12. Effective renal plasma flow

Question 1

Effective renal plasma flow ( eRPF) is ised to

Answer 1

calculate plasma flow (RPF) and hence estimate renal function

Question 2

Effective renal plasma flow (eRPF) can be estimating using … (why)

Answer 2

para- aminohippurinc ( PAH) clearance because filtration and secretion is nearly 100% excretion of all PAH that enter kidney

Question 3

Effective renal plasma flow ( eRPF) calculation

Answer 3

para- aminohippurinv ( PAH ) clearance = Upah x V/ Ppah

Question 4

renal blood flow (RBF) is the

Answer 4

volume of blood delivered to the kidneys per unit time

Question 5

renal blood flow (RBF) - calculation

Answer 5

renal blood flow (RBF) = RPF / (1-Hct)

1-Hct =plasma

Question 6

eRPG vs true renal plasma flow

Answer 6

eRPF underestimates true renal plasma flow slightly

Question 7

• What substance’s clearance is used to estimate renal plasma flow? Why?

Answer 7

Para-aminohippurate (PAH); 100% of PAH is actively secreted from the proximal tubule into the urine (the renal vein PAH concentration is 0)

Question 8

• What is the formula for estimating the effective renal plasma flow (ERPF) using para-aminohippuric acid (PAH)?

Answer 8

ERPF = urine PAH concentration times the urine flow rate divided by the plasma PAH concentration (UPAH × V/PPAH)

Question 9

• What is the formula for estimating renal blood flow (RBF) if renal plasma flow (RPF) is known?

Answer 9

RBF = RPF divided by (1 - the hematocrit) (normally, RBF is double the RPF)

Question 10

• You infuse para-aminohippuric acid (PAH) into a patient. An hour later, you still detect a trace of PAH in the patient’s blood. Why?

Answer 10

Ten percent of renal blood flow goes to the kidney parenchyma rather than through the glomerulus; PAH in this blood is not filtered

Question 11

• Effective renal plasma flow ____ (overestimates/underestimates) true renal plasma flow by approximately ____%.

Answer 11

Underestimates; 10 (because 10% of renal blood flow supplies the kidney parenchyma rather than passing through the glomerulus)

Question 12

• A man gets PAH. Plasma level (PPAH) is 2 mg/mL, urine level (UPAH) 550 mg/mL, urine flow (V) 4 mL/min. What is his ERPF (mL/min)?

Answer 12

ERPF = CPAH (mL/min) = (UPAH × V)/PPAH = (550 × 4)/2= 1100 mL/min

Question 13

• A man gets PAH; his ERPF is estimated to be 1000 mL/min. With a hematocrit (Hct) of 50%, what is his RBF (mL/min)?

Answer 13

RBF = RPF/(1 - Hct) = 1100 mL/min/(1 - 0.5) = 2200 mL/min (remember ERPF underestimates RPF by approximately 10%)