Download 7. URINE FORMATION Urine formation

glomerulus is excreted in the urine. The most obvious
element following this pattern is water (Fig. 7-2c).
7. URINE FORMATION
In order to form urine the kidneys have to carry out three
processes: glomerular filtration, tubular reabsorption and
finally tubular secretion (Fig. 7-1).
The last pattern consist in filtering a determined quantity in
the glomerulus and then further have a net addition to the
filtrate by tubular secretion of the same element, thus more
Urine formation
• Filtration
–Crossing of filtration membrane
–Formation of filtrate
Glomerular
filtration
• Tubular reabsorption
Tubular
secretion
–Solutes and water cross from filtrate
into interstitial fluid and peritubular
capillaries
• Tubular secretion
–Solutes are secreted into filtrate
Tubular
reabsorption
• Filtrate = Urine
Figure 7-1. General processes involved in the formation of urine
The addition of the effects of each portion determines the
composition and concentration of the urine.
Different substances reach urine through different routes
within this general pattern of filtration, reabsorption and
secretion. Some materials are filtered in the glomerulus
and are not reabsorbed at all in the tubules, thus, all what
is filtered is excreted in urine.
Understandable, since these products are not required by
the organism, most waste, toxic materials follow this
pattern of secretion (Fig. 7-2a). The opposite situation
takes place with substances which are filtered in the
glomerulus are shortly after reabsorbed in their entirety
through the tubules, thus, urine under normal conditions
does not carry any of these elements. Glucose, amino
acids and small proteins follow this pattern of circulation
(Fig. 7-2 b).
of what was originally filtered appears in the urine (Fig. 72d).
A
C
B
D
Figure 7-2. Alternative paths taken by different
substances in the kidney. A =waste, B=glucose,
aa, small prot., C=water, D= ions.
Two other patterns of secretion can be identified in the
nephrons. Some substances are filtered in the glomerulus
but some portion of it is reabsorbed in the tubular
component, thus, less of what is initially filtrated in the
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Use of inulin to calculate GFR
GLOMERULAR FILTRATION
The initial filtrate is formed in the renal capsule and
consists in the passage of a large proportion of all the
component of plasma, from the circulation to the tubular
structure of the kidney. Here the filtration barrier
selectively permits the passage of water, solutes
(electrolytes), amino acids, glucose, waste products and
some small proteins, therefore the concentration of most
components of the filtrate are almost identical to that of the
plasma left in the capillaries. Large proteins, cellular
components and cells are retained within the capillaries
and do not enter the filtrate. Some material such as fatty
acids, steroids and calcium can be partially retained in the
capillaries because a fraction may be bound to proteins.
The rate at which the filtrate is formed is called the
glomerular filtration rate (GFR) (Fig. 7-3).
To be able to determine the GFR we have to be able to
determine the clearance of a substance that behaves as
water in the glomerulus. The most common is to inject the
plasma with inulin. Inulin is a natural polysaccharide
synthesized by many plants (Fig 7-4).
•
•
•
•
Natural polysaccharide
Produced in plants
Fibre (fructan)
5200 D MW
• Filters completely
• Not reabsorbed or secreted
Figure 7-4. Characteristics of Inulin
It is a type of fibre classified as a fructan with a molecular
weight of 5200 daltons.
Inulin has the special
characteristic that the kidney is capable of completely filter
it in the glomerulus but it does not get reabsorbed or
secreted by the tubules. For the purposes of estimating
GFR inulin behaves like water in the glomerulus.
Therefore, once injected in the plasma its rate of excretion
is directly proportional to the rate of filtration of water and
solutes across the filtration barrier.
Figure 7-3. Glomerular filtration rate
Replacing plasma and urinary concentration of inulin and
the urinary flow rate in the above equation we can
calculate GFR (Fig. 7-5).
