Questions are grouped by lecture but numbered continuously across the whole bank. Every item is written in board style — a clinical vignette, a single best answer, and four distractors that are each real entities chosen to be plausible. Headers are here so you can drill one lecture at a time — if you want the blind-mixed experience, scroll past the headers and work straight through. Renumber sequentially as you add more.
Lecture 01 · Randall
Anemias of Diminished Production
Questions 1–12
1A 58-year-old woman with a 12-year history of rheumatoid arthritis has a hemoglobin of 9.8 g/dL and an MCV of 84 fL. Serum iron is 28 µg/dL and transferrin saturation is 11%.Which additional laboratory result would best support anemia of chronic disease rather than iron deficiency?
- ASerum ferritin of 8 ng/mL
- BElevated free erythrocyte protoporphyrin
- CTotal iron binding capacity of 470 µg/dL
- DRed cell distribution width of 18.4%
- ESerum ferritin of 240 ng/mL
Answer
E
Both conditions produce a low serum iron and low saturation, so those values cannot separate them. Ferritin is the discriminator: it is elevated in ACD, where hepcidin traps iron inside macrophages and enterocytes, and low in IDA, where stores are genuinely gone. A points the opposite way. C is also an IDA pattern — TIBC rises in IDA and falls in ACD. D is nonspecific and, if anything, favors IDA, since a widened distribution reflects new small cells mixing with older normal ones. B is the most tempting: free erythrocyte protoporphyrin is elevated in both conditions, because in each case protoporphyrin is made but cannot be paired with iron, so it carries no discriminating value.
2A 44-year-old man with alcohol use disorder has a hemoglobin of 10.2 g/dL and an MCV of 78 fL. The peripheral smear shows coarse basophilic stippling, and a marrow iron stain demonstrates ringed sideroblasts.Which set of iron studies is most consistent with this diagnosis?
- ALow ferritin, high TIBC, low serum iron, low saturation
- BHigh ferritin, low TIBC, low serum iron, low saturation
- CNormal ferritin, high TIBC, normal serum iron, low saturation
- DHigh ferritin, low TIBC, high serum iron, high saturation
- ELow ferritin, low TIBC, low serum iron, normal saturation
Answer
D
Sideroblastic anemia is a failure of protoporphyrin synthesis, not of iron supply. Iron arrives normally and then has nowhere to go, so it accumulates — producing the one microcytic anemia with an iron overload profile: high ferritin, high serum iron, and high saturation, with a low TIBC. A is the classic iron deficiency pattern and B is anemia of chronic disease; both are microcytic, which is what makes them plausible here. C is the pattern described for pregnancy, where TIBC rises and saturation falls without true depletion. The ringed sideroblasts in the stem are iron-laden mitochondria — a direct visual statement that iron is present and unusable.
3A 4-year-old child living in pre-1978 housing has a microcytic anemia, abdominal pain, and coarse basophilic stippling on the peripheral smear. Marrow examination shows ringed sideroblasts.Which pair of enzymes is inhibited by the responsible agent?
- AALA synthase and ferrochelatase
- BALA synthase and ALA dehydratase
- CALA dehydratase and ferrochelatase
- DFerrochelatase and methionine synthase
- EALA synthase and methylmalonyl-CoA mutase
Answer
C
Lead inhibits ALA dehydratase — the cytoplasmic step immediately after ALA synthase — and ferrochelatase, the terminal step that inserts ferrous iron into protoporphyrin. A convenient way to hold it: lead hits the second and last enzymes of the pathway. A and B are the common errors, both substituting ALA synthase, which is the first and rate-limiting mitochondrial step and the one relevant to vitamin B6, since B6 is its cofactor. D and E append enzymes from B12 and folate metabolism, which belong to the macrocytic anemias and have no role in heme synthesis.
4A 34-year-old man being treated for latent tuberculosis develops a microcytic anemia. Marrow examination shows ringed sideroblasts. The anemia resolves after a vitamin supplement is added to his regimen.Which mechanism best explains the anemia?
- AThe drug chelates iron within the duodenal lumen, preventing absorption
- BThe drug inhibits ferrochelatase, blocking insertion of iron into protoporphyrin
- CThe drug depletes pyridoxine, the cofactor required by ALA synthase
- DThe drug suppresses renal erythropoietin production
- EThe drug induces an antibody against erythroid precursors
Answer
C
Isoniazid causes a pyridoxine (B6) deficiency, and B6 is the cofactor for ALA synthase, the rate-limiting first step of heme synthesis. Blocking that step gives a sideroblastic anemia that is treatable and preventable with B6, which is exactly what the stem describes when the anemia resolves after a vitamin is added. B describes the mechanism of lead, not isoniazid, and lead poisoning would not respond to a vitamin. D would produce a normocytic anemia of renal failure. A and E describe processes that do not produce ringed sideroblasts, which specifically indicate iron accumulating in mitochondria because protoporphyrin synthesis has failed.
5A 32-year-old woman with menorrhagia has a hemoglobin of 11.8 g/dL, an MCV of 88 fL, and a red cell distribution width of 17.5%. Serum ferritin is 11 ng/mL.Which best explains the elevated RDW despite a normal MCV?
- AReticulocytosis from marrow compensation is enlarging a subset of cells
- BOlder normal-sized cells still circulate alongside newly formed smaller cells
- CA coexisting folate deficiency is producing a macrocytic subpopulation
- DRed cell fragmentation is generating schistocytes of variable size
- ETarget cell formation artifactually widens the measured volume distribution
Answer
B
RDW measures variation in cell size, and it moves before the mean does. As iron runs out, the marrow begins releasing small cells while the normal-sized population made earlier is still in circulation for its remaining lifespan. Two populations coexist, so the distribution widens while the average stays inside the reference range — making a high RDW with a normal MCV an early sign of iron deficiency. The low ferritin confirms it. A, C, and D are all genuine causes of an elevated RDW in other settings, which is what makes them plausible, but none is supported here: there is no evidence of hemolysis, no macrocytic indices, and no schistocytes described.
6A 61-year-old woman has a macrocytic anemia with hypersegmented neutrophils and glossitis. Serum homocysteine is elevated.Which finding would most reliably establish vitamin B12 deficiency rather than folate deficiency as the cause?
- AAn MCV above 110 fL
- BHypersegmented neutrophils on the peripheral smear
- CA markedly elevated serum homocysteine level
- DAn elevated serum methylmalonic acid level
- EThe presence of glossitis
Answer
D
The hematologic pictures are identical — same macrocytosis, same hypersegmented neutrophils, same glossitis, same elevated homocysteine — because both vitamins feed the same DNA-synthesis step. That eliminates A, B, C, and E as discriminators, and every one of them is listed in the stem or is a feature of both deficiencies. Only two things separate the two conditions: methylmalonic acid and neurologic findings, both present in B12 deficiency and absent in folate deficiency.
7A student is asked why methylmalonic acid accumulates in vitamin B12 deficiency but remains normal in folate deficiency.Which explanation is correct?
- AB12 is required for methylmalonyl-CoA mutase, a reaction independent of folate
- BB12 is required for methionine synthase, a reaction that folate does not participate in
- CMethylmalonic acid is a direct byproduct of homocysteine remethylation
- DFolate deficiency accelerates renal clearance of methylmalonic acid
- EFolate is the cofactor for methylmalonyl-CoA mutase, and B12 is not
Answer
A
B12 has two jobs. As methyl-B12 it runs methionine synthase, and folate is a partner in that reaction — which is precisely why losing either vitamin raises homocysteine and impairs DNA synthesis, giving identical blood findings. As adenosyl-B12 it separately runs methylmalonyl-CoA mutase in the mitochondrion, a pathway folate has nothing to do with. Lose B12 and methylmalonyl-CoA backs up as MMA. B is the most attractive wrong answer because it names a real B12-dependent enzyme — but methionine synthase is the reaction folate does share, so it cannot explain a difference between the two. E inverts the cofactor relationship. The same mitochondrial pathway is implicated in the neurologic damage, which is why MMA and neurologic findings travel together.
8A 62-year-old woman has a hemoglobin of 8.9 g/dL with an MCV of 118 fL, elevated methylmalonic acid, elevated homocysteine, and impaired vibration sense in both feet.Which underlying process is the most likely cause?
- AProlonged inadequate dietary intake of leafy vegetables
- BAutoimmune destruction of gastric parietal cells
- CPrior resection of the terminal ileum
- DChronic pancreatic exocrine insufficiency
- EChronic infection with Diphyllobothrium latum
Answer
B
The elevated MMA and the neurologic findings establish B12 deficiency rather than folate, which eliminates A. Among causes of B12 deficiency, the question asks for the most likely, and the answer is pernicious anemia — autoimmune destruction of parietal cells eliminating intrinsic factor, which is required for B12 absorption in the small bowel. C, D, and E are all genuine causes of B12 deficiency and are the reason this question is not a giveaway: terminal ileal disease removes the absorptive site, pancreatic insufficiency impairs release of B12 from its carrier protein, and the fish tapeworm competes for the vitamin. All are real; none is common.
9A 29-year-old woman has a hemoglobin of 8.0 g/dL and a hematocrit of 24% (normal 45%). Her reticulocyte count is reported as 6%.Which conclusion about the marrow is best supported?
- AThe response is appropriate, indicating destruction or blood loss
- BThe response is inadequate, indicating a production defect
- CThe raw percentage cannot be interpreted without the MCV
- DThe result is diagnostic of aplastic anemia
- EThe result indicates an appropriate response only if the MCV is elevated
Answer
A
The raw percentage is misleading because the denominator has shrunk — in anemia, a fixed number of reticulocytes represents a larger fraction of a smaller total. Correcting for that gives RI = 6 × (24/45) ≈ 3.2. An index above 3 indicates the marrow is responding appropriately, which shifts the differential toward hemolysis or blood loss and away from the production failures. B describes an index below 2. C and E invoke the MCV, which classifies the anemia by size but plays no part in the correction, which uses hematocrit. D contradicts the calculation entirely, since aplastic anemia would show a profoundly suppressed index.
10A 66-year-old man with metastatic prostate carcinoma is found to have a hemoglobin of 8.4 g/dL, a white cell count of 3.1 ×10³/µL, and a platelet count of 74 ×10³/µL.Which peripheral smear finding would most support marrow infiltration as the cause?
- AHypersegmented neutrophils
- BProminent rouleaux formation
- CBite cells with Heinz bodies on supravital stain
- DNucleated red blood cells with immature granulocytes
- EAn unremarkable smear apart from the reduced cell numbers
Answer
D
Both aplastic anemia and a myelophthisic process produce pancytopenia, so the counts alone do not separate them — the smear does. When tumor physically crowds the marrow, precursors are squeezed into the circulation before they should be, giving nucleated red cells and immature granulocytes. E is the pattern of aplastic anemia, where the marrow is empty rather than crowded and the surviving cells look normal; it is the intended contrast. A points to megaloblastic anemia, C to oxidative hemolysis, and B to a paraproteinemia — each a real smear finding attached to a different disease.
11An 8-year-old with sickle cell anemia whose baseline hemoglobin is 8.5 g/dL presents with a hemoglobin of 4.2 g/dL and a reticulocyte count of 0.2%. Parvovirus B19 IgM is positive.Which best explains why this infection produced such a severe drop in hemoglobin?
- AThe virus directly lyses circulating mature red cells throughout the vasculature
- BThe virus triggers an autoimmune hemolytic process directed at red cell antigens
- CShortened red cell survival makes the patient dependent on continuous marrow output
- DConcurrent splenic sequestration is trapping red cells
- EThe virus induces hypersplenism with accelerated clearance of otherwise normal red cells
Answer
C
Parvovirus B19 infects erythroid precursors and shuts down production for a week or two — which the suppressed reticulocyte count confirms. In a normal host with a 120-day red cell lifespan, losing production briefly barely registers. In sickle cell disease the cells survive only days, and the patient is compensated only because the marrow runs at maximum output continuously. Remove production from a system with short cell survival and the hemoglobin falls almost immediately. A is wrong because the virus targets precursors, not circulating cells. B, D, and E all describe increased destruction, which the near-zero reticulocyte count argues against — this is a production failure.
12A 47-year-old woman reports difficulty swallowing solid food. Examination shows a smooth beefy-red tongue and spoon-shaped nails. Hemoglobin is 9.1 g/dL with an MCV of 71 fL.Which additional finding is most expected?
- AElevated serum methylmalonic acid
- BSerum ferritin of 340 ng/mL
- CSerum ferritin of 6 ng/mL
- DRinged sideroblasts on marrow iron stain
- EA normal RDW with an elevated red cell count
Answer
C
Dysphagia from esophageal webs, atrophic glossitis, and a microcytic anemia together are Plummer-Vinson syndrome, and the underlying lesion is iron deficiency — so ferritin is low. Koilonychia in the stem is a further iron-deficiency sign, as is pica. B would suggest anemia of chronic disease or an iron-overload state. A belongs to B12 deficiency, which is macrocytic. D indicates sideroblastic anemia, the microcytic anemia with high rather than low iron stores. E describes the thalassemia pattern — uniform microcytosis with plenty of cells — and is the single best distractor here, since thalassemia is the other microcytic anemia that presents with a low MCV in a young adult.
Lecture 02 · Tayal
Hemolytic Anemias, Part 1
Questions 13–23
13A 27-year-old man has a hemoglobin of 9.2 g/dL, a reticulocyte count of 8%, an elevated LDH, and an unconjugated bilirubin of 3.1 mg/dL.Which additional finding would best indicate that the hemolysis is intravascular rather than extravascular?
- ASplenomegaly with left upper quadrant tenderness
- BAn undetectable serum haptoglobin with hemoglobinuria
- CA further rise in serum lactate dehydrogenase
- DPolychromasia with nucleated red cells on the smear
- EA further rise in unconjugated bilirubin with scleral icterus
Answer
B
Everything in the stem — reticulocytosis, elevated LDH, unconjugated hyperbilirubinemia — proves hemolysis but does not localize it, because all of it occurs in both patterns. C, D, and E simply restate findings already present and shared. A points the other way, since splenomegaly is a feature of extravascular destruction. Haptoglobin is the localizer: it binds hemoglobin free in plasma, so it is consumed heavily only when cells lyse directly into the bloodstream. In extravascular hemolysis the contents are degraded inside a macrophage and haptoglobin is only modestly reduced. Hemoglobinuria follows the same logic, requiring free plasma hemoglobin to exceed haptoglobin's binding capacity.
14A 24-year-old man of northern European descent has intermittent jaundice, splenomegaly, and pigment gallstones. His MCV is 84 fL, MCHC is 36.8 g/dL, and RDW is 16.5%.Which best explains the elevated MCHC?
- AIncreased hemoglobin synthesis per cell
- BRed cell agglutination interfering with the automated measurement
- CReticulocytosis raising the mean corpuscular hemoglobin
- DRedundant membrane in target cells
- ECellular dehydration from loss of potassium and water
Answer
E
In hereditary spherocytosis the cell loses membrane and lipid bilayer, then loses potassium and water, so the same quantity of hemoglobin occupies a smaller volume. MCHC is a ratio, so that modest volume loss pushes it detectably above range, while the MCV — an absolute value with a wide 80–100 normal range — usually stays inside it. That is why HS is classified as normocytic even though the cells look small and dense on a smear. A is wrong because hemoglobin content per cell is not increased; the cell is smaller. C would raise the MCV, not the MCHC, since reticulocytes are large. D describes cells with excess membrane, which lowers MCHC. B is a genuine cause of a spuriously high MCHC, but it occurs with cold agglutinins, and nothing here suggests that.
15A 31-year-old woman has a hemolytic anemia with numerous spherocytes on the peripheral smear.Which test best distinguishes hereditary spherocytosis from warm autoimmune hemolytic anemia?
- ADirect antiglobulin test
- BOsmotic fragility testing
- CSerum haptoglobin
- DReticulocyte count
- EHemoglobin electrophoresis
Answer
A
Spherocytes appear in both conditions and are formed by the same mechanical route — membrane is lost and the cell rounds up — so morphology cannot separate them. The DAT asks the one question that does: is antibody or complement bound to the cell? Positive points to an immune cause; negative with spherocytes points to hereditary spherocytosis. B is the key distractor and a common error: osmotic fragility is abnormal in both, because it detects the shape rather than the cause — any spherocyte lyses early since it has no spare membrane to expand into. C and D confirm and quantify hemolysis without addressing mechanism. E detects abnormal hemoglobins, which are not involved in either disorder.
16A 22-year-old man of African ancestry develops jaundice and dark urine three days after starting trimethoprim-sulfamethoxazole. Hemoglobin is 8.1 g/dL. A G6PD enzyme assay performed that day returns within the normal range.Which best explains this result?
- AThe diagnosis is excluded, and an autoimmune hemolytic process should be pursued instead
- BA concurrent folate deficiency is masking the underlying enzyme deficiency
- CThe assay is unreliable in males because the responsible gene is X-linked recessive
- DSulfonamide metabolites directly interfere with the enzymatic assay itself
- EOlder, more deficient cells have already lysed, leaving reticulocytes with higher activity
Answer
E
The abnormal G6PD protein misfolds and is degraded over time, so enzyme activity falls as a cell ages and the oldest cells are the most deficient. Those are exactly the cells destroyed first during an oxidant challenge. What remains in circulation immediately afterward is a young, reticulocyte-rich population with higher-than-representative activity, which can push the assay into the normal range. Repeating it weeks later, once the population has aged, is the correct approach. A abandons a diagnosis the clinical picture supports strongly — the timing after a sulfa drug is classic. C inverts the genetics: X-linked recessive inheritance is why the disorder is expressed in males. D and B invoke interference and masking effects that do not occur.
17A 19-year-old man develops an acute hemolytic episode after eating fava beans. The smear shows bite cells, and supravital staining reveals intracellular inclusions.Which sequence best describes the mechanism of red cell injury?
- AReduced NADH allows methemoglobin to accumulate, progressively impairing oxygen delivery
- BReduced ATP causes membrane pump failure, cellular swelling, and eventual osmotic lysis
- CReduced NADPH depletes reduced glutathione, allowing peroxide to denature hemoglobin
- DGlutathione depletion permits complement activation and membrane attack complex formation
- EReduced pyruvate kinase activity lowers ATP and progressively impairs deformability
Answer
C
G6PD is the entry point to the hexose monophosphate shunt, which is the red cell's only source of NADPH. NADPH regenerates reduced glutathione, and reduced glutathione plus glutathione peroxidase is what removes hydrogen peroxide. Lose the enzyme and that entire antioxidant chain collapses, so oxidants denature hemoglobin into membrane-bound precipitates — Heinz bodies — which splenic macrophages pluck out, taking a piece of membrane and leaving a bite cell. B and E describe pyruvate kinase deficiency, the other red cell enzymopathy, where the failure is energetic rather than oxidative. A describes methemoglobinemia, a different consequence of oxidant stress. D inserts complement, which has no role here.
18A newborn screening result shows hemoglobin SS.Which molecular change is responsible?
- AA chain terminator mutation abolishing β-globin production
- BA point mutation at the 6th codon of β-globin substituting lysine for glutamate
- CDeletion of two of the four α-globin genes
- DA splice-site mutation reducing the quantity of β-globin produced
- EA point mutation at the 6th codon of β-globin substituting valine for glutamate
Answer
E
Sickle hemoglobin is HbS = α2βS2, produced by a point mutation in the 6th codon of the β-globin gene replacing glutamate with valine. The change in protein charge alters electrophoretic mobility, which is what makes hemoglobin electrophoresis diagnostic. B is the single best distractor because it is the correct codon and the correct gene but the wrong substitution — glutamate to lysine at that same position gives HbC, another β-globin disorder. C describes α-thalassemia. D and A are the β+ and β0 thalassemia mutations, which reduce the amount of a normal globin rather than producing an abnormal one — the definitional line between a thalassemia and a hemoglobinopathy.
19A 21-year-old with sickle cell disease is counseled that maintaining good hydration is part of routine management.Which best explains why dehydration worsens this disease?
- AIt raises the MCHC, favoring HbS polymerization
- BIt lowers hemoglobin F levels within the red cell
- CIt promotes splenic sequestration of circulating red cells
- DIt increases plasma viscosity independently of the red cell
- EIt raises the oxygen affinity of HbS at low tension
Answer
A
Sickling itself drives potassium and water out of the cell, raising the MCHC. A higher intracellular hemoglobin concentration makes polymerization more likely at the next deoxygenation, so each sickling event makes the next one easier — a self-amplifying cycle ending in irreversibly sickled cells. Any additional dehydration feeds directly into that loop, which is why avoiding dehydration is listed as a treatment goal alongside raising HbF and lowering the HbS percentage. B is wrong because HbF is determined by gene expression, not hydration status. D is superficially reasonable and does happen, but the mechanism that matters here is intracellular. E misstates the effect and would, if anything, reduce deoxygenation.
20A 19-year-old with sickle cell disease has Howell-Jolly bodies on the peripheral smear and a small, calcified spleen on imaging.Which organism poses the greatest risk to this patient?
- AEscherichia coli
- BStaphylococcus aureus
- CStreptococcus pneumoniae
- DPseudomonas aeruginosa
- EListeria monocytogenes
Answer
C
Repeated infarction leaves the spleen small, shrunken, fibrotic, and non-functional — autosplenectomy — and the Howell-Jolly bodies are the smear evidence, since those DNA remnants are normally pitted out by a working spleen. The spleen's specific role is clearing encapsulated organisms, so those are the threat: pneumococci and Haemophilus influenzae. The remaining choices are genuine pathogens in other clinical contexts, but none is an encapsulated organism whose clearance depends principally on splenic function, which is what this patient has lost.
21A patient with sickle cell disease is started on hydroxyurea to raise hemoglobin F.Which best explains why increasing HbF reduces sickling?
- AHbF binds oxygen more avidly, preventing deoxygenation of HbS
- BHbF stabilizes the red cell membrane against dehydration
- CHbF contains no β chains and cannot enter the HbS polymer
- DHbF inhibits the enzyme that catalyzes HbS polymerization
- EHbF competitively binds the abnormal valine residue at position 6
Answer
C
HbF is α2γ2 — it contains no β chains at all, so it cannot be incorporated into a polymer built from βS chains. Raising HbF dilutes the polymerizing species, which places it alongside the other treatment goals: reduce the percentage of HbS by transfusion, and reduce the amount of HbS per cell by avoiding dehydration. All three attack the same variable from different directions. This is also why newborns are naturally protected until HbF declines. D is wrong because polymerization is a physical aggregation process, not an enzymatic one, so there is no enzyme to inhibit. A, B, and E propose plausible-sounding but non-existent mechanisms.
22A 26-year-old woman of equatorial African ancestry has elliptocytes on a routine peripheral smear. Hemoglobin, bilirubin, and reticulocyte count are all normal, and she has no symptoms.Which statement about her condition is correct?
