Questions from every lecture pool here and are not grouped or labeled. Work out what you are looking at before opening the answer. Renumber sequentially as you add more.
1A 58-year-old woman with rheumatoid arthritis has a hemoglobin of 9.8 and an MCV of 84. Serum iron is low and % saturation is low.Which additional result would most strongly favor anemia of chronic disease over iron deficiency?
- AElevated free erythrocyte protoporphyrin
- BElevated ferritin with decreased TIBC
- CDecreased ferritin with increased TIBC
- DAn elevated RDW
- EA normocytic rather than microcytic MCV
Answer
B
Both conditions give low serum iron and low % saturation, so those cannot separate them. Ferritin is the discriminator: high in ACD because hepcidin traps iron inside macrophages where ferritin stores it, and low in IDA because the stores are genuinely empty. TIBC moves opposite — down in ACD, up in IDA. C is the IDA pattern. A appears in both. D favors IDA if anything. E is unhelpful, since both start normocytic and become microcytic with time.
2A 4-year-old living in older housing has a microcytic anemia. The smear shows coarse basophilic stippling; marrow shows ringed sideroblasts.Which enzymes are inhibited?
- AALA synthase only
- BPorphobilinogen deaminase and uroporphyrinogen decarboxylase
- CMethionine synthase and methylmalonyl-CoA mutase
- DALA dehydratase and ferrochelatase
- EFerroportin and transferrin
Answer
D
Lead poisoning inhibits ALA dehydratase and ferrochelatase — effectively the second step and the final step of heme synthesis. Because ferrochelatase is what inserts Fe2+ into protoporphyrin, iron accumulates in mitochondria of erythroid precursors, giving ringed sideroblasts, while damaged ribosomal RNA aggregates produce basophilic stippling in circulating cells. A is the congenital X-linked sideroblastic anemia and also the step needing B6. B causes acute intermittent porphyria and porphyria cutanea tarda. C is the B12-dependent pair.
3A 62-year-old man has a macrocytic anemia with hypersegmented neutrophils, glossitis and an elevated homocysteine. Methylmalonic acid is normal.Which is true?
- AFolate deficiency, and neurologic findings would not be expected
- BB12 deficiency, most likely from pernicious anemia
- CB12 deficiency, and dorsal column findings should be sought
- DEither deficiency, since the labs shown cannot distinguish them
- ESideroblastic anemia from vitamin B6 deficiency
Answer
A
Normal methylmalonic acid is the finding that makes this folate. Both deficiencies impair DNA synthesis and raise homocysteine, because both feed the folate/methionine cycle — so macrocytosis, hypersegmented neutrophils and glossitis are shared. Only B12 also serves as adenosyl-B12 for methylmalonyl-CoA mutase, so only B12 deficiency raises MMA, and only B12 deficiency produces neurologic symptoms. D is wrong precisely because MMA does distinguish them.
4A 12-year-old with sickle cell disease has a baseline hemoglobin of 8 that drops to 4 over a week, with a reticulocyte count near zero.What is the most likely cause?
- AAcute splenic sequestration
- BSuperimposed iron deficiency from chronic hemolysis
- CParvovirus B19 infection of erythroid precursors
- DMyelophthisic replacement of the marrow by metastatic tumor
- EFolate deficiency from increased cell turnover
Answer
C
Parvovirus B19 infects erythroid precursors and inhibits erythropoiesis, and the near-zero reticulocyte count localizes the problem to production rather than destruction. It matters far more in patients with underlying marrow stress — sickle cell disease, post-HSCT — because those patients have short red cell survival and depend on continuous high output, so even a brief production halt causes a crash. A normal person with a 120-day red cell lifespan would barely notice. D would be expected to give pancytopenia and nucleated RBCs with granulocyte precursors on smear. E is a real risk in hemolysis but develops over weeks to months and would be macrocytic.
5A patient on isoniazid for latent tuberculosis develops a microcytic anemia with ringed sideroblasts in the marrow.Which is true regarding the mechanism and management?
- AIsoniazid chelates iron in the duodenum; treat with oral ferrous sulfate
- BIsoniazid inhibits ferrochelatase directly; the drug must be stopped permanently
- CIsoniazid causes folate antagonism; treat with folinic acid
- DIsoniazid suppresses EPO production; treat with exogenous EPO
- EIsoniazid causes vitamin B6 deficiency, and B6 is the cofactor for ALA synthase; treat with B6
Answer
E
Vitamin B6 (pyridoxine) deficiency is a listed side effect of isoniazid, and B6 is the cofactor for ALA synthase — the rate-limiting first step of heme synthesis. Blocking it starves protoporphyrin production, giving a sideroblastic picture, and it is treatable with vitamin B6. B misassigns the enzyme: ferrochelatase inhibition is lead. Note this is the one sideroblastic cause with a clean, specific fix.
6A 44-year-old woman has fatigue, dysphagia, and a beefy red tongue. Hemoglobin 9.1, MCV 72.Which is true?
- AThe triad describes Plummer-Vinson syndrome, and ferritin should be low
- BThe triad describes pernicious anemia, and MCV should be above 100
- CGlossitis excludes iron deficiency and points to B12 deficiency
- DKoilonychia would be inconsistent with this diagnosis
- EEsophageal webs indicate an underlying myelophthisic process
Answer
A
Plummer-Vinson syndrome is iron deficiency anemia + esophageal webs (dysphagia) + atrophic glossitis (beefy red tongue), and the microcytic MCV fits — so expect low ferritin, high TIBC, low serum iron, low % saturation, high RDW. C is the trap: glossitis occurs in both iron deficiency and the megaloblastic anemias, so it does not discriminate. D is wrong — koilonychia and pica are the other classic iron deficiency signs.
7A 70-year-old man with known prostate cancer has pancytopenia. The peripheral smear shows nucleated red blood cells and immature granulocytes.What does the smear finding indicate?
- AAn appropriate marrow response to hemolysis
- BAplastic anemia with a marrow void of hematopoietic elements
- CMegaloblastic maturation arrest in the marrow
- DA myelophthisic process replacing the marrow space
- EParvovirus B19 infection of erythroid precursors
Answer
D
Myelophthisic processes replace the marrow space with something else — usually metastatic cancer — often producing pancytopenia, and characteristically forcing marrow elements not normally present in blood into circulation: nucleated RBCs and neutrophil precursors. That smear finding is what separates it from aplastic anemia, where the marrow is empty rather than crowded and the smear shows no such immature forms despite equally low counts.
8A patient with a hematocrit of 21% (normal 45%) has a reticulocyte count of 4%.How should this be interpreted?
- AThe reticulocyte index is about 1.9, indicating an inadequate marrow response
- BThe reticulocyte index is about 8.6, indicating brisk marrow response
- CThe raw 4% is appropriate and needs no correction
- DThe reticulocyte index cannot be calculated without a ferritin level
- EAn index in this range establishes hemolysis as the cause
Answer
A
RI = % reticulocytes × (actual Hct / normal Hct) = 4 × (21/45) ≈ 1.9. Since <2 is an inadequate response and >3 is appropriate, this marrow is underperforming — a production problem. The correction exists precisely because the raw percentage is a fraction of a shrunken denominator, so 4% looks reassuring but represents little absolute output. Note that at this degree of anemia the RPI would be the more accurate calculation, since it also corrects for the prolonged maturation time of prematurely released reticulocytes.
9A patient underwent partial gastrectomy two years ago and now has a microcytic anemia.Which mechanism best explains the iron deficiency?
- ALoss of intrinsic factor production by parietal cells
- BLoss of gastric acid, which normally maintains iron in the better-absorbed ferrous Fe2+ state
- CIncreased hepcidin from surgical inflammation blocking ferroportin
- DLoss of terminal ileal absorptive surface
- EDecreased transferrin synthesis by the liver
Answer
B
Stomach acid increases iron absorption by keeping iron reduced as ferrous Fe2+, which is absorbed better than oxidized ferric Fe3+. Remove the acid-producing stomach and absorption falls; duodenectomy does the same by removing the site of absorption itself. A is a real consequence of gastrectomy, but intrinsic factor loss causes B12 deficiency and a macrocytic anemia. D also causes B12 deficiency. Note that iron is absorbed in the duodenum and exported into blood by ferroportin.