HOW TO CALCULATE OR
ESTIMATE GFR
There are several techniques to measure or closely
estimate the GFR. Renal clearance rate of a substance
(Cs) can be established if we know the volume of urine
being produced (V), the concentration of the substance in
the urine (Us) and the concentration of the substance in the
plasma (Ps).
Plasma concentration of inulin (Pinu) = 0.1µmol/mL
Urinary concentration of inulin (Uinu) = 6µmol/mL
Urinary flow rate (V) = 2 mL/ min
Then GFR= C inu = V x Uinu/ Pinu = 2 x 6/0.1 = 120 mL/min
Cs= Us x V / Ps
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Use of PAH to estimate renal plasma
flow
Para-aminohippuric acid (PAH), a product of aromatic
amino acid metabolism can be used to estimate Renal
Plasma Flow (RPF). When a substance is completely
cleared from the plasma, then the clearance rate of such
substance is equivalent to the total renal plasma flow.
Such a substance does not exist but about 90 % of the
PAH is excreted. That means in the average normal kidney
the extraction rate of PAH is 90 % (Fig. 7-7).
• Plasma inulin (PInu) = 0.1 µmol/mL
• Urinary inulin (UInu) = 6 µmol/mL
• Urinary flow rate (V) = 2 mL/min
GFR = CInu = V x UInu/ PInu = 2 x 6/ 0.1=
120 mL/min
Figure 7-5. Estimating GFR using inulin
Use of creatinine to estimate GFR
Muscle metabolism generates a waste by-product called
creatinine (Fig. 7-6). This compound can be used to
estimate GFR. Unlike inulin, creatinine does not have to
be injected, thus makes the estimation techniques simpler.
Creatinine clears in the glomeulus as water but the tubules
also secret a small amount of creatinine into the filtrate,
generating an overestimation of the amount cleared.
Fortunately, there is also a similar overestimation in the
technique to measure creatinine, thus, the combination of
these two errors tend to balance out.
•
•
•
•
•
•
•
• Product of aromatic AA
metabolism
• Used to estimate RPF
• 90 % is filtered in in kidney
Figure 7-7. Characteristics of PAH
Remember that ideally, if a compound is totally cleared
from the plasma, the clearance rate of that compound
would be equal to the total amount of that compound in the
RPF. What gets in the kidneys (RPF x Ps) is the same
quantity that is eliminated in the urine (Us x V) (Fig. 7-8).
RPF = Us x V / Ps= Cs
Protein metabolism waste product
Fairly constant
Does not require injection
Clears glomerulus as water
Tubules secrete small amount
Technique to measure overestimate it
They balance out
But, since the extraction rate is only 90 % then
Clearance of PAH/ Extraction rate of PAH
RPF=
And the extraction rate of PAH (EPAH) is the difference
between the amount of PAH in the renal artery (APAH) and
the amount left in the renal venous return (VPAH) divided by
the concentration of PAH in the renal artery.
EPAH = APAH- VPAH /APAH
Figure 7-6. Estimating GFR using inulin
For example an animal with the following values: (Fig. 7-9)
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Furthermore, once we know the plasma flow rate, by
knowing the blood hematocrit value we can determine the
total blood flow through the kidney.
If we measured hematocrit value is 0.45 then the total
plasma flow would be 430 / (1 – 0.45) or 782 mL/min
Volume of plasma flowing through the
kidneys per minute
RPF = Us x V/Ps = Cs
Filtration fraction
• Since PAH extraction rate = 90%
• RPF = CPAH / EPAH
Of all the plasma that circulates through the glomerulus
about 19% is filtered into the tubular compartment. The
exact amount at any given time is determined by several
physiological factors. The hydrostatic and colloidal
osmotic forces across the filtration barrier and the capillary
filtration coefficient, which is in turn determined by the
permeability and the surface area of the capillaries,
determine the GFR.
and
• EPAH = APAH – VPAH / APAH
Figure 7-8. Characteristics of PAH
The volume of filtrate formed or filtrate fraction is
= GFR x renal plasma flow.
The components of the filtration barrier all contribute to
determine the filtering capacity of this structure (Fig. 7-10).
