Medical Laboratory Scientist (MLS) ASCP Practice Exam

Question: 1 / 400

What is the glomerular filtration rate for a patient with a serum creatinine of 2 mg/dL, if the urine creatinine was 124 mg/dL and the urine volume was 2.2 L/24 hrs?

80 mL/min

85 mL/min

90 mL/min

95 mL/min

To calculate the glomerular filtration rate (GFR) using the creatinine clearance method, we use the following formula:

\[ \text{GFR} = \left( \frac{\text{Urine Creatinine} \times \text{Urine Volume}}{\text{Serum Creatinine} \times 1440} \right) \]

Where:

- Urine Creatinine is in mg/dL,

- Urine Volume is in mL/24 hours (1 L = 1000 mL),

- Serum Creatinine is in mg/dL,

- 1440 is the number of minutes in 24 hours.

Given the data:

- Urine Creatinine = 124 mg/dL,

- Urine Volume = 2.2 L = 2200 mL,

- Serum Creatinine = 2 mg/dL.

Plugging in the values:

\[ \text{GFR} = \left( \frac{124 \, \text{mg/dL} \times 2200 \, \text{mL}}{2 \, \text{mg/dL} \times 1440} \right) \]

Calculating the values step-by-step:

1

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