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MLPAO Medical Laboratory Calculations Flashcards

7 cards from real MLPAO practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

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  1. A reagent requires a 1 in 50 dilution (1:50) to prepare a working solution. If you need 200 mL of working solution, how many mL of reagent concentrate do you need?

    Answer: 4 mL

    Volume of concentrate = Final volume ÷ Dilution factor = 200 ÷ 50 = 4 mL.

  2. The coefficient of variation (CV) for a QC material is calculated as (SD/Mean) × 100. If the SD is 0.05 mmol/L and the mean is 2.5 mmol/L, what is the CV?

    Answer: 2.0%

    CV = (SD / Mean) × 100 = (0.05 / 2.5) × 100 = 0.02 × 100 = 2.0%.

  3. A urine specific gravity is measured at 1.010 using a refractometer at a temperature of 27°C. The refractometer is calibrated at 20°C. Each 3°C above 20°C requires adding 0.001 to the reading. What is the corrected specific gravity?

    Answer: 1.012

    Temperature difference = 27 − 20 = 7°C. Correction = 7°C ÷ 3°C × 0.001 = 0.0023 ≈ 0.002. Corrected SG = 1.010 + 0.002 = 1.012.

  4. A patient's hemoglobin is 145 g/L and hematocrit (Hct) is 0.44 L/L. Calculate the Mean Corpuscular Hemoglobin Concentration (MCHC) using the formula: MCHC = Hgb (g/L) ÷ Hct (L/L).

    Answer: 329.5 g/L

    MCHC = Hgb ÷ Hct = 145 ÷ 0.44 = 329.5 g/L.

  5. You receive a 24-hour urine collection with a total volume of 1,440 mL. The urine creatinine is 9,504 μmol/L and serum creatinine is 88 μmol/L. Calculate the creatinine clearance in mL/min (use: CrCl = U × V / P, where V is in mL/min).

    Answer: 109 mL/min

    V = 1,440 mL ÷ 1,440 min = 1.0 mL/min. CrCl = (9,504 × 1.0) / 88 = 108 ≈ 109 mL/min.

  6. A laboratory's glucose analyzer has a reportable range of 1.0–30.0 mmol/L. A patient result reads 'HI' (above the linear range). The technologist dilutes the specimen 1:4 with saline and obtains a result of 9.8 mmol/L. What is the corrected patient glucose?

    Answer: 39.2 mmol/L

    Corrected result = Diluted result × Dilution factor = 9.8 × 4 = 39.2 mmol/L.

  7. A peripheral blood smear differential count shows: neutrophils 68%, lymphocytes 22%, monocytes 7%, eosinophils 2%, basophils 1%. The WBC count is 8.0 × 10⁹/L. What is the absolute neutrophil count (ANC)?

    Answer: 5.44 × 10⁹/L

    ANC = WBC × (Neutrophil% ÷ 100) = 8.0 × (68 ÷ 100) = 8.0 × 0.68 = 5.44 × 10⁹/L.