similar illness in a sibling who died of a painful crisis. Which of the following best describes the molecular basis of this disorder?

  1. APoint mutation causing premature stop codon in the beta-globin gene
  2. BDeletion of one or more alpha-globin gene copies
  3. CSubstitution of valine for glutamic acid at the sixth position of the beta-globin chain
  4. DExpansion of a CAG trinucleotide repeat in the beta-globin gene
Show answer
Answer: C. Substitution of valine for glutamic acid at the sixth position of the beta-globin chain

Sickle cell disease results from a single point mutation in the beta-globin gene (GAG to GTG) that substitutes valine for glutamic acid at position 6, producing hemoglobin S. Under hypoxic or acidotic conditions, HbS polymerizes into rigid fibers that distort red cells into a sickle shape, causing vaso-occlusion and hemolysis. The diagnosis is confirmed by hemoglobin electrophoresis, which shows HbS migrating differently from HbA, along with HbF and absent or reduced HbA in homozygotes. This mutation is common in tribal and certain regional populations of central India, including Chhattisgarh and Odisha.

  • A. A premature stop codon causing absent beta-chain production describes beta-zero thalassemia, not sickle hemoglobin.
  • B. Alpha-globin gene deletions cause alpha-thalassemia, not sickle cell disease.
  • D. Trinucleotide repeat expansion is the mechanism of disorders like Huntington disease, not hemoglobinopathies.
Educational objective: Sickle cell disease is caused by a single beta-globin point mutation (Glu6Val) producing HbS, confirmed by hemoglobin electrophoresis.
38

A 14-month-old infant presents with severe pallor, failure to thrive, and massive splenomegaly. The infant has required blood transfusions every 3-4 weeks since 6 months of age. Peripheral smear shows microcytic hypochromic red cells, numerous target cells, and nucleated red blood cells. Hemoglobin electrophoresis shows markedly elevated HbF (88%), elevated HbA2, and absent HbA. Based on this electrophoresis pattern, which diagnosis is most likely?

  1. AAlpha-thalassemia trait
  2. BHereditary persistence of fetal hemoglobin
  3. CSickle cell trait
  4. DBeta-thalassemia major
Show answer
Answer: D. Beta-thalassemia major

Beta-thalassemia major results from absent or severely reduced beta-globin chain synthesis due to homozygous beta-zero or beta-plus mutations, leading to near-total absence of HbA. Compensatory elevation of HbF and HbA2 occurs, and the electrophoresis pattern of absent HbA with predominant HbF and raised HbA2 is diagnostic. Clinically, affected children present around 6 months of age, as fetal hemoglobin declines, with severe anemia, hepatosplenomegaly, and transfusion dependence. Ineffective erythropoiesis also causes marrow expansion, producing characteristic skull and facial bone changes if untreated.

  • A. Alpha-thalassemia trait shows mild microcytosis with normal HbA2 and HbF, not absent HbA.
  • B. Hereditary persistence of fetal hemoglobin is usually asymptomatic with high HbF but normal or near-normal HbA and no anemia.
  • C. Sickle cell trait shows HbS alongside HbA, not absent HbA with elevated HbF.
Educational objective: Beta-thalassemia major shows absent HbA with markedly elevated HbF and HbA2 on electrophoresis, correlating with severe transfusion-dependent anemia from infancy.
39

A 45-year-old woman presents with fatigue and yellowish discoloration of eyes for 2 weeks. Examination shows pallor and mild splenomegaly. Hemoglobin is 7.8 g/dL, reticulocyte count 12%, and peripheral smear shows spherocytes and polychromasia without central pallor abnormalities typical of hereditary conditions. Serum indirect bilirubin is elevated and lactate dehydrogenase is raised. A test is performed by incubating the patient's washed red cells with anti-human globulin reagent, resulting in visible agglutination. What does this finding indicate?

