2.  GFR = Clearance of inulin (Cin ) = ( U x V ) / P

            (30 x 1) / 5 =  6 ml/min

Amount Excreted = Amount Filtered - Amount Reabsorbed

        1 ml/min  =  6 ml/min - amount reabsorbed
        amount reabsorbed = 5 ml;   % reabsorbed = 5/6 x 100% =  83.3%

3.  Cx  = (U x V) / P =  10 x 10 / 1 =  100 ml/min

        Cx  >> GFR, therefore must be secreted.  100 ml of plasma is being cleared of substance X, whereas, only 6 ml of plasma is being cleared of inulin in a minute.

4.  Solute A will begin to appear in the urine.  Below the Tm, all the solute will be reabsorbed.

5.  The longer the Loop of Henle, the greater the potential corticomedullary gradient.  The larger the corticomedullary gradient, the greater the potential to concentrate urine.  Desert animals have a problem with dehydration and must conserve water.  being able to excrete a concentrated, scant urine is beneficial.

6.  GFR = 120 ml/min.   V = 36/ 30 = 1.2 ml/min

        CNa = (U x V) / P =  (200 x 1.2) / 144  = 1.67 ml/min

Mass Filtered = 144 mmole/L (0.12 L/min) =  17.28 mmole/min
Mass Excreted = 200 mmole/L (0.0012 L/min) = 0.24 mmole/min

Mass Reabsorbed = 17.28 - 0.24 = 17.04  or 17.04/17.28 = 98.6% reabsorbed