Ask Yourself!

• Can it be filtered?• Is it reabsorbed?• Is it secreted?• What factors regulate

the amount filtered, reabsorbed, and secreted?– Size– Permeability (charge)– Surface area

5L/min

CO

1.25L/min

RBF

660 ml/min

RPF

125 ml/min

GFR

Filtration fraction: GFR/RPF

Urine formation 2ml/min

RBF (renal blood flow) ml/min

• RBF = ¼ CO

• RBF = ¼ (5L/min) = 1.25L/min= 1250ml/min

• Turns out, it really is a little less than this (some vasoconstriction, etc.) so we will use 1200ml/min

1.25-1.2L/min

RBF

Effective Renal Blood Flow (ERBF; ml/min)

• Only about 90% of the RBF actually reaches the nephrons at the glomerulus, therefore; this number indicates the estimated, or effective, blood flow that actually reaches the nephrons.

• ERBF = 0.9(RBF) • ERBF = 0.9(1200ml/min) = 1080

ml/min

1.08L/min

Renal Plasma Flow (RPF in ml/min)

• Since it is the plasma that is actually filtered in the kidney, it is important to know the amount of plasma that actually reaches the kidneys which is represented by this number.

• RPF = RBF(1 – HCT)• Normal HCT = .45 (men) • RPF = 1200 ml/min(1-.45) = 660

ml/min• RPF ≈ 660ml/min (Normal value used

in lab)

660 ml/min

Effective Renal Plasma Flow (ERPF in ml/min)

• As in the blood flow, only about 90% of the plasma that flows through the renal artery actually reaches the nephrons. This figure represents the estimated amount of plasma that reaches the glomerulus.

• ERPF = 0.9(RPF) • ERPF = 0.9(660 ml/min) = 594

ml/min• ERPF ≈ 594 ml/min (Normal value

used in lab)594 ml/min

Glomerular Filtration Rate (GFR in ml/min) • This figure refers to the amount of plasma that is

actually filtered into Bowman’s capsule. About 20% of the plasma that reaches the glomerulus is filtered but the lab manual figures this number by reducing the amount per day to amount per minute.

• About 180 L of plasma is filtered daily.• GFR = 180 L/day = .125 L/min = 125 ml/min 1440 min/day• GFR ≈ 125 ml/min (Normal value used in lab)

125 ml/min

GFR

Filtration fraction: GFR/RPF

Filtered Load (aka-”tubular load”)

• The total amount of any non-protein or non-protein bound substance filtered into bowman’s capsule. Each solute is independently calculated!

• Filtration Load or Tubular Loads = (GFR) x [ substance] plasma

• Example: glucose: 180 L/day x 1g/L

• Compare what is filtered to what is actually found in the urine…..tells you what is reabsorbed or secreted (think about this).

Excretion in mg/min

• This will determine the amount of a substance that is excreted in the urine in a specific time frame considering filtration, secretion, & reabsorption. (again, each substance/solute is independent)

• Excretions = filtrations + secretions - reabsorptions

• Be ready to rearrange the above equation to solve for other values!

• Excretion = (Urine Flow Rate)(Urine Concentrations) This can also be used for excretion

Clearance

• Renal clearance of a substance is the volume ofplasma completely cleared of a substanceper min by the kidneys.

• Clearance is a general concept that describes the rate at which substances are removed (cleared) from the plasma— units ml/min.

Renal Clearance Value (RCV in ml/min)

Renal clearance of a substance is the volume of plasma completely cleared of a substance per min.

RCVs = Us x R Ps

RCVs x Ps = Us x R

Where: RCVs = clearance of substance SPs = plasma conc. of substance SUs = urine conc. of substance SR = urine flow rate

How long would it take to clear a substance that is only filtered?

This is ERPF (594ml) arriving in the glomerulus with a substance dissolved in the plasma

This is GFR (125) with some of the substance entering the glomerular capsule (filtered load)

This is the 469ml of plasma that did not get filtered with the substance dissolved in it leaving the glomerulus to leave the kidney and recirculates through the body.

