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
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