10A 55-year-old woman has a macrocytic anemia, decreased vibration sense in both feet, and a spastic gait.Which is true regarding the most likely underlying cause?
- AMethylmalonic acid would be expected to be normal
- BBody stores are minimal, so the deficiency developed over weeks
- CAutoimmune destruction of gastric parietal cells causing intrinsic factor deficiency is the most common cause
- DHomocysteine would be expected to be normal
- EDietary insufficiency is the most common cause in non-vegans
Answer
C
Neurologic findings make this B12, and pernicious anemia — autoimmune destruction of parietal cells → intrinsic factor deficiency → failure of B12 absorption in the small bowel — is the most common cause of B12 deficiency. The signs map to the tracts: dorsal columns give the lost proprioception and vibration sense, lateral corticospinal tracts give the spastic paresis. A and D are inverted; both MMA and homocysteine rise in B12 deficiency. B describes folate, whose stores are minimal — B12 has large hepatic stores, so dietary deficiency takes years and vegans are the exception.
11A patient started on a new medication develops pancytopenia. Marrow biopsy shows a hypocellular space largely replaced by fat, with no infiltrate.Which is true?
- AThis is aplastic anemia; management includes stopping the causative drug, supportive care, and HSCT in some cases
- BThis is a myelophthisic process and warrants a search for occult malignancy
- CExogenous EPO alone would be expected to correct the pancytopenia
- DIron studies would show an overload pattern
- EThe peripheral smear should show nucleated red cells and granulocyte precursors
Answer
A
Aplastic anemia is damage to stem cells from medications, infections, or autoimmune/abnormal T-cell activity, with some idiopathic cases. The marrow is void of hematopoietic elements, giving pancytopenia, and treatment is to discontinue the causative medication, provide supportive care, and consider HSCT. B and E describe the myelophthisic alternative — but the biopsy here shows an empty marrow, not one crowded out by tumor, and it is that distinction the question turns on. C is wrong: EPO drives erythropoiesis only, and the problem is the stem cell compartment across all lineages.
12Which sequence correctly describes the progression of untreated iron deficiency?
- ASerum iron falls first, then ferritin, with a microcytic anemia from the outset
- BTIBC falls as stores deplete, and the anemia is macrocytic early
- CFerritin depletes first with a rise in TIBC, then serum iron falls and % saturation drops, and the anemia is normocytic early before becoming microcytic and hypochromic
- DAll iron studies change simultaneously once anemia is established
- E% saturation rises early as transferrin increases
Answer
C
Stores go first: ferritin depletes and TIBC rises (TIBC measures transferrin molecules, which the body upregulates to scavenge harder). Then serum iron falls and % saturation drops — % saturation being the fraction of transferrin molecules actually carrying iron, so it falls both because iron is scarce and because the denominator grew. Only then does anemia appear, and it is normocytic early, becoming microcytic and hypochromic as hemoglobin synthesis fails. E is backwards for the same reason: more transferrin with less iron means a lower saturation.
13A patient with metastatic cancer has a normocytic anemia. Serum iron is low, ferritin is elevated, and TIBC is low.Which mechanism accounts for the low serum iron?
- ADepleted total body iron stores from occult blood loss
- BFailure of ferrochelatase to incorporate iron into protoporphyrin
- CImpaired transferrin synthesis reducing iron transport capacity
- DHepcidin binding ferroportin on enterocytes and macrophages, trapping iron intracellularly
- EAutoimmune destruction of gastric parietal cells
Answer
D
This is anemia of chronic disease, and inflammation raises hepcidin, an acute phase reactant. Hepcidin binds ferroportin — the only export channel for iron — on intestinal cells and macrophages, so absorbed and recycled iron cannot reach the blood. Serum iron falls while ferritin rises, because the iron is stranded in storage. A would give low ferritin. B is sideroblastic anemia, which produces an iron overload pattern with high serum iron. Note her slide also credits hepcidin with suppressing EPO production, a second contributor to the anemia.
14A 19-year-old with a family history of anemia and gallstones has hemoglobin 11.2, MCV 82.6, MCHC 36.5 (high), RDW 16.5 (high). The smear shows small dense red cells lacking central pallor.Which best explains the elevated MCHC?
- AIncreased hemoglobin synthesis per cell driven by high erythropoietin
- BExtra marrow cell divisions producing undersized cells
- CRelative cell dehydration from loss of K+ and water, concentrating the same hemoglobin in a smaller volume
- DPrecipitation of denatured globin within the cell
- EReticulocytosis, since reticulocytes carry more hemoglobin
Answer
C
This is hereditary spherocytosis, and her slide gives the mechanism directly: relative cell dehydration — the cell loses K+ and water, so the same amount of hemoglobin occupies a smaller volume, raising the concentration. Note the MCV of 82.6 is normal, which is the point: MCHC is a ratio and moves detectably, while MCV has a wide 80–100 range and stays inside it. B is the mechanism of iron deficiency, where cells are built undersized because hemoglobin is scarce — that lowers MCHC rather than raising it. D describes Heinz bodies in G6PD deficiency.
15Two patients both have hemolytic anemia with numerous spherocytes on smear.Which test best distinguishes hereditary spherocytosis from warm autoimmune hemolytic anemia?
- AOsmotic fragility test
- BReticulocyte count
- CSerum haptoglobin
- DDirect antiglobulin (Coombs) test
- EHemoglobin electrophoresis
Answer
D
Her HS slide states explicitly that spherocytes are seen in HS and WAHA, so morphology cannot separate them. The DAT asks the one question that does: is antibody or complement bound to the red cell surface? Positive in WAHA, negative in HS. The reason both produce spherocytes is mechanical — splenic macrophages remove part of the membrane, whether it was weakened by a protein defect or coated with autoantibody, and the cell rounds up. A is abnormal in both, since it detects the spherical shape rather than its cause. B and C confirm hemolysis without localizing the mechanism.
16In the osmotic fragility test, red cells are incubated in saline solutions of progressively decreasing concentration.Why do spherocytes lyse at higher salt concentrations than normal cells?
- AHaving lost membrane, a sphere has no surface-area reserve to accommodate the water it takes up
- BSpherocytes have abnormally high intracellular sodium, drawing in water faster
- CSpherocytes lack the ATP needed to maintain membrane integrity
- DAntibody bound to the spherocyte surface fixes complement in hypotonic media
- ESpherocytes are larger, so they reach their lytic volume sooner
Answer
A
A normal biconcave disc is, as her early slide puts it, a compromise that provides surface area beyond the minimum needed to enclose its volume — so in hypotonic saline it can swell toward a sphere before the membrane is stressed. A spherocyte has already lost that reserve; it is at minimum surface area for its contents, so any water entry immediately raises tension and it ruptures. Hence her stated principle: spherocytes are more sensitive to osmotic lysis and lyse sooner than normal RBCs. E is wrong on the facts — spherocytes are not larger; MCV is low-to-normal.
17A patient with hereditary spherocytosis undergoes splenectomy. Hemoglobin improves, and the smear now shows small round dark inclusions within occasional red cells.What are these, and why did they appear?
- AHeinz bodies, from oxidative denaturation of hemoglobin
- BHowell-Jolly bodies — DNA remnants, no longer removed because the spleen is gone
- CBasophilic stippling, from aggregates of damaged ribosomal RNA
- DRinged sideroblasts, from iron-laden mitochondria
- ESchistocytes, from shear injury during surgery
Answer
B
Howell-Jolly bodies are remnants of DNA, and they appear after splenectomy because the spleen is what normally pits such inclusions out of circulating cells. Their presence is effectively a marker of asplenia or hyposplenism from any cause — which is why they also show up in sickle cell disease after autosplenectomy. Note the therapeutic logic: splenectomy is often helpful in HS because it increases the lifespan of the abnormal cells — you are removing the organ doing the destroying, not fixing the membrane. A and C are the inclusions of G6PD deficiency and sideroblastic anemia respectively.