  1. AThe patient's red cells have an inherited membrane protein defect
  2. BFree antibody against red cells is circulating in the patient's serum
  3. CAntibody or complement is bound to the patient's red cell surface in vivo
  4. DThe patient's red cells are deficient in glucose-6-phosphate dehydrogenase
Show answer
Answer: C. Antibody or complement is bound to the patient's red cell surface in vivo

The direct antiglobulin (Coombs) test detects antibody (usually IgG) or complement (C3d) already bound to the surface of circulating red cells in vivo. Anti-human globulin reagent cross-links these bound antibodies or complement, causing visible agglutination, which confirms an immune-mediated hemolytic process. This is the single best test to diagnose autoimmune hemolytic anemia and distinguishes it from hereditary or enzymatic causes of hemolysis. A positive direct Coombs test with evidence of hemolysis (elevated LDH, indirect bilirubin, reticulocytosis) confirms the autoimmune etiology.

  • A. An inherited membrane defect would not produce a positive direct Coombs test, since no antibody is involved.
  • B. Free circulating antibody in serum is detected by the indirect Coombs test, not the direct test.
  • D. G6PD deficiency is an enzyme defect unrelated to antibody binding and would not cause agglutination with anti-human globulin.
Educational objective: A positive direct Coombs test confirms antibody or complement bound to red cells in vivo, establishing the diagnosis of autoimmune hemolytic anemia.
40

A 19-year-old man presents with easy bruising, recurrent gum bleeding, and fatigue for 3 weeks. He gives a history of taking chloramphenicol for typhoid fever one month ago. Complete blood count shows hemoglobin 6.5 g/dL, total leukocyte count 1,800/microL, and platelet count 15,000/microL, with normal red cell indices and no abnormal cells on peripheral smear. Which bone marrow biopsy finding would be most consistent with this presentation?

  1. AMarrow fibrosis with teardrop-shaped cells in the biopsy
  2. BMarrow infiltration with abnormal plasma cells exceeding 10%
  3. CHypercellular marrow with blast cells exceeding 20% of nucleated cells
  4. DHypocellular marrow with increased fat spaces replacing hematopoietic tissue
Show answer
Answer: D. Hypocellular marrow with increased fat spaces replacing hematopoietic tissue

Aplastic anemia is characterized by pancytopenia with a hypocellular bone marrow in which hematopoietic tissue is replaced by fat, reflecting failure of the stem cell compartment rather than infiltration or fibrosis. Chloramphenicol is a well-recognized idiosyncratic cause of aplastic anemia, acting through direct marrow toxicity independent of dose. Other causes include idiopathic (most common), radiation, benzene exposure, and certain viral infections. The diagnosis requires a bone marrow biopsy showing less than 25% cellularity along with peripheral pancytopenia and no evidence of infiltrative or dysplastic disease.

  • A. Marrow fibrosis with teardrop cells is typical of primary myelofibrosis, not aplastic anemia.
  • B. Plasma cell infiltration suggests multiple myeloma, not marrow failure from drug toxicity.
  • C. Blast cell excess defines acute leukemia, not aplastic anemia.
Educational objective: Aplastic anemia shows a hypocellular, fat-replaced bone marrow with pancytopenia, and chloramphenicol is a classic idiosyncratic causative drug.

WBC Disorders & Lymphoid Neoplasms

7 questions
41

A 4-year-old boy presents with fever, pallor, and bone pain for 2 weeks. Examination shows multiple enlarged cervical lymph nodes and hepatosplenomegaly. Complete blood count shows hemoglobin 7 g/dL, total leukocyte count 42,000/microL with 70% blasts, and platelet count 30,000/microL. The blasts are small with scant cytoplasm and high nuclear-cytoplasmic ratio. Which of the following markers, if positive on these blasts, would be most specific in confirming the lineage of this leukemia?

