Transcript renal clearance

RENAL CLEARANCE AND
RENAL BLOOD FLOW
Dr.Mohammed Sharique Ahmed
Quadri
Al Maarefa Colege
Objective
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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 .
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)x
Where :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)
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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
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CLEARANCE RATIO
 Clearance of any substance (x) compared with clearance of
Inulin = C x
( glomerular marker)
C inulin
 Cx
= 1 (filtered & neither reabsorbed nor secreted)
C inulin
 Cx
< 1 (substance is not filtered/filtered & reabsorbed)
C inulin
 Cx
> 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
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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 secreted
and is almost completely cleared from the renal plasma
1. amount enter kidney =
RPF x PPAH
2. amount entered =~ amount excreted
3. ERPF x Ppah
ERPF =
= UPAH x V
UPAH x V
PPAH
ERPF = Clearance PAH
~ 10 % PAH
remains
To Calculate Actual RPF , One Must Correct
for Incomplete Extraction of PAH( extraction
ratio)
APAH =1.0
EPAH =
APAH - VPAH
APAH
= 1.0 – 0.1 = 0.9
1.0
normally, EPAH = 0.9
i.e PAH is 90 % extracted
RPF =
ERPF
EPAH
VPAH = 0.1
Renal Clearance gives an indication of
the functioning of the kidneys.
Name
Equation
Units
Clearance
Ux V
Px
U inulin x V
P inulin
Cx
C inulin
U (PAH) x V
P (PAH)
mL/min
GFR
Clearance Ratio
Effective Renal
Plasma Flow
mL/min
None
mL/min
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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
R
The 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 .
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AUTOREGULATION
BLOOD FLOW
1.
Myogenic theory
OF
RENAL
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
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