Download Renal Physiology

Dr J. Mohan
Renal Physiology
April, 2011
J. Mohan, PhD.
Lecturer,
Physiology Unit,
Faculty of Medical Sciences,
U.W.I., St Augustine.
Office : Room 105, Physiology Unit.
References:
Koeppen B.E. & Stanton B.A. (2010). Berne & Levy
Physiology. 6th Edition. Mosby, Elsevier.
Costanzo L.S. (2006) Physiology. 3rd Edition, Elsevier,
Saunders.
Marieb, E. & Hoehn, K. (2010). Human Anatomy &
Physiology. 8th Edition, Pearson, Benjamin Cummings.
Stanfield, C.L. & Germann W.J. (2008). Principles of
Human Physiology. 3rd Edition, Pearson, Benjamin
Cummings.
Hall, J.E. (2011). Guyton and Hall Textbook of Medical
Physiology. 12th Edition, Elsevier, Saunders.
April 18, 2011
1
Dr J. Mohan
Physiology Objectives
1.
Explain, with the use of equations, the principles of clearance,
renal plasma flow, renal blood flow and urinary excretion.
2.
Define renal plasma flow (RPF) and renal blood flow (RBF).
3.
Justify the use of creatinine clearance over that of inulin in the
estimation of GFR.
4.
Explain the relationship between clearance and excretion for
inulin, glucose, urea and penicillin.
5.
Define tubular transport maximum (Tm), renal threshold and
renal splay.
6.
Explain, with the use of graphs, how renal glucose-handling
varies with plasma glucose concentration.
Today’s Topics
• Sample calculation of Urinary Excretion.
• Sample calculations of Renal Clearance.
– Some calculations of clearance of various substances.
• Measurement of Renal Plasma Flow and Renal Blood Flow.
– Sample calculation of Renal Plasma Flow and Renal
Blood Flow.
• Renal handling of glucose.
April 18, 2011
2
Dr J. Mohan
Excretion of a Substance
• Excretion by the kidneys = elimination of solute and water
from the body in the form of urine
• For any substance, the amount of a substance that is
excreted over a period of time is determined by this simple
rule: the amount of substance that enters the renal tubule is
excreted unless it is reabsorbed
• What determines entry into the renal tubule? Filtration or
Secretion
• Therefore : amount of substance excreted (E) = amount
filtered (F) + amount secreted (S) – amount reabsorbed (R)
•
i.e.
E =F +S - R
Excretion of a Substance
Example:
• Amount of substance excreted (E) = amount filtered (F) +
amount secreted (S) – amount reabsorbed (R)
•
i.e.
E =F +S - R
• If the GFR= 150 ml/min, the plasma concentration of a solute
is 0.08 mmol/ml. Assume that the solute is freely filterable is
secreted at a rate of 3 mmol/min and reabsorbed at a rate of 6
mmol/min. What is the excretion rate of the solute?
F
April 18, 2011
= GFR
x [Plasma] of solute
= 150 ml/min x 0.08 mmol/ml
= 12 mmol/min
3
Dr J. Mohan
Excretion of a Substance
Example:
• Amount of substance excreted (E) = amount filtered (F) +
amount secreted (S) – amount reabsorbed (R)
•
i.e.
E =F +S - R
F = 12 mmol/min
S = 3 mmol/min
R = 6 mmol/min
Therefore, solving for E = 12 + 3 - 6 = 9 mmol/min
Renal Clearance
• Recall :
• Principles of clearance
from lecture 1
and how to calculate
clearance of a
substance x as given in
the example
• That clearance of
creatinine and inulin can
be used to determine
GFR
Figure 32.12; Koeppen & Stanton, 2010
April 18, 2011
4
Dr J. Mohan
Renal Clearance
Cx =
Ux X V
_______________
Px
ml/min
• represents a volume of plasma from which all of
substance x has been removed and excreted into urine
per unit time
• i.e. rate of removal of substance x from plasma by
kidneys
• Note : ratio of the amount of x excreted in urine to the
amount of x in plasma
Renal Clearance
• Cx =
Ux X V
___________
Pa x
ml/min
• E.g. if Ux = 100 mg/ml; V= 1 ml/min
• Then excretion rate of x = 100 mg/ml x 1 ml/min
= 100 mg/min
• If Px = 1 mg.ml , then Cx = 100 mg/min
_________________________
1 mg/ml
= 100 ml/min
April 18, 2011
5
Dr J. Mohan
Renal Clearance
• Example 2 : A urine sample revealed that 450 ml of urine
was collected in 1 hour and this had a concentration of
Na+ of 15 mmol/L. The plasma concentration of Na+
was 145 mmol/L. What was the Na+ clearance?
