MLPAO MLPAO Medical Laboratory Calculations 3

Question 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?

Accepted Answer

4 mL

Answer

10 mL

Answer

2 mL

Answer

50 mL

Question 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?

Accepted Answer

2.0%

Answer

5.0%

Answer

0.2%

Answer

20.0%

Question 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?

Accepted Answer

1.012

Answer

1.010

Answer

1.013

Answer

1.009

Question 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).

Accepted Answer

329.5 g/L

Answer

63.8 g/L

Answer

145.0 g/L

Answer

300.0 g/L

Question 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).

Accepted Answer

109 mL/min

Answer

98 mL/min

Answer

144 mL/min

Answer

95 mL/min

Question 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?

Accepted Answer

39.2 mmol/L

Answer

9.8 mmol/L

Answer

2.45 mmol/L

Answer

44.0 mmol/L

Question 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)?

Accepted Answer

5.44 × 10⁹/L

Answer

6.80 × 10⁹/L

Answer

0.68 × 10⁹/L

Answer

54.4 × 10⁹/L