Lecture 5: How do we concentration urine

Question 1

What kind of environment is the renal medulla?

Answer 1

Dry hyperosmotic

Question 2

What must happen to the last portion of the tubule?

Answer 2

It has to go back through the hyperosmotic environment in the renal medulla which we have created

Question 3

What happens if no AQP are present in the collecting duct?

Answer 3

no water will be reabsorbed.

Question 4

if no water is reabsorbed how much of GFR can be lost?

Answer 4

up to ~15%

1L/h of H2O in urine

Question 5

What if there are AQP?

Answer 5

Water will move out of the collecting duct via AQP into the hypertonic environment if there are aquaporins in the tube’s epithelial cells

Question 6

Describe the process if antidiuretic hormone (vasopressin) is released

Answer 6

Ultimately water is reabsorbed.

1. Osmoreceptors in hypothalamus detect lower body fluids than normal
2. ADH released from posterior pituitary
3. ADH binds to V2 receptor in collecting duct
4. AQP are inserted in lumenal side
5. Max: reduce water loss from 1L/h ->15mL/h

Question 7

What do long descending thin limbs (of the loop of henle) have, ONLY in the outer medulla?

Answer 7

aquaporins.

short descending thin limbs have no detectable AQP

Question 8

What do short descending thin limbs have?

Answer 8

lots of chloride channels

Question 9

what channels/transporters etc do the thick ascending limb of the loop of henle have?

Answer 9

Na2ClK pump, Na K ATPase
Cl channel, K channel

ROMK (renal outer medullary potassium channel)

Question 10

What does ROMK do?

Answer 10

it recirculates potassium

Question 11

What does Furosemide do?

Answer 11

it blocks the 2Cl- from entering the thick ascending limb

Question 12

What happens to the remaining K and surplus Cl left from the proximal tubule?

Answer 12

they are reabsorbed, through the two Cl- and K+ channels

Question 13

What is the maximum achieveable gradient across the thick ascending limb known as ?

Answer 13

the single effect

Question 14

What is this maximum achieveable gradient across the TAL?

Answer 14

Tubular lumen has osmolarity of 290mosm/L
the Basolateral ECF has an osmolarity of 310mosmol/L

there is a 20mosmol/L gradient from tubular lumen to ECF

Question 15

Describe the short loops and the countercurrent multiplier they have

Answer 15

• No aquaporins
• NaCl pumped out of thck ascending limb -> ECF -> thin descending limb
• filtrate with higher NaCl concentration moves down the thin descneding limb

Question 16

What does optimum exchange require?

Answer 16

slow flow in order to achieve near equilibrium

Question 17

Why would a decrease in medullary blood flow be required?

Answer 17

E.g. in volume depletion, to improve the efficiency of the countercurrent exchange by allowing more time for blood in the ascending vasa recta to achieve osmotic equilibration.

Question 18

Where might an increase in medullary blood flow be required?

Answer 18

e.g. in osmotic diuresis, a decrreased urine concentrating ability and it will impair the efficiency of countercurrent exchange.

Question 19

Why is there no active mechanism in the Thick ascending limb of the long loops?

Answer 19

they do not extend into the outer medulla

Question 20

What is the current best guess of mechanism of concentration in the outer medulla?

Answer 20

the passive hypothesis

Question 21

How much of the total loops do the long loops of henle make up?

Answer 21

~15%

Question 22

The long loops which extend into inner medulla are

Answer 22

not the same as the short loops

Question 23

What happens in the thin descending limb to concentrate the urine?

Answer 23

AQP are in the apical and basolateral membranes

ECF gradient is created by the thick ascending limb and short loops

Net water absorption occurs

The filtrate volume is reduced. This causes a reduced flow to the outer medulla which increases the filtrate

Question 24

What happens if antidiuresis is present?

Answer 24

Dehydration will activate antidiuresis.

ADH inserts AQP2 channels in collecting duct
Urea concentration in collecting duct will rise due to water being reabsorbed

But there will also be some Urea channels open. Urea reabsorbed with the water ONLY IN THE INNER MEDULLA

Question 25

What causes NaCl to leave the thin descending limb and the thin ascending limb?

Answer 25

The reabsorption of water and urea dilutes the ECF sodium concentration

This gradient is facilitaed by the ClCK1

The total inner medulla osmolarity will rise due to the urea and Sodium
This osmolarity will be 1200mosmol/L cosisting of 50% urea and 50% NaCl

Question 26

What happens to the urea?

Answer 26

it goes round and round

Question 27

How much of tubular fluid is left at the end of the loop of henle?

Answer 27

~15% of glomerular filtrate

Question 28

What is the osmolarity at the end of the loop of Henle?

Answer 28

It is less than ECF, ~100mosmol/L

Question 29

What is the concentration at the hairpin bend of the loop?

Answer 29

4x that of anywhere else.

~1200mosmol/L

Question 30

Why is Urea surprisingly critical?

Answer 30

At the hairpin end during antidiuresis, the osmolaltiy of 1200-1400mosmol is ~ 50% NaCl and 50% Urea

Patients on low protein diets have a low urea production and therefore a low concentration ability

Urea can be restored rapidly by giving intravenous urea