- AMost patients are not anemic and require no treatment
- BSplenectomy should be performed to prevent future hemolytic crises
- CThis finding confers increased susceptibility to Plasmodium falciparum
- DThe autosomal recessive inheritance predicts progressive hemolysis
- EThe elliptocytes indicate coexisting iron deficiency
Answer
A
Hereditary elliptocytosis is usually autosomal dominant, most often from α-spectrin gene mutations, and the majority of patients are not anemic — the elliptocytes are an incidental finding and treatment is unnecessary, since HE is uncommonly associated with hemolysis. B follows from that: with no meaningful hemolysis there is nothing for splenectomy to accomplish. C inverts a real association — like G6PD deficiency, HE is common in equatorial Africa because it confers resistance to falciparum malaria. D misassigns the inheritance; the recessive form is hereditary pyropoikilocytosis, the severe variant that genuinely causes hemolysis and anemia, illustrating the general rule that one working allele usually leaves enough normal protein to keep cells usable.
23A patient is described as having compensated hemolysis.Which finding is most consistent with this description?
- AA normal hemoglobin with an elevated reticulocyte count and elevated LDH
- BA low hemoglobin with a suppressed reticulocyte count
- CA normal hemoglobin with a normal reticulocyte count and normal LDH
- DA low hemoglobin with normal haptoglobin and normal bilirubin
- EA normal hemoglobin with an elevated ferritin and low TIBC
Answer
A
Hemolysis is defined as destruction of red cells with release of hemoglobin — and notably, anemia is not part of that definition. If increased marrow production keeps pace with the loss, the process is compensated: hemoglobin stays normal while the markers of accelerated turnover, such as reticulocytosis and elevated LDH, remain abnormal. That combination is what A describes. C describes a patient with no hemolysis at all. B is the pattern of a production failure, since the marrow is not responding. D lists laboratory findings that argue against hemolysis in a patient who is nonetheless anemic. E describes an iron-handling pattern unrelated to red cell destruction.
Lecture 03 · Tayal
Hemolytic Anemias, Part 2
Questions 24–35
24A 26-year-old man of Mediterranean ancestry is evaluated for a mild anemia. Hemoglobin is 11.2 g/dL, MCV is 63 fL, red cell count is 6.2 ×10&sup6;/µL, and RDW is 12.8%.Which is the most likely diagnosis?
- AIron deficiency anemia
- Bβ-thalassemia minor
- CAnemia of chronic disease
- DSideroblastic anemia
- ELead poisoning
Answer
B
Three numbers here point away from iron deficiency and toward thalassemia. The RDW is normal, because the genetic defect is present in every cell from the start, so the population is uniformly small — whereas iron deficiency produces a mixed population and a high RDW. The red cell count is increased: the marrow makes plenty of cells, they are simply underfilled, while in iron deficiency the count falls. And an MCV of 63 is lower than iron deficiency usually reaches; below 67 should raise thalassemia specifically. C, D, and E are all microcytic anemias, which is why they are listed, but none produces an elevated red cell count with a normal RDW.
25A 30-year-old woman has microcytosis with a normal RDW and an elevated red cell count. Hemoglobin electrophoresis shows HbA2 of 6.2%.Which is the most likely diagnosis?
- Aα-thalassemia trait
- BHemoglobin H disease
- CIron deficiency anemia
- Dβ-thalassemia trait
- ESilent carrier α-thalassemia
Answer
D
HbA2 is α2δ2 — it requires α chains but not β chains. When β production falls, δ chains take up some of the slack and the HbA2 percentage rises, which is why elevated HbA2 (roughly 3.5–8%) is characteristic of β-thal trait. In α-thalassemia the missing chain is the one HbA2 also needs, so HbA2 is not increased — eliminating A, B, and E despite all three being genuine α-thalassemia states with compatible red cell indices. C is excluded by the normal RDW and elevated red cell count, and iron deficiency does not raise HbA2.
26A 14-year-old with transfusion-dependent β-thalassemia is found to have a mutation producing no β-globin at all.Which mutation type most commonly produces this phenotype?
- ASplicing mutations
- BPromoter methylation silencing transcription
- CLarge gene deletions
- DMissense substitutions altering globin charge
- EChain terminator mutations
Answer
E
The β0 phenotype means zero β-globin output, and the characteristic lesion is a chain terminator mutation — a premature stop codon from a nonsense or frameshift change halts translation, and the truncated globin is degraded. It is an all-or-nothing mutation producing an all-or-nothing result. A produces the β+ phenotype instead, and the reasoning is worth holding: splicing is competitive, so a weakened or alternative site is used most of the time but the correct site is still used occasionally, leaving a minority of normal mRNA and therefore reduced but non-zero output. C is the mechanism of α-thalassemia, the standard contrast. D describes the hemoglobinopathy mutations such as HbS and HbC.
27A 16-year-old with β-thalassemia major who has never been transfused has a ferritin of 2,100 ng/mL, and a liver biopsy shows heavy iron deposition on Prussian blue staining.Which best explains the iron overload?
- AHemolysis releases iron that has no route of excretion
- BRepeated transfusion has delivered an excess iron load
- CIncreased hepcidin production traps iron within hepatocytes
- DIneffective erythropoiesis signals increased intestinal iron absorption
- EPrecipitated α-globin aggregates chelate iron within the liver
Answer
D
The stem specifically states the patient has never been transfused, which removes the mechanism most students reach for first and is the reason B is listed. The remaining explanation is the important one: ineffective erythropoiesis means the marrow churns through erythroid precursors that die in place, and that futile activity signals for more iron, upregulating absorption even though iron was never the problem. Transfusion, when it occurs, then adds a second independent source — which is why chelation is a standing requirement in these patients. C inverts the physiology: high hepcidin would reduce absorption. A and E propose iron-handling mechanisms that do not drive the overload.
28A radiograph of the skull in a child with untreated β-thalassemia major shows perpendicular bony radiations from the outer table, described as a crewcut appearance.Which mechanism is responsible?
- AExtramedullary hematopoiesis arising within the periosteum
- BOsteoblastic reaction to infiltration of the marrow space
- CIron deposition within the diploic space of the skull
- DSecondary hyperparathyroidism complicating chronic systemic illness
- EMarrow expansion driven by erythropoietin in response to anemia
Answer
E
The chain runs: anemia → tissue hypoxia → increased erythropoietin → marrow expansion → skeletal deformity. The expanding marrow cavity pushes outward, and new bone laid down on the outer table produces the radiating pattern. This is also why transfusion helps structurally as well as hematologically — it reduces the anemia and therefore removes the erythropoietin drive. A names a real phenomenon in these patients, but extramedullary hematopoiesis occurs in liver and spleen, not periosteum, and is not what produces this radiographic finding. C misattributes the change to the iron overload, which affects heart, liver, and pancreas. D and B describe bone processes that do not arise from thalassemia.
29A fetus with hydrops fetalis is found to have deletion of all four α-globin genes.Which best explains why this condition is incompatible with life?
- AThe γ4 tetramers are entirely incapable of binding oxygen
- BThe γ4 tetramers precipitate and destroy the red cell membrane from within
- CThe γ4 tetramers bind oxygen with high affinity and do not release it
- Dβ4 tetramers form instead and are inherently unstable in the fetus
- EAbsence of α chains prevents heme synthesis in erythroid precursors
Answer
C
With no α chains available, whatever non-α chain is most abundant pairs with itself. In the fetus that is γ, giving γ4 = Hb Bart. The lethal property is its high oxygen affinity: it binds oxygen and will not release it to tissues, so the fetus is profoundly hypoxic despite having hemoglobin. Severe anemia plus hypoxia drives high-output cardiac failure and total body edema. D describes HbH disease, where three genes are deleted and the surviving fetus makes β4 after birth — the same principle applied to a different developmental stage, which is what makes it the strongest distractor. A is wrong because the problem is oxygen release, not binding. E is wrong because heme synthesis is independent of globin availability.
30A 34-year-old man has hemolytic anemia, recurrent unprovoked venous thromboses, and pancytopenia. The direct antiglobulin test is negative.Which underlying defect is most likely?
- AAnkyrin deficiency destabilizing the membrane skeleton
- BA somatic PIGA mutation preventing GPI anchor formation
- CAn autoantibody directed against CD59
- DDeficiency of complement factor H
- EReduced ADAMTS13 activity
Answer
B
The triad of chronic hemolysis, thrombosis, and pancytopenia with a negative DAT is paroxysmal nocturnal hemoglobinuria. The lesion is a somatic PIGA mutation in a hematopoietic stem cell, which prevents formation of the GPI anchor. Proteins that require that anchor cannot attach — functionally the critical two are CD55 and CD59, both complement brakes — leaving cells defenseless against ongoing low-level complement activation. Thrombosis is the most common cause of death. A is hereditary spherocytosis. C is a deliberate near-miss: the outcome resembles PNH but the defect is a failure to express the protein, not an antibody against it. D relates to complement dysregulation in atypical HUS, and E to TTP.
31A patient with paroxysmal nocturnal hemoglobinuria is described as having a non-immune hemolytic anemia despite complement-mediated cell destruction.Which best justifies that classification?
- AComplement regulatory proteins are absent, rather than antibody being bound
- BComplement is activated only after the cells have already begun to lyse
- CThe responsible IgM antibody elutes from the cell during sample processing
- DThe GPI anchor is required before any antibody binding can occur
- EDestruction of the cells in PNH is entirely extravascular
Answer
A
The distinction rests on what initiates the destruction. In the immune hemolytic anemias an antibody binds the red cell and the process follows from there. In PNH there is no antibody: the cell is destroyed because it has lost CD55 and CD59, so ordinary background complement activation proceeds unchecked. Complement is involved, but as the effector, not as the consequence of an immune recognition event — which is why PNH is grouped with the acquired non-immune disorders alongside microangiopathic and mechanical hemolysis. C describes the situation in cold agglutinin syndrome, where IgM does elute and leaves complement behind. B, D, and E misstate the biology.
32A 68-year-old man develops anemia and acrocyanosis of the fingers during winter, two weeks after an episode of atypical pneumonia. The peripheral smear shows clumped red cells.Which direct antiglobulin test result is most expected?
- APositive for IgG only
- BPositive for both IgG and complement in equal proportion
- CPositive for complement only
- DNegative
- EPositive for IgA
Answer
C
Cold agglutinin syndrome following Mycoplasma pneumoniae is mediated by an IgM autoantibody, usually against the I antigen. The DAT detects complement rather than immunoglobulin, and the reason is mechanical: IgM binds in the cool periphery and fixes complement there, then elutes as the blood rewarms on its way back to the core, leaving the deposited complement behind on the cell. A is the pattern of warm autoimmune hemolytic anemia, where IgG stays bound at body temperature. D would suggest a non-immune process such as PNH or hereditary spherocytosis. The red cell clumping in the stem is a further clue: agglutination points cold, since IgG does not agglutinate efficiently at 37°C.
33A 71-year-old man with back pain, renal insufficiency, and a monoclonal gammopathy has a peripheral smear showing red cells arranged in stacked columns.Which best describes this finding?
- AAgglutination from a cold-reacting antibody
- BTarget cells from redundant membrane
- CSpherocyte formation from partial membrane removal
- DSchistocytes from microvascular shearing
- ERouleaux formation from increased plasma paraproteins
Answer
E
Rouleaux are red cells stacked like coins, produced when high concentrations of monoclonal protein reduce the charge repulsion that normally keeps cells apart — the classic setting is plasma cell myeloma, which the back pain, renal insufficiency, and gammopathy describe. A is the intended contrast and the reason this question exists: agglutinates also look like red cells sticking together but are irregular clumps produced by an antigen–antibody reaction, pointing to cold agglutinin syndrome rather than a paraproteinemia. C, D, and B describe alterations in the shape of individual cells rather than in how cells associate with one another.
34A 6-year-old develops dark red urine several hours after playing outdoors on a cold day, two weeks after a viral illness. Serum drawn during the episode appears red.Which antibody is responsible?
- AAn IgM antibody against the I antigen active in the cold
- BAn IgM antibody against the i antigen arising after infection
- CAn IgG antibody against Rh antigens reacting maximally at body temperature
- DAn IgG antibody against the P antigen, binding cold and lysing on rewarming
- EAn IgG antibody against platelet glycoprotein IIb-IIIa complexes
Answer
D
This is paroxysmal cold hemoglobinuria, mediated by the Donath-Landsteiner antibody — an IgG directed against the P antigen that is biphasic: antibody and early complement components bind at low temperature in the periphery, then terminal components assemble on rewarming and the cell lyses. Because lysis completes inside the vessel, PCH is one of the few genuinely intravascular immune hemolytic anemias, which is why the serum is red and the urine dark. It classically follows viral infection in children. A and B are cold agglutinin antibodies, which are IgM and cause agglutination rather than biphasic intravascular lysis. C describes warm AIHA. Note also that corticosteroids are not helpful here, a deliberate contrast with warm disease.
35A patient taking quinine for nocturnal leg cramps develops acute hemolysis. Laboratory work shows an antibody that binds red cells only when the drug is present in the reaction mixture.Which mechanism is responsible?
- AHapten adsorption of drug to the red cell membrane
- BComplement lysis from deficient CD55 and CD59
- CTrue autoantibody formation
- DImmune complex (ternary complex) formation
- EOxidative injury from a drug metabolite
Answer
D
Quinine and quinidine are the standard examples of the immune complex or ternary complex mechanism, mediated by IgM: antibody forms on exposure, and on further exposure the drug–antibody complex adsorbs onto the red cell. A is the closest distractor and describes a genuinely different mechanism — in hapten adsorption the drug binds the membrane first and antibody forms against the drug–membrane complex, with penicillin and cephalosporins as the examples. Both A and D require the drug to be present, so hemolysis stops on withdrawal. C is the mechanism of α-methyldopa, where a true self-directed antibody is created that can persist after the drug is stopped — the clinically important difference. B is PNH and E is the G6PD mechanism.
Lecture 04 · Tayal
Bleeding Disorders, Part 1
Questions 36–47
36A 9-year-old boy develops palpable purpura over the buttocks and lower extremities along with abdominal pain and arthralgias two weeks after an upper respiratory infection. Platelet count is 284 ×10³/µL, PT is 12.4 seconds, and aPTT is 29 seconds.Which is the most likely diagnosis?
- AImmune thrombocytopenic purpura
- BDisseminated intravascular coagulation
- Cvon Willebrand disease
- DHemophilia A
- EHenoch-Schönlein purpura
Answer
E
Hemostasis has three arms — the vessel, the platelet, and the clotting factors. The platelet count tests the second and PT/aPTT test the third, so a patient who bleeds with a completely normal panel should direct attention to the arm nothing in the routine workup examines: the vessel wall. Henoch-Schönlein purpura is an immune complex vasculitis and is one of the non-thrombocytopenic purpuras. A is excluded by the normal platelet count, and D and B by the normal clotting times — DIC would also lower the platelet count. C is the best of the remaining distractors because vWD does cause mucocutaneous bleeding with a normal platelet count, but it does not produce palpable purpura with abdominal pain and arthritis.
37A 41-year-old woman undergoing evaluation for fatigue is incidentally found to have a platelet count of 62 ×10³/µL. She reports no bleeding, and examination shows no petechiae or ecchymoses.Which is the most appropriate initial management?
- ATransfuse one unit of apheresis platelets prophylactically
- BBegin high-dose corticosteroid therapy
- CObserve, since bleeding is not expected at this count
- DAdminister intravenous immunoglobulin over two days
- EProceed to elective splenectomy
Answer
C
The reference range begins at 150, but the threshold at which bleeding begins is far lower. Counts between 50 and 100 produce no clinical findings at all; bruising with minor trauma appears around 30–50, spontaneous bruising and menorrhagia at 10–30, and spontaneous bleeding at roughly 10. A count of 62 in an asymptomatic patient therefore requires evaluation of the cause but no immediate intervention. A, B, D, and E are all real treatments for thrombocytopenia, and each becomes appropriate at a lower count or in a bleeding patient — the error the question targets is attaching a treatment decision to a number simply because it falls below the reference range.
38A 28-year-old woman has a platelet count of 14 ×10³/µL with a normal hemoglobin and white cell count. The smear shows large platelets, and marrow examination demonstrates increased megakaryocytes.Which best explains these findings?
- AMarrow failure with ineffective and disordered megakaryopoiesis throughout
- BPeripheral destruction with a marrow response releasing young platelets
- CSplenic sequestration of platelets that were produced normally
- DDilution following large-volume crystalloid and colloid resuscitation
- EConsumption of platelets within widespread microvascular thrombi
Answer
B
The two findings point the same direction. The marrow is busy, not failing, so the problem lies in the periphery. And the platelets are large because they are young — the marrow is compensating by releasing newly formed platelets. That combination of large platelets with a hypercellular megakaryocyte compartment is the signature of a destructive thrombocytopenia, and in an otherwise healthy patient with isolated thrombocytopenia, immune thrombocytopenic purpura is the most common cause. A is the direct opposite, and a production failure would show small platelets and a hypocellular marrow. C, D, and E are genuine causes of thrombocytopenia but none produces increased marrow megakaryocytes with large circulating platelets.
39A 33-year-old woman with immune thrombocytopenic purpura has a platelet count of 4 ×10³/µL. She has no active bleeding.Which best explains why prophylactic platelet transfusion is not indicated?
- AThe autoantibody will destroy transfused platelets as rapidly as the patient's own
- BTransfusion would increase the risk of thrombosis in this setting
- CTransfusion risks HLA alloimmunization that would complicate later care
- DThe marrow will restore the count within a few hours without intervention
- ETransfused platelets cannot reach the microcirculation where they are needed
Answer
A
The rule is that there is no role for prophylactic platelet transfusion in ITP, even at a count of zero, in a patient who is not bleeding — and the reason is mechanistic. The circulating autoantibody recognizes a platelet surface glycoprotein such as GPIIb-IIIa, and it does not distinguish the patient's platelets from donor platelets. Transfusing is feeding the same machine. Transfusion is reserved for active, serious bleeding. B is a true statement about a different disorder — it is the reason platelets are avoided in HIT, where the state is actively prothrombotic. C is a real concern but not the governing reason. D and E are false.
40A 5-year-old boy develops petechiae and epistaxis ten days after a viral upper respiratory illness. Platelet count is 18 ×10³/µL, hemoglobin and white cell count are normal, and he is otherwise well.Which is the most appropriate management?
- ASplenectomy after failure of medical therapy
- BObservation, with spontaneous remission expected
- CLong-term corticosteroid therapy with a taper
- DRituximab given as four weekly infusions
- EA thrombopoietin receptor agonist such as eltrombopag
Answer
B
Acute immune thrombocytopenia in children is clinically a different disease from the chronic adult form. It has an abrupt onset, usually after a viral illness, does not usually require steroids, and undergoes spontaneous and permanent remission. The adult form, by contrast, is typically chronic with relapses, which is where the escalating treatment ladder becomes relevant. A, C, D, and E are all legitimate therapies for chronic adult ITP, arranged along that ladder from immune suppression through Fc receptor blockade, splenectomy, and increased production — and each is inappropriate here, where the natural history is resolution.
41A 64-year-old man receiving unfractionated heparin after orthopedic surgery has a platelet count fall from 245 to 88 ×10³/µL on day 7. He develops a new proximal deep venous thrombosis.Which best explains the thrombosis despite the falling platelet count?
- AHeparin-PF4 complexes directly activate factor X on the platelet surface
- BHeparin directly inhibits antithrombin, producing a rebound prothrombotic state in the vasculature
- CDestroyed platelets release tissue factor from activated splenic macrophages
- DThe antibody activates complement directly on the endothelial surface
- EAntibody binding to platelet FcγRIIa causes activation and microparticle release
Answer
E
Every other cause of thrombocytopenia in this material lowers the count by removing platelets. HIT lowers it by activating them first. IgG antibodies form against heparin–PF4 complexes, bind the platelet FcγRIIa receptor, and drive the platelet to release procoagulant microparticles including thrombin. Only afterward are the activated, antibody-coated platelets consumed and destroyed. Because activation precedes destruction, the falling count is a marker of consumption rather than a bleeding risk, and the clinical danger is thrombosis. The remaining options invoke mechanisms that do not occur in HIT, and B in particular inverts heparin's actual interaction with antithrombin.
42A patient with suspected heparin-induced thrombocytopenia has had all heparin discontinued, including line flushes. The platelet count is 62 ×10³/µL and a new thrombosis has been documented.Which is the most appropriate next step?
- AObserve, since removal of heparin is sufficient
- BResume heparin at a reduced dose
- CTransfuse platelets before initiating anticoagulation
- DBegin warfarin as sole anticoagulation
- EBegin argatroban
Answer
E
Stopping heparin removes the trigger but does nothing about the thrombin burst that has already occurred — which is why A is the trap this question is built around. A direct thrombin inhibitor such as argatroban or bivalirudin is required to neutralize the thrombin already generated. C is worse than unhelpful: platelet transfusion may increase thrombotic risk, since it adds substrate to an actively prothrombotic state. B reintroduces the antigen driving the process. D is the subtlest error — warfarin does not address the immediate thrombin excess, and starting it alone in the acute phase is specifically avoided.
43A patient begins heparin therapy and the platelet count falls from 220 to 132 ×10³/µL on day 2. No thrombosis occurs, and the count returns to baseline over the following days while heparin is continued.Which best describes this event?
- AImmune thrombocytopenic purpura triggered by heparin
- BType II heparin-induced thrombocytopenia, immune
- CType I heparin-induced thrombocytopenia, non-immune
- DDilutional thrombocytopenia
- EEarly disseminated intravascular coagulation
Answer
C
Three features identify Type I: the fall is mild (100,000–150,000), the onset is rapid at 1–2 days, and it resolves despite continued heparin, with no thrombosis. The mechanism is non-immune. Type II is the dangerous entity — severe (below 100,000), delayed to days 5–10, persistent until heparin is stopped, and associated with thromboembolic complications — and its timing is scored in the 4 Ts as one of four elements alongside the magnitude of the fall, thrombosis, and the absence of other causes. A, D, and E do not fit a count that recovers spontaneously while the drug continues.
44A term newborn delivered to a healthy primigravida develops petechiae and a platelet count of 22 ×10³/µL on the first day of life. The mother's platelet count is 258 ×10³/µL.Which is the most likely diagnosis?
- ARh hemolytic disease of the newborn
- BMaternal immune thrombocytopenic purpura with transplacental autoantibody
- CCongenital amegakaryocytic thrombocytopenia
- DNeonatal sepsis
- ENeonatal alloimmune thrombocytopenia
Answer
E
Two features settle this. The mother's platelet count is normal, which is exactly what distinguishes an alloantibody from an autoantibody: she lacks the HPA-1a antigen, so her antibody attacks fetal platelets but has nothing to attack on her own. In maternal ITP the autoantibody targets an antigen she also carries, so her count would be low — making B the intended contrast. The second feature is that this is a first pregnancy, and NAIT can affect the first child, unlike Rh disease, which requires a prior sensitizing pregnancy. Importing the Rh timing into the platelet disease is the classic error, which is why A is included.
