To review:
The type of inhibition by A and B inhibitor and the kind of affinity they possess for ES versus E.
Introduction:
Enzyme inhibitors are the molecules that are able to bind to enzyme and reduce the activity. Enzyme inhibitors work by several ways such as preventing the substrate from entering the active site of an enzyme or hinder in catalyzing the reaction. There are three types of enzyme inhibitors, namely, competitive, noncompetitive, and uncompetitive.
Explanation of Solution
The velocity of the reaction at different substrate concentrations in presence of A inhibitor and B inhibitor are given in the question. For calculating, the Vmax (maximum
[S] in mM (millimolar) |
V | |
|
1.3 | 1.17 | 0.76 | 0.855 |
2.6 | 2.10 | 0.38 | 0.476 |
6.5 | 4.00 | 0.15 | 0.25 |
13.0 | 5.70 | 0.077 | 0.175 |
26.0 | 7.20 | 0.038 | 0.139 |
The following table shows the values of [S], V,
[S] in mM (millimolar) |
V | |
|
1.3 | 0.62 | 0.76 | 1.613 |
2.6 | 1.42 | 0.38 | 0.704 |
6.5 | 2.65 | 0.15 | 0.377 |
13.0 | 3.12 | 0.077 | 0.321 |
26.0 | 3.58 | 0.038 | 0.28 |
Plotting the graph between
According, to the Michelis Menten equation, the value of y-intercept represents
From the graph, it can be observed that the value of y-intercept is 0.10 mM-1 sec. y-Intercept represents inverse of Vmax.
KM can be calculated by putting the value of Vmax in the equation Slope=KM/Vmax
So, the KM for the reaction is 9.8and the Vmax is 10 mM. sec-1. In presence of inhibitor A, KM is changed but Vmax is unchanged. So, it is a competitive inhibitor.
According, to the Michelis Menten equation, the value of y-intercept represents
From the graph, it can be observed that the value of y-intercept is 0.25 mM-1 sec. y-Intercept represents inverse of Vmax.
KM can be calculated by putting the value of Vmax in the equation Slope=KM/Vmax
So, the KM for the reaction is 5.68 and the Vmax is4 mM. sec-1. In presence of inhibitor B, KM andVmax both are changed. So, it is anuncompetitive inhibitor. A inhibitor is a competitive inhibitor so it can bind to free enzyme and has no affinity for ES. B inhibitor is an uncompetitive inhibitor, so it can bind to ES as well as E.
Therefore, it can be concluded that inhibitor A is competitive as Vmax remain unchanged whereas, inhibitor B is uncompetitive inhibitor. A can bind to free enzyme and B can bind to E as well as ES.
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Chapter 6 Solutions
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