Water balance + water (R)

Question 1

Water balance: input = output

Answer 1

• For maintenance of water homeostasis = amount of water we consume must equal the amount of water we excrete
• Typically daily intake: ~2.5 litres H2O/day
  
  • food & drink: ~ 2.2 litres water/day
  • aerobic metabolism produces ~ 300ml/day
• Daily output = ~2.5 litres /day
  
  • urine 1.5 L/day
  • faeces 0.1 L/day
  • insensible loss 0.9 L/day (sweat, water from breath)
• Normal urine output can vary from 420 ml to 20 litres/day
• Also water intake is stimulated by thirst

Question 2

How do we reduce urine volume?

Answer 2

• Recall: 180 L/day filtered at glomerulus.
• 65% (117 L) is (R) in PT + 15% (27 L) is (R) in LoH = this is constitutive (R)
• The remaining 20% (36 L) reaches the DT, + urine vol is ultimately determined by how much amount water gets (R) in the DT + CD
  - high levels of water (R) from the DT & CD will reduce urine vol, while less water (R) will increase urine vol
• The water permeability of the DT + CD depends on vasopressin (ADH)
• In the absence of vasopressin, the membrane is impermeable to water.
• This is because vasopressin regulates the presence of aquaporins (water channels) in the apical membrane of the CD

Question 3

Vasopressin controls water (R) in the DT + CD by regulating aquaporin-2:

Answer 3

• Aquaporin 2 is regulated by vasopressin (ADH)
• The stimuli for secretion are low blood volume, low BP & high [Na+] blood.

What happens:

1. vasopressin binds to receptors on the basolateral membrane of wall of CD
2. production of cAMP initiating “second messenger system”
3. aquaporin-2 (stored in vesicles) are inserted into apical membrane
4. membrane becomes permeable to water, allowing water to be (R)
5. water is (R) via osmosis, increasing [urine] up to 1200 mOsmol.

Question 4

The special role of the juxtamedullary nephron LoH in concentrating urine

Answer 4

• Normal blood plasma & IF have a concentration of 300 mOsmol.
• The IF in the renal cortex has a concentration of 300 mOsmol
• In the renal medulla there is a vertical gradient of increasing solute concen, up to 1200 mOsmol.
• = allows water to be (R) from the TF in the CD up to this concen
• This is achieved by:
  (i) massive NaCl (R) from the LoH (thick ascending limb, 750 mOsmol),
  (ii) urea recycling (450 mOsmol)

Question 5

(i) 750 mOsmol is generated by NaCl (R) (countercurrent multiplication)

Answer 5

• The ascending limb is impermeable to water but permeable to salts.
• In particular, the Na+,K+,2Cl-pump can pump out NaCl to a 200 mOsmol concen gradient
  
  • the K+ recycles back into the TF
• This achieves 2 objectives:
  (i) The TF is dilute when it reached the DT (can make dilute urine).
  (ii) A high [NaCl] is generated in the interstitium of the renal medulla (up to 750 mOsmol).
• The [NaCl] of the interstitial fluid of the renal medulla starts to increase
• The high concen draws water from the descending limb
  → the concen of the TF increase
• And the cycle repeats itself…

Question 6

(ii) An additional 450 mOsmol is generated by urea recycling

Answer 6

• Thick ascending limb, DT & most of CD impermeable to urea
• By the time the TF reaches the end of the CD, the relative concen of urea has increases because of (R) of water
• the more water (R), the higher the [urea].
• The papillary duct is permeable to urea.
  ∴urea is (R) due to conc. gradient
• The thin ascending part of the LoH is also permeable to urea.
  ∴urea is secreted into loop due to conc. gradient.
• The urea then cycles around the end of the nephron tubule, repeating the process.
• This is termed urea recycling.
• The effect is to create a pool of concentrated urea in the IF at the level of the renal papilla