Free ASCP MLT Practice Test: 100 Questions With Answers

Answer: A. Group O, D-positive

No agglutination with anti-A or anti-B means the cells carry neither A nor B antigen. Plasma that agglutinates both A1 and B cells contains anti-A and anti-B, the expected reverse grouping for group O. The 3+ reaction with anti-D makes the patient D-positive.

Why the other choices don't fit:

• B. Group AB cells would react with both antisera, and AB plasma would react with neither reagent cell.
• C. Group A needs a positive anti-A reaction and plasma that reacts only with B cells.
• D. This ignores the 3+ anti-D reaction.

Answer: B. Anti-Jka

Kidd antibodies are a classic cause of delayed hemolytic transfusion reactions. They often fall below detectable levels between exposures, so a pretransfusion screen can be negative, then rise quickly after re-exposure. They also show dosage: stronger reactions with homozygous cells.

Why the other choices don't fit:

• A. Usually a naturally occurring IgM cold antibody of little clinical significance; it doesn't fit an anamnestic rise a week after transfusion.
• C. It can show dosage, but it is usually a cold-reacting IgM antibody that is rarely implicated in hemolytic reactions.
• D. Usually a naturally occurring IgM cold antibody that is rarely clinically significant.

Answer: D. Anti-Fya

Proteolytic enzymes such as ficin and papain destroy Duffy (Fya, Fyb), M, and N antigens, and S variably. An antibody that disappears after enzyme treatment points to one of those systems.

Why the other choices don't fit:

• A. Rh antibodies typically react more strongly with enzyme-treated cells.
• B. Kidd antibodies are typically enhanced by enzyme treatment.
• C. Kell antigens are not destroyed by ficin or papain, so anti-K would still react.

Answer: B. ABO hemolytic disease of the fetus and newborn

Group O mothers can make IgG anti-A,B that crosses the placenta and coats a group A or B baby's red cells, giving a positive DAT. Her antibody screen stays negative because screening cells are group O and carry no A or B antigen.

Why the other choices don't fit:

• A. The mother is D-positive and her screen shows no unexpected alloantibody.
• C. It doesn't produce a positive DAT with anti-IgG.
• D. It can cause neonatal jaundice, but the DAT is negative.

Answer: B. 20–24 °C with continuous gentle agitation

Standard room-temperature platelets are stored at 20–24 °C with continuous gentle agitation to support gas exchange and maintain acceptable pH. Cold-stored platelets are a separate, specifically labeled option used for active bleeding when conventional platelets are unavailable or impractical.

Why the other choices don't fit:

• A. 1–6 °C applies to specifically manufactured cold-stored platelets, not standard room-temperature platelet components.
• C. That is frozen storage for plasma and cryoprecipitated AHF.
• D. Cold-stored platelets are a separate product pathway; this stem asks about standard room-temperature platelets.

Source: FDA, room-temperature platelet bacterial-risk guidance and FDA cold-stored platelet guidance

Answer: A. 2

Estimated fetal whole-blood volume = 0.8% × 5,000 mL = 40 mL. A 300 µg dose covers about 30 mL of fetal whole blood (15 mL fetal red cells), so 40 ÷ 30 = 1.33 doses and at least 2 full doses are needed. Product-specific instructions should be followed when dosing Rh immune globulin.

Why the other choices don't fit:

• B. This forgets the extra vial.
• C. This rounds up a decimal below .5 before adding the extra vial.
• D. This adds more than one extra vial.

Source: FDA RhoGAM and MICRhoGAM prescribing information

Answer: C. Cryoprecipitated AHF

Cryo is the cold-insoluble fraction of plasma and is rich in fibrinogen, factor VIII, von Willebrand factor, and factor XIII, so it corrects low fibrinogen in a small volume. Plasma also contains fibrinogen, but far less per milliliter.

Why the other choices don't fit:

• A. They treat thrombocytopenia, and this count is near normal.
• B. They restore oxygen-carrying capacity, not clotting factors.
• D. It is a volume expander with no fibrinogen.

Answer: D. Washed red blood cells

The reaction is driven by the patient's anti-IgA against IgA in donor plasma. Washing removes most plasma proteins, including IgA.

Why the other choices don't fit:

• A. Irradiation prevents transfusion-associated graft-versus-host disease.
• B. Leukoreduction reduces febrile reactions, HLA alloimmunization, and CMV transmission risk.
• C. Fresh units address storage concerns, not plasma protein exposure.

Answer: C. The negative result is invalid; the test must be repeated

Check cells (Coombs control cells) should agglutinate in every negative antiglobulin tube. If they don't, active anti-human globulin wasn't present, often because inadequate washing left free globulin that neutralized it, or the reagent was omitted. The original negative can't be trusted.

Why the other choices don't fit:

• A. Check cells exist to catch false negatives; failure to agglutinate means the result isn't valid.
• B. Check-cell failure is a test-system problem, not a patient finding.
• D. Each negative tube needs its own check-cell confirmation.

Answer: B. Saline replacement

High plasma protein causes rouleaux, which mimics agglutination. Removing the plasma and replacing it with saline disperses rouleaux, while true agglutination remains.

Why the other choices don't fit:

• A. Prewarming addresses cold-reacting antibodies, not protein-driven rouleaux.
• C. Enzymes change antigen reactivity and won't remove excess plasma protein.
• D. Elution recovers antibody from coated cells; it doesn't address rouleaux in plasma tests.

Answer: D. Anti-I

Adult red cells are rich in I antigen, while cord cells carry mostly i and very little I. A cold antibody that reacts with all adult cells, including the patient's own, but spares cord cells fits anti-I.

Why the other choices don't fit:

• A. Anti-i would react strongly with cord cells, the opposite pattern.
• B. Many adult cells are P1-negative, so anti-P1 wouldn't react with every panel cell.
• C. Lewis antigens are absent from cord cells too, but anti-Lea wouldn't react with all adult cells or with a Le(a−) patient's own cells.

