A 1:200 dilution of a urine specimen yields a creatinine result of 0.8 mg/dL; the serum value is 1.2 mg/dL. What is the final creatinine value in mg/dL in the undiluted urine sample?

• A. 160
• B. 0.8
• C. 960
• D. 860

Answer: (A)

When a sample is diluted, the concentration in the diluted solution is the original concentration divided by the dilution factor. To undo the dilution, you multiply the measured concentration by the dilution factor.

Here, the urine was diluted 1:200, so the dilution factor is 200. The creatinine in the diluted urine is 0.8 mg/dL. Multiply by 200 to get the undiluted value: 0.8 × 200 = 160 mg/dL.

The serum creatinine value isn’t needed for this calculation; it’s extra information that doesn’t affect the calculation of the urine concentration.