Cause they have no clue where the hell we were talking and we

1. RENAL SYSTEM
Renal Lecture 6
Dr. Janan Alrefaee

2. Renal clearance: renal clearance of a
substance is the volume of plasma that is
completely cleared of the substance by the
kidneys per unit time.
*Renal clearance is useful in measuring (1) the
excretory function of the kidneys (2) renal blood
flows rate (3) the basic functions of the kidneys:
glomerular filtration, tubular reabsorption, and
tubular secretion.

3. Cs= (US*V)/Ps
Where Cs is the clearance rate of a substance
s, Us is the urine concentration of that substance,
and V is the urine flow rate. Ps is the plasma
concentration of the substance.

4. *If a substance is freely filtered and is not
reabsorbed or secreted by the renal tubules, then
GFR can be calculated as the clearance of this
substance as follows:
GFR = (US*V)/Ps= Cs

5. Substances that fits these criteria and used to
measure GFR:
Inulin, is a polysaccharide which not produced
in the body, is found in the roots of certain plants
and must be administered intravenously to a
patient.
Creatinine is used clinically, it is a by-product
of muscle metabolism and is cleared from the
body fluids almost entirely by glomerular
filtration (more suitable because no need for
intravenous infusion as in inulin).

6. Renal Plasma Flow
Para-aminohippuric acid (PAH) clearance can
be used to estimate renal plasma flow.
Theoretically, if a substance is completely cleared
from the plasma, the clearance rate of that
substance is equal to the total renal plasma flow.

7. Filtration Fraction (FF)
FF is the fraction of plasma that filters through
the glomerular membrane. The filtration fraction
is calculated as follows:
Filtration fraction = GFR (inulin clearance)
/Renal plasma flow (PAH clearance).
As the normal plasma flow through both
kidneys is 650 ml/min and the normal GFR is 125
ml/min or 180 L/day in the average adult human,
the filtration fraction 0.19 = 0.2 (20%).
FF = 125 ml/min/650 ml/min = 0.19

8. Calculation of tubular reabsorption or
secretion from renal clearances
If the filtered load of substance (GFR x plasma
concentration of substances) and renal excretion of
a substance (urine concentration of substance x
urine flow rate) are known, one can calculate
whether there is a net reabsorption or a net
secretion of that substance by the renal tubules.

9. If the rate of excretion of the substance is less than
the filtered load of the substance, then some of the
substance must have been reabsorbed from the
renal tubules. Conversely, if the excretion rate of
the substance is greater than its filtered load, then
there is tubular secretion of the substance, and so
can be calculated both of them.

10. Renal mechanisms for controlling urine
concentration
Antidiuretic hormone (ADH) (vasopressin): when
there is a deficit of water and osmolarity of the body
fluids increases above normal, the posterior pituitary
gland secretes more ADH, which increases the
permeability of the distal tubules and collecting ducts to
water and decreases urine volume without obviously alter
the rate of renal excretion of the solutes.

11. When ingestion excess water or when there is
excess water in the body and extracellular fluid
osmolarity is reduced, the secretion of ADH by the
posterior pituitary decreases, thereby reducing the
permeability of the distal tubule and collecting
ducts to water, which causes excreting large
amounts of dilute urine.

12. The kidneys conserve water by excreting
concentrated urine
The kidney excretion a small volume of concentrated
urine minimizes the intake of fluid required to maintain
homeostasis, especially in short water supply. This
happen by continuing to excrete solutes while
increasing water reabsorption and decreasing the
volume of urine formed. The human kidney can produce
a maximal urine concentration of 1200 to 1400
mOsm/L.

13. Obligatory urine volume
It is the minimal volume of urine that must be
excreted; a normal 70-kilogram human must
excrete about 600 milliosmoles of solute each day.
If maximal urine concentrating ability is 1200
mOsm/L, the obligatory urine volume, can be
calculated as:
600 (mOsm /day) / 1200 (mOsm/ L) = 0.5 L /
day

14. The limited ability of the human kidney to concentrate
the urine to a maximal concentration of 1200 mOsm/L
explains why severe dehydration occurs if one attempts to
drink seawater. Because for every liter of seawater drunk
(1200 mOsm/L), there is a net fluid loss of 1 liter (in
addition to that the kidney must also excrete other solutes,
especially urea, which contribute about 600 mOsm/L, so
urine volume should be more) explaining the rapid
dehydration that occurs in shipwreck victims who drink
seawater.

15. Requirements for excreting concentrated
urine
(1) A high level of ADH, (2) a high osmolarity
of the renal medullary interstitial fluid (which
provides the osmotic gradient necessary for water
reabsorption to occur in the presence of high
levels of ADH).

16. Note: The countercurrent mechanism:
depends on the special anatomical arrangement
of the loops of Henle (countercurrent multiplier)
and the vasa recta (countercurrent exchange), in
addition to the collecting ducts (discuses later).

17. The process by which renal medullary
interstitial fluid becomes hyperosmotic:
1- The countercurrent multiplier
2-Recirculation of urea from medullary
collecting duct to Loop of Henle

18. 1- The countercurrent multiplier steps are:
[1]The operation start at loop of Henle assume that it
is filled with fluid with a concentration of 300 mOsm/L.
[2] Active transport of Na and CI out of thick ascending
limb from the tubular lumen to the interstium & increase
interstium osmolarity. This creates a high osmotic
gradient between the interstial fluid and the fluid in the
thin descending loop of Henle which is permeable to
water.

19. [3] Movement of water by osmosis from the thin
descending loop of Henle to the interstial fluid
leading to increase the osmolarity of tubular fluid
equal to interstium osmolarity. [4] The steps are
repeated over and over, so the continuous inflow
of isotonic tubular fluid from proximal tubule with
outflow hypotonic tubular fluid into the distal
tubule occurs.

20. With sufficient time, this process gradually
traps solutes in the medulla more than water
and multiplies the concentration gradient
along the medulla, eventually raising the
interstitial fluid osmolarity to 1200 to 1400
mOsm/L in medulla pelvic tip.

22. 2-Recirculation of urea from medullary collecting
duct to Loop of Henle
When there is water deficit and blood concentrations of
ADH are high, water is reabsorbed rapidly from the
cortical collecting tubule and inner medullary collecting
ducts, causing a higher concentration of urea in the tubular
fluid. This lead to large amount of urea is passively
reabsorbed from the inner medullary collecting ducts into
interstitial fluid.

23. This urea diffuses from interstitial fluid into the thin loop
of Henle, and then passes through the distal tubules, and
finally passes back into the collecting duct and so. The
recirculation of urea through these terminal parts of the
tubular system several times before it is excreted, trap
urea in the renal medulla leading to renal medulla
hyperosmolarity. This is essential to save body fluid when
water shortage

25. Urea contributes about 40 to 50 % of the renal
medullary interstitium osmolarity (500-600
mOsm/L) when the kidney is forming maximally
concentrated urine.

26. The people, who eat a high-protein diet, give
up large amounts of urea as a nitrogenous “waste”
product, can concentrate their urine much better
than people whose protein intake and urea
production are low.
Malnutrition is associated with great
impairment of urine concentrating ability.

27. Note: About one half of the filtered urea is
excreted. Urea excretion rate is determined by
two factors: (1) the concentration of urea in the
plasma and (2) the glomerular filtration rate.