45A 42-year-old woman presents with confusion, fever, and petechiae. Platelet count is 22 ×10³/µL, the smear shows schistocytes, LDH is markedly elevated, and creatinine is 1.5 mg/dL. PT is 12.8 seconds, aPTT is 30 seconds, and fibrinogen is 320 mg/dL.Which is the most likely diagnosis?
- ADisseminated intravascular coagulation
- BImmune thrombocytopenic purpura
- CTypical hemolytic uremic syndrome
- DThrombotic thrombocytopenic purpura
- EHeparin-induced thrombocytopenia
Answer
D
The pentad — thrombocytopenia, microangiopathic hemolysis, renal dysfunction, neurologic disturbance, and fever — is present, and the discriminating laboratory finding is that PT, aPTT, and fibrinogen are all normal. That is what separates TTP from DIC, which shares the low platelets, schistocytes, and microvascular thrombi but prolongs both clotting times and consumes fibrinogen. The logic is about what is being consumed: TTP consumes platelets on uncleaved vWF multimers while leaving the coagulation factors untouched. C is plausible but the renal impairment here is mild and neurologic findings dominate, the reverse of typical HUS. B produces isolated thrombocytopenia without schistocytes or organ dysfunction.
46A patient with thrombotic thrombocytopenic purpura is treated with therapeutic plasma exchange rather than plasma infusion alone.Which best explains this choice?
- AExchange delivers a higher concentration of ADAMTS13 than infusion can achieve
- BExchange removes the autoantibody and multimers while supplying enzyme
- CInfused plasma does not contain ADAMTS13
- DExchange carries a lower risk of volume overload
- EExchange physically removes schistocytes from the circulation
Answer
B
Plasma exchange performs two jobs in one procedure. Infusing plasma alone would supply ADAMTS13 but leave the IgG autoantibody in place to neutralize it — and would also leave the accumulated ultra-large vWF multimers circulating. Exchange removes the patient's plasma, carrying away both the antibody and the toxic product, and replaces it with donor plasma containing working enzyme. It attacks the cause and the consequence simultaneously, which is why it is the emergency intervention rather than an adjunct. C is false, and it is precisely because plasma does contain ADAMTS13 that it serves as the replacement fluid. A and D are secondary considerations, and E misdescribes the purpose.
47A 5-year-old develops bloody diarrhea followed by oliguria. Creatinine is 3.9 mg/dL, platelet count is 58 ×10³/µL, and schistocytes are present. ADAMTS13 activity is normal.Which is the most appropriate management?
- ATherapeutic plasma exchange
- BCorticosteroids with rituximab
- CEculizumab
- DSupportive care without specific therapy
- EPlatelet transfusion to a target above 50 ×10³/µL
Answer
D
Typical hemolytic uremic syndrome follows infection with Shiga toxin-producing E. coli, and its distinguishing features against TTP are prominent acute renal failure, less prominent neurologic involvement, a childhood predominance with frequent recovery, and normal ADAMTS13. Treatment follows the lesion: the toxin has already injured the endothelium and is gone, so there is nothing for exchange to remove — hence supportive care only. A is correct for TTP, where an antibody and an accumulated substrate can both be cleared. C is correct for atypical HUS, driven by unregulated complement. B targets autoantibody production, which is not the mechanism here.
Lecture 05 · Tayal
Bleeding Disorders, Part 2
Questions 48–59
48A 7-year-old boy has had three episodes of painful knee swelling following minor falls. He has never had epistaxis, gum bleeding, or petechiae.Which category of defect does this bleeding pattern suggest?
- AA qualitative platelet disorder
- BA clotting factor deficiency
- CA vessel wall abnormality
- DA quantitative platelet disorder
- EA disorder of fibrinolysis
Answer
B
The type of bleed is the single most useful historical discriminator, and it maps directly onto the underlying defect. Platelets plug tiny capillary leaks continuously, so losing them produces diffuse, superficial, small-vessel bleeding — petechiae, mucosal ooze, epistaxis — appearing immediately. Clotting factors stabilize a plug that has already formed, so losing them allows the initial plug to form and then fail later under pressure, giving bleeding into deep spaces: joints and soft tissue, often with a delay. Hemarthrosis is a factor disease until proven otherwise. A and D both describe platelet problems, which would present with the mucocutaneous pattern the stem explicitly excludes. C also produces petechiae and purpura.
49A 58-year-old man bleeds excessively after a dental extraction. He takes aspirin daily. Platelet count is 244 ×10³/µL, PT is 12.1 seconds, and aPTT is 28 seconds.Which test is most likely to reveal the abnormality?
- AA mixing study with normal plasma
- BFibrinogen level and thrombin time
- CFactor VIII and factor IX activity assays
- DD-dimer and fibrin degradation products
- EAutomated platelet function analysis
Answer
E
Qualitative platelet defects are invisible to every routine screening test. The count is normal because all the platelets are present; PT and aPTT are normal because the clotting factors are intact. The defect is in platelet function, and only a function test will show it — the automated analyzer exposes platelets to agonists such as collagen, epinephrine, and ADP and is specifically good at detecting antiplatelet medications, von Willebrand disease, and inherited function disorders. Aspirin acts by inhibiting cyclooxygenase and therefore thromboxane A2 production. A, B, D, and C all interrogate the coagulation cascade or its products, which the normal PT and aPTT have already shown to be intact. Note that bleeding time is no longer performed.
50A 22-year-old woman has had lifelong mucosal bleeding requiring occasional platelet transfusion. Platelet count and platelet size are normal. Aggregometry shows absent aggregation with ADP, collagen, and epinephrine, but a normal response to ristocetin.Which is the most likely diagnosis?
- ABernard-Soulier syndrome
- BHemophilia A
- Cvon Willebrand disease, type 1
- DStorage pool disease
- EGlanzmann thrombasthenia
Answer
E
The agonist that fails identifies the receptor that is missing. ADP, collagen, and epinephrine all funnel into activation of GPIIb-IIIa, the fibrinogen receptor that cross-links platelets to one another — so failure across all of them indicates a defect in aggregation, which is Glanzmann thrombasthenia. Ristocetin is the exception because it works by promoting vWF binding to GP1b, testing adhesion instead; a normal ristocetin response therefore exonerates that axis, eliminating A and C. Glanzmann is autosomal recessive, has a normal platelet count and size, and is the most common inherited disorder of platelet function. B is excluded by the mucosal rather than deep bleeding pattern and by the normal aggregometry that hemophilia would not affect.
51A 17-year-old has mild thrombocytopenia with strikingly large platelets on the smear. Aggregometry shows absent aggregation with ristocetin but normal responses to ADP, collagen, and epinephrine.Which receptor is deficient?
- AGPIIb-IIIa, the fibrinogen receptor
- BGP1b, the von Willebrand factor receptor
- CP2Y12, the ADP receptor
- DThe thromboxane A2 receptor
- EGPVI, the collagen receptor
Answer
B
This is the mirror image of the preceding pattern. Ristocetin tests the adhesion axis by promoting vWF binding to GP1b, so isolated failure of ristocetin aggregation localizes the defect there — and the giant platelets with mild thrombocytopenia identify Bernard-Soulier syndrome specifically rather than von Willebrand disease. A would give the opposite aggregometry result, failing with ADP, collagen, and epinephrine while responding to ristocetin. C and D name receptors targeted by clopidogrel and inhibited downstream of aspirin respectively, both of which are acquired rather than inherited and neither of which produces giant platelets.
52A 26-year-old woman with menorrhagia is found to have absent ristocetin-induced platelet aggregation. Her platelet count and platelet size are normal.Which additional result would best distinguish von Willebrand disease from Bernard-Soulier syndrome?
- Avon Willebrand factor antigen level
- BProthrombin time
- CPlatelet count and mean platelet volume
- DFactor IX activity
- ED-dimer
Answer
A
Both disorders lie on the same pathway and both abolish ristocetin aggregation, because the test requires vWF to bridge to GP1b and either partner can be the missing one. The way to separate them is to measure the ligand directly: vWF antigen is low in vWD and normal in Bernard-Soulier, where the deficiency is in the platelet receptor. C is a reasonable instinct, since Bernard-Soulier characteristically shows giant platelets with mild thrombocytopenia — but the stem states the count and size are normal, deliberately removing that clue. B, D, and E examine the coagulation cascade and fibrinolysis, neither of which is the site of the lesion.
53A hospitalized patient who has been receiving broad-spectrum antibiotics and taking nothing by mouth for eight days has a PT of 19.2 seconds, an aPTT of 33 seconds, and a platelet count of 214 ×10³/µL.Which best explains why the PT is prolonged before the aPTT?
- AVitamin K deficiency spares the factors measured by the aPTT
- BFactor VII does not require vitamin K for its activity
- CThe PT is intrinsically more sensitive than the aPTT to any factor deficiency
- DFactor VII is consumed first once the tissue factor pathway is triggered
- EFactor VII has the shortest half-life of the vitamin K-dependent factors
Answer
E
Vitamin K is the cofactor for gamma-carboxylation of factors 2, 7, 9, and 10, and all four fall when it is deficient — but not at the same rate. A factor's level declines in proportion to how fast it clears, and factor 7 has the shortest half-life of the four. Factor 7 is also the only one of them in the extrinsic limb, which is what the PT reads. The result is an isolated PT prolongation first, with the aPTT following as 9, 10, and 2 deplete — and the same logic explains why PT/INR is used to monitor warfarin. B and A are factually wrong. C misattributes the finding to assay sensitivity rather than factor kinetics. D invokes consumption, which is not the mechanism in a deficiency state.
54A 54-year-old man with sepsis has a PT of 23 seconds, an aPTT of 51 seconds, a platelet count of 64 ×10³/µL, a fibrinogen of 88 mg/dL, and a markedly elevated D-dimer.Which is the most likely diagnosis?
- AVitamin K deficiency
- BChronic liver disease
- Cvon Willebrand disease, type 3
- DHemophilia A with an acquired inhibitor
- EDisseminated intravascular coagulation
Answer
E
DIC is the only disorder in these lectures that strikes all three arms of hemostasis at once, and the combination is effectively diagnostic: widespread thrombin generation consumes platelets so the count falls, consumes factors so PT and aPTT both rise, and converts fibrinogen to fibrin so fibrinogen falls — after which fibrinolysis clears that fibrin and releases D-dimer. B is the closest competitor and shares the low platelets and prolonged times, but liver disease does not characteristically produce a markedly elevated D-dimer, and the fibrinogen here is very low. A produces a prolonged PT with a normal platelet count. D and C are single-lesion disorders that cannot move every parameter.
55A 31-year-old woman with a lifelong history of easy bruising and menorrhagia has a von Willebrand factor antigen level of 71% and a ristocetin cofactor activity of 19%.Which is the most likely diagnosis?
- Avon Willebrand disease, type 2
- Bvon Willebrand disease, type 1
- Cvon Willebrand disease, type 3
- DAcquired von Willebrand syndrome from aortic stenosis
- EBernard-Soulier syndrome
Answer
A
The discriminator is antigen versus activity. Types 1 and 3 are quantity problems: the protein that is present functions normally, so antigen and activity fall together. Type 2 is a quality problem: plenty of protein is produced but it is structurally abnormal and cannot interact properly with platelet GP1bα and factor 8, so antigen substantially exceeds activity — which is exactly the discrepancy shown here. Type 2 accounts for 20–35% of cases, most often from missense substitutions causing defective multimer assembly, with type 2A the most common variant. B would show both numbers modestly and proportionally reduced; C would show both near zero. E is a platelet receptor defect and would not lower vWF antigen.
56A patient with type 3 von Willebrand disease is scheduled for a dental procedure. Desmopressin is considered and rejected.Which best explains why it would be ineffective?
- ADesmopressin acts through V1 receptors, which are absent in type 3
- BType 3 patients have essentially no stored von Willebrand factor available for release
- CThe von Willebrand factor released in type 3 is structurally abnormal and non-functional
- DDesmopressin requires intact platelet GP1b to exert its effect
- EType 3 patients develop neutralizing antibodies against released factor
Answer
B
DDAVP works by emptying a storage depot — it acts on endothelial V2 receptors to release vWF already made and stored in Weibel-Palade bodies. That requires there to be something in the depot. Type 1 patients make normally functioning vWF, just too little, so a reserve exists and the released protein works, which is why DDAVP is the type 1 treatment. Type 3 patients make almost none, or none that is secreted, so there is nothing to mobilize and they require concentrate instead. C is the intended near-miss: it is the correct reasoning for type 2, where the released protein is defective. A misnames the receptor, and D and E describe mechanisms that do not apply.
57A patient with von Willebrand disease is noted to have a prolonged aPTT.Which best explains this finding?
- Avon Willebrand factor acts as a cofactor within the intrinsic cascade
- BImpaired platelet adhesion secondarily prolongs the aPTT
- Cvon Willebrand factor directly activates factor IX in plasma
- DUnbound factor VIII is rapidly cleared, lowering factor VIII levels
- EvWF deficiency causes secondary malabsorption of vitamin K
Answer
D
von Willebrand factor is not a clotting factor and has no place in the cascade, so on its own it should not affect the aPTT — which is why A and C are wrong despite sounding reasonable. It does so indirectly: factor 8 binds vWF with high affinity, and vWF stabilizes it, so unbound factor 8 is rapidly cleared and vWF deficiency drags factor 8 down with it. This explains two further observations: the aPTT in vWD is normal unless factor 8 falls significantly, and severe type 3 disease can produce hemarthrosis mimicking hemophilia, because the patient functionally has both an adhesion defect and a factor 8 deficiency. B is wrong because the aPTT is a plasma assay indifferent to platelet adhesion.
58A 74-year-old man with no personal or family history of bleeding develops extensive spontaneous soft tissue hematomas. The aPTT is 68 seconds and does not correct when his plasma is mixed with normal plasma.Which is the most likely diagnosis?
- AHemophilia A presenting late in life
- BAn acquired factor VIII inhibitor
- CFactor IX deficiency
- Dvon Willebrand disease
- EVitamin K deficiency
Answer
B
The mixing study is the decisive test, and its logic is direct: if the patient is simply missing a factor, the normal plasma supplies it and the clotting time corrects. If the patient has an inhibiting antibody, that antibody attacks the factor in the donated plasma too, so the aPTT stays prolonged. Corrects equals deficiency; does not correct equals inhibitor. Here the failure to correct, combined with an elderly patient and no prior bleeding history, indicates an acquired autoantibody against factor 8. A, C, D, and E are all deficiency states, and every one of them would correct on mixing — which is the point of the question. Inhibitor strength is quantified by the Bethesda titer.
59A patient with hemophilia A and a high-titer factor VIII inhibitor is bleeding and is given recombinant factor VIIa.Which best describes how this agent works?
- AIt supplies factor VIII in a form the antibody cannot recognize
- BIt neutralizes the circulating inhibitory antibody
- CIt activates factor X directly on the platelet surface, downstream of the blockade
- DIt inhibits conversion of plasminogen to plasmin, stabilizing existing clot
- EIt replaces all four vitamin K-dependent factors simultaneously
Answer
C
A factor 8 inhibitor blocks the intrinsic route to factor X. Recombinant 7a activates X directly on the platelet surface, entering the cascade downstream of the blockade entirely and producing a thrombin burst — so the antibody becomes irrelevant. That is what bypassing agent means: the antibody is not overcome, it is routed around. A describes a strategy that does not exist, though giving large doses of factor 8 to overwhelm the inhibitor is a separate real option. B misstates the mechanism. D describes the antifibrinolytics such as aminocaproic acid and tranexamic acid. E describes prothrombin complex concentrate, which is used for warfarin reversal, though activated PCC is also used in this setting.
Lecture 06 · Tayal
Blood Transfusion
Questions 60–71
60Ten minutes into a red cell transfusion, a patient develops fever, rigors, hypotension, severe lumbar back pain, and dark red urine.Which underlying failure most likely produced this event?
- ADonor HLA antibodies reacting with the recipient's neutrophils
- BFailure of the antibody screen to detect a Kell system alloantibody
- CBacterial contamination of the unit during collection or processing
- DMisidentification of the patient, giving ABO-incompatible blood
- EInfusion of a volume exceeding the patient's limited cardiac reserve
Answer
D
This is an acute hemolytic transfusion reaction, and the emphasis is deliberate: these are traced to patient misidentification — the wrong blood given to the wrong patient — rather than to a failure of serology. The most common single scenario is group A blood given to a group O patient. It is why mislabeled tubes are never accepted under any circumstances. B describes a delayed rather than acute mechanism and would not produce this presentation. C would cause high fever with profound hypotension but not back pain and hemoglobinuria. A is TRALI and E is TACO, both of which present with respiratory rather than hemolytic findings. The first step in any suspected reaction is to stop the transfusion.
61A student is asked why ABO incompatibility produces immediate intravascular hemolysis while Rh incompatibility does not.Which explanation is correct?
- AABO antigens are present at far higher copy number on the red cell surface
- BABO antibodies require prior sensitization and are therefore of higher affinity
- CAnti-A and anti-B are IgG and are therefore cleared rapidly by splenic macrophages
- DAnti-A and anti-B are IgM and fix complement efficiently to C5b-9
- EABO antibodies activate the coagulation cascade directly
Answer
D
Anti-A and anti-B are naturally occurring and preformed — the expected antibodies, present without any prior transfusion or pregnancy, which makes B wrong. They are IgM, which is pentameric and therefore an extremely efficient complement fixer: a single molecule can bridge the C1q arms and drive the cascade to C5b-9, punching holes in the cell inside the vessel. Anti-D, by contrast, is IgG, requires exposure, and fixes complement poorly, so those cells are removed slowly by splenic macrophages — extravascular. C states the IgG mechanism but attaches it to the wrong antibody. The general principle carries over from hemolytic anemia: IgM to complement to intravascular; IgG to spleen to extravascular.
62An Rh(D)-negative woman in her first pregnancy is carrying an Rh(D)-positive fetus and is given Rho(D) immune globulin.What is the purpose of this intervention?
- ATo neutralize preformed maternal anti-D of the IgM class
- BTo prevent ABO hemolytic disease of the newborn
- CTo prevent maternal formation of anti-D IgG
- DTo suppress fetal erythropoiesis and reduce antigen load
- ETo prevent maternal alloimmunization to Kell antigens
Answer
C
Rh antibodies are caused by exposure — transfusion, pregnancy, or transplant — rather than being naturally present, which is what makes A wrong. RhoGAM prevents the mother from forming anti-D after exposure to fetal red cells. The reason this matters is that anti-D is IgG, and only IgG crosses the placenta, so in a later pregnancy it would cross, bind fetal red cells, and destroy them. That single fact is why anti-D endangers a fetus while anti-A and anti-B, being IgM, largely do not, so B is incorrect. It is also the identical mechanism behind neonatal alloimmune thrombocytopenia, where maternal IgG against a fetal platelet antigen crosses to destroy fetal platelets.
63A blood bank technologist finds that a patient's red cells fail to react with anti-A, anti-B, or anti-D reagents. The patient's plasma agglutinates group A cells, group B cells, and group O cells.Which best explains these results?
- AThe patient is group O, Rh(D) negative
- BThe sample was drawn after a recent large-volume transfusion
- CThe patient is group AB, Rh(D) negative
- DThe patient has the Bombay phenotype
- EA cold agglutinin is interfering with the reverse typing
Answer
D
The front type looks like group O — carrying neither A nor B antigen and D negative — and would ordinarily be confirmed by a back type showing anti-A and anti-B. The critical abnormality is that the plasma also agglutinates group O cells, which a true group O patient's plasma would not, since group O cells carry the H antigen that everyone else has too. The H antigen (fucose) is the backbone that must be present before either ABO antigen can be built; individuals lacking it entirely have the Bombay phenotype and make anti-H, rendering them incompatible with essentially all ordinary donors, including group O. C is excluded because AB cells would react with both reagents. This case illustrates why front and back typing exist as reciprocal checks: when they disagree, something is wrong.
64A group AB patient requires urgent red cell transfusion, and no group AB units are available.Which group should be selected next?
- AGroup O only, since it is the universal donor
- BGroup A or B only after a full crossmatch and antibody identification
- CGroup A
- DNo transfusion should be given until AB units arrive
- EGroup O, but only if the patient is Rh(D) negative
Answer
C
The selection rule is that a recipient may receive any group whose antigens they will not attack. A group AB patient has neither anti-A nor anti-B, which is what makes AB the universal recipient, so an AB recipient takes AB units first, then group A or group B units, with group O last. A is a common reflex error: group O is the universal red cell donor because O cells carry no A or B antigen, but that makes it the last choice for an AB patient rather than the first, since better-matched options exist and the supply of O should be conserved. D is unsafe in an urgent setting when compatible alternatives exist. E introduces an Rh condition that does not govern the ABO selection sequence.
65A 24-year-old man with severe hemophilia A presents with a spontaneous hemarthrosis.Which product should be administered?
- AFactor VIII concentrate
- BFresh frozen plasma
- CCryoprecipitate
- DWhole blood
- EPlatelet concentrate
Answer
A
Cryoprecipitate does contain factor 8 and von Willebrand factor, and it is still the wrong answer for von Willebrand disease and for hemophilia A — this is stated explicitly. Those patients receive specific factor concentrates, recombinant or plasma-derived, which are purer, virally inactivated, and dose-calculable. Cryoprecipitate is a fibrinogen product, indicated for severe hypofibrinogenemia below 100 mg/dL in massive transfusion and DIC, for dysfibrinogenemia, and for factor 13 deficiency. B would deliver factor 8 only in dilute form along with every other factor. The principle worth carrying: the presence of an ingredient does not make a product the right treatment.
66A trauma patient who has received twelve units of red cells continues to bleed diffusely. The fibrinogen level is 62 mg/dL.Which product is most appropriate?
- AAdditional red blood cells
- BPlatelet concentrate
- CFactor IX concentrate
- DCryoprecipitate
- EAlbumin
Answer
D
This is the indication cryoprecipitate exists for: severe hypofibrinogenemia below 100 mg/dL in the setting of massive transfusion or DIC, with a post-transfusion target of 100–150 mg/dL. Cryoprecipitate is prepared from FFP and concentrates fibrinogen, factor 8, von Willebrand factor, and factor 13 into a small volume, which is what makes it the efficient way to raise fibrinogen. B addresses a different component and no platelet count is given. C treats hemophilia B rather than hypofibrinogenemia. A would worsen the dilutional coagulopathy without addressing it. E provides oncotic support but no hemostatic factors at all.
67A patient with thrombotic thrombocytopenic purpura undergoes therapeutic plasma exchange using fresh frozen plasma as the replacement fluid.Which property of FFP makes it the appropriate choice?
- AIt supplies ADAMTS-13
- BIt supplies fibrinogen at high concentration
- CIt supplies factor VIII and von Willebrand factor
- DIt provides oncotic volume expansion
- EIt supplies immunoglobulin that neutralizes the autoantibody
Answer
A
FFP is non-cellular plasma containing all clotting factors including fibrinogen, and its listed indications follow from that breadth: replacing all factors in DIC, liver disease, and dilutional coagulopathy from massive transfusion; Coumadin reversal, where it supplies factors 2, 7, 9, and 10; and TTP, where the relevant content is ADAMTS-13. In TTP the deficiency is of that specific metalloprotease, so the replacement fluid must restore it while the exchange simultaneously removes the autoantibody and the accumulated ultra-large vWF multimers. B and C name genuine constituents of plasma that are not the reason for its use here — and if fibrinogen alone were the goal, cryoprecipitate would be the more concentrated product. E misattributes the benefit to immunoglobulin content.
