qqqqqqqqq9 - yousmle99

[ArchivalUser]

The formula for calculating free-water clearance(C H20 ) is: C H20 = V - Cosm, where V = the volume
of urine in ml/min, and where Cosm = (Uosm x V)/Posm. If a patient has a V of 10 ml/min, a Uosm of
900 mOsm/kg, and a Posm of 300 mOsm/kg, you would conclude that

(A) the patient is excreting excess free water
(B) in this patient the diluting segment of the ascending limb is defective
© in this patient antidiuretic hormone (AD H) has been suppressed
(D) the patient must have hypernatremia
(E) the patient must be concentrating urine

[ArchivalUser]

The water clearance is decreased.. SIADH?
ADH is increased, causing hyponatremia and concentrated urine.

[ArchivalUser]

can you elaborate little more? how can you say Ch2O is decreased?

[ArchivalUser]

because form my calculation the free water clearance is -20.

Free water clearance= V { 1-(Uosm/Posm)}

10{1-(900/300)}

10(-2) = -20. so i concluded its decreased hence the negative

[ArchivalUser]

to be honest i was looking at B and E.
The descending loop can be hypertonic especially in a dehydrated state, and the ascending becoming hypotonic.
Not sure if its a defective problem or excess ADH produced.

Just my thinking, but you can rectify the thinking.

[ArchivalUser]

B and E