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USMLE Step 1 · United States

USMLE Step 1 hematology & oncology questions

50 original USMLE Step 1-style questions on hematology & oncology, 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
Sections15
Options per question5
Time≈ 75 min
USMLE Step 1 Hematology & oncology: practice quizPick an answer to see the correct option and the explanation at once. Your best score is saved in this browser.

Red Cell Disorders - Microcytic Anemia

4 questions
1

A 34-year-old woman complains of fatigue and brittle nails for 3 months. She has heavy menstrual bleeding requiring 6-7 pads daily for the past year. Vital signs are normal. Examination shows pale conjunctivae and spoon-shaped nails. Laboratory studies show hemoglobin 9.2 g/dL, mean corpuscular volume 68 fL, serum ferritin 8 ng/mL (reference 20-200 ng/mL), total iron-binding capacity 480 mcg/dL (reference 250-450 mcg/dL), and transferrin saturation 9% (reference 20-50%). Which of the following best explains this combination of findings?

  1. ADecreased transferrin synthesis by the liver secondary to malnutrition
  2. BChronic inflammation causing sequestration of iron within macrophages
  3. CIncreased hepcidin production blocking ferroportin-mediated iron export
  4. DDecreased body iron stores with compensatory upregulation of transferrin synthesis
  5. EDefective iron utilization by erythroid precursors due to mitochondrial dysfunction
Show answer
Answer: D. Decreased body iron stores with compensatory upregulation of transferrin synthesis

Low ferritin reflects depleted iron stores, which is the hallmark of true iron deficiency. The liver responds to low iron stores by increasing transferrin synthesis, raising TIBC, while the low circulating iron relative to this increased transferrin produces a low transferrin saturation. This pattern (low ferritin, high TIBC, low saturation) is diagnostic of iron deficiency anemia, commonly caused by chronic blood loss such as menorrhagia. Chronic disease anemia instead shows low TIBC and low-normal to high ferritin because hepcidin traps iron in macrophages.

  • A. Decreased transferrin synthesis would lower TIBC, the opposite of what is seen here.
  • B. Inflammation-mediated iron sequestration causes anemia of chronic disease, which shows low TIBC and normal to high ferritin, not this pattern.
  • C. Hepcidin elevation is the mechanism of anemia of chronic disease, which produces low TIBC, not elevated TIBC as seen here.
  • E. Sideroblastic anemia involves normal or increased iron stores with high ferritin, not low ferritin.
Educational objective: Iron deficiency anemia is characterized by low ferritin, high TIBC, and low transferrin saturation, distinguishing it from anemia of chronic disease.
2

A 19-year-old man of Mediterranean descent is found to have mild anemia on a routine pre-employment physical. He is asymptomatic and has no history of abnormal bleeding. Physical examination is unremarkable. Laboratory studies show hemoglobin 11.8 g/dL, mean corpuscular volume 62 fL, red blood cell count 5.9 million/mcL, and red cell distribution width 12.5% (reference 11.5-14.5%). Serum ferritin and iron studies are within normal limits. Which of the following is the most likely diagnosis?

  1. ASideroblastic anemia
  2. BAnemia of chronic kidney disease
  3. CHereditary spherocytosis
  4. DBeta-thalassemia minor
  5. EIron deficiency anemia
Show answer
Answer: D. Beta-thalassemia minor

The Mentzer index (MCV divided by RBC count) here is 62/5.9 = 10.5, which is below 13 and suggests thalassemia trait rather than iron deficiency. A normal RDW with a disproportionately high RBC count relative to the degree of anemia and microcytosis, plus normal iron studies, further supports beta-thalassemia minor, in which ineffective erythropoiesis produces numerous small, relatively uniform red cells. Iron deficiency anemia typically produces an elevated RDW from anisocytosis and a lower RBC count for the same degree of microcytosis.

  • A. Sideroblastic anemia shows elevated iron stores and ringed sideroblasts, not normal iron studies with this RBC pattern.
  • B. Chronic kidney disease causes normocytic anemia from decreased erythropoietin, not microcytosis.
  • C. Hereditary spherocytosis causes normocytic anemia with spherocytes and increased osmotic fragility, not microcytosis.
  • E. Iron deficiency would show low ferritin and an elevated RDW with a Mentzer index typically above 13.
Educational objective: A Mentzer index (MCV/RBC) below 13 with normal iron studies and a near-normal RDW favors thalassemia trait over iron deficiency anemia.
3

A 2-day-old male infant of Southeast Asian descent is found to have generalized edema, hepatosplenomegaly, and severe pallor at birth. He was delivered stillborn-appearing but resuscitated; he dies shortly after birth despite intervention. Hemoglobin electrophoresis of cord blood shows a predominant hemoglobin tetramer composed entirely of gamma chains. Which of the following best explains this infant's condition?

  1. ADeletion of a single alpha-globin gene copy
  2. BDeletion of two alpha-globin gene copies in cis
  3. CDeletion of three of four alpha-globin gene copies
  4. DDeletion of all four alpha-globin gene copies
  5. EDeletion of two alpha-globin gene copies in trans
Show answer
Answer: D. Deletion of all four alpha-globin gene copies

Deletion of all four alpha-globin genes prevents any alpha-globin synthesis, so excess gamma chains form tetramers called hemoglobin Barts, which has very high oxygen affinity and cannot deliver oxygen to tissues, causing hydrops fetalis and death in utero or shortly after birth. Three-gene deletion causes hemoglobin H disease, a moderately severe hemolytic anemia with beta-chain tetramers (HbH) but compatible with life. One or two-gene deletions cause silent carrier state or mild microcytic anemia (alpha-thalassemia trait), respectively, without hydrops.

  • A. A single gene deletion causes a silent carrier state with no clinical findings.
  • B. Two-gene deletion in cis causes alpha-thalassemia trait with mild microcytic anemia, not hydrops fetalis.
  • C. Three-gene deletion causes hemoglobin H disease, a moderate hemolytic anemia compatible with survival, not fatal hydrops fetalis.
  • E. Two-gene deletion in trans also causes mild alpha-thalassemia trait, not a fatal condition.
Educational objective: Deletion of all four alpha-globin genes causes hemoglobin Barts and hydrops fetalis, which is fatal, whereas three-gene deletion causes the less severe hemoglobin H disease.
4

A 6-year-old boy is evaluated for fatigue and pallor. He has no history of toxin exposure or pica. Laboratory studies show hemoglobin 8.9 g/dL, mean corpuscular volume 68 fL, serum iron 180 mcg/dL (reference 50-150 mcg/dL), ferritin 310 ng/mL (reference 20-200 ng/mL), and total iron-binding capacity 220 mcg/dL (reference 250-450 mcg/dL). Bone marrow aspirate with Prussian blue stain shows erythroid precursors with iron granules forming rings around the nucleus. Genetic testing reveals a mutation in the gene encoding erythroid-specific delta-aminolevulinic acid synthase. Which of the following best describes the biochemical consequence of this mutation?

  1. AImpaired dehydration of two molecules of delta-aminolevulinic acid
  2. BImpaired oxidation of coproporphyrinogen III to protoporphyrinogen IX
  3. CImpaired condensation of glycine and succinyl-CoA in mitochondria
  4. DImpaired incorporation of iron into protoporphyrin IX
  5. EImpaired conversion of porphobilinogen to hydroxymethylbilane
Show answer
Answer: C. Impaired condensation of glycine and succinyl-CoA in mitochondria

ALAS2 (erythroid-specific delta-aminolevulinic acid synthase) catalyzes the first and rate-limiting step of heme synthesis, condensing glycine and succinyl-CoA using vitamin B6 as a cofactor to form delta-aminolevulinic acid in mitochondria. Loss-of-function mutations impair this initial step, reducing heme production and causing iron to accumulate in mitochondria, visualized as ringed sideroblasts on Prussian blue stain. This produces a microcytic anemia with elevated iron stores, distinguishing it from iron deficiency, in which stores are depleted.

  • A. ALA dehydratase catalyzes this step; it is inhibited in lead poisoning and ALA dehydratase deficiency porphyria, not by ALAS2 mutation.
  • B. Coproporphyrinogen oxidase catalyzes this step; its deficiency causes hereditary coproporphyria.
  • D. Ferrochelatase catalyzes iron incorporation into protoporphyrin IX; this is inhibited in lead poisoning, not ALAS2 deficiency.
  • E. Porphobilinogen deaminase catalyzes this step; its deficiency causes acute intermittent porphyria.
Educational objective: ALAS2 mutations impair the first step of heme synthesis (glycine + succinyl-CoA condensation), causing X-linked sideroblastic anemia with ringed sideroblasts and elevated iron stores.

Red Cell Disorders - Macrocytic Anemia

3 questions
5

A 68-year-old man presents with a 4-month history of progressive gait instability and tingling in both feet. He underwent total gastrectomy for gastric cancer 3 years ago. Examination shows decreased vibration and proprioception in the lower extremities, a positive Romberg sign, and bilateral extensor plantar responses with hyperreflexia at the knees but absent ankle reflexes. Laboratory studies show hemoglobin 9.8 g/dL, mean corpuscular volume 112 fL, and serum vitamin B12 110 pg/mL (reference 200-900 pg/mL). Which of the following metabolic disturbances is most directly responsible for this patient's neurologic findings?

  1. AImpaired conversion of uridine monophosphate to deoxythymidine monophosphate
  2. BDecreased regeneration of tetrahydrofolate from dihydrofolate
  3. CImpaired conversion of homocysteine to cystathionine
  4. DDecreased synthesis of S-adenosylmethionine from homocysteine
  5. EAccumulation of methylmalonyl-CoA impairing myelin synthesis
Show answer
Answer: E. Accumulation of methylmalonyl-CoA impairing myelin synthesis

Vitamin B12 is a cofactor for methylmalonyl-CoA mutase, which converts methylmalonyl-CoA to succinyl-CoA. B12 deficiency causes methylmalonyl-CoA accumulation, which is thought to disrupt normal fatty acid synthesis and myelin formation in the dorsal columns and lateral corticospinal tracts, producing subacute combined degeneration with combined posterior column and corticospinal signs. B12 is also required to regenerate methionine (and thus S-adenosylmethionine) from homocysteine via methionine synthase, but this pathway is more directly linked to the macrocytic anemia and megaloblastic changes rather than the specific myelin pathology.

  • A. This step requires thymidylate synthase and folate, not B12, and explains megaloblastic changes, not the neurologic findings.
  • B. Tetrahydrofolate regeneration from dihydrofolate requires dihydrofolate reductase, not B12, and relates to folate metabolism, not this neurologic syndrome.
  • C. Cystathionine synthesis requires vitamin B6, not B12, and its impairment is not the mechanism of subacute combined degeneration.
  • D. Decreased SAM synthesis contributes to impaired methylation but is more closely linked to the hematologic than the specific demyelinating findings.
Educational objective: Vitamin B12 deficiency causes subacute combined degeneration through methylmalonyl-CoA accumulation, which impairs myelin synthesis in the dorsal columns and corticospinal tracts.
6

A 52-year-old woman with a history of chronic alcohol use disorder presents with fatigue. She reports poor dietary intake over the past several months. She has no paresthesias, gait abnormalities, or sensory deficits. Examination is unremarkable except for mild pallor. Laboratory studies show hemoglobin 10.1 g/dL, mean corpuscular volume 108 fL, and hypersegmented neutrophils on peripheral smear. Serum homocysteine is elevated. Which of the following laboratory results would best help confirm the most likely underlying cause of her anemia rather than an alternative vitamin deficiency? Serum methylmalonic acid: normal

  1. ADecreased serum haptoglobin level
  2. BElevated serum methylmalonic acid level
  3. CNormal serum methylmalonic acid level
  4. DElevated serum indirect bilirubin level
  5. EElevated serum lactate dehydrogenase level
Show answer
Answer: C. Normal serum methylmalonic acid level

Both folate and vitamin B12 deficiency cause megaloblastic macrocytic anemia with hypersegmented neutrophils and elevated homocysteine, because both vitamins are needed for homocysteine remethylation to methionine. However, methylmalonyl-CoA mutase requires only B12, not folate, to convert methylmalonyl-CoA to succinyl-CoA. A normal methylmalonic acid level in the setting of megaloblastic anemia and elevated homocysteine indicates that B12-dependent metabolism is intact, pointing to folate deficiency as the cause rather than B12 deficiency.

  • A. Decreased haptoglobin reflects hemolysis, which occurs in both B12 and folate deficiency from intramedullary destruction of defective erythroid precursors.
  • B. An elevated methylmalonic acid level would indicate B12 deficiency, not support a diagnosis of isolated folate deficiency.
  • D. Elevated indirect bilirubin reflects hemolysis from ineffective erythropoiesis, seen in both B12 and folate deficiency, so it does not distinguish between them.
  • E. Elevated LDH reflects increased cell turnover from ineffective erythropoiesis but occurs in both B12 and folate deficiency, so it does not distinguish between them.
Educational objective: A normal methylmalonic acid level with elevated homocysteine in megaloblastic anemia points to folate deficiency rather than vitamin B12 deficiency.
7

A 61-year-old woman presents with fatigue, a smooth, beefy-red tongue, and numbness in both feet for 2 months. She has a history of vitiligo and autoimmune thyroiditis. Laboratory studies show hemoglobin 8.4 g/dL, mean corpuscular volume 118 fL, and serum vitamin B12 95 pg/mL (reference 200-900 pg/mL). Serum gastrin is markedly elevated. Upper endoscopy shows atrophic mucosa limited to the body and fundus of the stomach, with antral sparing. Which of the following is the most likely underlying mechanism of this patient's disease?

  1. AAutoantibodies against gastric G cells
  2. BSurgical resection of the terminal ileum
  3. CChronic Helicobacter pylori infection of the gastric antrum
  4. DAutoantibodies against the thyrotropin receptor
  5. EAutoantibodies against gastric parietal cells and intrinsic factor
Show answer
Answer: E. Autoantibodies against gastric parietal cells and intrinsic factor

Pernicious anemia results from autoimmune destruction of gastric parietal cells and antibodies against intrinsic factor, leading to achlorhydria, loss of intrinsic factor, and impaired B12 absorption in the terminal ileum. The resulting hypochlorhydria removes negative feedback on gastrin-secreting G cells, causing hypergastrinemia, and the body/fundus-predominant atrophic gastritis with antral sparing is characteristic. Patients with pernicious anemia have an increased risk of gastric adenocarcinoma and are often associated with other autoimmune conditions such as vitiligo and autoimmune thyroiditis.