68A patient with heparin-induced thrombocytopenia and a new thrombosis has a platelet count of 34 ×10³/µL and is not bleeding.What is the appropriate approach to platelet transfusion?
- AAvoid transfusion, which may increase thrombotic risk
- BTransfuse only if the count falls below 10 ×10³/µL
- CTransfuse to maintain a count above 50 ×10³/µL
- DTransfuse concurrently with discontinuation of heparin
- ETransfuse before initiating a direct thrombin inhibitor
Answer
A
The reflex of low platelets, therefore give platelets is the trap; the correct first question is always why the count is low. In DIC, ITP, TTP, and HIT the low count reflects ongoing consumption or immune destruction rather than failed production, so transfused platelets are destroyed just as quickly. In HIT it is worse than futile: the underlying state is actively prothrombotic, driven by platelet activation and release of procoagulant microparticles, so adding platelets supplies further substrate for thrombosis. C, B, D, and E all propose thresholds or sequencing that would be reasonable in a production failure such as chemotherapy-induced thrombocytopenia, where the standard stable trigger is around 10,000.
69A 78-year-old woman with a history of heart failure receives two units of red cells and three units of plasma over four hours while also receiving maintenance intravenous fluids. She develops dyspnea, jugular venous distension, and bilateral crackles, and improves substantially after furosemide.Which is the most likely diagnosis?
- ATransfusion-related acute lung injury
- BFebrile nonhemolytic transfusion reaction
- CAnaphylactic transfusion reaction
- DTransfusion-associated circulatory overload
- EAcute hemolytic transfusion reaction
Answer
D
Both TRALI and TACO present as respiratory distress during or shortly after transfusion, and three features here point to volume rather than lung injury: a large cumulative volume on top of maintenance fluids, a patient with limited cardiac reserve, and findings of cardiogenic congestion that respond to diuretics. TRALI is non-cardiogenic lung injury from donor antibodies, does not respond to diuretics, and requires aggressive respiratory support. C would produce urticaria and bronchospasm without fever and typically without volume findings. B is fever and rigors only and is not life-threatening. E would present with fever, hypotension, back pain, and hemoglobinuria within minutes.
70A previously healthy 40-year-old man develops acute dyspnea, hypoxemia, and fever one hour into a plasma transfusion and requires intubation. Chest imaging shows bilateral infiltrates. Central venous pressure is normal and he does not improve with diuresis.Which mechanism is responsible?
- ADonor HLA or granulocyte antibodies activating recipient neutrophils
- BRecipient IgE antibodies directed against a donor plasma protein
- CVolume overload in a patient with reduced cardiac reserve
- DCytokines within the stored unit acting on the hypothalamic thermoregulatory center
- EBacterial endotoxin present in the transfused product
Answer
A
TRALI is lung injury rather than fluid overload, which the normal filling pressure and the failure to respond to diuresis establish, eliminating C. The majority of cases are associated with granulocyte or HLA antibodies in the donor that react with the patient's white cells; the resulting neutrophil activation injures the pulmonary capillary bed. Treatment is aggressive respiratory support in the ICU. It was formerly the most commonly reported cause of transfusion-related death. B describes an allergic reaction, which produces urticaria and wheezing without infiltrates. D is the febrile nonhemolytic mechanism, and E would give profound hypotension with high fever.
71Blood centers now restrict plasma donation to men, women who have never been pregnant, and women who have tested negative for HLA antibodies.What is the rationale for this policy?
- APlasma from female donors contains higher concentrations of inflammatory cytokines
- BPregnancy is the most common route to HLA and granulocyte alloimmunization
- CMale donors provide larger plasma volumes per collection
- DPreviously pregnant donors have a higher rate of bacterial contamination
- EFemale donors more frequently carry anti-IgA antibodies
Answer
B
The antibodies that cause TRALI reside in the donor plasma, not in the patient, and a person acquires HLA and granulocyte antibodies only through alloimmunization — the commonest route being pregnancy, where the mother is exposed to paternal HLA antigens on fetal cells. Restricting plasma donation removes the population most likely to be carrying those antibodies, which is why an epidemiologic intervention solved an immunologic problem and why TRALI risk fell substantially. A, C, and D propose differences between donor groups that are not the basis of the policy. E names a real antibody, but anti-IgA causes anaphylaxis in IgA-deficient recipients and is unrelated to donor sex or pregnancy history.
Lecture 07 · Randall
White Blood Cells & Reactive Changes
Questions 72–83
72A patient has a total white cell count of 2.2 ×10³/µL. The differential shows 78% lymphocytes, 15% neutrophils, and 7% monocytes.Which best describes this patient's status?
- ALymphocytosis
- BPancytopenia
- CNeutropenia with a normal absolute lymphocyte count
- DLeukocytosis with lymphocyte predominance
- EA normal differential in a patient with hemodilution
Answer
C
Always work from the absolute count rather than the percentage. Here the absolute neutrophil count is 2.2 × 0.15 = 0.33, which is well below the 1.4–6.5 reference range and in fact below the 0.5 threshold at which opportunistic infection risk climbs steeply. The absolute lymphocyte count is 2.2 × 0.78 = 1.7, comfortably within the normal 1.2–3.4 range. So the striking lymphocyte percentage is a disguised neutropenia, not a lymphocytosis — relative counts shift automatically whenever any single line moves. A and D misread the percentage as a real increase. B requires anemia and thrombocytopenia as well, neither of which is given.
73A 34-year-old man who has been taking clozapine for six weeks presents with fever and pharyngitis. His absolute neutrophil count is 0.2 ×10³/µL. Hemoglobin and platelet count are normal.Which mechanism is most likely responsible?
- ADose-dependent suppression of marrow progenitor cells
- BIncreased peripheral consumption from an occult infection
- CSequestration of neutrophils within an enlarged spleen
- DReplacement of marrow space by an infiltrative process
- EIdiosyncratic immune-mediated destruction of neutrophils
Answer
E
Drug toxicity is the single most common cause of neutropenia and it comes in two forms. Dose-dependent toxicity is predictable, occurs with chemotherapeutic agents, and is planned around because the nadir is expected. Idiosyncratic toxicity is by definition unpredictable and is characteristic of certain antibiotics and psychiatric medications — presenting exactly as here, in a previously well patient on a stable dose who becomes abruptly and profoundly neutropenic. The management difference follows: support through the first, stop the drug in the second. A is the intended contrast and would not fit an antipsychotic at steady dose. D would ordinarily produce other cytopenias, which the normal hemoglobin and platelet count exclude.
74A 46-year-old woman with Cushing syndrome has a white cell count of 14.8 ×10³/µL with an absolute neutrophil count of 12.1 and an absolute lymphocyte count of 0.6 ×10³/µL. She has no fever and no localizing symptoms.Which best explains this pattern?
- AOccult bacterial infection with a stress-related lymphopenia
- BIncreased marrow production of neutrophils with immune destruction of lymphocytes
- CCytokine-mediated suppression of lymphopoiesis alone
- DMarrow infiltration that spares the granulocytic line
- EDemargination of neutrophils together with redistribution and apoptosis of lymphocytes
Answer
E
Cortisol appears on both the leukocytosis and the leukopenia lists, and it moves the two lines in opposite directions simultaneously. Neutrophils that were adherent to vessel walls — the marginating pool — detach into the circulating pool where they are counted, and the marrow storage pool is released; no new neutrophils are made. Lymphocytes are simultaneously driven out of the blood into lymphoid tissue and undergo apoptosis. A patient on steroids or with endogenous hypercortisolism who has a high white count and a low lymphocyte count is therefore not necessarily infected, which is what makes A the trap. B misdescribes the neutrophil arm as increased production. D and C account for only one of the two changes.
75A 22-year-old man has two weeks of fever, hepatosplenomegaly, and pancytopenia. Ferritin is 21,400 ng/mL, soluble IL-2 receptor is markedly elevated, and EBV PCR is positive. PT and aPTT are prolonged, fibrinogen is 92 mg/dL, and D-dimer is elevated.Which is the best unifying diagnosis?
- ADisseminated intravascular coagulation from occult sepsis
- BAcute leukemia with marrow replacement
- CHemophagocytic lymphohistiocytosis with secondary DIC
- DInfectious mononucleosis with hepatic involvement
- EAplastic anemia following viral infection
Answer
C
Fever, hepatosplenomegaly, cytopenias, a ferritin in the thousands, an elevated soluble IL-2 receptor, and an EBV trigger constitute hemophagocytic lymphohistiocytosis — other supporting criteria include decreased NK cell activity and hemophagocytosis. The coagulation abnormalities are the point of the question: DIC is listed as a complication of HLH, not as an alternative diagnosis, alongside hepatitis, organ failure, and death. Stopping at A means treating a consequence and missing the disease that requires immunosuppression and chemotherapy. D is insufficient because ordinary mononucleosis does not produce pancytopenia with a ferritin of this magnitude. B and E would not explain the ferritin or the soluble IL-2 receptor.
76A patient with hemophagocytic lymphohistiocytosis has pancytopenia despite a florid systemic inflammatory state.Which best explains the cytopenias?
- ACytokines suppress hematopoiesis while macrophages ingest blood cells
- BMarrow space is replaced by a histiocytic neoplasm
- CMassive splenomegaly sequesters all three cell lines
- DAutoantibodies are formed against each hematopoietic lineage
- EComplement-mediated lysis destroys circulating cells of every lineage present
Answer
A
This looks backwards at first, since a hyperinflammatory state might be expected to raise counts. Two mechanisms drive them down at once. The cytokine flood — particularly IFNγ and TNF, along with IL-6 and IL-12 released from activated macrophages and cytotoxic T cells — directly suppresses hematopoiesis at the marrow. Simultaneously, those activated macrophages are literally ingesting blood cells, the finding the disease is named for. Production falls while destruction rises, so every line drops. The same activated macrophages release ferritin as an acute phase reactant, which is why a very high ferritin is such a useful screening lab in a septic-appearing patient with falling counts. D, B, C, and E each propose a single mechanism that does not fit the syndrome.
77A 3-year-old with a two-week history of paroxysmal coughing has a white cell count of 32 ×10³/µL with 78% lymphocytes.Which mechanism accounts for the lymphocytosis?
- AReactive expansion of cytotoxic T cells against infected epithelium
- BA toxin that blocks lymphocyte egress from the circulation
- CIncreased lymphoid production driven by bacterial antigen
- DDemargination of lymphocytes from the vessel wall
- EClonal proliferation of a lymphoid population
Answer
B
Bacteria classically cause neutrophilia, so Bordetella pertussis breaking that rule is exactly why it is examinable. It produces lymphocytosis-promoting factor, or pertussis toxin, which prevents lymphocytes from leaving the circulation. The mechanism deserves care: this is blocked egress, not increased production. Lymphocytes that would normally home into lymphoid tissue are stuck in the blood, so the peripheral count climbs without any true expansion of the lymphoid mass — which is what makes C the intended wrong answer. A describes the mechanism of infectious mononucleosis. D is the mechanism by which cortisol raises the neutrophil count. E would indicate a lymphoproliferative neoplasm rather than an infection.
78A 19-year-old college student has fever, exudative pharyngitis, posterior cervical lymphadenopathy, and splenomegaly. The smear shows numerous large lymphocytes with abundant cytoplasm, and a heterophile antibody test is positive.What is the identity of the atypical lymphocytes?
- AEpstein-Barr virus-infected B lymphocytes
- BPlasma cells responding to viral antigen
- CCirculating lymphoblasts
- DMonocytes with reactive morphologic changes
- EReactive CD8-positive cytotoxic T lymphocytes
Answer
E
This is the most commonly inverted fact about infectious mononucleosis. EBV infects B cells — along with oropharyngeal epithelial cells, producing the pharyngitis, and hepatocytes, producing the hepatitis — but the cells filling the smear are reactive CD8-positive cytotoxic T cells mounting a response against those infected B cells, which makes A the trap. The same logic explains the organ findings: the splenomegaly is hypertrophy of the periarteriolar lymphoid sheath, the T-cell zone of the spleen. The entire clinical picture is the host T-cell response rather than the virus itself, which is also why it takes weeks to settle and why splenic rupture is a risk. C would indicate acute leukemia.
79A 20-year-old with four days of fever, sore throat, and cervical lymphadenopathy has a negative monospot test. The clinical suspicion for infectious mononucleosis remains high.Which is the most appropriate next step?
- ARepeat the monospot test again in six months
- BBone marrow aspiration and biopsy
- CSerology for EBV viral capsid antigen antibodies
- DExcisional biopsy of an enlarged cervical lymph node
- EFlow cytometry on a peripheral blood specimen
Answer
C
A negative monospot does not exclude mononucleosis, and there are two separate reasons that are tested as distinct stems. The illness may be caused by CMV rather than EBV, in which case heterophile antibodies are never produced. Or the patient may be in the window period — the first few days after infection, before heterophile antibodies appear — which fits this patient at day four. Either way the next step is EBV-specific serology for viral capsid antigen antibodies. A delays diagnosis pointlessly. B, E, and D are invasive investigations appropriate to suspected malignancy, and nothing here suggests that; note also that painful, tender nodes in an acute febrile illness favor an infectious process.
80A 54-year-old man with a perforated appendix has a white cell count of 68 ×10³/µL with numerous myelocytes and metamyelocytes.Which additional finding would most support chronic myeloid leukemia rather than a reactive process?
- AAccompanying monocytosis
- BProminent toxic granulation
- CDöhle bodies within neutrophils
- DAbsence of circulating nucleated red blood cells
- EBasophilia and eosinophilia
Answer
E
A leukemoid reaction is leukocytosis with circulating immature granulocytes that mimics CML, which is what the name records, and it is usually driven by severe bacterial infection. Three findings distinguish it: a leukemoid reaction lacks eosinophilia, lacks basophilia, and lacks nucleated red cells. Of these, basophilia is the most useful positive marker of CML, since basophilia is rare and points to a myeloproliferative process whenever it appears. B and C are toxic changes, which are often present in a reactive process and therefore argue the opposite way. D also favors a reactive picture. A is a common accompaniment of reactive leukocytosis and does not distinguish the two.
81A 61-year-old woman with metastatic breast carcinoma has a white cell count of 17 ×10³/µL with a left shift, circulating nucleated red blood cells, and teardrop-shaped red cells. No toxic granulation is seen.Which best describes this blood picture?
- AA leukemoid reaction to an occult infection
- BA leukoerythroblastic picture indicating a marrow process
- CChronic myeloid leukemia in chronic phase
- DAcute myeloid leukemia with monocytic differentiation
- EReticulocytosis from a compensated hemolytic anemia
Answer
B
A left shift alone says only that the marrow is being driven hard. Nucleated red blood cells in the peripheral blood mean something more specific: erythroid precursors must normally enucleate before they can exit, so their appearance indicates the marrow's architectural barrier has been breached — by tumor, fibrosis, or granuloma occupying marrow space. The absence of toxic changes points away from infection. So left shift plus nucleated RBCs without toxic changes means look at the marrow, whereas left shift plus toxic changes without nucleated RBCs means look for infection, which is A. This is the same myelophthisic process that produces myelophthisic anemia and extramedullary hematopoiesis, viewed from the smear.
82A lymph node biopsy from a patient with generalized lymphadenopathy shows enlarged follicles containing germinal centers that are polarized into dark and light zones, contain numerous tingible body macrophages, and are surrounded by intact mantle zones.Which is the correct interpretation?
- AParacortical hyperplasia
- BFollicular lymphoma
- CFollicular hyperplasia, a reactive pattern
- DSinus histiocytosis
- EMetastatic carcinoma within the node
Answer
C
All three named features indicate a reactive germinal center, and each is the direct opposite of what follicular lymphoma shows — which is precisely why these are the features to know. Polarization into dark and light zones reflects an organized selection program that a neoplastic follicle has lost. Tingible body macrophages are clearing apoptotic B cells that failed selection, so their presence proves apoptosis is still occurring, whereas their absence in lymphoma reflects the anti-apoptotic BCL2 block. An intact mantle zone shows the follicle still respects its boundaries. Read together they say the follicle is following the rules. Follicular hyperplasia occupies the B-cell zone and is driven by autoimmune disease, infection, or vaccination.
83A 63-year-old man has a firm, non-tender 4 cm supraclavicular lymph node that has enlarged gradually over three months. He reports a 6 kg weight loss.Which feature of this presentation is most concerning?
- AThe supraclavicular location alone
- BThe absence of overlying erythema
- CThe size exceeding 2 cm
- DThe absence of tenderness
- EThe absence of fluctuance
Answer
D
Pain is the highest-yield single discriminator, and it runs opposite to intuition about seriousness. Tender nodes are usually infectious and usually benign, because tenderness arises from rapid capsular stretch during acute inflammation. A painless node is the one that raises concern for malignancy, since a slowly infiltrating tumor expands the node too gradually to hurt — and the three-month course here reinforces that. C, B, and E are all consistent with the picture but are secondary: size and the absence of erythema or fluctuance simply reflect that this is not an acute suppurative process. A contributes real concern but is less discriminating than the character of the node itself.
Lecture 08 · Randall
Acute Leukemias
Questions 84–96
84A 63-year-old man has a white cell count of 8.2 ×10³/µL. Bone marrow examination shows 12% blasts with a full spectrum of maturing granulocytic forms.Which conclusion about classification is correct?
- AThis meets criteria for acute leukemia because blasts are identifiable
- BThe normal white cell count argues against any leukemic process
- CAcute and chronic are distinguished by the rapidity of symptom onset
- DThe blast percentage falls below the threshold, indicating a chronic process
- EClassification cannot be attempted without flow cytometry
Answer
D
Acute leukemia is defined by ≥20% blasts in the blood or bone marrow. At 12%, with a spectrum of maturing granulocytic forms, this describes a chronic myeloid process — chronic leukemias are composed of more mature hematopoietic cells, with CML specifically carrying <20% blasts. A mistakes the mere presence of blasts for the threshold; a small percentage is normal in marrow. B is wrong because the total count does not define the category — acute leukemia may present with leukocytosis, a normal count, or pancytopenia. C is the most instructive error: despite the names, acute and chronic are defined by the maturity of the malignant cells, not the tempo of the illness. E overstates the case, since the blast percentage alone settles this question.
85A 5-year-old girl is diagnosed with B-lymphoblastic leukemia.Which cytogenetic finding would predict the most favorable outcome?
- AHypodiploidy with 42 chromosomes
- Bt(9;22); BCR::ABL1
- CHyperdiploidy with 56 chromosomes
- Dt(9;11); KMT2A::MLLT3
- EA complex karyotype including del(7q)
Answer
C
Hyperdiploidy is defined as more than 50 chromosomes and is a good prognostic indicator in B-ALL, alongside t(12;21); ETV6::RUNX1 and age above 1 and below 10 — so this patient carries two favorable features. The pairing is easier to hold as a single idea: gaining chromosomes is good, losing them is bad. Hyperdiploid blasts behave as a less aggressive, more chemosensitive clone, while hypodiploid blasts have lost genetic material, and what tends to be lost includes tumor suppressor genes — which makes A unfavorable. B is the Philadelphia chromosome, poor in B-ALL. D is the one poor-risk lesion among the AML defining abnormalities. E describes the pattern of AML-MR and therapy-related disease.
86A 9-year-old boy is diagnosed with B-lymphoblastic leukemia. His white cell count at presentation is 142 ×10³/µL, and cytogenetics show t(9;22). Flow cytometry demonstrates CD10, CD19, TdT, and CD34.Which feature of this presentation is associated with a favorable outcome?
- AThe white cell count
- BThe patient's age
- CThe translocation
- DThe expression of CD34
- EThe expression of TdT
Answer
B
His age of 9 falls within the favorable window of above 1 and below 10 — note that the lower bound exists because infants under 1 do poorly. Both other prognostic features here are unfavorable: a white count above 100K at presentation is a recognized poor sign rather than an incidental marker of disease burden, and t(9;22); BCR::ABL1 is poor in B-ALL. That translocation is worth holding as a two-sided fact, since in CML the same Philadelphia chromosome is the defining and therapeutically targetable lesion. D and E name diagnostic markers with no prognostic meaning: TdT indicates a lymphoblast and CD34 indicates a stem or progenitor cell, and neither is favorable or unfavorable.
87A 15-year-old boy presents with dyspnea and superior vena cava syndrome. Imaging shows a large anterior mediastinal mass. Biopsy demonstrates sheets of blasts expressing CD3, CD7, TdT, and CD34.Which statement about this disease is correct?
- AIt represents roughly 85% of acute lymphoblastic leukemia
- BThe TdT expression establishes B-lineage derivation
- CIt carries a slightly better prognosis than B-ALL in children
- DThe CD34 expression indicates myeloid derivation
- EIt carries a slightly worse prognosis than B-ALL
Answer
E
An adolescent with a mediastinal or thymic mass is the classic T-ALL presentation — Teenagers with a Thymic mass — and its prognosis is slightly poorer than B-ALL, with roughly 70–80% cure. The reason for the mass is anatomic: T lymphoblasts mature in the thymus, so a T-cell clone expands in the anterior mediastinum and presents as a mass, whereas B lymphoblasts develop in the marrow and spill into blood. A describes B-ALL. B and D misread the markers: TdT is expressed by both B- and T-lymphoblasts and CD34 is positive in both ALL and AML, so neither establishes lineage — here CD3 and CD7 do. C inverts the prognostic comparison.
88Flow cytometry on a marrow aspirate demonstrates that the blast population expresses CD34.What does this finding establish?
- AThe blasts are of lymphoid lineage
- BThe blasts are of myeloid lineage
- CThe blasts are B-lineage specifically
- DThe blasts represent acute promyelocytic leukemia
- EThe blasts are immature, but lineage remains undetermined
Answer
E
CD34 is a stem cell antigen and is positive in both ALL and AML, so on its own it reports only that the cell is a stem or progenitor cell — it carries no lineage information at all. The markers that do assign lineage are CD10 and CD19 for B-lineage, the CD1a–CD8 series for T-lineage, and CD33, CD117, and MPO for myeloid. A companion fact worth pairing with this: TdT is expressed by both B- and T-lymphoblasts, so it establishes that a cell is a lymphoblast without telling you which kind. D is doubly wrong, since APL is characteristically CD34-negative despite being myeloid.
89A 58-year-old woman has 40% blasts in the marrow. Careful examination identifies a slender pink cytoplasmic inclusion within one blast.What does this finding establish?