18A man of Mediterranean descent develops acute hemolysis three days after starting a sulfonamide antibiotic.Which sequence describes the underlying mechanism?
- ADrug binds a membrane protein → IgG forms against the drug-membrane complex → splenic clearance
- BAbnormal globin polymerizes on deoxygenation → membrane damage → hemolysis
- CLoss of the GPI anchor → absent CD55/CD59 → complement-mediated lysis
- DFree α-globin chains precipitate → membrane damage → extravascular hemolysis
- EReduced NADPH → less reduced glutathione → H2O2 not cleared → Heinz bodies → membrane damage
Answer
E
This is G6PD deficiency, and the chain is worth reciting in order: reduced G6PD activity gives less NADPH, which means less conversion of oxidized to reduced glutathione; since removing H2O2 requires reduced glutathione plus glutathione peroxidase, the peroxide accumulates and denatures hemoglobin, which precipitates as membrane-bound Heinz bodies that damage the membrane. Result can be extravascular or intravascular hemolysis. The triggers to recognize: antimalarials (primaquine, chloroquine), sulfonamides, nitrofurantoins, fava beans, and infections. A is the hapten drug mechanism, an immune process with a positive DAT — a distinct scenario despite also being drug-triggered.
19A smear drawn after an oxidant exposure shows cells with semicircular defects at their periphery, and a supravital stain reveals intracellular precipitates.Which is true?
- AThe defects are the result of fibrin strand shearing
- BThe precipitates are visible on routine Wright-Giemsa staining
- CSplenic macrophages produced the defects by removing the inclusions, and the disorder is X-linked recessive
- DThe inheritance is autosomal dominant with 75% penetrance
- ETreatment consists of replacing the deficient enzyme
Answer
C
Bite cells form because splenic macrophages pick out the Heinz body inclusions, taking a bite of membrane with them, and G6PD deficiency is X-linked recessive — so clinically it presents in males. B is the trap: Heinz bodies require supravital staining and are not seen on a routine smear, which is why the question specifies it. E is wrong and worth remembering — treatment is preventive, avoiding oxidant exposure; there is no enzyme to replace. A describes schistocytes in MAHA, which are fragments rather than bitten cells.
20Routine screening in a patient of West African ancestry shows numerous elliptical red cells. Hemoglobin, reticulocyte count, bilirubin and LDH are all normal.What is the appropriate interpretation?
- AHereditary elliptocytosis, most often from α-spectrin mutations, autosomal dominant, and requiring no treatment
- BHereditary pyropoikilocytosis, which is autosomal dominant and requires transfusion support
- CEarly sickle cell disease, since irreversibly sickled cells can appear elliptical
- Dβ-thalassemia trait, given the ancestry and abnormal cell shape
- EAn artifact of slide preparation requiring a repeat smear
Answer
A
Hereditary elliptocytosis comes most commonly from mutations in the α-spectrin genes (SPTA1, SPTB), is autosomal dominant, and is common in equatorial Africa because it confers malaria resistance. The critical clinical point is that the majority of patients are NOT anemic — elliptocytes are an incidental finding and treatment is not necessary, since HE is uncommonly associated with hemolysis. The entirely normal hemolysis panel here fits that exactly. B inverts two facts: pyropoikilocytosis is the severe form, is autosomal recessive, and does cause hemolysis and anemia.
21Which statement about the sickle cell mutation is correct?
- AA deletion in the α-globin gene cluster on chromosome 16
- BA splice-site mutation reducing but not eliminating β-globin synthesis
- CA point mutation replacing valine with glutamate, altering oxygen affinity
- DA point mutation in the 6th codon of the β-globin gene replacing glutamate with valine, which changes protein charge and electrophoretic mobility
- EA somatic mutation in the PIGA gene acquired by a hematopoietic stem cell clone
Answer
D
Point mutation in the 6th codon of the β-globin gene, glutamate (glutamic acid) → valine. The direction matters — C has it backwards. The consequence worth linking: because glutamate is charged and valine is not, the protein's charge changes, which changes electrophoretic mobility and is precisely what makes hemoglobin electrophoresis diagnostic (SS = disease, AS = trait, AA = normal). Note also the contrast in category: sickle cell is a hemoglobinopathy — an abnormal globin — whereas B describes a thalassemia, which is decreased synthesis of a normal globin. A is α-thal and E is PNH.
22A 3-year-old with sickle cell disease presents with sudden abdominal distension, a rapidly enlarging spleen, hemoglobin falling from 8 to 4, and hypotension.Which is true?
- AThis is autosplenectomy, and pneumococcal prophylaxis is the priority
- BThis is splenic sequestration crisis, a potentially life-threatening event from trapping of RBCs with hypovolemia and circulatory collapse
- CThis is an aplastic crisis and the reticulocyte count will be elevated
- DThis is microangiopathic hemolysis and schistocytes should be sought
- EThe spleen in sickle cell disease is invariably small and fibrotic at this age
Answer
B
Splenic sequestration crisis: a sudden enlargement of the spleen due to trapping of RBCs, with a rapid drop in Hb/Hct plus hypovolemia and circulatory collapse — potentially life-threatening because blood volume is being lost into the spleen, not just red cell mass. The spleen is congested with sinuses dilated by RBCs. E and A capture the later stage: repeated infarction produces autosplenectomy — small, shrunken, fibrotic, non-functional — leaving susceptibility to encapsulated organisms (pneumococci, H. influenzae). The sequence is congestion and sequestration first, autoinfarction later, so a young child can still have a spleen large enough to sequester. C is wrong in two ways: aplastic crisis has no splenic enlargement, and the retic count would be low.
23Which pairing of a sickle cell treatment strategy with its rationale is correct?
- AIron supplementation — to correct the chronic anemia
- BMaintaining a mild metabolic acidosis — to reduce polymerization
- CHydroxyurea — to reduce HbS by suppressing β-globin transcription
- DTransfusion to keep HbS above 30% — to maintain oxygen delivery
- EMaintaining hydration — because dehydration raises MCHC and promotes polymerization
Answer
E
Avoid dehydration, which decreases MCHC — her stated goal of reducing HbS per cell. The logic is the self-amplifying cycle: sickling drives out K+ and water, raising intracellular hemoglobin concentration, and a higher concentration makes polymerization easier on the next deoxygenation. Keeping the cell hydrated dilutes the polymerizing species. C has the right drug and the wrong mechanism — hydroxyurea increases HbF, which contains no β chains and therefore cannot join HbS polymer (the same reason newborns are protected). D inverts the target: transfusion aims to keep HbS below 30%. B is backwards — acidosis promotes sickling, so intracellular pH should be maintained.
24Two patients are microcytic and hypochromic. Patient 1: MCV 62, RBC count elevated, RDW normal. Patient 2: MCV 76, RBC count low, RDW elevated.Which interpretation is correct?
- ABoth are iron deficient; patient 1 is simply further along
- BPatient 1 has iron deficiency and patient 2 has thalassemia trait
- CPatient 1 has thalassemia and patient 2 has iron deficiency, and the RDW plus RBC count are what distinguish them
- DNeither pattern is consistent with thalassemia, since thalassemia raises the RDW
- EOnly hemoglobin electrophoresis can distinguish these, since the indices overlap completely
Answer
C
Her thalassemia lab slide gives the pattern: RBC count increased (“lots of small cells”), MCV decreased — think thalassemia if MCV <67, MCH and MCHC decreased, and critically a NORMAL RDW. The RDW is normal because the genetic defect is present in every cell, so the population is uniformly small. In iron deficiency the RDW is high, because normal cells made before the iron ran out still circulate alongside new small ones, and the RBC count is low. Her β-thal minor slide adds the practical instruction: exclude iron deficiency. Electrophoresis is confirmatory (↑HbA2 in β-thal trait), but the CBC already separates them.
25Which correctly matches the β-thalassemia mutation type to its phenotype and explains why?
