Chapter 6, Problem 29RE

MATHEMATICAL The hydrolysis of a phenylalanine-containing peptide is catalyzed by $\alpha$-chymotrypsin with the following results. Calculate $K_{\text{M}}$ and $V_{\text{max}}$ for the reaction.

$\begin{array}{cc}\text{Peptide Concentration }\left(M\right)& \text{Velocity }\left(M{\min }^{-1}\right)\\ \begin{array}{l}2.5\times {10}^{-4}\\ 5.0\times {10}^{-4}\\ 10.0\times {10}^{-4}\\ 15.0\times {10}^{-4}\end{array}& \begin{array}{l}2.2\times {10}^{-6}\\ 5.8\times {10}^{-6}\\ 5.9\times {10}^{-6}\\ 7.1\times {10}^{-6}\end{array}\end{array}$

Interpretation Introduction

Interpretation:

The values of ${\text{K}}_{\text{M}}$ and ${\text{V}}_{\text{max}}$ of the hydrolysis reaction of phenylalanine-containing peptide, which is catalyzed by alpha-chymotrypsin, is to be determined.

Concept introduction:

In an enzymatic reaction, ${\text{V}}_{\text{max}}$ is maximum rate of the reaction, ${\text{K}}_{\text{M}}$ is substrate concentration of the reaction that reaches half of its maximum velocity.

The ${\text{V}}_{\text{max}}$ and ${\text{K}}_{\text{M}}$ of a reaction can be calculated with the help of the Lineweaver–Burk plot. In this plot, the substrate concentration, $\left(\frac{1}{\left[\text{S}\right]}\right)$, is along the X-axis and the velocity of the reaction $\frac{1}{\text{V}}$ is along the Y-axis.

Answer to Problem 29RE

The graph shows a straight line and on extrapolating the line, it intersects X-axis as well as Y-axis, and the desired value is determined.

The value of the Y-intercept is 450, which also represents $\left(\frac{1}{{\text{V}}_{\text{max}}}\right)$.

So, the value of ${\text{V}}_{\text{max}}$ is calculated as,

$\begin{array}{l}\left(\frac{1}{{\text{V}}_{\text{max}}}\right)=\left(\frac{1}{450}\right)\\ {\text{V}}_{\text{max}}=2.2\times {10}^{-3}\text{M}{\sec }^{-1}\end{array}$

Conversion of $\text{M}{\sec }^{-1}$ into $\text{M}{\min }^{-1}$.

$\begin{array}{c}60\sec =1\min \\ 1=\frac{1\min }{60\sec }\\ 2.2\times {10}^{-3}\frac{\text{M}}{\sec }=\frac{60\sec }{1\min }\times 2.2\times {10}^{-3}\frac{\text{M}}{\sec }\\ =1.32\times {10}^{-3}\text{M}{\min }^{-1}\end{array}$

The value of the X- intercept is -800, which represents the value of $-\left(\frac{1}{{\text{K}}_{\text{M}}}\right)$.

So, the value of ${\text{K}}_{\text{M}}$ is calculated as,

$\begin{array}{l}-\left(\frac{1}{{\text{K}}_{\text{M}}}\right)=-\left(\frac{1}{800}\right)\\ {\text{K}}_{\text{M}}=1.25\times {10}^{-3}\text{M}\end{array}$

Explanation of Solution

Given information:

$\begin{array}{cc}\begin{array}{l}\text{Peptide Concentration}\\ \text{ (M)}\end{array}& \begin{array}{l}\text{Velocity}\\ \left({\text{M min}}^{-1}\right)\end{array}\\ 2.5\times {10}^{-4}& 2.2\times {10}^{-6}\\ 5.0\times {10}^{-4}& 5.8\times {10}^{-6}\\ 10.0\times {10}^{-4}& 5.9\times {10}^{-6}\\ 15.0\times {10}^{-4}& 7.1\times {10}^{-6}\end{array}$

The reciprocal of the substrate concentration and the velocity is required to draw the respective plots. Before plotting the graph, the data must be converted. The reaction velocity data is given in mole/min. It needs to be converted into mole/sec by dividing by 60.

${\begin{array}{ccccc}\begin{array}{l}\text{peptide concentration}\\ \left(\text{M}\right)\end{array}& \begin{array}{l}\text{pepetide concentration}\\ \left(\frac{1}{\text{M}}\right)\end{array}& \begin{array}{l}\text{concentation in milimoles}\\ \left(\text{mmol}\right)\end{array}& \begin{array}{l}\text{velocity (V)}=\left(\frac{1}{\text{V}}\right)\\ \text{M}{\min }^{-1}\end{array}& \begin{array}{l}\text{velocity}\left(\frac{\text{1}}{\text{V}}\right)\\ \text{M}{\sec }^{-1}\end{array}\\ 2.5\times {10}^{-4}& 0.400\times {10}^4& 4000& 2.2\times {10}^{-6}=0.4545\times {10}^6& 7575.5\\ 5.0\times {10}^{-4}& 0.200\times {10}^4& 2000& 5.8\times {10}^{-6}=0.1724\times {10}^6& 2873.3\\ 10.0\times {10}^{-4}& 0.100\times {10}^4& 1000& 5.9\times {10}^{-6}=0.1694\times {10}^6& 2823.3\\ 15.0\times {10}^{-4}& 0.067\times {10}^4& 670& 7.1\times {10}^{-6}=0.1408\times {10}^6& 2346.6\end{array}}^1{{}^{}}^{}{{}^{}}^{}{}^{}$

In enzyme kinetics, the values of ${\text{K}}_{\text{M}}$ and ${\text{V}}_{\text{max}}$ of a reaction can be determined with the help of the Michaelis–Menten equation. The equation is as follows:

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

This equation is further modified so that it can represent in a straight line.

Modification of the equation is,

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

From this equation, the values of both ${\text{K}}_{\text{M}}$ and ${\text{V}}_{\text{max}}$ can be determined. In the given plot, the values are calculated as follows:

${\text{K}}_{\text{M}}=-\frac{1}{\text{value intersect at x-axis}}$

${\text{V}}_{\text{max}}=\frac{1}{\text{value intersect at Y-axis}}$

Conclusion

With the help of the Lineweaver–Burk plot, the values of both ${\text{K}}_{\text{M}}$ and ${\text{V}}_{\text{max}}$ is determined. The value of ${\text{V}}_{\text{max}}$ is $1.32\times {10}^{-3}{\text{ Mmin}}^{-1}$ and the value of ${\text{K}}_{\text{M}}$ is $1.25\times {10}^{-3}\text{M}$.