The fenestrations in the endothelial cells of the glomerular
capillaries are large enough to permit the passage of all
materials in blood except very large proteins and cells
such as erythrocytes. The endothelial cells are, however,
are densely loaded with negatively charged components.
PPAH = 0.015 mg/mL
UPAH = 5.8 mg/mL
V
= 1 mL/min
RPF = Us x V/Ps = Cs
• CPAH = 5.8 mg/mL x 1 mL/min
0.015 mg/mL
= 387 mL/min = 430 mL/min
0.9
Figure 7-9. Estimating renal plasma flow
Plasma concentration PAH of 0.015 mg/mL.
Urinary concentration of PAH of 5.8 mg /mL
Urinary flow rate of 1 mL/min
The clearance of PAH in urine would be 5.8 mg/mL x 1
mL/min divided by the plasma concentration of PAH or
0.015 mg/mL= 387mL/min.
If the extraction rate of PAH is 90% then the real plasma
flow rate (PFR) can be determined by dividing 387 mL/min
by 0.9 which would yield 430 mL/min.
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• Maximum size allowed through 7µm or 40 kD
• Most proteins, cell components and cells are retained in capillaries
• Small protein hormones cross
• Protein bound FA and steroids are retained
• Filtrate may contain 0.03% protein
Figure 7-10. Characteristics of the filtration
barrier
This feature prevents the passage or filtration of many
smaller proteins that would comfortably fit through the
fenestrations.
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The composition of the basement membrane determines
its filtering capacity. This membrane is primarily made of
proteoglycans and collagen chains interlocked, leaving
large spaces through which most solutes and water can
filter. Proteoglycans are charged with strong negative
charges, thus preventing the passage of most proteins that
could have crossed the fenestrations.
The interdigtation of their extension create a series of
filtration slits which further contribute to the filtration
process. These elements are also charged negatively,
thus further preventing the passage of proteins.
If the filtration barrier is normal, this permits the passage
of molecules smaller than 7µm in diameter or with a
molecular mass of about 40,000 daltons (Fig. 7-10).
Because most proteins are larger than these they are
unable to be filtered out and they remain within the
capillaries. Some proteins are, however, smaller than
these thus, they cross the filtration membrane. Many
protein hormones fall in this category. It is estimated that
the filtrate can have about 0.03% of proteins in the
glomerular space. Proteins are nonetheless, actively
reabsorbed in the proximal convoluted tubule to the point
that very little protein ever reach the urine.
The last component of the filtration barrier is the layer of
specialized epithelial cell that lay on top of the basement
membrane, the podocytes (Fig. 6-11).
For example an 80 kg hog under resting conditions will
have a cardiac output (CO) of approximately 5.7L/min. As
stated earlier the kidney receives approximately 20% of
the volume of the CO so, in our case, the kidneys would be
irrigated by about 1.14 L of blood/min that is the Renal
Blood Flow (Fig 7-12).
Volume of blood flowing through the
kidneys per minute
Cardiac output x % entering kidneys
•
•
•
•
80kg hog
CO = 5.7L/min
Renal fraction 20%
RBF = 5.7 L/min x 0.2 = 1.14 L/min
Figure 7-12. Calculation of renal blood flow
How much urine is formed?
Renal Blood Flow= CO X % Renal fraction
To understand the process of urine formation is necessary
to understand all the intermediary steps between blood
moving through the circulatory system and the total
amount of urine excreted by an animal. There are several
concepts (Fig. 7-11) which will be explained with a
mathematical example.
RBF = 5.7 L/min X 0.2 = 1.14 L/min
Of the volume of blood entering the kidney, only certain
percentage is plasma. The rest is made up principally by
hematocrites.
Although these values can vary
tremendously a standard percentage for a normal animal is
60% plasma (Fig 7-13).
