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Biochemistry · Clinical biochemistry: creatinine clearance
Sample itemA 24-hour urine collection has a volume of 1440 mL and urine creatinine of 90 mg/dL. Plasma creatinine is 1.5 mg/dL. What is the creatinine clearance?
Original practice item, not a recalled exam question. Nothing here is recorded.
The answer and the reasoning
A. 60 mL/min
Clearance = (urine concentration x urine flow rate) / plasma concentration. The flow rate is 1440 mL per 1440 minutes = 1 mL/min, so clearance = (90 x 1) / 1.5 = 60 mL/min. Creatinine clearance slightly overestimates GFR because a small amount is secreted by the tubule.
Option by option
A. 60 mL/min (correct)
Correct. (90 x 1) / 1.5 = 60 mL/min.
B. 120 mL/min
120 mL/min is not obtained with any correct use of the given values.
C. 30 mL/min
30 mL/min would need a urine creatinine of 45 mg/dL or a plasma creatinine of 3.0 mg/dL, and does not follow from the given values.
D. 90 mL/min
90 mL/min results from omitting division by the plasma creatinine.
Source note: Original NextPrep sample item — pending expert review. This item is original practice material awaiting independent expert review, so confirm important facts in a standard textbook.
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