Page 626 - First Aid for the USMLE Step 1 2020, Thirtieth edition [MedicalBooksVN.com]_Neat

582 SeCTIOn III Renal ` RENAL—PhysioLogy Renal ` RENAL—PhysioLogy

Renal clearance C = (U V)/P = volume of plasma from which C = clearance of X (mL/min).
x
x
x
x
the substance is completely cleared in the U = urine concentration of X (eg, mg/mL).
x
urine per unit time. P = plasma concentration of X (eg, mg/mL).
x
If C < GFR: net tubular reabsorption and/or V = urine flow rate (mL/min).
x
not freely filtered.
If C > GFR: net tubular secretion of X.
x
If C = GFR: no net secretion or reabsorption.
x
Glomerular filtration Inulin clearance can be used to calculate GFR 14
rate because it is freely filtered and is neither
reabsorbed nor secreted. 12
C = GFR = U × V/P
inulin inulin inulin 10
= K [(P – P ) – (π – π )]
f GC BS GC BS
(GC = glomerular capillary; BS = Bowman 8
space; π normally equals zero; K = filtration Plasma creatinine (mg/100 mL)
BS f
coefficient). 6
Normal GFR ≈ 100 mL/min. 4
Creatinine clearance is an approximate measure
of GFR. Slightly overestimates GFR because 2
creatinine is moderately secreted by renal
tubules.
25 50 75 100 125 150
Glomerular filtration rate
(mL/min)

Effective renal plasma Effective renal plasma flow (eRPF) can be estimated using para-aminohippuric acid (PAH)
flow clearance. Between filtration and secretion, there is nearly 100% excretion of all PAH that enters
the kidney.
eRPF = U PAH × V/P PAH = C PAH .
Renal blood flow (RBF) = RPF/(1 − Hct). Usually 20–25% of cardiac output.
eRPF underestimates true renal plasma flow (RPF) slightly.

FAS1_2019_14-Renal.indd 582 11/7/19 5:42 PM