RENAL CLEARANCE AND RENAL BLOOD FLOW

Dr.Mohammed Sharique Ahmed Quadri

Al Maarefa Colege

Objective

Describe the concept of renal plasma clearanceUse the formula for measuring renal clearanceUse clearance principles for inulin, creatinine etc. for determination of GFRUse PAH clearance for measuring renal blood flowOutlines the factors affecting the renal blood flow .

Plasma clearance Definition Volume of plasma completely cleared of a

particular substance by kidneys per minute

(not the amount of the substance removed)

Varies for different substances, depending on how the kidneys handle each substance

Unit- ml/min

EQUATION FOR RENAL CLEARANCE

Cx = (U)x X V (P)xWhere :- Cx = Clearance(ml/min)

(U)x = Urine concentration of the substance(mg/ml) V =Urine flow rate (ml/min) (P)x = Plasma conc of the

substance(mg/ml)

If substance is filtered , not reabsorbed or secreted , its plasma clearance rate equals GFR

Clearance of Various Substances Albumin – 0: normally albumin is not

filtered across the membrane Glucose – 0 :normally filtered glucose is

completely reabsorbed back into the blood stream

Inulin – is equal to the GFR.(Glomerular Marker) Inulin is a Fructose polymer ; freely filtered across the membrane and neither reabsorbed nor secreted.

PAH – Para Amino Hippuric Acid(& other organic acids).Has highest clearance since it is both filtered and secreted.

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

CLEARANCE RATIO Clearance of any substance (x) compared with

clearance of Inulin = C x ( glomerular marker)

C inulin C x = 1 (filtered & neither reabsorbed nor

secreted) C inulin C x < 1 (substance is not filtered/filtered &

reabsorbed) C inulin C x > 1 (substance is filtered as well as

secreted ) C inulin

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?

Renal plasma flow can be estimated from the clearance of Para-amminohippuric acid

(Ficks Principle) – The amount of a substance entering the organ is equal to the amount of substance leaving it( assuming that substance is neither synthesisized nor degraded by the organ ).

by measuring the amount of a given substance taken up per unit of time and dividing this value by the arteriovenous difference for the substance across the kidney.

Since the kidney filters plasma, the renal plasma flow equals the amount of a substance excreted per unit of time divided by the renal arteriovenous difference as long as the amount in the red cells is unaltered during passage through the kidney.

Any excreted substance can be used if its concentration in arterial and renal venous plasma can be measured and if it is not metabolized, stored, or produced by the kidney and does not itself affect blood flow.

PAH CLEARANCE PAH acid is an organic acid which is

almost 90% cleared in 1 circulation through the kidneys since it is both filtered and secreted by the tubules. – neither metabolized nor synthesized by

the kidneys.– PAH does not alter the RPF. – No other organ extracts PAH

Hence clearance of PAH gives the effective Renal Plasma Flow (EPRF)

Use of PAH Clearance to Estimate Renal Plasma Flow

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

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

To Calculate Actual RPF , One Must Correct for Incomplete Extraction of PAH( extraction ratio)

APAH =1.0

EPAH =APAH - VPAH

APAH

RPF =ERPF

EPAH

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

VPAH = 0.1

= 1.0 – 0.11.0

= 0.9

Renal Clearance gives an indication of the functioning of the kidneys.

Name Equation UnitsClearance Ux V

Px

mL/min

GFR U inulin x V P inulin

mL/min

Clearance Ratio

C x C inulin

None

Effective Renal Plasma Flow

U (PAH) x V P (PAH)

mL/min

REGULATION OF RENAL BLOOD FLOWRBF (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)

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