Question

Enzymes are often described as following the two-step mechanism: $$ \begin{aligned} \mathrm{E}+\mathrm{S} & \rightleftharpoons \mathrm{ES} \text { (fast) } \\ \mathrm{ES} & \ldots \mathrm{E}+\mathrm{P} \text { (slow) } \end{aligned} $$ 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 this enzyme-catalyzed reaction following the given two-step mechanism is: \(Rate = k[\mathrm{E}][\mathrm{S}]\), where \(k = k_2 K_\text{eq}\).

Step-by-step solution

Write the rate equation for the slow step.

Since the second step of the enzyme mechanism is the slow step and it involves the breakdown of the enzyme-substrate complex (ES), it can be assumed to be the rate-determining step. So, we can write the rate equation for this step as: \(Rate = k_2[\mathrm{ES}]\) Where \(k_2\) is the rate constant for the slow step, and [ES] represents the concentration of the enzyme-substrate complex.

Establish an equilibrium expression for the fast step.

Since the first step is an equilibrium, we can write the equilibrium constant, \(K_\text{eq}\), for the first reaction as: \(K_\text{eq}=\frac{[\mathrm{ES}]}{[\mathrm{E}][\mathrm{S}]}\)

Solve [ES] in terms of the equilibrium constant and enzyme, substrate concentrations.

Now, using the equilibrium expression, we can solve the concentration of the enzyme-substrate complex ([ES]) in terms of the enzyme concentration ([E]), the substrate concentration ([S]) and the equilibrium constant (\(K_\text{eq}\)). \([\mathrm{ES}] = K_\text{eq}[\mathrm{E}][\mathrm{S}]\)

Substitute [ES] in the rate equation from Step 1.

Substitute the expression for [ES] we found in Step 3 into the rate equation from Step 1: \(Rate = k_2(K_\text{eq}[\mathrm{E}][\mathrm{S}])\)

Modify the rate law equation.

We can further simplify the rate law equation by combining the rate constant for the slow step (\(k_2\)) and the equilibrium constant (\(K_\text{eq}\)) into a new constant, \(k\): \(Rate = k[\mathrm{E}][\mathrm{S}]\) Where \(k = k_2 K_\text{eq}\).

Result:

The rate law we can expect for this enzyme-catalyzed reaction following the given two-step mechanism is: \(Rate = k[\mathrm{E}][\mathrm{S}]\)

Key concepts

Rate Law for Enzyme-Catalyzed Reactions

Understanding how enzymes catalyze reactions is pivotal in biochemistry. An enzyme's efficiency plays a critical role in metabolic pathways. The rate law for enzyme-catalyzed reactions provides insight into this process.

When looking at enzyme kinetics, the rate law helps us predict the speed of the reaction under varying conditions of enzyme and substrate concentrations. It's derived from the premise that the reaction rate is proportional to the concentration of the reactants, in this case, the enzyme-substrate complex (ES).

The rate-determining step in an enzyme-catalyzed reaction, typically the slowest step, defines the rate law. This step usually involves the conversion of the ES complex to the product and unbound enzyme. The rate at which this occurs is expressed as the rate constant, denoted as k, multiplied by the concentration of the ES complex. The final form of the rate law equation, Rate = k[ES], succinctly reveals how the speed of the reaction is directly tied to the ES complex's concentration.

By evaluating the rate law, students and researchers can gain a deeper appreciation for the catalytic power of enzymes and predict how changes in conditions might affect the reaction kinetics.

Enzyme-Substrate Complex

The enzyme-substrate complex (ES) is a temporary molecular union where an enzyme binds to its substrate. It's akin to a lock and key, where the enzyme's active site (the lock) perfectly fits the substrate (the key). This specificity determines the enzyme's catalytic action and is fundamental to the biochemistry of life processes.

Formation of the ES complex is the initial step in enzymatic catalysts and occurs rapidly. The complex formation is reversible and leads either to the substrate’s transformation into the product with the enzyme remaining unchanged or to the dissociation back into the enzyme and unaltered substrate.

Understanding this complex offers enormous benefits, such as insights into drug design and metabolic control. It also provides crucial information to manipulate reaction rates by altering substrate concentration, which is especially relevant in industrial processes and therapeutic applications.

Equilibrium Constant

The equilibrium constant (Keq) is a value that reflects the ratio of the concentration of products to reactants at a state of equilibrium, for reversible reactions. In enzyme kinetics, Keq describes the relation between the concentrations of the enzyme-substrate complex, free enzyme, and free substrate.

In a chemical equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, which means no net change occurs in the concentration of reactants and products.

The equilibrium constant is a crucial factor in enzyme kinetics as it provides a quantitative measure of how easily the enzyme forms an ES complex with the substrate. This information, in turn, affects the rate law of an enzyme-catalyzed reaction. A high Keq indicates a strong affinity between enzyme and substrate, leading to a higher concentration of the ES complex, whereas a low Keq suggests a reduced enzyme-substrate affinity, resulting in less ES complex formation.

Enzymologists use this constant to predict reaction tendencies and to calculate the concentrations of different species present within an equilibrium mixture, enabling a deeper understanding of reaction dynamics.