Chapter 24, Problem 35IL

Interpretation Introduction

Interpretation:

The value of ${\text{Rate}}_{\text{max}}$ for the given reaction has to be determined.

Concept introduction:

For enzymatic reaction, the common equation is given below,

  $\text{E}+\text{S}\rightleftharpoons \text{ES}\rightleftharpoons \text{E}+\text{P}$

Here,

E is the enzyme.

S is the substrate.

ES is the enzyme substrate complex.

P is the product.

The general equation for Michaelis-Menten is dervied from above equation,

  ${\text{v}}_{\text{°}}={\text{V}}_{\max }\left(\frac{[\text{S}]}{[\text{S}]+K_{\text{m}}}\right)$                                                                                      (1)

Here,

${\text{v}}_{\text{°}}$ is the initial rate of the reaction.

${\text{V}}_{\text{max}}$ is the rate maximum of the reaction.

$\text{S}$ is the substrate concentration.

${\text{K}}_{\text{m}}$ is the Michaelis-Menten constant which is half of the maximum velocity of the reaction.

Take the Reciprocal of both the sides in equation (1),

  $\frac{1}{v_{°}}=\left(\frac{K_m}{V_{max}}\right)\left(\frac{1}{[S]}\right)+\frac{1}{V_{max}}$ (2)

This equation is called as Lineweaver-Burk equation. By using this equation the value of ${\text{Rate}}_{\text{max}}$ is calculated for the given reaction.

Explanation of Solution

The ${\text{Rate}}_{\text{max}}$ for the given reaction is calculated below,

Given:

The substare concentration and reaction rate are given.

By using the equation,

  $\frac{1}{v_{°}}=\left(\frac{K_m}{V_{max}}\right)\left(\frac{1}{[S]}\right)+\frac{1}{V_{max}}$

Calculate the value of $\frac{\text{1}}{{\text{v}}_{\text{°}}}$ and $\frac{\text{1}}{\text{[S]}}$,

The graph between $\frac{\text{1}}{{\text{v}}_{\text{°}}}$ and $\frac{\text{1}}{\text{[S]}}$ is plotted by taking the values from table (1).

The equation for the straight line obtained from the graph is $\text{y}=0.148\text{x}+0.00\text{92}$.

$\frac{\text{1}}{{\text{v}}_{\text{°}}}\times {\text{10}}^{-6}=\text{0}{\text{.148×10}}^{-\text{5}}\text{x}+\frac{\text{1}}{{\text{V}}_{\text{max}}}$

Rearrange the equation, in form of $\text{y = mx + c}$.

$\text{y}=\text{1}\text{.5x}+\text{9200}$

The value of intercept comes from the graph is $\text{9200}$.

$\frac{\text{1}}{{\text{Rate}}_{\text{max}}}=9200$

Rearrange for ${\text{Rate}}_{\max }$,

$\begin{array}{c}{\text{Rate}}_{\max }=\frac{\text{1}}{\text{9200}}\\ =\text{1}{\text{.1×10}}^{-\text{4}}\text{mol}{\text{.L}}^{-\text{1}}\cdot {\text{min}}^{-\text{1}}\end{array}$.

So, the value of ${\text{Rate}}_{\text{max}}$ is $1.1\times 10^{-4}mol.L^{-1}\cdot mi{n}^{-1}$.

The value of ${\text{Rate}}_{\text{max}}$ for the given reaction is $1.1\times 10^{-4}mol\cdot L^{-1}\cdot mi{n}^{-1}$.