• Express V in ml/min : 450 ml/60 mins = 7.5 ml/min
• U Na+ = 15 mmol/L; V= 7.5 ml/min; P Na+ = 145 mmol/L
• Clearance of Na+ = 15 mmol/L x 7.5 ml/min
_________________________
145 mmol/L
= 0.78 ml/min
Renal Clearance
• Example 2 : In a 24 hour period, 1.44.L of urine is
collected from a man receiving an infusion of inulin (the
fructose polymer used to estimate GFR). In his urine the
[inulin] is 150 mg/ml and in his plasma the [inulin] in 1
mg/ml. What is his inulin clearance?
• Express V in ml/min : 1.44 L /24 hours
= 1440 ml / 1440 min
= 1.0 ml /min
U inulin = 150 mg/ml; V= 1.0 ml/min; P inulin = 1 mg/ml
• Clearance of inulin = 150 mg/ml x 1.0 ml/min
_________________________
1 mg/ml
= 150 ml/min
April 18, 2011
6
Dr J. Mohan
Renal Clearance
• From previous examples, it can be seen that clearance can
be calculated for any substance
• Clearance an vary from zero to ~ 600 ml/min depending on
the renal handling of the substance
• A substance that is not normally filtered across the
glomerular capillaries will have a clearance of approx. zero
e.g. albumin
• Renal clearance of glucose = 0; glucose is filtered, unlike
albumin, but is completely reabsorbed back into blood
• Substances that are filtered but partially reabsorbed e.g.
Na+, urea, phosphate, Cl- have a clearance higher than
zero
Renal Clearance
April 18, 2011
•
Recall that the clearance of inulin, the fructose polymer, is
used to estimate GFR
•
Like creatinine, (the clearance of which is also used to
estimate GFR), inulin is freely filtered across the gloerular
capillaries, but is neither reabsorbed nor secreted, therefore its
clearance measures GFR
•
Organic acids e.g. para - aminohippuric acid (PAH) is both
filtered & secreted, therefore has the highest clearance values
•
Drugs such as penicillin is also secreted, therefore has a high
clearance
7
Dr J. Mohan
Renal Clearance
•
why is the use of creatinine preferred over that of inulin for the
estimation of GFR?
•
inulin is freely filtered across the glomerular capillaries, but is
neither reabsorbed nor secreted, therefore its clearance is the most
accurate measure of GFR available
•
however, the use of inulin clearance is limited in the clinical
situation because of several factors:
1. an intravenous injection followed by constant infusion is
required
2. complete emptying of the bladder is necessary before the
beginning of the clearance period, in order to remove all
urine not containing inulin, and again at the end of the
period in order to obtain all the urine produced during this
period
Renal Clearance
3. the urine flow must be high so that enough urine may be
obtained in a short period of time to permit analysis and
to reduce possible errors introduced by urine remaining in
the bladder at the beginning and end of the clearance
period
•
April 18, 2011
these requirements can be difficult to meet in a patient with
compromised renal function
8
Dr J. Mohan
Renal Clearance
•
the use of creatinine clearance overcomes some of these practical
problems, but creatinine is secreted into the PT, and certain drugs
e.g. cimetidine (organic cations) can inhibit tubular secretion of
creatinine, thereby reducing creatinine clearance (but not actual
GFR)
•
Otherwise :
1. Creatinine is an end product of protein metabolism and is
always present in the blood; its concentration remains fairly
constant over a 24 hour period
2. therefore there is no need for an intravenous infusion, and
a clearance period can extend for as long as 24 hours so
that adequate amounts of urine can be collected and the
frequency of bladder emptying is minimised
Renal Clearance
3. only 1 blood sample is needed and it can be taken during