  1. AGlycophorin A
  2. BCD14
  3. CTerminal deoxynucleotidyl transferase (TdT)
  4. DMyeloperoxidase
Show answer
Answer: C. Terminal deoxynucleotidyl transferase (TdT)

Terminal deoxynucleotidyl transferase (TdT) is a nuclear enzyme expressed in immature lymphoid precursor cells and is positive in the vast majority of acute lymphoblastic leukemia cases, confirming lymphoblastic lineage. ALL is the most common childhood malignancy, with peak incidence between 2 and 5 years of age, and presents with marrow failure features (anemia, thrombocytopenia), lymphadenopathy, and hepatosplenomegaly due to blast infiltration. TdT positivity, combined with CD19 and CD10 (common ALL antigen) expression, distinguishes ALL from acute myeloid leukemia, which is typically TdT negative. This immunophenotyping is essential since morphology alone cannot reliably distinguish lymphoblasts from myeloblasts.

  • A. Glycophorin A is an erythroid marker, used in erythroleukemia, not ALL.
  • B. CD14 is a monocytic marker, seen in monocytic leukemias, not lymphoblastic leukemia.
  • D. Myeloperoxidase is a marker of myeloid lineage, positive in AML, not ALL.
Educational objective: TdT positivity confirms lymphoblastic lineage in acute lymphoblastic leukemia, which peaks in incidence among children aged 2-5 years.
42

A 32-year-old woman presents with gum bleeding, easy bruising, and epistaxis for 5 days. Complete blood count shows hemoglobin 8 g/dL, total leukocyte count 3,200/microL, and platelet count 18,000/microL. Peripheral smear and bone marrow examination show abnormal promyelocytes with heavy cytoplasmic granulation and bundles of needle-shaped cytoplasmic inclusions. Coagulation studies show prolonged prothrombin time and low fibrinogen, consistent with disseminated intravascular coagulation. Which cytogenetic abnormality is most characteristic of this condition?

  1. At(8;14) involving MYC-IGH fusion
  2. Bt(15;17) involving PML-RARA fusion
  3. Ct(9;22) involving BCR-ABL1 fusion
  4. Dt(12;21) involving ETV6-RUNX1 fusion
Show answer
Answer: B. t(15;17) involving PML-RARA fusion

Acute promyelocytic leukemia (AML M3) is characterized by the translocation t(15;17), which creates the PML-RARA fusion gene and blocks promyelocyte differentiation at the promyelocyte stage. The abnormal promyelocytes contain bundles of multiple Auer rods, sometimes called faggot cells, and release procoagulant material that precipitates disseminated intravascular coagulation, a hallmark clinical complication. Recognizing this translocation is critical because it predicts dramatic response to all-trans retinoic acid, which induces terminal differentiation of the leukemic clone. Prompt initiation of ATRA improves survival and reduces the high early mortality from coagulopathy-related hemorrhage.

  • A. t(8;14) MYC-IGH is characteristic of Burkitt lymphoma.
  • C. t(9;22) BCR-ABL1 is the Philadelphia chromosome, characteristic of chronic myeloid leukemia.
  • D. t(12;21) ETV6-RUNX1 is a common favorable-prognosis translocation in childhood B-cell ALL.
Educational objective: Acute promyelocytic leukemia is defined by t(15;17) PML-RARA fusion, presenting with Auer rod-laden promyelocytes and DIC, and responds to ATRA therapy.
43

A 48-year-old man presents with fatigue and early satiety for 3 months. Examination reveals massive splenomegaly extending 10 cm below the left costal margin. Total leukocyte count is 2,10,000/µL with a full spectrum of granulocytic maturation including basophilia, and platelet count is 5,80,000/µL. Cytogenetic analysis shows a reciprocal translocation between chromosomes 9 and 22. Which of the following best explains the molecular consequence of this translocation that drives leukemogenesis?

  1. ATranslocation of MYC adjacent to immunoglobulin gene enhancers causing its overexpression
  2. BOverexpression of cyclin D1 due to juxtaposition with the immunoglobulin heavy chain enhancer
  3. CFormation of a PML-RARA fusion protein that blocks myeloid differentiation
  4. DFusion of ABL1 to BCR produces a chimeric protein with constitutive tyrosine kinase activity
Show answer
Answer: D. Fusion of ABL1 to BCR produces a chimeric protein with constitutive tyrosine kinase activity

The t(9;22)(q34;q11) translocation fuses the ABL1 gene on chromosome 9 to the BCR gene on chromosome 22, forming the Philadelphia chromosome. The resulting BCR-ABL1 fusion protein has constitutively active tyrosine kinase activity, driving uncontrolled proliferation of the granulocytic lineage independent of normal growth signals. This is the molecular basis for the clinical picture of chronic myeloid leukemia with marked leukocytosis and splenomegaly. Tyrosine kinase inhibitors such as imatinib specifically target this abnormal kinase.