Answer: B. To remove antibody from the patient's red cells so it can be identified

Elution frees antibody bound to red cells. Testing the eluate against panel cells tells you what is coating the cells, such as an alloantibody against transfused cells or an autoantibody.

Why the other choices don't fit:

• A. That is adsorption.
• C. That is titration.
• D. That describes enzyme or thiol treatment.

Answer: B. To detect ABO incompatibility between donor and recipient

Immediate spin mixes patient plasma with donor cells at room temperature and centrifuges right away. It is a check for ABO incompatibility, which IgM anti-A and anti-B detect immediately.

Why the other choices don't fit:

• A. IgG alloantibodies need the antiglobulin phase.
• C. Donor D typing is done separately.
• D. That is not the purpose of the crossmatch.

Answer: D. A patient receiving a directed donation from a first-degree relative

Irradiation prevents transfusion-associated graft-versus-host disease (TA-GVHD) by stopping donor lymphocytes from dividing. Blood from a close relative carries higher risk because shared HLA types can keep the recipient from recognizing and destroying donor lymphocytes.

Why the other choices don't fit:

• A. Leukoreduction, not irradiation, addresses febrile reactions.
• B. This patient needs K-negative units, not irradiation.
• C. Cryo is acellular and isn't irradiated for TA-GVHD prevention.

Answer: D. Direct antiglobulin test

A positive DAT on the post-transfusion specimen, especially one that was negative before transfusion, suggests transfused cells coated with antibody, the hallmark of an immune hemolytic reaction.

Why the other choices don't fit:

• A. Titers aren't part of the initial reaction workup.
• B. It quantifies fetal-maternal hemorrhage.
• C. It isn't an initial test for suspected hemolysis.

Answer: B. Transfusion-related acute lung injury (TRALI)

TRALI is acute hypoxemic, noncardiogenic pulmonary edema with onset during or within 6 hours of transfusion. Bilateral infiltrates without evidence that circulatory overload is the main cause fit TRALI better than TACO.

Why the other choices don't fit:

• A. TACO is cardiogenic, with volume overload and typically increased BNP.
• C. It causes fever and chills, not hypoxemia with bilateral infiltrates.
• D. It appears days later with falling hemoglobin, not acute lung injury.

Source: TRALI consensus redefinition

Answer: C. Bombay (Oh) phenotype with anti-H

Bombay individuals lack H antigen, so they can't make A or B antigens and type as O. They form anti-A, anti-B, and a potent anti-H that reacts with group O cells, which are richest in H. They can receive only Bombay blood.

Why the other choices don't fit:

• A. An autoantibody would also react with the patient's own cells, and group O cells would react with anti-H lectin.
• B. Anti-A1 doesn't react with group O cells.
• D. D typing has nothing to do with this ABO and H pattern.

Answer: C. Myoglobinuria

The blood pad detects the pseudoperoxidase activity of heme, so hemoglobin and myoglobin both react. Few red cells rule out hematuria. Clear plasma, a muscle injury, and very high CK point to myoglobin, which clears into urine quickly and doesn't tint plasma.

Why the other choices don't fit:

• A. It would show increased RBCs on microscopy (ASCP reference 0–3/HPF).
• B. It usually comes with pink or red plasma from intravascular hemolysis and low haptoglobin.
• D. Bilirubin doesn't react on the blood pad.

Source: ASCP BOC MLT Content Guideline (rev. June 9, 2026)

Answer: C. Acute pyelonephritis

Casts form in the renal tubules, so WBC casts place the inflammation in the kidney, not just the bladder. With bacteria, fever, and flank pain, that's an upper urinary tract infection.

Why the other choices don't fit:

• A. It causes WBCs and bacteria but not casts.
• B. It is associated with RBC casts and dysmorphic RBCs.
• D. It is associated with heavy proteinuria, oval fat bodies, and fatty casts.

Answer: D. A non-glucose reducing sugar such as galactose

Glucose oxidase strips react only with glucose; copper reduction detects any reducing substance. A positive copper test with a negative strip, with ascorbic acid excluded, suggests another reducing sugar. Galactose fits an infant with symptoms after milk feeds.

Why the other choices don't fit:

• A. Backwards: copper reduction is less sensitive to glucose than the strip.
• B. Ketones aren't reducing substances that explain this pattern.
• C. Protein doesn't explain a positive copper reduction result.

Answer: C. WBC 1,800/µL, mostly neutrophils; glucose 20 mg/dL; protein 220 mg/dL

Bacterial meningitis classically shows a high neutrophil-predominant count, low glucose, and high protein. ASCP's CSF reference ranges are WBC 0–5/µL, glucose 50–80 mg/dL, and protein 15–45 mg/dL.

Why the other choices don't fit:

• A. Lymphocytes with normal glucose fit viral meningitis.
• B. All within reference ranges.
• D. Lymphocytes with low glucose and high protein are more typical of tuberculous or fungal meningitis.

Source: ASCP BOC MLT Content Guideline (rev. June 9, 2026)

Answer: B. Cystine; an abnormal crystal associated with cystinuria

Hexagonal plates in acidic urine are the classic appearance of cystine, an abnormal crystal linked to an inherited amino acid transport defect and kidney stones. It is confirmed with the cyanide-nitroprusside test.

Why the other choices don't fit:

• A. Uric acid crystals are usually yellow-brown and rhomboid, barrel, or rosette-shaped.
• C. Calcium oxalate dihydrate looks like an envelope; the monohydrate form is oval or dumbbell-shaped.
• D. Triple phosphate forms in alkaline urine and looks like a coffin lid.

Answer: D. Enterococcus does not reduce nitrate to nitrite

The nitrite pad detects bacteria that convert dietary nitrate to nitrite, mainly gram-negative Enterobacterales. Enterococci and many other gram-positive organisms don't, so a negative nitrite doesn't rule out infection. Short bladder incubation can also cause false negatives.