Renal Plasma Flow = RBF X % of plasma in blood
RPF= 1.14 L/min X 0.6 = 0.684 L/min or 684 mL/min
Of the plasma which enters the kidneys only about 19%
goes through the filtration barrier. The rest leaves the
glomerulus through the efferent arteriole with all the
hematocrits and large proteins (Fig. 7-13).
• Renal blood flow
• Renal plasma flow
• Glomerular filtration rate
• Urine
• Daily urine volume
Glomerular Filtration Rate = RPF X % filtration fraction
GFR = 684 mL/min X 0.19 = 130 mL of filtrate/min
Of the volume that enters the capsular space only a very
small fraction is excreted in urine, the rest is all reabsorbed
Figure 7-11. Renal flow rates between
circulation and urine
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at different places in the nephron. The average volume
which is not reabsorbed under normal conditions is
onlyabout 0.8% of the filtrate (Fig. 7-14).
Volume of no reabsorbed filtrate
leaving the kidney per minute
GFR x % no reabsorbed filtrate
Volume of plasma flowing through
the kidneys per minute
• GFR =130 mL/min
• Non reabsorbed filtrate (NRF) = 0.8%
• Urine = 130 mL x 0.008 = 1.04 mL/min
Renal blood flow x % plasma (blood-hematocrit)
• RBF = 1.14 L/min
• % plasma in blood = 60%
• RPF = 1.14 L/min x 0.6 = 0.684 L/min
= 684 mL/min
Figure 7-15. Rate of urine formation
Figure 7-13. Calculation of renal plasma flow
• Urine = 1.04 mL/min
• Day = 60 min x 24 h= 1440 min/day
• L = 1000 mL
• Daily volume urine = urine x min / 1000
Volume of plasma (filtrate) entering
Bowman’s capsule per minute
• Volume of urine = 1.04 mL/min x 1.44 =
1.49 L/Day
Renal plasma flow x plasma entering the renal capsule
• RPF = 684 mL/min
• Filtration fraction (FF) = 19%
• GFR = 684 mL/min x 0.19 = 130 mL /min
Figure 7-16. Calculation of daily urine production
volume
Figure 7-14. Calculation of glomerular filtration
rate
Urine = GFR X % no reabsorbed filtrate
Urine = 130 m L/min X 0.008 = 1.04 mL/min (Fig. 7-15).
This volume per minute can be converted to L/day by
multiplying by 1.44 (Number of minutes in one day/1000
mL) (Fig 7-16).
Thus, this animal produces 1.04 mL/min X 1.44 = 1.4976
L/day
Minimum obligatory volume of urine
In order to eliminate all the solutes that are ingested over a
given period of time the animal has to produce an
obligatory volume of urine. The obligatory volume
depends on the ability of the animal to concentrate urine.
The more it can be concentrated the smaller is the
required volume of urine excreted. Some animals adapted
to life in the dessert have the ability to concentrate urine to
levels as high as 10,000.00 mOsm/L while animals living in
water can only concentrate urine to about 500 mOsm/L.
Let’s assume that our 80 Kg hog has a daily solute intake
of about 690 mOsm/day and it is only able to concentrate
urine to 200 mOsm/L (Fig. 7-17). Then the obligatory
volume for this animal will be:
690mOsm/day/1200mOsm/L = 575 mL/day
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The GFR is determined by the filtration pressure and the
filtration coefficient (Kf) (Fig. 7-19).
GFR = FP X Kf
• Smaller volume of urine
• As concentrated as possible
• Filtration pressure (PG‐PB‐πG+ πB)
ƒ Intake
= 690 mOsm/day
ƒ Max. concentration = 1200 mOsm/L
690 mOsm/day
= 575 mL/day
1200 mOsm/L
– Glomerular capillary pressure (PG or GCP)
– Capsular pressure (PB or CP)
– Blood colloid osmotic pressure (πG or BCOP)
– Capsular osmotic pressure (πB or COP)
Figure 7-17. Calculation of the minimum
obligatory daily volume of urine production
• Filtration coefficient (Kf)
This equation explains why animals, that in desperation
drink sea water, undergo rapid dehydration. The sea
water has a solute concentration of about 2400 mOsm/L.
