Chapter 6, Problem 28RE

MATHEMATICAL The enzyme $\text{β}$-methylaspartase catalyzes the deamination of $\text{β}$-methylaspartate

[V. Williams and J. Selbin, J. Biol. Chem. 239, 1636 (1964)]. The rate of the reaction was determined by monitoring the absorbance of the product at 240 nm $\left({\text{A}}_{240}\right)$. From the data in the following table, determine $K_{\text{M}}$ for the reaction. How does the method of calculation differ from that in Questions 26 and 27?

Interpretation Introduction

Interpretation:

The ${\text{K}}_{\text{m}}$ value for the reaction in which “enzyme b-methyl aspartate catalyzes the deamination of b-methyl aspartate” is to be determined.

Concept introduction:

In an enzymatic reaction, ${\text{V}}_{\text{max}}$ is the maximum rate of the reaction, ${\text{K}}_{\text{m}}$ is the substrate concentration at which the rate of the reaction is half of ${\text{V}}_{\text{max}}$. ${\text{K}}_{\text{m}}$ also defines substrate affinity of the enzyme.

To determine ${\text{V}}_{\text{max}}$ and ${\text{K}}_{\text{m}}$, kinetic data obtained from the reaction needs to be put into the Lineweaver–Burk plot.

To draw this plot, substrate concentration $\frac{1}{\left[\text{S}\right]}$ and its reciprocal velocity value $\frac{1}{\text{V}}$, needs to be plotted on the y- and x- axes.

The $\text{y}\text{intercept}$ will be the $\frac{1}{{\text{V}}_{\text{max}}}$ of the reaction. From ${\text{V}}_{\text{max}}$ value, ${\text{K}}_{\text{m}}$ can be determined. The $x\text{intercept}$ is $-\frac{1}{K_{\text{m}}}$. From that, ${\text{K}}_{\text{m}}$ can also be obtained.

Answer to Problem 28RE

Plotting the data gives the following straight line:

From this straight-line graph, the value of ${\text{K}}_{\text{m}}$ is $2.86\times {10}^{-2}\text{M}$.

Explanation of Solution

Given information:

Following data containing substrate concentrations and velocity of the reaction at that substrate concentration, is given.

$\begin{array}{cc}SubstrateConcentration\left(mmol/Lit\right)& Velocity(\Delta {\text{A}}_{240}{\min }^{-1})\\ 0.002& 0.045\\ 0.005& 0.115\\ 0.20& 0.285\\ 0.040& 0.380\\ 0.060& 0.460\\ 0.080& 0.475\\ 0.100& 0.505\end{array}$.

In enzyme kinetics, to determine ${\text{K}}_{\text{m}}$ and ${\text{V}}_{\text{max}}$ of a reaction, the Michaels–Menten equation can be used.

$V=\frac{K_{\text{m}}}{V_{\text{max}}}$ (This is the actual Michaelis–Menten equation.)

But ${\text{K}}_{\text{m}}$ and ${\text{V}}_{\text{max}}$ of a reaction cannot be accurately determined from this equation. To do that, this equation needs to be modified, so that it can be represented as a straight line and the straight-line equation, $y=\text{m}x+\text{b}$, can be used.

Modification of the equation is

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

This equation can be plotted on the graph, $\frac{1}{V}$ represents the y-axis and $\frac{K_{\text{m}}}{V_{\text{max}}}$ is the slope $m$. $\frac{1}{\left[\text{S}\right]}$ represents the x-axis and b is $\frac{1}{V_{\text{max}}}$. The y intercept is $\frac{1}{V_{\text{max}}}$. And the x-intercept is $-\frac{1}{K_{\text{m}}}$. This plot is known as the Lineweaver–Burk plot. From this plot, ${\text{K}}_{\text{m}}$ and ${\text{V}}_{\text{max}}$ can be determined easily.

$V_{\text{max}}=\frac{1}{y\text{intercept}}$ and $K_{\text{m}}=\frac{1}{x\text{intercept}}$

Conclusion

The enzyme, b-methyl aspartate, catalyzes the deamination of b-methyl aspartate. By plotting the given data into a Lineweaver–Burk plot, K_m for this reaction is 2.86 × 10^-2 mole. Although in this case, concentration cannot be determined directly. Absorbance values were used instead, as a matter of convenience.