Why the other choices don't fit:

• A. High specific gravity isn't the main reason nitrite misses enterococci.
• B. A negative nitrite doesn't disprove a heavy, pure culture.
• C. The two pads are independent tests.

Answer: A. Monosodium urate; consistent with gout

Needle-shaped, strongly negatively birefringent crystals are monosodium urate, the crystal of gout. They appear yellow when aligned parallel to the slow axis of the compensator.

Why the other choices don't fit:

• B. CPPD crystals are rhomboid or rod-shaped and weakly positively birefringent.
• C. Cholesterol crystals are flat plates with notched corners.
• D. Calcium oxalate isn't the crystal of acute gout.

Answer: A. 26 mmol/L, increased

138 − (100 + 12) = 26 mmol/L, well above 7–16. A high anion gap with low bicarbonate suggests unmeasured anions, as in lactic acidosis, ketoacidosis, kidney failure, or some poisonings.

Why the other choices don't fit:

• B. This adds potassium, which the stem told you not to use.
• C. This confuses the bicarbonate value with the gap.
• D. Right number, wrong interpretation.

Source: ASCP BOC MLT Content Guideline (rev. June 9, 2026)

Answer: C. 35 mOsm/kg

Calculated: 2(140) + 180/18 + 28/2.8 = 280 + 10 + 10 = 300 mOsm/kg. Gap = measured − calculated = 335 − 300 = 35. A gap this large suggests an unmeasured osmotically active substance such as ethanol, methanol, or ethylene glycol.

Why the other choices don't fit:

• A. That is the calculated osmolality, not the gap.
• B. Arithmetic slip.
• D. Arithmetic slip.

Source: ASCP BOC MLT Content Guideline (rev. June 9, 2026)

Answer: A. 140 mg/dL

LDL = total cholesterol − HDL − (triglycerides ÷ 5) = 220 − 50 − 30 = 140 mg/dL. The equation isn't valid when triglycerides are 400 mg/dL or higher.

Why the other choices don't fit:

• B. That is non-HDL cholesterol (220 − 50).
• C. Subtracts the wrong terms.
• D. Subtracts only the triglyceride term.

Source: ASCP BOC MLT Content Guideline (rev. June 9, 2026)

Answer: D. 3.0 mg/dL; prehepatic (hemolytic)

Indirect = total − direct = 3.4 − 0.4 = 3.0 mg/dL. A mostly unconjugated rise with high LD, very low haptoglobin (ASCP reference 30–200 mg/dL), and reticulocytosis points to red cell destruction.

Why the other choices don't fit:

• A. Adds instead of subtracting.
• B. Obstruction raises conjugated bilirubin, which is low here.
• C. Reports the direct value as the answer.

Source: ASCP BOC MLT Content Guideline (rev. June 9, 2026)

Answer: D. 100 mL/min

Convert the volume to a rate: 1,440 mL ÷ 1,440 min = 1.0 mL/min. Clearance = (U × V) ÷ P = (120 × 1.0) ÷ 1.2 = 100 mL/min.

Why the other choices don't fit:

• A. Misapplies the minute conversion.
• B. Arithmetic slip.
• C. Skips the minute conversion.

Source: ASCP BOC MLT Content Guideline (rev. June 9, 2026)

Answer: A. Metabolic acidosis, partially compensated

pH is low (ASCP reference 7.35–7.45), so the patient is acidotic. Bicarbonate is low (23–29 mmol/L), so the primary problem is metabolic. pCO₂ is low (35–44 mm Hg) because the lungs are blowing off CO₂ to compensate. The pH hasn't returned to normal, so compensation is partial.

Why the other choices don't fit:

• B. Respiratory acidosis would show a high pCO₂.
• C. Alkalosis would show a high pH.
• D. Full compensation would bring the pH inside the reference range.

Source: ASCP BOC MLT Content Guideline (rev. June 9, 2026)

Answer: A. Potassium

Red cells hold far more potassium than plasma does, so hemolysis releases it and falsely raises the result (LD and AST rise for the same reason). The right action is to recollect, not to report.

Why the other choices don't fit:

• B. Sodium is higher outside cells, so hemolysis doesn't raise it.
• C. Hemolysis doesn't raise glucose; delayed separation actually lowers it.
• D. Chloride is higher outside cells, so hemolysis doesn't raise it.

Answer: D. Lipase

Lipase comes mainly from the pancreas and remains increased for days after amylase has fallen, which makes it the more useful marker.

Why the other choices don't fit:

• A. Points to bone or hepatobiliary disease.
• B. Points to muscle or cardiac injury.
• C. A sensitive hepatobiliary marker.

Answer: B. 10%; iron deficiency

Transferrin saturation = (iron ÷ TIBC) × 100 = (45 ÷ 450) × 100 = 10%. Low iron, high TIBC, and low saturation is the classic iron-deficiency pattern. UIBC here would be 450 − 45 = 405 µg/dL.

Why the other choices don't fit:

• A. Iron is also low, but TIBC is normal or low, not high.
• C. The arithmetic is inverted; overload shows high saturation.
• D. Hemochromatosis shows high saturation.

Source: ASCP BOC MLT Content Guideline (rev. June 9, 2026)

Answer: A. Immediately before the next scheduled dose

The trough is the lowest concentration in the dosing interval, which occurs just before the next dose.

Why the other choices don't fit:

• B. That is peak-type timing; exact peak timing depends on the drug and route.
• C. A random level can't be compared with trough targets.
• D. That is peak-type timing, not a trough.

Answer: B. TSH increased, free T4 decreased

When the thyroid itself fails, free T4 falls and the pituitary responds by raising TSH.