If they can only eliminate 1200 mOsm for each litter of
urine, then, to eliminate the 2400 that they ingest with each
litter of sea water that they drink they need to eliminate
2400mOsm/ 1200 mOsm = 2 L of water. Since they only
ingested 1 litter the other litter has to come from internal
sources causing rapidly an unbalance resulting in
dehydration (Fig 7-18).
• Sea water
= 2400 mOsm/L
• Max concentration = 1200 mOsm/L
Figure 7-19. Factors influencing glomerular
filtration rate
The filtration coefficient is determined by the area of the
capillary and the permeability of the filtration barrier. As a
result there are many possible mechanisms to alter the
GFR. A change in the filtration coefficient will change the
GFR. These changes are, however, not responsible for
the fine tuning of the GFR. Changes in Kf usually are
observed under pathological conditions. A reduction in the
total capillary surface resulting from loss of nephrons will
reduce the Kf and in turn diminish the overall GFR.
Thickening of the filtration barrier, as a result of
hypertension, will also reduce the Kf and the GFR.
The rate at which the glomerular filtration takes place is
drastically influenced by the filtration pressure. The
filtration pressure is the balance between the forces trying
to push material out of the capillaries into the Bowman’s
space, against those forces trying to prevent it (Figs. 7-19,
7-20).
ƒ For each litter drank need
2400 mOsm = 2L to eliminate
1200 mOsm/L
Figure 7-18. Minimum obligatory volume of urine
required to eliminate all the ingested solutes
The above described example assumes constant ideal
conditions. In a living organism that is never the case.
There are always physiological changes that influence the
rate at which most processes take place. The glomerular
filtration rate is no exception.
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– Permeability FB
– Capillary area
GFR= Kf x (PG‐PB‐πG+ πB)
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The main force in favour of filtration is the glomerular
capillary pressure (GCP). This pressure is about 50 mm
Hg and it is generated by low resistance (vasodilation) of
the afferent capillary in conjunction with high resistance
(vasoconstriction) of the efferent arteriole. Modifications in
the resistance of these two arterioles can increase further
or drastically reduce the GCP with the consequent
increase or reduction in production of filtrate, respectively.
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The main forces opposing filtration are the capsular
pressure (CP) and the blood colloid osmotic pressure
(BCOP). The CP is generated by the physical pressure
exerted by the filtrate against the Bowman’s capsule. This
usually reaches values of about 10 mm Hg. The BCOP is
more variable and it is generated by the osmotic
concentration existing in the capillary due to the high
concentration of proteins that do not cross the filtration
barrier.
pressure, and as a result will increase the filtration
pressure, thus increasing the GFR.
The capsular pressure does not usually change much,
unless there is a pathological situation such as deposit of
calcium in the tubule or the formation of stones that
prevent emptying of the content in the Bowman’s capsule.
If this is the case the capsule pressure can increase to the
point of eliminating the filtration pressure thus stopping GF.
In order to try to maintain a somehow constant glomerular
filtration pressure or to modify it according to the needs of
the moment, the kidney resort to two mechanisms,
autoregulation and sympathetic stimulation (Fig. 7-21).
These mechanism permit that the animal maintains
relatively constant GFR when mean arterial blood pressure
oscillate within a large range (90 to 180 mm Hg). Dropping
below that range triggers a significant reduction in GFR.
Filtration pressure
• Autoregulation
Figure 7-20. Forces influencing the filtration
pressure. GCP=glomerular capillary pressure,
CP= capsular pressure, BCOP= blood colloid
osmotic pressure
Because of its nature, there is an increasing gradient
within the capillaries from the end leaving the afferent
arteriole towards the end of the capillary where it becomes
the efferent arteriole. The increase in osmotic pressure
takes place because as the blood flow within the capillary it
becomes more concentrated as water leaves the capillary.