  • A. G cells are not the target of autoimmune attack in pernicious anemia; their hyperplasia and increased gastrin secretion are a secondary consequence of achlorhydria.
  • B. Terminal ileum resection causes B12 deficiency from loss of the absorption site, but there is no history of surgery here, and it would not explain the atrophic gastritis and hypergastrinemia.
  • C. H. pylori-associated gastritis typically involves the antrum and does not classically spare it, and does not directly cause B12 deficiency through parietal cell autoimmunity.
  • D. Antibodies against the thyrotropin receptor cause Graves disease, which is a separate autoimmune condition that may coexist but does not cause this patient's anemia.
Educational objective: Pernicious anemia is caused by autoantibodies against parietal cells and intrinsic factor, producing body-predominant atrophic gastritis, achlorhydria, hypergastrinemia, and B12 deficiency, with increased gastric cancer risk.
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Red Cell Disorders - Hemolytic Anemia (Immune/Other)

5 questions
8

A 58-year-old woman with systemic lupus erythematosus presents with fatigue and dark urine for 1 week. Temperature is 37.1°C (98.8°F), pulse is 102/min, and blood pressure is 110/70 mm Hg. Examination shows scleral icterus and mild splenomegaly. Laboratory studies show hemoglobin 8.1 g/dL, total bilirubin 3.4 mg/dL (mostly indirect), lactate dehydrogenase 420 U/L, and haptoglobin below the detection limit. Peripheral blood smear shows numerous spherocytes and polychromasia without schistocytes. A direct antiglobulin test is positive for IgG. Which of the following best describes the mechanism responsible for this patient's anemia?

  1. AComplement-mediated intravascular lysis of red cells by IgM antibodies
  2. BOxidative denaturation of hemoglobin due to glucose-6-phosphate dehydrogenase deficiency
  3. CSplenic macrophage Fc receptors phagocytose antibody-coated red cells
  4. DOsmotic lysis of red cells due to a membrane protein deficiency
  5. EMechanical fragmentation of red cells by fibrin strands in small vessels
Show answer
Answer: C. Splenic macrophage Fc receptors phagocytose antibody-coated red cells

Warm autoimmune hemolytic anemia is caused by IgG autoantibodies that bind red cell surface antigens optimally at 37°C. Splenic macrophages recognize the Fc portion of IgG and phagocytose part or all of the red cell membrane, producing spherocytes and extravascular hemolysis predominantly in the spleen. This process is reflected by a positive direct antiglobulin (Coombs) test for IgG, elevated indirect bilirubin and LDH, and low haptoglobin. Warm AIHA is associated with autoimmune diseases such as SLE and lymphoproliferative disorders such as CLL.

  • A. IgM-mediated complement lysis describes cold agglutinin disease, not warm AIHA with IgG and spherocytes.
  • B. G6PD deficiency causes episodic hemolysis with Heinz bodies and bite cells after oxidant stress, not a Coombs-positive process.
  • D. A primary membrane protein defect describes hereditary spherocytosis, which would not have a positive Coombs test.
  • E. Fibrin strand shearing causes schistocytes in microangiopathic hemolytic anemia, not the spherocytes seen here.
Educational objective: Warm AIHA is an IgG-mediated, Coombs-positive, spherocytic extravascular hemolytic anemia caused by splenic Fc-receptor phagocytosis, often associated with SLE or CLL.
9

A 24-year-old man develops a nonproductive cough, fever, and malaise 10 days after returning from a college dormitory outbreak of respiratory illness. On examination he has blue-purple discoloration of the fingertips and ears that resolves with warming. Temperature is 38.0°C (100.4°F). Chest auscultation reveals scattered crackles. Laboratory studies show hemoglobin 9.8 g/dL, reticulocyte count 4.5%, and a peripheral smear with red cell agglutination that disperses on warming the slide. A direct antiglobulin test is positive for complement (C3) but negative for IgG. Which of the following is the most likely underlying cause of this patient's hemolysis?

  1. AADAMTS13 autoantibody formation reducing von Willebrand factor cleavage
  2. BMechanical destruction of red cells by a defective prosthetic heart valve
  3. CShiga toxin-mediated endothelial injury following enteric infection
  4. DIgG antibodies against red cell antigens triggered by a lupus flare
  5. EIgM antibodies against red cell antigens induced by Mycoplasma pneumoniae infection
Show answer
Answer: E. IgM antibodies against red cell antigens induced by Mycoplasma pneumoniae infection

Cold agglutinin disease is caused by IgM autoantibodies, frequently triggered by Mycoplasma pneumoniae infection (or lymphoproliferative disorders such as CLL in older patients), that bind red cell antigens at low temperatures in the peripheral circulation. IgM strongly activates complement, which deposits C3 on the red cell surface; the direct antiglobulin test is therefore positive for complement but negative for IgG, since IgM itself dissociates at warmer temperatures before testing. Acrocyanosis in extremities exposed to cold reflects agglutination-induced sludging in peripheral capillaries. Warming the blood causes the agglutinated cells to disperse, confirming the cold-dependent nature of the antibody.

  • A. ADAMTS13 deficiency causes TTP with schistocytes, not a complement-positive, IgG-negative agglutination picture.
  • B. Mechanical valve hemolysis produces schistocytes from shear stress, not temperature-dependent agglutination with acrocyanosis.
  • C. Shiga toxin causes HUS with schistocytes and thrombocytopenia, not IgM-mediated agglutination.
  • D. A lupus-associated IgG mechanism would produce a warm, Coombs-IgG-positive hemolysis with spherocytes, not cold-induced agglutination with acrocyanosis.
Educational objective: Cold agglutinin disease is caused by IgM autoantibodies, often following Mycoplasma pneumoniae infection, that fix complement on red cells and cause peripheral agglutination and acrocyanosis in the cold.
10

A 34-year-old woman is admitted with fever, confusion, and petechiae. Temperature is 38.6°C (101.5°F), pulse is 110/min, and blood pressure is 128/82 mm Hg. Laboratory studies show hemoglobin 7.9 g/dL, platelet count 18,000/mm³, creatinine 1.6 mg/dL, lactate dehydrogenase 680 U/L, and undetectable haptoglobin. Prothrombin time and partial thromboplastin time are normal. A peripheral blood smear shows numerous fragmented, helmet-shaped red cells and crescent-shaped cell fragments with no increase in spherocytes. Which of the following best explains the origin of the abnormal red cell fragments seen on this patient's smear?

  1. AOsmotic fragility due to a deficiency of membrane skeletal protein ankyrin
  2. BComplement-mediated lysis of red cells by cold-reactive IgM antibodies
  3. CMechanical shearing of red cells as they pass through vessels partially occluded by platelet microthrombi
  4. DOxidative damage to red cell membranes from an unstable hemoglobin variant
  5. EPhagocytosis of antibody-coated red cells by splenic macrophages
Show answer
Answer: C. Mechanical shearing of red cells as they pass through vessels partially occluded by platelet microthrombi

Schistocytes (helmet cells, fragmented red cells) arise when red cells are forced through small vessels narrowed or occluded by platelet-rich microthrombi, as occurs in thrombotic thrombocytopenic purpura, hemolytic uremic syndrome, and disseminated intravascular coagulation. The intravascular shear stress fragments the red cell membrane, producing the characteristic jagged fragments seen on smear, together with thrombocytopenia from platelet consumption and elevated LDH from cell lysis. A normal PT/PTT in this patient argues against DIC and supports TTP, but the smear mechanism (mechanical shear from microthrombi) is shared by all microangiopathic processes. Haptoglobin is consumed because it binds free hemoglobin released by intravascular hemolysis.

  • A. Ankyrin deficiency underlies hereditary spherocytosis, which produces spherocytes rather than fragmented cells.
  • B. Cold IgM-mediated lysis causes agglutination and acrocyanosis, not schistocyte formation.
  • D. Unstable hemoglobin causing oxidative damage produces Heinz bodies and bite cells, not helmet cells.
  • E. Splenic phagocytosis of antibody-coated cells produces spherocytes, not fragmented schistocytes, and is not consistent with this smear.
Educational objective: Schistocytes on a peripheral smear reflect mechanical red cell fragmentation as cells traverse vessels occluded by platelet microthrombi, as in TTP, HUS, and DIC.
11

A 29-year-old woman presents with fatigue, easy bruising, and confusion. Temperature is 38.3°C (100.9°F). Examination shows scattered petechiae and no lymphadenopathy. Laboratory studies show hemoglobin 8.0 g/dL, platelet count 15,000/mm³, creatinine 0.9 mg/dL, and lactate dehydrogenase 710 U/L. Peripheral smear shows schistocytes. Prothrombin time, partial thromboplastin time, and fibrinogen are normal. Plasma activity of a specific protease is found to be less than 10% of normal. Deficiency of this protease leads to this patient's disease by which of the following mechanisms?

  1. ADepletion of coagulation factors from widespread activation of the extrinsic pathway
  2. BAccumulation of unusually large von Willebrand factor multimers that promote platelet aggregation
  3. CLoss of protein C-mediated inhibition of factor Va and factor VIIIa
  4. DImpaired cleavage of fibrin degradation products leading to vascular fibrin deposition
  5. EUncontrolled activation of the complement alternative pathway on endothelial surfaces
Show answer
Answer: B. Accumulation of unusually large von Willebrand factor multimers that promote platelet aggregation

ADAMTS13 is a plasma metalloprotease that normally cleaves large von Willebrand factor (vWF) multimers released from endothelial cells into smaller, less adhesive forms. Severe deficiency of ADAMTS13 (due to autoantibody inhibition or, less commonly, a genetic defect) allows unusually large vWF multimers to persist in the circulation, where they spontaneously bind platelets and promote widespread platelet microthrombus formation in small vessels. This causes the classic findings of TTP: thrombocytopenia from platelet consumption, microangiopathic hemolytic anemia from shear-induced red cell fragmentation, and organ ischemia (notably neurologic and renal), typically with a normal PT/PTT/fibrinogen since the coagulation cascade is not primarily activated.

  • A. Extrinsic pathway consumption of coagulation factors describes DIC, which classically shows abnormal PT/PTT and low fibrinogen.
  • C. Loss of protein C function describes a hypercoagulable state unrelated to ADAMTS13 or vWF multimers.
  • D. Impaired fibrin degradation is not the mechanism of ADAMTS13 deficiency, which acts on vWF rather than fibrin.
  • E. Uncontrolled complement activation on endothelium underlies atypical HUS, not ADAMTS13 deficiency.
Educational objective: ADAMTS13 deficiency allows persistence of ultra-large vWF multimers that trigger platelet microthrombi, producing the thrombocytopenia and microangiopathic hemolysis of TTP.
12

A 4-year-old boy is brought to the emergency department with bloody diarrhea for 3 days after eating a hamburger at a family barbecue. Temperature is 37.8°C (100°F). Examination shows pallor and mild abdominal tenderness without peritoneal signs. Laboratory studies show hemoglobin 7.6 g/dL, platelet count 42,000/mm³, creatinine 2.1 mg/dL, and lactate dehydrogenase 650 U/L. Peripheral smear shows schistocytes. Stool culture grows a sorbitol-nonfermenting strain of Escherichia coli. Which of the following best describes the mechanism by which this organism produces this child's syndrome?

  1. AA toxin that ADP-ribosylates elongation factor-2, halting host protein synthesis in renal tubular cells
  2. BA toxin that activates adenylate cyclase, causing massive secretory diarrhea and volume depletion
  3. CA toxin that inhibits the 60S ribosomal subunit, causing endothelial cell injury and local thrombus formation
  4. DDirect bacterial invasion of the renal glomerular capillaries with complement deposition
  5. EAn exotoxin that cross-links MHC class II and the T-cell receptor, causing polyclonal T-cell activation
Show answer
Answer: C. A toxin that inhibits the 60S ribosomal subunit, causing endothelial cell injury and local thrombus formation

Enterohemorrhagic E. coli O157:H7 produces Shiga-like toxin, which inhibits the 60S ribosomal subunit and halts protein synthesis in target cells, particularly the glomerular endothelium, which is rich in the toxin's receptor (globotriaosylceramide, Gb3). Toxin-induced endothelial injury exposes subendothelial collagen and promotes local platelet activation and microthrombus formation, producing the triad of microangiopathic hemolytic anemia (schistocytes), thrombocytopenia, and acute kidney injury that defines HUS. This typically follows a prodrome of bloody diarrhea after ingestion of undercooked beef or contaminated produce, and antibiotics are generally avoided because bacterial killing can increase toxin release.

  • A. ADP-ribosylation of EF-2 describes diphtheria toxin and Pseudomonas exotoxin A, not Shiga toxin.
  • B. Adenylate cyclase activation describes cholera toxin and E. coli heat-labile toxin, which cause watery diarrhea, not HUS.
  • D. HUS results from a circulating toxin's endothelial effects, not direct bacterial invasion of the kidney.
  • E. Superantigen-mediated T-cell activation describes toxic shock syndrome toxins, not Shiga toxin.
Educational objective: Shiga toxin from EHEC O157:H7 inhibits the 60S ribosomal subunit in glomerular endothelial cells, triggering microthrombosis and the triad of HUS: hemolytic anemia, thrombocytopenia, and acute kidney injury.

Red Cell Disorders - Aplastic/Other

2 questions
13

A 19-year-old man presents with fatigue, gum bleeding, and recurrent fevers over the past 3 weeks. He reports no recent travel and takes no medications. Temperature is 38.4°C (101.1°F), pulse is 102/min. Examination shows pallor, scattered petechiae, and no lymphadenopathy or splenomegaly. Laboratory studies show hemoglobin 7.2 g/dL, white blood cell count 1,900/mm³ with absolute neutrophil count 600/mm³, and platelet count 22,000/mm³. Reticulocyte count is 0.3%. A bone marrow biopsy shows less than 10% cellularity with fatty replacement and no abnormal infiltrate. Which of the following is the most likely underlying mechanism of this patient's condition?