Why the other choices don't fit:

• A. That is primary hyperthyroidism.
• C. That suggests central (pituitary or hypothalamic) hypothyroidism.
• D. Euthyroid.

Answer: D. The previous 2–3 months

Glucose attaches to hemoglobin slowly and permanently, so A1c tracks the red cell life span of roughly 120 days, weighted toward the most recent weeks.

Why the other choices don't fit:

• A. That is the window for fructosamine (glycated serum proteins).
• B. Daily averages come from glucose monitoring, not A1c.
• C. Red cells don't survive that long.

Source: CDC, A1C Test for Diabetes and Prediabetes

Answer: B. Citrate-buffered glycolysis-inhibitor tube

A rapidly effective citrate-buffered glycolysis inhibitor minimizes the early fall in glucose when prompt cell separation is not possible. Sodium fluoride inhibits enolase later in the glycolytic pathway and should not be relied on by itself to prevent the initial glycolysis that occurs after collection.

Why the other choices don't fit:

• A. Heparin prevents clotting but does not rapidly stop cellular glycolysis.
• C. A clot activator does not prevent blood cells from consuming glucose before separation.
• D. Fluoride alone is a delayed glycolysis inhibitor; current laboratory guidance prefers a rapidly effective inhibitor such as citrate buffer when prompt separation cannot occur.

Source: ADA/AACC laboratory guidance for glucose specimens

Answer: A. Place on ice immediately and test promptly

Ammonia rises in standing blood as cells and amino acids break down, so it must be chilled and tested quickly. Hemolysis and patient smoking can also falsely raise results.

Why the other choices don't fit:

• B. Delay at room temperature falsely raises ammonia.
• C. Light protection matters for bilirubin, not ammonia.
• D. The delay makes the result unreliable.

Answer: A. Bilirubin

Bilirubin breaks down when exposed to light, giving falsely low results. Specimens, especially from newborns, are shielded with foil or amber tubes.

Why the other choices don't fit:

• B. Needs ice and prompt testing, not light protection.
• C. Affected by hemolysis, not light.
• D. Needs prompt testing because cells keep producing it.

Answer: B. 1.3

Globulin = total protein − albumin = 7.0 − 4.0 = 3.0 g/dL. A/G = 4.0 ÷ 3.0 = 1.3.

Why the other choices don't fit:

• A. Inverts the ratio (globulin ÷ albumin).
• C. Divides albumin by total protein.
• D. Divides total protein by albumin.

Source: ASCP BOC MLT Content Guideline (rev. June 9, 2026)

Answer: C. 900 mg/24 h

Convert the volume to deciliters: 1,800 mL = 18 dL. Then 50 mg/dL × 18 dL = 900 mg/24 h.

Why the other choices don't fit:

• A. Multiplies by liters instead of deciliters.
• B. Multiplies by mL ÷ 10 incorrectly.
• D. Divides the volume by the concentration.

Source: ASCP BOC MLT Content Guideline (rev. June 9, 2026)

Answer: D. Cardiac troponin

Cardiac troponin I or T is highly specific to heart muscle and rises with myocardial injury, which is why it replaced older markers.

Why the other choices don't fit:

• A. Less specific and less sensitive; largely replaced by troponin.
• B. Rises early but is found in all muscle, so it isn't specific.
• C. Present in many tissues and no longer used for this purpose.

Answer: C. Cholestatic (biliary obstruction)

ALP plus GGT rising together points to the biliary tree; GGT confirms the ALP is hepatobiliary rather than bone. A mild ALT rise means hepatocyte damage isn't the main process.

Why the other choices don't fit:

• A. Bone disease raises ALP but leaves GGT normal.
• B. Hepatitis causes a much larger ALT and AST rise than ALP rise.
• D. Doesn't produce this enzyme pattern.

Answer: D. The level is interpreted against hours since ingestion, starting at 4 hours, to assess liver-toxicity risk

For a single acute ingestion, the acetaminophen concentration is plotted on a time-based nomogram. A level drawn before 4 hours can't be interpreted reliably because absorption may be incomplete.

Why the other choices don't fit:

• A. Timing matters for clinical interpretation, not calibration.
• B. Timing documents the exposure, not specimen stability.
• C. There is no such reporting convention.

Source: FDA ACETADOTE prescribing information

Answer: C. Pseudohyponatremia from the electrolyte exclusion effect in the indirect method

Indirect ISE dilutes the sample and assumes a normal water fraction. Excess lipid (or protein) takes up volume and lowers the water fraction, so sodium is underestimated. Direct ISE measures activity in the water phase and isn't affected.

Why the other choices don't fit:

• A. Direct ISE is the method that avoids this error.
• B. Hemolysis doesn't explain a 15 mmol/L gap tied to lipemia.
• D. The pattern matches a known lipemia effect on indirect ISE.

Answer: B. Failure to suppress, supporting hypercortisolism (Cushing syndrome)

Dexamethasone normally suppresses ACTH and therefore cortisol. Cortisol that stays high despite dexamethasone suggests autonomous or excess cortisol production.

Why the other choices don't fit:

• A. A normal response is suppression of cortisol.
• C. Insufficiency is evaluated with stimulation, not suppression, and shows low cortisol.
• D. This test evaluates the adrenal axis.

Answer: B. 60 fL

MCV = (Hct % × 10) ÷ RBC (× 10⁶/µL) = (27 × 10) ÷ 4.5 = 60 fL, microcytic (ASCP reference 76–100 fL).

Why the other choices don't fit:

• A. That is the MCHC in g/dL: (8.1 ÷ 27) × 100 = 30.
• C. That is the MCH in pg: (8.1 × 10) ÷ 4.5 = 18.
• D. Drops a factor of 10.

Source: ASCP BOC MLT Content Guideline (rev. June 9, 2026)

Answer: B. 9.6 × 10³/µL

The counter includes nRBCs in the WBC count. Corrected WBC = (uncorrected WBC × 100) ÷ (100 + nRBCs) = (12.0 × 100) ÷ 125 = 9.6. ASCP expects this correction when more than 10 nRBCs per 100 WBCs are present.