any point in the collection period
4. plasma creatinine alone can be used to follow changes in
GFR in a patient with chronic renal disease
April 18, 2011
9
Dr J. Mohan
Measurement of Renal Plasma Flow and
Renal Blood Flow
• Recall from Lecture 1 the definition and regulation of
Renal Blood Flow
• Recall that red blood cells are not filtered, so the flow of
plasma that enters the kidney is Renal Plasma Flow
• Renal Plasma Flow (RPF) can be estimated from the
clearance of para - aminohippuric acid (PAH)
• Renal Blood Flow can then be calculated from RPF and
the haematocrit
Measurement of Renal Plasma Flow and
Renal Blood Flow
• Measuring True Renal Plasma Flow (RPF) by applying the
Fick Principle to the kidney :
• i.e. the amount of substance entering the kidney via the renal
artery = amount of substance leaving the kidney via the renal
vein + the amount excreted in the urine
• For PAH :
Amt. of PAH entering kidney= Amt. of PAH leaving kidney
Amt. of PAH entering kidney = [RA] PAH x RPF
Amt. of PAH leaving kidney = [RV] PAH x RPF + [U] PAH x V
April 18, 2011
10
Dr J. Mohan
Measurement of Renal Plasma Flow and
Renal Blood Flow
Amt. of PAH entering kidney= Amt. of PAH leaving kidney
[RA] PAH x RPF = [RV] PAH x RPF + [U] PAH x V
RPF =
[U] PAH x V
____________
[RA] PAH - [RV] PAH
where RPF = Renal Plasma Flow
[U] PAH
= [PAH] in urine
V
= urine flow rate
[RA]
= [PAH] in renal artery
[RV]
= [PAH] in renal vein
Measurement of Renal Plasma Flow and
Renal Blood Flow
• PAH is ideal for measuring RPF because :
1. PAH is neither used nor produced by the kidney
2. PAH does not alter RPF
3. The kidneys remove most of PAH from renal arterial blood by
filtration & secretion leaving very little in renal vein so [RV] is
nearly zero
4. No organ apart from the kidney removes PAH from blood so
[PAH] in the renal artery = [PAH] in any peripheral vein, which
is easier to sample from humans than the renal artery
April 18, 2011
11
Dr J. Mohan
Measurement of Renal Plasma Flow and
Renal Blood Flow
• The measurement of “True Renal Plasma Flow” involves the
infusion of PAH into the body, taking urine samples and blood
samples
• Certain simplifications can be made which estimate “Effective
Renal Plasma Flow” which approximates true RPF to 10%
1. [RV] PAH = assumed to be 0
2. [RA] PAH = [PAH] in any peripheral vein
Measurement of Renal Plasma Flow and
Renal Blood Flow
Therefore:
Effective RPF =
where
[U] PAH
V
[P]
April 18, 2011
[U] PAH x V
____________
[P] PAH
= Clearance of PAH
= [PAH] in urine
= urine flow rate
= [PAH] in plasma
12
Dr J. Mohan
Measurement of Renal Plasma Flow and
Renal Blood Flow
• Renal Blood Flow can then be calculated from RPF and
the haematocrit :
• RBF = RPF
_______
1 – Hct
Where : RBF = Renal Blood Flow (ml/min)
RPF = Renal Plasma Flow (ml/min)
Hct = haematocrit = fraction of blood volume
occupied by red blood cells
1-Hct = fraction of blood volume occupied by
plasma
Measurement of Renal Plasma Flow and
Renal Blood Flow
Example : A man with a urine flow rate of 1 ml/min has a plasma
concentration of PAH of 1 mg/dl and urine concentration of PAH
of 600 mg/dl, and a haematocrit of 0.45. What is his RBF?
Effective RPF = 600 mg/dl x 1 ml/min
____________
1 mg/dl
= Clearance of PAH
= 600 ml/min
RBF
= 600 ml/min
--------------1-0.45
=
April 18, 2011
1091 ml/min
13
Dr J. Mohan
Today’s Topics
• Sample calculation of Urinary Excretion.
• Sample calculations of Renal Clearance.
– Some calculations of clearance of various substances.
• Measurement of Renal Plasma Flow and Renal Blood Flow.
– Sample calculation of Renal Plasma Flow and Renal
Blood Flow.
• Renal handling of glucose.