Production of Glomerular Ultrafiltrate

Excretionrate = Filtration

rateSecretion

rateReabsorption

rate

A substance that is neither reabsorbed nor secreted can be used to determine the amount of filtrate produced (GFR) We use inulin or creatinine

Filtration

Excretion

+ -

Inulin

• Small polysaccharide (energy storage in some plants)

• Not digested

• Not reabsorbed or secreted

Amount of inulin excreted / min = amount of inulin filtered / min

Renal Clearance of Inulin: Calculating GFR

1. Inject inulin 2. Obtain blood sample and determine plasma inulin concentration (P i)

from Jung et al. (1992)

Filtration load = GFR X Pi

(mg/min) (ml/min) (mg/ml)

1. Inject inulin 2. Obtain blood sample and determine plasma inulin concentration (P i)3. Obtain urine sample and determine rate of inulin excretion

Excretion per minute = R X Ui

(mg/min) (ml/min) (mg/ml)

R = Urine flow rate Ui = inulin concentration in urine

Renal Clearance of Inulin: Calculating GFR

1. Inject inulin 2. Obtain blood sample and determine plasma inulin concentration (P)3. Obtain urine sample and determine rate of inulin excretion4. Solve for GFR

For inulin...

Thus,

Quantity filtered per minute (mg/min)

= Quantity excreted per minute (mg/min)

= GFR X P (ml/min) (mg/ml)

R X U(ml/min) (mg/ml)

GFR = R X Ui = RCVi

Pi

R = urine flow rateUi = inulin concentration in urinePi = inulin concentration in plasma

Renal Clearance of Inulin: Calculating GFR

ExampleA doctor suspects her patient has kidney damage and has inulin infused into his vein. She then recovered 30 mg/ml of inulin is his blood plasma and 0.5 mg/ml of inulin in his urine. If the rate of urine formation was 2 ml/min, what is the GFR of the patient?

GFR = R X Ui

Pi

GFR = 2 ml/min x 30 mg/ml = 120 ml/min 0.5 mg/ml

Stages of Kidney Disease

Stage Kidney damage GFR (ml/min)

1 Little (e.g. some proteinuria) 90 or above

2 Mild 60 to 89

3 Moderate 30 to 59

4 Severe 15 to 29

5 Kidney failure Less than 15

Can we use creatinine? What are its advantages? Disadvantages? Yes! Endogenous slight over-estimate of GFR

Theoretically, if a substance is completely cleared from the plasma, its clearance rate would equal effective renal plasma flow.

Use of Clearance to Estimate Renal Plasma Flow

Cx = renal plasma flow

Paraminohippuric acid (PAH) is freely filtered and secretedand is virtually completely cleared from the renal plasma

Excretionrate = Filtration

rateSecretion

rateReabsorption

rate+ -

Measuring Effective Renal Plasma Flow

[RA] PAH x ERPF

[RV] PAH x ERPF

[U] PAH x R

Amount of PAH entering the kidney = the amount leaving the kidney

[RA] PAH x ERPF =( [RV] PAH x ERPF) + ([U] PAH x R)

Solve for ERPF

ERPF = [U] PAH x R[RA] PAH - [RV] PAH

Because PAH is filtered and completely secreted, [RV] PAH is nearly zero!

SO.........

Use of PAH Clearance to Estimate Renal Plasma Flow

1. amount enter kidney =ERPF x PPAH

3. ERPF x Ppah = UPAH x R

ERPF = UPAH x R

PPAH

ERPF = Clearance PAH

2. amount entered = amount excreted~

Copyright © 2006 by Elsevier, Inc.

Clearances of Different Substances

Clearance of inulin (Cin) = GFRif Cx < Cin: indicates reabsorption of x

Clearance of PAH (Cpah) ~ effective renal plasma flow

Substance Clearance (ml/min inulin (used to estimate GFR) 125Creatinine (used to estimate GFR) 140 (slight secretion) PAH(used for ERPF/ RPF/ RBF) 594 glucose 0 sodium 0.9 urea 70

Clearance creatinine (Ccreat) ~ 140 (used to estimate GFR)if Cx > Cin: indicates secretion of x

Use the “long” excretion equation for calculation of tubular reabsorption

Reabsorption = Filtration -Excretion

Filt s = GFR x Ps (Ps = Plasma conc of s)

Excret s = Us x R Us = Urine conc of s R = urine flow rate

(when Excret s < Filt s)

Use the “long” excretion equation for calculation of tubular secretion

(when Excret s > Filt s)Secretion = Excretion - Filtration

Filt s = GFR x Ps

Excret s = Us x R