Question

In clinical studies, Inulin and para-amino hippuric acid given intraveneously until steady state is reached in a 70 kg man. Once steady state was reached, the following measurements were collected: arterial plasma concentration of inulin 0.18 mg/ml, of paraamino hippuric acid 0.21 mg/ml; renal venous plasma concentration of para-amino hippuric acid 0.0014 mg/ml; urine flow 2.2 ml/min; urinary concentration of inulin 11.0 mg/ml, of para-amino hippuric acid 5.5 mg/ml. What is Tm (transport maximum) for PAH secretion?

Total Body Water = 42L

Extracellular Fluid = 14L

Osmolarity = 300mOsmoles/L

Intracellular Fluid = 28L

Answer 1

We know that glomerular filtration rate = 125 mL/min and urine production rate = 2.2 mL/min

Given, plasma PAH concentration = 0.21 mg/mL. Then, rate of filtration of plasma PAH = 0.21 X 125 = 26.25 mg/min

Given, urine PAH concentration = 5.5 mg/mL. Then, rate of excretion of urine PAH = 2.2 X 5.5 = 12.1 mg/min

Therefore, from 26.25 mg/min PAH filtered at glomerulus, PAH is excreted out at a rate of 12.1 mg/min.

Hence, tubular maximum for PAH secretion is given by,

Tmax. = rate of plasma PAH filtration - rate of urine PAH excretion = 26.25 - 12.1 = 14.15 mg/min