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

  1. $\displaystyle\ce{S + E -> ES}$
  2. $\displaystyle\ce{ES -> EP}$
  3. $\displaystyle\ce{EP -> E + P}$

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?


2 Answers 2


The rate determining step doesn't have anything to do with any of the factors you mentioned. It's not even true that the rate determining step can be determined directly from the rate of $$\ce{\mathrm{ES}->\mathrm{EP}}$$ The correct definition is that it is the biggest barrier from a stable species (reactant or intermediate) to any subsequent transition state along the reaction coordinate.

Your comment about the apparent rate of the reaction is unrelated. The rate-determining step is an intrinsic property of the reaction and its mechanism. It is unaffected by concentration, though concentration effects may change which mechanism out of several is predominant.


To add onto Zhe's answer, consider the Michaelis-Menten model of enzyme kinetics.$$v=\frac{k_{cat}[E_0][S]}{K_M+[S]}$$The $k_{cat}[E_0]$ term is sometimes given as $V_{max}$ which is the maximum speed of the reaction while $k_{cat}$ itself is the rate at which the enzyme produces product molecules. $K_M$ is known as the Michaelis constant and behaves like an equilibrium constant. Thus, enzymes with a small $K_M$ are said to have a strong affinity for their substrate because at very low concentrations of substrate, the enzyme is bound.

To answer the question you ask about a large concentration of substrate, this is the case where,


In this case, we can rewrite our equation as,

$$v=k_{cat}[E_0]=V_{max}.$$Notice the cancellation of substrate concentration when $K_M$ is very small.

This means, as you point out, the reaction goes no faster for large concentrations of substrate because the enzyme is always completely saturated and able to find another substrate molecule immediately. Notice that what exactly is a large concentration, however, depends on the size of $K_M.$

The Michaelis-Menten equation does not make any assumptions about what is the rate determining step when being derived, so there is no reason to worry about that. There are limiting cases to think about though.

It's worth pointing out that in the other limiting case, we get a linear equation.


Your Answer

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

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