MEDizzy - In an experiment, the renal vein of a rat was cannulated,

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Renal and Urinary Systems

In an experiment, the renal vein of a rat was cannulated, and para-aminohippurate (PAH) was infused. Following a sufficient period of equilibration, plasma PAH concentration was 0.2 mg/mL and urinary concentration was 100 mg/ml. Urinary flow was 1 mL/min. Inulin clearance was found to be 100 mL/min. The fraction of filtered plasma at the glomerulus is which of the following?

ExplanationThe fraction of filtered plasma at the glomerulus, or filtration fraction (FF), equals glomerular filtration rate (GFR) divided by renal plasma flow (RPF), or inulin clearance divided by para-aminohippurate (PAH) clearance. Inulin clearance is given as 100 mL/min. PAH clearance can be calculated as 500 mL/min, when PAH urine concentration (100 mg/mL) is multiplied by urine flow (1 mL/min) and divided by PAH plasma concentration (0.2 mg/mL). GFR of 100 mL/min divided by RPF of 500 mL/min equals 0.2.