  1. AInfiltration of the marrow by malignant lymphocytes
  2. BClonal proliferation of myeloid blasts replacing normal marrow elements
  3. CMegaloblastic maturation arrest from impaired DNA synthesis
  4. DT-cell-mediated destruction of hematopoietic stem cells
  5. EAntibody-mediated destruction of mature circulating blood cells
Show answer
Answer: D. T-cell-mediated destruction of hematopoietic stem cells

Aplastic anemia results from immune-mediated, typically T-cell-mediated, destruction of hematopoietic stem cells, leading to pancytopenia with a hypocellular, fatty marrow and low reticulocyte count reflecting the inability to produce new blood cells. Known triggers include viral infections (parvovirus B19, particularly in patients with an underlying hemolytic condition), ionizing radiation, and drugs such as chloramphenicol and certain chemotherapeutics, although many cases are idiopathic and autoimmune in nature. The absence of blasts, abnormal infiltrate, or organomegaly distinguishes this from a leukemic or infiltrative process, and the low reticulocyte count with marrow fatty replacement is the key diagnostic finding.

  • A. Malignant lymphocytic infiltration describes lymphoma or lymphocytic leukemia involving the marrow, which would show cellular infiltration, not fatty hypocellularity.
  • B. Clonal blast proliferation describes acute leukemia, which would show a hypercellular marrow with blasts, not hypocellularity.
  • C. Megaloblastic maturation arrest occurs in vitamin B12 or folate deficiency, with a hypercellular marrow showing giant precursors, not marrow aplasia.
  • E. Antibody-mediated destruction of mature cells describes peripheral cytopenias such as ITP or AIHA, not a hypocellular marrow failure state.
Educational objective: Aplastic anemia is caused by T-cell-mediated destruction of hematopoietic stem cells, producing pancytopenia with a hypocellular, fatty bone marrow and low reticulocyte count.
14

A 61-year-old woman with long-standing rheumatoid arthritis presents for a routine visit with persistent fatigue. She has no gastrointestinal symptoms and no history of blood loss. Laboratory studies show hemoglobin 9.8 g/dL, mean corpuscular volume 82 fL, serum iron 32 µg/dL (normal 60-170), total iron-binding capacity 180 µg/dL (normal 250-450), ferritin 310 ng/mL (normal 20-200), and C-reactive protein 42 mg/L (normal <10). Which of the following best explains the laboratory findings in this patient?

  1. AIncreased hepcidin production limits iron release from macrophages and absorption from the gut
  2. BChronic occult gastrointestinal blood loss depleting total body iron stores
  3. CVitamin B12 malabsorption impairing DNA synthesis in erythroid precursors
  4. DAutoantibody-mediated destruction of erythroid precursors in the bone marrow
  5. EImpaired globin chain synthesis due to a point mutation in the beta-globin gene
Show answer
Answer: A. Increased hepcidin production limits iron release from macrophages and absorption from the gut

In anemia of chronic disease, chronic inflammation (as in rheumatoid arthritis) increases hepatic production of hepcidin, which degrades ferroportin on macrophages and enterocytes. This traps iron within macrophages and reduces dietary iron absorption, producing low serum iron despite normal or increased total body iron stores. Because iron is sequestered rather than depleted, ferritin (an acute phase reactant and marker of stored iron) is normal to high, and TIBC (transferrin) is low because transferrin synthesis is also downregulated by inflammation, distinguishing this pattern from iron deficiency anemia, where ferritin is low and TIBC is high.

  • B. Chronic GI blood loss would cause iron deficiency anemia with low ferritin and high TIBC, not the pattern seen here.
  • C. B12 deficiency causes megaloblastic anemia with macrocytosis, not the microcytic/normocytic picture with low TIBC seen here.
  • D. Autoantibody-mediated marrow destruction describes pure red cell aplasia, which does not explain this iron profile.
  • E. A beta-globin point mutation describes beta-thalassemia, which presents with microcytosis and a different iron profile from early childhood.
Educational objective: Anemia of chronic disease results from hepcidin-driven iron sequestration in macrophages, producing low serum iron and low TIBC with normal or high ferritin, in contrast to iron deficiency anemia.

WBC Disorders - Benign

1 questions
15

A 45-year-old man undergoes evaluation for fatigue and early satiety. Examination shows massive splenomegaly extending 10 cm below the left costal margin. Laboratory studies show white blood cell count 118,000/mm³ with a full spectrum of maturing granulocytic forms including myelocytes, metamyelocytes, and basophils, hemoglobin 10.1 g/dL, and platelet count 620,000/mm³. A similar degree of leukocytosis is also seen in a second patient, a 62-year-old woman with pneumococcal pneumonia and a white blood cell count of 32,000/mm³ with marked neutrophilia and toxic granulation but no basophilia or splenomegaly. Which of the following laboratory tests would best distinguish the underlying marrow process in these two patients?

  1. APeripheral blood lactate level
  2. BLeukocyte alkaline phosphatase score
  3. CErythrocyte sedimentation rate
  4. DSerum lactate dehydrogenase level
  5. ESerum uric acid level
Show answer
Answer: B. Leukocyte alkaline phosphatase score

Leukocyte alkaline phosphatase (LAP) score distinguishes a benign leukemoid reaction from chronic myeloid leukemia (CML). In a leukemoid reaction, driven by infection or inflammation, mature neutrophils express high levels of alkaline phosphatase, giving a high LAP score. In CML, the neoplastic granulocytes derived from the BCR-ABL1 fusion clone have low LAP activity, so the LAP score is characteristically low despite the markedly elevated white cell count. This distinction, together with basophilia and massive splenomegaly favoring CML, helps separate a reactive process from a myeloproliferative neoplasm, which can be confirmed with cytogenetic or molecular testing for the Philadelphia chromosome.

  • A. Lactate reflects tissue hypoperfusion or sepsis severity, not marrow clonality.
  • C. ESR is a nonspecific marker of inflammation elevated in infection but not helpful in distinguishing CML from a leukemoid reaction.
  • D. LDH is elevated in many high-turnover states, including both infection and malignancy, and does not reliably distinguish the two.
  • E. Uric acid rises with any high cell turnover state and does not differentiate a reactive from a neoplastic leukocytosis.
Educational objective: A low leukocyte alkaline phosphatase score points to CML, while a high score favors a benign leukemoid reaction, in a patient with marked neutrophilic leukocytosis.

Leukemia - Acute

4 questions
16

A 4-year-old boy with Down syndrome is brought to the pediatrician for easy bruising and fatigue over the past 3 weeks. Temperature is 99.9°F (37.7°C). Examination shows pallor, scattered petechiae on the lower extremities, and generalized nontender lymphadenopathy. Complete blood count shows hemoglobin 7.8 g/dL, platelet count 42,000/mm³, and white blood cell count 38,000/mm³ with 70% blasts. The blasts are small with scant cytoplasm and condensed nuclear chromatin. Immunohistochemical staining of the blasts is most likely to show positivity for which of the following markers?

  1. AMyeloperoxidase
  2. BTartrate-resistant acid phosphatase
  3. CTerminal deoxynucleotidyl transferase
  4. DCD15
  5. EAlpha-naphthyl butyrate esterase
Show answer
Answer: C. Terminal deoxynucleotidyl transferase

This child's presentation with pancytopenia-related symptoms, marked leukocytosis with blasts, and a background of Down syndrome is classic for acute lymphoblastic leukemia (ALL), the most common childhood malignancy and the most common cancer overall in children, with increased incidence in trisomy 21. ALL blasts are lymphoid precursors that express terminal deoxynucleotidyl transferase (TdT), a DNA polymerase found in pre-B and pre-T lymphoblasts but not in myeloid cells or mature lymphocytes, making it a key immunohistochemical marker distinguishing ALL from AML. TdT positivity, combined with small blasts having scant cytoplasm and condensed chromatin, supports a lymphoblastic rather than myeloblastic origin.

  • A. Myeloperoxidase is a marker of myeloid differentiation, positive in AML, not in lymphoblastic leukemia.
  • B. Tartrate-resistant acid phosphatase is characteristic of hairy cell leukemia, not ALL.
  • D. CD15 is expressed on myeloid cells and granulocytes, not lymphoblasts.
  • E. Alpha-naphthyl butyrate esterase is a marker for monocytic differentiation, as seen in acute monocytic leukemia.
Educational objective: Acute lymphoblastic leukemia, the most common childhood cancer and associated with Down syndrome, is characterized by TdT-positive lymphoblasts.
17

A 58-year-old man presents with a 2-week history of fatigue, gum bleeding, and low-grade fevers. Temperature is 100.4°F (38°C). Examination shows pallor, scattered ecchymoses, and gingival hypertrophy with bleeding. Complete blood count shows hemoglobin 8.1 g/dL, platelet count 28,000/mm³, and white blood cell count 54,000/mm³. Peripheral smear shows large myeloid blasts with fine, needle-like cytoplasmic inclusions. Cytochemical staining of the blasts with Sudan black and myeloperoxidase is strongly positive. Which of the following is the most likely diagnosis?

  1. AChronic myeloid leukemia, blast crisis
  2. BAcute myeloid leukemia
  3. CMyelodysplastic syndrome
  4. DChronic lymphocytic leukemia
  5. EAcute lymphoblastic leukemia
Show answer
Answer: B. Acute myeloid leukemia

This patient's presentation with cytopenias, gingival infiltration, and peripheral blasts containing needle-like cytoplasmic inclusions (Auer rods) that stain strongly for myeloperoxidase is diagnostic of acute myeloid leukemia (AML). Auer rods are crystalline aggregates of azurophilic granule material, and their presence confirms myeloid lineage. Myeloperoxidase positivity further supports granulocytic differentiation, distinguishing AML from lymphoblastic leukemias, which lack Auer rods and are myeloperoxidase-negative. Gum infiltration is particularly associated with the monocytic subtype of AML.

  • A. Blast crisis of CML can show myeloid or lymphoid blasts but occurs in a patient with a known antecedent chronic phase and typically massive splenomegaly, not described here.
  • C. Myelodysplastic syndrome shows dysplastic changes with cytopenias but blast counts below 20% in the bone marrow, not an acute leukemic blast population with Auer rods.
  • D. CLL presents with mature small lymphocytes and smudge cells in older adults, not blasts with Auer rods.
  • E. ALL blasts lack Auer rods and are myeloperoxidase-negative; they instead stain positive for TdT.
Educational objective: Auer rods are myeloperoxidase-positive cytoplasmic inclusions that confirm myeloid lineage in acute myeloid leukemia.
18

A 34-year-old woman presents with gum bleeding, easy bruising, and heavy menstrual bleeding for 1 week. Temperature is 98.9°F (37.2°C). Laboratory studies show hemoglobin 9.0 g/dL, platelet count 18,000/mm³, prothrombin time 19 seconds (control 12 seconds), fibrinogen 90 mg/dL (reference 200-400 mg/dL), and D-dimer markedly elevated. Bone marrow biopsy shows abnormal promyelocytes with numerous cytoplasmic granules and bundles of Auer rods. Cytogenetic analysis reveals a reciprocal translocation between chromosomes 15 and 17. Which of the following best describes the mechanism by which the drug used to treat this condition exerts its therapeutic effect?

  1. AIt blocks dihydrofolate reductase, halting purine synthesis in rapidly dividing promyelocytes
  2. BIt alkylates DNA within the abnormal promyelocytes, triggering apoptosis
  3. CIt inhibits the tyrosine kinase activity generated by the chromosomal fusion protein
  4. DIt inhibits topoisomerase II, preventing resealing of double-strand DNA breaks
  5. EIt binds the fusion protein and induces terminal differentiation of the abnormal promyelocytes
Show answer
Answer: E. It binds the fusion protein and induces terminal differentiation of the abnormal promyelocytes

Acute promyelocytic leukemia (APL) is caused by t(15;17), producing the PML-RARA fusion protein, which blocks myeloid differentiation at the promyelocyte stage and recruits corepressors that silence genes needed for maturation. The abnormal promyelocytes release procoagulant granule contents, causing disseminated intravascular coagulation, reflected here by prolonged PT, hypofibrinogenemia, thrombocytopenia, and elevated D-dimer. All-trans retinoic acid (ATRA) binds the retinoic acid receptor portion of the PML-RARA fusion protein, reversing the differentiation block and inducing the promyelocytes to mature into terminally differentiated neutrophils, which then undergo apoptosis. This differentiation mechanism distinguishes ATRA from cytotoxic chemotherapy.

  • A. Dihydrofolate reductase inhibition describes methotrexate's mechanism, which is not the targeted treatment for APL.
  • B. Alkylation is the mechanism of agents like cyclophosphamide, not ATRA.
  • C. Tyrosine kinase inhibition describes imatinib's mechanism in BCR-ABL-driven CML, not the mechanism of ATRA in APL.
  • D. Topoisomerase II inhibition describes drugs such as etoposide, not the mechanism used to treat APL specifically.
Educational objective: All-trans retinoic acid treats APL by binding the PML-RARA fusion protein and inducing terminal differentiation of malignant promyelocytes.
19

A 9-year-old boy with newly diagnosed Burkitt lymphoma begins induction chemotherapy. Thirty-six hours later, he develops muscle cramps and decreased urine output. Laboratory studies show potassium 6.8 mEq/L, phosphorus 8.9 mg/dL (reference 2.5-4.5 mg/dL), calcium 6.2 mg/dL (reference 8.5-10.5 mg/dL), and uric acid 14.2 mg/dL (reference 3.5-7.2 mg/dL). Serum creatinine is 2.4 mg/dL. Which of the following agents, if given prior to chemotherapy, would most likely have prevented this patient's acute kidney injury?

  1. ARasburicase
  2. BSodium bicarbonate infusion
  3. CAllopurinol
  4. DCalcium gluconate
  5. EFurosemide
Show answer
Answer: A. Rasburicase

This patient has tumor lysis syndrome, characterized by hyperkalemia, hyperphosphatemia, hypocalcemia (from calcium-phosphate precipitation), and markedly elevated uric acid leading to acute kidney injury, a known complication following initiation of chemotherapy in high-grade lymphomas such as Burkitt lymphoma with rapid cell turnover. Rasburicase is a recombinant urate oxidase that catalyzes the enzymatic breakdown of uric acid into allantoin, a much more water-soluble compound that is readily excreted, making it the most effective agent for rapidly lowering uric acid and preventing urate nephropathy in high-risk patients prior to chemotherapy. Allopurinol works by blocking new uric acid formation but does not reduce existing uric acid levels as rapidly or effectively as rasburicase in high-risk settings.

