In a steady state, para-aminohippurate plasma concentration is 0.02 mg/mL, urine concentration is 12 mg/mL, and urine flow is 1 mL/min. Simultaneously measured inulin clearance is 120 mL/min. What is the filtration fraction?
- A 0.15
- B 0.20 ✓
- C 0.25
- D 0.30
Correct answer: B. 0.20
Explanation
PAH clearance = (U x V)/P = (12 x 1)/0.02 = 600 mL/min, which approximates effective renal plasma flow. Filtration fraction = GFR/RPF = 120/600 = 0.20, the normal value of roughly one fifth of renal plasma flow being filtered. Option C would result if RPF were taken as 480 mL/min, a common arithmetic slip when dividing instead of multiplying in the clearance equation.
Reference: Ganong Review of Medical Physiology, 26th ed.
High-yield for: NEET PGINI-CETNExTFMGEUSMLEPLABMRCP
Written and medically reviewed by the StethoPrep medical team.