Problem 116 - Chemistry The Central Science

Question

Enzymes are often described as following the two-step mechanism: $$ \begin{array}{c}{\mathrm{E}+\mathrm{S} \rightleftharpoons \mathrm{ES} \text { (fast) }} \\ {\mathrm{ES} \longrightarrow \mathrm{E}+\mathrm{P} \quad(\text { slow })}\end{array}$$ 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.

Step-by-Step Solution

Answer

The rate law for the enzyme-catalyzed reaction following the given two-step mechanism is \(\text{Rate} = \alpha [\text{E}][\text{S}]\), where \(\alpha = \frac{k_1k_2}{k_{-1}}\). To account for the inhibition reaction, an additional elementary step is \(\text{E + I} \rightleftharpoons \text{EI} \quad (\text{rapid equilibrium})\), which represents the formation of the enzyme-inhibitor (EI) complex.

Step 1: Identify the Reaction Mechanism

The given two-step mechanism of the enzyme-catalyzed reaction is: \[ \begin{array}{c} \text{E + S} \rightleftharpoons \text{ES} \quad (\text{fast}) \\ \text{ES} \longrightarrow \text{E + P} \quad (\text{slow}) \end{array} \]

Step 2: Derive the Rate Law for the Reaction

Since the second step is the slow (rate-limiting) step, the overall rate of the reaction depends on the rate of the ES complex formation. The rate law for the second step can be written as follows: \[ \text{Rate} = k_2[\text{ES}] \] To determine the rate law for the overall reaction, we need to express [ES] in terms of [E] and [S]. To do this, we will look at the first step and assume that it is in a rapid equilibrium state, which means that the forward and reverse rates are equal: \[ k_1[\text{E}][\text{S}] = k_{-1}[\text{ES}] \] Solving for [ES], we get: \[ [\text{ES}] = \frac{k_1}{k_{-1}} [\text{E}][\text{S}] \] Now we can substitute this expression for [ES] in the rate law we derived before: \[ \text{Rate} = k_2\left(\frac{k_1}{k_{-1}} [\text{E}][\text{S}]\right) \] Finally, we define \(\alpha = \frac{k_1k_2}{k_{-1}}\), the rate law for the reaction is: \[ \text{Rate}=\alpha [\text{E}][\text{S}] \]

Step 3: Write an Additional Elementary Step for the Inhibition Reaction

Now we need to add an additional step accounting for the interaction of enzyme E with the inhibitor I. Since the inhibitor binds to the active site of the enzyme, it competes with the substrate, and the resulting enzyme-inhibitor complex EI is not converted into the product. The additional elementary step could be written as: \[ \text{E + I} \rightleftharpoons \text{EI} \quad (\text{rapid equilibrium}) \] With this additional step, the enzyme's interaction with the inhibitor is now incorporated into the mechanism.

Key Concepts

Reaction Mechanism

A reaction mechanism is a detailed, step-by-step description of how a chemical reaction occurs. In enzyme kinetics, the mechanism often involves multiple steps, including the formation and breakdown of enzyme-substrate complexes. For enzymes, a common model is the two-step mechanism:

• Formation of an enzyme-substrate (\( \text{ES} \)) complex: \( \text{E} + \text{S} \rightleftharpoons \text{ES} \) (fast step)
• Conversion of the \( \text{ES} \) complex into enzyme (\( \text{E} \)) and product (\( \text{P} \)): \( \text{ES} \rightarrow \text{E} + \text{P} \) (slow step)

The slow step, also known as the rate-limiting step, is crucial as it determines the overall reaction rate. Understanding each step helps in predicting the kinetics and behavior of the reaction under different conditions. Moreover, being aware of the mechanism helps to identify where inhibitors might interact, potentially altering the reaction rate.

Rate Law

The rate law in enzyme-catalyzed reactions describes how the concentration of reactants affects the speed of the reaction. It is often derived from the rate-limiting step of the mechanism. For our two-step mechanism:The rate-limiting step is:\[\text{ES} \longrightarrow \text{E} + \text{P}\]The rate law is then dependent on the concentration of \( \text{ES} \). Typically expressed as:\[\text{Rate} = k_2[\text{ES}]\]However, because \( \text{ES} \) is formed in the fast equilibrium step:\[\text{E} + \text{S} \rightleftharpoons \text{ES}\]We can express \( [\text{ES}] \) in terms of \( [\text{E}] \) and \( [\text{S}] \):\[[\text{ES}] = \frac{k_1}{k_{-1}} [\text{E}][\text{S}]\]Substituting this into the rate equation gives:\[\text{Rate} = \alpha [\text{E}][\text{S}]\]where\( \alpha = \frac{k_1k_2}{k_{-1}} \). This expression indicates that the rate depends linearly on both the enzyme and substrate concentrations.

Enzyme Inhibitors

Enzyme inhibitors are molecules that bind to enzymes and reduce their activity. They are crucial in regulating biological processes and are commonly used in drug development. There are several types of inhibitors, but in the context of our reaction mechanism, we are concerned with competitive inhibitors.Competitive inhibitors compete with the substrate for the enzyme's active site. This means they can bind to the enzyme (\( \text{E} \)) to form an enzyme-inhibitor complex (\( \text{EI} \)), which does not lead to product formation. This interaction can be represented by an additional step in the mechanism:

• \( \text{E} + \text{I} \rightleftharpoons \text{EI} \) (rapid equilibrium)

Since the inhibitor competes with the substrate, it effectively reduces the concentration of enzyme available for substrate binding, thus decreasing the reaction rate. Understanding how inhibitors work is vital for controlling enzymatic reactions both in vivo and in various industrial processes.

Enzyme-Substrate Complex

The enzyme-substrate complex (\( \text{ES} \)) is a temporary molecular structure formed when an enzyme binds to its substrate. This complex is an essential part of the catalytic process.In the two-step mechanism, the formation of the \( \text{ES} \) complex is the initial step:\[\text{E} + \text{S} \rightleftharpoons \text{ES}\]This step usually happens quickly and establishes rapid equilibrium, meaning the rate of formation equals the rate of dissociation. The stability and properties of the \( \text{ES} \) complex are vital for the following reaction steps because they lead to the conversion into products or provide information about the specificity and efficiency of the enzyme.By understanding the nature of the \( \text{ES} \) complex, scientists can learn how modifications in substrate or enzyme structure affect the overall reaction. This understanding is also crucial when engineering enzymes for improved activity or designing inhibitors that can effectively block enzyme function.