Why the other choices don't fit:

• A. Multiplies instead of dividing.
• C. Leaves the count uncorrected.
• D. Subtracts 25% of the count, which isn't the formula.

Source: ASCP BOC MLT Content Guideline (rev. June 9, 2026)

Answer: A. 120 × 10³/µL

Absolute retic = 0.040 × 3.00 × 10⁶/µL = 0.120 × 10⁶/µL = 120 × 10³/µL, above ASCP's reference of 20–115 × 10³/µL, consistent with increased marrow output.

Why the other choices don't fit:

• B. Misplaces the decimal.
• C. Divides instead of multiplying.
• D. Misplaces the decimal.

Source: ASCP BOC MLT Content Guideline (rev. June 9, 2026)

Answer: A. Beta-thalassemia minor

Marked microcytosis with a high red cell count, a normal RDW (ASCP reference 11.5–14.5%), normal ferritin, and target cells is the thalassemia-trait pattern. Hemoglobin electrophoresis or HPLC showing increased HbA2 would support it.

Why the other choices don't fit:

• B. Usually a low RBC count, high RDW, and low ferritin.
• C. Usually normocytic or mildly microcytic, and it doesn't raise the RBC count.
• D. Shows ringed sideroblasts in the marrow and increased iron.

Source: ASCP BOC MLT Content Guideline (rev. June 9, 2026)

Answer: C. Vitamin B12 or folate deficiency

Oval macrocytes plus hypersegmented neutrophils are the hallmark of megaloblastic anemia, caused by impaired DNA synthesis. Pancytopenia fits because all cell lines are affected.

Why the other choices don't fit:

• A. Causes round macrocytes and target cells, without hypersegmentation.
• B. Raises the MCV slightly but doesn't cause pancytopenia or hypersegmentation.
• D. Causes microcytosis.

Answer: C. Factor VIII deficiency

Correction on mixing means a factor is missing rather than inhibited, and lasting correction after incubation argues against a time-dependent factor VIII inhibitor. An isolated prolonged APTT points to the intrinsic pathway (factors VIII, IX, XI, XII).

Why the other choices don't fit:

• A. Would prolong the PT, not the APTT.
• B. Typically fails to correct on mixing.
• D. Prolongs the PT first.

Answer: B. Disseminated intravascular coagulation (DIC)

Widespread clotting consumes platelets and factors (low platelets, low fibrinogen, prolonged PT and APTT) while fibrinolysis raises D-dimer; fibrin strands shear red cells into schistocytes. Sepsis is a common trigger.

Why the other choices don't fit:

• A. Lowers platelets but leaves coagulation tests normal.
• C. Doesn't cause thrombocytopenia, consumption, or schistocytes.
• D. Causes thrombocytopenia and schistocytes, but PT, APTT, and fibrinogen are usually near normal.

Answer: B. Recollect in sodium citrate and recount, correcting for the citrate dilution per procedure

Clumping in EDTA (EDTA-dependent pseudothrombocytopenia) falsely lowers analyzer counts. Recollecting in another anticoagulant such as sodium citrate usually prevents clumping; the count is then corrected for the anticoagulant dilution.

Why the other choices don't fit:

• A. Reports a falsely low result.
• C. Acts on an artifact.
• D. Won't fix clumps caused by the EDTA itself.

Answer: A. Acute myeloid leukemia

Auer rods are fused azurophilic granules found in myeloid blasts, and myeloperoxidase positivity confirms myeloid lineage.

Why the other choices don't fit:

• B. Lymphoblasts lack Auer rods and are MPO-negative.
• C. A mature lymphoid neoplasm, not a blast proliferation.
• D. A mature lymphoid neoplasm, not a blast proliferation.

Answer: B. Warm the specimen to 37 °C and reanalyze

A cold agglutinin clumps red cells at room temperature; the analyzer counts clumps as single large cells, lowering the RBC count and raising MCV and MCHC. Warming disperses the agglutinates.

Why the other choices don't fit:

• A. That fixes EDTA-dependent platelet clumping.
• C. Plasma replacement is used for lipemia-type plasma interference.
• D. Dilution doesn't break up cold-induced clumps.

Answer: D. Bernard-Soulier syndrome

The platelets lack the GPIb-IX-V receptor that binds von Willebrand factor, so ristocetin can't induce aggregation even when normal plasma supplies VWF. Giant platelets and thrombocytopenia are typical.

Why the other choices don't fit:

• A. The opposite pattern: absent aggregation with ADP, epinephrine, and collagen, normal with ristocetin, and normal platelet size.
• B. The ristocetin defect corrects with normal plasma because the missing piece is VWF itself.
• C. Impairs aggregation with arachidonic acid, collagen, and epinephrine, not ristocetin alone.

Answer: B. Hemoglobin S is present, but trait and disease can't be distinguished without electrophoresis or HPLC

Reduced HbS is insoluble and turns the solution turbid. Both sickle cell trait (HbAS) and sickle cell disease (HbSS) test positive, so hemoglobin electrophoresis or HPLC is needed to identify the genotype.

Why the other choices don't fit:

• A. The screen can't separate trait from disease.
• C. The screen can't separate trait from disease.
• D. HbC doesn't produce a positive solubility test.

Answer: D. Iron, such as in Pappenheimer bodies and ringed sideroblasts

Prussian blue reacts with ferric iron, so it identifies iron-containing inclusions (siderotic granules) in red cells and ringed sideroblasts in the marrow.

Why the other choices don't fit:

• A. That is the MPO stain.
• B. That is the Kleihauer-Betke acid elution stain.
• C. Heinz bodies need a supravital stain.

