Glomerular Filtration

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Glomerular Filtration Factors Affecting
GFR A.A.J.Rajaratne
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Objectives

• Describe the glomerular membrane, interms of the
  major layers and its permeability characteristics
• Explain in terms of size and electrical charges
  of pores of the membrane and why the glomerular
  membrane has a high degree of selectivity.

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• State the glomerular pressure, Bowmans capsular
  pressure and colloid osmotic pressure in the
  glomerular capillaries and explain how these
  pressures cause filtration of fluid at the
  glomerulus
• State the composition of the glomerular filtrate
• Explain the terms of GFR and filtration fraction
  and give their normal values

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• Describe the effect of the following on the GFR
• Renal blood flow
• Afferent and efferent arteriolar constriction
• Sympathetic stimulation
• Outflow obstruction
• Recognize that GFR is kept constant with wide
  changes in arterial blood pressure by means of
  autoregulation

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GLOMERULAR FILTRATION
GFR Males 90-140 ml/min Females 80 -125
ml/min About 180 litres of plasma is filtered
per day
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Glomerular Filtrate
Devoid of cells and protein Concentration of
salts and organic molecules (glucose and amino
acids) are similar in the plasma and ultra
filtrate
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The Glomerulus

• Capillary endothelium - pores 100nM
• Basement membrane
• Tubular epithelium pores - 8nM

Filtrate plasma - protein
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EFP(PGC PBS) (COPGC-COPBS) GFR Kf .
EFP Kf Filtration coefficient ml/min/mmHg
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Starling Forces
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Starling Forces
For Filtration Hydrostatic Pressure in
Glomerulus (45 mm Hg) Colloid O.P. in tubule
(0 mmHg) Against Filtration Hydrostatic
Pressure in tubule (10 mm Hg) Colloid O.P.
in glomerular blood (25mm Hg)
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• GFR can be altered by changing
• Kf
• 2. Any of the Starling forces

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• Factors affecting glomerular filtration
• Role of hydrostatic and oncotic pressures
• 2. Role of capillaries
• 3. Role of mesangial cells

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Renal Blood Flow
1.25 L per minute 25 of the CO Indirectly
determine GFR Modify rate of water and solute
reabsorption by the PCT Participating in
concentration and dilution of urine Delivering
substrates for the excretion in the urine
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Autoregulation of RBF GFR
RBF GFR remains relatively constant between 90
180 mmHg
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Autoregulation of RBF GFR

1. Pressure sensitive - Myogenic mechanism
2. NaCl concentration dependent mechanism

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Autoregulation of RBF GFR

• Is absent below arterial pressures of 90mmHg
• Autoregulation is not perfect
• Despite autoregulation, RBF GFR can be changed
  under appropriate conditions by several hormones.

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Major Hormones that Influence GFR RBF
Vasoconstrictors Sympathetic nerves Angitensin
II Endothelin
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Major Hormones that Influence GFR RBF
Vasodilators Prostaglandin (PGE2, PGI2)Nitric
Oxide Bradykinin ANP
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Tubulo-glomerular Feedback

• SNGFR is greater in juxtamedullary nephrons 50
  nl/min (others 30 nl/min)
• The SNGFR is determined by composition of the
  fluid in the distal nephron.
• The mechanism is called tubulo-glomerular feed
  back.

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Estimation of glomerular filtration rate
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Clearance The clearance of a substance is the
volume of plasma from which the substance was
completely cleared by the kidneys per unit time,
(units vol. plasma/time),
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Clearance of X, CX GFR when the substance X
meets the following criteria, i. freely
filterable at the glomerulus ii. not reabsorbed
by tubules iii. not secreted by tubules iv. not
synthesised by tubules v. not broken-down by
tubules
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These criteria are met by the polysaccharide
inulin, so CIN GFR. However, inulin does not
occur naturally in the body and requires several
hours of infusion to reach steady state
concentration. Therefore, creatinine is used to
estimate GFR.
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Creatinine is formed from muscle creatine and
is released at approximately a constant rate.
Therefore blood Cr changes little per 24
hrs. However, Cr is secreted by the tubules and
overestimates GFR by small amount. for freely
filterable substances, when, a. CX lt CIN ? net
tubular reabsorption b. CX gt CIN ? net tubular
secretion
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Para-aminohippurate PAH Undergoes tubular
secretion and is freely filterable. At low
PAHpl, virtually all PAH escaping filtration is
secreted by the tubule. Therefore virtually all
plasma supplying secreting nephrons is cleared of
PAH. About 10-15 of total renal plasma flow
(TRPF) supplies non-secreting portions of the
kidney. Thus, CPAH actually measures effective
renal plasma flow (ERPF). This is 85-90 of
TRPF.
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This only applies at a low PAHpl, at higher
levels the TMax is exceeded. The radiographic
contrast Diodrast is handled similarly to
PAH. urea is freely filterable but 50 is
reabsorbed (R 40-60). The amount reabsorbed
depends on flow rate. Therefore urea is less
accurate than creatinine as an estimate of GFR.
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Fractional Excretion Fractional excretion, FEX ,
is the mass of "x" excreted as a fraction of the
total mass filtered,
and if, 1. FEX lt 1.0 ? net tubular
reabsorption 2. FEX gt 1.0 ? net tubular
secretion
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Estimation of glomerular filtration rate
This may be done using the clearance of a
substance present in plasma that is filtered
freely at the glomerulus, but is neither
reabsorbed nor secreted by the tubules. Inulin,
a polysaccharide of MW 5000, satisfies these
criteria.
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Inulin clearance
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Since inulin cannot be absorbed or secreted, all
inulin that is filtered at the glomerulus
must appear in the urine. In the steady state,
the rate of filtration (moles/min) of inulin must
be equal to its excretion in the urine
(moles/min). Rate of inulin filtration
(moles/min) Pinulin (moles/ml) x GFR
(ml/min) Rate of excretion of inulin (moles/min)
Uinulin (moles/ml) x V (ml/min) where Uinulin
and Pinulin represent the concentrations of
inulin in urine and plasma, respectively, and V
represents the rate of urine flow.
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2. inulin filtration rate inulin excretion rate
Pinulin x GFR Uinulin x V So that GFR
UinV/Pin 3. The right-hand-side of equation
the clearance of inulin. (The clearance of any
other substance is UV/P for that substance.) 4.
The GFR is equal to the clearance of inulin, or
of any other substance that is freely filtered,
but neither reabsorbed nor secreted. 5. The
"normal" value of GFR for a 70 kg human is about
125 ml/min.
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Creatinine clearance