  • B. Sodium bicarbonate was previously used to alkalinize urine but is no longer routinely recommended because it can promote calcium-phosphate crystallization.
  • C. Allopurinol inhibits xanthine oxidase to prevent new uric acid formation but acts more slowly and is less effective than rasburicase in high-risk tumor lysis prevention.
  • D. Calcium gluconate treats symptomatic hypocalcemia or hyperkalemia-induced arrhythmia but does not prevent the uric acid-mediated kidney injury.
  • E. Furosemide promotes diuresis but does not address the underlying hyperuricemia driving the kidney injury.
Educational objective: Rasburicase prevents tumor lysis syndrome-associated nephropathy by enzymatically converting uric acid to the more soluble allantoin.

Leukemia - Chronic

3 questions
20

A 71-year-old man undergoes a routine physical examination and is found to have painless, nontender lymphadenopathy in the cervical and axillary regions. He reports no fever, weight loss, or night sweats. Complete blood count shows hemoglobin 13.1 g/dL, platelet count 210,000/mm³, and white blood cell count 62,000/mm³ with 85% small, mature-appearing lymphocytes. Peripheral blood smear shows numerous fragile, ruptured lymphocytes with smeared nuclear material. Flow cytometry shows the circulating lymphocytes co-express CD5 and CD23 along with B-cell markers. Which of the following is the most likely diagnosis?

  1. AInfectious mononucleosis
  2. BChronic lymphocytic leukemia
  3. CHairy cell leukemia
  4. DFollicular lymphoma
  5. EMantle cell lymphoma
Show answer
Answer: B. Chronic lymphocytic leukemia

This elderly patient's presentation of asymptomatic lymphocytosis with small, mature-appearing lymphocytes, smudge cells on peripheral smear, and B cells co-expressing CD5 (normally a T-cell marker) along with CD23 is characteristic of chronic lymphocytic leukemia (CLL), the most common leukemia in older adults in the United States. Smudge cells result from the mechanical fragility of the malignant lymphocytes during slide preparation. CLL is typically an indolent disease often discovered incidentally, in contrast to more aggressive lymphoid malignancies, and the combination of CD5+/CD23+ B cells is a key immunophenotypic clue for diagnosis.

  • A. Infectious mononucleosis occurs in young patients with atypical reactive T lymphocytes, not smudge cells or a CD5+/CD23+ B-cell population in an elderly patient.
  • C. Hairy cell leukemia shows cells with fine cytoplasmic projections and is tartrate-resistant acid phosphatase positive, not CD5+/CD23+ smudge cells.
  • D. Follicular lymphoma is CD10-positive and associated with t(14;18), not the CD5+/CD23+ phenotype seen here.
  • E. Mantle cell lymphoma is also CD5+ but is typically CD23-negative and presents with lymphadenopathy and t(11;14), not smudge cells on peripheral smear.
Educational objective: Chronic lymphocytic leukemia, the most common leukemia in older adults, is characterized by smudge cells on smear and CD5+/CD23+ B lymphocytes.
21

A 45-year-old man presents with fatigue and early satiety over the past 2 months. Examination reveals a spleen palpable 8 cm below the left costal margin. Complete blood count shows white blood cell count 180,000/mm³ with a full spectrum of granulocytic maturation stages and basophilia. Cytogenetic analysis of bone marrow cells reveals a reciprocal translocation between the long arms of chromosomes 9 and 22. The patient is started on a medication that targets the abnormal protein product of this translocation. Which of the following best describes the mechanism of action of this medication?

  1. ACompetitive inhibition of ATP binding at the kinase domain of the fusion protein
  2. BInhibition of topoisomerase II, preventing resealing of double-strand DNA breaks in proliferating granulocytes
  3. CInhibition of dihydrofolate reductase, blocking purine synthesis in proliferating granulocyte precursors
  4. DMonoclonal antibody-mediated complement activation against the fusion protein on the cell surface
  5. EBinding to the fusion protein and inducing its terminal differentiation into mature granulocytes
Show answer
Answer: A. Competitive inhibition of ATP binding at the kinase domain of the fusion protein

This patient's markedly elevated white cell count with full granulocytic maturation, splenomegaly, and basophilia along with the t(9;22) translocation is classic for chronic myeloid leukemia (CML). This translocation fuses the BCR gene on chromosome 22 with the ABL1 gene on chromosome 9, creating the BCR-ABL fusion protein with constitutively active tyrosine kinase activity that drives uncontrolled myeloid proliferation. Imatinib is a tyrosine kinase inhibitor that works by competitively binding the ATP-binding site within the kinase domain of BCR-ABL, blocking its ability to phosphorylate downstream substrates and thereby halting the proliferative signaling cascade. This targeted mechanism distinguishes imatinib from cytotoxic chemotherapy and has transformed CML into a chronic, manageable disease for most patients.

  • B. Topoisomerase II inhibition describes agents such as etoposide, unrelated to BCR-ABL tyrosine kinase inhibition.
  • C. Dihydrofolate reductase inhibition describes methotrexate's mechanism, unrelated to tyrosine kinase inhibition in CML.
  • D. Monoclonal antibody mechanisms apply to agents like rituximab, not to imatinib, which is a small-molecule kinase inhibitor.
  • E. Differentiation induction describes the mechanism of all-trans retinoic acid in acute promyelocytic leukemia, not imatinib in CML.
Educational objective: Imatinib treats CML by competitively inhibiting ATP binding at the kinase domain of the constitutively active BCR-ABL fusion protein produced by t(9;22).
22

A 58-year-old man comes to the physician because of fatigue and easy bruising for the past 2 months. He reports no fevers or night sweats. Physical examination shows a palpable spleen tip 6 cm below the left costal margin; no lymph nodes are palpable. Laboratory studies show hemoglobin 9.8 g/dL, platelet count 85,000/mm3, and leukocyte count 3200/mm3 with circulating cells that have fine, hair-like cytoplasmic projections. Bone marrow aspirate is unsuccessful because of marrow fibrosis ("dry tap"), and biopsy shows cells diffusely infiltrating the marrow with abundant clear cytoplasm giving a "fried egg" appearance. Cytochemical staining of the circulating cells is positive for tartrate-resistant acid phosphatase. Which of the following is most likely to be identified on molecular analysis of these cells?

  1. ABCR-ABL1 fusion gene
  2. BJAK2 V617F mutation
  3. CFLT3-ITD mutation
  4. DPML-RARA fusion gene
  5. EBRAF V600E mutation
Show answer
Answer: E. BRAF V600E mutation

Hairy cell leukemia is a B-cell neoplasm characterized by TRAP-positive cells with fine cytoplasmic projections, marrow fibrosis causing a dry tap, splenomegaly without lymphadenopathy, and pancytopenia. Nearly all cases harbor an activating BRAF V600E point mutation, which constitutively activates the RAF-MEK-ERK signaling pathway and drives proliferation; this finding is used diagnostically and as a target for BRAF inhibitor therapy (e.g., vemurafenib). The other mutations are characteristic of different hematologic malignancies and are not associated with this clinical and cytochemical picture.

  • A. BCR-ABL1 is the hallmark of chronic myeloid leukemia, which presents with marked leukocytosis and basophilia, not TRAP-positive hairy cells.
  • B. JAK2 V617F is seen in myeloproliferative neoplasms such as polycythemia vera, not hairy cell leukemia.
  • C. FLT3-ITD is an acute myeloid leukemia mutation associated with poor prognosis, unrelated to this indolent B-cell leukemia.
  • D. PML-RARA defines acute promyelocytic leukemia, which presents with blasts containing azurophilic granules and DIC risk, not hairy cells.
Educational objective: Hairy cell leukemia is a B-cell leukemia with TRAP-positive hairy cells, marrow fibrosis causing a dry tap, and splenomegaly without lymphadenopathy, nearly always driven by an activating BRAF V600E mutation.

Lymphoma

5 questions
23

A 24-year-old woman comes to the physician because of a painless mass in her neck that she noticed 3 weeks ago. She also reports intermittent fevers and a 5-kg (11-lb) weight loss over the past month. Physical examination shows a firm, nontender, enlarged cervical lymph node. Excisional biopsy shows scattered large binucleate cells with prominent eosinophilic nucleoli surrounded by a mixed inflammatory infiltrate of lymphocytes, eosinophils, and histiocytes. Immunohistochemical staining of the large cells is positive for CD15 and CD30 and negative for CD20. Which of the following is the most likely diagnosis?

  1. APeripheral T-cell lymphoma
  2. BHodgkin lymphoma
  3. CDiffuse large B-cell lymphoma
  4. DAnaplastic large cell lymphoma
  5. EBurkitt lymphoma
Show answer
Answer: B. Hodgkin lymphoma

Hodgkin lymphoma is characterized by Reed-Sternberg cells, large binucleate ("owl's eye") cells that are CD15+ and CD30+ but typically CD20-negative, set within a reactive background of lymphocytes, eosinophils, plasma cells, and histiocytes. It shows a bimodal age distribution with peaks in young adults and older adults, and the classic/mixed cellularity subtype has a strong association with EBV infection. The disease classically presents with painless lymphadenopathy, often cervical, with contiguous spread and systemic B symptoms such as fever and weight loss.

  • A. Peripheral T-cell lymphoma lacks the characteristic CD15+/CD30+ Reed-Sternberg cells and reactive mixed infiltrate.
  • C. Diffuse large B-cell lymphoma shows sheets of large atypical B cells that are CD20-positive, not scattered Reed-Sternberg cells in a reactive background.
  • D. Anaplastic large cell lymphoma is CD30+ but is a T-cell lineage neoplasm that is ALK-positive in many cases and lacks the classic reactive background and CD15 positivity of Hodgkin lymphoma.
  • E. Burkitt lymphoma shows a monotonous population of medium-sized blasts with a starry sky pattern, not binucleate giant cells.
Educational objective: Hodgkin lymphoma is defined by CD15+/CD30+ Reed-Sternberg cells in a reactive inflammatory background, with a bimodal age distribution and association with EBV infection.
24

An 8-year-old boy living in rural Uganda is brought to a clinic because of rapid swelling of the right side of his jaw over the past 3 weeks. Examination shows a firm, nontender mass involving the mandible. Serologic testing is positive for Epstein-Barr virus. Biopsy of the mass shows a monotonous population of intermediate-sized lymphoid cells with high mitotic activity interspersed with pale macrophages containing cellular debris, giving a "starry sky" pattern. Cytogenetic analysis of the tumor cells shows a reciprocal translocation between chromosomes 8 and 14. Which of the following best describes the molecular consequence of this translocation?

  1. AFusion of the PAX5 gene to a partner locus, impairing B-cell differentiation
  2. BFusion of the CCND1 gene to the immunoglobulin heavy chain locus, increasing cyclin D1 activity
  3. CJuxtaposition of the MYC gene to the immunoglobulin heavy chain enhancer, driving its overexpression
  4. DFormation of a BCR-ABL1 fusion protein with constitutive tyrosine kinase activity
  5. EFusion of the BCL2 gene to the immunoglobulin heavy chain locus, blocking apoptosis
Show answer
Answer: C. Juxtaposition of the MYC gene to the immunoglobulin heavy chain enhancer, driving its overexpression

Burkitt lymphoma is driven by a t(8;14) translocation that relocates the MYC proto-oncogene from chromosome 8 to a position adjacent to the immunoglobulin heavy chain enhancer on chromosome 14, causing constitutive overexpression of MYC and unchecked cellular proliferation. This produces the characteristic starry sky histology, where high proliferation and apoptosis create a monotonous sheet of lymphoid cells interspersed with tingible-body macrophages. The endemic (African) form is strongly associated with EBV infection and classically presents as a jaw or facial bone mass in children.

  • A. PAX5 rearrangements are not the defining molecular lesion of Burkitt lymphoma; the hallmark abnormality is MYC dysregulation.
  • B. This describes the t(11;14) translocation of mantle cell lymphoma, which overexpresses cyclin D1, not MYC.
  • D. BCR-ABL1 is the product of the t(9;22) Philadelphia chromosome seen in chronic myeloid leukemia, not Burkitt lymphoma.
  • E. This describes the t(14;18) translocation of follicular lymphoma, which overexpresses the anti-apoptotic BCL2 protein rather than MYC.
Educational objective: Burkitt lymphoma results from a t(8;14) translocation that places MYC under control of the immunoglobulin heavy chain enhancer, producing constitutive MYC overexpression and the starry sky histologic pattern.
25

A 62-year-old woman comes to the physician because of painless swelling in her neck and groin for the past 4 months. She feels well otherwise. Physical examination shows multiple nontender, rubbery lymph nodes in the cervical and inguinal regions, each measuring 1 to 2 cm. Lymph node biopsy shows effacement of normal architecture by numerous abnormal follicle-like nodules composed of small lymphoid cells with irregular, cleaved nuclei. Cytogenetic studies show a translocation involving chromosomes 14 and 18. Which of the following best explains how this genetic abnormality contributes to lymphomagenesis in this patient?

  1. AConstitutive activation of a tyrosine kinase that promotes proliferation
  2. BLoss of function of a tumor suppressor that normally promotes p53-mediated apoptosis
  3. COverexpression of a transcription factor that drives unchecked cell cycle entry via MYC
  4. DOverexpression of an anti-apoptotic protein that inhibits mitochondrial outer membrane permeabilization
  5. EOverexpression of a cyclin that drives unchecked progression through the G1-S checkpoint
Show answer
Answer: D. Overexpression of an anti-apoptotic protein that inhibits mitochondrial outer membrane permeabilization

Follicular lymphoma is characterized by a t(14;18) translocation that juxtaposes the BCL2 gene with the immunoglobulin heavy chain locus, resulting in BCL2 overexpression. BCL2 is an anti-apoptotic protein that resides in the mitochondrial outer membrane and inhibits pro-apoptotic proteins such as BAX and BAK, preventing mitochondrial outer membrane permeabilization and the release of cytochrome c. This blocks the intrinsic apoptotic pathway, allowing abnormal lymphoid cells to accumulate and form the characteristic neoplastic follicles seen on biopsy.

