The document discusses glomerular filtration, detailing the structure and function of the glomerular membrane, and the factors influencing glomerular filtration rate (GFR). It covers the roles of pressures, renal blood flow, and hormonal influences, as well as methods for estimating GFR using inulin and creatinine clearance. Additionally, it explains concepts like fractional excretion and effective renal plasma flow, providing normal values and implications for kidney function assessment.

1. Glomerular Filtration
Glomerular Filtration
&
&
Factors Affecting GFR
Factors Affecting GFR

2. Objectives
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.

3. •State the glomerular pressure, Bowman’s 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

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

5. GFR –
Males 90-140 ml/min
Females 80 -125 ml/min
About 180 litres of plasma is
filtered per day
GLOMERULAR FILTRATION

6. 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

9. • Capillary endothelium - pores 100nM
• Basement membrane
• Tubular epithelium pores - 8nM
The Glomerulus
Filtrate = plasma - protein

10. EFP=(PGC – PBS) – (COPGC-COPBS)
GFR = Kf . EFP
Kf – Filtration coefficient ml/min/mmHg

12. Starling Forces
Starling Forces

13. 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)
Starling Forces
Starling Forces

15. GFR can be altered by changing
1. Kf
2. Any of the Starling forces

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

22. 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
Renal Blood Flow

25. Autoregulation of RBF & GFR
RBF & GFR remains relatively
constant between 90 – 180 mmHg

26. Autoregulation of RBF & GFR
1. Pressure sensitive - Myogenic
mechanism
2. NaCl concentration dependent
mechanism

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

29. Major Hormones that Influence
GFR & RBF
Vasoconstrictors
Sympathetic nerves
Angiotensin II
Endothelin

30. Major Hormones that
Influence GFR & RBF
Vasodilators
Prostaglandin
(PGE2, PGI2)
Nitric Oxide
Bradykinin
ANP

32. Tubulo-glomerular Feedback
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.

33. Estimation of
glomerular filtration
rate

34. 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),
CX = UX V
PX

35. 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

36. 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.

37. 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 < CIN  net tubular reabsorption
b. CX > CIN  net tubular secretion

38. Para-aminohippurate PAH
Undergoes tubular secretion and is freely
filterable. At low [PAH]pl, 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.

39. This only applies at a low [PAH]pl, 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.

40. Fractional Excretion:
Fractional excretion, FEX , is the mass of "x"
excreted as a fraction of the total mass filtered,
where FEX = UX ´V
GFR ´PX
and if,
1. FEX < 1.0  net tubular
reabsorption
2. FEX > 1.0  net tubular secretion

41. 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.
Estimation of glomerular filtration
rate:

42. PA(inulin)x RPF PV(inulin)x RPF
Filtered = Pin
X GFR
Excreted =
Uin X V
Filtered = Excreted
Pin X GFR = Uin X V
No reabs; No
secretion.
Inulin clearance

43. 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.

44. 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.
2. inulin filtration rate = inulin
excretion rate

45. In clinical practice creatinine clearance may
be used to estimate GFR for the
following reasons.
1. 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.
Creatinine clearance:

46. Creatinine clearance - contd:
However, in humans, a small amount of
creatinine is secreted into the urine in
the proximal tubules.
Consequently, 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%.

47. 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.
Creatinine clearance - contd:

48. PA
cr x RPF PV
crx RPF
Filtered =
PcrX GFR
Excreted =
Ucr X V
Filtered = Excreted
Pcr X GFR = Ucr X V
No reabs; No
secretion.
Small amount is
secreted

49. 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.
Estimation of renal plasma flow rate:

50. 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

51. PA
PAH x RPF PV
PAHx RPF = 0
Near-complete
extraction in one
passage through
kidney.
UPAH V

52. 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.

53. 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%.

54. 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.

55. 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.