Twenty-four hour urine sample: Total volume = 1350 ml Inulin concentration = 1.8
ID: 208137 • Letter: T
Question
Twenty-four hour urine sample: Total volume = 1350 ml
Inulin concentration = 1.86 mg/ml, Creatinine concentration = 3.6 mg/ml
Sodium concentration = 0.17 mEq/ml, Urea concentration = 24.0 mg/ml
Potassium concentration = 0.12 mEq/ml, PAH concentration = 10 mg/ml
Plasma sample taken at the midpoint during urine collection: Hematocrit = 43%
Inulin concentration = 0.013 mg/ml, Creatinine concentration = 0.019 mg/ml
Sodium concentration = 0.13 mEq/ml, Urea concentration = 0.28 mg/ml
Potassium concentration = 0.008 mEq/ml, PAH concentration = 0.014 mg/ml
What is FF exp (filtration fraction experimental) %?
When I worked it out, I got 18.03%. Is this correct?
Explanation / Answer
The experimental FF (filtration fraction) is the portion of the plasma which is filtered by the kidneys and is calculated from the diagnostic measured values as GFR/RPF (Glomerular filtration rate/Renal plasma flow). It other words it is the part of the fluid reaching the kidney which is passed on to the renal tubules, and is usually 20%.
GFR = (urine creatinine/plasma creatinine)*flow = (3.6/0.019)*flow = 189.47 *flow
RPF = (urine PAH/plasmaPAH)*flow = (10/0.014)*flow = 714 *flow
Considering the flow rate to be same (1ml/mm)
FF = GFR/RPF = 189.47/714 = 26.5%.
The FF is higher than normal conditions.
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