Michaelis-Menten equation

The Michaelis–Menten equation describes how the rate of an enzyme-catalysed reaction increases with substrate concentration and eventually plateaus at a maximum value ( $v_{max}$ ) due to enzyme saturation.

Consider an enzyme $E$ binding reversibly to a substrate $S$ to form the Michaelis complex $ES$ , which then produces product(s) $P$ and regenerates the enzyme:

$E+S\overset{k_{1}}{\underset{k_{-1}}{\rightleftharpoons}}ES\xrightarrow{k_2} E + P$

where $k_1$ , $k_{-1}$ and $k_2$ are rate constants.

The rate of formation and breakdown of $ES$ is given by:

$\frac{d[ES]}{dt}=k_1[E][S]-(k_{-1}+k_2)[ES]$

Let’s assume:

1. The initial rate method is used to measure the reaction.
2. The initial substrate concentration $[S]_0$ is much greater than the initial enzyme concentration $[E]_0$ , so that the free substrate concentration $[S]$ is approximately equal to $[S]_0$ .
3. Initial product formation is negligible and the step $ES\rightarrow E+P$ is irreversible.
4. Total enzyme concentration is conserved: $[E]_0=[E]+[ES]$ .

Therefore, the enzyme–substrate complex builds up rapidly during an initial transient phase and then reaches a constant concentration (see diagram above). During this period, $\frac{d[ES]}{dt}=0$ . This is known as the steady-state approximation, with

$\frac{[E][S]}{[ES]}=K_M\;\;\;\;\;\;\;\;80$

where $K_M=\frac{k_{-1}+k_2}{k_1}$ is the Michaelis constant.

Substituting $[E]=[E]_0-[ES]$ into eq80 gives:

$[ES](K_M+[S])=[E]_0[S]$

which rearranges to:

$[ES]=\frac{[E]_0}{1+K_M/[S]}$

Since $[S]\approx [S]_0$ ,

$[ES]=\frac{[E]_0}{1+K_M/[S]_0}\;\;\;\;\;\;\;\;81$

Because product formation is relatively slow initially, $ES\rightarrow E+P$ becomes the rate determining step, with the initial reaction velocity $v$ of the overall reaction being:

$v=\frac{d[P]}{dt}=k_2[ES]\;\;\;\;\;\;\;\;82$

Substituting eq81 into eq82 yields the Michaelis-Menten equation:

$v=\frac{v_{max}}{1+K_M/[S]_0}\;\;\;\;\;\;\;\;83$

where $v_{max}=k_2[E]_0$ is a constant for a fixed $[E]_0$ .

When $[S]_0\gg K_M$ , we have $v\approx v_{max}$ (see above diagram). In other words, the rate of the enzyme-catalysed reaction reaches a maximum when the substrate concentration is in excess. Here, the reaction is zero-order with respect to the substrate concentration.

Rearranging eq83 to $v=\frac{v_{max}[S]_0}{[S]_0+K_M}$ shows that at low initial substrate concentration ( $[S]_0\ll K_M$ ), the reaction approximates first-order kinetics for a fixed $[E]_0$ : $v=\frac{v_{max}}{K_M}[S]_0\propto [S]_0$ .

When $[S]_0=K_M$ , we have $v=\frac{1}{2}v_{max}$ .

At a specific temperature and pH, $K_M$ is characteristic for a given enzyme-substrate pair. It can be interpreted as a measure of the affinity of the enzyme for its substrate because it can be rewritten as:

$K_M=K_S+\frac{k_2}{k_1}$

Here, $K_S=\frac{k_{-1}}{k_1}$ is the equilibrium constant for $ES\rightleftharpoons E+S$ . A smaller $K_S$ (and hence a smaller $K_M$ ) corresponds to greater the affinity of the enzyme for its substrate, since the enzyme–substrate complex is less likely to dissociate back into free enzyme and substrate. Additionally, because $K_M$ is equal to the substrate concentration at which the reaction rate is half-maximal, a smaller $K_M$ means the enzyme can achieve this rate at lower substrate concentrations and therefore operates more efficiently.

Question

What is the definition of catalytic efficiency?

Answer

Catalytic efficiency is defined as $\frac{v_{max}}{[E]_0K_M}$ , which is equivalent to $\frac{k_2}{K_M}$ because $v_{max}=k_2[E]_0$ . The term $\frac{v_{max}}{[E]_0}$ corresponds to the maximum number of reactions each enzyme molecule can catalyse per unit time. Furthermore, a smaller $K_M$ indicates a higher affinity of the enzyme for its substrate. Therefore, $\frac{v_{max}}{[E]_0K_M}$ captures both the catalytic speed and substrate-binding efficiency of the enzyme, with a larger value indicating higher catalytic efficiency.

To determine $K_M$ experimentally, $v=\frac{v_{max}[S]_0}{[S]_0+K_M}$ is rearranged into a Lineweaver-Burk form:

$\frac{1}{v}=\frac{1}{v_{max}}+\frac{K_M}{v_{max}}\frac{1}{[S]_0}$

A plot of $\frac{1}{v}$ against $\frac{1}{[S]_0}$ for different initial substrate concentrations results in a straight line with intercept $\frac{1}{v_{max}}$ and slope $\frac{K_M}{v_{max}}$ , from which $K_M$ can be determined (see diagram below).