USMLE forum
 
USMLE Forum
Step 1
Step 2 CK
Step 2 CS
Matching & Residency
Step 3
Classifieds
 
Archives
 
 
  <<   < *  Step 1   *  >   >>  

* qqqqqqqqq9
 #848158  
  yousmle99 - 02/13/18 15:38
 
  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
(C) in this patient antidiuretic hormone (AD H) has been suppressed
(D) the patient must have hypernatremia
(E) the patient must be concentrating urine
 
Report Abuse

 
 

* Re:qqqqqqqqq9
#3360717
  crossover - 02/13/18 18:02
 
  The water clearance is decreased.. SIADH?
ADH is increased, causing hyponatremia and concentrated urine.
 
Report Abuse

* Re:qqqqqqqqq9
#3360758
  yousmle99 - 02/14/18 06:28
 
  can you elaborate little more? how can you say Ch2O is decreased?  
Report Abuse

* Re:qqqqqqqqq9
#3360824
  crossover - 02/14/18 16:57
 
  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

 
Report Abuse

* Re:qqqqqqqqq9
#3360831
  crossover - 02/14/18 17:12
 
  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.
 
Report Abuse

* Re:qqqqqqqqq9
#3360844
  meenugal - 02/14/18 19:18
 
  B and E  
Report Abuse

          Page 1 of 1          

[<<First]   [<Prev]  ... Message ...  [Next >]   [Last >>]

 
Logon to post a new Message/Reply
 
 
 
 

 

 

Google
  Web USMLEforum.com
 

Step 1 Step 2 CK Step 2 CS Matching & Residency Step 3 Classifieds
LoginUSMLE LinksHome