Calculating Renal BF:

Renal Measurements:

Renal Blood Flow

GFR

Renal processes

Filtration

Reabsorption

Secretion

To calculate Clearance & RBF,

Inject known amount of a solute, i.v.

Wait a few minutes

Measure P_A in any artery

Measure P_V in RENAL VEIN

Measure Rate of urine formation

Measure [solute]_urine

Renal clearance

volume of plasma per minute needed to excrete the quantity of solute appearing in the urine in a minute

Renal Clearance:

The hypothetical volume of plasma from which a substance is completely removed per minute in one pass thorough the kidney

If there were 1 mg of Z in 100 ml of plasma, and

You found 0.5 mg of Z appearing in the urine/ min, then

the clearance of Z would = 50 ml of plasma

Clearance

C = U x V / P_A

U = [solute] in urine (mg/ml)

V = volume of urine/min (ml/min)

P_A = [solute] in arterial plasma (mg/100 ml plasma

Calculate Clearance

P_A = 1.0 mg/100 ml plasma

U = 0.1 mg/ml

V = 1.0 ml/min

C = 0.1mg/ml x 1.0 ml/min

1.0 mg / 100 ml

C = 10 ml/min

Understanding Clearance:

If in this example, 0.1 mg of solute appears in urine / min, in how much plasma was that 0.1 mg delivered if P_A = 1.0 mg/100 ml?

Answer = 10 ml/min

In other words, that 0.1 mg of solute that appeared in the urine/min was dissolved in 10 ml of plasma. Thus C = 10 mls/min.

Extraction Ratio (E)

The E of a solute is equal to the fraction of the substance that is removed from the plasma in one pass through the kidney:

E = P_A - P_V / P_A

Calculate E

P_A = 0.2 mg/100 ml

P_V = 0.1 mg/100 ml

E = 0.2 - 0.1 / 0.2

E = 0.5

Renal Plasma Flow

RPF = C / E

RPF = 10 ml/min / 0.5

RPF = 20 ml / min

Renal Blood Flow

RBF = RPF x 1 / 1- hct

If hct = 0.5,

RBF = 20 ml/min x 1 / 1 - 0.5

RBF = 40 ml / min

Calculating Renal BF:

X 1.0 mg/100ml 0.8/100 0.2 mg/ml 1 ml/min

Y 1.0 0.3 0.7 1

Z 1.0 0.9 0.1 1

Hct = 0.50

To calculate RBF, first calculate C and E of substance with largest E

E_Y = 0.7

C_Y = 0.7 mg/ml x 1 ml/min

1.0 mg / 100 mls

C_Y = 70 ml / min

From C and E, calculate RPF:

RPF = C / E

RPF = 70 ml / min

0.7

RPF = 100 ml/ min

From RPF & Hct, calculate RBF:

RBF = RPF x 1

1 - Hct

RBF = 100 ml / min x 1

0.5

RBF = 200 ml / min

GFR = C of solute which is ONLY FILTERED, e.g., inulin.


WHY?

Inferences from GFR

If C_solute = GFR, solute is only filtered

If C_solute > GFR, solute is filtered and secreted

If C_solute < GFR, solute is filtered and reabsorbed

Filtration Fraction

FF = fraction of plasma that is made into filtrate

Usually = 0.2 in humans

Calculate:

FF = GFR / RPF

Inferences from FF:

If E_solute = FF, solute is only filtered

If E_solute > FF, solute is filtered and secreted

If E_solute < FF, solute is filtered and reabsorbed

Assume X from data set is only filtered

Then, GFR = C_X = 0.2 x 1

1.0 / 100

GFR = 20 ml / min

FF = GFR / RPF

FF = 20 ml/min / 100 ml/min

FF = .2

Calculate Amount of a solute filtered / min:

Amount Filtered / min = P_A x GFR

For Z, Amt Filt =

1.0mg/100 mls x 20 ml / min

Amt Filt = 0.2 mg / min

Amount of a Solute Appearing in Urine / min

Amt Urine = U x V

For Solute Z, Amt Urine =

0.1 mg / ml x 1 ml / min

Amt in Urine / min = 0.1 mg / min

Amount reabsorbed / min

= Amt filtered - Amt in Urine

= (GFR x P_A) - (U x V)

= (20 ml/min x 1.0 / 100) -

(0.1 mg/ml x 1 ml/min)

= (0.2 mg/min) - (0.1 mg / min)

= 0.1 mg / min is reabsorbed

Amount Secreted / min

= Amt in Urine - Amt Filtered

= (U x V) - (P_A x GFR)

= (0.7 mg/ml x 1 ml/min) -

(1.0 mg/ 100 ml x 20 ml/min)

= (0.7 mg/min) - (0.2 mg/min)

= 0.5 mg/min

Inferences:

If UV = GFR P_A , solute is only filtered

If UV > GFR P_A , solute is filtered and secreted

If UV < GFR P_A , solute is filtered and reabsorbed