- AChain terminator mutations cause β0, because a premature stop codon yields a truncated, degraded globin with no usable product; splicing mutations cause β+, because the correct splice site is still used part of the time
- BChain terminator mutations cause β+, because translation of a shortened chain still yields partially functional globin
- CSplicing mutations cause β0, because all transcripts are aberrantly spliced and degraded
- DBoth types cause β0; the β+ phenotype results only from large gene deletions
- ENeither type applies to β-thalassemia, which is caused predominantly by gene deletions
Answer
A
Chain terminator → β0; splicing → β+. A premature stop codon is all-or-nothing: the truncated globin cannot assemble into hemoglobin and is degraded, so nothing usable comes off that allele. Splicing is competitive: the weakened or alternative site is used most of the time, but the correct site is still used some of the time, leaving a minority of normal mRNA — hence reduced but present output. E states the contrast backwards: β-thal is mostly point mutations, while α-thal is mostly gene deletions.
26A child with β-thalassemia major who has never been transfused is found to have iron deposition in the liver on a Prussian blue stain.What is the mechanism?
- AHepcidin-mediated trapping of iron within hepatic macrophages
- BFailure of ferrochelatase to incorporate iron into protoporphyrin
- CRepeated intravascular hemolysis delivering free hemoglobin to the liver
- DIneffective erythropoiesis driving increased dietary iron absorption
- ECoexistent hereditary hemochromatosis, which is required to produce overload without transfusion
Answer
D
On her Cooley's anemia pathogenesis figure, the arrow from ineffective erythropoiesis leads to increased iron absorption from the stomach, then to iron deposition in heart, liver and pancreas — systemic iron overload (secondary hemochromatosis). Because free α chains precipitate and kill erythroblasts, most die in the marrow; that futile erythroid activity signals for more iron even though iron was never the problem. Transfusion then adds a second, independent iron burden, which is why chronically transfused patients need chelation — but the question specifies no transfusions, and overload still occurs. B is sideroblastic anemia. A describes anemia of chronic disease, where serum iron is low.
27A newborn is severely anemic with generalized edema, ascites and marked hepatosplenomegaly. Hemoglobin electrophoresis shows tetramers of γ globin.Which is true?
- AThree of four α-globin genes are deleted, producing HbH disease
- BAll four α-globin genes are deleted; the γ4 tetramers have high oxygen affinity, and death in utero is usual
- CThis is β-thalassemia major, and HbA2 would be markedly elevated
- DTwo α-globin genes are deleted, giving thalassemia trait with mild anemia
- EThe tetramers are composed of excess β globin, which cannot bind oxygen
Answer
B
Hb Bart hydrops fetalis syndrome — all four α genes deleted (––/––), with tetramers of excess γ globin having high O2 affinity, giving severe anemia, generalized edema, ascites, marked hepatosplenomegaly, skeletal and cardiovascular malformations, and usually death in utero. The high oxygen affinity is the lethal detail: γ4 binds oxygen but will not release it to tissues, so the fetus is hypoxic despite carrying hemoglobin. The tetramer identity tracks gene count — 3 genes deleted gives HbH disease with β4 tetramers (A and E describe that instead), because after birth β is the abundant unpaired chain, while in the fetus it is γ.
28A microcytic patient has an elevated HbA2 of 6% on electrophoresis.What does this indicate?
- Aα-thalassemia trait, since HbA2 rises with any reduction in α chains
- BIron deficiency anemia, which characteristically elevates HbA2
- Cβ-thalassemia major, in which HbA2 is the predominant hemoglobin
- DSickle cell trait, since HbS migrates with HbA2
- Eβ-thalassemia trait or intermedia; HbA2 is not increased in α-thalassemia
Answer
E
Her lab slide states it directly: increased HbA2 in β-thal trait and β-thal intermedia, but NOT in α-thal, with the β-thal minor range given as HbA2 4–8% against a normal 1.5–4%. This is what makes electrophoresis useful for sorting the two thalassemias, since both give the same microcytic CBC. C is the exception worth noting on her electrophoresis table: in β-thal major, HbA2 is only trace — HbF at 70–90% dominates the percentages. So elevated HbA2 points to the heterozygous or intermediate states, not major.
29A 30-year-old has chronic hemolytic anemia, pancytopenia, and a deep vein thrombosis. Flow cytometry shows red cells and granulocytes lacking CD55 and CD59.Which is true?
- AA somatic PIGA mutation prevents GPI anchor formation, thrombosis is the most common cause of mortality, and eculizumab reduces hemolysis
- BThe disorder is inherited in an X-linked recessive pattern and treated by avoiding oxidants
- CThe direct antiglobulin test should be strongly positive, since the hemolysis is complement-mediated
- DCorticosteroids and rituximab are first-line, as in warm autoimmune hemolytic anemia
- EThe defect is in spectrin and ankyrin, and splenectomy is curative
Answer
A
PNH: a somatic mutation in PIG-A, required to form the red cell GPI anchor, so anchor-dependent proteins including CD55 (decay accelerating factor) and CD59 (membrane inhibitor of reactive lysis, which inhibits C3 convertase) cannot attach — leaving cells sensitive to complement-mediated lysis. The triad is chronic hemolytic anemia, thrombosis (the most common cause of mortality), and pancytopenia from marrow failure. Treatment is marrow transplant or anti-C5 antibody (eculizumab); median survival 10 years. C is the instructive distractor: the hemolysis involves complement but no antibody, so the DAT is negative — PNH is classified as a non-immune, acquired clonal disorder.
30An older adult develops hemolytic anemia and acrocyanosis of the fingers in cold weather. The smear shows clumped red cells, and the DAT is positive for complement but negative for IgG.Which is true?
- AThis is warm autoimmune hemolytic anemia; the clumping reflects IgG cross-linking at 37°C
- BThe clumping is rouleaux, indicating a paraprotein from plasma cell myeloma
- CCold agglutinin syndrome from an IgM antibody, often anti-I; a workup for lymphoma or monoclonal gammopathy is warranted
- DParoxysmal cold hemoglobinuria from an IgG anti-P antibody
- EHereditary spherocytosis, since cold exposure precipitates splenic trapping
Answer
C
Cold agglutinin syndrome — usually IgM, often against I or i antigens, with the thermal amplitude of the antibody determining severity, hence symptoms in fingers, toes and cold-exposed areas. The DAT pattern is the tell: it detects complement rather than IgG, because IgM elutes as the blood rewarms while the complement it deposited remains bound. In an older adult, her slide directs you toward monoclonal gammopathies — plasma cell myeloma, lymphomas; the infection-associated form (Mycoplasma pneumoniae, EBV, CMV, influenza, HIV) is usually self-limited. A is excluded because in WAHA the cells are NOT stuck together. B describes the other cause of cells adhering — rouleaux, stacked from paraproteins rather than clumped by antibody.
31A child develops cola-colored urine after playing outside in the cold, several weeks after a viral illness. Serum testing shows an antibody that binds red cells at low temperature and lyses them on warming.Which is true?
- AThe antibody is IgM against the I antigen, and corticosteroids are first-line therapy
- BThis is warm autoimmune hemolytic anemia and splenectomy should be considered
- CThe hemolysis is extravascular, occurring in splenic macrophages
- DThis is the Donath-Landsteiner biphasic hemolysin — IgG against the P antigen — causing intravascular hemolysis; it is self-limited and corticosteroids are not helpful
- EThis is microangiopathic hemolysis and schistocytes will dominate the smear
Answer
D
Paroxysmal cold hemoglobinuria, mediated by the Donath-Landsteiner antibody — an IgG autoantibody against the P antigen, historically linked to syphilis and now typically following viral infection in children. It is a biphasic hemolysin: antibody and early complement components bind at low temperature, then terminal complement components form on warming and lyse the cells intravascularly — which is why the presentation is hemoglobinuria, and why C is wrong. Management is the deliberate contrast with WAHA: self-limited, avoid cold, corticosteroids are not helpful. A confuses it with cold agglutinin syndrome, which is IgM anti-I.
