2
$\begingroup$

From Fersht, Enzyme Structure and Mechanism p. 87:

The Michaelis-Menten mechanism assumes that the enzyme-substrate complex is in thermodynamic equilibrium with free enzyme and substrate.

In my understanding what this means is that the (E-, S- and ES-concentration dependent) rates of association and dissociation have equated. So we're kind of in this situation: enter image description here

where "product" would be referring to [ES] and "reactant" to [E] and [S]. Does that make sense?

$\endgroup$
  • $\begingroup$ Product is still $P$; $ES$ is the transient Enzyme-Substrate complex. If you mean product in the sense of product of a reaction then yes but it is a confusing usage. So stick to calling $ES$ as a transient state $\endgroup$ – WYSIWYG Jun 3 '14 at 9:41
  • $\begingroup$ @tmottm Can you please accept answers, which answer your question satisfactory? This is a nice gesture towards the people who write the answers. You can find all your questions in your User menu. $\endgroup$ – Chris Jun 13 '14 at 10:27
2
$\begingroup$

Thermodynamic equilibrium means that:

Rate of forward reaction = Rate of backward reaction

In this case:

$E + S\xrightleftharpoons[k2]{k1} ES\\ \ \\ \ \\ \\ at\ equilibrium:\\ \ \ \\ \ k1.[E][S] \tiny{\ (forward\ rate)}\normalsize= k2.[ES] \tiny\ (backward\ rate) $

This was the initial assumption in the Michaelis-Menten model.

Later on this was improvised by assuming pseudo-steady state of ES complex. This means that $[ES]$ does not change over time, which is both as a result of its production by the reversible reaction: $E + S \leftrightharpoons\ ES$ and consumption by the irreversible reaction $\ ES\ \xrightarrow{k3} E+P$

that is:

$k1[E][S]\tiny\ (production)\normalsize=(k2+k3)[ES]\tiny\ (consumption)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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