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Discussion Starter · #1 ·
A 45 y/o woman is hospitalized after an automobile accident. The physician collects 1.44L of urine from the patient in a 24hr period. The clinical lab returns the following result.

Plasma creatinine con : 2.0 mg/ml
Plasma urea con : 15 umol/ml
Urine creatinine con : 100 mg/ml
Urine urea con : 160 umol/ml

What is the approx glomerular filtration Rate (GFR) in ml/min of this patient?

A) 10
B) 25
C) 50
D) 75
E) 100

....?
 

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I think 50. I did it like this:
1.44L/24h is equivalent at 1440ml/24h wich is equivalent at 60ml/h wich is the same as saying 1ml/min. That's the V (urine flow).

Then you do the formula of Clearance of creatinine.

Cl: VxUrine concentration of creatinine
_________________________________
Plasma concentration of creatinie

Cl: 1x100
______
2

Answer is 50.
 

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Discussion Starter · #5 ·
I think 50. I did it like this:
1.44L/24h is equivalent at 1440ml/24h wich is equivalent at 60ml/h wich is the same as saying 1ml/min. That's the V (urine flow).

Then you do the formula of Clearance of creatinine.

Cl: VxUrine concentration of creatinine
_________________________________
Plasma concentration of creatinie

Cl: 1x100
______
2

Answer is 50.
Thanks EVAVAR, Your explanation is very impressive. The answer is C) 50 ml/min.
 
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