MLT GlobeASCP exam preparation (USA · MLS / MLT) – page 50
1200 practice MCQs for the ASCP medical laboratory exam. Level: Advanced.
A European donor's red cells type D+, C−, E−, c+, e+. What is the most probable genotype?
C and E are absent, so both haplotypes are ce. At least one carries D. Because r is very common in Europeans and R0 is rare, R0r is more probable than R0R0.
A pregnant woman types as serologic weak D. RHD genotyping shows weak D type 2. How should she be managed?
People with weak D types 1, 2 or 3 do not make alloanti-D, so they can be managed as D-positive. RHD genotyping avoids unnecessary RhIG and saves D-negative units.
Red cells type D-negative by direct agglutination and by the indirect antiglobulin test, but D is shown by adsorption–elution. This phenotype is:
Del cells carry so few D sites that only adsorption–elution detects them. Del is relatively common among D-negative East Asians.
With low-protein monoclonal anti-D, a separate negative control is required when the patient types as:
A negative anti-A or anti-B tube serves as a control for spontaneous agglutination. In group AB, D-positive samples every tube is positive, so a separate control is needed.
A plasma antibody reacts with D+C− cells and D−C+ cells, but not D−C− cells. Adsorption–elution studies show one antibody reacting with both. The specificity is:
G antigen is present on almost all cells carrying D or C. Anti-G mimics anti-D + anti-C; adsorption–elution separates them. The distinction matters because true anti-D shows RhIG failure.
An antibody reacts strongly with D-positive and weakly with D-negative adult cells. It does not react with DTT-treated cells but is not reduced by ficin. The most likely specificity is:
LW antigens are stronger on D-positive cells and are destroyed by DTT but resist enzymes. Rh antigens like D are not affected by DTT.
Two weeks after transfusion, a patient has a new anti-Jkb and a positive DAT, but hemoglobin is stable and there are no signs of hemolysis. This is best classified as:
A new alloantibody with a positive DAT but no clinical or laboratory hemolysis is a delayed serologic transfusion reaction. A delayed hemolytic reaction requires evidence of hemolysis.
A patient with normal IgA has anaphylaxis during plasma transfusion. Deficiency of which plasma protein, with antibodies to it, is a known cause, especially in East Asian patients?
Haptoglobin deficiency with anti-haptoglobin antibodies is a recognized cause of transfusion anaphylaxis, mainly in East Asian populations. It is an alternative to IgA deficiency.
Why can an immunocompetent patient develop TA-GVHD after receiving blood from a relative?
If the donor is homozygous for an HLA haplotype the recipient also carries, the recipient sees donor lymphocytes as self and does not reject them. The donor cells then attack the recipient.
In metabolic acidosis, Winter's formula predicts expected pCO2 = 1.5 × HCO3− + 8 (±2). A patient has HCO3− 12 mmol/L and pCO2 32 mmHg. This indicates:
Expected pCO2 = 1.5 × 12 + 8 = 26 (24–28) mmHg. The measured 32 mmHg is higher than expected, so ventilation is inadequate, meaning an additional respiratory acidosis.
Which is the main effect of temperature correction if a hypothermic patient's blood gas is analysed at 37 °C?
Gas solubility increases as blood cools, so at 37 °C the analyser reports higher partial pressures than exist in the cold patient, and a lower pH.
Glucose oxidase reacts only with beta-D-glucose. Which enzyme is added to some reagents to speed up conversion of alpha-D-glucose to the beta form?
Mutarotase speeds the conversion of alpha- to beta-D-glucose so all glucose is measured quickly. Hexokinase is the enzyme of a different glucose method.
Na 138, K 4.0, Cl 104, HCO3 24 mmol/L, albumin 20 g/L (2.0 g/dL). The anion gap without K is 10 mmol/L. What is the albumin-corrected anion gap?
Corrected AG = AG + 2.5 × (40 − albumin g/L)/10 = 10 + 2.5 × 2 = 15 mmol/L. Low albumin lowers the measured gap and can hide a high-anion-gap acidosis.
An unexpectedly low or negative anion gap may be caused by:
Cationic IgG paraproteins add unmeasured cations and lower the gap; hypoalbuminemia and bromide interference also do so. The other options raise the anion gap.
Total calcium is 1.90 mmol/L (7.6 mg/dL) and albumin 25 g/L (2.5 g/dL). Using the common correction formula, the adjusted calcium is about:
Adjusted Ca (mg/dL) = measured + 0.8 × (4.0 − albumin g/dL) = 7.6 + 0.8 × 1.5 = 8.8 mg/dL (2.20 mmol/L). Ionised calcium measurement is preferred when accuracy matters.
Chloride measured with an ISE may be falsely increased in a patient taking which substance?
Chloride electrodes respond to other halides; bromide and iodide give positive interference and a falsely low anion gap. Lithium does not affect chloride ISEs.
On a Lineweaver–Burk plot, a noncompetitive inhibitor changes which value?
A noncompetitive inhibitor binds outside the active site and lowers Vmax, raising 1/Vmax (y-intercept) while Km stays the same. An unchanged y-intercept is the pattern for competitive inhibition.
Apolipoprotein E is important in lipid metabolism because it:
Apo E binds hepatic receptors (LDL receptor and LRP) so remnants are cleared. Apo(a), not apo E, binds apo B-100 in Lp(a).
A patient has cholesterol and triglycerides both near 400 mg/dL, palmar xanthomas and a broad beta band on lipoprotein electrophoresis. The most likely disorder is:
Type III results from defective remnant clearance, usually apo E2/E2, giving remnant accumulation seen as a broad beta band, with palmar xanthomas. Familial hypercholesterolemia raises LDL with normal triglycerides.
A patient receiving intravenous glycerol-containing fluids has a markedly high triglyceride result by a routine enzymatic method. Why?
Enzymatic methods hydrolyse triglyceride and measure the released glycerol, so free glycerol falsely raises results unless a glycerol-blanked method is used.