Answer: B. A supravital stain such as crystal violet or new methylene blue

Heinz bodies are clumps of denatured hemoglobin. They aren't visible with Wright stain and need a supravital stain applied to living cells.

Why the other choices don't fit:

• A. Heinz bodies don't show up with Wright stain.
• C. Detects iron, not denatured hemoglobin.
• D. Detects glycogen and isn't used for Heinz bodies.

Answer: B. 7.5 × 10³/µL

Cells/µL = (cells counted × dilution) ÷ (area counted in mm² × depth in mm) = (150 × 20) ÷ (4 × 0.1) = 3,000 ÷ 0.4 = 7,500/µL, or 7.5 × 10³/µL.

Why the other choices don't fit:

• A. Leaves out the depth correction.
• C. Leaves out the volume counted.
• D. Leaves out the area counted.

Source: ASCP BOC MLT Content Guideline (rev. June 9, 2026)

Answer: C. Increased plasma fibrinogen

Fibrinogen and other acute-phase proteins reduce the charge that keeps red cells apart, promoting rouleaux. Stacked cells settle faster.

Why the other choices don't fit:

• A. More red cells slow settling and lower the ESR.
• B. Abnormal shapes resist rouleaux and lower the ESR.
• D. Abnormal shapes resist rouleaux and lower the ESR.

Answer: B. Chronic lymphocytic leukemia

CD5 and CD23 co-expression on a clonal B-cell population with dim surface immunoglobulin is the typical CLL immunophenotype, and smudge cells are a common smear finding.

Why the other choices don't fit:

• A. Usually CD5+ but CD23−.
• C. Expresses CD11c, CD25, and CD103, usually without CD5.
• D. A T-cell blast population, not mature CD19+ B cells.

Answer: D. Polycythemia vera

JAK2 V617F drives growth-factor-independent signaling and is present in the large majority of polycythemia vera cases. It also appears in some essential thrombocythemia and primary myelofibrosis cases.

Why the other choices don't fit:

• A. Defined by BCR::ABL1, the Philadelphia chromosome.
• B. Defined by PML::RARA.
• C. A lymphoid neoplasm without a defining JAK2 mutation.

Answer: B. t(15;17), PML::RARA

Abnormal promyelocytes with bundled Auer rods (faggot cells) and DIC point to acute promyelocytic leukemia, defined by the t(15;17) PML::RARA fusion. Recognizing it quickly matters because the coagulopathy can be life-threatening.

Why the other choices don't fit:

• A. The Philadelphia chromosome of chronic myeloid leukemia.
• C. Associated with myeloproliferative neoplasms.
• D. Not the defining abnormality of this presentation.

Answer: C. PT reported as INR

Warfarin lowers the vitamin K–dependent factors II, VII, IX, and X. The PT is sensitive to these changes, and the INR standardizes PT results across reagents.

Why the other choices don't fit:

• A. Used for unfractionated heparin, not routine warfarin monitoring.
• B. Measures fibrin breakdown, not anticoagulant effect.
• D. Doesn't measure warfarin's effect.

Answer: A. Antibodies against platelet factor 4/heparin complexes

Heparin-induced thrombocytopenia (HIT) is caused by antibodies against PF4/heparin complexes that activate platelets, causing thrombocytopenia and paradoxical clotting, typically 5–10 days after heparin starts.

Why the other choices don't fit:

• B. Doesn't test for the heparin-dependent antibody.
• C. Evaluates prolonged clotting times, not HIT.
• D. Unrelated to platelet destruction.

Source: American Society of Hematology, heparin-induced thrombocytopenia

Answer: A. Falsely prolonged PT and APTT from excess citrate

Coagulation tubes depend on a 9:1 blood-to-citrate ratio. An underfilled tube has too much citrate for the plasma, which binds the calcium added during testing and prolongs clotting times. Recollect with a properly filled tube.

Why the other choices don't fit:

• B. Excess anticoagulant prolongs, not shortens, the results.
• C. Mixing doesn't correct the ratio.
• D. Fill volume doesn't raise the platelet count.

Answer: C. Immunity from vaccination

The vaccine contains only surface antigen, so vaccinated people make anti-HBs but never anti-HBc. Anti-HBs alone, with negative HBsAg and anti-HBc, fits vaccine immunity in CDC's interpretation table.

Why the other choices don't fit:

• A. Would also show a positive total anti-HBc.
• B. Would show a positive HBsAg.
• D. Would show a positive HBsAg.

Source: CDC, Clinical Testing and Diagnosis for Hepatitis B

Answer: A. A false-positive nontreponemal result

Nontreponemal tests such as RPR detect antibodies that aren't specific to Treponema pallidum and can react with other infections, autoimmune conditions, pregnancy, and older age. CDC guidance says a reactive nontreponemal test must be confirmed with a treponemal test; a nonreactive TP-PA doesn't confirm syphilis.

Why the other choices don't fit:

• B. Would usually show a reactive treponemal test.
• C. Treponemal tests usually stay reactive for life after infection.
• D. Typically produces high nontreponemal titers with reactive treponemal tests.

Source: CDC STI Treatment Guidelines: Syphilis

Answer: A. Serum sickness

Antigen–antibody complexes deposit in vessels, joints, and kidneys and activate complement.

Why the other choices don't fit:

• B. Type I (IgE-mediated).
• C. Type II (antibody against cell-surface antigens).
• D. Type IV (T cell–mediated, delayed).

Answer: A. IgG

IgG is actively transported across the placenta, which is why maternal IgG can cause hemolytic disease of the fetus and newborn and also gives newborns passive protection.

Why the other choices don't fit:

• B. A large pentamer that doesn't cross; fetal IgM points to infection in the fetus itself.
• C. Doesn't cross in meaningful amounts.
• D. Doesn't cross in meaningful amounts.

Answer: A. Anti-double-stranded DNA

ANA is sensitive but not specific. Anti-dsDNA (along with anti-Smith) is highly specific for SLE and helps confirm the diagnosis.

