Download Kidney 1

Kidney 1
Renal Functions
• Regulate volume/composition of extracellular
fluid
• Contribute to acid/base balance by modulating
excretion of fixed acids (i.e. anything but CO2).
• Excretion of metabolites and foreign substances
• Endocrine functions: secretion of renin and
erythropoietin; conversion of calcitriol (Vitamin
D) to active or inactive forms.
The concept of renal clearance
• Clearance of a solute X is the volume of
plasma totally cleared of that substance
per unit time
• Cx = ([X]urine)(Furine)/[X]plasma
• For example, if [creatinine]plasma = 4 mg/liter, and
• Total loss of creatinine in urine/day is 720 mg, then
• The clearance of creatinine is 720mg/24 hr/4 mg/L = 180
liters/24 hours.
Gross Structure
There are two
basic types of
nephrons: cortical
or superficial ones
with short Loops of
Henle, and
juxtamedullary
ones with loops
that descend into
the inner medulla.
In humans, only
about 15% of the
nephrons are
juxtamedullary
ones.
The nephron is the
basic functional unit of
the kidney
The major parts of the
nephron are
Glomerulus
Proximal tubule
Loop of Henle
Distal tubule
Collecting
duct
The first step in urine formation is glomerular
filtration
• Glomerular capillaries are
100X more leaky than ordinary
capillaries
• Glomerular filtrate contains all
of the small solutes of plasma,
but much less protein.
• The driving force for
glomerular filtration falls from
about 17 mmHg at the head of
the glomerular capillary bed to
about 8 mmHg at the tail end –
this is mostly due to an
increasing concentration of
plasma proteins.
The actual filter is the basement membrane of the
glomerular capillary
Glomerular Filtration Rate
• GFR is typically 90-140 ml/min in adult males,
80-125 ml/min in adult females. Or roughly 180
liters/day.
• Of total renal plasma flow RPF, 20-25% gets
filtered – this is the filtered fraction FF.
• So, FF=GFR/RPF
• Therefore, if cardiac output is about 7,200
liters/day, RPF is about 900 liters/day, and total
renal blood flow is about 1,800 liters/day.
Measurement of GFR
• To measure GFR, a marker substance is chosen
that is freely filtered across glomeruli and
neither reabsorbed nor secreted in the
subsequent segments of the nephron.
• Classically the substance used in experimental
physiology was inulin, a plant polysaccharide.
This has to be injected. Clinically, creatinine is
used, although there is a small secretory
component, because it is normally present in the
blood.
Clearance effects of reabsorption
and secretion
• If a substance is both filtered and
reabsorbed, its clearance is less than
GFR; if a substance is both filtered and
secreted, its clearance is greater than
GFR.
• Typically, both water and Na+ handling by
the kidney are characterized by high levels
of reabsorption: 95-98% of filtered water
and Na+ are reabsorbed.
Measurement of RPF
• A marker substance is chosen that is both
filtered and also so effectively secreted that
essentially all plasma that passes through the
kidney is cleared of the substance.
• The substance classically used for these
measurements is para-amino hippuric acid, a
product of metabolism of aromatic amino acids.
• Remember PAH clearance measures renal
plasma flow, not the volume flow of whole blood.
To get whole-blood flow, divide the PAH
clearance by the hematocrit.
Functions of the Proximal Tubule
• Secreted:
–
–
–
–
–
–
–
–
PAH
Organic acids and bases
Thiamine
Guanidine
Penicillin
Choline
Sulfonamides
H+ (exchanged for Na+)
• Reabsorbed:
–
–
–
–
–
–
–
Na+
K+
Glucose
L amino acids
Some vitamins
Ca++, PO43About 2/3 of filtered water,
driven by net solute uptake.
Tubular fluid leaving the PT
is isosmotic with plasma.
The concepts of filtered load and tubular
maximum (Tmax) The filtered load of a
substance is the rate
of delivery to the renal
tubules of that
substance. The
tubular maximum is
the maximum rate of
reabsorption of that
substance. For
glucose is about 375
mg/min. If the filtered
load exceeds this
value, the additional
glucose will “spill
over” into the urine.
A thought question
• What would be the relationship between
plasma concentration and excretion rate
for a substance that is both filtered and
secreted?
Independent regulation of salt
and water
The Loop of Henle and DCT: Independent
regulation of salt and water excretion
• A regulatory problem is posed by the fact that
loss and gain of salt and water do not occur in
an isosmotic way.
• The combination of transport processes in the
DCT and Loop of Henle allow the production of
either concentrated or dilute urine depending on
the homeostatic need. This is done by
modulating the rates of reabsorption of salt and
water independently.
What happens in the Loop of Henle
• Tubular fluid enters the Loop at about 300
mOsM.
• As the fluid passes through the thin descending
limb of the Loop, it passes through regions of
steadily ascending ISF concentration.
• Because the thin descending limb is much more
permeable to water than to salt, water leaves the
tubule, concentrating NaCl within the tubule
Ascending the Loop
• As the tubular fluid enters the thin ascending
limb of the Loop, it passes into a part of the
tubule that is much more permeable to salt than
to water. NaCl diffuses out of the tubule, leaving
water behind.
• As the tubular fluid passes through the thick
ascending limb, Na+, Cl- and K+ are actively
reabsorbed by the NK2C cotransporter.
• The tubular fluid leaves the Loop more dilute
than it was when it entered. However, because
of NaCl reabsorption, urea has replaced NaCl as
the major solute.
So, how are we going to get a
concentrated urine?
• The Loop of Henle creates an osmotic
driving force that can be used if necessary
for water recovery and urine
concentration. How does it do it? The key
lies in the active reabsorption of NaCl in
the thick ascending limb of the Loop of
Henle and distal convoluted tubule,
together with the loop shape of the
nephron.
The Two-solute Hypothesis
Permeability Properties
Ascending
thin limb
urea
NaCl
Medullary ISF
urea
urea
NaCl
Collecting
Duct
urea
NaCl
NaCl
H2O
H2O
H2O
H2O
Driving Forces (remember that, along the way between ATL
and CD, urea replaced NaCl as the major solute)
ATL
Medullary ISF
urea
UREA
NaCl
H2O
NaCl
H2O
Collecting
Duct
UREA
NaCl
H2O
H2O
Imagine starting the situation with equal osmotic concentrations in each
compartment. The two solute gradients would both tend to drive solute into the
ISF. Water can follow, but only from the collecting duct. The collecting duct
urine therefore becomes concentrated. Urea can reenter the thin ascending
limb, so on the average, each urea molecule may circle around several times
before being excreted, carrying a proportionate amount of water back into the
renal ISF each time.
Summary of concentrated urine production
What if a dilute urine is needed?
ADH could be thought of as the hormone of
water conservation
• To produce a concentrated urine and thus recover the
maximum amount of water, a high collecting duct water
permeability is required. This process is controlled by
antidiuretic hormone (ADH; arginine vasopressin in
humans), a posterior pituitary neuropeptide.
• In the presence of adequate ADH, water channels
(aquaporins) are inserted into the collecting duct plasma
membranes, enabling CD urine to become osmotically
concentrated.
• In the absence of ADH, aquaporins are recovered into
the interior of the CD cells, and the dilute urine that
enters the collecting duct passes through without much
further modification.