Given the following data: multiple choice questions. please help me
with to answer the following questions.
 
Plasma Creatinine  0.07 mg/ml  Plasma X 0.1 mg/ml 
Urine Creatinine  9 mg/ml  Urine X 3 mg/ml 
Plasma PAH   0.03 mg/ml  Plasma Y 0.04 
Urine PAH   17 mg/ml  Urine Y 12 mg/ml  
Urine Flow   1 ml/min  Hematocrit 0.53 
 
1. ??  Effective Renal Plasma Flow is: 
 
A. 427 ml/min 
B. 524 ml/min 
C. 567 ml/min 
D. 604 ml/min 
E. 728 ml/min 
 
2. ??  Effective Renal Blood Flow is: 
 
A.  998 ml/min 
B. 1115 ml/min 
C. 1206 ml/min 
D. 1285 ml/min 
E. 1549 ml/min 
 
3. ??  Glomerular Filtration Rate is: 
 
A. 113 ml/min 
B. 119 ml/min 
C. 122 ml/min 
D. 129 ml/min 
E. 133 ml/min 
 
4. ??  The filtration fraction is: 
 
A. 0.23 
B. 0.21 
C. 0.19 
D. 0.17 
E. 0.14 
 
 
 
  
5. ??  Substance X is filtered at a rate of: 
 
A. 12.9 mg/min 
B. 13.4 mg/min 
C. 15.8 mg/min 
D. 16.7 mg/min 
E. Insufficient data 
 
6. ??  Substance X is excreted at a rate of: 
 
A. 1 mg/min 
B. 2 mg/min 
C. 3 mg/min 
D. 4 mg/min 
E. Insufficient data 
 
7. ??  Substance X is probably: 
 
A. Filtered only 
B. Filtered and reabsorbed 
C. Filtered and secreted 
D. Synthesized by the kidney 
E. Insufficient data 
 
8. ??  Substance Y is filtered at a rate of: 
 
A. 3 mg/min 
B. 4 mg/min 
C. 5 mg/min 
D. 6 mg/min 
E. 7 mg/min 
 
9. ??  Substance Y is excreted at a rate of: 
 
A. 8 mg/min 
B. 9 mg/min 
C. 10 mg/min 
D. 11 mg/min 
E. 12 mg/min 
 
10. ??  Substance Y is probably: 
 
A. Filtered only 
B. Filtered and reabsorbed 
C. Filtered and secreted 
D. Synthesized by the kidney 
E. Insufficient data

1)  Renal plasma flow can be measured by measuring the clearance of a
substance that is completely removed in one pass through the kidney
(like PAH).  The amount of substance in the urine thus reflects the
renal plasma flow.  The equation is:

RPF = ( UPAH* V) / PPAH = CPAH  
    = ( 17mg/mL * 1mL/min) / 0.03mg/mL = 567 mL/min

2)  Renal plasma flow can be used to estimate renal blood flow if the
hematocrit is known - the hematocrit is the fraction of blod which is
cellular, the remainder is the plasma.  Thus, the total blood flow can
be calculated using the formula

RBF = RPF x  1 / 1- hct
    = 567mL/min / (1 - 0.53) = 1206 mL/min

3)  Glomerular filtration rate can be estimated by measuring the
clearance of a substance which is only filtered (like creatinine)


GFR = ( UCR* V) / PCR = CCR
    = ( 9 mg/mL * 1 mL/min ) / 0.07 mg/mL = 129 mL/min

4) The filtered fraction is the percentage of renal plasma flow which
is filtered:

FF = GFR / RPF 
   = 129 mL/min / 567 mL/min = 0.23 = 23%

5)  The amount of substance that is filtered is calculated by
measuring the amount in the plasma times the filtration rate:

amount filtered = PX * GFR 
                = 0.1mg/mL * 129mL/min = 12.9mg/min

6)  The amount of substance appearing in the urine is calculated by
multiplying the urine concentration by the flow rate:

amount in urine = UX * V
                = 3mg/mL * 1mL/min = 3mg/min

7) Since the amount excreted, 3mg/min is less than the amount
filtered, 12.9mg/min, this substance must be filtered and resorbed.

8) Repeating 5-7 for substance Y

amount filtered = PY * GFR
                = 0.04 mg/mL * 129 mL/min = 5 mg/min

9) 
 
amount in urine = UY * V
                = 12mg/mL * 1mL/min = 12mg/min

10) Since the amount in the urine, 12mg/min, is greater than the
amount being filtered (5mg/min) - this substance must be filtered and
secreted


A page with information of how to answer all of these questions
http://www.life.umd.edu/classroom/bsci440/higgins/10_Renal_clearance.htm

Thank you