MLT GlobeASCP exam preparation (USA · MLS / MLT) – page 56
1200 practice MCQs for the ASCP medical laboratory exam. Level: Advanced.
Which of these Rh haplotypes is the rarest?
ry (dCE) is extremely rare in all populations. r′ and r″ are each about 1% in Europeans, and R0 is common in people of African descent.
Anti-f (anti-ce) reacts with red cells that carry:
f is a compound antigen made when c and e are in cis. R1R2 cells have c and e in trans and are f-negative.
A Kleihauer-Betke test gives an unexpectedly high fetal cell count, but flow cytometry with anti-D shows very few D-positive cells. Which maternal condition best explains this?
Maternal cells containing HbF resist acid elution and are counted as fetal cells. Anti-D flow cytometry counts only the D-positive fetal cells.
A rosette test on a D-negative mother's postpartum sample is falsely positive most often when:
Maternal weak D cells bind anti-D and form rosettes with indicator cells throughout the sample. A weak D baby tends to cause a false negative instead.
The E and e antigens differ by:
E/e is due to Pro226Ala in RhCE. C/c differ by several amino acids (mainly Ser103Pro) on the same RhCE protein.
An R2R2 patient has anti-e. About what proportion of donors of European descent are e-negative?
The e antigen is present in about 98% of Europeans, so only about 2% of donors are e-negative, making compatible units hard to find.
Blood gas: pH 7.40, pCO2 60 mmHg, HCO3− 36 mmol/L. How is this best interpreted?
A normal pH with both pCO2 and HCO3− markedly abnormal in opposite directions indicates two primary disorders. Compensation rarely returns pH fully to 7.40, so this is not simple compensation.
A 20-year-old has muscle cramps and dark urine after exercise. In a forearm exercise test, lactate fails to rise while ammonia rises normally. The most likely disorder is:
Without muscle phosphorylase, glycogen cannot be broken down for glycolysis, so lactate does not rise. In myoadenylate deaminase deficiency the pattern is reversed: lactate rises but ammonia does not.
In the o-cresolphthalein complexone method for calcium, 8-hydroxyquinoline is added to the reagent to:
o-Cresolphthalein complexone also binds magnesium; 8-hydroxyquinoline chelates Mg so that only calcium forms the red complex. The alkaline pH is provided by a separate buffer.
A patient with chronic lymphocytic leukemia (WBC 250 × 10^9/L) has K 2.9 mmol/L on a sample left at room temperature for 6 hours; a promptly separated sample shows K 4.1 mmol/L. The low result is due to:
Large numbers of active leukocytes continue to take up potassium in vitro, causing pseudohypokalemia, especially at warm temperatures. Hemolysis would raise, not lower, potassium.
A patient has mild hypercalcemia with high-normal PTH. The calcium/creatinine clearance ratio is 0.005. The most likely diagnosis is:
A calcium/creatinine clearance ratio below 0.01 suggests FHH, caused by calcium-sensing receptor mutations. Primary hyperparathyroidism usually gives a ratio above 0.02.
A patient has high free T4 and free T3 with a normal TSH (not suppressed). Assay interference has been excluded. The most likely cause is:
High thyroid hormones should suppress TSH. An inappropriately normal or high TSH suggests a TSH-secreting adenoma or thyroid hormone resistance. Graves disease and T4 overdose suppress TSH.
An uncompetitive inhibitor binds only to the enzyme–substrate complex. What is its effect on the kinetic constants?
By removing ES complex, an uncompetitive inhibitor lowers Vmax and also lowers apparent Km, giving parallel lines on a Lineweaver–Burk plot. Km rising with Vmax unchanged is competitive inhibition.
An NADH-linked assay gives ΔA/min of 0.050 at 340 nm. Total volume 1.0 mL, sample 0.05 mL, light path 1 cm, NADH molar absorptivity 6220 L·mol⁻¹·cm⁻¹. What is the enzyme activity?
U/L = (ΔA/min × total volume × 10⁶) ÷ (ε × path × sample volume) = (0.050 × 1.0 × 10⁶) ÷ (6220 × 1 × 0.05) = 50 000 ÷ 311 ≈ 161 U/L.
An elderly woman without chest pain has CK-MB by immunoinhibition of 60 U/L with total CK 90 U/L. CK-MB mass by immunoassay is normal. The most likely cause is:
Immunoinhibition treats all non-M activity as MB, so CK-BB in a macro-complex gives a falsely high CK-MB, often a large share of total CK. A mass assay using MB-specific antibodies is not affected.
A person of blood group O who is a secretor has a mildly raised ALP after a fatty meal; a fasting repeat sample is normal. The extra ALP most likely came from:
Intestinal ALP enters the blood after fatty meals, especially in group O and B secretors. For this reason ALP is best measured on a fasting sample.
The CDC reference method for LDL cholesterol (beta-quantification) is based on:
Ultracentrifugation at density 1.006 g/mL floats off VLDL; the bottom fraction is precipitated to remove LDL and Lp(a), leaving HDL. LDL cholesterol = bottom-fraction cholesterol − HDL cholesterol.
An infant with prolonged conjugated jaundice has a very faint alpha-1 band on electrophoresis. Liver biopsy shows PAS-positive, diastase-resistant globules in hepatocytes. The most likely genotype is:
The Z variant of alpha-1 antitrypsin misfolds and is trapped in hepatocytes, causing neonatal cholestasis and low serum levels. PiMM is normal; PiMS and PiSS give only mild reductions.
In the Jendrassik–Grof method, alkaline tartrate is added at the end of the reaction to:
Alkaline tartrate shifts azobilirubin to a blue color with maximum near 600 nm, away from hemoglobin and carotene interference. Ascorbic acid is the reagent that destroys excess diazo reagent.
A serum stored for several days shows a smaller beta-2 band than when it was fresh. Which protein is most likely responsible?
C3 is unstable and breaks down during storage into fragments that move differently, so the beta-2 band shrinks. Fresh serum should be used for electrophoresis.