32A patient with sepsis develops anemia, thrombocytopenia, and numerous red cell fragments on smear.Which best describes the mechanism and the expected laboratory pattern?
- ASplenic macrophages partially removing antibody-coated membrane, producing spherocytes with a positive DAT
- BRed cells sheared on dense fibrin strands, producing schistocytes with markedly elevated LDH, very low haptoglobin, and hemoglobinuria
- COxidative denaturation of hemoglobin producing Heinz bodies and bite cells
- DComplement-mediated lysis from absent CD55 and CD59
- EPrecipitation of unpaired globin chains within erythroblasts causing marrow death
Answer
B
Microangiopathic hemolytic anemia — red cells are damaged on contact with dense fibrin strands, giving schistocytes (red cell fragments). Causes: DIC (the fit here, with sepsis and thrombocytopenia), TTP, and HUS. Because the mechanism is shearing inside the vessel rather than phagocytosis, the whole cytoplasm spills into plasma — hence the intravascular pattern of markedly high LDH, very low haptoglobin, and hemoglobinuria. The useful morphologic rule: fragmented cell means shear; rounded cell means spleen. A is WAHA, C is G6PD, D is PNH, E is thalassemia.
33Three patients develop immune hemolytic anemia on medication: one on penicillin, one on quinidine, one on α-methyldopa.Which statement is correct?
- AAll three act by the hapten mechanism, in which the drug binds a membrane protein
- BAll three generate IgM antibodies requiring the drug to be present for hemolysis to occur
- CQuinidine acts by autoantibody formation with apparent specificity for Rh antigens
- DPenicillin acts by immune complex formation, with the drug-antibody complex adsorbing to red cells
- EPenicillin is hapten/drug adsorption, quinidine is immune complex (ternary), and α-methyldopa induces true autoantibodies like those in WAHA
Answer
E
Her three mechanisms, matched to their drugs: hapten/drug adsorption — the drug binds an RBC membrane protein and IgG forms against the drug-membrane complex (penicillin, cephalosporins, tetracycline); immune complex / ternary complex — an IgM antibody forms on exposure and on further exposure the drug-antibody complex adsorbs to RBCs (quinidine, quinine, chlorpropamide); and autoantibody formation — true autoantibodies (IgG) like those in WAHA, with apparent specificity for Rh antigens (α-methyldopa). Treatment across all three is identifying and stopping the offending drug, and note her warning that drugs can cause severe intravascular hemolysis. The conceptual difference: the first two need the drug present, while the third has created a genuinely self-directed antibody that can persist.
34A 9-year-old develops palpable purpura over the buttocks and lower legs two weeks after an upper respiratory infection, along with abdominal pain and joint pain. Platelet count 280,000, PT and aPTT normal.Which best explains the bleeding?
- AAn IgG autoantibody against platelet GPIIb-IIIa
- BA qualitative platelet function defect
- CImmune complex–mediated vessel wall injury
- DConsumption of clotting factors and platelets
- EDeficiency of a vWF-cleaving metalloprotease
Answer
C
This is Henoch-Schönlein purpura, an immune complex vasculitis and one of the “non-thrombocytopenic purpuras.” The diagnostic signature is purpura with a completely normal panel — normal platelet count and normal PT/aPTT — because no routine test evaluates the vessel wall. A is ITP (would give thrombocytopenia). B is possible in principle but does not explain the post-infectious vasculitic syndrome with abdominal and joint findings. D is DIC (PT and aPTT would be prolonged). E is TTP (would give thrombocytopenia and schistocytes).
35A patient's routine CBC shows a platelet count of 62,000/µL. He reports no bruising, no bleeding, and no petechiae.Which is true?
- AThe absence of symptoms is inconsistent with this count and the result is likely spurious
- BThis is the expected finding, since counts in the 50,000–100,000 range are typically asymptomatic
- CHe should be transfused prophylactically to prevent spontaneous hemorrhage
- DSpontaneous bruising and menorrhagia are expected at this count
- EBleeding risk correlates with platelet function rather than platelet count in all cases
Answer
B
Her thresholds are worth memorizing: 50,000–100,000 → nothing; 30,000–50,000 → bruising with minor trauma; 10,000–30,000 → spontaneous bruising and menorrhagia; <10,000 → spontaneous bleeding into gums and gut, nosebleeds. D describes the 10,000–30,000 range. C is wrong — prophylactic transfusion is not indicated, and in ITP specifically it has no role at any count in a non-bleeding patient. E overstates a real point; count and function both matter.
36A 28-year-old woman has isolated thrombocytopenia of 22,000 with petechiae. Hemoglobin and white count are normal, PT and aPTT are normal, and the smear shows large platelets. A marrow exam is performed.What finding is expected, and why?
- ADecreased megakaryocytes, because the autoantibody suppresses megakaryocyte development
- BA hypocellular marrow largely replaced by fat
- CMegakaryocytes replaced by an infiltrating process
- DIncreased megakaryocytes, because the defect is peripheral destruction and the marrow is compensating
- ENormal megakaryocytes with ringed sideroblasts in the erythroid line
Answer
D
ITP is a destructive thrombocytopenia — IgG coats platelet glycoproteins and macrophages remove them — so the marrow responds by increasing megakaryocytes. The large platelets on smear reflect the same thing: young platelets released under drive are large. The marrow's other purpose here is exclusion — ruling out infiltration, leukemia, lymphoma, and MDS, which matters because ITP is a diagnosis of exclusion. B is aplastic anemia, C is a myelophthisic process, E belongs to sideroblastic anemia.
37A patient with known chronic ITP has a platelet count that has fallen to essentially zero. He has no bleeding of any kind.Regarding platelet transfusion, which is correct?
- AThere is no role for prophylactic transfusion, because the autoantibody will destroy transfused platelets as rapidly as the patient's own
- BTransfusion is mandatory at counts below 10,000 regardless of bleeding
- CTransfusion is indicated because it will suppress autoantibody production
- DTransfusion is contraindicated because it will precipitate thrombosis
- ETransfusion should be given together with heparin to prevent clotting
Answer
A
Her slide states it explicitly: no role for prophylactic platelet transfusion, even with ZERO platelets, in a non-bleeding patient. The mechanism explains the rule — the circulating autoantibody targets a glycoprotein present on donor platelets too, so they are cleared just as fast. Transfusion is reserved for active serious bleeding. D describes the reasoning in HIT and TTP, where transfusion may worsen thrombosis — a different mechanism, and worth keeping separate. Treatment here is steroids, rituximab, IVIG or anti-D, splenectomy, or a TPO receptor agonist.
38A 5-year-old develops sudden petechiae and a platelet count of 18,000 ten days after a viral illness. He is otherwise well, with no organomegaly.Which is true regarding this presentation compared with the adult form?
- AIt is more likely to become chronic with relapses than the adult form
- BCorticosteroids are mandatory to prevent intracranial hemorrhage
- CSplenectomy is the treatment of choice in this age group
- DOnset is characteristically insidious rather than abrupt
- EOnset is abrupt, typically follows a viral illness, and spontaneous permanent remission is expected
Answer
E
Acute childhood ITP is clinically a different disease from the adult chronic form: abrupt onset, usually following a viral illness, usually managed by observation without steroids, and characterized by spontaneous and permanent remission. The contrast she draws is with adults, who often have chronic disease with relapses — so A and D invert the picture, and B and C over-treat a self-limited illness.
39A postoperative patient on unfractionated heparin has a platelet count fall from 240,000 to 85,000 on day 7, and develops a new deep vein thrombosis.Which best explains why this patient clots rather than bleeds?
- AHeparin directly cross-links platelets, forming aggregates
- BThe antibody binds the platelet FcγRIIa receptor and activates platelets, which release procoagulant microparticles including thrombin, before they are consumed
- CThe falling count reflects marrow suppression, and thrombosis is coincidental
- DUltra-large vWF multimers accumulate and aggregate platelets in the microcirculation
- EAntithrombin is consumed, leaving thrombin unopposed
Answer
B
The order of events is the whole answer in HIT. IgG against heparin-PF4 binds FcγRIIa and activates platelets; activated platelets release procoagulant microparticles, including thrombin; only then does the count fall as platelets are consumed and destroyed. Because activation precedes destruction, the thrombocytopenia is a marker of consumption, not a bleeding risk. Roughly 1/3 develop thrombosis, and 1/3 of those are amputated or die. D is the TTP mechanism.
