A renal function test uses inulin clearance. Over a 30‑minute collection period the patient produces 36 mL of urine. The inulin concentration in plasma is (0.9\,\text{mg\,mL}^{-1}\) and in the collected urine it is (2.7\,\text{mg\,mL}^{-1}\).

(a) Determine the urine flow rate (in mL min$^{-1}$).
(b) Calculate the glomerular filtration rate (GFR) using the clearance formula $GFR = \frac{U_{\text{inulin}}\times V}{P_{\text{inulin}}}$.
(c) Compare the calculated GFR with the normal adult value of about 125 mL min$^{-1}$ and state what the result indicates about kidney function.

Question

A renal function test uses inulin clearance. Over a 30‑minute collection period the patient produces 36 mL of urine. The inulin concentration in plasma is (0.9\,\text{mg\,mL}^{-1}\) and in the collected urine it is (2.7\,\text{mg\,mL}^{-1}\).

(a) Determine the urine flow rate (in mL min$^{-1}$).
(b) Calculate the glomerular filtration rate (GFR) using the clearance formula $GFR = \frac{U_{\text{inulin}}\times V}{P_{\text{inulin}}}$.
(c) Compare the calculated GFR with the normal adult value of about 125 mL min$^{-1}$ and state what the result indicates about kidney function.

Accepted Answer

Urine flow rate = 1.2 mL min⁻¹; GFR = 3.6 mL min⁻¹, which is markedly below the normal ~125 mL min⁻¹, indicating severe renal impairment.

Igcse Model Questions Vol1 2020 Biology Question 2

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