Renal Equations

Question 1

Renal clearance

Answer 1

volume of plasma from which the substance is completely cleared per unit time

Cx = UxV/Px

Cx = clearance of X 
Ux = urine concentration of X 
Px = plasma concentration of X 
V = urine flow rate 

If Cx GFR: net tubular *secretion of X.
If Cx = GFR: *no net secretion or reabsorption.

Question 2

Glomerular filtration rate

Answer 2

*Inulin clearance can be used to calculate GFR because it is freely filtered and is *neither reabsorbed nor secreted.

Normal GFR ≈ 100 mL/min

GFR = Uinulin × V/Pinulin = CInulin

Uinulin=urine concentration
V=flow rate
Pinulin=plasma conc
Cinulin=clearance

GFR= Kf [(PGC – PBS) – (πGC – πBS)]
(GC = glomerular capillary; BS = Bowman space.)
πBS normally equals zero; Kf = filtration constant.

Question 3

Effective renal plasma flow (eRPF)

Answer 3

can be estimated using para-aminohippuric acid (PAH)
clearance because between filtration and secretion there is nearly *100% excretion of all PAH that enters the kidney.

eRPF = UPAH × V/PPAH = CPAH.

Renal blood flow (RBF) = RPF/(1 − Hct)

Plasma = 1 − hematocrit.

eRPF underestimates true renal plasma flow (RPF) slightly.

Question 4

Filtration fraction (FF)

Answer 4

Filtration fraction (FF) = GFR/RPF

Normal FF = 20%.

Question 5

Filtered load

Answer 5

GFR × Px

Px=Plasma conc

Question 6

Excretion rate

Answer 6

V × Ux

V=Urine flow rate
Ux=Urine conc

Question 7

Reabsorption

Answer 7

filtered – excreted

Question 8

Secretion

Answer 8

excreted – filtered