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Biochemistry · Clinical biochemistry: colorimetry calculation
Sample itemIn a colorimetric glucose assay, a 200 mg/dL standard gives an absorbance of 0.40 at the same path length. A patient sample gives an absorbance of 0.25. Assuming the Beer-Lambert law applies, what is the sample glucose?
Original practice item, not a recalled exam question. Nothing here is recorded.
The answer and the reasoning
B. 125 mg/dL
By the Beer-Lambert law absorbance is directly proportional to concentration at a fixed path length. Sample concentration = (absorbance of test / absorbance of standard) x standard concentration = (0.25 / 0.40) x 200 = 0.625 x 200 = 125 mg/dL. The relationship holds only in the linear range, so highly concentrated samples are diluted and re-run.
Option by option
A. 80 mg/dL
80 mg/dL results from multiplying the standard concentration by 0.40 instead of by the absorbance ratio.
B. 125 mg/dL (correct)
Correct. (0.25 / 0.40) x 200 = 125 mg/dL.
C. 160 mg/dL
160 mg/dL would need a sample absorbance of 0.32, and does not follow from the given values.
D. 320 mg/dL
320 mg/dL arises from multiplying by the inverted ratio of 1.6.
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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