17 9 points Recall that the renal clearance of substance X i

17.) (9 points) Recall that the renal clearance of substance X is calculated using

                                           C_X = (U_X)(V)/P_X

Inulin and paraaminohippurate (PAH) were infused intravenously into a human subject. The following measurements were made:

Plasma inulin concentration = 1 mg/100ml

Urine inulin concentration = 100mg/100ml

Plasma PAH concentration =60 mg/100ml

Urine PAH concentration= 12670 mg/100ml

The flow of urine (V) was 1.2 ml/min

Calculate the glomerular filtration rate

Calculate the clearance of PAH

What can you conclude about the fate of PAH in the kidney?

Solution

1. Glomerular filteration rate is calculated as,

GFR= U_insulin × V ÷ P_{insulin =} C_insulin

_{= 100×1.2÷1}

_{= 120 mg/100 ml.}

_{2.Clearance of PAH is calculated as,}

Clearance=U_{PAH ÷} P_PAH × V

= 12670÷60×1.2

= 211.16×1.2

= 253.39 mg/100 ml.

3.PAH undergoes both glomerular filteration and tubular secretion. By this we can conclude that the rate at which the kidneys can clear PAH from the blood reflects total renal blood flow.