40Two patients on heparin develop thrombocytopenia. Patient 1: count falls to 130,000 on day 2 and recovers while heparin continues, with no thrombosis. Patient 2: count falls to 55,000 on day 7 and does not recover until heparin is stopped.Which interpretation is correct?
- ABoth represent Type II HIT, differing only in severity
- BPatient 1 has Type II HIT and requires a direct thrombin inhibitor
- CPatient 2 has Type I HIT and heparin may be safely continued
- DPatient 1 has non-immune Type I HIT; Patient 2 has immune Type II HIT
- ENeither is HIT, since immune HIT always presents within 48 hours
Answer
D
Type I is non-immune: mild (100,000–150,000), rapid onset at 1–2 days, resolves despite continued heparin, and no thrombosis. Type II is immune (heparin-PF4 IgG): severe (<100,000), delayed onset at 5–10 days, persists until heparin is discontinued, and carries thromboembolic complications. The timing and the behavior on continued heparin are the two discriminators, and they are also two of the 4 Ts.
41A patient with a high 4 Ts score and a positive heparin-PF4 immunoassay has all heparin discontinued, including flushes. Her platelet count is 48,000.What else is required?
- ANothing further; stopping heparin removes the trigger and is sufficient
- BProphylactic platelet transfusion to raise the count above 100,000
- CA non-heparin anticoagulant such as argatroban or bivalirudin, and avoidance of platelet transfusion
- DImmediate transition to warfarin monotherapy
- ETherapeutic plasma exchange to remove the heparin-PF4 antibody
Answer
C
Her slide is emphatic that stopping heparin is NOT ENOUGH — the thrombin already generated persists, so a direct thrombin inhibitor (argatroban, bivalirudin) is required to neutralize it. And platelet transfusions should be avoided, since they may increase thrombotic risk in an actively prothrombotic state. B does exactly the wrong thing. E is the treatment for TTP, not HIT.
42Two patients develop thrombocytopenia on medication. Patient 1 is on quinidine; Patient 2 is receiving cytotoxic chemotherapy.Which correctly pairs the mechanisms?
- APatient 1 — drug-dependent antibody reacting with platelet membrane glycoproteins; Patient 2 — direct toxic effect on the marrow
- BPatient 1 — direct marrow toxicity; Patient 2 — drug-dependent antibody
- CBoth are immune-mediated, differing only in the target glycoprotein
- DBoth reflect splenic sequestration
- EPatient 1 — antibody against a drug-platelet factor 4 complex; Patient 2 — dilutional
Answer
A
Quinine, quinidine, and sulfonamide antibiotics cause an immune thrombocytopenia — antibodies react with platelet membrane glycoproteins only when the drug is present, and coated platelets are removed. Chemotherapeutic agents are the classic non-immune cause: direct marrow toxicity and failed production. E confuses this with HIT, where the antibody targets a heparin-PF4 complex specifically. Note the cross-lecture link: quinine and quinidine are also the ternary complex drugs in immune hemolytic anemia.
43A term newborn has a platelet count of 14,000 and scattered petechiae. The mother's platelet count is 265,000. This is her first pregnancy.Which is true?
- AThis is transplacental passage of maternal autoantibody from maternal ITP
- BA normal maternal count excludes an antibody-mediated process
- CLike Rh disease, this requires a prior sensitizing pregnancy, so an alloimmune cause is unlikely
- DThe mechanism is fetal marrow failure from an intrauterine infection
- EMaternal IgG alloantibody against a fetal platelet antigen the mother lacks, most often HPA-1a, with the first child often affected
Answer
E
NAIT. Two features nail it. First, the mother's platelet count is normal — the antibody is an alloantibody against HPA-1a, an antigen she does not possess, so her own platelets are untouched. In maternal ITP (option A) the antibody is an autoantibody and the mother would be thrombocytopenic too. Second, the first child is often affected, in deliberate contrast to Rh disease of the newborn, which classically requires a second pregnancy — which is why C is wrong. Detected by ELISA; treated with IVIG ± corticosteroids and platelet transfusion, and subsequent pregnancies are high-risk.
44A 38-year-old woman presents with fever, confusion, a platelet count of 18,000, hemoglobin 7.8 with numerous schistocytes, LDH 1,400, and a creatinine of 1.6.Which mechanism underlies this presentation?
- AShiga-like toxin injuring endothelium
- BAutoantibody against ADAMTS13, so ultra-large vWF multimers are not cleaved and aggregate platelets in the microcirculation
- CLoss of CD55 and CD59 permitting complement-mediated lysis
- DWidespread tissue factor exposure consuming factors and fibrinogen
- EInherited mutations in complement regulatory genes
Answer
B
The pentad — thrombocytopenia, MAHA, renal dysfunction, neurologic disturbance, fever — with only a mild creatinine rise is TTP, and the acquired form is caused by an IgG autoantibody against ADAMTS13. Without that metalloprotease, sticky ultra-large vWF multimers persist and drive platelet activation and aggregation in small vessels. A is typical HUS (where renal failure would be prominent and ADAMTS13 normal). C is PNH. D is DIC. E is aHUS.
45Two patients each have thrombocytopenia, anemia, and schistocytes on smear. Patient 1: PT and aPTT normal, fibrinogen normal. Patient 2: PT and aPTT both prolonged, fibrinogen low, D-dimer markedly elevated.Which interpretation is correct?
- ABoth are DIC; the coagulation times simply reflect different stages
- BPatient 1 has DIC and Patient 2 has TTP
- CThe smear findings distinguish them, and the coagulation studies add nothing
- DPatient 1 has a thrombotic microangiopathy consuming platelets only; Patient 2 has DIC consuming factors and fibrinogen as well
- EPatient 1 has DIC, because normal coagulation times exclude a microangiopathy
Answer
D
This is the most testable contrast across the two bleeding lectures. Both produce microvascular thrombi and shear red cells, so the smear cannot separate them. The difference is what gets consumed. In TTP, platelets are consumed on vWF strands while the cascade is untouched — so PT and aPTT are normal (unless DIC supervenes) and fibrinogen is normal. In DIC, the cascade itself is activated, consuming factors and fibrinogen — so both times prolong, fibrinogen falls, and secondary fibrinolysis raises D-dimer.
46A patient with TTP is started on therapeutic plasma exchange rather than plasma infusion alone.What is the rationale?
- AIt removes schistocytes, preventing further hemolysis
- BIt replaces platelets that have been consumed
- CIt removes the autoantibody and the accumulated ultra-large multimers while supplying functional ADAMTS13
- DIt removes complement components responsible for lysis
- EIt corrects the prolonged PT and aPTT characteristic of TTP
Answer
C
Exchange does two jobs in one procedure, exactly as her slide annotates it: removes the autoantibody and provides ADAMTS13. Infusion alone would supply enzyme but leave the antibody behind to neutralize it — which is why exchange, not infusion, is the emergency intervention. Adjuncts reduce antibody production (steroids, rituximab) or block vWF (caplacizumab). E is wrong on its premise: PT and aPTT are normal in TTP.
47A 4-year-old develops bloody diarrhea after a family barbecue, then anemia with schistocytes, a platelet count of 60,000, and a creatinine of 3.8 requiring dialysis. ADAMTS13 activity is normal.Which is true?
- AShiga-like toxin from enterohemorrhagic E. coli disrupted endothelium; plasma exchange is not the treatment
- BNormal ADAMTS13 activity excludes a microangiopathy
- CPlasma exchange should be started urgently, as in TTP
- DAn inherited complement regulatory mutation is the most likely cause
- ENeurologic involvement is expected to dominate the presentation
Answer
A
Typical HUS. Three features separate it from TTP: prominent acute renal failure rather than a mild creatinine rise, NORMAL ADAMTS13, and neurologic symptoms less common than in TTP (so E is wrong). It follows food contaminated with enterohemorrhagic E. coli producing Shiga-like toxins that disrupt endothelium and activate platelets. It is NOT treated by plasma exchange — there is no autoantibody to remove and no missing enzyme to supply, so care is supportive. Children often recover. D describes aHUS.