  • A. MYC translocation with immunoglobulin genes, t(8;14), is the mechanism in Burkitt lymphoma.
  • B. Cyclin D1 overexpression via IGH juxtaposition is the mechanism in mantle cell lymphoma, t(11;14).
  • C. PML-RARA fusion is characteristic of acute promyelocytic leukemia, t(15;17).
Educational objective: The Philadelphia chromosome t(9;22) creates a BCR-ABL1 fusion gene encoding a constitutively active tyrosine kinase that causes chronic myeloid leukemia.
44

A 68-year-old man is found to have an elevated white cell count on a routine health check. He is asymptomatic, and examination reveals mild, painless bilateral cervical lymphadenopathy. Total leukocyte count is 62,000/µL with 85% mature-appearing small lymphocytes. The peripheral smear shows numerous fragmented, disrupted lymphocytes resulting from preparation artifact. Flow cytometry of the lymphocytes shows co-expression of CD5, CD23, and dim CD20. Which of the following is the most likely diagnosis?

  1. AHairy cell leukemia
  2. BChronic lymphocytic leukemia
  3. CFollicular lymphoma
  4. DMantle cell lymphoma
Show answer
Answer: B. Chronic lymphocytic leukemia

Chronic lymphocytic leukemia presents in older adults with an incidental lymphocytosis of small, mature-looking B lymphocytes that are fragile and rupture during smear preparation, producing characteristic smudge cells. The diagnostic immunophenotype is co-expression of CD5 (normally a T-cell marker) with CD23 and dim CD20 on the neoplastic B cells. This combination distinguishes CLL from other CD5-positive lymphoid neoplasms such as mantle cell lymphoma, which is typically CD23-negative. The clinical picture of painless lymphadenopathy with marked lymphocytosis in an elderly patient is classic.

  • A. Hairy cell leukemia shows cells with fine cytoplasmic projections and is positive for CD11c, CD25, and CD103, not CD5/CD23.
  • C. Follicular lymphoma cells are typically CD10 and bcl-2 positive but CD5-negative.
  • D. Mantle cell lymphoma is also CD5-positive but is characteristically CD23-negative and associated with cyclin D1 overexpression.
Educational objective: CLL is diagnosed by smudge cells on smear and a CD5+/CD23+ B-cell immunophenotype.
45

A 24-year-old woman presents with a painless, enlarging mass in the lower neck for two months, without fever, weight loss, or night sweats. Excisional lymph node biopsy shows scattered large binucleate cells with prominent eosinophilic nucleoli, set within a background of broad collagen bands dividing the node into nodules containing abundant lymphocytes and eosinophils. Immunohistochemistry on the large cells is positive for CD15 and CD30. Which of the following subtypes is most likely, and carries the best prognosis among Hodgkin lymphoma subtypes?

  1. ALymphocyte-rich type
  2. BLymphocyte-depleted type
  3. CMixed cellularity type
  4. DNodular sclerosis type
Show answer
Answer: D. Nodular sclerosis type

The described broad bands of collagen dividing the lymph node into nodules, along with lacunar-type Reed-Sternberg cells, are classic for nodular sclerosis Hodgkin lymphoma, the most common subtype overall and the typical presentation in young women with cervical or mediastinal masses. Reed-Sternberg cells characteristically express CD15 and CD30 but are usually negative for CD45 and CD20, distinguishing Hodgkin lymphoma from non-Hodgkin lymphomas. Although lymphocyte-predominant type has the single best prognosis among all Hodgkin subtypes overall, within classical Hodgkin lymphoma nodular sclerosis carries an excellent prognosis and is the subtype matching this histology. Prognosis in Hodgkin lymphoma correlates more with clinical stage than histologic subtype, but nodular sclerosis is traditionally taught as having a favorable outcome.

