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MCB 450
Problem Set 2  Answers
1.
From the following data on an enzymic reaction, determine (a) the type of
inhibition, (b) K
M
for the substrate, (c) V
max
, and (d) K
i
, assuming that the
inhibitor is present at a final concentration of 5mM in the reaction mixture.
s
v
v inhib
1/s
1/v
1/v inhib
2
139
88
0.5 0.007194 0.011364
3
179
121
0.333333 0.005587 0.008264
4
213
149
0.25 0.004695 0.006711
10
313
257
0.1 0.003195 0.003891
15
370
313
0.066667 0.002703 0.003195
Using the graphing program at:
http://www.changbioscience.com/stat/ec50.html
we then make plots of 1/v vs. 1/s for both the uninhibited and inhibited cases, and
get the following values:
Inhibited:
Results:
Model: Linear
a x + b
a = 0.018798 ( +/ 1.015E4 )
b = 0.001985 ( +/ 3.005E5 )
Uninhibited:
Results:
Model: Linear
a x + b
a = 0.010269 ( +/ 2.154E4 )
b = 0.002107 ( +/ 6.376E5 )
Calculation of V
max
The yintercept in each case is approximately 0.002. V
max
= 1/0.002 = 500
μ
g/hr
This is a reasonable answer if one eyeballs the numbers in the uninhibited case.
Inhibition is competitive since plots intersect on the yaxis.
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View Full DocumentCalculation of K
M
Slope of the uninhibited plot = K
M
/V
max
0.010269 = K
M
/500
K
M
= 5.13 mM
This is a reasonable answer. K
M
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 Summer '09
 MINTEL

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