48An adult has recurrent episodes of microangiopathic hemolysis, thrombocytopenia, and progressive renal failure. There is no diarrheal prodrome, ADAMTS13 is normal, and a mutation in a complement regulatory gene is identified.Which is true regarding treatment?
- APlasma exchange is definitive, as in TTP
- BAntibiotics directed at Shiga toxin–producing organisms
- CDirect thrombin inhibition with argatroban
- DNo therapy exists, and the course is uniformly self-limited
- EA complement inhibitor such as eculizumab, since the lesion is excessive complement activation
Answer
E
Atypical HUS — non-infective, inherited (complement component gene mutations) or sporadic, chronic and recurring with >50% progressing to end-stage renal failure and 10–25% early mortality. Because the lesion is excessive complement activation, patients respond to complement inhibitors (eculizumab). Note the cross-lecture symmetry: eculizumab also treats PNH, where cells lack CD55 and CD59 and complement is therefore unopposed — too little regulation in one disease, too much activation in the other, same drug target.
49A patient bleeds excessively after a dental extraction. Platelet count 310,000, PT normal, aPTT normal. He takes aspirin daily.Which is true?
- AThe normal panel excludes a hemostatic defect, so the bleeding is purely surgical
- BAspirin blocks the P2Y12 receptor, preventing ADP-mediated activation
- CA prolonged aPTT would be expected if the defect were significant
- DAspirin inhibits cyclooxygenase and thus thromboxane A2 production; qualitative platelet defects are invisible to the routine screening panel
- EBleeding time should be measured, as it remains the standard test of platelet function
Answer
D
Cyclooxygenase inhibitors (aspirin, other NSAIDs) block thromboxane A2 production, and TxA2 stimulates platelets. The key teaching point is that qualitative defects give a completely normal screening panel — the count is normal because the platelets are all present, and PT/aPTT are normal because the factors are intact. You need a platelet function analyzer to see it. B describes clopidogrel and ticlopidine, not aspirin. E is wrong: bleeding time is not done anymore — difficult and not reproducible.
50Two patients have lifelong mucocutaneous bleeding. Patient 1: normal platelet count and size, no aggregation to ADP, collagen, or epinephrine, but normal ristocetin response. Patient 2: giant platelets with mild thrombocytopenia, absent aggregation to ristocetin.Which interpretation is correct?
- APatient 1 has Bernard-Soulier syndrome; Patient 2 has Glanzmann thrombasthenia
- BPatient 1 lacks GPIIb-IIIa and cannot aggregate; Patient 2 lacks GP1b and cannot adhere
- CBoth lack the platelet vWF receptor, differing in severity
- DPatient 1 has storage pool disease; Patient 2 has von Willebrand disease
- EBoth are autosomal dominant, and carriers are symptomatic
Answer
B
Match the agonist to the receptor. Ristocetin tests vWF binding to GP1b — the adhesion receptor — so absent ristocetin aggregation plus giant platelets and mild thrombocytopenia is Bernard-Soulier (GP1BA/GP1BB). ADP, collagen, and epinephrine all converge on GPIIb-IIIa and fibrinogen cross-linking — the aggregation step — so failure to those with preserved ristocetin response is Glanzmann thrombasthenia (ITGA2B/ITGB3), the most common inherited platelet function disorder. E is wrong: both are autosomal recessive with asymptomatic carriers.
51A hospitalized, poorly nourished patient on broad-spectrum antibiotics has a PT of 19 seconds (INR 1.7) with a normal aPTT and a normal platelet count.Which best explains the isolated PT prolongation?
- AVitamin K deficiency, and factor 7 has the shortest half-life of the vitamin K–dependent factors
- BVitamin K deficiency, and factor 7 is the only vitamin K–dependent factor
- CAn acquired factor 8 inhibitor
- DConsumption of fibrinogen by intravascular thrombin
- EHeparin contamination of the specimen
Answer
A
Malnutrition plus decreased gut flora is the classic setup, and PT increases first because of the short half-life of factor 7. All four vitamin K–dependent factors (2, 7, 9, 10) require gamma-carboxylation and all will fall, but F7 falls fastest and is the only one of them read by the PT — so B is wrong on the facts even though it reaches the right diagnosis. Treatment is vitamin K, with response in 24–48 hours; urgent reversal in a bleeding patient uses FFP (all factors) or PCC (the four vitamin K–dependent factors). C and E would prolong the aPTT.
52A woman with menorrhagia and easy bruising has a normal platelet count, a normal aPTT, vWF activity (ristocetin cofactor) of 22%, and a vWF antigen of 78%.What does the antigen–activity discrepancy indicate?
- AType 1 vWD, since antigen and activity both fall in quantitative disease
- BType 3 vWD, since the activity is markedly reduced
- CType 2 vWD — a qualitative defect, with structurally abnormal vWF present in adequate amount but unable to function
- DHemophilia A, since vWF stabilizes factor 8
- EAn acquired vWF inhibitor, since a normal aPTT excludes inherited disease
Answer
C
Antigen measures how much protein is there; activity measures whether it works. In Types 1 and 3 the problem is quantity — what is present functions normally, so antigen and activity fall together. In Type 2 the problem is quality — mostly missense mutations causing defective multimer assembly, producing abnormal vWF that cannot interact with GP1bα and F8 — so antigen exceeds activity, exactly as here. Type 2A is the most common variant, and the platelet count is normal except in type 2B. Note that vWF activity is low in all types; it is the discrepancy that localizes the defect.
53Two patients with vWD need coverage for surgery. Patient 1 has Type 1 disease with vWF activity of 30%. Patient 2 has Type 3 disease with essentially undetectable vWF.Which approach is correct?
- ADDAVP for both, since it stimulates vWF synthesis
- BFactor 8 concentrate alone for both, since F8 is the deficient factor
- CAntifibrinolytics alone are sufficient in both
- DPlatelet transfusion for both, since the defect is one of platelet adhesion
- EDDAVP for Patient 1, since it releases stored vWF from endothelial Weibel-Palade bodies; VWF:F8 concentrate for Patient 2, who has nothing to release
Answer
E
DDAVP empties a depot; it does not create protein. It acts on vasopressin V2 receptors to release vWF from endothelial Weibel-Palade bodies, so it works in Type 1, where the vWF made is normal in function, just reduced in amount — and her slide adds the practical caveat to confirm it works in the individual patient. In Type 3 there is almost no vWF, or none that is secreted, so there is nothing to mobilize; replacement with VWF:F8 concentrate (or recombinant vWF) is required, as it is in Type 2. Antifibrinolytics (Amicar, tranexamic acid) block plasminogen→plasmin and are adjuncts, not primary therapy.
54A 22-year-old man with lifelong spontaneous hemarthroses has an aPTT of 68 seconds and a normal PT and platelet count. Mixing his plasma 1:1 with normal plasma normalizes the aPTT.Which is true?
- ACorrection indicates an inhibitor, and the Bethesda titer should be measured
- BCorrection indicates a factor deficiency; a specific F8 or F9 assay is needed, since hemophilia A and B are clinically indistinguishable
- CThe normal PT excludes a clotting factor disorder
- DHemarthrosis points to a platelet disorder rather than a factor deficiency
- EHemophilia B is more common than hemophilia A, so factor 9 deficiency is most likely
Answer
B
Mixing study corrects = deficiency. Does not correct = inhibitor. Normal plasma supplies the missing factor, so the time normalizes; an antibody would inactivate the factor in the added plasma too and the aPTT would stay long (that is when the Bethesda titer applies, making A backwards). Both hemophilias are X-linked recessive and clinically indistinguishable, so specific factor assays are required. A is 1 in 5,000 males; B (“Christmas disease”) is 1 in 30,000, so E inverts the frequencies. And hemarthrosis is the factor-deficiency pattern — petechiae would suggest platelets.