Why the other choices don't fit:

• B. Associated with limited cutaneous systemic sclerosis (CREST).
• C. Associated with primary biliary cholangitis.
• D. Associated with rheumatoid arthritis.

Answer: C. Prozone: antibody excess prevented lattice formation

When antibody far outnumbers antigen, each antigen particle is coated by separate antibody molecules and visible cross-linking doesn't form. Diluting the serum brings the two into a better ratio, and the reaction appears.

Why the other choices don't fit:

• A. Postzone is antigen excess, which dilution of the serum wouldn't fix.
• B. Dilution reveals the true reaction; it doesn't create one.
• D. Hemolysis doesn't explain the dilution-dependent pattern.

Answer: C. C1q

C1q binds the Fc portion of IgM or clustered IgG in immune complexes, activating C1r and C1s and launching the classical pathway.

Why the other choices don't fit:

• A. Central to all pathways but isn't the classical initiator.
• B. Part of the alternative pathway.
• D. Starts the lectin pathway.

Answer: D. Staphylococcus aureus

Catalase separates staphylococci (positive) from streptococci (negative), and coagulase separates S. aureus from the coagulase-negative staphylococci.

Why the other choices don't fit:

• A. Coagulase-negative.
• B. Coagulase-negative.
• C. Catalase-negative and grows in chains.

Answer: B. Streptococcus pyogenes

Group A strep is beta-hemolytic and PYR-positive, and the throat is its classic source.

Why the other choices don't fit:

• A. PYR-negative and CAMP-positive.
• C. PYR-positive but bile esculin–positive and usually not beta-hemolytic.
• D. Alpha-hemolytic and optochin-susceptible.

Answer: D. Neisseria meningitidis

Glucose plus maltose utilization identifies N. meningitidis, and CSF is its classic source.

Why the other choices don't fit:

• A. Uses glucose only.
• B. Asaccharolytic: it doesn't use these sugars.
• C. A gram-negative coccobacillus, not a diplococcus.

Answer: C. Haemophilus influenzae

H. influenzae needs both X and V factors. Chocolate agar supplies both; on blood agar it grows only near S. aureus, which releases V factor (satellitism).

Why the other choices don't fit:

• A. Needs V factor only.
• B. Needs X factor only and is associated with genital ulcers.
• D. Isn't identified by X and V requirements.

Answer: D. Salmonella species

Black centers on HE agar mean H₂S production, which most Salmonella make; blue-green colonies mean no lactose fermentation.

Why the other choices don't fit:

• A. Also a non-fermenter but doesn't produce H₂S (green colonies without black centers).
• B. Ferments lactose (yellow-orange colonies on HE).
• C. Oxidase-positive and needs microaerophilic incubation, typically at 42 °C, on selective media.

Answer: C. 85,000 CFU/mL

Colonies × (1 ÷ loop volume) = 85 × 1,000 = 85,000 CFU/mL.

Why the other choices don't fit:

• A. Misapplies the multiplier (a 0.01 mL loop would use × 100).
• B. Misapplies the multiplier.
• D. Misapplies the multiplier.

Answer: C. Decontamination: killing normal flora that would overgrow mycobacteria

Mycobacteria tolerate NaOH better than most normal respiratory flora, so it reduces overgrowth during weeks-long incubation. N-acetyl-L-cysteine (NALC) is the mucolytic, and a phosphate buffer is added afterward to dilute and neutralize.

Why the other choices don't fit:

• A. That is the buffer's job.
• B. That is NALC's job.
• D. Done later with Ziehl-Neelsen, Kinyoun, or fluorochrome stains.

Answer: A. Stop further workup, keep work in a biosafety cabinet, and contact the LRN reference laboratory for referral

Nonhemolytic, nonmotile, catalase-positive large gram-positive rods with ground-glass, tenacious colonies mean Bacillus anthracis has not been ruled out. Sentinel protocols say to stop and refer to the Laboratory Response Network (LRN) reference laboratory, where confirmation happens.

Why the other choices don't fit:

• B. Sentinel bench cards warn that automated systems can misidentify these organisms; confirmation belongs to the LRN.
• C. B. cereus is usually beta-hemolytic and motile.
• D. Ignores findings that require referral.

Source: APHL Sentinel Laboratory Biothreat Bench Cards

Answer: D. Request a new specimen because bacterial overgrowth can falsely increase the colony count

Bacteria multiply in urine left at room temperature, so a delayed, unpreserved specimen can give a falsely high count. Specimens that can't be processed promptly should be refrigerated or collected in a preservative tube, following the laboratory's acceptance criteria.

Why the other choices don't fit:

• A. There is no valid correction for overgrowth.
• B. Delay at room temperature directly affects counts.
• C. Refrigeration after the fact doesn't undo growth that already occurred.

Answer: A. Clostridium perfringens

A double zone of hemolysis (an inner zone of complete hemolysis and an outer zone of partial hemolysis) and boxcar-shaped rods, with spores rarely seen, are classic for C. perfringens, a cause of gas gangrene.

Why the other choices don't fit:

• B. A cause of antibiotic-associated diarrhea, not gas gangrene with double-zone hemolysis.
• C. Grows well in air, is usually motile, and doesn't give a double zone of hemolysis.
• D. Small, club-shaped aerobic rods without double-zone hemolysis.

Answer: C. Pseudomonas aeruginosa

Oxidase positivity, nonfermentation, grape-like odor, blue-green pyocyanin pigment, and growth at 42 °C together identify P. aeruginosa.

Why the other choices don't fit:

• A. Oxidase-negative and ferments lactose.
• B. Oxidase-negative, ferments lactose, and forms mucoid colonies.
• D. Oxidase-negative and swarms on blood agar.