- AThe leukemia is of myeloid lineage
- BThe leukemia carries a favorable cytogenetic abnormality
- CThe blasts express CD34
- DThe diagnosis is acute promyelocytic leukemia
- EFlow cytometry is no longer necessary for classification
Answer
A
The inclusion is an Auer rod, which is nothing more exotic than crystallized myeloperoxidase aggregating in the cytoplasm. Because only myeloid cells make MPO, an Auer rod is a visible confession of lineage — Auer rods are present only in myeloid blasts, so finding one eliminates the lymphoid possibilities without flow cytometry. It establishes nothing further. B is wrong because prognosis in AML derives from karyotype and molecular testing. C is wrong and is a useful reminder that APL is MPO-positive but CD34-negative. D overreaches, since APL requires t(15;17); PML::RARA. E is wrong because subclassification still requires flow, cytogenetics, and NGS.
90A 40-year-old man with newly diagnosed acute myeloid leukemia develops diffuse oozing from venipuncture sites. PT and aPTT are prolonged, fibrinogen is 74 mg/dL, and D-dimer is markedly elevated. Cytogenetics show t(15;17).Which mechanism links this leukemia to the coagulopathy?
- ABlasts consume platelets directly within marrow sinusoids
- BThe PML::RARA fusion protein cleaves fibrinogen
- CCytoplasmic granules within the abnormal promyelocytes are rich in procoagulants
- DThrombocytopenia from marrow replacement fully accounts for the picture
- EAn autoantibody against factor VIII has developed as a paraneoplastic phenomenon
Answer
C
This is acute promyelocytic leukemia, defined by t(15;17); PML::RARA. The fusion produces a retinoic acid receptor that no longer responds to physiologic ligand, so the clone arrests at the promyelocyte stage — and the promyelocyte is precisely the stage packed with primary granules. Those granules contain procoagulants, so when the cells lyse they release that material into the circulation and trigger DIC, which is what the prolonged times, low fibrinogen, and elevated D-dimer describe. D is insufficient because isolated thrombocytopenia does not prolong PT and aPTT or consume fibrinogen. A, B, and E propose mechanisms that are not features of this disease.
91A patient with acute promyelocytic leukemia is treated with an agent that targets the underlying molecular lesion.Which best describes how that agent works?
- AIt inhibits the fusion tyrosine kinase produced by the translocation
- BIt supplies ligand that overcomes the receptor defect
- CIt inhibits terminal complement, protecting the cells from lysis
- DIt depletes CD20-positive cells that sustain the clone
- EIt directly inhibits thrombin generated by the granule contents
Answer
B
RARA is the retinoic acid receptor alpha, and it normally drives promyelocytes to differentiate into mature granulocytes. Fusion to PML yields a receptor that no longer responds to physiologic retinoic acid, so the cell arrests. Because the problem is a receptor-sensitivity defect rather than a missing receptor, giving pharmacologic doses of all-trans retinoic acid overwhelms it, forces differentiation, and the cells mature and die on schedule. This is differentiation therapy — one of the few malignancies treated by making the cells grow up rather than by killing them. A describes imatinib in CML, D rituximab, C eculizumab, and E a direct thrombin inhibitor.
92Flow cytometry on a marrow aspirate shows a blast population that is positive for MPO, CD33, and CD117 but negative for CD34.Which interpretation is most appropriate?
- AThe CD34 negativity excludes acute myeloid leukemia
- BThis pattern is expected in acute promyelocytic leukemia
- CThis pattern favors acute megakaryoblastic leukemia
- DThe MPO result is likely artifactual given the absent CD34
- EThis phenotype is characteristic of B-lymphoblastic leukemia
Answer
B
APL blasts are CD34 negative while remaining positive for MPO, CD33, and CD117 — the exception to the usual AML immunophenotype, and the logic is worth holding rather than memorizing. CD34 is a stem cell antigen, and APL is arrested at the promyelocyte stage, which lies further along the maturation sequence than a stem cell, so the antigen has already been lost. The phenotype is telling you where the block sits. A is wrong for exactly that reason. E is wrong because B-ALL blasts express CD10, CD19, TdT, and CD34 and are MPO-negative. C is wrong because megakaryoblastic leukemia is MPO-negative, and MPO is present here.
93A 55-year-old woman has a white cell count of 180 ×10³/µL and markedly swollen, infiltrated gingivae. Marrow blasts are MPO negative and express monocytic markers.Which conclusion follows?
- AThe MPO negativity establishes lymphoid lineage
- BThe gingival findings indicate a leukemoid reaction to periodontal infection
- CThis is acute monocytic leukemia, a myeloid neoplasm classified by differentiation
- DAuer rods should be identifiable given the very high white cell count
- EThis presentation is most characteristic of Down syndrome
Answer
C
Acute monocytic or monoblastic leukemia presents with a high white count and gingival involvement, and its monoblasts are MPO-negative. It belongs to the AML defined by differentiation group — the wastebasket for AMLs not otherwise defined by clinical history or cytogenetics, comprising eight subtypes analogous to the historic FAB M0–M7 system. A is the central trap: MPO-negative does not mean lymphoid, because MPO is made by the granulocytic arm and monocytes branch away from it while remaining firmly myeloid. B ignores the blast population. D is impossible, since Auer rods are aggregates of MPO and cannot form in an MPO-negative blast. E describes megakaryoblastic leukemia.
94A 3-year-old child with Down syndrome develops an acute leukemia. The blasts are MPO negative, express megakaryocytic markers, and lack lymphoid antigens.Which is the most likely diagnosis?
- AB-lymphoblastic leukemia
- BAcute monocytic leukemia
- CAcute promyelocytic leukemia
- DAcute megakaryoblastic leukemia
- EAcute myeloid leukemia with t(8;21)
Answer
D
Acute megakaryoblastic leukemia features MPO-negative megakaryoblasts, belongs to the AML defined by differentiation group, and is associated with Down syndrome in early childhood — the megakaryocytic markers settle it. A is the strongest distractor on epidemiology alone, since ALL is the most common childhood cancer and children with Down syndrome are at increased risk of leukemia generally — but B-ALL blasts express CD10, CD19, TdT, and CD34, and the stem states lymphoid antigens are absent. B is the other MPO-negative subtype, but it is identified by monocytic markers and gingival involvement. C is MPO-positive and defined by t(15;17).
95A 61-year-old man treated six years ago with an alkylating agent for lymphoma now presents with acute myeloid leukemia. The karyotype is complex and includes del(7q).Which statement is most accurate?
- AThis represents transformation of the original lymphoma rather than a new neoplasm
- BTherapy-related disease is characteristically associated with an isolated balanced translocation
- CTopoisomerase inhibitors are the only drug class implicated in this complication
- DThis is secondary AML, which most often falls into the AML-MR category and carries a very poor prognosis
- EThe presence of a defining cytogenetic abnormality places this in the favorable risk group
Answer
D
Secondary, therapy-related AML follows cytotoxic chemotherapy for a solid or hematopoietic tumor, most often falls into the AML-MR category, shows a complex karyotype, and carries a very poor prognosis. E misreads defining abnormality as favorable, but a complex karyotype with del(7q) is an MDS-defining, poor-risk pattern — quite unlike the favorable single lesions t(8;21), inv(16), t(15;17), and mutated NPM1. The underlying principle is that both AML-MR and therapy-related AML arise from a marrow that was already genomically damaged: one driver lesion is treatable, a shattered genome is not. B inverts the genetics, and C is too narrow, since alkylating agents are implicated as well. A is wrong because this is a new myeloid neoplasm arising from mutagenized stem cells.
96A 68-year-old man with acute myeloid leukemia has a white cell count of 84 ×10³/µL, of which 90% are blasts. His absolute neutrophil count is 0.4 ×10³/µL. He presents with fever and hypotension.Which best explains his susceptibility to infection?
- ABlasts consume complement components, impairing opsonization of bacteria
- BThe mature neutrophil count is low and blasts are non-functional
- CLeukostasis impairs perfusion of lymphoid organs
- DThe blasts secrete immunosuppressive cytokines
- EChemotherapy has not yet been administered
Answer
B
A markedly elevated white cell count can be reassuring in a way that is entirely misleading. The count is high because blasts are spilling into the blood, not because mature neutrophils are being produced — and blasts cannot perform chemotaxis, phagocytosis, or killing. The absolute neutrophil count of 0.4 confirms it, sitting below the 0.5 threshold at which opportunistic infection risk climbs steeply. The patient is functionally neutropenic despite a white count of 84, which is why infection is a leading cause of death in acute leukemia. The practical lesson is to look at the differential rather than the total. A, C, D, and E propose mechanisms that are not the explanation for the infection risk here.
Lecture 09 · Randall
Myelodysplastic Neoplasms / Syndromes
Questions 97–104
97A 71-year-old man is found on a routine CBC to have a hemoglobin of 9.4 g/dL, a white cell count of 2.8 ×10³/µL, and a platelet count of 88 ×10³/µL.Which marrow finding would best support myelodysplastic neoplasm over aplastic anemia?
- AA markedly hypocellular marrow with normal-appearing residual cells and no dysplasia
- BA hypercellular marrow with dysplastic changes in multiple lineages
- CA marrow replaced by fibrous tissue and collagen
- DA marrow infiltrated by metastatic carcinoma
- EA normocellular marrow with increased iron stores
Answer
B
Both diseases present with cytopenias, so the blood count cannot separate them — the marrow points in opposite directions. MDS is a disorder of ineffective hematopoiesis: cells are produced in abundance but carry maturation defects and die before reaching the circulation, so the marrow is hypercellular with dysplasia while the blood is empty. A is the aplastic anemia pattern, where the marrow is genuinely empty and the remaining cells look normal. C and D describe myelophthisic processes, which would also produce a leukoerythroblastic smear. E fits no cause of pancytopenia in this material.
98A marrow aspirate from a patient with suspected MDS shows dysplastic erythroid precursors.What proportion of cells in a lineage must be dysplastic for that lineage to count as significant?
- AMore than 2%
- BMore than 5%
- CMore than 10%
- DMore than 20%
- EMore than 50%
Answer
C
Dysplasia requires greater than 10% dysplastic cells in a respective lineage to be significant. The threshold exists because occasional dysplastic-looking cells are found in normal marrow and in reactive states such as nutritional deficiency, so a cutoff is needed to separate incidental atypia from a clonal disorder. Do not confuse this figure with the blast thresholds, which are entirely separate: ≥5% blasts in marrow or ≥2% in blood counts as increased blasts and worsens the prognostic score, and ≥20% means the disease is now AML.
99A peripheral smear from a 68-year-old with pancytopenia shows neutrophils with bilobed, spectacle-shaped nuclei and markedly reduced cytoplasmic granules.How are these findings best described?
- AToxic granulation with Döhle bodies
- BHypersegmentation from megaloblastic maturation
- CPseudo-Pelger-Huët change with hypogranulation
- DAuer rods within maturing granulocytes
- ELeukoerythroblastic change from marrow infiltration
Answer
C
These are the two classic features of granulocytic dysplasia: hypolobation, described as pseudo-Pelger-Huët or “pelgeroid” change and sometimes producing unilobate forms, together with hypogranulation. Both record a cell that began a maturation program and failed to complete it — a neutrophil should segment its nucleus and fill with granules, and this one did neither. A describes reactive changes seen in infection and would argue for a leukemoid reaction. B is a feature of megaloblastic anemia, though hypersegmentation is also a less common dysplastic change. D indicates myeloid blasts in acute leukemia.
100A patient with MDS has 7% blasts in the bone marrow and 3% blasts in the peripheral blood.How should these findings be interpreted?
- ABlasts are increased by both criteria, worsening the prognostic score
- BBlasts are increased in the marrow only
- CBlasts are increased in the blood only
- DBlasts are not increased by either criterion
- EThe disease now meets the diagnostic criteria for acute myeloid leukemia rather than MDS
Answer
A
Increased blasts means ≥5% in the marrow and/or ≥2% in the peripheral blood, and this patient exceeds both. That matters because the prognostic score in MDS is assigned from cytogenetic abnormalities, blast percentage, and degree of cytopenia, so increased blasts worsen the outlook and raise the risk of progression. E is wrong because ≥20% in either compartment is required to call AML. Note that the marrow threshold is higher than the blood threshold, because a small blast population is normal in marrow and abnormal in blood.
101Cytogenetic analysis is performed on a newly diagnosed case of myelodysplastic neoplasm.Which abnormality is among the most commonly identified?
- At(9;22)(q34.1;q11.2); BCR::ABL1
- BMonosomy 7 or deletion of 7q
- CJAK2 V617F
- Dt(15;17)(q24;q21); PML::RARA
- Et(8;14); IGH::MYC
Answer
B
The recurrent cytogenetic abnormalities in MDS are −5/del(5q), −7/del(7q), and complex karyotype defined as ≥3 aberrations. Note the pattern: these are losses of genetic material and chaotic karyotypes rather than clean balanced translocations, which is why MDS carries the prognostic profile it does and why there is no single driver mutation present across the board. A is the defining lesion of CML, B of acute promyelocytic leukemia, C of the myeloproliferative neoplasms, and E of Burkitt lymphoma — all specific lesions defining specific diseases, in contrast to the pattern here.
102Two patients are diagnosed with MDS on the same day. One has de novo disease; the other developed MDS after cytotoxic chemotherapy for breast cancer.Which statement about the therapy-related case is correct?
- AIt carries a substantially lower risk of progression to acute myeloid leukemia
- BIt is not associated with cytogenetic abnormalities
- CIt has the most aggressive course and the highest rate of progression to AML
- DIt responds better to hypomethylating agents
- EIt should be reclassified as acute myeloid leukemia
Answer
C
MDS after cytotoxic therapy has the highest risk of progression to AML and the most rapid clinical course. The reasoning parallels therapy-related AML from Lecture 08: the marrow stem cell pool was mutagenized before the disease began, so the clone that emerges arises from a background of widespread genomic damage rather than a single clean lesion — which is what a complex karyotype reports. A inverts the relationship. B is wrong because these cases characteristically show complex cytogenetics. E is wrong because classification still requires ≥20% blasts to call AML.
103A 52-year-old man with high-risk MDS is otherwise fit and has an HLA-matched sibling.Which treatment offers the possibility of cure?
- AHypomethylating agent therapy
- BA BCL2 inhibitor
- CAllogeneic stem cell transplantation
- DErythropoietin with G-CSF support
- EThalidomide-like therapy
Answer
C
In younger patients with high-risk disease, allogeneic stem cell transplant is potentially curative — it is the only option listed that replaces the abnormal clone rather than managing its consequences. The remaining choices are all genuine MDS therapies, which is why they are listed, but they belong to the other arm of the algorithm: in older patients or those who cannot undergo transplant, treatment is aimed at improving cytopenias and preventing complications using hypomethylating agents, BCL2 inhibitors, thalidomide-like therapies, and growth factor support. That distinction — curative intent versus supportive intent — is driven by patient fitness as much as by disease risk.
104A hematologist is distinguishing among the chronic myeloid neoplasms.Which feature best separates a myelodysplastic neoplasm from a myeloproliferative neoplasm?
- AMarrow cellularity
- BThe peripheral blood count
- CThe presence of organomegaly
- DPatient age at presentation
- EThe percentage of blasts in the marrow
Answer
B
The three categories of chronic myeloid neoplasm are distinguished first by the peripheral blood count, then further defined by marrow morphology and genetics. MPN produces cytoses; MDS produces cytopenias. A is the trap and the reason this question exists: both have a hypercellular marrow, so cellularity is precisely the feature that fails to separate them. The explanation is that in MPN the cells successfully reach the blood, whereas in MDS they carry maturation defects and die before release. C and D overlap substantially between the two, and E is not increased in either by definition, since ≥20% blasts would make the disease acute.
Lecture 10 · PALM 820
Myeloproliferative Neoplasms
Questions 105–116
105A 54-year-old man has a white cell count of 78 ×10³/µL with myelocytes and metamyelocytes on the smear. There is no toxic granulation.Which additional finding would most support chronic myeloid leukemia?
- AProminent Döhle bodies within neutrophils
- BAn accompanying monocytosis above 10%
- CAbsolute basophilia with eosinophilia
- DAbsence of circulating nucleated red cells
- EA platelet count below the reference range
Answer
C
Absolute basophilia and eosinophilia are characteristic of CML chronic phase, and basophilia is the most useful single positive marker because basophils are rare in reactive states. A leukemoid reaction, by contrast, lacks basophilia, lacks eosinophilia, and lacks nucleated red cells, while showing toxic granulation and Döhle bodies — so A argues the opposite way. B misstates the CML pattern, which includes an absolute monocytosis with a relative percentage under 3%. D is characteristic of a reactive process. E is wrong because platelets in CML are normal to increased.
106A 58-year-old woman is being evaluated for a suspected myeloproliferative neoplasm.Which test should be performed first?
- ABCR::ABL1 testing by FISH or PCR
- BCALR sequencing
- CMPL sequencing
- DJAK2 V617F analysis
- EBone marrow biopsy with reticulin staining
Answer
A
The diagnostic approach runs clinical suspicion → BCR::ABL1 testing → JAK2, then CALR and MPL → hematology/oncology referral for marrow biopsy. BCR::ABL1 comes first because it is present in 100% of CML and its presence or absence partitions the entire differential in a single step; every other MPN requires that CML be excluded. D, B, and C are the correct second step, and notably JAK2, CALR and MPL are shared between ET and PMF, so they cannot separate those two. E is essential for primary myelofibrosis specifically, but it follows rather than precedes the molecular testing.
107A 62-year-old man has a hemoglobin of 19.2 g/dL and a hematocrit of 57%. He reports headaches and generalized itching after hot showers.Which laboratory finding would best support polycythemia vera rather than a secondary erythrocytosis?
- AA subnormal serum erythropoietin level
- BAn elevated serum erythropoietin level
- CAn elevated red cell distribution width
- DAn elevated serum ferritin level
- EA normal serum erythropoietin level
Answer
A
In secondary erythrocytosis — hypoxia, altitude, an erythropoietin-secreting tumor — the marrow is responding appropriately to a signal, so EPO is normal or high. In polycythemia vera the JAK2-mutated progenitor no longer needs the signal; it makes red cells autonomously, and the body responds by suppressing EPO. A subnormal EPO alongside a high hematocrit is physiologically impossible unless production has become independent of regulation, which is why it serves as the minor diagnostic criterion. B and E describe secondary causes. D is not part of the criteria. E would not distinguish the two conditions.
108A 60-year-old woman with polycythemia vera develops abdominal pain, ascites, and tender hepatomegaly. Imaging shows hepatic vein thrombosis.Which complication does this represent?
- APortal vein thrombosis
- BBudd-Chiari syndrome
- CMesenteric venous thrombosis
- DSplenic vein thrombosis
- EPost-polycythemic myelofibrosis
Answer
B
Budd-Chiari syndrome is hepatic vein thrombosis, and it is specifically named among the thrombotic complications of polycythemia vera. Thrombosis occurs in 20–25% of patients and may be arterial or venous, with a striking predilection for mesenteric, portal, and splenic veins in addition to the hepatic veins — which is why A, C, and D are all genuine PV complications and are listed here as plausible alternatives, distinguished only by the vessel involved. E is the spent phase, a late marrow event rather than a thrombotic one. Note that Budd-Chiari also occurs in essential thrombocythemia.
109A 55-year-old woman has a persistent platelet count of 780 ×10³/µL. Hemoglobin and white cell count are normal. The marrow shows increased large megakaryocytes with hyperlobated nuclei.Which is the most likely diagnosis?
- APolycythemia vera
- BEssential thrombocythemia
- CChronic myeloid leukemia
- DPrimary myelofibrosis, overt fibrotic stage
- EReactive thrombocytosis
Answer
B
Essential thrombocythemia is sustained thrombocytosis above 450 ×10&sup9;/L predominantly involving the megakaryocyte lineage, with increased large to giant megakaryocytes having hyperlobated, “staghorn-like” nuclei and no significant erythroid or granulocytic increase — which the normal hemoglobin and white count confirm. A and C would show expansion of other lineages. D is possible, since PMF can present with isolated thrombocytosis, but the overt fibrotic stage requires grade 2 or 3 reticulin fibrosis, which is not described. E remains a formal exclusion in the criteria but does not explain the abnormal megakaryocyte morphology.
110A 68-year-old man has anemia, a palpable spleen extending to the pelvis, and a smear showing nucleated red cells, immature granulocytes, and teardrop-shaped red cells.Which is the most likely diagnosis?
- APrimary myelofibrosis
- BPolycythemia vera, polycythemic phase
- CChronic myeloid leukemia, chronic phase
- DEssential thrombocythemia
- EMyelodysplastic neoplasm
Answer
A
The combination of massive splenomegaly, a leukoerythroblastic smear, and teardrop cells is primary myelofibrosis. The mechanism explains all three at once: collagen replacing marrow space squeezes precursors out prematurely, giving the leukoerythroblastic picture and deforming red cells into teardrops as they exit, while hematopoiesis relocates to the fetal sites, producing extramedullary hematopoiesis in spleen and liver and therefore enormous organomegaly. This is the same myelophthisic mechanism as marrow-infiltrating tumor from Lecture 01. D and B are other MPNs without fibrosis at presentation. E would not produce this degree of splenomegaly.
111A bone marrow biopsy in chronic myeloid leukemia is examined.Which megakaryocyte morphology is expected?
- ALarge, giant forms with hyperlobated staghorn nuclei
- BPleomorphic and hypersegmented, in loose clusters near trabeculae
- CMarkedly hypolobated with separate nuclei in pawn-ball arrangement
- DSmall, hyposegmented “dwarf” forms
- EClustered and hyperchromatic with bare nuclei
Answer
D
Megakaryocyte morphology is one of the few things that differs by disease in a group where cell morphology is otherwise normal, so it is worth holding as a set. CML gives smaller than normal, hyposegmented “dwarf” megakaryocytes. A describes essential thrombocythemia, B describes polycythemia vera, and E describes the atypical megakaryocytes of primary myelofibrosis, which also show clustering and marked size variation adjacent to trabeculae or sinuses. C is megakaryocytic dysplasia in MDS, a different disease category entirely.
112A patient with chronic myeloid leukemia on imatinib develops fatigue and bone pain. Marrow examination now shows 34% blasts.Which statement is correct?
- AThis remains chronic phase with adverse risk features
- BThis represents blast phase, usually a myeloblast crisis
- CThis represents blast phase, usually a lymphoblast crisis
- DThis indicates transformation to a myelodysplastic neoplasm
- EThis is within the expected range for treated chronic phase
Answer
B
Blast phase is diagnosed at ≥20% blasts in peripheral blood or bone marrow, and it is typically a myeloblast crisis, with a minority having a lymphoblast crisis — which makes C the near-miss. A is wrong because the adverse-risk band in chronic phase is 10–19% blasts, along with ≥20% basophils, new chromosome abnormalities, and TKI resistance; 34% is past that. E is wrong because chronic phase requires blasts under 5%. Note that the ≥20% threshold is identical to the definition of acute leukemia — blast phase is a chronic myeloid neoplasm becoming acute, and it carries a poor prognosis. Extramedullary blast proliferation may involve skin, lymph nodes, bone, and CNS.
113Molecular testing on a patient with a myeloproliferative neoplasm returns a JAK2 V617F mutation.What does this result establish?