55A patient with severe hemophilia A who has received many factor 8 concentrate exposures now bleeds despite full replacement dosing. His mixing study does not correct.Which treatment addresses the problem, and why?
- ADDAVP, to release stored factor 8 from endothelial cells
- BFresh frozen plasma, since it contains all clotting factors
- CVitamin K, since factor 8 requires gamma-carboxylation
- DRecombinant factor 7a, which activates X to Xa on the platelet surface and so enters the cascade downstream of the inhibited step
- EPlasma exchange, which is the definitive therapy for all acquired inhibitors
Answer
D
A non-correcting mixing study identifies an F8 inhibitor — an alloantibody in a treated hemophiliac, or an autoantibody in an older patient without hemophilia; both bleed severely, and strength is graded by the Bethesda titer. Recombinant 7a (NovoSeven) is a bypassing agent: it activates X to Xa on the platelet surface, producing a “thrombin burst” from a point downstream of the blockade, so the antibody becomes irrelevant. Other options are large doses of F8 to overwhelm the inhibitor, or activated PCC. C is wrong on its premise — the vitamin K–dependent factors are 2, 7, 9, 10, not 8.
56A patient with gram-negative sepsis develops oozing from IV sites. Platelets 42,000, PT prolonged, aPTT prolonged, fibrinogen 90 mg/dL, D-dimer markedly elevated.Which best accounts for this combination?
- AWidespread thrombin generation consuming platelets and factors, thrombin-induced conversion of fibrinogen to fibrin, and secondary fibrinolysis releasing degradation products
- BAn IgG autoantibody against platelet glycoproteins
- CDeficient gamma-carboxylation of factors 2, 7, 9, and 10
- DFailure to cleave ultra-large vWF multimers
- ESplenic sequestration of platelets with hypersplenism
Answer
A
DIC — a “consumption coagulopathy.” It is the one disorder in these lectures that moves every number, and each moves for a stated reason: platelets down (thrombin activation, destruction in the microvasculature, endotoxemia), PT up (decreased factor V), aPTT up (decreased V and VIII), fibrinogen down (thrombin-induced conversion to fibrin), D-dimer up (secondary lysis of intravascular fibrin). Management is treat the underlying disease, supportive care, serial labs, and transfusion of red cells, plasma, platelets, or cryoprecipitate if bleeding. D is TTP, where fibrinogen and the clotting times are normal.
57Which statement about the factor VIII–vWF complex is correct?
- AFactor 8 is synthesized by hepatocytes, so liver disease is the leading cause of factor 8 deficiency
- BvWF and factor 8 circulate independently, and their levels are unrelated
- CFactor 8 is made by endothelial cells rather than hepatocytes, and because vWF stabilizes it, vWF deficiency produces a secondary decrease in factor 8
- DFactor 8 stabilizes vWF, so factor 8 deficiency lowers vWF levels
- EvWF is a clotting factor within the intrinsic pathway, which is why vWD prolongs the aPTT directly
Answer
C
F8 is produced by endothelial cells in the liver and elsewhere — NOT hepatocytes, making A a deliberate trap. F8 binds vWF with high affinity, so when vWF is normal all F8 circulates bound to it; unbound F8 is cleared rapidly. Since vWF stabilizes F8, the dependency runs vWF → F8, not the reverse (so D is inverted). That is also why E is wrong in its mechanism: vWF is not a cascade factor, so the aPTT in vWD is prolonged only indirectly, and only when F8 falls significantly — as in severe Type 3, which can produce hemarthrosis resembling hemophilia.
58 A patient receives a transfusion of packed red blood cells. Thirty minutes later she develops fever, chills, back pain, hypotension, and hemoglobinuria. What is the most likely diagnosis, and what is the immediate next step?
Answer
Acute hemolytic transfusion reaction (AHTR) due to ABO incompatibility. Immediate next step: stop the transfusion. Then maintain urine output with IV fluids and diuretics; cardiovascular support in the ICU. Root cause is almost always patient misidentification.
59 A stable oncology patient on uncomplicated chemotherapy has a platelet count of 8,000/mm³ and is not actively bleeding. At what platelet count threshold does prophylactic transfusion become indicated?
Answer
10,000/mm³. Stable thrombocytopenic oncology patients on uncomplicated chemotherapy can tolerate counts of 5,000–10,000 without spontaneous bleeding, but 10,000 is the accepted prophylactic trigger. (Spontaneous bleeding is rare above 20,000.)
60 An immunosuppressed cancer patient receives a blood transfusion from a relative. Nine days later he develops fever, a diffuse rash spreading from the face and trunk, mucositis, hepatitis, and pancytopenia. What is the diagnosis and how is it prevented?
Answer
Transfusion-associated graft-versus-host disease (TA-GVHD). Donor T-lymphocytes proliferate and attack the immunosuppressed recipient; risk is higher with related donors. Prevention: irradiation of cellular blood products (crosslinks T-lymphocyte DNA). Irradiation is 100% effective. TA-GVHD is rare but nearly always fatal.
61 During a massive transfusion, a patient with DIC is found to have fibrinogen of 60 mg/dL. Which blood product is most appropriate, and which product should NOT be used to treat the fibrinogen deficiency?
Answer
Use cryoprecipitate (contains fibrinogen, F8, vWF, F13); target post-transfusion fibrinogen >100–150 mg/dL. Do NOT use cryoprecipitate as a substitute for specific factor concentrates in hemophilia A or vWD — it is a fibrinogen replacement product, not a hemophilia treatment.
62 A patient with TTP requires plasma exchange. Why is fresh frozen plasma used as the replacement fluid?
Answer
FFP contains functional ADAMTS-13. Plasma exchange removes the patient's anti-ADAMTS-13 antibodies (pathologic plasma) and replaces with FFP, simultaneously removing the antibody and replenishing the enzyme that cleaves ultra-large vWF multimers.
63 A patient receiving a platelet transfusion develops pruritus and urticaria but no fever. What type of transfusion reaction is this, and what is the mechanism?
Answer
Allergic reaction. Recipient IgE antibodies recognize an allergen contained in the transfused unit. No fever distinguishes it from febrile nonhemolytic reactions. Treatment: antihistamines. Anaphylactic reactions are rare but can occur in patients with IgA deficiency.
64 Why are platelets at higher risk of causing septic transfusion reactions compared to red blood cells?
Answer
Platelets are stored at room temperature (to maintain function), while RBCs are refrigerated at 4°C. Room temperature storage supports bacterial growth far more readily. Platelets are now tested or treated for bacterial contamination to reduce this risk.
65 An AB-positive patient needs an emergency RBC transfusion but AB blood is unavailable. In order, what are the next acceptable choices?
Answer
AB patients are universal recipients. If AB is unavailable: 1st alternative = A, 2nd = B, 3rd = O. For Rh, Rh-positive patients can receive Rh-positive or Rh-negative blood.
66 Why do ABO-incompatible transfusions cause immediate intravascular hemolysis, whereas Rh-incompatible transfusions typically cause a delayed extravascular reaction?
Answer
ABO antibodies (anti-A, anti-B) are preformed IgM — they exist without prior exposure and activate complement rapidly → C5b-9 MAC → immediate intravascular lysis. Rh antibodies (anti-D) are IgG, formed only after prior sensitization. IgG-coated cells are cleared extravascularly by splenic macrophages, causing a slower delayed hemolytic reaction.
67 A patient develops acute respiratory distress with hypoxemia requiring intubation during a plasma transfusion. Chest X-ray shows bilateral pulmonary infiltrates but no evidence of fluid overload. What is the diagnosis, and what is the underlying mechanism?
Answer
Transfusion-related acute lung injury (TRALI). Donor granulocyte or HLA antibodies react with patient white blood cells → neutrophil activation → lung injury (non-cardiogenic pulmonary edema). Treatment is aggressive respiratory support in the ICU. Risk has decreased by restricting plasma donations to males and tested females.