Transcript kidney 6

‫بسم هللا الرحمن الرحيم‬
‫﴿و ما أوتيتم من العلم إال قليال﴾‬
‫صدق هللا العظيم‬
‫االسراء اية ‪58‬‬
By
Dr. Abdel Aziz M. Hussein
Lecturer of Medical Physiology
Measurement of :
1. Reabsorption rate = filtered load – excretion rate
2. Secretion rate = excretion rate – filtered load
3. Excretion fraction (fraction excretion) (Fex):
•
It is the fraction of the filtered load of substance that's
excreted
Measurement of :
4. Clearance ratio;
• It is the ratio between clearance of a substance x and
clearance of inulin
1. If Clearance ratio = 1  substance is not reabsorbed
nor secreted i.e. substance reacts as inulin).
2. If the ratio is < 1 substance is partially reabsorbed
as urea.
3. If ratio equals zero  substance is filtered but
completely reabsorbed e.g. glucose when Pglucose <
renal threshold.
4. If ratio is > 1  substance is secreted e.g. PAH.
125 ml/min
124 ml/min
1 ml/min
Clearance ratio is = 1
125 ml/min
124 ml/min
1 ml/min
Clearance ratio less than 1
125 ml/min
124 ml/min
1 ml/min
Clearance more than 1
125 ml/min
124 ml/min
1 ml/min
Clearance = zero
Transport Maximum (Tm
Def.,
• It is the maximal amount of substance transported by
renal tubules per min
Types:
• 2 types;
1. Reabsorptive Tm e.g. Tm for glucose (TmG)
2. Secretory Tm e.g. Tm for PAH (TmPAH)
Filtered load = 3
Normal Plasma Concentration
Reabsorption rate = 3
Filtered load = 3
Normal Plasma Concentration
↑ Filtered load = 6
High Plasma Concentration
Reabsorption rate = 4
(Tm)
Filtered load = 6
Excretion rate = 2
Transport Maximum (TmG)
Def.,
• It is the maximum amount of glucose reabsorbed by
the renal tubules/min
Types:
• It is 375 mg/min in male and 300 mg/min in female
• It occurs when plasma glucose concentration [PG] 300
mg/dl.
Def.,
• It is the plasma concentration of substance above it;
the filtered substance begins to appear in urine.
Value:
• It is 180 mg/dl for glucose
Cause:
• When PG is below 180 mg/dl, any filtered load of
glucose is completely reabsorbed
• If PG increase above 180 mg/dl  filtered load
exceeds reabsorptive capacity of renal tubules 
glucose starts to appear in urine.
Secretion rate = 4
Filtered load = 3
Excretion Rate = 7
Normal Plasma Concentration
Secretion rate = 7
Filtered load = 4
Excretion Rate = 11
High Plasma Concentration
Transport Maximum (TmPAH)
Def.,
• It is the maximum amount of PAH secreted by the
renal tubules/min.
Types:
• It is 80 mg/min corresponding to [PPAH] of 20 mg/dl
• Below [PG] 180 mg/dl: all filtered glucose will be
reabsorbed and so, no glucose will appear in urine and
hence the clearance of glucose = zero
• Increase PG above 180 mg/dl, glucose starts to appear in
urine, and so its clearance starts.
• Further increase of PG above 180 mg/dl will be
accompanied by increase in glucose clearance
Excreted glucose = filtered glucose – reabsorbed glucose
UG . Vo = GFR.PG – TG
• By dividing the equation by [PG] the result will be:
Gradually decrease as long as [PG] is increasing till
theoretically reach very low value and so can be
neglected and  CG will equal Cin.
Excreted PAH = filtered PAH + secreted PAH
UPAH. Vo = GFR. PPAH + TPAH
By dividing the equation by [PPAH] the result will be
gradually decrease with any increase in PPAH till by time
it will be much decreased to the degree that can be
neglected and  CPAH will equal Cin.
Clearance (ml/min)
CPAH
CG
Plasma concentration
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• Glomerular filtration rate would be increased by
:
a) constriction of the afferent arteriole
b) a decrease in afferent arteriolar pressure
c) compression of the renal capsule
d) a decrease in the concentration of plasma
protein
e) a decrease in renal blood flow
• The volume of plasma needed each minute to
supply a substance at the rate at which it is
excreted in the urine is known as the :
a) diffusion constant of the substance
b) clearance of the substance
c) extraction ratio of the substance
d) tubular mass of the substance
e) filtration rate of the substance .
