Question

Enzymes are often described as following the two-step mechanism: $$ \begin{aligned} \mathrm{E}+\mathrm{S} & \rightleftharpoons \mathrm{ES} \text { (fast) } \\ \mathrm{ES} &-\rightarrow \mathrm{E}+\mathrm{P} \text { (slow) } \end{aligned} $$ Where \(\mathrm{E}=\) enzyme, \(\mathrm{S}=\) substrate, and \(\mathrm{P}=\) product. If an enzyme follows this mechanism, what rate law is expected for the reaction?

Short answer

The expected rate law for the given two-step enzyme mechanism is: \[ \text{rate} = k'[\mathrm{E}][\mathrm{S}] \] where k' represents the product of the rate constant and the equilibrium constant.

Step-by-step solution

Identify the rate-determining step

The rate-determining step is the slowest step in a reaction mechanism. In this case, the second step is slower than the first one. Therefore, the rate-determining step is: \[ \mathrm{ES} \rightarrow \mathrm{E} + \mathrm{P} \]

Write the rate expression for the rate-determining step

The rate law for a reaction is the product of the concentrations of the reactants raised to their stoichiometric coefficients in the rate-determining step. Since the rate-determining step involves the conversion of ES to E and P, the rate law for the reaction can be written as: \[ \text{rate} = k[\mathrm{ES}] \] where k is the rate constant for this step.

Establish an expression for the concentration of ES

The formation and dissociation of the enzyme-substrate complex (ES) in the fast equilibrium step can be described by an equilibrium constant, K: \[ \frac{[\mathrm{ES}]_{eq}}{[\mathrm{E}]_{eq}[\mathrm{S}]_{eq}} = K \] Expressing the concentration of the ES complex: \[ [\mathrm{ES}]_{eq} = K[\mathrm{E}]_{eq}[\mathrm{S}]_{eq} \]

Substitute the expression for ES into the rate expression

Now we will substitute the expression for ES from step 3 into the rate expression from step 2: \[ \text{rate} = kK[\mathrm{E}]_{eq}[\mathrm{S}]_{eq} \]

Simplify the rate expression

Combining the rate constant (k) and the equilibrium constant (K) into a new constant (k'): \[ k' = kK \] The rate law is then expressed as: \[ \text{rate} = k'[\mathrm{E}][\mathrm{S}] \] The expected rate law for the reaction, assuming it follows the given two-step mechanism, is: \[ \text{rate} = k'[\mathrm{E}][\mathrm{S}] \] where k' represents the product of the rate constant and the equilibrium constant.

Key concepts

Reaction Mechanism

A reaction mechanism describes the sequence of elementary steps that make up a complex reaction. In the case of enzymes, the mechanism often begins with the enzyme (E) binding to a substrate (S) to form an enzyme-substrate complex (ES). This occurs quickly and is considered a fast reversible step, represented by the equation:

• \[ \mathrm{E} + \mathrm{S} \rightleftharpoons \mathrm{ES} \]

Once the complex is formed, the slow step follows, where the ES complex breaks down into the product (P) and releases the enzyme. This can be denoted as:

• \[ \mathrm{ES} \rightarrow \mathrm{E} + \mathrm{P} \]

Understanding this reaction mechanism is crucial because it explains how the reactants turn into products and helps determine the kinetics of the reaction.

Rate Law

The rate law of a reaction provides the relationship between the reaction rate and the concentration of its reactants. For enzymatic reactions following the two-step mechanism mentioned above, the rate law can be derived from the rate-determining step. This step is the slow step, characterized by the breakdown of the enzyme-substrate complex (ES) into the enzyme (E) and the product (P).

The rate law for this reaction can be expressed as:

• \[ \text{rate} = k[\mathrm{ES}] \]

Here, \( k \) is the rate constant that applies to this specific step. To connect ES to measurable concentrations, we use the fast step's equilibrium to redefine the concentration of ES in terms of E and S, ultimately simplifying the rate law to:

• \[ \text{rate} = k'[\mathrm{E}][\mathrm{S}] \]

where \( k' \) is the combined product of the rate and equilibrium constants.

Rate-Determining Step

The rate-determining step (RDS) in a reaction mechanism is the slowest step, and it controls the overall rate of the reaction. In enzymatic reactions, this step is critical because it signals the point where the reaction sequence bottlenecks. In our enzyme example, the conversion of the enzyme-substrate complex (ES) to the product (P) is the rate-determining step:

• \[ \mathrm{ES} \rightarrow \mathrm{E} + \mathrm{P} \]

This particular step is slower than the initial formation of the ES complex, and because it is slow, it dictates the speed at which the whole reaction can proceed.

Focusing on the rate-determining step allows chemists to understand better which factors most significantly affect the reaction speed. By focusing on this step, strategies can be developed to increase reaction rates, such as optimizing conditions that favor the transition from ES to the product.

Enzyme-Substrate Complex

The enzyme-substrate complex (ES) plays a pivotal role in enzyme kinetics. It forms when an enzyme binds its substrate in a lock-and-key fit, initiating the reaction process. The formation of this complex is typically fast and establishes an equilibrium with the individual enzyme and substrate:

• \[ \mathrm{E} + \mathrm{S} \rightleftharpoons \mathrm{ES} \]

Once the ES complex is formed, the conversion to product usually follows, often being the slow, rate-determining step. The concentration of this complex, hence, significantly influences the reaction's rate law.

By recognizing and establishing the concentration of the ES complex, the rate of the reaction can be expressed as:

• \[ \text{rate} = k'[\mathrm{E}][\mathrm{S}] \]

This expression shows that the reaction speed directly depends on how frequently the enzyme and substrate collide effectively to form ES. The study of enzyme-substrate interactions is fundamental in enzyme kinetics, as it provides deep insight into how enzymes work and how they can be manipulated in various scientific applications.