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ASCP exam preparation (USA · MLS / MLT) – page 53

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

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Q1041HardLab math

Arterial blood has bicarbonate 24 mmol/L and pCO2 40 mm Hg. Using pH = 6.1 + log [HCO3⁻/(0.03 × pCO2)], the pH is:

Answer: C. 7.40

0.03 × 40 = 1.2 mmol/L; 24/1.2 = 20; log 20 = 1.30; 6.1 + 1.30 = 7.40.

ID MG-LOP-0044 · Found a mistake? Report it
Q1042HardLab math

How many grams of calcium chloride (CaCl2, MW 111 g/mol) are needed to prepare 1 L of a solution containing 100 mEq/L?

Answer: B. 5.55 g

Ca²⁺ has valence 2, so 100 mEq = 50 mmol. 50 mmol × 111 mg/mmol = 5550 mg = 5.55 g. 11.1 g treats mEq as mmol.

ID MG-LOP-0046 · Found a mistake? Report it
Q1043HardLab math

A WBC count uses a 1:20 dilution. A total of 100 cells are counted in the four large corner squares (each 1 mm², depth 0.1 mm). The WBC count is:

Answer: B. 5.0 × 10⁹/L

Volume counted = 4 × 1 × 0.1 = 0.4 µL. Cells/µL = (100 × 20)/0.4 = 5000/µL = 5.0 × 10⁹/L.

ID MG-LOP-0047 · Found a mistake? Report it
Q1044HardLab safety

The OSHA permissible exposure limit for formaldehyde as an 8-hour time-weighted average is:

Answer: B. 0.75 ppm

OSHA 29 CFR 1910.1048 sets the formaldehyde PEL at 0.75 ppm (8-h TWA), a short-term limit of 2 ppm and an action level of 0.5 ppm.

ID MG-LOP-0063 · Found a mistake? Report it
Q1045HardPhlebotomy & specimen handling

A patient has a hematocrit of 60%. Using C = 1.85 × 10⁻³ × (100 − Hct) × V, how much 3.2% citrate is needed for 4.5 mL of blood?

Answer: B. 0.33 mL

C = 0.00185 × (100 − 60) × 4.5 = 0.00185 × 40 × 4.5 = 0.333 mL. The standard 0.5 mL would be too much citrate because plasma volume is reduced.

ID MG-LOP-0071 · Found a mistake? Report it
Q1046HardQuality control & QA

Four consecutive results for one control level are +1.4, +1.6, +1.2 and +1.8 SD. Which Westgard rule is violated?

Answer: B. 4-1s

Four consecutive values beyond the same 1 SD limit violate 4-1s, which signals a small systematic error. No value exceeds 2 SD, so 2-2s is not violated.

ID MG-LOP-0086 · Found a mistake? Report it
Q1047HardQuality control & QA

Five control results are 98, 100, 102, 100 and 100 mg/dL. What is the sample standard deviation?

Answer: C. 1.41 mg/dL

Mean = 100. Squared deviations sum to 4+0+4+0+0 = 8. Sample SD = √(8/(n−1)) = √(8/4) = 1.41. Dividing by n (5) gives 1.26, the population SD, which is not used for QC.

ID MG-LOP-0091 · Found a mistake? Report it
Q1048HardQuality control & QA

A lab verifies a manufacturer's reference interval using 20 healthy local subjects. The interval is accepted if no more than how many results fall outside it?

Answer: B. 2

Under CLSI EP28, if 2 or fewer of 20 results (≤10%) fall outside the proposed limits, the interval can be transferred. If more fall outside, a further 20 are tested or a new interval is set.

ID MG-LOP-0100 · Found a mistake? Report it
Q1049HardQuality control & QA

Which method validation experiment is designed to estimate proportional systematic error?

Answer: C. Recovery experiment

In a recovery study, known amounts of analyte are added to patient samples; incomplete recovery that grows with concentration shows proportional error. Interference studies estimate constant systematic error.

ID MG-LOP-0103 · Found a mistake? Report it
Q1050HardAnaerobes

An anaerobic gram-negative rod from an abdominal abscess is resistant to vancomycin 5 µg, kanamycin 1 mg and colistin 10 µg special-potency disks. It grows on Bacteroides bile esculin agar with blackening. It belongs to the:

Answer: D. Bacteroides fragilis group

The B. fragilis group resists all three disks, grows in 20% bile and hydrolyzes esculin (black colonies). Fusobacterium is kanamycin and colistin susceptible, and pigmented Prevotella and Porphyromonas are bile sensitive.