- AAn MPN is supported, but the specific entity is not determined
- BThe diagnosis is essential thrombocythemia
- CThe diagnosis is primary myelofibrosis
- DChronic myeloid leukemia is confirmed
- EThe diagnosis is polycythemia vera rather than another myeloid neoplasm
Answer
A
JAK2 is shared across polycythemia vera, essential thrombocythemia, and primary myelofibrosis — present in over 95% of PV and in 50–60% of both ET and PMF, which also share CALR at 30% and MPL. The mutation confirms a clonal myeloproliferative process but cannot say which one, which is precisely why the WHO criteria for each disease include a clause requiring that the criteria for the others are not met, and why a marrow biopsy is essential for primary myelofibrosis. D is excluded because CML is defined by BCR::ABL1, present in 100% of cases and tested for first.
114A patient with a ten-year history of polycythemia vera no longer requires phlebotomy. Hemoglobin is now 9.8 g/dL and the spleen has enlarged further.Which best explains this change?
- AThe disease has entered the spent phase with marrow fibrosis
- BThe patient has developed iron deficiency from prior phlebotomy
- CThe disease has entered remission
- DA secondary acute leukemia has developed
- EHydroxyurea has produced an expected therapeutic response
Answer
A
The spent phase, or post-polycythemic myelofibrosis, is marked by cytopenias and ineffective hematopoiesis: the red cell mass normalizes and then decreases, producing the loss of the phlebotomy requirement, while progressive reticulin and collagenous fibrosis drive a leukoerythroblastic smear with teardrop cells and progressive organomegaly. C is the trap this question is built around — not needing phlebotomy sounds like improvement but signals a failing, fibrosing marrow. D is possible in PV, particularly after cytotoxic agents, but would require increased blasts, which are not described.
115A patient with a myeloproliferative neoplasm undergoes bone marrow examination.Which combination of findings is expected?
- AHypocellular marrow, maturation present, blasts not increased
- BHypercellular marrow, maturation blocked, blasts above 20%
- CHypercellular marrow, maturation present, blasts not increased
- DHypercellular marrow, maturation defective, prominent dysplasia
- ENormocellular marrow with fibrosis and no cellular expansion
Answer
C
The shared MPN profile is increased cellularity, maturation present, normal cell morphology, and blasts typically not increased. This follows directly from the molecular lesion: a constitutively active tyrosine kinase delivers a permanent growth signal but does not break the maturation program, so cells proliferate and still mature normally. B describes acute leukemia, where maturation is blocked at the blast stage. D describes MDS, where maturation is defective and cells die in the marrow. A and E fit no chronic myeloid neoplasm. The one morphologic exception within MPN is the megakaryocyte, whose appearance differs by disease.
116A 61-year-old man with polycythemia vera has an MCV of 74 fL and hypochromic red cells. Marrow iron staining shows absent stainable iron.Which best explains these findings?
- ACoexisting thalassemia trait
- BAnemia of chronic disease from the underlying neoplasm
- CSideroblastic change from clonal evolution
- DIron consumption by the expanded erythron and repeated phlebotomy
- EOccult gastrointestinal blood loss unrelated to the neoplasm
Answer
D
It looks contradictory to find microcytic hypochromic cells and absent marrow iron in a disease of too many red cells, and that tension is the point. The expanded erythron burns through iron stores making all those cells, and therapeutic phlebotomy removes more iron with every unit taken. The patient becomes iron deficient while remaining polycythemic — which is why the lecture explicitly lists both normocytic normochromic or microcytic hypochromic red cells and absent marrow iron as expected findings. B would show increased stores, since iron is trapped rather than depleted. A and E propose unrelated diagnoses, and C would show ring sideroblasts with iron present.
Lecture 11 · Randall
B-cell Lymphomas
Questions 117–130
117A 68-year-old man is found on a routine CBC to have a white cell count of 32 ×10³/µL with 82% small mature lymphocytes. Numerous smudge cells are present.Which statement about this condition is correct?
- AIt arises from a germinal center B cell
- BThe smudge cells represent a distinct neoplastic population
- CIt is the same disease as small lymphocytic lymphoma
- DIt is aggressive and requires immediate intensive chemotherapy
- EIt is curable with combination chemotherapy in most patients
Answer
C
CLL and SLL are the same disease, distinguished only by where it is found — blood and marrow only is CLL, which is the majority, and tissue only is SLL. Many patients have both. A is wrong because the cell of origin is a naïve B cell. B is a common misconception: smudge cells are an artifact, produced when fragile CLL lymphocytes rupture as the smear is spread — a diagnostic clue rather than a cell type. D and E both misstate the behavior, since CLL/SLL is indolent and incurable, and small B-cell lymphomas respond poorly to intensive chemotherapy because of their low proliferative rate.
118A 70-year-old woman with a five-year history of CLL/SLL develops rapidly enlarging cervical lymph nodes, fever, and weight loss over three weeks.Which development is most likely?
- ARichter syndrome with transformation to diffuse large B-cell lymphoma
- BAutoimmune hemolytic anemia
- CProlymphocytic progression
- DProgression to acute lymphoblastic leukemia
- EDevelopment of hypogammaglobulinemia with recurrent bacterial infection
Answer
A
Richter syndrome is transformation of CLL/SLL to diffuse large B-cell lymphoma, and it is characterized clinically by rapidly enlarging lymph nodes in a patient with known CLL/SLL — precisely the vignette. C is the other recognized transformation, occurring in 10–30%, but prolymphocytic progression is a change in the circulating population rather than an abrupt nodal mass. B and E are genuine complications of CLL/SLL and are listed for that reason, but neither produces rapidly enlarging nodes: autoimmune hemolytic anemia causes anemia and jaundice, and hypogammaglobulinemia causes infections. D is not a recognized transformation of CLL.
119A 57-year-old man has painless generalized lymphadenopathy. Node biopsy shows a nodular proliferation of small cleaved cells. FISH demonstrates t(14;18).Which gene is placed under IGH promoter control?
- ACCND1
- BMALT1
- CMYC
- DBCL2
- EBIRC3
Answer
D
t(14;18) moves BCL2 from chromosome 18 to the IGH locus on chromosome 14, causing overexpression of BCL2, an anti-apoptotic molecule that is normally NOT expressed in germinal centers. This is follicular lymphoma, the second most common lymphoma after DLBCL, arising from a germinal center B cell. A is the partner in t(11;14), defining mantle cell lymphoma through cyclin D1. C is the partner in t(8;14), defining Burkitt lymphoma. B and E are the partners in t(11;18) BIRC3::MALT1, associated with gastric MALT lymphoma. All four translocations share the theme of an oncogene driven by an immunoglobulin promoter.
120A pathologist compares a reactive germinal center with a follicle from a patient with follicular lymphoma.Which feature is characteristic of the neoplastic follicle?
- APolarization into dark and light zones
- BNumerous tingible body macrophages
- CAn intact surrounding mantle zone
- DA high proportion of cells undergoing apoptosis
- EAbsence of tingible body macrophages
Answer
E
The germinal center normally works by deliberately killing most of the cells in it — B cells mutate their antibody genes at random, most produce a worse or autoreactive antibody, and those are driven into apoptosis. Tingible body macrophages are the cells clearing that debris, so their presence proves apoptosis is occurring. In follicular lymphoma, t(14;18) switches BCL2 back on permanently, apoptosis is blocked, and the macrophages disappear. A, B, C, and D are all features of a reactive follicle from Lecture 07, each reflecting an organized selection program the neoplastic follicle has lost. The same failure to die explains why follicular lymphoma is indolent yet incurable.
121A 66-year-old man has widespread lymphadenopathy, marrow involvement, and polypoid lesions throughout the colon. Tumor cells are CD5-positive and strongly express cyclin D1.Which statement about this lymphoma is correct?
- AIt is the aggressive exception among small B-cell lymphomas
- BIt commonly undergoes Richter transformation to DLBCL
- CIt carries t(14;18) involving BCL2
- DIt arises from a germinal center B cell
- EIt is indolent, like other small B-cell lymphomas
Answer
A
Cyclin D1 expression with a CD5-positive phenotype and lymphomatoid polyposis identifies mantle cell lymphoma, defined by t(11;14) placing CCND1 under IGH control. It is the EXCEPTION to the indolent nature of small B-cell lymphomas — incurable and aggressive, yet still too slowly growing to respond to intensive chemotherapy, with roughly 3–5 year survival. That combination is the worst of both worlds and is exactly why it is singled out, making E incorrect. B is wrong because mantle cell shows no Richter transformation; aggressive variants remain mantle cell. C names the follicular lymphoma translocation, and D is wrong because the cell of origin is a naïve, pre-germinal center B cell.
122A 64-year-old woman with a gastric MALT lymphoma is treated with H. pylori eradication therapy. The lymphoma fails to regress. FISH shows t(11;18).Which best explains the treatment failure?
- AThe translocation indicates transformation to diffuse large B-cell lymphoma
- BThe translocation prevents antibiotics from reaching the gastric mucosa
- CThe translocation renders the clone independent of antigenic stimulation
- DThe organism was never present in this patient
- EThe translocation confers resistance to the antibiotics themselves
Answer
C
Gastric MALT begins as a chronically antigen-driven proliferation — there is no MALT in the normal stomach, and H. pylori recruits it. Persistent stimulation keeps the B cells dividing until a clone emerges that is still dependent on that stimulus, which is why eradication is often curative. Once t(11;18) (BIRC3::MALT1) occurs, the clone acquires an autonomous proliferative signal and no longer needs the antigen, so removing the organism accomplishes nothing. The translocation marks the transition from antigen-dependent to antigen-independent growth. A is a different event entirely, and B, D, and E propose mechanisms with no basis.
123A 72-year-old man has fatigue, blurred vision, and mucosal bleeding. Serum viscosity is markedly elevated and protein electrophoresis shows an IgM monoclonal spike. Marrow shows a lymphoplasmacytic infiltrate.Which is the most appropriate immediate treatment for his visual symptoms?
- AHigh-dose corticosteroids
- BPlasmapheresis
- CRed cell transfusion
- DRituximab monotherapy
- EAllogeneic stem cell transplantation
Answer
B
This is Waldenström macroglobulinemia, defined as lymphoplasmacytic lymphoma in the marrow plus an IgM monoclonal gammopathy plus hyperviscosity syndrome, and hyperviscosity is treated with plasmapheresis. The mechanism explains the choice: IgM is a large pentamer that raises serum viscosity directly, so physically removing it relieves symptoms within hours, whereas therapies directed at the clone take weeks. C would be actively harmful, since adding red cells raises viscosity further. D and A may have a role in treating the underlying lymphoma but do not address the immediate emergency.
124A marrow biopsy shows a lymphoplasmacytic infiltrate. Molecular testing is performed to support the diagnosis.Which mutation is present in more than 90% of cases?
- ABRAF V600E
- BBCL2 rearrangement
- CJAK2 V617F
- DMYD88 L265P
- EMYC rearrangement
Answer
D
Lymphoplasmacytic lymphoma has no recurrent chromosomal abnormalities, which distinguishes it from most of the small B-cell lymphomas, but MYD88 L265P is present in over 90% of cases and therefore carries the diagnostic weight a translocation would carry elsewhere. A is the mutation of hairy cell leukemia, present in essentially 100% of those cases, and also appears in Langerhans cell histiocytosis. C defines the myeloproliferative neoplasms. B indicates follicular lymphoma and E indicates Burkitt lymphoma, both structural rearrangements rather than point mutations.
125A 61-year-old man has massive splenomegaly and pancytopenia. He has no palpable lymphadenopathy. Bone marrow aspiration yields a dry tap, and the core biopsy shows small lymphocytes with abundant pale cytoplasm and reticulin fibrosis.Which additional finding would best support the diagnosis?
- AAn absolute monocytosis
- BA t(11;14) translocation
- CGeneralized lymphadenopathy on imaging
- DAn absolute monocytopenia
- EAn IgM monoclonal gammopathy
Answer
D
Hairy cell leukemia is defined as much by absences as by findings: it involves marrow, blood, and spleen but is not associated with lymphadenopathy, and patients present with splenomegaly and pancytopenia including monocytopenia — a specific and unexpected cytopenia that makes A exactly wrong. The dry tap follows from the reticulin fibrosis, which is why diagnosis rests on core biopsy, and the marrow shows “fried egg” morphology. C contradicts the disease. B indicates mantle cell lymphoma and E indicates lymphoplasmacytic lymphoma. Essentially 100% of cases carry BRAF V600E, and treatment uses purine analogues such as cladribine.
126A 9-year-old boy in Kenya presents with a rapidly enlarging jaw mass. Biopsy shows sheets of intermediate-sized cells with interspersed pale macrophages containing apoptotic debris.Which cytogenetic abnormality is expected?
- At(8;14); IGH::MYC
- Bt(14;18); IGH::BCL2
- Ct(11;14); IGH::CCND1
- Dt(11;18); BIRC3::MALT1
- Et(15;17); PML::RARA
Answer
A
The “starry sky” appearance and an African child with a jaw mass identify endemic Burkitt lymphoma, defined by t(8;14) placing MYC under IGH control. The morphology and the biology are the same fact: Burkitt has one of the highest proliferation rates of any human tumor, so cells die as fast as they divide and macrophages full of apoptotic debris appear as pale “stars” against the dark “sky” of tumor. The same kinetics make it rapidly fatal untreated, highly curable with intensive chemotherapy, and a very high risk for tumor lysis syndrome. The other translocations define follicular, mantle cell, gastric MALT, and acute promyelocytic leukemia.
127A clinician is asked to describe the three clinical settings in which Burkitt lymphoma occurs.Which pairing is correct?
- AEndemic disease presents in adults with a mesenteric mass
- BSporadic disease presents in African children with a jaw mass
- CImmunodeficiency-related disease arises only after transplantation and is always EBV-negative
- DEndemic disease presents in African children with a jaw mass and is EBV-positive
- ESporadic disease occurs only in patients with HIV infection
Answer
D
The three flavors are endemic — Africa, children, jaw mass, EBV-positive; sporadic — adults, mesenteric mass, ± EBV; and immunodeficiency-related — HIV or post-transplant, ± EBV. A and B swap the endemic and sporadic descriptions, which is the most common error. C is wrong because immunodeficiency-related cases are ± EBV rather than uniformly negative. E is wrong because sporadic disease occurs in immunocompetent people, with HIV belonging to the separate immunodeficiency-related category. Note that EBV also drives post-transplant lymphoproliferative disorders.
128A 70-year-old man presents with a rapidly enlarging neck mass over six weeks. Biopsy shows sheets of large atypical B cells.Which statement about this lymphoma is correct?
- AIt arises only de novo and never from a lower-grade process
- BIt is indolent and requires only observation
- CIt is incurable regardless of therapy
- DIt is the most common lymphoma worldwide and is potentially curable
- EIt is defined by the presence of a t(14;18) translocation
Answer
D
Diffuse large B-cell lymphoma is the most common lymphoma in the world and, like other high-grade lymphomas, is aggressive and fatal if untreated but potentially curable with intensive chemotherapy — roughly 70% remission and 40% cure, improved by the addition of rituximab, an anti-CD20 antibody. B and C invert the central prognostic principle, which is that aggressive lymphomas are curable and indolent ones are not, because chemotherapy kills dividing cells. A is wrong because DLBCL can arise by transformation from CLL/SLL or follicular lymphoma. E describes follicular lymphoma.
129Flow cytometry is performed on a lymph node suspension to determine whether a B-cell population is neoplastic.Which finding would best support a clonal process?
- AA mixture of kappa- and lambda-expressing cells in roughly equal numbers
- BRestriction of surface light chain expression to kappa only
- CExpression of CD19 and CD20 by the majority of cells
- DThe presence of numerous T cells alongside the B cells
- EExpression of CD45 by all lymphoid cells
Answer
B
Neoplastic processes are monoclonal: a single cell produces an expansile clone with an identical antigen receptor rearrangement, independent of antigen stimulation. Because each B cell expresses either kappa or lambda but not both, a clone shows light chain restriction — the practical flow cytometry readout of clonality. A describes the opposite and indicates a polyclonal, reactive population responding to various antigens. C, D, and E describe normal B-cell, T-cell, and leukocyte markers that carry no information about clonality. Note the caveat that not all clonal processes are neoplastic, especially in T cells, and that T cells have no light chain equivalent.
130A student asks why low-grade lymphomas are incurable while high-grade lymphomas can be cured.Which explanation is correct?
- AHigh-grade lymphomas are more likely to carry targetable translocations
- BLow-grade lymphomas express drug efflux pumps that high-grade lymphomas lack
- CHigh-grade lymphomas are detected at a much earlier clinical stage than low-grade lymphomas
- DLow-grade lymphomas involve the marrow, which shields them from chemotherapy
- ELow-grade lymphomas have a low proliferative rate, so most cells escape cytotoxic therapy
Answer
E
Cytotoxic chemotherapy kills dividing cells. An aggressive lymphoma is dividing furiously, so nearly the whole tumor is exposed at once and can be eradicated. An indolent lymphoma has a low proliferative rate, so at any moment most of the clone sits quietly out of reach — the tumor shrinks and then returns, and the patient lives a long time with the disease without being rid of it. C is factually backwards, since low-grade lymphomas such as follicular are typically widespread at diagnosis yet still indolent. B, D, and A propose mechanisms that are not the explanation. Mantle cell lymphoma is the worst of both — aggressive yet still too slow-growing to cure.
Lecture 12 · Torres
Plasma Cell Neoplasms
Questions 131–139
131A 75-year-old African American man presents with back pain, fatigue, and recurrent infections. Laboratory studies show a normocytic normochromic anemia, thrombocytopenia, and elevated creatinine and calcium.Which set of findings does this presentation represent?
- AThe CRAB criteria of multiple myeloma
- BThe diagnostic triad of Waldenström macroglobulinemia
- CThe criteria for monoclonal gammopathy of undetermined significance
- DThe features of AL amyloidosis without myeloma
- EThe presentation of a solitary plasmacytoma
Answer
A
CRAB stands for hyperCalcemia, Renal insufficiency, Anemia, and Bone lesions, and this patient has all four elements — the calcium and creatinine are elevated, the anemia is present, and the back pain reflects multifocal osteolytic lesions. Note the demographics, which fit: myeloma is male predominant at 1.5:1, twice as frequent in African Americans, and peaks at 65–70 years. B describes an IgM-associated hyperviscosity syndrome, a different disease. C is excluded because MGUS is asymptomatic by definition. D and E are related plasma cell disorders lacking this constellation.
132A 68-year-old asymptomatic woman is found to have an IgG monoclonal spike of 1.8 g/dL. Marrow shows 6% clonal plasma cells. Calcium, creatinine, hemoglobin, and skeletal survey are all normal.Which is the correct diagnosis?
- AMonoclonal gammopathy of undetermined significance
- BSmoldering myeloma requiring immediate therapy
- CMultiple myeloma
- DWaldenström macroglobulinemia
- EAL amyloidosis
Answer
A
MGUS requires three conditions simultaneously: an M-spike under 3 g/dL, fewer than 10% monoclonal plasma cells in the marrow, and no symptoms — meaning no CRAB findings. This patient satisfies all three. MGUS is a pre-neoplastic condition and about 1% progress to plasma cell myeloma, so surveillance rather than treatment is appropriate, making B wrong. C is excluded by the absence of CRAB findings and the low plasma cell percentage. D would require an IgM spike. The discriminating step in any such vignette is to check for hypercalcemia, renal insufficiency, anemia, and lytic lesions rather than react to the M-spike alone.
133A 73-year-old man has an M-spike identified on serum protein electrophoresis. The paraprotein is IgM, and he has hyperviscosity symptoms but no lytic bone lesions.Which diagnosis does the isotype favor?
- AMultiple myeloma with an unusual IgM paraprotein secretion pattern
- BHeavy chain disease
- CLight chain only myeloma
- DLymphoplasmacytic lymphoma / Waldenström macroglobulinemia
- EMonoclonal gammopathy of undetermined significance
Answer
D
The isotype decides. Myeloma secretes IgG > IgA > light chain only; an IgM-secreting lymphoproliferative disorder is Waldenström macroglobulinemia, which is lymphoplasmacytic lymphoma. The clinical consequences follow the molecule: IgM is a large pentamer that raises serum viscosity and produces hyperviscosity syndrome treated with plasmapheresis, whereas myeloma produces lytic bone lesions and CRAB findings. The absence of lytic lesions here supports that split. C describes a myeloma variant with little or no serum spike. E is excluded by the presence of symptoms.
134A patient with multiple myeloma has a serum creatinine of 3.2 mg/dL. Urine electrophoresis demonstrates a monoclonal light chain.Which term describes the urinary protein, and what is the mechanism of renal injury?
- ATamm-Horsfall protein; normal tubular protein is overproduced
- BBence-Jones protein; whole immunoglobulin is filtered and obstructs glomeruli
- CM-protein; immune complexes deposit along the basement membrane
- DAmyloid; light chains form fibrils exclusively within glomeruli
- EBence-Jones protein; light chains deposit in tubules causing injury
Answer
E
Immunoglobulin eliminated in the urine is the Bence-Jones protein, and in myeloma kidney the free light chains excreted by the kidney deposit in the tubules and lead to kidney injury. The reason light chains and not whole immunoglobulin do this is size: a free light chain is small enough to be filtered by the glomerulus, whereas intact IgG is not, so only the light chain reaches the tubule — making B mechanistically wrong. D describes AL amyloidosis, the other light-chain deposition disorder, in which free light chains deposit in tissues in the form of amyloid. C and E misidentify the protein.
135A 69-year-old woman with multiple myeloma has multiple well-circumscribed lucent lesions on skeletal survey. A radionuclide bone scan is reported as unremarkable.Which best explains the discrepancy?
- AMyeloma lesions are osteoblastic and require different tracer
- BBone scans cannot detect lesions of the axial skeleton
- CThe lesions are too small to be resolved by conventional radionuclide imaging
- DHypercalcemia suppresses radionuclide uptake by bone
- EMyeloma lesions are purely lytic with little osteoblastic activity to detect
Answer
E
Myeloma cells release MIP1α, TNF, and IL-1β, which activate osteoclasts via RANK while osteoblast activity is suppressed, so bone is resorbed with no attempt at repair. A bone scan works by detecting osteoblastic activity, so lesions that provoke none are effectively invisible — which is why myeloma is imaged with a skeletal survey. A states the opposite of the biology. B, C, and D propose technical explanations that do not apply. The same unopposed resorption dumps calcium into the blood, producing the C of CRAB with its neurologic manifestations and kidney injury.
136A peripheral smear from a patient with a large M-spike shows red cells arranged in stacks resembling columns of coins.Which finding is described, and what causes it?
