7
$\begingroup$

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.

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

1 Answer 1

Reset to default
5
$\begingroup$

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 $[E]_0 = [E] + [ES]$.

Improve this answer
$\endgroup$
10
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.