Chapter 6, Problem 25RE

MATHEMATICAL For an enzyme that displays Michaelis–Menten kinetics, what is the reaction velocity, V (as a percentage of $V_{\text{max}}$), observed at the following values?

(a) $\left[\text{S}\right]=K_{\text{M}}$

(b) $\left[\text{S}\right]=0.5K_{\text{M}}$

(c) $\left[\text{S}\right]=0.1K_{\text{M}}$

(d) $\left[\text{S}\right]=2K_{\text{M}}$

(e) $\left[\text{S}\right]=10K_{\text{M}}$

Interpretation Introduction

Interpretation:

The reaction velocity V (as the percentage of V_max) of an enzyme that displays Michaelis–Menten equation is to be calculated.

Concept introduction:

In 1913, Leonor Michaelis and Maud Menten discovered an equation that proved to be an eminent model for the enzyme kinetics. To measure the velocity or rate of a reaction, the Michaelis–Menten equation is used. The equation is as follows:

$\text{V }=\frac{{\text{V}}_{\max }\left[\text{S}\right]}{{\text{K}}_{\text{m}}+\left[\text{S}\right]}$

Answer to Problem 25RE

V = 0.5 ${\text{V}}_{\text{max}}$

V = 0.33 ${\text{V}}_{\text{max}}$

V = 0.09 ${\text{V}}_{\text{max}}$

V = 0.67 ${\text{V}}_{\text{max}}$

V = 0.91 ${\text{V}}_{\text{max}}$

Explanation of Solution

$\left[S\right]{\text{ = K}}_{\text{M}}$ is given and calculating the reaction velocity (V) on the basis of the Michaelis and Menten equation,

$\text{V }=\frac{{\text{V}}_{\max }\left[\text{S}\right]}{{\text{K}}_{\text{m}}+\left[\text{S}\right]}$

Since $\left[S\right]{\text{ = K}}_{\text{M}}$, substituting the value in the above equation,

$\begin{array}{c}\text{V }=\frac{{\text{V}}_{\text{max}}{\text{ K}}_{\text{M}}}{{\text{K}}_{\text{M}}+{\text{ K}}_{\text{M}}}\\ \text{V }=\frac{{\text{V}}_{\text{max}}\cancel{{\text{K}}_{\text{M}}}}{2\cancel{{\text{K}}_{\text{M}}}}\\ \text{V }=\frac{{\text{V}}_{\text{max}}}{2}\\ \text{V }=0.5{\text{ V}}_{\text{max}}\end{array}$

$\left[S\right]=0.5{\text{ K}}_{\text{M}}$ is given and calculating of reaction velocity (V) on the basis of Michaelis and Menten equation,

$\text{V }=\frac{{\text{V}}_{\max }\left[\text{S}\right]}{{\text{K}}_{\text{m}}+\left[\text{S}\right]}$

Since $\left[S\right]=0.5{\text{ K}}_{\text{M}}$, substituting the value in the above equation,

$\begin{array}{c}\text{V }=\frac{{\text{V}}_{\text{max}}0.5{\text{ K}}_{\text{M}}}{{\text{K}}_{\text{M}}+0.5{\text{ K}}_{\text{M}}}\\ \text{V }=\frac{{\text{V}}_{\text{max}}0.5\cancel{{\text{K}}_{\text{M}}}}{1.5\cancel{{\text{K}}_{\text{M}}}}\\ \text{V }=\frac{{\text{V}}_{\text{max}}}{3}\\ \text{V }=0.33{\text{ V}}_{\text{max}}\end{array}$

$\left[S\right]=0.1{\text{ K}}_{\text{M}}$ is given and calculating of reaction velocity (V) on the basis of Michaelis and Menten equation,

$\text{V }=\frac{{\text{V}}_{\max }\left[\text{S}\right]}{{\text{K}}_{\text{m}}+\left[\text{S}\right]}$

Since $\left[S\right]=0.1{\text{ K}}_{\text{M}}$, substituting the value in the above equation,

$\begin{array}{c}\text{V }=\frac{{\text{V}}_{\text{max}}0.1{\text{ K}}_{\text{M}}}{{\text{K}}_{\text{M}}+0.1{\text{ K}}_{\text{M}}}\\ \text{V }=\frac{{\text{V}}_{\text{max}}0.1\cancel{{\text{K}}_{\text{M}}}}{1.1\cancel{{\text{K}}_{\text{M}}}}\\ \text{V }=\frac{{\text{V}}_{\text{max}}}{11}\\ \text{V }=0.09{\text{ V}}_{\text{max}}\end{array}$

$\left[S\right]=2{\text{ K}}_{\text{M}}$ is given and calculating of reaction velocity (V) on the basis of Michaelis and Menten equation,

Since $\left[S\right]=2{\text{ K}}_{\text{M}}$, substituting the value in the above equation,

$\begin{array}{c}\text{V }=\frac{{\text{V}}_{\text{max}}{\text{ 2 K}}_{\text{M}}}{{\text{K}}_{\text{M}}+{\text{ 2 K}}_{\text{M}}}\\ \text{V }=\frac{{\text{V}}_{\text{max}}\text{ 2}\cancel{{\text{K}}_{\text{M}}}}{3\cancel{{\text{K}}_{\text{M}}}}\\ \text{V }=\frac{{\text{2 V}}_{\text{max}}}{3}\\ \text{V }=0.67{\text{ V}}_{\text{max}}\end{array}$

$\left[S\right]={\text{ 10 K}}_{\text{M}}$ is given and calculating of reaction velocity (V) on the basis of Michaelis and Menten equation,

Since $\left[S\right]={\text{ 10 K}}_{\text{M}}$, substituting the value in the above equation,

$\begin{array}{c}\text{V }=\frac{{\text{V}}_{\text{max}}{\text{ 10 K}}_{\text{M}}}{{\text{K}}_{\text{M}}+{\text{ 10 K}}_{\text{M}}}\\ \text{V }=\frac{{\text{V}}_{\text{max}}\text{ 10}\cancel{{\text{K}}_{\text{M}}}}{11\cancel{{\text{K}}_{\text{M}}}}\\ \text{V }=\frac{{\text{10 V}}_{\text{max}}}{11}\\ \text{V }=0.91{\text{ V}}_{\text{max}}\end{array}$