  • A. Lymphocyte-rich type shows abundant lymphocytes without the broad fibrous bands and nodularity, and is less common.
  • B. Lymphocyte-depleted type is rare, seen in older or HIV-positive patients, and carries the worst prognosis.
  • C. Mixed cellularity type lacks the broad sclerotic collagen bands and nodular pattern described here.
Educational objective: Nodular sclerosis is the most common Hodgkin lymphoma subtype, showing CD15+/CD30+ lacunar Reed-Sternberg cells within collagen-banded nodules, and carries a favorable prognosis.
46

An 8-year-old boy from a malaria-endemic region presents with a rapidly enlarging left mandibular mass over three weeks. Biopsy shows a diffuse infiltrate of medium-sized lymphoid cells with a very high mitotic rate, interspersed with pale-staining macrophages that have ingested apoptotic tumor cells, giving a characteristic microscopic appearance. Cytogenetic studies reveal a translocation involving chromosome 8 and chromosome 14. Which of the following best describes the molecular mechanism underlying this lymphoma?

  1. ADeletion of the RB1 tumor suppressor gene on chromosome 13
  2. BTranslocation placing BCL2 under the control of the immunoglobulin heavy chain enhancer
  3. CTranslocation placing MYC under the control of the immunoglobulin heavy chain enhancer
  4. DPoint mutation in the RAS oncogene leading to constitutive signaling
Show answer
Answer: C. Translocation placing MYC under the control of the immunoglobulin heavy chain enhancer

The t(8;14) translocation characteristic of Burkitt lymphoma juxtaposes the MYC oncogene on chromosome 8 with the immunoglobulin heavy chain locus on chromosome 14, placing MYC under the control of a highly active enhancer and causing its overexpression. This drives a very high proliferative rate, evident histologically as the starry sky pattern of numerous apoptotic tumor cells being phagocytosed by benign macrophages against a dark background of dense lymphoid cells. The endemic African form is strongly associated with Epstein-Barr virus infection and typically presents as a jaw mass in children, as in this case. MYC dysregulation distinguishes Burkitt lymphoma from other translocation-driven lymphomas such as follicular lymphoma, which involves BCL2.

  • A. RB1 deletion is associated with retinoblastoma and osteosarcoma, not Burkitt lymphoma.
  • B. BCL2 under IGH enhancer control, t(14;18), is the mechanism of follicular lymphoma, which has an indolent course.
  • D. RAS mutations are not the characteristic driver of Burkitt lymphoma.
Educational objective: Burkitt lymphoma is driven by a MYC-immunoglobulin gene translocation, most commonly t(8;14), producing the starry sky histologic pattern.
47

A 62-year-old woman presents with low back pain and fatigue for two months. Investigations show hemoglobin 8.9 g/dL, serum calcium 11.8 mg/dL, and serum creatinine 2.1 mg/dL. Serum protein electrophoresis reveals a monoclonal spike of 4.2 g/dL in the gamma region. Bone marrow aspirate shows 35% plasma cells with abnormal morphology. Skeletal survey shows multiple well-defined lucent lesions in the skull and pelvis. Urine electrophoresis demonstrates monoclonal free kappa light chains. Which of the following best accounts for the renal impairment in this patient?

