Question

The next two questions refer to the following data.

	Plasma concentration	Urine concentration	Excretion rate
Glucose	20 mg/L	0.025 mg/L	0.0001 mg/min
Inulin	15 mg/L	750 mg/L	3 mg/min

14) calcute The glomerular filtration rate

a. 1 ml/min b. 200 ml/min c. 0.01 ml/min d. 0.0002 ml/min e. 10000 ml/min

15) what is the rate at which glucose passes into the kidney fluid is…

a. 0.2 mg/min b. 0.25 mg/min c. 1 mg/min d. 4 mg/min e. 200 mg/min

Answer 1

14)b.200ml/min

Gfr can be calculated from inulin clearance.

Gfr=(Urinary inulin/plasma inulin)×urine flow rate

Here 1st we have to calculate urine flow rate. Here excretion rate of inulin is 3mg in 1 minute. Also urine contains 750 mg inulin in 1000ml.so 3mg inulin will be present in 1000/250 =4ml urine. So urine flow rate is 4ml/minute.

So gfr=(750/15)×4=200ml/minute.

15)d.4mg/minute.

Here we have to calculate how much glucose passes from plasma to kidney fluid in one minute.

200ml of plasma is filtered per minute. And plasma contains 20mg glucose in 1000ml.so glucose passing =20/5=4mg/minute.