  • A. Constitutive tyrosine kinase activation describes BCR-ABL1 in chronic myeloid leukemia, not the BCL2-driven mechanism of follicular lymphoma.
  • B. Follicular lymphoma results from a gain-of-function translocation causing BCL2 overexpression, not from loss of a tumor suppressor.
  • C. MYC-driven unchecked cell cycle entry describes the mechanism of Burkitt lymphoma, not follicular lymphoma.
  • E. Overexpression of cyclin D1 driving G1-S progression describes the mechanism of mantle cell lymphoma, not follicular lymphoma.
Educational objective: Follicular lymphoma arises from a t(14;18) translocation causing BCL2 overexpression, which blocks the intrinsic apoptotic pathway and allows survival of abnormal lymphoid cells.
26

A 68-year-old man comes to the physician because of fatigue and abdominal fullness for the past 2 months. Physical examination shows diffuse nontender lymphadenopathy in the cervical, axillary, and inguinal regions and a palpable spleen. Colonoscopy performed for evaluation of occult blood in the stool shows multiple small polyp-like lesions scattered throughout the colonic mucosa. Biopsy of a lymph node shows a monotonous population of small to medium-sized lymphocytes with irregular nuclear contours expanding the mantle zone around residual reactive germinal centers. Cytogenetic analysis shows a translocation between chromosomes 11 and 14. Overexpression of which of the following proteins is responsible for the pathogenesis of this lymphoma?

  1. ACyclin D1
  2. BCyclin E
  3. CBCL6
  4. DMYC
  5. EBCL2
Show answer
Answer: A. Cyclin D1

Mantle cell lymphoma is caused by a t(11;14) translocation that places the CCND1 gene, encoding cyclin D1, next to the immunoglobulin heavy chain locus, resulting in cyclin D1 overexpression. Cyclin D1 binds CDK4/6 and promotes phosphorylation of the retinoblastoma protein, driving cells through the G1-S checkpoint in an unregulated manner. This produces a monotonous proliferation of small lymphocytes that expand the mantle zone, as seen in this biopsy, and the disease can present with multiple lymphomatous polyposis of the gastrointestinal tract.

  • B. Cyclin E is not the product implicated in the characteristic translocation of mantle cell lymphoma; cyclin D1 is.
  • C. BCL6 dysregulation is associated with some diffuse large B-cell lymphomas, not the t(11;14) translocation described here.
  • D. MYC overexpression from t(8;14) is the mechanism of Burkitt lymphoma, not mantle cell lymphoma.
  • E. BCL2 overexpression from t(14;18) is the mechanism of follicular lymphoma, not mantle cell lymphoma.
Educational objective: Mantle cell lymphoma results from a t(11;14) translocation causing cyclin D1 overexpression, which drives unchecked progression through the G1-S cell cycle checkpoint.
27

A 70-year-old man comes to the physician because of a rapidly enlarging neck mass over the past 6 weeks accompanied by fevers and a 7-kg (15-lb) weight loss. Physical examination shows a firm, nontender right cervical lymph node measuring 4 cm. Excisional biopsy shows diffuse effacement of nodal architecture by large atypical lymphoid cells with vesicular nuclei and prominent nucleoli, arranged in sheets without a follicular pattern. Immunohistochemistry shows the large cells are positive for CD20 and negative for CD15. Which of the following is the most likely diagnosis?

  1. AHodgkin lymphoma
  2. BDiffuse large B-cell lymphoma
  3. CMantle cell lymphoma
  4. DFollicular lymphoma
  5. EChronic lymphocytic leukemia/small lymphocytic lymphoma
Show answer
Answer: B. Diffuse large B-cell lymphoma

Diffuse large B-cell lymphoma is the most common non-Hodgkin lymphoma in adults and presents as a rapidly growing, often extranodal or nodal mass with constitutional symptoms. Histology shows diffuse (non-follicular) effacement of the normal architecture by large atypical B cells expressing CD20, with a high proliferation rate accounting for the aggressive clinical course. Unlike Hodgkin lymphoma, the neoplastic cells are CD20-positive and CD15-negative, and unlike follicular or mantle cell lymphoma, the growth pattern is diffuse rather than nodular.

  • A. Hodgkin lymphoma shows CD15+/CD30+ Reed-Sternberg cells in a reactive background, not diffuse CD20+ large cell sheets.
  • C. Mantle cell lymphoma shows small to medium lymphocytes expanding the mantle zone and is associated with t(11;14), not large diffuse sheets of atypical cells.
  • D. Follicular lymphoma shows a nodular (follicular) growth pattern of small cleaved lymphocytes, an indolent course, and is associated with t(14;18), not a diffuse sheet of large cells.
  • E. Chronic lymphocytic leukemia/small lymphocytic lymphoma shows small, mature-appearing lymphocytes with an indolent course, not large atypical cells with an aggressive presentation.
Educational objective: Diffuse large B-cell lymphoma is the most common non-Hodgkin lymphoma in adults, presenting aggressively with diffuse sheets of CD20-positive large atypical B cells.

Plasma Cell & Related Disorders

4 questions
28

A 66-year-old man comes to the physician because of persistent lower back pain and fatigue for the past 2 months. Laboratory studies show hemoglobin 9.1 g/dL, calcium 11.4 mg/dL (reference range 8.5-10.5 mg/dL), and creatinine 1.9 mg/dL. Serum protein electrophoresis shows a sharp monoclonal spike in the gamma globulin region. Peripheral blood smear shows red cells arranged in stacked, chain-like aggregates. A skeletal survey shows multiple well-defined lytic lesions in the skull and vertebral bodies. Bone marrow biopsy shows greater than 10% clonal plasma cells. Which of the following mechanisms best explains the development of the lytic bone lesions in this patient?

  1. ASecretion of RANKL and IL-6 by marrow stromal cells, leading to increased osteoclast activation
  2. BOsteoblast-stimulating factor secretion causing excess sclerotic bone formation
  3. CDecreased renal activation of vitamin D leading to secondary hyperparathyroidism
  4. DComplement-mediated lysis of osteocytes by circulating immune complexes
  5. EDirect invasion and destruction of bone cortex by amyloid fibril deposition
Show answer
Answer: A. Secretion of RANKL and IL-6 by marrow stromal cells, leading to increased osteoclast activation

In multiple myeloma, clonal plasma cells in the bone marrow stimulate stromal cells and osteoblasts to increase expression of RANKL while decreasing osteoprotegerin (a RANKL decoy receptor), and plasma cells also secrete IL-6, which further promotes osteoclast differentiation and activity. The resulting increase in osteoclastic bone resorption relative to osteoblastic bone formation produces the characteristic punched-out lytic lesions seen on imaging and contributes to hypercalcemia. This mechanism, along with anemia from marrow crowding, renal injury, and the monoclonal (M) spike on SPEP, forms the basis of the CRAB criteria used to diagnose symptomatic myeloma.

  • B. Myeloma bone lesions are lytic, not sclerotic; osteoblastic activity is suppressed, not stimulated, in this disease.
  • C. Secondary hyperparathyroidism from vitamin D deficiency is not the mechanism of myeloma bone disease, which is driven by local cytokine-mediated osteoclast activation.
  • D. Lytic lesions in myeloma result from cytokine-driven osteoclast activation, not complement-mediated lysis of bone cells.
  • E. Amyloid deposition can occur in myeloma-associated AL amyloidosis but does not cause the classic punched-out lytic bone lesions; the mechanism is osteoclast activation via RANKL.
Educational objective: In multiple myeloma, plasma cell-driven RANKL and IL-6 secretion activates osteoclasts, producing lytic bone lesions and hypercalcemia as part of the CRAB criteria.
29

A 64-year-old woman with a known history of multiple myeloma comes to the emergency department because of decreased urine output and worsening leg swelling over the past week. Laboratory studies show creatinine 4.2 mg/dL (baseline 1.0 mg/dL) and urinalysis shows no significant proteinuria on urine dipstick but a markedly positive result on urine protein electrophoresis for free light chains. Renal biopsy shows eosinophilic, fractured casts within the distal tubules and collecting ducts surrounded by multinucleated giant cells and associated tubular epithelial injury. Which of the following best explains the mechanism of renal injury in this patient?

  1. AObstruction and direct tubular toxicity from free light chains binding Tamm-Horsfall protein and precipitating in the distal nephron
  2. BHyperuricemia from tumor cell turnover causing urate crystal deposition in the renal tubules
  3. CHypercalcemia-induced nephrogenic diabetes insipidus from downregulation of aquaporin-2 channels
  4. DImmune complex deposition in the mesangium triggering a proliferative glomerulonephritis
  5. EDeposition of amyloid light chain fibrils in the glomerular basement membrane causing nephrotic-range proteinuria
Show answer
Answer: A. Obstruction and direct tubular toxicity from free light chains binding Tamm-Horsfall protein and precipitating in the distal nephron

Myeloma cast nephropathy results from excess monoclonal free light chains (Bence Jones proteins) filtered by the glomerulus that bind to Tamm-Horsfall protein in the distal tubule and collecting duct, forming insoluble, eosinophilic, fractured casts that obstruct the tubule lumen and incite a surrounding giant cell reaction. The light chains are also directly toxic to tubular epithelial cells, causing apoptosis and further impairing renal function. Because the excreted protein is primarily light chains rather than albumin, a standard urine dipstick (which detects mainly albumin) can be negative or only mildly positive despite a markedly positive result on urine protein electrophoresis.

  • B. Tumor lysis syndrome with urate nephropathy is not characteristic of myeloma and does not explain the light chain casts seen on biopsy.
  • C. Hypercalcemia can contribute to renal dysfunction in myeloma via vasoconstriction and volume depletion, but it does not explain the fractured casts and giant cell reaction seen on this biopsy.
  • D. Immune complex-mediated glomerulonephritis is not the mechanism of myeloma kidney; the injury is tubular and related to light chain cast formation, not glomerular immune complex deposition.
  • E. AL amyloidosis causes glomerular deposition and nephrotic syndrome with heavy proteinuria, not the obstructive tubular cast pattern with fractured casts and giant cells described here.
Educational objective: Myeloma cast nephropathy results from free light chains binding Tamm-Horsfall protein in the distal nephron, forming obstructive casts and causing direct tubular toxicity.
30

A 68-year-old man comes to the physician because of blurred vision, headache, and fatigue for the past month. He also reports numbness and tingling in his feet. Examination shows hepatosplenomegaly. Funduscopic examination reveals dilated, segmented retinal veins. Laboratory studies show hemoglobin 9.8 g/dL, platelet count 210,000/mm3, and total protein 11 g/dL. Serum protein electrophoresis shows a tall, narrow spike in the gamma region. Skeletal survey shows no lytic lesions. Bone marrow biopsy shows infiltration by small lymphocytes with plasmacytoid features. Which of the following is the most likely diagnosis?

  1. AChronic lymphocytic leukemia
  2. BMonoclonal gammopathy of undetermined significance
  3. CMultiple myeloma
  4. DWaldenstrom macroglobulinemia
  5. EAmyloid light-chain amyloidosis
Show answer
Answer: D. Waldenstrom macroglobulinemia

This patient has a monoclonal IgM gammopathy with hyperviscosity symptoms (visual changes, headache, retinal vein dilation) and peripheral neuropathy but no lytic bone lesions, consistent with Waldenstrom macroglobulinemia, a lymphoplasmacytic lymphoma that secretes large pentameric IgM molecules. The large size and pentameric structure of IgM markedly raise serum viscosity, producing the hyperviscosity syndrome. Lytic bone lesions are characteristic of multiple myeloma (IgG or IgA) but are absent in Waldenstrom because the malignant cells rarely cause significant osteoclast activation. Bone marrow shows lymphoplasmacytic infiltration rather than the sheets of plasma cells seen in myeloma.

  • A. CLL presents with lymphocytosis and lymphadenopathy, not a monoclonal IgM spike causing hyperviscosity.
  • B. MGUS is asymptomatic by definition and does not cause hyperviscosity symptoms or organomegaly.
  • C. Multiple myeloma classically causes lytic bone lesions, hypercalcemia, and renal failure, which are absent here.
  • E. Amyloidosis presents with organ infiltration by light chains, not a hyperviscosity syndrome from a macroglobulin.
Educational objective: Waldenstrom macroglobulinemia is a lymphoplasmacytic lymphoma producing monoclonal IgM that causes hyperviscosity syndrome without lytic bone lesions.
31

A 72-year-old woman has a routine serum protein electrophoresis performed during a workup for mild anemia that shows a monoclonal IgG spike of 2.1 g/dL. She has no bone pain, and skeletal survey shows no lytic lesions. Serum calcium is 9.4 mg/dL (normal 8.5-10.2), creatinine is 0.9 mg/dL, and hemoglobin is 11.8 g/dL. Bone marrow biopsy shows 4% plasma cells. She is asymptomatic and reports no fatigue or weight loss. Which of the following best explains why this patient does not meet diagnostic criteria for multiple myeloma?

  1. AAbsence of end-organ damage and bone marrow plasma cells below 10%
  2. BAge under 75 years, which excludes a diagnosis of plasma cell malignancy
  3. CAbsence of a detectable monoclonal protein on electrophoresis
  4. DPresence of polyclonal rather than monoclonal plasma cell proliferation
  5. ENormal serum free light chain ratio excluding any plasma cell dyscrasia
Show answer
Answer: A. Absence of end-organ damage and bone marrow plasma cells below 10%

Monoclonal gammopathy of undetermined significance (MGUS) is defined by serum M protein less than 3 g/dL, bone marrow clonal plasma cells less than 10%, and absence of CRAB features (hypercalcemia, renal insufficiency, anemia, bone lesions) attributable to the plasma cell disorder. This patient's M spike, plasma cell percentage, and normal calcium, creatinine, and lack of lytic lesions all fall below myeloma thresholds. Multiple myeloma requires clonal plasma cells of 10% or greater or a biopsy-proven plasmacytoma plus at least one myeloma-defining event. MGUS carries a roughly 1% per year risk of progression to myeloma and requires monitoring rather than treatment.

  • B. Age is not a diagnostic criterion for distinguishing MGUS from multiple myeloma.
  • C. A monoclonal protein is present here; its presence does not distinguish MGUS from myeloma, only its magnitude and associated findings do.
  • D. The plasma cells in MGUS are clonal, not polyclonal; monoclonality is shared with myeloma.
  • E. Free light chain ratio can be abnormal in MGUS as well and alone does not define the distinction from myeloma.
Educational objective: MGUS is distinguished from multiple myeloma by an M spike under 3 g/dL, under 10% clonal bone marrow plasma cells, and absence of end-organ damage.

