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QCHP exam preparation (Qatar) – page 25

749 practice MCQs for the QCHP medical laboratory exam. Level: Intermediate.

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Q481MediumNeisseria & fastidious

A patient with severe pneumonia has a negative Legionella urinary antigen test, but Legionella is later grown on BCYE. The best explanation is that the urinary antigen test:

Answer: C. Detects only L. pneumophila serogroup 1

Most urinary antigen kits detect only L. pneumophila serogroup 1. Infections due to other serogroups or species are missed, so culture on BCYE remains important.

ID MG-BAC-0157 · Found a mistake? Report it
Q482MediumNon-fermenters & other GNR

An oxidase-positive non-fermenter produces a yellow-green pigment that fluoresces under UV light, but no blue pigment. It does not grow at 42 °C. The most likely organism is:

Answer: D. Pseudomonas fluorescens

Fluorescent pseudomonads other than P. aeruginosa produce pyoverdin but not pyocyanin and fail to grow at 42 °C. P. aeruginosa grows at 42 °C and usually makes pyocyanin. Acinetobacter and Stenotrophomonas are oxidase negative.

ID MG-BAC-0163 · Found a mistake? Report it
Q483MediumNon-fermenters & other GNR

A man with liver cirrhosis develops septic shock and haemorrhagic bullae on his legs one day after eating raw oysters. Blood cultures grow a curved, halophilic, lactose-fermenting gram-negative rod. The most likely organism is:

Answer: B. Vibrio vulnificus

V. vulnificus causes rapidly fatal septicaemia with bullous skin lesions in patients with liver disease or iron overload. It is the lactose-positive vibrio. V. parahaemolyticus mainly causes gastroenteritis and is lactose negative.

ID MG-BAC-0166 · Found a mistake? Report it
Q484MediumNon-fermenters & other GNR

In a rapid urease test on a gastric biopsy, the phenol red indicator turns from yellow to pink when Helicobacter pylori is present. The colour change is caused by:

Answer: B. Ammonia raising the pH

H. pylori urease splits urea into ammonia and CO2; ammonia makes the gel alkaline, turning phenol red pink. CO2 alone does not cause this colour change.

ID MG-BAC-0171 · Found a mistake? Report it
Q485MediumBlood & tissue protozoa

A thin blood film shows normal-sized red cells containing band-shaped trophozoites across the cell and a schizont with 8 merozoites arranged in a rosette around pigment. The species is:

Answer: B. Plasmodium malariae

Band forms and rosette schizonts with about 8 merozoites in normal or small red cells suggest P. malariae. P. vivax enlarges red cells and has 12–24 merozoites.

ID MG-PAR-0006 · Found a mistake? Report it
Q486MediumLab diagnosis & methods

On a thick film, 400 asexual parasites are counted against 200 WBCs. The patient's WBC count is 8 × 10^9/L (8000/µL). The parasite density is:

Answer: C. 16,000 parasites/µL

Parasites/µL = (parasites counted ÷ WBCs counted) × WBC/µL = (400 ÷ 200) × 8000 = 16,000. When the WBC count is unknown, 8000/µL is often assumed.

ID MG-PAR-0009 · Found a mistake? Report it
Q487MediumIntestinal protozoa

Round oocysts 8–10 µm that stain variably with modified acid-fast stain and show blue autofluorescence under UV light are:

Answer: B. Cyclospora cayetanensis

Cyclospora oocysts are about twice the size of Cryptosporidium, stain unevenly acid-fast, and autofluoresce. Cryptosporidium does not autofluoresce.

ID MG-PAR-0022 · Found a mistake? Report it
Q488MediumLab diagnosis & methods

Which method is most sensitive for detecting Strongyloides larvae in stool?

Answer: B. Agar plate culture looking for bacterial tracks

Larvae crawl across nutrient agar carrying bacteria and leave visible tracks, making agar plate culture more sensitive than direct examination.

ID MG-PAR-0041 · Found a mistake? Report it
Q489MediumTrematodes (flukes)

For the best recovery of Schistosoma haematobium eggs, urine should be collected:

Answer: A. Around midday, including the last portion after exercise

Egg excretion in urine peaks between about 10:00 and 14:00, and the terminal portion contains the most eggs. Urine is then sedimented or filtered.

