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ASCP exam preparation (USA · MLS / MLT) – page 55

1200 practice MCQs for the ASCP medical laboratory exam. Level: Advanced.

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Q1081HardOther body fluids (synovial, serous, semen, amniotic, fecal)

A man has azoospermia, semen volume 0.6 mL, acidic pH and absent fructose. The most likely cause is:

Answer: B. Congenital absence of vas deferens and seminal vesicles, or ejaculatory duct obstruction

Fructose and most semen volume come from the seminal vesicles; their absence or obstruction gives low-volume, acidic, fructose-negative semen with no sperm. Testicular failure gives normal volume and fructose.

ID MG-UBF-0036 · Found a mistake? Report it
Q1082HardUrine chemical (dipstick)

A diabetic patient in severe ketoacidosis has only a small positive urine ketone result. The best explanation is:

Answer: D. Most ketones are present as beta-hydroxybutyrate, which the pad does not detect

In severe ketoacidosis with poor tissue perfusion, the ratio of beta-hydroxybutyrate to acetoacetate rises. The nitroprusside test therefore underestimates total ketosis; a blood beta-hydroxybutyrate assay is better.

ID MG-UBF-0048 · Found a mistake? Report it
Q1083HardUrine chemical (dipstick)

To help distinguish myoglobinuria from hemoglobinuria, the most useful finding is:

Answer: C. Pink or red colour of the patient's plasma

Free hemoglobin binds haptoglobin and colours the plasma pink-red in hemolysis. Myoglobin is small and cleared quickly, so plasma stays clear while CK is very high.

ID MG-UBF-0050 · Found a mistake? Report it
Q1084HardUrine chemical (dipstick)

A urine specimen has refractometer specific gravity 1.040 but reagent strip specific gravity 1.015. The patient recently had an intravenous pyelogram. The best explanation is:

Answer: A. Radiographic contrast raises refractometer reading but not the ionic strip pad

Non-ionic contrast media increase refractive index but do not affect the ionic strip pad. Refractometer values above about 1.040 in a normal urine suggest contrast or other high-molecular-weight substances.

ID MG-UBF-0057 · Found a mistake? Report it
Q1085HardUrine microscopy (casts, crystals)

Yellow-brown spheres with spiky projections ('thorny apples') are seen in an old, alkaline urine specimen. They are most likely:

Answer: A. Ammonium biurate

Ammonium biurate crystals form in alkaline, often old specimens and look like thorny apples. They convert to uric acid when acetic acid is added. Leucine spheres have concentric rings and no spikes.

ID MG-UBF-0076 · Found a mistake? Report it
Q1086HardABO system

Red cells do not agglutinate with anti-A or anti-A,B, but anti-A can be adsorbed onto and eluted from them. The plasma lacks anti-A, and secretor saliva contains A and H. The subgroup is:

Answer: B. Am

Am cells are not agglutinated but adsorb and elute anti-A, the plasma lacks anti-A, and secretors have normal A in saliva. Ael differs because its saliva contains H only.

ID MG-BBK-0326 · Found a mistake? Report it
Q1087HardABO system

The anti-A,B present in group O plasma is best described as:

Answer: C. An antibody to a structure shared by A and B antigens

Anti-A,B is a cross-reacting antibody: after adsorption with A cells, the eluate still reacts with B cells. This is why anti-A,B detects some weak A subgroups better than anti-A.

ID MG-BBK-0328 · Found a mistake? Report it
Q1088HardABO system

A group O patient with a myelodysplastic neoplasm has persistent polyagglutination, weak mixed-field reactions with some anti-A reagents, and reactivity with Salvia sclarea lectin. Most likely:

Answer: A. Tn activation

Tn is a persistent somatic change in hematopoietic cells that exposes GalNAc, which can react like a weak A. T activation is transient and linked to infection.

ID MG-BBK-0346 · Found a mistake? Report it
Q1089HardAntibody screen & identification

For allogeneic adsorption in a recently transfused patient with a warm autoantibody, adsorbing cells such as R1R1, R2R2 and rr are chosen so that:

Answer: D. Each common significant antigen (K, Jka, Jkb, S, s) is lacking on one cell

Each adsorbed aliquot then keeps any alloantibody whose antigen is absent from that cell, so alloantibodies are not removed with the autoantibody.