- AAgglutination; an IgM autoantibody cross-links red cells
- BSchistocytes; shearing within a fibrin mesh
- CRouleaux; high paraprotein reduces the charge repulsion between red cells
- DSpherocytes; partial membrane removal by macrophages within the splenic cords
- ETarget cells; excess membrane relative to hemoglobin content
Answer
C
Rouleaux are red cells stacked like coins, formed when a high concentration of monoclonal protein reduces the charge repulsion that normally keeps cells apart — a characteristic peripheral blood finding in plasma cell myeloma. A is the deliberate contrast carried over from Lecture 03: agglutinates also look like red cells sticking together, but they are irregular clumps produced by an antigen–antibody reaction and point to cold agglutinin syndrome. The distinction matters because rouleaux reflect plasma composition while agglutination reflects an immune process. B, D, and E describe changes in individual cell shape rather than in how cells associate.
137A patient with multiple myeloma and a total serum protein of 11 g/dL suffers recurrent pneumococcal pneumonia.Which best explains this susceptibility?
- AThe paraprotein directly inhibits neutrophil function
- BThe monoclonal protein is functionally useless and normal antibody production is suppressed
- CHypercalcemia impairs lymphocyte activation
- DThe elevated total serum protein concentration reflects preserved and adequate humoral immunity
- ERenal loss of immunoglobulin produces hypogammaglobulinemia
Answer
B
Myeloma produces an enormous quantity of immunoglobulin while leaving the patient immunodeficient, and the resolution is that the paraprotein is monoclonal — a single specificity directed at nothing useful — while the expanding clone suppresses normal plasma cells, so functional polyclonal antibody falls. Patients suffer infections due to immune deficiency, a listed clinical feature. This is the same pattern as hypogammaglobulinemia in CLL and functional neutropenia in AML: a high number of a useless product alongside a deficiency of the working version. D mistakes quantity for function, and A, C, and E propose mechanisms that are not the explanation.
138A 71-year-old woman with a monoclonal gammopathy develops nephrotic-range proteinuria, macroglossia, and restrictive cardiomyopathy. Congo red staining of a fat pad biopsy shows apple-green birefringence.Which process is responsible?
- AFree light chains depositing in tissues as amyloid
- BFree light chains obstructing renal tubules
- CWhole immunoglobulin depositing along basement membranes
- DOsteoclast activation mediated by RANK
- EPlasma cell infiltration of the affected organs
Answer
A
AL amyloidosis occurs when free light chains circulate in serum and deposit in tissues in the form of amyloid, giving primary amyloidosis. The misfolded light chains form fibrils that accumulate systemically, which is why the presentation spans kidney, tongue, and heart rather than a single organ. B describes the other light chain deposition disorder, myeloma kidney, where the same free light chains deposit as casts in the tubules — the deliberate contrast, since one protein causes two different diseases depending on where and how it deposits. D is the mechanism of the lytic bone lesions, and C and E describe processes that are not responsible here.
139Which cytokine is described as promoting the expansion and survival of myeloma cells?Select the correct mediator.
- AInterferon gamma
- BInterleukin-6
- CInterleukin-2
- DTumor necrosis factor alpha
- EGranulocyte colony-stimulating factor
Answer
B
IL-6 promotes expansion and survival of myeloma cells, and it sits alongside the other pathogenetic factors listed — exposure to toxins and radiation, chronic antigenic stimulation, and IGH gene rearrangement. D is a genuine participant but in a different role: TNF, along with MIP1α and IL-1β, is produced by the myeloma cells to activate osteoclasts via RANK, driving the bone disease rather than sustaining the clone. A, C, and E are cytokines with no described role in this disease. Keeping IL-6 attached to survival of the clone and the MIP1α/TNF/IL-1β group attached to bone destruction separates the two arms of the pathogenesis.
Lecture 13 · Howell
T-cell Lymphomas
Questions 140–150
140Flow cytometry on a lymph node shows a T-cell population that is CD2, CD3, CD5, and CD4 positive and TdT negative.What does the TdT result establish?
- AThe cells are of B-cell rather than T-cell lineage
- BThe cells are precursor thymic lymphoblasts
- CThe result cannot be interpreted without CD1a staining
- DThe cells are undergoing active receptor rearrangement
- EThe cells are mature, post-thymic T cells
Answer
E
TdT is the enzyme that inserts random nucleotides during antigen receptor gene rearrangement, so it is expressed only while a lymphocyte is building its receptor — in marrow and thymus. Once the receptor is complete and the cell leaves for the periphery, TdT is switched off permanently, making it a timestamp: TdT-positive means precursor and therefore lymphoblastic leukemia/lymphoma, while TdT-negative means post-thymic and therefore one of the mature peripheral T-cell lymphomas. B and D describe the immature compartment, which would also express CD1a and cytoplasmic CD3 and be double positive or double negative for CD4/CD8. A is wrong because TdT marks both B- and T-lymphoblasts and says nothing about lineage.
141A pathologist notes that B-cell lymphomas are classified by stage of development while T-cell lymphomas are classified differently.On what basis are T-cell lymphomas classified?
- AClinical presentation, as cutaneous, leukemic, extranodal, or nodal
- BCell size, as small, intermediate, or large
- CThe specific chromosomal translocation identified on karyotype or FISH
- DSurface light chain restriction pattern
- EDegree of TdT expression
Answer
A
T-cell lymphomas are classified according to clinical presentation — cutaneous, disseminated/leukemic, extranodal, and nodal — because the usual classification tools fail. Morphology shows striking variability and extensive overlap; the immunophenotype offers no clear surface marker of clonality and usually no disease-specific phenotype; and only a few entities have a defining lesion, such as t(2;5) in anaplastic large cell lymphoma. D is the specific point of failure: light chain restriction is a B-cell tool with no T-cell equivalent, so clonality requires T-cell receptor gene rearrangement studies. B is the framework used for B-cell lymphomas.
142A 58-year-old man has a ten-year history of scaly patches progressing to plaques on the trunk. Biopsy shows small cerebriform lymphocytes infiltrating the epidermis singly and in small intraepidermal collections.Which immunophenotypic finding would best support the diagnosis?
- ACD4 positive with retained CD7
- BCD4 positive with loss of CD7
- CCD8 positive with loss of CD4
- DCD30 positive with ALK expression
- ECD56 positive with EBV positivity
Answer
B
Mycosis fungoides arises from a mature CD4-positive epidermotropic T cell, and the neoplastic population is positive for CD2, CD3, CD5, CD4 and CLA while being negative for CD7, CD8 and ALK. The loss of CD7 is the diagnostic lever, because a normal mature T cell carries the full panel — an aberrant phenotype, meaning a population missing a marker it should have, is the closest thing T-cell pathology has to light chain restriction, making A wrong. The intraepidermal collections are Pautrier microabscesses, highly characteristic but present in a minority of cases. D describes anaplastic large cell lymphoma and E describes extranodal NK/T-cell lymphoma.
143A 67-year-old man has diffuse erythroderma, generalized lymphadenopathy, and 3,200 circulating atypical lymphocytes per mL with cerebriform nuclei. Bone marrow examination is nearly normal.Which best explains the sparing of the bone marrow?
- AThe cells express skin homing receptors and traffic between skin, blood, and nodes
- BMarrow involvement occurs only in the terminal phase of all T-cell lymphomas
- CThe tumor cells lack the adhesion molecules needed to enter marrow
- DCirculating cells in this disease are reactive rather than neoplastic
- EMarrow sampling error is typical in erythrodermic disease
Answer
A
This is Sézary syndrome, defined by the triad of erythroderma, lymphadenopathy, and circulating Sézary cells exceeding 1000 per mL. Despite being leukemic, the bone marrow is remarkably spared, with involvement described as sparse and usually interstitial. The explanation is the skin homing receptor CCR4 and cutaneous lymphocyte antigen, which program these cells to traffic skin to blood to lymph node — the blood is their transit route rather than their destination. That is why Sézary is grouped with the cutaneous lymphomas and why the triad describes three stops on one circuit. Flow typically shows a CD4/CD8 ratio above 10:1 with increased CD4+CD7− cells.
144A 49-year-old man from Japan presents with generalized lymphadenopathy, hepatosplenomegaly, skin lesions, leukocytosis, and a calcium of 13.8 mg/dL. The smear shows lymphocytes with markedly polylobated nuclei.Which virus is implicated?
- AEpstein-Barr virus
- BHuman T-cell leukemia virus 1
- CHuman herpesvirus 8
- DHuman immunodeficiency virus
- EHepatitis C virus
Answer
B
Adult T-cell leukemia/lymphoma is caused by HTLV-1, is endemic in Japan, the Caribbean, and Central Africa, has a median age of 47, and shows characteristic “flower cells” with polylobated nuclei. The hypercalcemia with or without lytic bone lesions is a feature of the acute variant and reflects the general T-cell mechanism of osteoclast activating factor. Note that HTLV-1 alone is not sufficient — additional genetic hits are needed after a long latency following infection very early in life via breast milk, sexual intercourse, or blood products. A drives extranodal NK/T-cell lymphoma and Burkitt lymphoma, and B drives primary effusion lymphoma.
145A patient with adult T-cell leukemia/lymphoma dies of Pneumocystis pneumonia and cryptococcal meningitis.Which feature of the neoplastic cell best explains this vulnerability?
- AThe cells express CD25 and consume available interleukin-2
- BThe cells secrete immunoglobulin that is functionally inactive
- CThe cells infiltrate and destroy the bone marrow
- DThe cells are derived from a CD4-positive regulatory T cell
- EThe virus directly infects and lyses neutrophils
Answer
D
The cell of origin is a CD4-positive peripheral regulatory T cell, whose normal role is to suppress immune responses. Expanding that population enormously produces profound immunosuppression, which is why the listed causes of death are the same organisms seen in advanced HIV — Pneumocystis, cryptococcal meningitis, disseminated herpes zoster — alongside hypercalcemia, rather than complications of tumor bulk. A names a genuine marker, since CD25 is the IL-2 receptor alpha chain and a normal regulatory T-cell marker, but the immunosuppression follows from the regulatory function rather than from cytokine consumption. B describes a plasma cell neoplasm.
146A lymph node from a 12-year-old shows sheets of large bizarre cells, some with eccentric horseshoe-shaped nuclei and a prominent paranuclear eosinophilic Golgi zone. The cells are uniformly and strongly CD30 positive.Which genetic abnormality is most likely?
- At(2;5)(p23;q35) producing an NPM-ALK fusion
- Bt(8;14) producing an IGH::MYC fusion
- Ct(14;18) producing an IGH::BCL2 fusion
- Dt(11;14) producing an IGH::CCND1 fusion
- Et(15;17) producing a PML::RARA fusion
Answer
A
The cells described are hallmark cells — large cells with eccentric horseshoe- or kidney-shaped nuclei and a prominent paranuclear eosinophilic Golgi region — and with strong uniform CD30 in a child they identify ALK-positive anaplastic large cell lymphoma. 75–80% of cases carry t(2;5)(p23;q35), joining the nucleophosmin gene at 5q35 to the ALK gene at 2p23. The ALK staining pattern reports the genetics: cytoplasmic and nuclear staining indicates t(2;5), because nucleophosmin normally shuttles to the nucleus and carries the fusion protein with it. The remaining options are B-cell and myeloid translocations.
147A 14-year-old with ALK-positive anaplastic large cell lymphoma is being counseled about prognosis.Which statement is accurate?
- APrognosis depends entirely on the presence of B symptoms
- BIt has a five-year survival under 20%, typical of T-cell lymphomas
- CIt is incurable but indolent, with survival measured in decades
- DPrognosis is identical to peripheral T-cell lymphoma, not otherwise specified
- EIt has a five-year survival of 80–90%, unusually good for a T-cell lymphoma
Answer
E
ALK-positive ALCL has a five-year survival of 80–90%, which stands out sharply against the general rule that T-cell lymphomas are aggressive with poorer survival — worse than B-cell lymphoma and Hodgkin lymphoma. It is explicitly one of the exceptions to that rule, which is what makes it examinable. It represents 3% of all non-Hodgkin lymphoma but 10–30% of childhood lymphomas, occurs in the first three decades with a 3:1 male predominance, and presents with advanced disease in 70% and B symptoms — so a good outcome despite advanced stage is exactly the point. B and D apply the general T-cell rule without recognizing the exception.
148A 44-year-old man of East Asian descent has a destructive ulcerating mass of the nasal cavity and hard palate. Biopsy shows extensive necrosis with atypical lymphoid cells invading vessel walls.Which combination of findings is expected?
- ACD56 positive, EBV positive, cytotoxic granule proteins positive
- BCD30 positive, ALK positive, EBV negative
- CCD4 positive, CD7 negative, CLA positive
- DCD4 positive, CD25 positive, HTLV-1 integrated
- ETdT positive, CD1a positive, cytoplasmic CD3 positive, CD4/CD8 double negative
Answer
A
Extranodal NK/T-cell lymphoma, nasal type involves the upper aerodigestive tract, is angiodestructive with necrosis and vascular destruction, and occurs in Asians and indigenous populations of Mexico, Central and South America. It is EBV-positive, and the majority are of NK-cell lineage (66–75%) expressing CD2, cytoplasmic CD3, CD56, and cytotoxic markers TIA-1, granzyme B, and perforin. Those cytotoxic proteins are working weapons rather than mere markers, which is why the tumor destroys vessels and produces the classic destructive midline facial lesion. B describes ALCL, C mycosis fungoides, D adult T-cell leukemia/lymphoma, and E a precursor T-lymphoblastic process.
149Molecular studies are ordered to establish clonality in a suspected T-cell lymphoma.Which test is appropriate?
- ASurface light chain restriction by flow cytometry
- BImmunoglobulin heavy chain gene rearrangement
- CKaryotype analysis for a complex karyotype
- DJAK2 V617F mutation analysis
- ET-cell receptor gene rearrangement
Answer
E
T-cell receptor gene rearrangement studies establish clonality in T-cell neoplasms, because there is no clear surface marker of clonality for T cells — A is precisely the tool that does not exist on the T side, since light chain restriction is a B-cell readout, and B is the corresponding B-cell molecular test. D belongs to the myeloproliferative neoplasms. C is unhelpful because complex karyotypes occur in mycosis fungoides and Sézary syndrome with no specific changes, so they cannot define a disease. Note the important caveat that a clonal T-cell population is not necessarily neoplastic.
150A dermatopathologist finds a clonal T-cell receptor gene rearrangement in a skin biopsy from a patient with a chronic inflammatory dermatosis.How should this result be interpreted?
- AIt establishes a diagnosis of cutaneous T-cell lymphoma
- BIt excludes a reactive process
- CIt indicates that the process will inevitably progress to lymphoma
- DIt must be interpreted alongside morphology and clinical context
- EIt is a technical artifact and should be disregarded
Answer
D
The governing principle is that while all neoplasms are clonal, not all clonal processes are neoplastic — especially in T cells. A vigorous reactive T-cell response can generate a detectable clone without being a lymphoma, so a rearrangement result is supporting evidence rather than proof and must be read alongside morphology and clinical presentation. This is one concrete reason the entire family is classified by clinical presentation rather than by laboratory markers. A, B, and C all treat the finding as diagnostic, which overstates it, and E dismisses a genuine result. Note that mycosis fungoides is notoriously difficult to separate from chronic dermatitis for exactly this reason.
Lecture 14 · Bhagavathi
Hodgkin Lymphoma
Questions 151–159
151A lymph node biopsy from a 30-year-old with painless cervical lymphadenopathy shows scattered large binucleate cells within an abundant background of lymphocytes, macrophages, eosinophils, and plasma cells.Which immunophenotype is expected in the large cells?
- ACD30 positive, CD15 negative, PAX5 negative, CD20 positive
- BCD30 negative, CD15 negative, PAX5 positive, CD20 positive
- CCD30 positive, CD15 positive, PAX5 positive, CD20 negative
- DCD30 negative, CD15 positive, PAX5 negative, CD20 negative
- ECD30 positive, CD15 positive, PAX5 negative, CD20 positive
Answer
C
The large binucleate cells are Reed-Sternberg cells, and classical Hodgkin lymphoma is PAX5 positive, CD30 positive, CD15 positive, and CD20 negative. The pattern encodes real biology: the RS cell arises from a germinal center B cell but has lost its B-cell program, which is why CD20 is negative and why the cell looks so morphologically strange. The one B-cell marker that survives is PAX5, a B-cell transcription factor emphasized as the marker “that's not lost,” and it is used to prove B-cell origin. B is the NLPHL panel, which is essentially the opposite. The abundant inflammatory background constitutes 90% of the tumor cellularity.
152A 26-year-old man has an isolated enlarged cervical node. Biopsy shows nodules containing scattered large cells with multilobated nuclei resembling popcorn, in a background of follicular dendritic cells and reactive B cells.Which is the most likely diagnosis?
- AClassical Hodgkin lymphoma, nodular sclerosis type
- BClassical Hodgkin lymphoma, mixed cellularity type
- CNodular lymphocyte-predominant Hodgkin lymphoma
- DFollicular lymphoma, grade 1
- EDiffuse large B-cell lymphoma
Answer
C
NLPHL is uncommon and typically presents in a young male with cervical or axillary lymphadenopathy, with frequent L&H or “popcorn” cell variants in a background of follicular dendritic cells and reactive B cells. Its phenotype is the mirror image of classical disease: PAX5 positive, CD20 positive, OCT2 positive, CD30 negative, CD15 negative — it has retained the B-cell program that classical HL loses. A and B are classical subtypes defined by Reed-Sternberg cells rather than popcorn cells. Note that NLPHL may evolve to T-cell/histiocyte-rich large B-cell lymphoma.
153A clinician is asked how Hodgkin lymphoma differs from non-Hodgkin lymphoma in its behavior.Which statement is correct?
- AIt typically presents with widespread noncontiguous lymphadenopathy at diagnosis
- BIt is distinguished by the absence of an inflammatory infiltrate
- CIt rarely involves lymph nodes and presents at extranodal sites
- DIt is defined by a characteristic chromosomal translocation
- EIt arises in a single node or chain and spreads to contiguous lymphoid tissue
Answer
E
Hodgkin lymphoma arises in a single node or chain of nodes and spreads first to anatomically contiguous lymphoid tissue, so it mimics a carcinoma rather than producing the generalized lymphadenopathy of non-Hodgkin lymphoma — making A the intended contrast. That predictable behavior has a practical consequence: because the extent of disease could be mapped and encompassed in a radiation field, Hodgkin lymphoma became the first cancer successfully treated with radiation and chemotherapy, and stage rather than histologic subtype is now the dominant prognostic variable. D is wrong because HL is defined by the presence of Reed-Sternberg cells, and B inverts the characteristic morphology.
154A student asks why the neoplastic cells make up only a small fraction of a Hodgkin lymphoma mass.Which explanation is correct?
- AThe neoplastic cells are destroyed by the host immune response as fast as they arise
- BThe neoplastic cells secrete cytokines and chemokines that attract inflammatory cells
- CThe inflammatory cells are themselves part of the neoplastic clone
- DSampling artifact causes the neoplastic cells to be underrepresented
- EThe neoplastic cells divide slowly compared with reactive lymphocytes
Answer
B
The defining morphologic feature is rare neoplastic cells in an abundant inflammatory background of lymphocytes, macrophages and granulocytes that constitutes 90% of the tumor cellularity, and the mechanism is that the transformed B cells secrete cytokines, chemokines and other factors that attract inflammatory cells. The upstream driver is activation of the transcription factor NF-κB, by several mechanisms including EBV infection — the same signal that keeps the crippled germinal center cell alive also makes it pour out chemokines. This explains why B symptoms are cytokine-driven and why identification of RS cells and variants is crucial for diagnosis. C is wrong because the background is reactive, not clonal.
155A pathologist reviews the distribution of classical Hodgkin lymphoma subtypes.Which subtype accounts for approximately 70% of cases?
- AMixed cellularity
- BLymphocyte-rich
- CLymphocyte-depleted
- DNodular lymphocyte-predominant
- ENodular sclerosis
Answer
E
Nodular sclerosis accounts for 70% of Hodgkin lymphoma cases, making it by far the most common subtype. The four classical subtypes are nodular sclerosis, mixed cellularity, lymphocyte-rich, and lymphocyte-depleted, with the latter two described as uncommon. D is not a classical subtype at all — nodular lymphocyte-predominant Hodgkin lymphoma is classified separately and has a different immunophenotype and behavior. Clinically, nodular sclerosis and lymphocyte predominance usually present at stage I–II free of systemic symptoms, whereas mixed cellularity and lymphocyte-depleted more often present with disseminated disease and B symptoms.
156Two patients with Hodgkin lymphoma are compared: one has nodular sclerosis at stage IIA, the other has mixed cellularity at stage IIA.Which factor most influences prognosis with current treatment protocols?
- AThe tumor stage
- BThe histologic subtype
- CThe presence of Reed-Sternberg variants
- DThe CD15 staining intensity
- EThe patient's sex
Answer
A
With current treatment protocols, tumor stage rather than histologic type is the important prognostic variable. This represents a genuine historical shift — subtype once mattered a great deal, but effective combined-modality therapy has largely erased those differences, which is why B is the trap. The stage-based figures are worth carrying: cure rate for stage I and stage IIA is 90%, and even in advanced disease at stage IVA or IVB, 60–70% five-year disease-free survival is common. Subtype still correlates with how patients present — nodular sclerosis and lymphocyte predominance at low stage without symptoms — but that is a matter of presentation rather than independent prognostic weight.
157A 47-year-old woman treated 15 years ago for Hodgkin lymphoma with alkylating chemotherapy and mantle radiotherapy now presents with pancytopenia. Marrow examination shows 28% blasts and a complex karyotype.Which best explains this development?
- ARelapsed Hodgkin lymphoma with marrow involvement
- BAn aplastic anemia caused by late radiation effects
- CA myelodysplastic neoplasm unrelated to prior therapy
- DTransformation of Hodgkin lymphoma to a T-cell lymphoma
- ETherapy-related acute myeloid leukemia
Answer
E
Long-term survivors of Hodgkin lymphoma treated with alkylating chemotherapy and radiotherapy have an increased risk of developing a secondary malignancy, and the marrow findings here — ≥20% blasts with a complex karyotype — identify therapy-related AML. This connects directly to Lecture 08, where alkylating agents and topoisomerase inhibitors are named as the culprits and secondary AML is described as showing a complex karyotype with a very poor prognosis. The underlying principle is that the marrow stem cells were mutagenized by the curative therapy. C is contradicted by the blast count, which defines AML rather than MDS, and A and D are not supported by a myeloid blast population.
158A 34-year-old man with Hodgkin lymphoma reports fevers, drenching night sweats, and a 12 kg weight loss.Which pattern of disease is most consistent with these symptoms?
- AMixed cellularity at stage III or IV
- BLymphocyte predominance at stage II
- CNodular sclerosis at stage I
- DAny subtype at stage I
- ENodular lymphocyte-predominant disease at stage I
Answer
A
B symptoms — fever, night sweats, and weight loss — are most characteristic of patients with disseminated disease at stage III–IV and of the mixed cellularity and lymphocyte-depleted types. By contrast, nodular sclerosis and lymphocyte predominance usually present at clinical stage I–II and are free of systemic symptoms, which excludes C, B, D, and E. The mechanism ties back to the pathogenesis: the cytokines and chemokines secreted by the transformed B cells that recruit the inflammatory background are also what produce the constitutional symptoms, so B symptoms track cytokine output and disease burden rather than tumor mass alone.