• The glomerular filtration barrier is composed
of all the following except :
•
a) fenestrated capillary endothelium .
•
b) macula densa .
•
c) basement membrane .
•
d) podocytes .
•
e) mesangial cells .
•
‫موقف على الدهر‬
‫• سالما ً شباب النيل في كل‬
‫ٍ‬
‫يجلب الفخرا‬
‫المجدَ أو‬
‫يجني‬
‫ُ‬
‫ٌ‬
‫بكور‬
‫عيون فإننا بكرنا‬
‫شباب اذا نامت‬
‫•‬
‫ٌ‬
‫َ‬
‫الطير نستقبل الفجرا‬
‫ِ‬
‫مصنع يدر‬
‫• تعالوا نشيّ ْد مصنعا ً رب‬
‫ٍ‬
‫صناعنا المغن َم الوفرا‬
‫على ُ‬
‫شباب نزلنا حومةَ المج ِد كلنا َ ومن‬
‫•‬
‫ٌ‬
‫ع النصرا‬
‫يغتدي للنصر ينتز ُ‬
1. Reabsorption:
•
Transport of materials from tubular lumen (filtrate) to
peritubular space or interstitial fluid to peritubular
capillaries (PTC).
2. Secretion:
•
Transport of materials from blood in PTC to interstitial
fluid to tubular lumen.
1) Surface area
Is very large since
every tubule
receives only 60
nl/min (SNGFR).
2) Flux
is the rate of
transport/unit
time/unit
surface area
1) Properties of the
membrane of epithelial
cells
•
The difference between the
apical and basolateral
membranes properties
account for the transepithelial
transport of all solutes
Apical
membrane
Basolateral membrane
2) Tight junctions
•
Attach cells at their apical
borders
• The tight junction may be;
a. Loose tight junction (allows
passage of solutes and
water) or
b. Tight junction (doesn't'
allow transport or difficult
transport).
Tight junction
Paracellular space
3) Establishment of transport maximum (Tm) & development
of concentration gradient between the lumen and peritubular
space
• This allow back diffusion of substance across the leaky tight
junction. Both characters limit the transport process.
4) Metabolic state and vitality of the epithelium:
• Any transport whether active or passive needs energy.
• O2 is needed for reabsorption of Na+ which is
responsible for reabsorption of other. So, O2 lack
affects tubular transport.
5) Hormonal & chemical substance as epinephrine,
aldosterone, parathormone calcitonin, AII, ADH, ANP,
PG and diuretics.
6) Removal of reabsorbed materials by PTC:
• The dynamics of reabsorption in PTC is determined by Starling
forces:
• a) Capillary hydrostatic pressure (13 mmHg).
• b) Capillary oncotic pressure of plasma proteins (32 mmHg)
• c) Peritubular interstitial pressure (6 mmHg)
• d) Interstitial oncotic pressure (15 mmHg)
• If the algebraic sum of forces  favoring reabsorption  it
will be faster.
• If not, back diffusion to tubular lumen will occur and
reabsorption will be decreased.
• Net reabsorbing force: = (32+ 6 ) – (13+ 15) = 10 mmHg
1. Cuboidal in shape.
2. Have two surfaces: luminal and basolateral surfaces:
a. Luminal surface:
• has microvilli (making the brush border) that increases
its surface area.
• Held together by leaky tight Junction
b. Basolateral surface:
• Has many process that interdigitate each other
• Are separated by paracellular spaces
3.The cells have abundant mitochondria especially
near basolateral border supplying energy for Na+K+ pump.
4. The cells are rich in intracellular and brush border
carbonic anhydrous enzyme.
5. The proximal tubule is divided into two segments,
convoluted and straighted segments;
Convoluted segment
Straighted segment
(pars convolute)
Represents segments 1 & 2.
Early 2/3 which lies in the
cortex.
Convoluted.
Have greater surface area due
to numerous microvilli.
Preferential reabsorption of
essential nutrients as
glucose, amino acids &
HCO3-.
Transepithelial potential
difference (TEPD) is -4 m.v.
(pars recta)
Segment 3.
Late 1/3 which lies in cortex
and outer medulla.
Straighted.
Smaller surface area.
Preferential Cl- reabsorption
TEPD is +4 m.v.
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