The average value for the BCOP is about 30 mm Hg.
Because the concentration of proteins in the filtrate is so
low the capsular osmotic pressure is negligible, thus for
calculations is considered to be zero.
Therefore the filtration pressure can be equated as:
FP= GCP- CP- BOCP or 50 - 10 - 30 = 10 mm Hg
More subtle changes in GFR can be observed as a result
of changes in the parameters that determine the filtration
pressure. For example a small decrease in resistance by
the afferent arteriole or a small increase in resistance by
the efferent arteriole will increase the capillary hydrostatic
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– Modify afferent and / or efferent arteriole resistance
– Tubuloglomerular feedback
• Initiated by macula densa detecting flow
• Activate renin‐angiotensin system
• Sympathetic stimulation
– Norepinephrine causes afferent arteriole vasoconstriction
• Effective under severe stress or shock
Figure 7-21. Mechanisms used to maintain
glomerular filtration rate semi constant
Autoregulation.
The mechanism of autoregulation
consists in modifying the resistance in the afferent and / or
efferent arterioles. If the systemic mean arterial blood
pressure increases, the afferent arteriole vasocontricts and
or the efferent arteriole vasodilates reducing the pressure
in the glomerular capillaries thus maintaining constant
filtration pressure and GFR. The opposite takes place
when the systemic mean arterial blood pressure
decreases.
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This auto regulatory mechanism can also be triggered by
flow rate of filtrate through the distal convoluted tubule.
This is described as tubuloglomerular feedback.
If the cells in the macula densa detect a large change in
the flow of filtrate they convey this information to the
juxtaglomerular apparatus which then triggers a
vasoconstriction or vasodilation in the afferent arteriole to
reduce or increase filtration pressure, thus maintaining
GFR constant.
The signalling for this regulatory
mechanism is not completely understood but it is
suggested that it is implemented in the following manner.
A decrease in GFR will result in a slower rate of passage
of the filtrate though the loop of Henle. As a result there
will be more time to reabsorb solutes from the filtrate. The
filtrate leaving the loop of Henle and reaching the cells in
the macula densa will have lower concentration of NaCl.
This will be detected by the cells of the macula densa
which will send information to the afferent arteriole to
vasodilate, increasing the filtration pressure and the GFR.
This signal also triggers the release of renin by the
juxtaglomerular cells. Renin elevation translates in
elevation of angiotensin II and, since this is a powerful
vasoconstrictor for the efferent arteriole, it elevates
filtration pressure and brings back the GFR to the normal
rate (Fig. 7-21).
Sympathetic stimulation. Sympathetic innervations
uses norepinephrine as their neurotransmitter and this
catecholamine has an effect in the afferent arterioles
causing their vasoconstriction. This mechanism has a
negligible effect when the stimulation is mild because the
autoregulation mechanism takes over, but it may have a
drastic effect when the stimulation is very strong. In
situations of severe stress or shock, there is a drastic
reduction in the irrigation of the kidney and if maintained
for a long period it may cause kidney damage. A short
term reduction is not harmful but provide the resources to
maintain irrigated other vital organs (heart, brain, liver),
and maintain homeostasis.
Changes from filtrate to urine
The second step in the process of making urine is to
reduce the volume of the filtrate by removing most of their
content, leaving the undesirable compounds to be voided
in a small volume of fluid. This reduction is initiated in the
tubular component of the nephron with the reabsorption of
a large volume of water. The majority of the filtrate is
reabsorbed in the proximal convoluted tubule. About 65%
of the water is reabsorbed in this section. Another 15 % is
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reabsorbed in the loop of Henle and the final 19 % is
reabsorbed in the collecting tubes leaving only about 1 %
of the volume to be secreted as urine (Fig. 7-22).
