mahendra
Forum Guru
Topics: 173
Posts: 419
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A 45-year-old woman is hospitalized after an automobile accident.
The physician collects 1.44 L of urine from the patient in a
24-hour period. The clinical lab returns the following results:
Plasma creatinine concentration 2.0 mg/mL
Plasma urea concentration 15 mmol/mL
Urine creatinine concentration 100 mg/mL
Urine urea concentration 160 mmol/mL
What is the approximate glomerular filtration rate (GFR; in mL/min)
of this patient?
A. 10
B. 25
C. 50
D. 75
E. 100
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| asmi
Forum Hero
Topics: 1043
Posts: 4,609
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E :?:
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| Idiopathic
Forum Guru
Topics: 19
Posts: 641
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It should be D...U times V over P
1.44 times 100 divided by 2 is ~ 75
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| Idiopathic
Forum Guru
Topics: 19
Posts: 641
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crap...forgot it was in a day.
1.44L/day X 100 mg/mL / 2.0 mg/mL = 72L/day = 72000mL/day
72000mL/1440min = 50mL/min
:oops:
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| Idiopathic
Forum Guru
Topics: 19
Posts: 641
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ANother trick is this:
serum creatinine should be 1. For every doubling in creatinine, the GFR gets cut in half. If you couldnt figure it out, you could use this formula.
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| asmi
Forum Hero
Topics: 1043
Posts: 4,609
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thanks 
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