Answer: D. Escherichia coli

A lactose-fermenting, oxidase-negative, indole-positive gram-negative rod from urine is presumptively E. coli, the most common cause of urinary tract infection.

Why the other choices don't fit:

• A. Indole-negative and usually mucoid.
• B. Doesn't ferment lactose and is indole-negative.
• C. Indole-negative.

Answer: C. Cefoxitin disk

Cefoxitin is a strong inducer of the mecA-encoded altered penicillin-binding protein, so it detects methicillin (oxacillin) resistance more reliably than oxacillin disks.

Why the other choices don't fit:

• A. Disk diffusion isn't reliable for vancomycin in staphylococci.
• B. Detects beta-lactamase-mediated penicillin resistance, not mecA.
• D. Used with clindamycin to detect inducible clindamycin resistance.

Source: CLSI 2026 AST meeting summary, cefoxitin surrogate testing

Answer: C. Resistant, because inducible clindamycin resistance is present

The flattened zone shows that erythromycin induced resistance to clindamycin. Such isolates can fail clindamycin therapy, so clindamycin is reported as resistant.

Why the other choices don't fit:

• A. The D-zone overrides the zone size.
• B. The finding requires a resistant report.
• D. Clindamycin is reported for staphylococci after D-testing.

Answer: B. Cryptococcus neoformans

The halo is the polysaccharide capsule of Cryptococcus. Urease positivity and CSF involvement fit C. neoformans; cryptococcal antigen testing is the more sensitive test.

Why the other choices don't fit:

• A. Not encapsulated and typically urease-negative; forms germ tubes.
• C. A small intracellular yeast, not a capsulated yeast in India ink.
• D. A mold with septate hyphae, not budding yeast.

Answer: D. Plasmodium falciparum

Multiple delicate rings per cell, high parasitemia, normal-sized infected cells, and banana-shaped gametocytes are the classic picture of P. falciparum, the most dangerous malaria species.

Why the other choices don't fit:

• A. Infected cells are enlarged, often with Schüffner dots.
• B. Shows band forms and rosette schizonts.
• C. Can show tetrads (Maltese cross) and has no banana-shaped gametocytes.

Answer: A. Giardia duodenalis (lamblia)

Giardia trophozoites are pear-shaped and bilaterally symmetrical with two nuclei; the cysts are oval with up to four nuclei.

Why the other choices don't fit:

• B. An amoeba with a single nucleus in the trophozoite and up to four in round cysts.
• C. A urogenital flagellate without a cyst stage.
• D. Small round oocysts seen with modified acid-fast stain.

Answer: C. To help distinguish true bacteremia from skin contamination

An organism such as a coagulase-negative staphylococcus in only one of several sets often reflects skin contamination, while the same organism in multiple sets from separate sites supports true infection. Multiple sets also improve detection.

Why the other choices don't fit:

• A. Blood culture bottles aren't used for cell counts.
• B. Susceptibility testing needs an isolate first.
• D. A set already pairs aerobic and anaerobic bottles; the reason for separate sites is contamination.

Answer: A. 75 mg/dL

Absorbance is proportional to concentration: C(unknown) = (A(unknown) ÷ A(standard)) × C(standard) = (0.300 ÷ 0.400) × 100 = 75 mg/dL.

Why the other choices don't fit:

• B. Inverts the ratio.
• C. Skips the standard concentration.
• D. Misplaces the decimal.

Source: ASCP BOC MLT Content Guideline (rev. June 9, 2026)

Answer: A. 450 mg/dL

Dilution = sample ÷ total volume = 0.2 ÷ (0.2 + 0.8) = 1:5. Multiply by the dilution factor: 90 × 5 = 450 mg/dL.

Why the other choices don't fit:

• B. Treats it as 1:4 (sample-to-diluent instead of sample-to-total).
• C. Divides instead of multiplying.
• D. Uses a factor of 6.

Answer: D. 2:2s violation; reject the run and look for a systematic cause

Two control measurements exceeding the same 2 SD limit (here, +2 SD) violate the 2:2s rule. A shift in the same direction typically points to systematic error, such as calibration or reagent problems.

Why the other choices don't fit:

• A. 1:2s is a warning that triggers checking the other rules, and the 2:2s rule is broken here.
• B. R:4s requires one control above +2 SD and another below −2 SD.
• C. 1:3s requires a control beyond 3 SD.

Source: Westgard QC, "Westgard Rules"

Answer: D. 90%

PPV = true positives ÷ all positive test results = 72 ÷ (72 + 8) = 90%.

Why the other choices don't fit:

• A. That is sensitivity: 72 ÷ (72 + 18).
• B. That is specificity: 102 ÷ (102 + 8).
• C. That is negative predictive value: 102 ÷ (102 + 18).

Answer: B. Flush the eyes at the eyewash station for at least 15 minutes, then report and seek medical evaluation

Immediate, prolonged flushing removes the chemical before it causes more damage; 15 minutes is the standard minimum. Reporting and medical follow-up come after flushing.

Why the other choices don't fit:

• A. Neutralizing agents can cause further injury; use water or eyewash.
• C. Flushing comes first; the SDS can be checked afterward.
• D. Far too short, and the exposure must be reported.

Source: OSHA eyewash guidance

Answer: A. 3%

CV = (SD ÷ mean) × 100 = (6 ÷ 200) × 100 = 3%. CV expresses imprecision relative to the mean, so you can compare it across concentrations.

Why the other choices don't fit:

• B. Divides the mean by the SD.
• C. Forgets to multiply by 100.
• D. Reports the SD as a percentage.

Answer: C. 5.85 g

Moles needed = molarity × liters = 0.2 mol/L × 0.5 L = 0.1 mol. Grams = 0.1 mol × 58.5 g/mol = 5.85 g.

Why the other choices don't fit:

• A. Calculates for 1 L instead of 500 mL.
• B. Calculates for 0.4 M in 1 L.
• D. Calculates for 250 mL.