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Saudi Laboratory Specialist Licensure Examination (SLLE): free practice test

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Practice test questions with answers and explanations

Try the questions in the test above first, then open each one here to see the correct answer and why every other option is wrong.

01Forward grouping of Mr. Turki gives no reaction with anti-A or anti-B but a 1+ reaction with anti-A,B. In reverse grouping, B cells react 4+ and A1 cells react 1+. Which additional result would best support a weak A subgroup rather than group O?

Practice question Unit: Blood Bank

  1. AHis red cells react strongly with anti-H lectin

    Strong H reactivity is seen in both group O and weak A subgroups, so it does not separate them.

  2. BAnti-A can be adsorbed onto his red cells and eluted ✓ Correct answer

    Adsorption-elution recovers anti-A from red cells that carry a small amount of A antigen. Group O cells would give no eluted anti-A.

  3. CHis direct antiglobulin test and autocontrol are negative

    A negative DAT is expected in both situations and says nothing about the A antigen.

  4. DHis serum agglutinates group O reagent red cells

    This would suggest anti-H, as in the Bombay phenotype, and is not a feature of weak A subgroups.

Reference: SCFHS SLLE Content Guideline, Blood Bank: Blood Group Nomenclature and Antigens: ABO, p. 4; Harmening, Modern Blood Banking & Transfusion Practices, 6th ed.

02Mr. Fahad empties his bladder at 07:00 on Monday to begin a 24-hour urine collection. How should the 07:00 Monday urine and the 07:00 Tuesday urine be handled?

Practice question Unit: Urinalysis and Other Body Fluids

  1. AThe Monday urine is discarded and the Tuesday urine is collected ✓ Correct answer

    The first void only empties urine formed before the start time, while the final void at the end time completes exactly 24 hours of formation.

  2. BBoth urines are discarded

    Discarding the final void loses urine produced in the last hours and makes the total volume and results too low.

  3. CBoth urines are collected

    Keeping the first void adds urine formed before the timed period, so the collection exceeds 24 hours and results are falsely high.

  4. DThe Monday urine is collected and the Tuesday urine is discarded

    This reverses the correct order; the start void is discarded and the end void is kept.

Reference: SCFHS SLLE Content Guideline, Urinalysis and Other Body Fluids: Urinalysis (Urine Specimen: Types, Collection, Preservation), p. 5; Brunzel, Fundamentals of Urine and Body Fluid Analysis, 4th ed.

03A laboratory finds that HbA1c results in patients carrying common hemoglobin variants are unreliable on charge-based methods. Which method separates glycated hemoglobin by a property that does not depend on hemoglobin charge?

Practice question Unit: Clinical Chemistry

  1. ABoronate affinity chromatography ✓ Correct answer

    Boronate affinity binds the cis-diol groups of glucose bound to hemoglobin, so it measures total glycation regardless of hemoglobin charge and is not affected by common variants.

  2. BCation-exchange liquid chromatography

    Cation-exchange separates hemoglobins by charge, so variants such as HbS or HbC can co-elute or distort the glycated fraction.

  3. CIsoelectric focusing

    Isoelectric focusing separates by isoelectric point, which is a charge property that variants alter.

  4. DElectrophoresis at alkaline pH

    Electrophoresis separates by net charge and migration, so variant hemoglobins interfere.

Reference: SCFHS SLLE Content Guideline, Clinical Chemistry: Carbohydrates: Glycated hemoglobin, p. 6; Bishop, Fody & Schoeff, Clinical Chemistry: Principles, Techniques, and Correlations, 8th ed.

04A blood sample has a bicarbonate concentration of 24 mmol/L and a pCO2 of 80 mmHg. Using the Henderson-Hasselbalch equation (pK 6.1; CO2 solubility coefficient 0.03 mmol/L per mmHg), what is the calculated pH?

Practice question Unit: Clinical Chemistry

  1. A7.00

    This value would need a bicarbonate to dissolved CO2 ratio of about 8:1, which means a pCO2 near 100 mmHg, not this case.

  2. B7.20

    This value would need a ratio of about 12.6:1, which means a pCO2 near 63 mmHg with the same bicarbonate.

  3. C7.10 ✓ Correct answer

    Dissolved CO2 = 0.03 x 80 = 2.4 mmol/L; the ratio 24 / 2.4 = 10, its log is 1, so pH = 6.1 + 1 = 7.10.

  4. D7.40

    This is the normal pH at a 20:1 ratio (for example pCO2 40 mmHg), but a pCO2 of 80 lowers the ratio to 10:1.

Reference: SCFHS SLLE Content Guideline, Clinical Chemistry: Acid Base, p. 6; Bishop, Clinical Chemistry: Principles, Techniques, and Correlations, 8th ed.

05A CBC on a visibly hemolyzed sample from Mr. Fahad shows hemoglobin 13.0 g/dL, Hct 33% and MCHC 39 g/dL. What is the best interpretation?

Practice question Unit: Hematology

  1. AAll values are accurate and the high MCHC is genuine

    Incorrect: in vitro hemolysis makes RBC and Hct lower than the truth, so the calculated MCHC is falsely high.

  2. BHemoglobin is falsely low because free hemoglobin is not measured

    Incorrect: the analyzer lyses cells to measure hemoglobin anyway, so free hemoglobin is included in the value.

  3. CRBC and Hct falsely decreased, hemoglobin accurate, MCHC falsely increased ✓ Correct answer

    Correct: total hemoglobin is measured after lysis and stays accurate, while counted red cells and Hct fall, raising MCHC (13.0 ÷ 33 ≈ 39 g/dL).

