Chapter 6, Problem 66TQ

Summary Introduction

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 V_max (maximum rate of reaction)of the enzyme in presence of inhibitors, graph between $\frac{\text{1}}{\text{[S]}}$ and $\frac{\text{1}}{\text{V}}$, where [S] is the substrate concentration and V is the velocity or the rate of reaction needs to be plotted. The following table shows the values of [S], V, $\frac{\text{1}}{\text{[S]}}$, and $\frac{\text{1}}{\text{V}}$ in presence of inhibitor A.

[S] in mM (millimolar)	V	$\frac{\text{1}}{\text{[S]}}$ in mM-1	$\frac{\text{1}}{\text{V}}$ (mM-1 sec)
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, $\frac{\text{1}}{\text{[S]}}$, and $\frac{\text{1}}{\text{V}}$ in presence of inhibitor B.

[S] in mM (millimolar)	V	$\frac{\text{1}}{\text{[S]}}$ in mM-1	$\frac{\text{1}}{\text{V}}$ (mM-1 sec)
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 $\frac{\text{1}}{\text{V}}$ and $\frac{\text{1}}{\text{[S]}}$ in presence of inhibitor A and inhibitor B.

According, to the Michelis Menten equation, the value of y-intercept represents $-\frac{1}{{\text{V}}_{\max }}$. It can be observed from the graph ( $\frac{\text{1}}{\text{V}}$ versus $\frac{\text{1}}{\text{[S]}}$ ) that the value of y-intercept in presence of A is 0.10 mM^-1 sec. The slope of the graph reflects $\frac{{\text{K}}_{\text{M}}}{{\text{V}}_{\text{max}}}$, where K_M is the Michelis Menten constan, thich can be calculated by using the following formula:

$\begin{array}{c}\text{Slope,}\frac{{\text{K}}_{\text{M}}}{{\text{V}}_{\text{max}}}=\frac{\text{y2}-\text{y1}}{\text{x2}-\text{x1}}\\ =\frac{\text{0}\text{.476}-\text{0}\text{.25}}{\text{0}\text{.38}-\text{0}\text{.15}}\\ =\frac{\text{0}\text{.226}}{\text{0}\text{.23}}\\ \frac{{\text{K}}_{\text{M}}}{{\text{V}}_{\text{max}}}=\text{0}\text{.98}\end{array}$

From the graph, it can be observed that the value of y-intercept is 0.10 mM^-1 sec. y-Intercept represents inverse of V_max.

$\begin{array}{c}y\text{-intercept}=\frac{\text{-1}}{{\text{V}}_{\text{max}}}\\ \text{0}\text{.10}=\frac{\text{-1}}{{\text{V}}_{\text{max}}}\\ {\text{V}}_{\text{max}}={\text{10 mM sec}}^{\text{-1}}\end{array}$

K_M can be calculated by putting the value of V_max in the equation Slope=K_M/V_max

$\begin{array}{l}\frac{K_{\text{M}}}{{\text{V}}_{\text{max}}}=\text{0}\text{.98}\\ \frac{K_{\text{M}}}{10\text{mM}{\sec }^{-1}}=\text{0}\text{.98}\\ K_{\text{M}}=\text{0}\text{.98}\times 10\\ K_{\text{M}}=9.8\end{array}$

So, the K_M for the reaction is 9.8and the V_max is 10 mM. sec^-1. In presence of inhibitor A, K_M is changed but V_max is unchanged. So, it is a competitive inhibitor.

According, to the Michelis Menten equation, the value of y-intercept represents $-\frac{1}{{\text{V}}_{\max }}$. It can be observed from the graph ( $\frac{\text{1}}{\text{V}}$ versus $\frac{\text{1}}{\text{[S]}}$ ) that the value of y-intercept in presence of B is 0.25 mM^-1 sec. The slope of the graph reflects $\frac{K_{\text{M}}}{{\text{V}}_{\text{max}}}$, where K_M is the Michelis Menten constan, thich can be calculated by using the following formula:

$\begin{array}{c}\text{Slope,}\frac{K_{\text{M}}}{{\text{V}}_{\text{max}}}=\frac{\text{y2}-\text{y1}}{\text{x2}-\text{x1}}\\ =\frac{\text{0}\text{.704}-\text{0}\text{.377}}{\text{0}\text{.38}-\text{0}\text{.15}}\\ =\frac{\text{0}\text{.326}}{\text{0}\text{.23}}\\ \frac{K_{\text{M}}}{{\text{V}}_{\text{max}}}=1.42\end{array}$

From the graph, it can be observed that the value of y-intercept is 0.25 mM^-1 sec. y-Intercept represents inverse of V_max.

$\begin{array}{c}y\text{-intercept}=\frac{\text{-1}}{{\text{V}}_{\text{max}}}\\ \text{0}\text{.25}=\frac{\text{-1}}{{\text{V}}_{\text{max}}}\\ {\text{V}}_{\text{max}}=4{\text{ mM sec}}^{\text{-1}}\end{array}$

K_M can be calculated by putting the value of V_max in the equation Slope=K_M/V_max

$\begin{array}{l}\frac{K_{\text{M}}}{{\text{V}}_{\text{max}}}=1.42\\ \frac{K_{\text{M}}}{-4mM{\sec }^{-1}}=1.42\\ K_{\text{M}}=5.68\end{array}$

So, the K_M for the reaction is 5.68 and the V_max is4 mM. sec^-1. In presence of inhibitor B, K_M andV_max 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.

Conclusion

Therefore, it can be concluded that inhibitor A is competitive as V_max remain unchanged whereas, inhibitor B is uncompetitive inhibitor. A can bind to free enzyme and B can bind to E as well as ES.