  1. ADeposition of free light chains causing obstructive cast nephropathy in the distal tubules
  2. BImmune complex-mediated membranoproliferative glomerulonephritis
  3. CAscending bacterial pyelonephritis secondary to immunosuppression
  4. DDiffuse glomerular amyloid deposition causing nephrotic-range proteinuria
Show answer
Answer: A. Deposition of free light chains causing obstructive cast nephropathy in the distal tubules

This patient meets diagnostic criteria for multiple myeloma with bone marrow plasma cells over 10% (here 35%), a monoclonal (M) spike on serum protein electrophoresis, lytic bone lesions, hypercalcemia, and Bence Jones proteinuria (monoclonal free light chains in urine). The renal impairment in myeloma most commonly results from excess free light chains being filtered by the glomerulus and precipitating with Tamm-Horsfall protein in the distal tubules and collecting ducts, forming obstructive casts that cause tubular injury, a process known as myeloma cast nephropathy. This is distinct from AL amyloidosis, which can also occur in myeloma but typically causes nephrotic syndrome rather than the acute kidney injury pattern seen here. The combination of hypercalcemia, light chain toxicity, and dehydration commonly contributes to renal failure in myeloma patients.

  • B. Immune complex glomerulonephritis is not the typical mechanism of renal injury in myeloma.
  • C. There is no indication of infection in this presentation, and pyelonephritis is not the characteristic renal lesion of myeloma.
  • D. Amyloid deposition causes a nephrotic picture with heavy proteinuria rather than the tubular cast nephropathy typical of this presentation.
Educational objective: Multiple myeloma is diagnosed by the combination of marrow plasmacytosis, M spike, lytic lesions, and Bence Jones proteinuria, and renal failure typically results from light chain cast nephropathy.

Cardiovascular Pathology

2 questions
48

A pathology study examines the earliest grossly visible lesion of atherosclerosis in the aorta of a young adult who died of trauma. The lesion appears as a flat, yellow, irregular area that does not protrude significantly into the lumen. Microscopically, it consists of lipid-laden macrophages within the intima. Which of the following best describes the initiating step in the pathogenesis leading to the formation of this lesion?

  1. ARupture of the internal elastic lamina with medial smooth muscle proliferation
  2. BThrombus organization following platelet aggregation on an intact endothelium
  3. CEndothelial injury and dysfunction leading to increased permeability and leukocyte adhesion
  4. DDystrophic calcification of pre-existing fibrous plaques
Show answer
Answer: C. Endothelial injury and dysfunction leading to increased permeability and leukocyte adhesion

The lesion described is the fatty streak, the earliest grossly visible lesion of atherosclerosis, composed of intimal foam cells, which are macrophages that have ingested oxidized LDL. According to the response-to-injury hypothesis, atherogenesis begins with chronic endothelial injury or dysfunction caused by factors such as hemodynamic stress, hyperlipidemia, or toxins, which increases endothelial permeability and promotes adhesion of leukocytes, particularly monocytes. These monocytes migrate into the intima, differentiate into macrophages, and ingest lipoproteins to become foam cells, forming the fatty streak. This lesion can progress through intermediate stages to the mature fibrofatty atheromatous plaque, but it begins with endothelial dysfunction, not with medial injury, calcification, or thrombosis on intact endothelium.

  • A. Medial smooth muscle changes and elastic lamina disruption occur in later, more advanced atherosclerotic lesions, not the initiating fatty streak.
  • B. Significant thrombus formation on intact endothelium is not the mechanism of fatty streak formation; it requires endothelial injury.
  • D. Calcification is a feature of advanced, complicated atherosclerotic plaques, not the earliest lesion.
Educational objective: The fatty streak, composed of lipid-laden macrophages, is the earliest visible atherosclerotic lesion and begins with chronic endothelial injury and dysfunction.
49

A 9-year-old girl presents with migratory polyarthritis involving the knees and ankles for one week, along with a low-grade fever. She had a sore throat three weeks ago that resolved without treatment. Examination reveals a new pansystolic murmur at the apex and subcutaneous nodules over the extensor surfaces of both elbows. Anti-streptolysin O titer is elevated. According to the Jones criteria, which combination of findings supports the diagnosis of acute rheumatic fever in this patient, and which cardiac valve is most commonly affected in resulting chronic rheumatic heart disease?