ID MG-PAR-0047 · Found a mistake? Report it
Q490MediumTrematodes (flukes)

A small (about 30 µm) operculated egg with prominent 'shoulders' around the operculum and a small knob at the other end is typical of:

Answer: B. Clonorchis sinensis

The Chinese liver fluke egg is small with a shouldered operculum. Fasciola eggs are very large (130–150 µm) with an indistinct operculum.

ID MG-PAR-0048 · Found a mistake? Report it
Q491MediumHepatitis viruses

A screening anti-HCV test is reactive. What is the recommended next test?

Answer: C. HCV RNA by nucleic acid test

A reactive antibody result is followed by HCV RNA testing to tell current infection from past infection or a false-positive. Genotyping is done only if treatment is planned.

ID MG-VIR-0010 · Found a mistake? Report it
Q492MediumGeneral virology & lab methods

Why should calcium alginate or wooden-shaft swabs be avoided for viral specimens?

Answer: A. They can inactivate viruses and inhibit PCR

Calcium alginate and substances in wood can be toxic to viruses and inhibit amplification. Synthetic swabs such as flocked nylon or polyester with plastic shafts are used.

ID MG-VIR-0030 · Found a mistake? Report it
Q493MediumGeneral virology & lab methods

In real-time PCR, sample A has a cycle threshold (Ct) of 18 and sample B has a Ct of 32 for the same target. What does this mean?

Answer: C. Sample A contains much more target nucleic acid

Ct is the cycle at which fluorescence crosses the threshold; more starting target gives an earlier (lower) Ct. Each cycle difference is about a twofold difference in target.

ID MG-VIR-0032 · Found a mistake? Report it
Q494MediumArboviruses & zoonotic viruses

A patient with encephalitis in summer is suspected of West Nile virus infection. Which test is most useful?

Answer: A. IgM antibody to West Nile virus in CSF

IgM does not cross an intact blood–brain barrier, so WNV IgM in CSF shows central nervous system infection. Viremia is short, so culture and PCR are often negative.

ID MG-VIR-0041 · Found a mistake? Report it
Q495MediumArboviruses & zoonotic viruses

A livestock worker in Pakistan has fever and bleeding after a tick bite. Crimean-Congo hemorrhagic fever is suspected. What is the key laboratory precaution?

Answer: B. Treat samples as highly infectious; testing requires high-containment (BSL-4) facilities or approved safety procedures

CCHF virus spreads through blood and body fluids and is a risk group 4 pathogen. Samples need strict precautions, and culture is done only in maximum containment laboratories.

ID MG-VIR-0042 · Found a mistake? Report it
Q496MediumABO system

How does treatment of red cells with ficin affect their agglutination by anti-A and anti-B?

Answer: C. Reactions are enhanced

ABH antigens are carbohydrates and are not destroyed by proteases. Enzymes remove sialic acid and lower surface charge, so ABO reactions become stronger.

ID MG-BBK-0329 · Found a mistake? Report it
Q497MediumABO system

A healthy, never-transfused donor has a mixed-field reaction with anti-A. The two cell populations type as A and O. Her plasma has anti-B but no anti-A, and her twin brother is group A. The most likely cause is:

Answer: A. Blood group chimerism

Twin chimerism gives two separate red cell populations, and the person is tolerant to both, so no anti-A is made. A3 cells do not separate into cells typing as normal A and O.

ID MG-BBK-0340 · Found a mistake? Report it
Q498MediumABO system

A group A2 patient has anti-A1 that reacts at 37 °C and in the antiglobulin phase. Which red cells are the best choice?

Answer: A. Group O or A2 units

Anti-A1 active at 37 °C may shorten survival of A1 cells, so group O or A2 cells are given. AB and B units carry antigens the plasma reacts with.

ID MG-BBK-0348 · Found a mistake? Report it
Q499MediumABO system

The H (FUT1) and secretor (FUT2) genes are located on which chromosome?

Answer: D. 19

FUT1 and FUT2 are on chromosome 19. The ABO gene is on chromosome 9.

ID MG-BBK-0357 · Found a mistake? Report it
Q500MediumAntibody screen & identification

A patient has anti-E and anti-K. About 70% of donors are E-negative and 91% are K-negative. About how many random units must be tested to find 2 compatible units?

Answer: B. 4

0.70 × 0.91 = 0.64 of units are compatible. 2 ÷ 0.64 = 3.1, so about 4 units must be screened.

ID MG-BBK-0374 · Found a mistake? Report it
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