Myeloproliferative Disorders

3 questions
32

A 58-year-old man comes to the physician because of itching after hot showers and a burning, red discoloration of his hands and feet that improves with cooling. He has a history of a deep venous thrombosis 1 year ago. Examination shows facial plethora and splenomegaly. Laboratory studies show hemoglobin 19.2 g/dL, hematocrit 58%, and leukocyte count 13,200/mm3 with platelet count 620,000/mm3. Serum erythropoietin level is low. Which of the following is the most likely underlying mechanism of this patient's condition?

  1. AAutoantibody-mediated activation of the erythropoietin receptor
  2. BChronic hypoxia leading to compensatory renal erythropoietin overproduction
  3. CReciprocal translocation producing a constitutively active fusion tyrosine kinase
  4. DPoint mutation causing constitutive activation of a tyrosine kinase downstream of the erythropoietin receptor
  5. EEctopic erythropoietin secretion from a renal cell carcinoma
Show answer
Answer: D. Point mutation causing constitutive activation of a tyrosine kinase downstream of the erythropoietin receptor

This patient has erythromelalgia, aquagenic pruritus, splenomegaly, thrombosis, and markedly elevated hemoglobin/hematocrit with a low erythropoietin level, classic for polycythemia vera. PV is caused in most cases by a point mutation (V617F) in JAK2, a cytoplasmic tyrosine kinase that normally transmits signaling from the erythropoietin receptor; the mutation causes constitutive, erythropoietin-independent activation of downstream JAK-STAT signaling, producing autonomous erythrocytosis. Because red cell production is independent of erythropoietin, the appropriately low EPO level (suppressed by negative feedback from the elevated hematocrit) distinguishes PV from secondary polycythemia, in which EPO is elevated. This also explains the frequent accompanying leukocytosis and thrombocytosis, since JAK2 signaling affects multiple myeloid lineages.

  • A. Polycythemia vera is not caused by an autoantibody against the erythropoietin receptor.
  • B. Chronic hypoxia with compensatory EPO elevation describes secondary polycythemia, which would show a high, not low, EPO level.
  • C. A fusion tyrosine kinase from a translocation describes chronic myeloid leukemia (BCR-ABL1), not polycythemia vera.
  • E. Ectopic EPO secretion also causes secondary polycythemia with elevated EPO, not the low EPO seen in this patient.
Educational objective: Polycythemia vera results from a JAK2 V617F point mutation causing EPO-independent erythrocytosis with a characteristically low serum erythropoietin level.
33

A 61-year-old woman comes to the physician because of a burning sensation and redness of her fingertips for the past 2 weeks. She has no history of smoking. Vital signs are normal. Examination shows erythema and warmth of the fingertips without ulceration. Laboratory studies show hemoglobin 13.5 g/dL, leukocyte count 9,800/mm3, and platelet count 980,000/mm3 (normal 150,000-400,000). Peripheral smear shows abundant large platelets without blasts. Bone marrow biopsy shows increased numbers of large, hyperlobulated megakaryocytes without significant fibrosis. A mutation in which of the following genes is most likely responsible for this patient's condition?

  1. AJAK2
  2. BNPM1
  3. CFLT3
  4. DRUNX1
  5. EBCR-ABL1
Show answer
Answer: A. JAK2

This patient has marked, isolated thrombocytosis with megakaryocytic hyperplasia on bone marrow biopsy and digital erythromelalgia from microvascular platelet aggregation, findings consistent with essential thrombocythemia. About half of ET cases carry the JAK2 V617F mutation, which activates the JAK-STAT pathway downstream of the thrombopoietin receptor (MPL) and drives thrombopoietin-independent megakaryocyte proliferation; a smaller subset instead carries CALR or MPL mutations. The massively elevated, dysfunctional platelet mass paradoxically causes both thrombotic complications (erythromelalgia, stroke, DVT) and, at very high counts, acquired bleeding from consumption of von Willebrand factor.

  • B. NPM1 mutations are characteristic of a subtype of acute myeloid leukemia with a relatively favorable prognosis, not a myeloproliferative disorder.
  • C. FLT3 mutations are associated with acute myeloid leukemia and confer a worse prognosis, not essential thrombocythemia.
  • D. RUNX1 mutations are seen in acute myeloid leukemia and myelodysplastic syndromes, not essential thrombocythemia.
  • E. BCR-ABL1 is the fusion kinase of chronic myeloid leukemia, which presents with marked leukocytosis and basophilia, not isolated thrombocytosis.
Educational objective: Essential thrombocythemia is commonly driven by a JAK2 mutation causing thrombopoietin-independent megakaryocyte proliferation and marked thrombocytosis with thrombotic risk.
34

A 65-year-old man comes to the physician because of fatigue, early satiety, and unintentional 7-kg (15-lb) weight loss over 4 months. Examination shows massive splenomegaly extending below the umbilicus. Laboratory studies show hemoglobin 9.0 g/dL, leukocyte count 11,500/mm3, and platelet count 140,000/mm3. Peripheral blood smear shows teardrop-shaped red cells, nucleated red blood cells, and immature myeloid precursors. Bone marrow aspiration yields a dry tap. Bone marrow biopsy shows extensive replacement of hematopoietic tissue by dense collagen fibrosis with clusters of atypical megakaryocytes. Which of the following is the most likely diagnosis?

  1. APrimary myelofibrosis
  2. BAcute myeloid leukemia
  3. CChronic myeloid leukemia
  4. DMyelodysplastic syndrome
  5. EHairy cell leukemia
Show answer
Answer: A. Primary myelofibrosis

This patient has massive splenomegaly, constitutional symptoms, a dry bone marrow tap, and biopsy showing extensive marrow fibrosis with atypical megakaryocyte clusters, diagnostic of primary myelofibrosis. The leukoerythroblastic peripheral smear, with teardrop cells (dacryocytes), nucleated red cells, and immature myeloid forms, reflects extramedullary hematopoiesis in the spleen and liver as the fibrotic marrow fails to support normal hematopoiesis. Most cases are driven by JAK2, CALR, or MPL mutations causing megakaryocyte hyperplasia, which release fibrogenic cytokines such as platelet-derived growth factor and transforming growth factor-beta that stimulate fibroblast proliferation and marrow collagen deposition.

  • B. Acute myeloid leukemia presents with circulating blasts exceeding 20% of marrow cellularity, not a leukoerythroblastic smear with teardrop cells.
  • C. CML typically shows marked leukocytosis with a full spectrum of granulocytic maturation and basophilia, not a fibrotic dry tap with teardrop cells.
  • D. Myelodysplastic syndrome shows dysplastic changes and cytopenias but not the dense marrow fibrosis and dry tap characteristic of myelofibrosis.
  • E. Hairy cell leukemia causes splenomegaly and pancytopenia with cells showing hairy cytoplasmic projections, not marrow fibrosis with teardrop cells.
Educational objective: Primary myelofibrosis is diagnosed by marrow fibrosis with atypical megakaryocytes on biopsy, a dry tap, and a leukoerythroblastic peripheral smear with teardrop cells from extramedullary hematopoiesis.

Hemostasis - Platelet Disorders

4 questions
35

A 7-year-old girl is brought to the physician because of scattered small bruises and pinpoint red spots on her legs for the past 3 days. Two weeks ago she had an upper respiratory infection that resolved without treatment. She appears well and has no fever, lymphadenopathy, or hepatosplenomegaly. Laboratory studies show hemoglobin 12.8 g/dL, leukocyte count 7,200/mm3, and platelet count 18,000/mm3. Peripheral smear shows normal red cell and white cell morphology with a few large platelets and no schistocytes. Prothrombin time and partial thromboplastin time are normal. Which of the following is the most likely underlying mechanism of this patient's thrombocytopenia?

  1. AIgG autoantibodies against platelet glycoprotein IIb/IIIa causing splenic platelet destruction
  2. BHeparin-dependent antibody formation against platelet factor 4
  3. CDecreased platelet production from megakaryocyte hypoplasia in the bone marrow
  4. DDeficiency of ADAMTS13 causing accumulation of ultra-large von Willebrand factor multimers
  5. EMicrovascular thrombi formation causing platelet consumption
Show answer
Answer: A. IgG autoantibodies against platelet glycoprotein IIb/IIIa causing splenic platelet destruction

This child has isolated thrombocytopenia with petechiae and purpura following a viral illness, normal coagulation studies, and no schistocytes or organomegaly, consistent with immune thrombocytopenic purpura (ITP). In ITP, IgG autoantibodies, most commonly directed against platelet surface glycoprotein IIb/IIIa, bind circulating platelets, which are then opsonized and phagocytosed by splenic macrophages via Fc receptors, shortening platelet lifespan. Megakaryocytes in the bone marrow are typically increased, not decreased, as a compensatory response. Coagulation times remain normal because the clotting cascade is intact; only platelet number is affected.

  • B. Heparin-induced thrombocytopenia requires prior heparin exposure, which this patient does not have.
  • C. Bone marrow megakaryocytes are increased, not decreased, in ITP as a compensatory response to peripheral platelet destruction.
  • D. ADAMTS13 deficiency causes TTP, which presents with schistocytes, renal dysfunction, and neurologic symptoms, not seen here.
  • E. Microvascular thrombi with platelet consumption describes thrombotic microangiopathies such as TTP or HUS, which would show schistocytes and often renal or neurologic findings.
Educational objective: Immune thrombocytopenic purpura is caused by IgG autoantibodies against platelet glycoproteins (commonly GPIIb/IIIa) leading to splenic macrophage-mediated platelet destruction.
36

A 4-year-old boy is brought to the physician because of recurrent epistaxis and easy bruising since infancy. His parents are first cousins. Examination shows scattered ecchymoses and petechiae. Laboratory studies show platelet count 95,000/mm3 (normal 150,000-400,000) and prolonged bleeding time. Prothrombin time and partial thromboplastin time are normal. Peripheral blood smear shows platelets that are markedly enlarged, approaching the size of red blood cells. Platelet aggregation studies show normal aggregation in response to adenosine diphosphate, epinephrine, and collagen, but absent aggregation in response to ristocetin that is not corrected by adding normal plasma. Which of the following is the most likely defective platelet receptor?

  1. AGlycoprotein VI
  2. BGlycoprotein Ib
  3. CP2Y12 receptor
  4. DThromboxane A2 receptor
  5. EGlycoprotein IIb/IIIa
Show answer
Answer: B. Glycoprotein Ib

This child has thrombocytopenia with giant platelets, a prolonged bleeding time, and ristocetin-induced aggregation that fails to correct with mixing normal plasma, consistent with Bernard-Soulier syndrome, an autosomal recessive disorder (consistent with the consanguineous parents) caused by deficiency of glycoprotein Ib. GPIb is the platelet receptor for von Willebrand factor and is essential for platelet adhesion to exposed subendothelial collagen at sites of vascular injury. Because the defect is intrinsic to the platelet receptor rather than a deficiency of plasma von Willebrand factor, adding normal plasma does not correct the ristocetin aggregation defect, distinguishing it from von Willebrand disease. Aggregation to ADP, epinephrine, and collagen remains normal because those pathways act through other receptors.

  • A. GPVI is a collagen receptor; its deficiency would impair collagen-induced aggregation but not primarily cause the ristocetin defect described.
  • C. P2Y12 is the ADP receptor targeted by clopidogrel; its defect impairs ADP-induced aggregation specifically, not ristocetin response.
  • D. Thromboxane A2 receptor defects impair platelet activation amplification but do not cause the giant platelets or ristocetin defect seen here.
  • E. GPIIb/IIIa deficiency (Glanzmann thrombasthenia) causes defective platelet-platelet aggregation to ADP, epinephrine, and collagen, but ristocetin-induced agglutination is typically preserved.
Educational objective: Bernard-Soulier syndrome results from glycoprotein Ib deficiency, impairing platelet adhesion to von Willebrand factor and producing giant platelets with an uncorrectable ristocetin aggregation defect.
37

A 6-year-old girl is brought to the physician because of frequent nosebleeds and gum bleeding after tooth brushing for the past year. She is the child of first-cousin parents. Examination shows mucosal petechiae and a healing gingival laceration. Laboratory studies show a platelet count of 310,000/mm3 and a prolonged bleeding time. Prothrombin time and partial thromboplastin time are normal. Peripheral blood smear shows platelets of normal size and number that fail to form clumps or aggregates near sites of vessel injury. Platelet aggregation studies show normal agglutination with ristocetin but absent aggregation in response to adenosine diphosphate, epinephrine, and collagen. Which of the following is the most likely mechanism underlying this patient's bleeding disorder?

  1. ADeficiency of von Willebrand factor in plasma and platelet alpha granules
  2. BDeficiency of the platelet receptor that binds von Willebrand factor at the subendothelium
  3. CDeficiency of a plasma coagulation factor required for fibrin clot formation
  4. DDeficiency of the platelet receptor that binds fibrinogen to link adjacent platelets
  5. EAutoantibody-mediated destruction of circulating platelets
Show answer
Answer: D. Deficiency of the platelet receptor that binds fibrinogen to link adjacent platelets

This child has a normal platelet count with a prolonged bleeding time, normal ristocetin-induced agglutination, but absent aggregation to ADP, epinephrine, and collagen, consistent with Glanzmann thrombasthenia, an autosomal recessive disorder caused by deficiency of glycoprotein IIb/IIIa. This integrin receptor on activated platelets binds fibrinogen, which cross-links adjacent platelets to form platelet-platelet aggregates during primary hemostasis. Because the defect is in platelet-platelet linkage rather than platelet-vessel wall adhesion, ristocetin-induced agglutination (which depends on GPIb-von Willebrand factor interaction) remains normal, distinguishing this condition from Bernard-Soulier syndrome. Platelet count and coagulation times are normal because platelet number and the secondary coagulation cascade are unaffected.

