MLT (ASCP) Practice Questions (2026) — Free Sample Questions with Answers

6 free MLT (ASCP) questions from our 25-question bank, grouped by content domain. Each answer explains why the right option is right — and why every wrong option is wrong.

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Blood Banking

Question 1

A patient's forward grouping shows anti-A 0, anti-B 4+, anti-D 3+. Reverse grouping shows A1 cells 4+ and B cells 0. The patient needs fresh frozen plasma. Which plasma groups are ABO-compatible for this patient?

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Answer: B

Group B or group O plasma: Group O plasma contains anti-A and anti-B; its anti-B would attack the patient's group B red cells, so it is not compatible.

Group B or group AB plasma: Correct. The patient is group B. Plasma must lack anti-B: group B plasma has only anti-A, and group AB plasma has neither antibody.

Group O plasma only: This reverses the rule for red cells. Group O is the universal red-cell donor, but O plasma carries anti-B and is the least compatible plasma choice.

Group A or group AB plasma: Group A plasma contains anti-B, which would react with the patient's B antigen. Only the AB part of this choice is acceptable.

Plasma compatibility is the reverse of red-cell compatibility: give plasma that lacks antibodies to the recipient's antigens. Group AB plasma is the universal plasma donor.

Urinalysis and Other Body Fluids

Question 2

A urine sample is negative for protein on the reagent strip, but sulfosalicylic acid (SSA) precipitation is clearly positive. Which substance most likely explains this discrepancy?

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Answer: A

Immunoglobulin light chains (Bence Jones protein): Correct. The strip's protein-error-of-indicators pad reacts mainly with albumin. SSA precipitates all proteins, including light chains.

A large amount of albumin from glomerular injury: Albumin is exactly what the strip detects best, so heavy albuminuria would give a positive strip, not a negative one.

Highly alkaline, strongly buffered urine: Very alkaline urine causes a false-positive strip protein, the opposite of the pattern described.

A high concentration of urine glucose: Glucose does not precipitate with SSA and does not cause this strip-negative, SSA-positive pattern.

Strip protein is albumin-selective; SSA detects all proteins. Strip negative with SSA positive points to non-albumin protein such as Bence Jones light chains.

Chemistry

Question 3

Arterial blood gas results: pH 7.30, pCO2 60 mm Hg, HCO3⁻ 29 mEq/L. Which interpretation is correct?

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Answer: A

Partially compensated respiratory acidosis: Correct. Low pH with high pCO2 is respiratory acidosis; the raised HCO3⁻ shows renal compensation, but pH is still out of range.

Partially compensated metabolic alkalosis: The pH is acidotic (below 7.35), so the primary disorder cannot be an alkalosis.

Uncompensated metabolic acidosis: Metabolic acidosis would show a low bicarbonate; here HCO3⁻ is high and pCO2 is the abnormal value driving the pH.

Fully compensated respiratory acidosis: Full compensation would bring the pH back into the 7.35–7.45 range; a pH of 7.30 is still abnormal.

Match pH to the value moving the same way: high pCO2 with low pH means respiratory acidosis. Compensation is partial while pH is still outside 7.35–7.45.

Hematology

Question 4

A patient has an RBC count of 4.5 × 10¹²/L, hemoglobin 12.0 g/dL and hematocrit 36%. What is the mean corpuscular volume (MCV)?

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Answer: C

26.7 fL: This is the MCH (Hgb × 10 ÷ RBC = 120 ÷ 4.5), which is measured in picograms, not the MCV.

33.3 fL: 33.3 is the MCHC (Hgb ÷ Hct × 100) in g/dL, not the MCV.

80 fL: Correct. MCV = Hct (%) × 10 ÷ RBC (× 10¹²/L) = 360 ÷ 4.5 = 80 fL, at the lower edge of normocytic.

8.0 fL: This leaves out the ×10 factor (36 ÷ 4.5). Hematocrit as a percentage must be multiplied by 10 to give femtoliters.

MCV = Hct × 10 / RBC (fL); MCH = Hgb × 10 / RBC (pg); MCHC = Hgb / Hct × 100 (g/dL).

Immunology

Question 5

A health care worker's hepatitis B panel shows HBsAg negative, anti-HBs positive and total anti-HBc negative. How should these results be interpreted?

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Answer: C

Immune because of a past natural infection: Past natural infection leaves anti-HBc positive; a negative anti-HBc rules this out.

Chronic hepatitis B infection: Chronic infection requires a positive HBsAg for at least 6 months, which is not present here.

Immune because of hepatitis B vaccination: Correct. The vaccine contains only HBsAg, so vaccinated people develop anti-HBs without anti-HBc.

Acute infection in the window period: The window period shows anti-HBc (IgM) positive with HBsAg and anti-HBs both negative.

Anti-HBs alone = vaccine immunity; anti-HBs + anti-HBc = immunity from past infection; HBsAg positive = current infection.

Laboratory Operations

Question 6

In today's run, the level 1 control is +2.3 SD and the level 2 control is +2.4 SD from their means. Which Westgard rule is violated, and what type of error does it suggest?

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Answer: C

1-2s; a warning only, so accept the run: One control beyond 2 SD is a warning, but here two controls exceed 2 SD on the same side, which is a rejection rule.

R-4s; random error: R-4s requires one control above +2 SD and another below −2 SD (a range over 4 SD); both controls here are on the positive side.

2-2s; systematic error: Correct. Two control results beyond 2 SD on the same side of the mean violate 2-2s and indicate a shift (systematic error); reject the run.

1-3s; random error: 1-3s requires a single control beyond 3 SD; neither result exceeds 3 SD.

Random error rules: 1-3s and R-4s. Systematic error rules: 2-2s, 4-1s and 10x. Rejection rules mean patient results are not reported until the problem is fixed.

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