Reabsorption of filtrate
Figure 7-22. Areas of the nephron where
different volumes of filtrate are reabsorbed
Reabsorption of the different components of the filtrate is
achieved by active and passive mechanisms. The active
mechanisms involve the expenditure of energy in the form
of ATP to drive the movement of sodium against a
concentration gradient from the intracellular space into the
interstitial space through the basal membrane or the basolateral membrane of the tubular cells. This generates a
Na+ gradient in the tubular cell with respect to the lumen of
the tubule where the filtrate is located. The passive
mechanisms involve diffusion, facilitated diffusion, cotransport or symport, and antiport.
Proximal convoluted tubule
Tubular cells are equipped with a variety of proteins,
normally incorporated to the apical, lateral or basal
membrane, which participate in the movement of ions and
other molecules from the filtrate to the tubular cell and then
to the interstitial fluid, so they can be eventually
translocated to the capillaries and to circulations. These
proteins make ATPases, symports, antiports, as well as,
leek channels (Fig. 7-23).
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Materials that are transported through this mechanism into
the cell are, among others, amino acids, small proteins,
glucose, fructose, Cl-, Ca+, K+, HCO3- and of course Na+.
Water, the most abundantly reabsorbed molecule simply
diffuses into the cell by osmosis.
Once inside the cell most of the molecules are discarded
into the interstitial fluid by facilitated diffusion. Amino acids,
glucose, fructose, small proteins and K+ are all discarded
through facilitated diffusion. Cl- is co transported with K+
and HCO3- is also co transported with Na+ into the
interstitial fluid. Water follows by osmosis. From here all
the solutes are uptaked by the peritubular capillaries and
carried towards the interlobular and arcuate vein
respectively.
Figure 7-23. Components of a tubular cell of the
proximal convoluted tubule
In the proximal convoluted tubule, most of the reabsorption
is dependent on the creation and maintenance of a low
intracellular concentration of Na+. A low intracellular Na+
concentration generates a gradient respect to the filtrate in
the tubule. The creation of this gradient is achieved by
actively transporting Na+ out of the cell into the interstitial
space using a pump driven by ATP (ATPase) (Fig. 7-24).
In this process, for each Na+ which is removed from the
cell, one K+ is incorporated into the cell from the interstitial
space. This process is great to create a deficiency of Na+
but, at the same time, it accumulates K+ in the cell.
After a lot of K+ is accumulated inside the cell most of it is
discarded into the interstitial space by facilitated diffusion
or leak channels.
Once the Na+ gradient is created, a large variety of cotransporters located in the apical membrane of the tubular
cell became active. These symports are proteins capable
of binding a variety of solutes and molecules as well as
Na+. They bind the substance first and then Na+. When
Na+ binds then the protein rotate, propelled by the energy
of the Na+ attraction to the inside of the cell. As it rotates
the protein brings into the cell the co-transported material.
Once inside the cell both Na+ and the accompanying
solute or molecules are released and the protein counter
rotates back to the original position, again exposing the
binding sites to the lumen of the tubule.
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The net result is that large reduction I filtrate volume (about
65%). Because the cells of the convoluted tubule are so
permeable to water, still at this point the filtrate maintains
the same osmolal concentration than the interstitial fluid of
about 300 mOsm/kg.
Loop of Henle
Depending on the type of nephron the loop of Henle may
only descend a little bit into the medulla (cortical
nephrons), or, in the case of Juxtamedullary nephrons,
extend the loop of Henle deep into the medulla (Fig. 6-9).
The osmolal concentration of the interstitial fluid gradually
increases towards the medulla, reaching values of 1200
mOsm/kg in the area close to the tip of the renal pyramid
(Fig. 7-24). This feature permits progressively increase
removal of water from the filtrate. The thin descending
segment of the loop of Henle is very permeable to water
and less permeable to other substances as urea, sodium
and other ions. As the filtrate moves in the descending
limb of the loop of Henle an additional 15 % of the water is
removed. Since not many solutes are removed in this
section the filtrate becomes more concentrated.