  • A. Von Willebrand factor deficiency impairs ristocetin-induced aggregation, which is normal in this patient, making this mechanism inconsistent.
  • B. Deficiency of the GPIb receptor for von Willebrand factor causes Bernard-Soulier syndrome, which would show giant platelets and absent ristocetin aggregation, not the pattern described here.
  • C. A plasma coagulation factor deficiency would prolong PT or PTT, both of which are normal here, pointing to a primary platelet defect instead.
  • E. Autoantibody-mediated platelet destruction describes ITP, which causes thrombocytopenia, whereas this patient has a normal platelet count.
Educational objective: Glanzmann thrombasthenia results from glycoprotein IIb/IIIa deficiency, impairing fibrinogen-mediated platelet-platelet aggregation while sparing ristocetin-induced agglutination.
38

A 58-year-old man is started on unfractionated heparin for a deep vein thrombosis after knee replacement surgery. On day 7 of therapy, his platelet count falls from 230,000/mm3 to 68,000/mm3. He now has a cool, mottled right foot with absent pedal pulses. Doppler ultrasound confirms arterial occlusion. Which of the following best explains this patient's current presentation?

  1. AIgG antibodies against platelet factor 4-heparin complexes activate platelets and promote thrombin generation
  2. BHeparin triggers complement-mediated lysis of platelets with release of procoagulant microparticles
  3. CAnti-platelet antibodies bind glycoprotein Ib, causing splenic sequestration and consumption
  4. DHeparin directly cross-links platelet glycoprotein IIb/IIIa receptors causing irreversible aggregation
  5. EHeparin depletes antithrombin III, leading to unopposed factor Xa activity and clot formation
Show answer
Answer: A. IgG antibodies against platelet factor 4-heparin complexes activate platelets and promote thrombin generation

Heparin-induced thrombocytopenia (HIT) occurs when heparin binds platelet factor 4 (PF4), creating a neoepitope that provokes IgG antibody formation. The resulting immune complexes bind platelet Fc receptors, causing widespread platelet activation, aggregation, and consumption (thrombocytopenia) along with release of procoagulant platelet microparticles. This paradoxically causes thrombosis rather than bleeding despite the low platelet count, producing arterial or venous occlusions such as the ischemic limb seen here. Heparin must be stopped immediately and a non-heparin anticoagulant (e.g., argatroban) started.

  • B. HIT is not primarily complement-mediated lysis; it is Fc-receptor-mediated platelet activation.
  • C. HIT involves PF4-heparin antibodies, not anti-GPIb antibodies or splenic sequestration as the primary mechanism.
  • D. Heparin does not directly cross-link GPIIb/IIIa; this is not the mechanism of HIT.
  • E. Heparin potentiates antithrombin III rather than depleting it; this is heparin's therapeutic mechanism, not HIT.
Educational objective: Heparin-induced thrombocytopenia results from IgG antibodies against platelet factor 4-heparin complexes that activate platelets, causing thrombocytopenia with paradoxical thrombosis.

Hemostasis - Coagulation Factor Disorders

3 questions
39

A 6-year-old boy is brought in for evaluation of easy bruising and a swollen, painful left knee after a minor fall. He has no history of mucosal or skin bleeding episodes. Laboratory studies show platelet count 280,000/mm3, PT 12 seconds (reference 11-13.5 seconds), and PTT 58 seconds (reference 25-35 seconds). The PTT corrects to 32 seconds after mixing the patient's plasma 1:1 with normal plasma. Which of the following is the most likely underlying abnormality in this patient?

  1. ADeficiency of a clotting factor in the intrinsic pathway
  2. BCirculating inhibitor against a clotting factor
  3. CDeficiency of von Willebrand factor
  4. DQualitative platelet function defect
  5. EDeficiency of vitamin K-dependent clotting factors
Show answer
Answer: A. Deficiency of a clotting factor in the intrinsic pathway

An isolated elevated PTT with normal PT and normal platelet count in a male child with hemarthrosis suggests a deficiency of an intrinsic pathway factor (factor VIII in hemophilia A or factor IX in hemophilia B). The mixing study corrects the PTT because normal plasma supplies the missing factor, confirming a factor deficiency rather than an inhibitor. If an inhibitor (such as a factor VIII autoantibody) were present, the PTT would fail to correct because the inhibitor would also neutralize the factor in the added normal plasma. Specific factor assays are needed to distinguish factor VIII deficiency (hemophilia A) from factor IX deficiency (hemophilia B).

  • B. An inhibitor would prevent correction of the PTT on mixing study; here the PTT corrected.
  • C. Von Willebrand disease typically also prolongs bleeding time and is not X-linked; this boy's isolated PTT correction pattern fits a factor deficiency.
  • D. Platelet function defects affect bleeding time, not PTT, and platelet count here is normal.
  • E. Vitamin K-dependent factor deficiency would also prolong the PT, which is normal here.
Educational objective: A corrected PTT on 1:1 mixing study indicates a factor deficiency (as in hemophilia A or B), whereas failure to correct indicates a factor inhibitor.
40

A 19-year-old woman comes to the clinic because of heavy menstrual bleeding and easy bruising since menarche. She also reports prolonged bleeding after a tooth extraction last year. Platelet count is 260,000/mm3, PT is 12 seconds (reference 11-13.5 seconds), and PTT is 42 seconds (reference 25-35 seconds). Bleeding time is prolonged. Which of the following tests would best confirm the suspected diagnosis?

  1. AProthrombin time with factor VII assay
  2. BFibrinogen level and thrombin time
  3. CPlatelet aggregation with ADP and collagen
  4. DBone marrow aspirate for megakaryocyte number
  5. ERistocetin-induced platelet aggregation assay
Show answer
Answer: E. Ristocetin-induced platelet aggregation assay

The combination of mucocutaneous bleeding, prolonged bleeding time, and mildly prolonged PTT (due to reduced factor VIII, which von Willebrand factor normally stabilizes) suggests von Willebrand disease (vWD). Ristocetin induces vWF binding to platelet GPIb, and in vWD this agglutination is reduced or absent, directly testing vWF function and confirming the diagnosis. vWF also mediates platelet adhesion to subendothelial collagen, so its deficiency impairs primary hemostasis, explaining the prolonged bleeding time. Desmopressin, which releases stored vWF from endothelial cells, is a mainstay of treatment for mild type 1 disease.

  • A. PT and factor VII are unaffected in vWD; PT is typically normal.
  • B. Fibrinogen and thrombin time assess the common pathway and fibrinogen, not vWF function.
  • C. This nonspecific platelet aggregation test does not directly assess vWF-GPIb interaction as ristocetin does.
  • D. Megakaryocyte number is normal in vWD; this is not a platelet production disorder.
Educational objective: The ristocetin-induced platelet aggregation test assesses von Willebrand factor function and is used to confirm von Willebrand disease.
41

A 34-year-old woman develops fever, hypotension, and oozing from venipuncture sites 2 days after a complicated vaginal delivery complicated by retained placental fragments and endometritis. Temperature is 102.6°F (39.2°C), blood pressure is 88/54 mmHg, and heart rate is 118/min. Laboratory studies show platelet count 42,000/mm3, PT 19 seconds (reference 11-13.5 seconds), PTT 52 seconds (reference 25-35 seconds), fibrinogen 68 mg/dL (reference 200-400 mg/dL), and D-dimer markedly elevated. Which of the following best explains this constellation of findings?

  1. ASelective deficiency of von Willebrand factor triggered by endothelial injury
  2. BPrimary fibrinolysis without antecedent coagulation activation
  3. CSystemic activation of coagulation with widespread fibrin deposition and secondary consumption of platelets and clotting factors
  4. DIsolated increased platelet destruction mediated by autoantibodies
  5. EDecreased hepatic synthesis of clotting factors due to sepsis-related liver dysfunction
Show answer
Answer: C. Systemic activation of coagulation with widespread fibrin deposition and secondary consumption of platelets and clotting factors

Sepsis from endometritis triggers widespread release of tissue factor, which activates the extrinsic coagulation pathway and causes systemic thrombin generation with diffuse fibrin deposition in the microvasculature. This consumes platelets and clotting factors (causing thrombocytopenia, prolonged PT/PTT, and low fibrinogen) while plasmin simultaneously degrades the fibrin, producing markedly elevated D-dimer. The combination of low platelets, low fibrinogen, prolonged PT/PTT, and elevated D-dimer in a septic patient is characteristic of disseminated intravascular coagulation (DIC). Treatment focuses on addressing the underlying trigger and supportive transfusion of blood products as needed.

  • A. Isolated vWF deficiency does not explain the low fibrinogen, prolonged PT, or elevated D-dimer seen here.
  • B. Primary fibrinolysis is rare and would not typically cause the concurrent thrombocytopenia and factor consumption seen with an infectious trigger.
  • D. Isolated immune thrombocytopenia would not explain the concurrent coagulation factor consumption, low fibrinogen, and elevated D-dimer.
  • E. Liver dysfunction alone would not cause this degree of platelet consumption or the markedly elevated D-dimer reflecting active fibrin turnover.
Educational objective: DIC is a consumptive coagulopathy triggered by conditions such as sepsis, presenting with thrombocytopenia, prolonged PT/PTT, low fibrinogen, and elevated D-dimer from systemic fibrin formation and breakdown.

Hemostasis - Hypercoagulable States

3 questions
42

A 29-year-old woman presents with swelling and pain in her left calf. Ultrasound confirms a deep vein thrombosis. She reports that her mother and maternal aunt both had unprovoked venous thromboses before age 45. Genetic testing reveals a point mutation in the gene encoding a coagulation factor. Which of the following best explains the mechanism by which this mutation predisposes to thrombosis?

  1. AThe mutation results in a prothrombin variant with increased plasma concentration
  2. BThe mutation causes overproduction of antithrombin III inhibitor
  3. CThe mutation impairs hepatic synthesis of protein S
  4. DThe mutated factor is resistant to cleavage and inactivation by activated protein C
  5. EThe mutated factor binds tissue factor pathway inhibitor with higher affinity
Show answer
Answer: D. The mutated factor is resistant to cleavage and inactivation by activated protein C

Factor V Leiden results from a point mutation in the factor V gene (Arg506Gln) that eliminates the cleavage site normally recognized by activated protein C. This renders factor Va resistant to inactivation, allowing it to persist in an active procoagulant form and sustain thrombin generation, leading to a hypercoagulable state. It is the most common inherited thrombophilia in people of European descent and should be suspected in patients with a strong family history of unprovoked venous thromboembolism. Testing can be done via activated protein C resistance assay or genetic analysis.

  • A. This describes the prothrombin G20210A mutation, a separate inherited thrombophilia, not factor V Leiden.
  • B. Antithrombin III levels are not increased in factor V Leiden; this describes a different mechanism unrelated to the mutation.
  • C. Factor V Leiden does not affect protein S synthesis; protein S deficiency is a distinct, separate thrombophilia.
  • E. Factor V Leiden's defect is resistance to protein C cleavage, not altered binding to tissue factor pathway inhibitor.
Educational objective: Factor V Leiden is caused by a mutation that makes factor Va resistant to cleavage by activated protein C, making it the most common inherited thrombophilia.
43

A 31-year-old woman has had three spontaneous pregnancy losses, each occurring after 10 weeks of gestation. She now presents with acute left leg swelling and is found to have a femoral vein thrombosis on ultrasound. Laboratory studies show PTT of 48 seconds (reference 25-35 seconds) that fails to correct with 1:1 mixing study, but corrects with addition of excess phospholipid. Which of the following best explains the mechanism underlying both her laboratory finding and her clinical presentation?

  1. AA deficiency of factor XII causes prolonged PTT without clinical bleeding or thrombotic risk
  2. BAn antibody binds phospholipid-protein complexes, interfering with the in vitro clotting assay while promoting platelet activation and thrombin generation in vivo
  3. CAn antibody directly neutralizes antithrombin III, mimicking heparin's anticoagulant effect in vitro
  4. DConsumption of fibrinogen by ongoing microvascular thrombosis prolongs clotting-based assays
  5. EA mutation in the prothrombin gene increases circulating prothrombin, prolonging clotting assays while promoting thrombosis
Show answer
Answer: B. An antibody binds phospholipid-protein complexes, interfering with the in vitro clotting assay while promoting platelet activation and thrombin generation in vivo

Antiphospholipid syndrome involves antibodies (lupus anticoagulant, anticardiolipin, anti-beta2-glycoprotein I) directed against phospholipid-binding proteins. In vitro, these antibodies interfere with phospholipid-dependent clotting assays, prolonging the PTT, and this fails to correct with normal plasma (mixing study) because the antibody, not a factor deficiency, is responsible; adding excess phospholipid overwhelms the antibody and corrects the PTT, confirming a lupus anticoagulant. Paradoxically, in vivo these antibodies promote platelet activation, endothelial injury, and thrombin generation, causing a hypercoagulable state with recurrent arterial/venous thrombosis and placental infarction leading to late pregnancy loss. This dissociation between prolonged in vitro clotting time and in vivo thrombosis is the classic diagnostic paradox of the syndrome.

  • A. Factor XII deficiency prolongs PTT but is not associated with thrombosis or pregnancy loss; it is typically asymptomatic.
  • C. Antithrombin III neutralization is not the mechanism; lupus anticoagulant interferes with phospholipid-dependent assays directly, and antithrombin deficiency would not explain the mixing study correction pattern.
  • D. DIC-type fibrinogen consumption would also show low fibrinogen and elevated D-dimer, and would not selectively correct with excess phospholipid.
  • E. The prothrombin G20210A mutation increases thrombosis risk but does not prolong PTT via phospholipid interference or explain this mixing study pattern.
Educational objective: Antiphospholipid syndrome features antibodies against phospholipid-binding proteins that prolong phospholipid-dependent clotting assays in vitro while paradoxically promoting thrombosis in vivo.
44

A 56-year-old woman with newly diagnosed deep vein thrombosis is started on warfarin without bridging heparin. On day 3 of therapy, she develops painful, well-demarcated purple patches with hemorrhagic bullae on her abdomen and thighs that progress to necrosis. She has no prior personal history of thrombosis, but her father died of a massive pulmonary embolism at age 40. Which of the following best explains the pathogenesis of her skin findings?

