According to my chemistry book, there are 3 simplified steps within an enzyme catalysed reaction:

  1. S + E → ES
  2. ES → EP
  3. EP → E + P

Where S is substrate, E is enzyme, ES is enzyme substrate complex, EP is enzyme product complex and P is product.

It is my understanding that at first, when substrate concentration is low the rate determining step is step 1, as this is the slowest step in the mechanism.

When you increase substrate concentration to a certain level, where there are no free enzymes remaining to bond with them, adding more substrate will not make the reaction faster, so the order has become 0 with respect to substrate conc.

Is the rate determining step now 2 or 3? I thought it would be 3 as that is the step that frees up more enzymes which could go on to catalyse more reactions. My book says that it is step 2, as "it depends on how fast the ES can convert into EP". But surely this step won't free up more enzymes to catalyse reaction further?

Is the book correct? Can anyone help me understand this?


1 Answer 1


This is enzyme and substrate concentration specific, since the reaction rate depends on the reaction rate constant k that depends on the enzyme. However if there is a large amount of P present in the solution the EP -> E + P reaction slows down so this can be the rate determining step. On the other hand if you dynamically remove the P molecules from the system the ES->EP can be the rate determining step.

  • $\begingroup$ Is it enzyme specific, or is it related to the amount of P in solution? Your first sentence and the second two seem to be be giving two separate answers. -- FWIW, I agree with the answer of "enzyme specific". Even in the absence of appreciable amounts of product there are enzymes which have a slow off rate of product release, and it's this product release which is the rate limiting step. $\endgroup$
    – R.M.
    Oct 29, 2016 at 19:03
  • $\begingroup$ @R.M. You are right, my explanation was poorly constructed. I wanted to say that it is both enzyme specific and depends on the concentration since the reaction speed depends on a k constant and on the concentrations of the reactants. This constant depends on the enzyme. $\endgroup$
    – fazekaszs
    Oct 29, 2016 at 19:19
  • $\begingroup$ You can (and should) edit your answers to add additional information and to clarify things that might be confusing. (Edits are not just for typo correction.) Just click the "edit" link in the bottom left of your answer. $\endgroup$
    – R.M.
    Oct 29, 2016 at 19:35

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