  4. DHereditary spherocytosis, so the sample is acceptable and the results are genuine

    Incorrect: the high MCHC here is due to visible hemolysis and requires rejecting the sample and recollecting before any disease interpretation.

Reference: SCFHS SLLE Content Guideline, Hematology: Sampling and Acceptance Criteria (Integrity of samples), p. 7; Keohane et al., Rodak's Hematology, 5th ed.

06Ahmed, 8 years old, has had two episodes of meningococcal meningitis (Neisseria meningitidis). Which complement abnormality is most likely?

Practice question Unit: Immunology and Serology

  1. ADeficiency of C1 esterase inhibitor

    C1 inhibitor deficiency causes hereditary angioedema with swelling attacks, not recurrent neisserial infection.

  2. BDeficiency of C5 to C9 ✓ Correct answer

    C5 to C9 form the membrane attack complex (MAC) that lyses Neisseria. Their deficiency causes recurrent neisserial infection.

  3. CDeficiency of C1q in the classical pathway

    C1q deficiency is linked to lupus-like autoimmune disease because immune complexes are poorly cleared.

  4. DDeficiency of CD55 and CD59 on cells

    Loss of CD55 and CD59 leaves red cells open to complement lysis, as in paroxysmal nocturnal hemoglobinuria (PNH).

Reference: SCFHS SLLE Content Guideline, Immunology and Serology: Immunity: Physiology of the Immune System, p. 9; Todd, Spickett & Fairclough, Lecture Notes: Immunology, 7th ed.

07A breast core biopsy for ER, PR and HER2 testing stays on the operating-room counter for 3 hours before it is placed in formalin. Which preanalytic variable has been exceeded?

Practice question Unit: Histotechnology and Cytotechnology

  1. AFixation time

    Fixation time starts only when the tissue is in formalin. This specimen was not in formalin yet.

  2. BDecalcification time

    A breast core has no bone and needs no decalcification, so this step does not apply.

  3. CCold ischemia time ✓ Correct answer

    Cold ischemia is the interval between removal from the patient and immersion in fixative. Guidelines set one hour at most for breast biomarkers, because longer delays degrade antigens and cause false-negative results.

  4. DParaffin infiltration time

    Infiltration happens later in the processor and is not related to the delay before fixation.

Reference: SCFHS SLLE Content Guideline, Histo- and Cyto Techniques: Sample Preparation and Fixation, p. 9; Bancroft & Gamble, Theory and Practice of Histological Techniques, 6th ed.

08Ahmed, a 4-year-old boy, develops bloody diarrhea after eating undercooked chicken. A stool Gram stain shows curved gram-negative rods. Which incubation conditions are best for isolating the most likely pathogen?

Practice question Unit: Microbiology

  1. AAmbient air at 35°C

    Ambient air contains more oxygen than Campylobacter tolerates, so it grows poorly.

  2. BAnaerobic jar at 35°C

    Campylobacter is not an obligate anaerobe, and complete absence of oxygen does not suit it.

  3. C5% CO2 incubator at 25°C

    25°C is too low, and CO2 alone is not what this organism needs.

  4. DMicroaerophilic atmosphere (about 5% O2, 10% CO2) at 42°C ✓ Correct answer

    The picture (bloody diarrhea after chicken, curved rods) points to Campylobacter jejuni, which is microaerophilic and grows best at 42°C on selective medium.

Reference: SCFHS SLLE Content Guideline, Microbiology: General Principles of Medical and Clinical Microbiology, Microbial Taxonomy, Structure, Metabolism, Genetics, and Host Interaction, p. 10; Procop & Koneman, Koneman's Color Atlas and Textbook of Diagnostic Microbiology, 7th ed.

09A new screening test is evaluated in 600 patients. Among 200 patients with the disease, 180 test positive. Among 400 patients without the disease, 40 test positive. What is the sensitivity of the test?

Practice question Unit: Laboratory Operations

  1. A82%

    180/220 = 82% is the positive predictive value (true positives among all positive results), not the sensitivity.

  2. B10%

    10% is the false-negative rate (20/200), the complement of sensitivity.

  3. C95%

    About 95% is the negative predictive value, 360/380; sensitivity uses only the diseased group.

  4. D90% ✓ Correct answer

    Sensitivity = TP/(TP+FN) = 180/200 = 90%.

Reference: SCFHS SLLE Content Guideline, Laboratory Operations: Pre-Analytical, Analytical, Post-Analytical, p. 13–14; Bishop, Fody, Schoeff, Clinical Chemistry: Techniques, Principles, Correlations, 6th ed.

10Mrs. Noura, a licensed laboratory specialist, stopped practicing for three consecutive years for family reasons, not for study or training in her field. What must she do before returning to practice?

Practice question Unit: Patient Safety and Professionalism

  1. AResume immediately, since a license does not lapse through absence

    Incorrect; an absence of two consecutive years requires renewal before returning.

  2. BInform the new employer only, then resume

    Informing an employer does not replace the legal renewal of the license.

  3. CRepeat the full mandatory internship

    The Law does not require repeating the internship; it requires renewing the license.

  4. DRenew her license before resuming practice ✓ Correct answer

    A practitioner who has been away for two consecutive years, other than for study or training in the field, may return only after renewing the license.

Reference: SCFHS SLLE Content Guideline, Patient Safety and Professionalism: Saudi Law for Health Practitioners, p. 15; Law of Practicing Healthcare Professions (Royal Decree M/59, 1426H), Article 3

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