159A clinician summarizes the epidemiology of Hodgkin lymphoma.Which statement is accurate?
- AIt occurs almost exclusively in immunosuppressed patients
- BIt is a disease of the elderly with a median age above 70
- CIt accounts for approximately 30% of all new cancers
- DIt is uniformly fatal without stem cell transplantation
- EIt has an average age of onset of 32 and is curable in most cases
Answer
E
Hodgkin lymphoma has an average age of 32, making it one of the most common lymphomas of young adults, and it is curable in most cases — it was the first cancer to be successfully treated with radiation and chemotherapy. It accounts for 0.7% of all new cancers in the United States with about 8,000 new cases each year, so C overstates its frequency by a wide margin. B misstates the age distribution. D contradicts the cure rates of 90% in stage I–IIA and 60–70% five-year disease-free survival even in stage IV. A is wrong, although EBV is one mechanism of NF-κB activation in pathogenesis.
Lecture 15 · Torres
Langerhans Cell Histiocytosis
Questions 160–166
160A bone lesion biopsy shows large cells with grooved and folded nuclei and abundant pink, finely granular cytoplasm.Which immunophenotype would confirm the diagnosis?
- AS100 positive, CD1a positive, langerin positive
- BCD30 positive, ALK positive, EBV negative
- CCD15 positive, CD30 positive, PAX5 positive
- DCD56 positive, granzyme B positive, EBV positive
- ECD19 positive, CD10 positive, TdT positive
Answer
A
Langerhans cell histiocytosis is a clonal neoplastic process of Langerhans-type cells, and the defining phenotype is S100 positive, CD1a positive, and CD207 (langerin) positive, with CD4 also expressed. The described morphology — large cells with grooved and folded nuclei and abundant pink, finely granular cytoplasm — is the light microscopic counterpart. Langerhans cells are dendritic cells of the innate immune system that derive from myeloid stem cells via common dendritic cell progenitors and process antigens for presentation to T cells. B describes anaplastic large cell lymphoma, C classical Hodgkin lymphoma, D extranodal NK/T-cell lymphoma, and E B-lymphoblastic leukemia.
161Electron microscopy of a lesional cell demonstrates racket-shaped pentalaminar cytoplasmic structures.What are these structures, and what do they contain?
- ADense granules, containing ADP and calcium
- BAuer rods, containing myeloperoxidase
- CWeibel-Palade bodies, containing von Willebrand factor
- DBirbeck granules, containing langerin
- EPrimary granules, containing procoagulants
Answer
D
Birbeck granules are racket-shaped pentalaminar structures identifiable only by electron microscopy, and they contain langerin. They are pathognomonic for the Langerhans lineage because they are the physical footprint of a protein only these cells express: langerin binds carbohydrate and is endocytosed, and its structure forces the internalized membrane into a rigid five-layered rod with a vesicular dilation at one end — a rod plus a bulb, hence the tennis racket description. B is the aggregated myeloperoxidase of myeloid blasts, C stores vWF in endothelium and is the depot released by DDAVP, and E describes the promyelocyte granules that cause DIC in APL.
162Molecular testing is performed on a case of Langerhans cell histiocytosis.Which mutation is most frequently identified?
- ANPM1 mutation
- BMYD88 L265P
- CJAK2 V617F
- DKIT D816V
- EBRAF p.V600E
Answer
E
BRAF p.V600E is the most frequent genetic abnormality in LCH, present in 55–60% of cases. It is an activating valine-to-glutamate substitution at residue 600 and is conveniently detectable by immunohistochemical stain, so it does not require sequencing. Less frequent mutations include TP53, RAS, and MET. The same BRAF V600E is found in essentially 100% of hairy cell leukemia, which is worth pairing since both are treatable with BRAF inhibitors. B belongs to lymphoplasmacytic lymphoma, C to the myeloproliferative neoplasms, D to systemic mastocytosis, and A to a favorable-risk subset of AML.
163An 18-month-old presents with a seborrheic-appearing skin eruption, hepatosplenomegaly, lymphadenopathy, multiple lytic bone lesions, and pancytopenia.Which form of Langerhans cell histiocytosis is this?
- AMultifocal multisystem LCH, formerly Letterer-Siwe disease
- BMultifocal unisystem LCH
- CUnifocal eosinophilic granuloma involving the skeletal system
- DPulmonary Langerhans cell histiocytosis
- EHand-Schüller-Christian disease
Answer
A
Multifocal multisystem LCH, formerly Letterer-Siwe disease, is a malignant proliferation with an aggressive course affecting children younger than 2. It involves the skin resembling a seborrheic eruption, with hepatosplenomegaly, lymphadenopathy, lung and osteolytic bone lesions, and bone marrow failure — which accounts for the pancytopenia. It is rapidly fatal if untreated, with 50% five-year survival after chemotherapy, making it the form with the worst prognosis. C and B are the benign, indolent forms grouped as eosinophilic granuloma. D occurs in adult smokers. E names the triad seen in multifocal unisystem disease rather than this multisystem presentation.
164A 6-year-old with multiple erosive skull lesions develops polyuria and polydipsia. Serum sodium is elevated and urine is inappropriately dilute.Which best explains the endocrine finding?
- AHypercalcemia from bone destruction causing nephrogenic diabetes insipidus
- BMarrow failure with secondary adrenal insufficiency
- CAnterior pituitary destruction causing panhypopituitarism
- DInvolvement of the posterior pituitary and hypothalamus causing diabetes insipidus
- ERenal tubular injury from light chain deposition
Answer
D
In multifocal unisystem LCH, lesions sometimes involve the posterior pituitary and hypothalamus, causing diabetes insipidus in 50% of cases. The anatomy explains it: LCH has a marked predilection for the calvarium and skull base, and a lesion eroding through the sella reaches the structures that make and release ADH. This is central rather than nephrogenic diabetes insipidus, which is why A is wrong. The same anatomic logic produces the Hand-Schüller-Christian triad — calvarial bone lesions, diabetes insipidus, and exophthalmos — where the exophthalmos comes from an orbital lesion pushing the globe forward. E belongs to plasma cell myeloma.
165A 38-year-old woman who smokes one pack daily has bilateral upper-zone pulmonary nodules and cysts. Biopsy shows CD1a-positive cells with grooved nuclei.Which is the most appropriate initial management?
- AImmediate multi-agent chemotherapy
- BSmoking cessation
- CAllogeneic stem cell transplantation
- DSystemic corticosteroids for life
- ESurgical resection of all involved lung
Answer
B
Pulmonary LCH is most often seen in adult smokers and regresses spontaneously upon smoking cessation, which makes cessation the correct first intervention and one of the few instances in oncology where removing an exposure treats an established clonal lesion. About 40% of pulmonary cases carry a BRAF mutation, somewhat lower than the 55–60% seen overall. A and C would be appropriate for multifocal multisystem disease (Letterer-Siwe), which is aggressive and rapidly fatal untreated. D and E are disproportionate to a condition with a strong tendency toward spontaneous regression once the stimulus is removed.
166A 22-year-old man has an isolated painful lytic lesion of the femur. Biopsy shows sheets of Langerhans cells admixed with numerous eosinophils.Which statement about this lesion is correct?
- AIt is associated with diabetes insipidus in most cases
- BIt carries a 50% five-year survival even with chemotherapy
- CIt is indolent and may spontaneously regress
- DIt requires urgent multi-agent chemotherapy
- EIt typically arises in children younger than 2 years
Answer
C
This is unifocal LCH, also called eosinophilic granuloma — a benign and indolent proliferation of LCH cells admixed with inflammatory cells, predominantly eosinophils, which is where the name comes from. It commonly involves bones such as the calvarium, ribs and femur, occurs more frequently in the skeletal system of older children and young adults, may be asymptomatic or cause bone pain or pathologic fracture, and can spontaneously regress. B and E describe multifocal multisystem disease (Letterer-Siwe), which affects children under 2 and is aggressive. A applies to multifocal lesions in young children, where posterior pituitary involvement causes diabetes insipidus in 50%.
Lecture 16 · Howell
Spleen
Questions 167–174
167A pathologist examines a normal spleen and identifies the periarteriolar lymphoid sheath surrounding a central artery.Which cells predominate in this compartment?
- AMacrophages
- BT lymphocytes
- CB lymphocytes
- DPlasma cells
- EErythroid precursors
Answer
B
The white pulp is the immune compartment, and it is organized into two parts: the periarteriolar lymphoid sheath (PALS) contains T cells, while the lymphoid follicles contain B cells. Together they carry out immune reaction and antibody production, functioning much like a lymph node. A describes the splenic cords of Billroth in the red pulp, the filtration compartment. This anatomy has a direct clinical payoff carried over from Lecture 07: the splenomegaly of infectious mononucleosis is hypertrophy of the PALS, which makes sense because the disease is driven by a massive reactive CD8-positive T-cell response.
168A student asks how the spleen removes abnormal red cells from the circulation.Which mechanism is responsible?
- AAntibody-mediated opsonization within the white pulp follicles
- BComplement fixation on the sinusoidal endothelium
- CEnzymatic digestion of aged membranes within the central artery
- DDirect lysis of aged cells by splenic natural killer cells
- EBlood leaving capillaries into the cords, then squeezing into sinusoids
Answer
E
Filtration is accomplished through the “open” circulation: blood leaves the capillaries, passes through the cords of Billroth, and enters the venous sinusoids. The mechanism is purely mechanical — to re-enter the sinusoids a red cell must squeeze between endothelial cells through slits narrower than itself, so a deformable cell passes while a rigid one is stranded among the cord macrophages. This single fact underlies the removal of aged red cells, red cell inclusions, and cells with membrane abnormalities, and it explains why hereditary spherocytosis is a splenic hemolysis, why Howell-Jolly bodies appear after splenectomy, and why bite cells form in G6PD deficiency.
169A patient with cirrhosis has a spleen measuring 22 cm and a platelet count of 68 ×10³/µL. Marrow examination is normal and no antiplatelet antibody is detected.Which best explains the thrombocytopenia?
- AImmune destruction by an undetected autoantibody
- BDecreased thrombopoietin production by the cirrhotic liver alone
- CSplenic sequestration of a large fraction of the platelet mass
- DConsumption within microvascular thrombi
- EMarrow suppression from portal hypertension
Answer
C
Normally one third of the platelet mass is in the spleen; with splenomegaly, 80–90% of platelets may be sequestered. The platelets are not destroyed, they are pooled — which is why the marrow is normal, no antibody is found, and the count corrects after splenectomy. This is hypersplenism, a mechanism of cytopenia distinct from destruction and underproduction, and it is why the correct first question about any low platelet count is why it is low. A describes ITP, which would show an increased marrow megakaryocyte population, and D describes DIC or TTP, which would show schistocytes. B contributes in cirrhosis but does not account for the splenic findings.
170A 58-year-old man with cirrhosis has splenomegaly. The spleen shows an expanded, beefy red pulp with loss of white pulp and fibrosis of vessels and sinusoids.Which mechanism produced these changes?
- AInfiltration by a low-grade lymphoma
- BExtramedullary hematopoiesis replacing normal parenchyma
- CObstruction of venous outflow with red pulp congestion
- DStorage material accumulating within cord macrophages
- ERepeated infarction with fibrous replacement
Answer
C
Congestive splenomegaly results from obstruction of venous outflow, producing red pulp congestion, and the described morphology is exactly that — red pulp expanded and beefy red with loss of white pulp, and fibrosis of vessels and sinusoids when long-standing. The causes are cirrhosis of the liver, portal or splenic vein thrombosis, and cardiac failure, all of which impede outflow. B occurs in myeloproliferative neoplasms and chronic anemia, A in lymphoma, and D in storage diseases such as Gaucher and the mucopolysaccharidoses — all genuine causes of splenomegaly, but none produces this specific congestive picture in a cirrhotic patient.
171A 19-year-old college student with infectious mononucleosis is counseled to avoid contact sports for several weeks.What is the rationale for this advice?
- AExercise accelerates viral replication within lymphoid tissue
- BThe spleen has enlarged rapidly, predisposing it to rupture
- CPhysical activity precipitates airway obstruction from tonsillar swelling
- DExertion worsens the associated hemolytic anemia
- EContact sport increases the risk of transmitting the virus
Answer
B
Trauma is the most common cause of splenic rupture, and “spontaneous rupture” occurs where a predisposing condition has caused rapid splenic enlargement — the listed examples being infectious mononucleosis, malaria, typhoid fever, and lymphoid neoplasms. The reason rapidity matters is that the capsule is stretched thin without time to remodel, and the enlarged organ descends below the costal margin where the ribs no longer shield it. By contrast, a slowly enlarging spleen develops a thickened, fibrotic capsule and may be far less fragile despite being much larger. A, C, D, and E are not the basis of the recommendation.
172A 24-year-old with sickle cell disease has acute left upper quadrant pain. Imaging shows a wedge-shaped peripheral hypodensity in the spleen.Which process does this represent?
- ASplenic abscess
- BSplenic infarct from vascular occlusion
- CSubcapsular hematoma from occult trauma
- DA littoral cell angioma
- EFocal extramedullary hematopoiesis
Answer
B
A wedge-shaped pale infarct is the classic morphology of splenic infarction, defined as an ischemic insult to the splenic parenchyma due to vascular occlusion. The listed causes are arterial thromboembolism, disruption of blood supply, sickle cell anemia, myeloproliferative neoplasms with extensive extramedullary hematopoiesis, vasculitis, and hypercoagulable states — and this patient has one of them. The wedge shape reflects the territory of an occluded end artery, which is why it is peripheral and based on the capsule. Repeated infarction in sickle cell disease is what ultimately produces autosplenectomy and the appearance of Howell-Jolly bodies. D is a rare vascular neoplasm rather than an ischemic lesion.
173A splenectomy specimen from a patient with isolated splenomegaly and no lymphadenopathy shows a lymphoid infiltrate.Which lymphoma most commonly arises primarily in the spleen?
- ASplenic marginal zone lymphoma
- BFollicular lymphoma
- CMantle cell lymphoma
- DBurkitt lymphoma
- EClassical Hodgkin lymphoma
Answer
A
Among splenic lymphomas, the primary entity is splenic marginal zone lymphoma, while secondary involvement by lymphoma arising elsewhere is more common overall. Marginal zone lymphoma has nodal, splenic, and extranodal varieties, the extranodal form being MALT lymphoma. Note that another lymphoid neoplasm centred on the spleen is hairy cell leukemia, which involves the red pulp with blood lakes and characteristically presents with splenomegaly, pancytopenia and monocytopenia without lymphadenopathy. B, C, D, and E all involve the spleen but characteristically present with lymphadenopathy and are not primary splenic diseases.
174A clinician lists the functions of the spleen for a teaching session.Which function explains the vulnerability of asplenic patients to specific pathogens?
- ASequestration of one third of the platelet mass
- BExtramedullary hematopoiesis in chronic anemia
- CAntibody production within the periarteriolar lymphoid sheath
- DPhagocytosis providing defense against encapsulated bacteria
- ERemoval of aged red cells from the circulation
Answer
D
The four splenic functions are antibody production, phagocytosis of blood cells and particulate matter — including defense against encapsulated bacteria, hematopoiesis in fetal life and as extramedullary hematopoiesis later, and sequestration of blood elements. It is specifically the phagocytic defense against encapsulated organisms that is lost after splenectomy or autosplenectomy, which is why those patients are at risk from pneumococcus and Haemophilus influenzae and why vaccination is required. A, B, C, and E are all genuine splenic functions, but none accounts for the specific organism susceptibility.
Lecture 17 · Randall
Thymus
Questions 175–182
175A neonate has hypocalcemic tetany, a conotruncal cardiac defect, and markedly decreased circulating T cells. Imaging shows an absent thymic shadow.Which genetic abnormality is expected?
- A11q23 rearrangement
- B5q deletion
- CTrisomy 21
- D22q11 deletion
- E22q11 duplication
Answer
D
This is DiGeorge syndrome, caused by a 22q11 deletion and characterized by thymic and parathyroid aplasia or severe hypoplasia with markedly decreased T cells, variable defects involving the heart and great vessels, and defects in cell-mediated immunity. The findings track the embryology: the thymus arises from the 3rd branchial pouch, and the parathyroids and cardiac outflow tract derive from adjacent structures, so one developmental field failure produces immunodeficiency, hypocalcemia, and a cardiac defect together. DiGeorge is the aplasia end of the spectrum; dysplasia occurs in SCID, ataxia-telangiectasia, and incomplete DiGeorge. B belongs to MDS and A to KMT2A-rearranged leukemias.
176A small, lymphocyte-depleted thymus is examined. The pathologist must decide between thymic dysplasia and acute thymic involution.Which feature best distinguishes them?
- AThe overall weight of the gland
- BThe number of B-cell follicles present
- CThe ratio of CD4 to CD8 thymocytes
- DThe presence of fatty replacement
- EThe presence of well-formed Hassall corpuscles
Answer
E
Dysplasia or aplasia is distinguished from acute thymic involution by the absence of well-formed Hassall corpuscles. The reasoning is developmental: in acute involution from stress or infection the organ developed normally and then shrank, so the medullary architecture was built and Hassall corpuscles remain. In dysplasia or aplasia the organ never formed properly, so there are no cortical and medullary zones and no Hassall corpuscles. That one structure separates a reversible stress response from a primary immunodeficiency. A fails because both are small. D describes normal age-related involution after puberty. B relates to a different entity entirely.
177A 34-year-old woman with fatigable ptosis and diplopia undergoes thymectomy. The gland contains secondary B-cell follicles with germinal centers.Which condition is most associated with this finding?
- ADiGeorge syndrome
- BMyasthenia gravis
- CSevere combined immunodeficiency
- DThymic carcinoma
- ETrue thymic hyperplasia
Answer
B
Thymic follicular hyperplasia is defined by the presence of secondary B-cell follicles with germinal center formation and is seen in 65% of myasthenia gravis patients, as well as in systemic lupus erythematosus, Graves disease, and rheumatoid arthritis. The finding is abnormal because the thymus is a T-cell organ with a cortex and medulla, not follicles — germinal centers are structures of a B-cell response, and their presence means autoantibody is being generated inside the organ responsible for enforcing self-tolerance. E is the deliberate contrast: true thymic hyperplasia is enlargement beyond normal weight for age with normal microscopy. A and C produce hypoplastic glands without follicles.
178A pathologist distinguishes true thymic hyperplasia from thymic follicular hyperplasia.Which statement is correct?
- ATrue thymic hyperplasia is defined by germinal center formation on routine microscopy
- BFollicular hyperplasia is defined by increased gland weight alone
- CTrue thymic hyperplasia is enlargement beyond normal weight with normal microscopy
- DBoth are defined by the presence of Hassall corpuscles
- EFollicular hyperplasia is never seen in children
Answer
C
True thymic hyperplasia is enlargement beyond the upper limits of normal weight for age, with normal microscopy, possibly reflecting failed involution. Thymic follicular hyperplasia, by contrast, is defined by microscopic architecture — secondary B-cell follicles with germinal center formation — rather than by weight, so A and C swap the two definitions. E is wrong because a few follicles are normal in children, which matters practically: the finding is significant only when follicles are numerous or occur in an adult. D is wrong because Hassall corpuscles are a normal medullary structure and do not define either form of hyperplasia.
179An anterior mediastinal mass is resected from a 15-year-old boy. It consists of sheets of immature lymphoid cells that are TdT positive, CD1a positive, and CD7 positive.Which is the most likely diagnosis?
- AThymoma
- BThymic carcinoma
- CPrimary mediastinal large B-cell lymphoma
- DT-lymphoblastic lymphoma
- EMediastinal teratoma
Answer
D
TdT and CD1a positivity identify a precursor T cell, and in an adolescent with an anterior mediastinal mass this is T-lymphoblastic lymphoma — the Teenagers with a Thymic mass pattern from Lecture 08. The reason it arises here is anatomic: the thymic cortex normally contains immature thymocytes that are TdT positive, CD1a positive, and double positive or negative for CD4/CD8, so the neoplasm appears where its normal counterpart lives. A and B arise from thymic epithelial cells rather than lymphoid cells. C and E are also anterior mediastinal masses, which is why they are listed, but neither would show this precursor T-cell phenotype.
180A 52-year-old man with a thymic mass is evaluated.Which presentation accounts for approximately 40% of thymomas?
- ADiscovery during evaluation of recurrent infection
- BPresentation with pure red cell aplasia
- CDiscovery during evaluation of hypercalcemia
- DPresentation with superior vena cava obstruction alone
- EDiscovery during evaluation of myasthenia gravis
Answer
E
Thymoma presentation splits three ways: 40% with symptoms related to impingement on mediastinal structures, 40% during evaluation of myasthenia gravis, and 20% incidental. So both the impingement group and the myasthenia group are 40%, and among the options given only the myasthenia route is stated correctly — D names a single specific impingement syndrome rather than the category. B is a genuine association, since patients may present with other autoimmune disorders or pure red cell aplasia, but it is not one of the three main routes. Roughly 40% of patients with thymoma have myasthenia gravis, which is the reciprocal of this statistic.
181A thymic epithelial tumor is diagnosed as squamous cell carcinoma of the thymus.Which statement is correct?
- AIt is associated with myasthenia gravis in about 40% of cases
- BIt represents the majority of thymic epithelial neoplasms
- CIt has a five-year survival exceeding 90%
- DIt is not associated with myasthenia gravis or autoimmune conditions
- EIt is best classified as a minimally invasive thymoma
Answer
D
Thymic carcinoma is not associated with myasthenia gravis or autoimmune conditions, which breaks the reflex link between the thymus and myasthenia that holds for thymoma. The difference tracks how much organ function is retained: a thymoma preserves enough thymic architecture and immature T cells to keep generating an aberrant autoimmune response, whereas a carcinoma has abandoned thymic organization entirely. B is wrong because thymic carcinoma is only 5% of thymic epithelial neoplasms, and it is usually squamous cell carcinoma among more than ten subtypes. C contradicts its 18 month median survival, and E misclassifies a frankly malignant tumor.
182A thymoma is found to extend through its capsule into surrounding mediastinal fat and pleura at multiple sites.Which prognosis is expected?
- AClinically benign behavior with no risk of recurrence
- BGreater than 90% five-year survival
- CA median survival of 18 months
- DLess than 50% five-year survival
- EPrognosis identical to a non-invasive thymoma
Answer
D
Thymomas are graded by invasion rather than by cytology. A non-invasive thymoma is cytologically benign and clinically benign. An invasive thymoma shows extension through the surrounding capsule, and outcome depends on extent: minimally invasive tumors have greater than 90% five-year survival, while extensively invasive tumors have less than 50%. Involvement of multiple sites beyond the capsule describes the extensive category, so B understates the risk and A and E ignore the invasion entirely. C is the 18 month median survival of thymic carcinoma, a different and more aggressive entity accounting for 5% of thymic epithelial neoplasms.