Question

Calculate ${\mathbf{K}}_{\mathbf{M}}$and ${\mathbf{V}}_{\mathbf{max}}$ from the following data:

$\left[S\right]\left(\mathrm{\mu M}\right)$	${\mathbf{v}}_{\mathbf{0}}\left({\mathrm{mMs}}^{‐1}\right)$
0.1	0.34
0.2	0.53
0.4	0.74
0.8	0.91
1.6	1.04

Short answer

The values of ${\mathrm{K}}_{\mathrm{M}}$and ${\mathrm{V}}_{\max }$are $0.25\mathrm{\mu M}$and $1.25{\mathrm{mMs}}^{‐1}$.

Step-by-step solution

Introduction

The enzyme kinetics investigates the chemical processes that these enzymes catalyze. The rate at which the enzyme's substrate reacts with the enzyme and creates the products is used to calculate the reaction rate. ${\mathrm{K}}_{\mathrm{M}}$ stands for the Michaelis-Menten constant.

The value is not a direct measure of the enzyme affinity for the substrate but rather the concentration of $\mathrm{S}$, where the maximum velocity is half. ${\mathrm{V}}_{\max }$is the greatest rate at which an enzyme catalyzes a reaction.

Lineweaver-Burk plot

The Michaelis-Menten equation is

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

By taking the reciprocal of both sides of the Michaelis-Menten equation, the Lineweaver-Burk plot (double reciprocal equation) is obtained.

$\frac{1}{\mathrm{V}}=\frac{{\mathrm{K}}_{\mathrm{M}}}{{\mathrm{V}}_{\max }}\frac{1}{\left[\mathrm{S}\right]}+\frac{1}{{\mathrm{V}}_{\max }}$

Where

$\mathrm{V}=$Reaction velocity

localid="1649227403811" ${\mathrm{K}}_{\mathrm{M}}=$Michaelis-Menten constant

${\mathrm{V}}_{\max }=$Maximum reaction velocity

$\left[\mathrm{S}\right]=$Substrate concentration

Tabulation and graphical representation for the Lineweaver-Burk plot

$\frac{1}{\left[\mathrm{S}\right]}\left({\mathrm{\mu M}}^{‐1}\right)$	$\frac{1}{{\mathrm{v}}_0}\left({\mathrm{mM}}^{‐1}\mathrm{s}\right)$
10	2.9
5	1.8
2.5	1.3
1.25	1.0
0.625	0.9

Calculating KM and Vmax

$$\begin{gathered} {\mathrm{K}}_{\mathrm{M}}=\frac{‐1}{\mathrm{x}\mathrm{intercept}} \\ =\frac{‐1}{‐4{\mathrm{\mu M}}^{‐1}} \\ =0.25\mathrm{\mu M} \end{gathered}$$

Similarly,

$$\begin{gathered} {\mathrm{V}}_{\max }=\frac{‐1}{\mathrm{y}\mathrm{interpret}} \\ =\frac{‐1}{‐0.8{\mathrm{mM}}^{‐1}\mathrm{s}} \\ =1.25{\mathrm{mMs}}^{‐1} \end{gathered}$$