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DHA exam preparation (Dubai) – page 33

700 practice MCQs for the DHA medical laboratory exam. Level: Basic to intermediate.

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Q641MediumLab math

What volume of a 1 mol/L stock solution is needed to prepare 100 mL of a 0.1 mol/L solution?

Answer: A. 10 mL

Using C1V1 = C2V2: 1 × V1 = 0.1 × 100, so V1 = 10 mL, made up to 100 mL.

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Q642MediumInstrumentation & methods

Which grade of laboratory water, with the highest purity, is used for sensitive tests such as trace analysis and molecular methods?

Answer: A. Type I (clinical laboratory reagent water)

Type I (CLRW) water has the highest resistivity and lowest contamination; Type III is used for rinsing glassware.

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Q643HardNematodes (roundworms)

A soil-dwelling larva with a closed oesophagus and a POINTED tail penetrates the skin of a barefoot farmer. This infective stage belongs to:

Answer: D. Hookworm (filariform larva)

Hookworm filariform (L3) larvae have a pointed tail and infect through skin. Strongyloides filariform larvae have a notched tail, and rhabditiform larvae are non-infective feeding stages.

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Q644HardNematodes (roundworms)

Which nematode is described as ovoviviparous, its eggs hatching in the intestinal mucosa so larvae rather than eggs appear in stool?

Answer: D. Strongyloides stercoralis

Strongyloides eggs hatch almost immediately in the mucosa, so rhabditiform larvae are passed. Trichinella and Dracunculus are viviparous (release live larvae), while Enterobius is oviparous.

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Q645HardMycology

Lactophenol cotton blue mount of a mould from a BAL sample shows columnar conidial heads with phialides only on the upper part of the vesicle. Likely identity?

Answer: A. Aspergillus fumigatus

A. fumigatus has uniseriate phialides on the upper two-thirds of the vesicle, forming columnar heads; it is the commonest cause of invasive aspergillosis. A. niger has biseriate, radiate black heads; Rhizopus has sporangia.

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Q646HardBlood & tissue protozoa

Ring forms in which the chromatin dot sits inside the ring, giving a 'bird's-eye' appearance, are described for:

Answer: A. Plasmodium malariae

The bird's-eye ring is a described feature of P. malariae, which also shows band-form trophozoites and rosette schizonts.

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Q647HardIntestinal protozoa

In current protist classification, Trichomonas vaginalis belongs to which group?

Answer: A. Parabasalia (Metamonada)

Trichomonads are anaerobic flagellates with hydrogenosomes and parabasal bodies, grouped as Parabasalia. The older phylum Sarcomastigophora is outdated.

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Q648HardImmune system principles

To identify an immunoglobulin's isotype by sequencing, which part of the molecule is most informative?

Answer: C. Heavy-chain constant region at the carboxy end

Isotype is defined by the heavy-chain constant region, toward the C-terminus. Variable regions at the N-terminus determine specificity.

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Q649HardImmune system principles

Regarding experimental tolerance induction, which statement is accurate?

Answer: C. T cells become tolerant more easily (lower doses, longer) than B cells

T cells are tolerised at lower antigen doses and remain tolerant longer than B cells; both can be tolerant. Simple, soluble, non-aggregated antigens induce tolerance more readily.

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Q650HardUrine chemical (dipstick)

A dipstick protein pad is negative, but sulfosalicylic acid (SSA) precipitation is positive. The most likely explanation is:

Answer: C. A protein other than albumin, such as Bence Jones protein

The protein pad uses the 'protein error of indicators' and reacts mainly with albumin. SSA precipitates all proteins, including immunoglobulin light chains. Alkaline urine gives false positives on the dipstick, not false negatives.

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Q651HardAntibody screen & identification

A patient on high-dose intravenous penicillin has a positive DAT (IgG). The eluate does not react with untreated panel cells. The next useful test is:

Answer: A. Test the eluate with penicillin-treated cells

Penicillin binds to red cells, and antibody to the drug reacts only with drug-coated cells (drug adsorption mechanism). A nonreactive eluate with a positive DAT in a treated patient suggests this.

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Q652HardDonor selection & processing

In acute normovolemic hemodilution, units collected before surgery are reinfused in which order?

Answer: B. Reverse order of collection, so the first unit is given last

The first unit collected has the highest hematocrit and most platelets and clotting factors. It is saved for last, when bleeding has stopped and it gives the most benefit.

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Q653HardCarbohydrates & diabetes

A patient has fasting hypoglycemia with high insulin, high C-peptide and high proinsulin. Before diagnosing insulinoma, which test is essential?

Answer: A. Screen for sulfonylurea/meglitinide drugs

Sulfonylureas stimulate beta cells and give the same biochemical pattern as insulinoma, so a drug screen is needed. Anti-insulin antibodies alone do not exclude drug use.

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Q654HardProteins & electrophoresis

In a patient with normal renal function, the usual reference interval for the serum free kappa/lambda light chain ratio is about:

Answer: B. 0.26–1.65

The commonly used reference range for the serum free kappa/lambda ratio is 0.26–1.65. A ratio outside this range suggests a clonal plasma-cell process; renal impairment widens the expected range.

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Q655HardTumor markers & iron studies

A man taking finasteride (5-alpha-reductase inhibitor) for 1 year has a PSA of 2.0 ng/mL. How is this usually interpreted?

Answer: A. It is about half the true value; roughly double it

5-alpha-reductase inhibitors lower PSA by about 50% after 6–12 months. Doubling the value is a common way to compare with usual reference limits.

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Q656HardTumor markers & iron studies

A non-pregnant woman has a persistently low positive serum hCG, but urine hCG is negative, and results change with different assays. The likely cause is:

Answer: D. Heterophile antibody interference

Heterophile (anti-animal) antibodies bridge assay antibodies in serum and give 'phantom hCG'. They do not pass into urine, so urine hCG is negative.

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Q657HardHematology methods & instruments

For a manual platelet count, blood is diluted 1:100 and 250 platelets are counted in the central large squares of both sides of the hemacytometer (total 2 mm², depth 0.1 mm). What is the platelet count?

Answer: C. 125 × 10^9/L

Platelets/µL = 250 × 100 ÷ (2 × 0.1) = 125,000/µL, or 125 × 10^9/L. 250 × 10^9/L is obtained by counting only one side's area (1 mm²) in the formula.

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Q658HardHemolytic anemias & hemoglobinopathies

A man from Malaysia has mild or no anemia. His smear shows rigid oval cells, many with one or two transverse slits. The underlying defect is most likely:

Answer: B. A 27-base-pair deletion in the band 3 gene

Southeast Asian ovalocytosis results from a band 3 deletion producing rigid stomatocytic ovalocytes, often with little hemolysis and some protection against malaria. Alpha-spectrin mutations cause ordinary elliptocytosis with thin, non-slit elliptocytes.

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Q659HardRBC indices & anemias

A patient has a reticulocyte count of 6% and a hematocrit of 25%. Using a maturation (shift) factor of 2.0 and normal Hct of 45%, what is the reticulocyte production index (RPI)?

Answer: A. 1.7

Corrected retic = 6 × 25/45 = 3.3%; RPI = 3.3 ÷ 2.0 = 1.7. The shift factor corrects for early release of stress reticulocytes that circulate longer.

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Q660HardInstrumentation & methods

At high absorbance values, a spectrophotometer's standard curve bends toward the concentration axis and loses linearity. The most likely instrumental cause is:

Answer: C. Stray light reaching the detector

Stray light adds unabsorbed light at the detector, making absorbance read falsely low; the effect is largest in highly absorbing samples, causing negative deviation from Beer's law.

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