  1. AOne major and two minor criteria with evidence of preceding streptococcal infection; aortic valve most commonly affected
  2. BTwo major criteria (carditis, polyarthritis) with evidence of preceding streptococcal infection; mitral valve most commonly affected
  3. CTwo major criteria (subcutaneous nodules, polyarthritis) with evidence of preceding streptococcal infection; tricuspid valve most commonly affected
  4. DTwo major criteria (carditis, polyarthritis) with evidence of preceding streptococcal infection; pulmonary valve most commonly affected
Show answer
Answer: B. Two major criteria (carditis, polyarthritis) with evidence of preceding streptococcal infection; mitral valve most commonly affected

This patient has two major Jones criteria: carditis, indicated by the new pansystolic murmur, and polyarthritis, along with subcutaneous nodules which would count as a third major criterion. Evidence of preceding group A streptococcal infection is supported by the recent sore throat and elevated ASO titer. The Jones criteria require either two major criteria or one major plus two minor criteria, together with evidence of preceding streptococcal infection, to diagnose acute rheumatic fever. In chronic rheumatic heart disease, the mitral valve is most commonly and severely affected, typically producing mitral stenosis or regurgitation, because it bears the highest hemodynamic stress among the four valves.

  • A. This patient has two clear major criteria (carditis and polyarthritis, plus nodules), not just one major criterion; also, the aortic valve is affected less often than the mitral valve.
  • C. Subcutaneous nodules and polyarthritis are both major criteria, but the murmur indicates carditis, also major; the tricuspid valve is rarely the most affected valve in rheumatic heart disease.
  • D. The valve most commonly affected in rheumatic heart disease is the mitral valve, not the pulmonary valve, which is rarely involved.
Educational objective: Diagnosis of acute rheumatic fever by Jones criteria requires two major or one major plus two minor manifestations with evidence of prior streptococcal infection, and the mitral valve is most commonly affected in resulting chronic rheumatic heart disease.

GI & Liver Pathology

1 questions
50

A 58-year-old man undergoes colonoscopy for occult blood in stool and is found to have a 2.5 cm tubulovillous adenoma in the sigmoid colon. Molecular studies on the polyp show biallelic inactivation of a tumor suppressor gene on chromosome 5q21, which is considered the earliest genetic event in the pathway leading to malignant transformation of such polyps. Loss of function of this gene primarily promotes tumorigenesis by which of the following mechanisms?

  1. ALoss of p53-mediated cell cycle arrest and apoptosis in response to DNA damage
  2. BDefective repair of DNA mismatches, resulting in microsatellite instability
  3. CFailure to degrade beta-catenin, leading to unchecked Wnt signaling and increased cell proliferation
  4. DConstitutive activation of RAS-MAPK signaling due to a point mutation in codon 12
Show answer
Answer: C. Failure to degrade beta-catenin, leading to unchecked Wnt signaling and increased cell proliferation

The APC (adenomatous polyposis coli) gene on chromosome 5q21 normally forms a complex with axin and GSK-3beta that phosphorylates beta-catenin, targeting it for ubiquitin-mediated degradation. Biallelic loss of APC prevents this degradation, so beta-catenin accumulates, translocates to the nucleus, and acts as a transcriptional co-activator with TCF to drive expression of proliferative genes such as cyclin D1 and MYC. This is the initiating event in the classical (chromosomal instability) adenoma-carcinoma sequence, occurring before KRAS activation and p53 loss. Germline APC mutations cause familial adenomatous polyposis, and sporadic APC mutation or promoter hypermethylation is found in the majority of sporadic colorectal adenomas and carcinomas.

  • A. p53 loss is a later event in the adenoma-carcinoma sequence, associated with progression from large adenoma to carcinoma, not the initiating step.
  • B. This describes the mismatch repair pathway (HNPCC/Lynch syndrome), a distinct microsatellite instability route, not the APC-driven chromosomal instability pathway.
  • D. KRAS mutation is an intermediate event that occurs after APC loss, promoting growth of the adenoma rather than initiating it.
Educational objective: APC gene inactivation on chromosome 5q21 is the earliest event in the classical adenoma-carcinoma sequence, causing beta-catenin accumulation and unchecked Wnt pathway activation.

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