Question

Enzymes are often described as following the two-step mechanism: $$ \begin{aligned} &\mathrm{E}+\mathrm{S} \rightleftharpoons \mathrm{ES} \quad \text { (fast) } \\\ &\mathrm{ES} \longrightarrow \mathrm{E}+\mathrm{P} \quad \text { (slow) } \end{aligned} $$ where \(\mathrm{E}=\) enzyme, \(\mathrm{S}=\) substrate, \(\mathrm{ES}=\) enzyme-substrate complex, and \(\mathrm{P}=\) product. (a) If an enzyme follows this mechanism, what rate law is expected for the reaction? (b) Molecules that can bind to the active site of an enzyme but are not converted into product are called enzyme inhibitors. Write an additional elementary step to add into the preceding mechanism to account for the reaction of \(E\) with \(I\), an inhibitor.

Short answer

The short answer to the problem is as follows: (a) The expected rate law for the enzyme reaction is: Rate = k' [E] [S], where k' is the overall rate constant for the enzymatic reaction. (b) The additional elementary step for enzyme inhibition is: E + I ⇌ EI, where EI represents the enzyme-inhibitor complex.

Step-by-step solution

Write the given enzyme mechanism

The enzyme mechanism involves two elementary steps: 1. E + S ⇌ ES (fast) 2. ES → E + P (slow)

Identify the Rate Determining Step (RDS)

The slow step in the mechanism is the RDS, which is the conversion of the enzyme-substrate complex (ES) to the enzyme (E) and product (P): ES → E + P

Express the rate of the RDS

The rate of the RDS is given by: Rate = k [ES], where k is the rate constant.

Express [ES] in terms of [E] and [S]

For the fast equilibrium step, E + S ⇌ ES, we can write the equilibrium expression as: K_eq = [ES] / ([E] [S]) Now solve for [ES]: [ES] = K_eq [E] [S]

Substitute [ES] expression in the RDS rate

Substitute the expression for [ES] from step 4 into the rate equation from step 3: Rate = k (K_eq [E] [S]) = k' [E] [S], where k' = k K_eq is the overall rate constant for the enzymatic reaction. So, the expected rate law for the enzyme reaction is: Rate = k' [E] [S] #Part B: Introduce the Enzyme Inhibition Step

Write the elementary step for enzyme inhibition

In the presence of an inhibitor (I), the additional elementary step is the reaction of the enzyme (E) with the inhibitor (I): E + I ⇌ EI In this step, EI represents the enzyme-inhibitor complex. Now, the complete enzyme mechanism with inhibition becomes: 1. E + S ⇌ ES (fast) 2. ES → E + P (slow) 3. E + I ⇌ EI (inhibition)

Key concepts

Enzyme-Substrate Complex

Enzymes play a pivotal role in facilitating biochemical reactions, acting as catalysts that speed up the reaction without being consumed in the process. At the heart of enzymatic activity is the formation of an enzyme-substrate complex. This complex forms when the enzyme (E) binds specifically to a substrate (S), which is the molecule upon which the enzyme acts. The affinity between an enzyme and its substrate is crucial for the enzyme's catalytic activity.

When E and S bind together, they form the ES complex, a temporary association that is essential for the transformation of the substrate into the product (P). This process typically involves a fast equilibrium where E and S quickly form ES, which is then converted to P in a slower, rate-determining step. The formation of the ES complex ensures that the substrate is correctly positioned and oriented within the enzyme's active site, facilitating the chemical reactions necessary to form the product. Understanding this interaction is vital for studying enzyme kinetics and developing drugs that can affect enzyme activity.

Rate Law in Enzymatic Reactions

Understanding the rate of enzymatic reactions is crucial for controlling and predicting biochemical processes. The rate law expresses the relationship between the rate of a reaction and the concentration of its reactants. In the case of enzyme-catalyzed reactions, the rate law can be influenced by various factors, but it typically depends on the concentration of the enzyme-substrate complex (ES).

The rate-determining step (RDS), usually the slowest step of the mechanism, dictates the overall reaction rate. For the provided enzymatic reaction, the RDS involves the conversion of ES to E and P. By applying the equilibrium constant (\(K_{eq}\)) for the fast step where E and S form ES, the concentration of the ES complex can be related to the concentrations of E and S. By substituting \[ES\] with \[K_{eq} \left[ E \right]\left[ S \right]\], we derive the rate of the reaction as a function of these concentrations. This application of the rate law in enzymatic reactions is fundamental to predicting how changes in substrate or enzyme concentration will affect the speed of product formation.

Enzyme Inhibition

Enzyme inhibition is a critical concept in both biochemistry and pharmacology, as it can dramatically influence the rate of enzymatic reactions. An inhibitor (\(I\)) is a molecule that binds to an enzyme and decreases its activity. The inhibitor may bind to the active site, blocking substrate access, or to another site on the enzyme (allosteric site), altering the enzyme's shape and function.

When studying enzyme inhibition, we consider how the inhibitor interacts with the enzyme to form an enzyme-inhibitor complex (EI). The formation of EI competes with substrate binding and can be temporary (reversible inhibition) or permanent (irreversible inhibition). Adding the reaction \[E + I ⇌ EI\] to the enzymatic mechanism introduces a new layer of complexity, as it competes with the formation of the ES complex, thereby impacting the overall reaction rate and the efficiency of the enzyme. Understanding enzyme inhibition is crucial for drug design, as inhibitors can be used therapeutically to slow down or halt the progression of certain diseases by targeting specific enzymes.