Renal handling of glucose
• Recall from Lectures 1& 2 that glucose is freely filtered
across glomerular capillaries and reabsorbed in the 1st half
of PT
• Recall the reabsorption of glucose involves :
– active transport of Na+ at basolateral membrane via Na+/K+ATPase
– Na+ enters PT cell at apical membrane with glucose (Na+ glucose symport or Na+ - glucose cotransport) on the Na+ glucose transporter (SGLUT)
– glucose leaves the cell at basolateral membrane via passive
mechanisms (facilitated diffusion) on the glucose transporters
GLUT 1 & GLUT 2
• Because of limited number of glucose transporters, the
mechanism is saturable i.e. it has a transport maximum Tm
April 18, 2011
14
Dr J. Mohan
Na+ Reabsorption in 1st half of PT
Figure 33.1, Koeppen & Stanton, 2010
Filtration, Reabsorption & Excretion of glucose
“true”
“theoretical”
renal threshold
splay
Figure 19.16 ; Germann & Stanfield.
April 18, 2011
15
Dr J. Mohan
Renal handling of glucose
• Graph shows the relationship between plasma glucose
concentration and glucose reabsorption. For comparison,
the filtered load of glucose and the excretion rate of glucose
are also shown
• Known as “glucose titration curve” which is obtained
experimentally by infusing glucose and measuring its
plasma concentration and the rate of glucose reabsorption
• Filtration : glucose is freely filtered across glomerular
capillaries; filtered load is calculated as GFR x plasma
concentration of glucose [P glucose]
• As plasma concentration of glucose increases, filtered load
increases linearly (see graph)
Renal handling of glucose
•
Reabsorption :
•
Tm for glucose reabsorption = 375 mg/min = maximum concentration
of glucose that can be transported per unit time from the tubular fluid
back into the blood
– the amount of glucose transported is proportional to the amount
present in tubular fluid (amount filtered) up to the Tm for glucose
– a higher concentrations, the Tm is saturated and there is no
increase in the amount transported
•
Normal plasma glucose level = 80 -100 mg/dL
– when plasma conc = 100 mg/dL, assuming GFR = 125 ml/min
then filtered load = GFR
x plasma concentration of
glucose
= 1.25 dL/min x 100 mg/dL
= 125 mg/min
– 125 mg/min, which is < Tm, so all glucose is reabsorbed & none
excreted in urine
April 18, 2011
16
Dr J. Mohan
Renal handling of glucose
•
glucose level eventually the amount of glucose in the filtrate > the
capacity for reabsorption & some glucose will be excreted in the urine
•
‘theoretical” renal threshold = 300 mg/dL = plasma concentration at
which the amount of glucose in the filtrate exceeds the transport
maximum and at which glucose appears in the urine (see graph)
•
can be calculated as follows :
GFR x renal threshold = transport maximum
Renal threshold = transport maximum
_________________
GFR
= 375 mg/min
------------------1.25 dL /min
= 300 mg /dL
Renal handling of glucose
• however, “true” renal threshold = 160 -180 mg/dL ; at this
plasma concentration of glucose, the filtered load of glucose
= 225 mg/min (see graph) which is < Tm, but glucose starts
to appear in the urine at this plasma concentration - this
creates a rounded curve in the graph called “splay”
– Splay is the portion of the graph where reabsorption is
approaching saturation , but is not fully saturated;
because of splay, glucose appears in the urine before
reabsorption levels off at the Tm value
April 18, 2011
17
Dr J. Mohan
Renal handling of glucose
• Explanations for splay :
– Low affinity of Na+ glucose transporter for glucose
• as the glucose concentration in the tubular fluid rises,
if glucose molecules detach from their transporter
protein SGLT, those glucose molecules will be
excreted because of the few remaining “free”
transporters to which they may re-attach
– Heterogeneity of nephrons
• Tm represents the Tm for the whole kidney
• all the nephrons in the kidney do not have the same
Tm; some will reach Tm at lower plasma
concentrations than others and glucose will be
excreted in the urine in those nephrons before others
Renal handling of glucose
• at normal plasma glucose concentrations, 80 – 100 mg/dL,
all of filtered glucose is reabsorbed and none is excreted
(see graph)
• in uncontrolled diabetes mellitus, the plasma glucose
concentration > Tm and glucose in excreted in the urine
• during pregnancy, GFR is , filtered load of glucose
so that it may > reabsorption of glucose
• congenital abnormalities of the Na+- glucose cotransporter
Tm glucose excreted in urine
April 18, 2011
18