Furthermore, towards the end of the thin segment of the
loop of Henle, some solutes are incorporated into the
filtrate making it more concentrated. Once the lowest point
is reached, the characteristics of the thin segment changes
in such a way that the thin ascending segment of the loop
of Henle is totally impermeable to water but permeable to
solutes. In this section, then, no further water is removed
but many solutes diffuse out of the filtrate in the lumen into
the interstitial fluid and then into the vasa recta making the
filtrate less concentrated. The thick segment of the
ascending limb of the loop of Henle is not permeable to
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water or to solutes but some solutes (Na+, K+ and Cl-) are
symported from the filtrate to the inside of the tubular cells
and once in the cell they are discarded towards the
interstitial fluid by facilitated diffusion. The fact that only
solutes are removed in this section make the filtrate reach
concentrations of only 100 mOsm/kg as it reaches the
distal convoluted tubule while the interstitial fluid at this
point in the cortex is about 300 mOsm/kg (Fig. 7-24).
The Starling forces are a combination of hydrostatic and
oncotic forces determining the net filtration of fluids across
the capillary membranes (Fig. 7-25).
These can be represented by the following simplified
equation:
Jv = Kf ([Pc – Pi] – [πc –πi])
where
Jv represents the net movement of fluids between inside
and outside the capillaries. It is agreed that a positive
value of Jv indicates that blood is leaving the capillary while
a negative value indicates that fluid is entering the
capillary.
Kf represents the filtration coefficient which in turn depends
of the capillary surface area and the permeability.
Pc represents the hydrostatic pressure of the capillary.
Pi represents the hydrostatic pressure of the interstitial
space
Figure 7-24. Concentration of the filtrate and the
interstitial fluid at different depth of the
glomerulus
The movement of fluid (mainly water) between the
interstitial space and the interior of the capillaries in the
pericapillary bed, as well as, in the vasa recta is
determined by the balance of Starling forces between
these two mediums.
• Hydrostatic and oncotic forces
• Determine net filtration of fluids across
a capillary membrane
•
•
•
•
•
•
•
πi represents the osmotic pressure of the interstitial fluid
Distal convoluted tubule
In the distal convoluted tubule and first section of the
collecting duct there is further removal of solutes. Cl- is
symported with Na+ through the apical membrane into the
tubular cells and discarded into the interstitial fluid through
the basal membrane by symport with K+. All of this is
possible because the intracellular concentration of Na+ is
maintained very low by the ATPase pump (Fig. 7-26).
Collecting Duct
Jv = Kf([Pc-Pi] – [πc – πi])
Jv = Net movement of fluid
Kf = Filtration coefficient
Pc = Hydrostatic pressure of capillaries
Pi = Hydrostatic pressure interstitial space
πc = Osmotic pressure of capillaries
πi = Osmotic pressure of interstitial space
Fig 7-25. Forces influencing the passage of fluid
through capillaries
V BS 122 Physiology II
πc represents the capillary osmotic pressure
57
As the collecting tube enters the medulla of the kidney the
interstitial fluid becomes again more and more
concentrated. This concentration gradient will facilitate
initially the diffusion of water from the filtrate towards the
interstitial fluid. However, the permeability of the collecting
duct cells to water becomes hormonal dependent. In the
collecting duct the final removal of water takes place
leaving the filtrate converted into urine which can be
carried towards the ureter and bladder
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Fate of other solutes
In the initial filtrate there are many other solutes not
mentioned so far. Urea, a product of gluconeogenesis
from amino acids is one of them. Creatinine, sulphates,
nitrates and phosphates are other compounds that all
filtered in the renal corpuscle but eventually only partially
reabsorbed. This leaves a large proportion of these
materials dissolved in the urine, destined for excretion. As
the filtrate reaches the collecting duct these materials are
very concentrated in urine. Between 40 and 60 % of the
urea in the filtrate is passively reabsorbed. The rest is
concentrated and eliminated.
Distal convoluted tubule
Fig 7-26. The role of these cells is to further
remove solutes and water. No removal of
glucose or amino acids takes place in here
V BS 122 Physiology II
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