enzymes and substrates
overall reaction: S -> P
step 1: E + S -> ES fast
step 2: ES -> E + S fast
step 3: ES -> E + P slow
what is the order of the reaction in E and what is the order of reaction in S?
To determine the order of the reaction in E and the order of the reaction in S, we need to examine the rate equation for the overall reaction (S -> P) and the rate equations for the individual steps of the reaction.
The rate equation for the overall reaction is typically given by: Rate = k[E]^a[S]^b, where k is the rate constant, [E] represents the concentration of the enzyme, [S] represents the concentration of the substrate, and a and b are the orders of the reaction with respect to E and S, respectively.
Looking at the individual steps of the reaction:
Step 1: E + S -> ES (fast reaction)
This step involves the binding of the enzyme (E) and the substrate (S) to form an enzyme-substrate complex (ES).
Step 2: ES -> E + S (fast reaction)
This step involves the dissociation of the enzyme-substrate complex (ES) back into free enzyme (E) and free substrate (S).
Step 3: ES -> E + P (slow reaction)
This step involves the conversion of the enzyme-substrate complex (ES) into product (P) along with the release of the enzyme (E).
Since the slowest step in a reaction usually determines the overall rate, step 3 is often the rate-determining step.
Now, let's analyze the rate equation for the overall reaction:
Rate = k[E]^a[S]^b
In the rate equation, we assume that the concentrations of the enzyme and substrate are much greater than the concentrations of the enzyme-substrate complex (ES). Therefore, we can assume that the concentration of the enzyme is approximately constant, such that [E] ≈ [E]_total.
Since step 3 is the slow step and involves the conversion of ES to E and P, the reaction rate depends on the concentration of ES. Therefore, the order of the reaction in E is often referred to as zero-order, meaning that the concentration of E does not significantly affect the reaction rate.
On the other hand, step 3 also involves the decomposition of ES into E and P, indicating that the rate of the reaction is directly influenced by the concentration of ES. Therefore, the order of the reaction in S is often referred to as first-order, meaning that the reaction rate is directly proportional to the concentration of S.
In summary:
Order of reaction in E: Zero-order
Order of reaction in S: First-order