ID MG-BBK-0366 · Found a mistake? Report it
Q1090HardAntibody screen & identification

A child on ceftriaxone develops sudden intravascular hemolysis. The DAT is strongly positive with anti-C3 and weak with anti-IgG. The most likely mechanism is:

Answer: A. A drug-dependent antibody reacting with cells only when drug is present

Ceftriaxone causes drug-dependent antibodies that activate complement, giving C3 on cells and severe intravascular hemolysis. The penicillin type gives IgG coating and slower extravascular hemolysis.

ID MG-BBK-0368 · Found a mistake? Report it
Q1091HardAntibody screen & identification

Plasma reacts with all cells of one manufacturer's LISS-suspended panel, but not when the same cells are washed and resuspended in saline. The most likely cause is:

Answer: D. Antibody to a reagent additive or preservative

Reactivity that disappears once the reagent diluent is washed away points to an antibody against a diluent component. A red cell antibody would still react after washing.

ID MG-BBK-0381 · Found a mistake? Report it
Q1092HardAntibody screen & identification

Plasma contains anti-Fya and a second antibody that is hard to identify. A useful way to separate them is to:

Answer: A. Adsorb with Fy(a+) cells that lack the other suspected antigen

Adsorbing with cells carrying only one antigen removes that antibody and leaves the other for identification.

ID MG-BBK-0386 · Found a mistake? Report it
Q1093HardBlood components & storage

Under AABB standards, a leukocyte-reduced whole-blood-derived platelet concentrate must contain fewer than:

Answer: D. 8.3 × 10^5 leukocytes

Leukoreduced whole-blood-derived platelets must contain < 8.3 × 10^5 leukocytes per unit. The limit of < 5 × 10^6 applies to leukoreduced red cells and apheresis platelets.

ID MG-BBK-0401 · Found a mistake? Report it
Q1094HardBlood components & storage

After leukocyte reduction by filtration, a red cell unit must still contain at least what proportion of the original red cells?

Answer: C. 85%

AABB standards require that leukoreduced red cells retain at least 85% of the original red cell content. The 75% figure refers to the 24-hour in vivo recovery criterion for stored red cells.

ID MG-BBK-0402 · Found a mistake? Report it
Q1095HardBlood components & storage

Red cells frozen by the low-glycerol (about 20%) method need rapid freezing and storage at:

Answer: A. −120 °C or colder

The low-glycerol method gives less protection, so cells are frozen rapidly in liquid nitrogen and stored at −120 °C or colder (nitrogen vapour). High-glycerol (40%) units are stored at −65 °C or colder.

ID MG-BBK-0416 · Found a mistake? Report it
Q1096HardBlood components & storage

During deglycerolization of a frozen red cell unit, severe hemolysis occurs, and the unit clogs the washing set. The donor most likely has:

Answer: C. Sickle cell trait

Hemoglobin S cells gel and lyse in the hypertonic solutions used for deglycerolization, so units from donors with sickle trait need a modified procedure. G6PD deficiency does not cause this effect.

ID MG-BBK-0418 · Found a mistake? Report it
Q1097HardRh system

A pregnant woman types weak D. RHD genotyping shows weak D type 1. How should she be managed?

Answer: D. As D-positive; RhIG is not needed

People with weak D types 1, 2 or 3 do not make allo-anti-D, so they can receive D-positive blood and do not need RhIG. This is the AABB/CAP recommended approach.

ID MG-BBK-0446 · Found a mistake? Report it
Q1098HardOther blood group systems

The Lutheran glycoprotein functions as a receptor for:

Answer: C. Laminin

Lutheran/BCAM binds laminin, and this may increase adhesion of sickle red cells to vessel walls.

ID MG-BBK-0500 · Found a mistake? Report it
Q1099HardOther blood group systems

Anti-Fy3, made by some Fy(a−b−) patients, differs from anti-Fya because it:

Answer: D. Still reacts with enzyme-treated Fy-positive cells

Fy3 lies on a part of the Duffy protein not cleaved by ficin or papain, so anti-Fy3 reacts with enzyme-treated cells, while Fya and Fyb are destroyed.

ID MG-BBK-0514 · Found a mistake? Report it
Q1100HardRh system

The regulator type of Rhnull is caused by mutations in which gene?

Answer: D. RHAG

RhAG is needed to carry Rh proteins to the membrane, so RHAG mutations give the regulator Rhnull type with normal RH genes. The amorph type is due to inactive RHCE with RHD deleted.

ID MG-BBK-0516 · Found a mistake? Report it
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