Chapter 6, Problem 61RE

MATHEMATICAL For the following aspartase reaction (see Question 28) in the presence of the inhibitor hydroxymethylaspartate, determine $K_{\text{M}}$ and whether the inhibition is competitive or noncompetitive.

$\begin{array}{ccc}\begin{array}{l}\left[s\right]\\ \left(\text{molarity}\right)\end{array}& \begin{array}{l}V,\text{ no Inhibitor}\\ \left(\text{arbitrary units}\right)\end{array}& \begin{array}{l}V,\text{ Inhibitor Present }\\ \left(\text{same arbitrary units}\right)\end{array}\\ \begin{array}{l}1\times {10}^{-4}\\ 5\times {10}^{-4}\\ 1.5\times {10}^{-3}\\ 5\times {10}^{-3}\\ 11.0\end{array}& \begin{array}{l}0.026\\ 0.092\\ 0.136\\ 0.165\\ 9.52\end{array}& \begin{array}{l}0.010\\ 0.040\\ 0.086\\ 0.142\\ 7.60\end{array}\end{array}$

Interpretation Introduction

Interpretation:

The ${\text{K}}_{\text{M}}$ of both the inhibited and the non-inhibited reaction is to be determined. The type of inhibition, either competitive or noncompetitive is to be also identified.

Concept introduction:

In an enzymatic reaction, the inhibitors are the molecules that bind with the enzymes and block their activity. Enzymatic inhibition can be of many types. If a substrate and an inhibitor compete with each other to bind with the enzyme and both have similar structures, then the inhibition is known as competitive. On the other hand, if an inhibitor does not compete with the substrate and both of them can bind to an enzyme at their respective specific sites, then it is known as non-competitive inhibition.

Answer to Problem 61RE

Plot the reciprocal substrate concentration on the x-axis and two reaction velocities (both in the presence and absence of inhibitors) on the y-axis.

Before drawing the plot, the data needs to be modified so that it can be aligned in the Lineweaver–Burk plot. The modification of the data is as follows:

$\begin{array}{cccccc}\begin{array}{l}\left[\text{S}\right]\\ \text{molarity}\end{array}& \frac{1}{\left[\text{S}\right]}& \begin{array}{l}\text{V, no inhibitor}\\ \left(\text{arbituary units}\right)\end{array}& \begin{array}{l}\frac{\text{1}}{\text{V}}\\ \text{no inhibitor}\end{array}& \begin{array}{l}\text{V, inhibitor}\\ \left(\text{same arbituary units}\right)\end{array}& \begin{array}{l}\frac{\text{1}}{\text{V}}\\ \text{inhibitor}\end{array}\\ 1\times {10}^{-4}& \frac{1}{1\times {10}^{-4}}=\text{100000}& 0.026& \frac{1}{0.026}=38.46& 0.010& \frac{1}{0.010}=100\\ 5\times {10}^{-4}& \frac{1}{5\times {10}^{-4}}=20000& 0.092& \frac{1}{0.092}=10.86& 0.040& \frac{1}{0.040}=25\\ 1.5\times {10}^{-3}& \frac{1}{1.5\times {10}^{-3}}=666.66& 0.136& \frac{1}{0.136}=7.35& 0.086& \frac{1}{0.086}=11.62\\ 2.5\times {10}^{-3}& \frac{1}{2.5\times {10}^{-3}}=4000& 0.150& \frac{1}{0.150}=6.66& 0.120& \frac{1}{0.120}=8.33\\ 5\times {10}^{-3}& \frac{1}{5\times {10}^{-3}}=2000& 0.165& \frac{1}{0.165}=6.06& 0.142& \frac{1}{0.142}=7.04\end{array}$

The Lineweaver–Burk plot from above data is:

From this plot ${\text{K}}_{\text{M}}$ can be determined for both of the reactions.

In an uninhibited reaction, the $\frac{-1}{{\text{K}}_{\text{M}}}$ is -1/15000

Therefore, the value of ${\text{K}}_{\text{M}}$ is $6.5\times {10}^{-4}$

$\frac{-1}{{\text{K}}_{\text{M}}}$ value for an inhibited reaction -1/5000

Therefore, the value of ${\text{K}}_{\text{M}}$ is $2\times {10}^{-4}$.

By comparing the Lineweaver–Burk plot of both reactions, it can be said that the inhibition is competitive in nature

Explanation of Solution

Given information:

$\begin{array}{ccc}\begin{array}{l}\left[\text{S}\right]\\ \text{molarity}\end{array}& \begin{array}{l}\text{ V, no inhibitor}\\ \left(\text{arbituary units}\right)\end{array}& \begin{array}{l}\text{ V, inhibitor}\\ \left(\text{same arbituary units}\right)\end{array}\\ {\text{1×10}}^{\text{-4}}& \text{0}\text{.026}& \text{0}\text{.010}\\ {\text{5×10}}^{\text{-4}}& \text{0}\text{.092}& \text{0}\text{.040}\\ \text{1}{\text{.5×10}}^{\text{-3}}& \text{0}\text{.136}& \text{0}\text{.086}\\ \text{2}{\text{.5×10}}^{\text{-3}}& \text{0}\text{.150}& \text{0}\text{.120}\\ {\text{5×10}}^{\text{-3}}& \text{0}\text{.165}& \text{0}\text{.142}\end{array}$

In competitive inhibition, the substrate and the inhibitor have a similar structure. Hence, like the substrate, the inhibitor has also got nearly same affinity toward the active site of an enzyme and is able to bind with the enzyme. As a result, the enzyme’s affinity towards the substrate changes (it inhibits the substrate to bind with the enzyme). In other words, the ${\text{K}}_{\text{M}}$ value gets increased. In this case, the reaction is altered by an inhibitor. High ${\text{K}}_{\text{M}}$ value shows that the enzyme’s affinity with the substrate is decreased.

Also, in Lineweaver–Burk plot the ${\text{V}}_{\text{Max}}$ value (x-intercept) stays the same but the ${\text{K}}_{\text{M}}$ value (y-intercept) changes in presence of inhibitors. So, this is a competitive inhibition.

Conclusion

The reaction (without inhibitor) has a low value of ${\text{K}}_{\text{M}}$ and shows high affinity between the substrate and the enzyme. The reaction (with inhibitor) has a high value of ${\text{K}}_{\text{M}}$ and shows low affinity between the enzyme and the substrate. This is probably due to the involvement of an inhibitor.