RENAL BLOOD FLOW AND CLEARANCE

Dr. Shaikh Mujeeb AhmedAssistant ProfessorAlMaarefa College

URINARY BLOCK 313

Objectives

• Describe the concept of renal plasma clearance• Use the formula for measuring renal clearance• Use clearance principles for inulin, creatinine

etc. for determination of GFR• Use PAH clearance for measuring renal blood

flow• Outlines the factors affecting the renal blood

flow .

• Although kidneys constitute less than 0.5% of total body mass, they receive 20-25% of resting cardiac output

• Left and right renal artery enters kidney

• Branches into segmental, interlobar, arcuate, interlobular arteries

• Each nephron receives one afferent arteriole

• Divides into glomerulus – capillary ball

• Reunite to form efferent arteriole (unique)

• Divide to form peritubular capillaries or some have vasa recta

• Peritubular venule, interlobar vein and renal vein exits kidney

RENAL BLOOD SUPPLY

RENAL BLOOD SUPPLY

4

Renal blood flow and Oxygen consumption

• Oxygen consumption per 100 g of renal tissue is 5 ml/min. which is next highest to the heart (8ml/min).

• Total renal blood flow is ≈ 1200/min.• RBF per 100 g tissue is ≈ 400 mL/min which is

disproportionately high as compared to heart which is only 80 mL/min.

• A-V O2 difference is 1.5ml O2/100ml of blood flow. In heart it is ≈ 10 ml O2/100ml.

Renal blood flow and Oxygen consumption

• 90 % of RBF goes to the renal cortex, • 9% goes to outer medulla and • 1% to the inner medulla

REGULATION OF RENAL BLOOD FLOW

RBF (Q)is directly proportional to the pressure gradient

(ΔP) between the renal artery and the renal vein

Is inversely proportional to the resistance(R) of the renal vasculature

(Q) = Δ P RThe major mechanism of changing Renal blood

flow is by changing Afferent or Efferent Arteriolar resistance.

1) SYMPATHETIC NERVES AND CIRCULATING CATACHOLAMINES

Both afferent and efferent arterioles are innervated by sympathetic nerves that act via α1 receptors to cause vasoconstriction.

However ,since far more α1 receptors are present on Afferent arterioles, increased sympathetic stimulation will cause a decrease in both RBF & GFR.

2) ANGIOTENSIN II This is a potent vasoconstrictor. However Efferent

arteriole is more sensitive to Angiotensin II. Hence low levels of Angiotensin II causes increase in GFR while high levels of Angiotensin II will decrease GFR. RBF is decreased.

3) PROSTAGLANDINS PGE 2, PGI 2 are produced locally in the kidneys –

cause vasodilation of both afferent & efferent arterioles.

This effect is protective for renal blood flow , it modulates the vasoconstriction produce by sympathetic & angiotensin-II

  4)DOPAMINE At low levels Dopamine dilates Cerebral, Cardiac,

Splanchnic & Renal arterioles and constricts Skeletal Muscle and Cutaneous arterioles. Hence low dose Dopamine can be used in the treatment of hemorrhage .

AUTOREGULATION OF RENAL BLOOD FLOW

  1. Myogenic theory 2. Tubuloglomerular feedback by Juxta

Glomerular Apparatus (JGA)

Assessing Kidney Function

• Albumin excretion (microalbuminuria)• Plasma concentration of waste products

(e.g. BUN, creatinine)• Urine specific gravity, urine concentrating ability• Imaging methods (e.g. MRI, PET, arteriograms,

iv pyelography, ultrasound etc)• Isotope renal scans• Biopsy• Clearance methods (e.g. 24-hr creatinine clearance)• etc

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.

Clearance Technique

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

Cs = Us x V Ps

Cs x Ps = Us x V

Where: Cs = clearance of substance SPs = plasma conc. of substance SUs = urine conc. of substance SV = urine flow rate

INULINGlucose

UreaPAH

For a substance that is freely filtered, but not reabsorbed or secreted (inulin, 125 I-iothalamate, creatinine), renal clearanceis equal to GFR

Use of Clearance to Measure GFR

amount filtered = amount excreted

GFR x Pin = Uin x V

GFR = Pin

Uin x V

Calculate the GFR from the following data:

Pinulin = 1.0 mg / 100mlUinulin = 125 mg/100 mlUrine flow rate = 1.0 ml/min

GFR =125 x 1.0

1.0= 125 ml/min

GFR = Cinulin =Pin

Uin x V

• Clinically it is not convenient to use inline clearance – to maintain a constant plasma concentration it must be infused continuously throughout measurement

• Creatinine clearance is used as a rough estimate of GFR

• Normal plasma serum creatinine level is 1mg/100ml.

Effect of reducing GFR by 50 % on serum creatinine

concentration and creatinine excretion rate

Figure 27-18;Guyton and Hall

Steady-state relationship

betweenGFR and serum

creatinine concentration

Figure 27-19;Guyton and Hall

Sample problem

In a 24hr period, 1.44 L of urine is collected from a man receiving an infusion of inulin.

In his urine, the [inulin] is 150mg/ml, and [Na+] is 200 mEq/L.

In his plasma, the [inulin] is 1mg/mL, and the [Na+] is 140mEq/L

What is the clearance ratio for Na+, and what is the significance of its value?

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

Use of Clearance to Estimate Renal Plasma Flow

Cx = renal plasma flow

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

Use of PAH Clearance to Estimate Renal Plasma Flow

1. amount enter kidney =RPF x PPAH

3. ERPF x Ppah = UPAH x V

ERPF = UPAH x V

PPAH

ERPF = Clearance PAH

2. amount entered = amount excreted~

~ 10 % PAHremains

Copyright © 2006 by Elsevier, Inc.

PAH clearance example

Calculate the RPF from the following data:

PPAH = 0.01 mg / 100mlUPAH = 5.85 mg/100 mlUrine flow rate = 1.0 ml/min

ERPF =5.85 x 1.0

0.01= 585 ml/min

ERPF = CPAH =PPAH

UPAH x V

To Calculate Actual RPF , One Must Correct for Incomplete Extraction of PAH

EPAH =APAH - VPAH

APAH

RPF =ERPF

EPAH

normally, EPAH = 0.9i.e PAH is 90 % extracted

APAH =1.0

VPAH = 0.1

= 1.0 – 0.11.0

= 0.9

To Calculate Actual RPF , One Must Correct for Incomplete Extraction of PAH

EPAH =APAH - VPAH

APAH

normally, EPAH = 0.9i.e PAH is 90 % extracted

APAH =1.0

VPAH = 0.1

= 1.0 – 0.11.0

= 0.9

RPF =585

0.9= 650

RBF =RPF

1 - Hct

RBF =650

0.55 = 1182

RBF =650

1 - 0.45

Calculation of Tubular Reabsorption

Reabsorption = Filtration -Excretion

Filt s = GFR x Ps

Excret s = Us x V

Calculation of Tubular Secretion

Secretion = Excretion - Filtration

Filt s = GFR x Ps

Excret s = Us x V

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 125 PAH 600 glucose 0 sodium 0.9 urea 70

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

References

• Human physiology by Lauralee Sherwood, seventh edition

• Text book of physiology by Linda .s contanzo,third edition

• Text book physiology by Guyton &Hall,11th edition