• In clinical practice creatinine clearance may be
  used to estimate GFR for the following reasons.
• Inulin is not produced endogenously.
• Therefore, it must be infused intravenously if
  it is to be used in renal function tests.
• It is much more convenient to use a substance
  that is normally present in plasma that is freely
  filtered, but neither secreted, nor reabsorbed.
• 2. Creatinine, a normal breakdown product of
  creatine, is an endogenous compound that fulfils
  these criteria.

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Creatinine clearance - contd However, in
humans, a small amount of creatinine is secreted
into the urine in the proximal tubules. Consequen
tly, the rate of excretion of creatinine exceeds
its rate of filtration by 5 to 10. The
clearance of creatinine thus exceeds the true GFR
by 5 to 10.
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Creatinine clearance - contd
All that is required is a single plasma sample
and 24 hour urine collection. In most renal
diseases the GFR is substantially reduced. This
is detected by a diminished creatinine clearance
or, more frequently, by elevated Pcreatinine.
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PAcr x RPF
PVcrx RPF
No reabs No secretion.
Filtered PcrX GFR
Filtered Excreted Pcr X GFR Ucr X V
Small amount is secreted
Excreted Ucr X V
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Estimation of renal plasma flow rate
Certain substances are extracted from peritubular
capillary plasma and secreted into the proximal
tubular fluid in large amounts. The
concentration of such substances in renal venous
plasma is therefore, much less than in renal
arterial plasma. In such cases, the direct
Fick Principle (blood flow rate of
excretion/(A-V) concentration difference) can
be used to estimate renal plasma flow rate. PAH
(para-aminohippuric acid) might be used for this
purpose.
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1. In the steady state, the rate at which PAH
enters the kidney (moles/min) in the renal
arterial plasma is equal to the rate at which PAH
leaves the kidney in the urine and in renal
venous blood. PAH (moles/min) entering kidney in
renal arterial plasma Pa PAH x RPF PAH
(moles/min) leaving kidney in renal venous plasma
Pv PAH x RPF PAH (moles/min) leaving kidney in
the urine UPAH V
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2. The rate at which PAH enters the kidney is
equal to the rate that it leaves. Then,
3. Note that this last expression is the Direct
Fick formula for measuring plasma flow the blood
flow is equal to the rate of consumption
(excretion) divided by the arterio-venous
concentration difference.
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4. It is common to make the approximation that
the concentration of PAH in renal venous blood is
zero. Substituting zero for venous PAH, allows
us to compute a quantity called the effective
renal plasma flow (ERPF).
5. Note that the ERPF is equal to the clearance
of PAH and that it underestimates the true renal
plasma flow by approximately 10.
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6. The "normal value" of ERPF in a 70 kg human is
about 625 ml/min. Since the PAH clearance
underestimates renal plasma flow by about 10,
the true renal plasma flow (RPF) is about 700
ml/min. 7. The renal blood flow (RBF) is
then RBF RPF/1 - Hct ml blood/min, so that if
the hematocrit is 45, and RPF 700 ml/min, then
RBF is 1273 ml/min. This is 20 to 25 of cardiac
output.
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8. The filtration fraction is defined as the
ratio of GFR/RPF. If GFR 125 ml/min and RPF
700 ml/min, then the filtration fraction is
0.179. The filtration fraction is typically
between 0.15 and 0.20.