ID MG-BAC-0025 · Found a mistake? Report it
Q1051HardAnaerobes

An anaerobic gram-positive coccus from a wound is inhibited by a sodium polyanethol sulfonate (SPS) disk. It is most likely:

Answer: B. Peptostreptococcus anaerobius

P. anaerobius is characteristically susceptible to SPS, giving a zone of 12 mm or more. Other anaerobic gram-positive cocci are resistant, and Veillonella is a gram-negative coccus.

ID MG-BAC-0035 · Found a mistake? Report it
Q1052HardCulture media, specimens & AST

A continuously monitored blood culture bottle signals positive, but the Gram stain shows no organisms. What is the most appropriate next step?

Answer: C. Perform an acridine orange stain and subculture the broth

Low numbers of organisms, or organisms that stain poorly, may be missed on Gram stain; acridine orange is more sensitive, and subculture confirms growth. Very high white cell counts can also give false signals.

ID MG-BAC-0041 · Found a mistake? Report it
Q1053HardCulture media, specimens & AST

Enterococcus faecalis ATCC 29212 gives a very small zone around trimethoprim-sulfamethoxazole on a new lot of Mueller-Hinton agar. The most likely problem is:

Answer: B. Excess thymidine in the agar

Thymidine lets bacteria bypass folate inhibition, falsely lowering trimethoprim-sulfonamide activity. E. faecalis ATCC 29212 is used to check that thymidine content is acceptably low. pH 7.2–7.4 is the normal range.

ID MG-BAC-0049 · Found a mistake? Report it
Q1054HardCulture media, specimens & AST

Pseudomonas aeruginosa ATCC 27853 shows aminoglycoside zones below the QC range, while other drugs are in range. The most likely medium problem is:

Answer: A. Excess calcium and magnesium ions

High divalent cation levels reduce aminoglycoside uptake by P. aeruginosa, giving smaller zones (false resistance). Low cations have the opposite effect. Low zinc affects carbapenem testing, not aminoglycosides.

ID MG-BAC-0050 · Found a mistake? Report it
Q1055HardCulture media, specimens & AST

On a gradient diffusion strip, the edge of the ellipse of inhibition meets the strip between the 0.19 and 0.25 µg/mL marks. What MIC should be reported?

Answer: C. 0.25 µg/mL

Gradient strips have values between standard two-fold dilutions. The reading is taken where the ellipse meets the strip and is rounded up to the next two-fold dilution, 0.25 µg/mL, before interpretation.

ID MG-BAC-0057 · Found a mistake? Report it
Q1056HardCulture media, specimens & AST

A laboratory wants to change routine AST QC from daily to weekly testing. According to CLSI, which result supports this change?

Answer: A. 15-replicate (3 × 5-day) plan with ≤1 out-of-range result per drug/strain

CLSI allows weekly QC after a 15-replicate plan (or the 20/30-day plan) with acceptable results for each drug and QC strain. Five days is too few, and proficiency testing does not replace QC.

ID MG-BAC-0059 · Found a mistake? Report it
Q1057HardCulture media, specimens & AST

A Staphylococcus aureus has a penicillin zone of 30 mm. The edge of the zone is sharp and clear-cut ('cliff') rather than fuzzy ('beach'). How should penicillin be reported?

Answer: C. Resistant, because a sharp edge indicates beta-lactamase

CLSI recommends the penicillin zone-edge test for S. aureus with large zones. A sharp 'cliff' edge shows beta-lactamase production, so the isolate is reported resistant despite the large zone. A fuzzy 'beach' edge suggests no beta-lactamase.

ID MG-BAC-0062 · Found a mistake? Report it
Q1058HardEnterobacterales

Which E. coli pathotype is often non-motile, lactose non-fermenting or late, lysine decarboxylase negative, and causes dysentery like Shigella?

Answer: C. Enteroinvasive E. coli

EIEC invades colonic epithelium like Shigella and shares many of its biochemical features, which can cause misidentification.

ID MG-BAC-0075 · Found a mistake? Report it
Q1059HardGram-positive cocci

An Enterococcus faecium from a rectal screening swab is highly resistant to both vancomycin and teicoplanin. Which gene is most likely responsible?

Answer: A. vanA

vanA gives high-level resistance to both vancomycin and teicoplanin and is the commonest VRE gene. vanB strains usually remain teicoplanin susceptible.

ID MG-BAC-0112 · Found a mistake? Report it
Q1060HardGram-positive cocci

A catalase-negative gram-positive coccus in pairs and short chains is intrinsically vancomycin resistant and produces gas from glucose in MRS broth. It is most likely:

Answer: C. Leuconostoc

Leuconostoc is intrinsically vancomycin resistant and produces gas from glucose. Pediococcus is also vancomycin resistant but forms tetrads and produces no gas.

ID MG-BAC-0122 · Found a mistake? Report it
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