Question

The next two questions refer to the following data. You make the following measurements:

14) The glomerular filtration rate is

a. 1 ml/min

b. 200 ml/min

c. 0.01 ml/min

d. 0.0002 ml/min

e. 10000 ml/min

15) 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

explain the answer too.

	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

Answer 1

1 Ans B: 200ml/min

clearance of inulin = GFR

excretion rate = U x V

clearance of substace (inulin) = excretion rate / plasma concentration

= 3 mg/min / 15 mg /L

= (3/15) L/min

= 3 x 1000 ml /15

= 3000/15

= 200 ml/min

2. Ans D

given glucose values

	Plasma concentration	Urine concentration	Excretion rate
Glucose	20 mg/L	0.025 mg/L	0.0001 mg/min

from inulin GFR = 200ml/min

excretion rate = U x V

V = U / Excretion rate

GFR = U x V / Plasma con

200= U x V / 20

200 x 20 /1000 = U xV

4 mg /min = U x V

U x V excretion of glucose from kidney = 4 mg/min