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Enzymatic action may be described as follows:

$$\ce{Enzyme + Substrate <=>[k_1] ES complex ->[k_\mathrm{2}] Enzyme + Product}$$

The initial rate of enzyme-catalyzed reactions can be described by the Michaelis-Menten equation:

$$\mathrm{rate} = \frac{V_\mathrm{max}[\ce{S}]}{K_\mathrm{M} + [\ce{S}]} = \frac{k_\mathrm{cat}[\ce{E}][\ce{S}]}{K_\mathrm{M} + [\ce{S}]}$$

where $V_\mathrm{max}$ is the maximum rate, $[\ce{S}]$ the substrate concentration, $[\ce{E}]$ is the enzyme concentration, $K_\mathrm{M}$ is the Michaelis constant and $k_\mathrm{cat}$ is the number of catalytic cycles per second.

It is known that

$$V_\mathrm{max} = k_2[E]_0$$

and by inspecting the equation above we can deduce that

\begin{align} V_\mathrm{max} &= k_\mathrm{cat}[\ce{E}]\\ \implies k_2[\ce{E}]_0 &= k_\mathrm{cat}[\ce{E}] \end{align}

How, and why, does this equality hold?

How do we know when to use ${[\ce{E}]_0}$ and $[\ce{E}]$ in rate equations, and what are the implications of using either?

${[\ce{E}]_0}$ refers to the enzyme concentration at the start of the reaction, and [E] refers to the concentration of enzyme at any point in time during the course of the reaction.

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    $\begingroup$ Can you define with $[\ce{E}]_0$ and $[\ce{E}]$ are, just to make sure everybody is on the same page $\endgroup$
    – orthocresol
    Sep 1, 2017 at 6:25
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    $\begingroup$ Something seems wrong, what is $\mathrm{k_2}$? $\endgroup$
    – getafix
    Sep 1, 2017 at 12:45
  • $\begingroup$ I think you are getting messed up by adding in extra constants. What you have listed in the reaction as $k_2$ is $k_{cat}$ and $V_{max}=k_{cat}[E]_0$ not $[E]$. On the Wikipedia page, they go through the derivation which explains why the equation depends only the initial enzyme concentration. $\endgroup$
    – Tyberius ♦
    Sep 2, 2017 at 3:12

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Your equation

$$\ce{Enzyme + Substrate <=>[k_1] ES complex ->[k_2] Enzyme + Product}$$

contains only two rate constants, $k_1$ and $k_2$, but not $k_{cat}$ you refer to.

The correct schema for the Michaelis-Menten kinetics would be

$$\ce{Enzyme + Substrate <=>[k_1][k_{-1}] ES complex ->[k_2 = k_{cat}] Enzyme + Product}$$

Note that $k_{cat} = k_2$ is the rate constant for the decay of the ES complex in the second step.

From this you can derive the Michaelis-Menten rate law as described in the Wikipedia article you refer to.

$[E]_0$ is the initial concentration of the enzyme. During the reaction the enzyme exists as free enzyme ($[E]$) and bound in the enzyme substrate complex $[ES]$. These concentrations are related by the equation $[E]_0 = [E] + [ES]$.

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  • $\begingroup$ If k2 = kcat, then wouldn't that yield [E] nought = [E] ? $\endgroup$
    – Jonathan Smith
    Sep 2, 2017 at 14:43
  • $\begingroup$ $[E] = [E_0]$ holds exactly only when $[ES] = 0$, e.g. at the (idealized) beginning of the reaction. $\endgroup$
    – aventurin
    Sep 2, 2017 at 14:55
  • $\begingroup$ I'm slightly confused now. By that logic, at any point in time during the course of the reaction, [E] does not equal [E] nought, then k2 cannot equal to kcat, otherwise the equation $k_2[E]_0 = k_{cat}[E] $ will not hold. $\endgroup$
    – Jonathan Smith
    Sep 2, 2017 at 15:02
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    $\begingroup$ Your equation is wrong. One reason for this is your assumption $V_{max}=k_{cat} [E]$ which is wrong since $[E]$ usually is not constant over time. Another reason is that your $k_{cat}$ came from nowhere. $\endgroup$
    – aventurin
    Sep 2, 2017 at 15:28
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    $\begingroup$ @Jonathan Smith yes, in the rate equation you wrote, the [E] should be $[E]_0$ and in your equation for $V_{max}$ it should equal $k_{cat}[E]_0$. $\endgroup$
    – Tyberius ♦
    Sep 2, 2017 at 20:31

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