  1. ACross-reactive antibodies against warfarin-platelet factor 4 complexes cause microvascular thrombosis
  2. BWarfarin accumulation in subcutaneous fat causes local vasospasm and ischemic necrosis
  3. CInherited deficiency of protein C causes a transient hypercoagulable state when warfarin rapidly depletes this short half-life anticoagulant protein before depleting procoagulant factors
  4. DRapid depletion of fibrinogen by warfarin promotes microvascular thrombosis in the skin
  5. EWarfarin directly damages dermal capillary endothelium through a cytotoxic mechanism independent of coagulation factors
Show answer
Answer: C. Inherited deficiency of protein C causes a transient hypercoagulable state when warfarin rapidly depletes this short half-life anticoagulant protein before depleting procoagulant factors

Protein C has a short plasma half-life (about 8 hours), similar to factor VII, so warfarin rapidly lowers protein C levels within the first days of therapy before the longer half-life procoagulant factors (II and X) are sufficiently depleted. In a patient with an underlying inherited protein C (or protein S) deficiency, this creates a transient relative hypercoagulable state, leading to microvascular thrombosis in the skin and warfarin-induced skin necrosis. This is why heparin bridging is used when initiating warfarin in patients at risk, to maintain anticoagulation while protein C levels fall. The family history of early unprovoked fatal pulmonary embolism suggests an underlying inherited thrombophilia such as protein C deficiency.

  • A. This describes heparin-induced thrombocytopenia, which is unrelated to warfarin-induced skin necrosis and involves a different antibody target.
  • B. Warfarin does not cause vasospasm through fat accumulation; the mechanism is procoagulant-anticoagulant imbalance.
  • D. Fibrinogen is not a vitamin K-dependent factor and is not preferentially depleted by warfarin.
  • E. Warfarin's skin necrosis is mediated through disrupted coagulation factor balance, not direct endothelial cytotoxicity.
Educational objective: Warfarin-induced skin necrosis results from a transient hypercoagulable state caused by rapid depletion of protein C (which has a short half-life) relative to procoagulant factors, especially in patients with underlying protein C deficiency.

Anticoagulant & Antiplatelet Pharmacology

4 questions
45

A 58-year-old man is started on a continuous intravenous infusion for treatment of an acute pulmonary embolism. His activated partial thromboplastin time (aPTT) is monitored every 6 hours to adjust the infusion rate, with a target of 60-80 seconds (control 25-35 seconds). Which of the following best describes the mechanism of action of this drug?

  1. AInhibits vitamin K epoxide reductase, reducing synthesis of clotting factors II, VII, IX, and X
  2. BBinds antithrombin III and accelerates its inhibition of thrombin and factor Xa
  3. CBinds and irreversibly inactivates thrombin directly without a cofactor
  4. DDegrades fibrin by activating plasminogen to plasmin
  5. EDirectly inhibits factor Xa independent of antithrombin
Show answer
Answer: B. Binds antithrombin III and accelerates its inhibition of thrombin and factor Xa

Unfractionated heparin is a glycosaminoglycan that binds antithrombin III through a unique pentasaccharide sequence, inducing a conformational change that accelerates antithrombin's inhibition of thrombin (factor IIa) and factor Xa by roughly 1000-fold. Because unfractionated heparin has variable bioavailability and a short half-life, its anticoagulant effect is monitored using the aPTT, which reflects the intrinsic pathway. This contrasts with low-molecular-weight heparins, which preferentially inhibit factor Xa and generally do not require routine monitoring. Protamine sulfate can reverse unfractionated heparin by binding it directly and neutralizing its negative charge.

  • A. This describes warfarin's mechanism, which is monitored by PT/INR, not aPTT.
  • C. Direct thrombin inhibitors like argatroban or bivalirudin act this way, without requiring antithrombin as a cofactor.
  • D. This describes thrombolytics such as alteplase, which activate the fibrinolytic system rather than inhibiting coagulation factors.
  • E. This describes direct Xa inhibitors such as rivaroxaban or apixaban, not heparin.
Educational objective: Unfractionated heparin potentiates antithrombin III to inhibit thrombin and factor Xa, and its effect is monitored via the aPTT.
46

A 65-year-old man develops severe oozing from his sternotomy incision shortly after cardiopulmonary bypass surgery, during which he received a large dose of unfractionated heparin. His activated partial thromboplastin time is markedly prolonged. An intravenous drug is administered that rapidly normalizes his clotting times. Which of the following best describes the mechanism of this reversal agent?

  1. AA monoclonal antibody fragment that binds and inactivates circulating dabigatran
  2. BA recombinant factor Xa decoy that sequesters circulating direct Xa inhibitors
  3. CA synthetic vitamin K antagonist that competitively displaces heparin from antithrombin
  4. DA vitamin K analog that restores synthesis of clotting factors II, VII, IX, and X
  5. EA positively charged protein that binds and neutralizes negatively charged heparin molecules
Show answer
Answer: E. A positively charged protein that binds and neutralizes negatively charged heparin molecules

Protamine sulfate is a highly positively charged (cationic) protein, originally derived from salmon sperm, that binds to the strongly negatively charged heparin molecule through ionic interactions, forming a stable inactive complex that is cleared from circulation. This neutralizes heparin's anticoagulant activity within minutes and rapidly normalizes the aPTT, making it the treatment of choice for heparin-induced bleeding or heparin overdose, including after cardiopulmonary bypass. Protamine has a lesser and less predictable effect on low-molecular-weight heparins since smaller fragments bind less completely. Rapid protamine administration can itself cause hypotension or anaphylactoid reactions, particularly in patients previously exposed to protamine-containing insulin (e.g., NPH insulin).

  • A. This describes idarucizumab, the reversal agent for dabigatran.
  • B. This describes andexanet alfa, used to reverse direct factor Xa inhibitors.
  • C. Heparin does not act through antithrombin displacement, and no such vitamin K antagonist mechanism applies here.
  • D. Vitamin K reverses warfarin, not heparin, and acts over hours to days, not minutes.
Educational objective: Protamine sulfate, a cationic protein, rapidly reverses heparin by ionic binding and neutralization of the anionic heparin molecule.
47

A 60-year-old man with a history of myocardial infarction takes a daily low-dose oral medication for secondary prevention. Platelet function studies show markedly reduced thromboxane A2 production that persists for the lifespan of circulating platelets (approximately 7-10 days) despite a single dose. Which of the following best explains this prolonged effect?

  1. AReversible competitive inhibition of cyclooxygenase-1 that outlasts the drug's plasma half-life
  2. BCovalent blockade of the platelet P2Y12 receptor preventing ADP binding
  3. CIrreversible inhibition of thromboxane synthase in platelet mitochondria
  4. DPermanent inhibition of platelet glycoprotein IIb/IIIa receptor conformational activation
  5. EIrreversible acetylation of a serine residue on cyclooxygenase-1, and platelets lack nuclei to resynthesize the enzyme
Show answer
Answer: E. Irreversible acetylation of a serine residue on cyclooxygenase-1, and platelets lack nuclei to resynthesize the enzyme

Aspirin irreversibly acetylates a serine residue in the active site of cyclooxygenase-1 (COX-1), permanently blocking the enzyme's ability to convert arachidonic acid to prostaglandin H2, the precursor of thromboxane A2. Because mature platelets are anucleate and cannot synthesize new COX-1 protein, this inhibition persists for the entire 7-10 day lifespan of the affected platelets, even though aspirin itself has a plasma half-life of only about 20 minutes. This is why aspirin's antiplatelet effect lasts far longer than its presence in the circulation and why a single low dose produces a sustained reduction in thromboxane A2-mediated platelet aggregation. New, unaffected platelets released from the bone marrow gradually restore overall thromboxane production over several days.

  • A. Reversible inhibition would wear off as drug is cleared and would not explain a 7-10 day effect.
  • B. This describes clopidogrel's mechanism, not aspirin's.
  • C. Aspirin targets cyclooxygenase-1, not thromboxane synthase, and platelets lack mitochondria-based protein synthesis relevant here.
  • D. This describes GPIIb/IIIa inhibitors such as abciximab, not aspirin.
Educational objective: Aspirin irreversibly inhibits COX-1 via serine acetylation, and because platelets cannot synthesize new enzyme, the antiplatelet effect lasts the platelet's entire lifespan.
48

A 55-year-old woman is started on an oral antiplatelet agent after placement of a coronary drug-eluting stent. The drug is a prodrug that requires hepatic conversion by cytochrome P450 enzymes to its active metabolite. Which of the following best describes the mechanism by which the active metabolite inhibits platelet aggregation?

  1. ADirect inhibition of thrombin-mediated platelet activation via protease-activated receptor-1
  2. BIrreversible acetylation of cyclooxygenase-1, decreasing thromboxane A2 synthesis
  3. CReversible antagonism of glycoprotein IIb/IIIa, blocking fibrinogen cross-linking
  4. DInhibition of phosphodiesterase, increasing intracellular cyclic AMP in platelets
  5. EIrreversible blockade of the platelet ADP receptor P2Y12, preventing ADP-induced platelet activation
Show answer
Answer: E. Irreversible blockade of the platelet ADP receptor P2Y12, preventing ADP-induced platelet activation

Clopidogrel is a thienopyridine prodrug that is converted by hepatic cytochrome P450 enzymes (primarily CYP2C19) to an active metabolite, which irreversibly binds the platelet P2Y12 receptor. This blocks ADP-induced activation of the glycoprotein IIb/IIIa receptor complex, preventing fibrinogen binding and platelet aggregation. Because the inhibition is irreversible, the antiplatelet effect lasts for the lifespan of the affected platelets, similar to aspirin. Genetic polymorphisms in CYP2C19 can reduce conversion to the active metabolite, leading to clopidogrel resistance and reduced antiplatelet efficacy in some patients.

  • A. This describes direct thrombin inhibitors, not P2Y12 inhibitors.
  • B. This describes aspirin's mechanism, not clopidogrel's.
  • C. This describes drugs such as eptifibatide or tirofiban, not clopidogrel.
  • D. This describes cilostazol or dipyridamole, not clopidogrel.
Educational objective: Clopidogrel's active metabolite irreversibly blocks the platelet P2Y12 ADP receptor, preventing ADP-induced platelet aggregation.

Transfusion Medicine

2 questions
49

A 70-year-old woman with a history of multiple prior blood transfusions receives a unit of packed red blood cells. Thirty minutes into the transfusion, she develops a temperature of 101.3°F (38.5°C) and chills, without hypotension, rash, or respiratory distress. The transfusion is stopped, and laboratory evaluation shows a negative direct antiglobulin test, normal plasma haptoglobin, and no hemoglobinuria. Which of the following best explains the mechanism of this reaction?

  1. APreformed recipient IgM antibodies against donor ABO antigens activate complement
  2. BIgE cross-linking on recipient mast cells triggers degranulation and histamine release
  3. CBacterial contamination of the blood product causes septic shock
  4. DRecipient antibodies directed against donor leukocyte antigens react with residual donor white blood cells, releasing pyrogenic cytokines
  5. EDonor antibodies against recipient neutrophil antigens cause neutrophil sequestration in the pulmonary capillaries
Show answer
Answer: D. Recipient antibodies directed against donor leukocyte antigens react with residual donor white blood cells, releasing pyrogenic cytokines

Febrile non-hemolytic transfusion reactions are the most common transfusion reaction and result from recipient antibodies against donor human leukocyte antigens (HLA) or other leukocyte antigens, which react with residual white blood cells and platelets in the blood product, or from accumulated cytokines (such as IL-1, IL-6, and TNF-alpha) released by leukocytes during storage of the blood product. This triggers fever and chills without hemolysis, so hemoglobinuria, decreased haptoglobin, and a positive direct antiglobulin test are absent. It is more common in patients with prior transfusions or pregnancies due to prior sensitization to leukocyte antigens. Leukoreduction of blood products reduces the incidence of this reaction.

  • A. This describes an acute hemolytic transfusion reaction, which would show hemolysis markers and hypotension.
  • B. This describes an anaphylactic reaction, which presents with hypotension, urticaria, or bronchospasm.
  • C. Bacterial contamination typically causes high fever with hypotension and signs of sepsis, and would show positive blood cultures from the unit.
  • E. This describes transfusion-related acute lung injury, which causes respiratory distress, not isolated fever.
Educational objective: Febrile non-hemolytic transfusion reactions are caused by recipient anti-leukocyte antibodies or stored cytokines reacting with donor white blood cells, without hemolysis.
50

A 50-year-old woman develops acute shortness of breath, hypoxemia with oxygen saturation of 84% on room air, and bilateral crackles beginning 2 hours after receiving a unit of fresh frozen plasma. Chest X-ray shows bilateral pulmonary infiltrates. Her blood pressure is 110/70 mm Hg, and jugular venous pressure is normal with no evidence of fluid overload or elevated B-type natriuretic peptide. Which of the following best describes the underlying mechanism of this reaction?

  1. AIgE-mediated mast cell degranulation causes bronchospasm and laryngeal edema
  2. BVolume overload from rapid infusion increases pulmonary capillary hydrostatic pressure
  3. CDonor antibodies against recipient leukocyte antigens activate neutrophils, which become sequestered and degranulate in the pulmonary capillaries
  4. DPreformed recipient IgM antibodies against donor red cell antigens activate complement, causing intravascular hemolysis
  5. ERecipient antibodies against donor leukocytes release pyrogenic cytokines causing fever without lung injury
Show answer
Answer: C. Donor antibodies against recipient leukocyte antigens activate neutrophils, which become sequestered and degranulate in the pulmonary capillaries

Transfusion-related acute lung injury (TRALI) occurs when donor plasma contains antibodies against recipient human leukocyte antigens (HLA) or neutrophil-specific antigens. These antibodies bind recipient neutrophils, activating and sequestering them in the pulmonary microvasculature, where they degranulate and damage the capillary endothelium, causing noncardiogenic pulmonary edema. This presents within 6 hours of transfusion with acute hypoxemia and bilateral infiltrates, mimicking acute respiratory distress syndrome, but with normal cardiac filling pressures and normal B-type natriuretic peptide, distinguishing it from transfusion-associated circulatory overload. TRALI is a leading cause of transfusion-related mortality, and management is supportive with oxygen and ventilatory support as needed, since it is typically self-limited.

  • A. This describes an anaphylactic transfusion reaction, not TRALI.
  • B. This describes transfusion-associated circulatory overload, which would show elevated BNP and signs of volume overload.
  • D. This describes an acute hemolytic transfusion reaction, which causes hemolysis, not lung injury.
  • E. This describes a febrile non-hemolytic transfusion reaction, which does not cause hypoxemia or pulmonary infiltrates.
Educational objective: TRALI results from donor anti-leukocyte antibodies activating recipient neutrophils in the pulmonary capillaries, causing noncardiogenic pulmonary edema with normal cardiac filling pressures.

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