Question

A patient’s GFR is 125 ml/min, and his urine is produced at a rate of 1.25 ml/min. (A) By what factor is the inulin concentrated in his urine. (B) The concentration of glucose in his plasma is 5 mmol/l. His renal reabsorption of glucose is completely inhibited. What would be the concentration of glucose in his urine.

Answer 1

Here we will use a formula c=u×v/p

c-clearence-of a substance is the amount of plasma that is cleared of a substance per unit time also called renal clearence.

v-volume of urine flow=1.25ml/min(given)

u-concentration of substance in urine

p-concentration of substance in plasma.

For substance which are neither reabsorbed nor secreted in the tubules of kidney clearence is equal to the GFR(glomerular filtration rate) like inulin ,creatinin.

A.Here c=gfr=125ml/min

v=1.25ml/min

thus applying c=UV/p,

u/p=c/v

=125/1.25=100

u=100p

thus concentration of inulin in urine is 100 times that of concentration of inulin in plasma.( factor is 100)

B. Given glucose reabsorption is completely inhibited which refers c=gfr.=125ml/min

as per question,v=1.25nl/min and

p=5mmol/l

thus applying c=UV/p

u=cp/v

=[125ml/min×5mmol/l]÷[1.25ml/min]=500mmol/l

thus concentration of glucose in urine found tobe 100 times that of plasma that is 500 mmol/l.