Kidneys effect the separation of solute and H2O; free H2O is excreted = dilute urine; reabsorbed = concentrated urine
Determined by the maintenance of H2O homeostasis. Solute in = Solute out
Osmolar Clearance
The excretion of the daily solute load per unit time in a vol isotonic to plasma: True clearance term.
Cosm = (Uosm x V [in mL/min]) / Posm
Normal Cosm =2mL/min and remains stable for a given person with daily solute load (900mOsm) in balance (solute in = out)
In a concentrated urine: Cosm = (1200 mOsm/L)(0.5mL/min) / 300 mOsm/L = 2mL/min
In a dilute urine: Cosm = (100 mOsm/L) (6mL/min) / 300 mOsm/L = 2mL/min
In both situations the daily filtered load is the same Posm x Cosm = 300 mOsm/L x 0.002 L/min = 0.6 mOsm/min = 900 mOsm/day; therefore, in = out
When solute is in balance, filtered load = ultrafiltrate solute concentrationx GFR = plasma solute concentration x Cosm
filtered load = excretion
When urine is isotonic to plasma (Uosm = Posm), Cosm = (Uosm x V) / Posm = V
When urine is not isotonic then a given volume of urine can be considered to have two components: V = Cosm (isotonic urine) + CH2O (free water clearance)
Free Water
Free Water Clearance: CH2O
Volume of H2O that must be removed per unit time from a dilute final urine in order to render it isotonic with plasma, or alternatively volume of H2O that must be added per unit time to isotonic tubule fluid to produce the final dilute urine
C H2O = V – Cosm
I.e.: 1L of 150 mOsm/kg urine = 0.5L isotonic (300mOsm/kg) and 0.5L pure H2O. If excreted in 30 min then C H2O = (V – Cosm )/30 min = 500/30 mL/min = 16.7mL/min
Negative Free Water Clearance: (Tc H2O)
tubular conservation of water
Volume of H2O that must be added to a hypertonic urine per unit time to make it isotonic to urine, or alternatively volume of H2O that must be removed from the isotonic tubule fluid to make the final urine hypertonic to plasma
Tc
H2O = -C H2O = Cosm – V
i.e.: 0.5L of 600mOsm/kg urine = 1.0L of isotonic (300mOsm/kg) urine minus 0.5 l of pure H2O. if excreted in 1 Hr, then: Tc H2O = (1.0-0.5)L/Hr = 0.5L/Hr = 8.3mL/min
the purpose of CH2O and Tc H2O is to maintain normal pOsm compensating for fluctuations in body content, tonicity and solute load
Neither are true clearance terms
Neither are the same as Uosm
Factors that impair free water clearance
GFR is reduced (e.g. renal failure)
Decreased delivery of tubular fluid to diluting segment (Ý proxm tub reabsorption of filtrate seen in vol depletion)
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solute reabsorption in the diluting segment (Thick Ascending Limb and early convoluted tubule); seen with diuretics
Inability to suppress ADH secretion
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Solute load ( total amount of free water produced is directly proportional to the solute load )
Factors necessary for Tc H2O:
Adequate delivery of fluid to segments that separate solute and H2O: Mainly Thick ascending Limb
NaCl reabsorption must be normal
Hyperosmotic intersitium must be present: depends on first 2 and urea accumulation which depends on pp intake
Maximal ADH must be present and collecting ducts must respond normally to ADH