MLPAO MLPAO Medical Laboratory Calculations 2 — Questions and Answers
Question 1: A patient's serum creatinine is 132 μmol/L. Convert this value to mg/dL. (Molecular weight of creatinine = 113.12 g/mol)
- 1.49 mg/dL (Correct answer)
- 1.32 mg/dL
- 0.13 mg/dL
- 13.2 mg/dL
Correct answer: 1.49 mg/dL
Convert μmol/L to mg/dL: divide by 10 to get mmol/L (0.132), then multiply by MW (113.12): 0.132 × 113.12 / 10 = 1.49 mg/dL. Alternatively, μmol/L ÷ 88.4 ≈ mg/dL.
The standard conversion for creatinine is: mg/dL = μmol/L ÷ 88.4. Therefore: 132 ÷ 88.4 = 1.493 ≈ 1.49 mg/dL. This falls in the slightly elevated range for adults (normal: 0.7–1.2 mg/dL for males, 0.5–1.0 for females). In Ontario, creatinine is reported in μmol/L per SI units. The eGFR calculation (CKD-EPI or Cockcroft-Gault) uses the creatinine value and is automatically calculated by most Ontario laboratory information systems (LIS).
Question 2: A Levey-Jennings chart shows five consecutive QC values all falling above the mean on the same side. Which Westgard rule is violated?
- 10x rule (Correct answer)
- 2s rule
- R4s rule
- 1₃s rule
Correct answer: 10x rule
The 10x rule (10ₓ) is violated when 10 consecutive control values fall on the same side of the mean, indicating a systematic shift or bias.
Westgard rules help identify systematic and random errors in QC data. The 10ₓ rule flags a systematic shift when 10 consecutive control results fall on the same side of the mean (all above or all below), regardless of whether they exceed 1 SD. This suggests reagent drift, calibration shift, or equipment deterioration. The 1₃s rule rejects when one control exceeds mean ± 3 SD (random error). The R4s rule detects random error when the range between two controls exceeds 4 SD. Ontario labs must document corrective action when any rejection rule is triggered per ISO 15189 and IQMH accreditation requirements.
Question 3: You are preparing a working solution of NaCl at 0.9% (w/v). How many grams of NaCl are needed to prepare 500 mL of this solution?
- 4.5 g (Correct answer)
- 9.0 g
- 0.9 g
- 45.0 g
Correct answer: 4.5 g
0.9% w/v means 0.9 g per 100 mL. For 500 mL: 0.9 g × (500/100) = 0.9 × 5 = 4.5 g.
A weight/volume (w/v) percentage solution means grams of solute per 100 mL of solution. 0.9% NaCl (normal saline) = 0.9 g NaCl per 100 mL. To prepare 500 mL: Mass = (0.9 g/100 mL) × 500 mL = 4.5 g NaCl. Dissolve in approximately 450 mL of distilled water, then bring to a final volume of 500 mL. In Ontario clinical labs, reagent preparation must be documented in the reagent preparation log, including lot numbers, expiry, and preparer initials per IQMH standards.
Question 4: A urine microscopy report shows 15–20 RBCs per high-power field (HPF). This finding is best described as:
- Microscopic hematuria (Correct answer)
- Gross hematuria
- Normal finding
- Pseudohematuria
Correct answer: Microscopic hematuria
More than 3–5 RBCs/HPF is defined as microscopic hematuria. 15–20 RBCs/HPF is a clearly abnormal finding requiring clinical follow-up.
Normal urine contains 0–3 RBCs/HPF. Microscopic hematuria is defined as ≥3–5 RBCs/HPF in a centrifuged specimen (or ≥5 RBCs/μL in uncentrifuged). A count of 15–20 RBCs/HPF is significantly elevated and suggests urinary tract pathology such as UTI, kidney stones, glomerulonephritis, or malignancy. Gross hematuria is visible to the naked eye. Pseudohematuria is a red/brown urine discoloration without true RBCs (e.g., from myoglobin, hemoglobin, beets, rifampin). In Ontario, microscopic hematuria findings must be reported to the physician for follow-up per IQMH guidelines.
Question 5: A laboratory receives a hemolyzed specimen for potassium analysis. The measured potassium is 6.8 mmol/L. What is the most appropriate course of action?
- Reject the specimen, document hemolysis, and request a recollection (Correct answer)
- Report the result with a delta check notation
- Dilute the specimen 1:2 and multiply the result by 2
- Accept the result as potassium is stable in hemolyzed samples
Correct answer: Reject the specimen, document hemolysis, and request a recollection
Hemolysis releases intracellular potassium (up to 30× higher than plasma), causing falsely elevated results. The specimen must be rejected and recollected.
Intracellular potassium concentration is approximately 100–150 mmol/L, compared to plasma levels of 3.5–5.0 mmol/L. Even mild hemolysis significantly elevates measured potassium. The degree of interference depends on the hemoglobin concentration in the specimen. In Ontario labs, hemolysis is graded using a hemolysis index (HI), and analytes sensitive to hemolysis (K⁺, LDH, AST) are flagged or rejected based on the HI threshold. Per IQMH and CSMLS standards, the specimen must be rejected, the ordering physician notified, and a recollection requested with instructions to minimize venipuncture trauma.
Question 6: Using the formula: Clearance (mL/min) = (U × V) / P, where U = urine creatinine (8,800 μmol/L), V = urine volume per minute (1.0 mL/min), and P = plasma creatinine (88 μmol/L), calculate the creatinine clearance.
- 100 mL/min (Correct answer)
- 88 mL/min
- 1000 mL/min
- 8.8 mL/min
Correct answer: 100 mL/min
Clearance = (U × V) / P = (8,800 × 1.0) / 88 = 8,800 / 88 = 100 mL/min.
The creatinine clearance formula is: CrCl (mL/min) = (Urine creatinine × Urine flow rate) / Plasma creatinine. All units must be consistent. Here: CrCl = (8,800 μmol/L × 1.0 mL/min) / 88 μmol/L = 100 mL/min. This result is within the normal range for adults (85–125 mL/min for males, 75–115 mL/min for females). A 24-hour urine collection is typically used, converting total volume to mL/min: V = total volume (mL) / 1,440 min. In Ontario, creatinine clearance is used to estimate GFR and adjust drug dosing in renal impairment.
Question 7: A technologist needs to dilute a patient sample that has a potassium result of 9.2 mmol/L (above linearity). After a 1:2 dilution with saline, the re-measured result is 4.8 mmol/L. What is the corrected patient potassium?
- 9.6 mmol/L (Correct answer)
- 4.8 mmol/L
- 2.4 mmol/L
- 7.2 mmol/L
Correct answer: 9.6 mmol/L
Multiply the diluted result by the dilution factor: 4.8 × 2 = 9.6 mmol/L.
When a specimen is diluted 1:2 (1 part sample + 1 part diluent = 2 parts total), the measured analyte concentration is halved. To correct back to the original concentration: Original value = Measured value × Dilution factor = 4.8 mmol/L × 2 = 9.6 mmol/L. This is critically elevated (normal: 3.5–5.0 mmol/L) and constitutes a critical value per Ontario lab critical value policies. The reporting technologist must immediately notify the ordering physician or nurse and document the communication per IQMH/CSMLS requirements. The original out-of-range result (9.2 mmol/L) had prompted the dilution, and the corrected value confirms severe hyperkalemia.
MLPAO MLPAO Medical Laboratory Calculations 2
A patient's serum creatinine is 132 μmol/L.
Convert this value to mg/dL. (Molecular weight of creatinine = 113.12 g/mol)