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Studies of the inhibition of enzymes by various compounds often provide information about the nature of the binding site and the mechanism. Competitive inhibition is when the inhibitor, I, competes with the substrate for the catalytic site. This mechanism can be written as $$ \begin{gathered} \mathrm{E}+\mathrm{S} \rightleftharpoons \mathrm{ES} \rightarrow \mathrm{E}+\mathrm{P} \\ \mathrm{E}+\mathrm{I} \rightleftharpoons \mathrm{EI} \end{gathered} $$ Derive the steady-state rate law for this mechanism and show that it follows Michaelis-Menten kinetics when the inhibitor concentration is constant. Assume the inhibitor concentration is much greater than the enzyme concentration.

Short Answer

Expert verified
The rate law is \( v = \frac{k_2 [E]_0 [S]}{K_M(1 + \frac{[I]}{K_i}) + [S]} \), which follows Michaelis-Menten kinetics with modified \(K_M\).

Step by step solution

01

Understanding the Mechanism

In competitive inhibition, the enzyme (E) can bind to either the substrate (S) to form an enzyme-substrate complex (ES) or to the inhibitor (I) to form an enzyme-inhibitor complex (EI). Notably, the inhibitor competes for the active site, preventing substrate binding.
02

Write the Reaction Rates

The rate of formation for the ES complex is given by: \[ v = k_1 [E][S] - k_{-1} [ES] \]. Similarly, for EI complex formation: \[ v_{EI} = k_3 [E][I] - k_{-3} [EI] \]. These equations form the basis to derive the steady-state rate law.
03

Assume Steady-State for ES

In the steady-state, the rate of formation of ES is equal to its rate of breakdown: \[ k_1 [E][S] = k_{-1} [ES] + k_2 [ES] \], where \(k_2\) represents the conversion of ES to product (P). This means the net rate of change of [ES] is zero.
04

Relate the Total Enzyme Concentration

The total enzyme concentration can be expressed as: \[ [E]_0 = [E] + [ES] + [EI] \]. This allows us to express free enzyme concentration \([E]\) in terms of total enzyme concentration.
05

Substitute and Simplify Equations

Substitute \([E] = [E]_0 - [ES] - [EI]\) into the steady-state expressions. This allows us to express rates based on known concentrations and constants.
06

Derive the Rate Law under Inhibition

The rate of product formation \(v\) is given by: \[ v = k_2 [ES] \]. Substitute \([ES]\) using the steady-state expression and simplify to \[ v = \frac{k_2 [E]_0 [S]}{K_M(1 + \frac{[I]}{K_i}) + [S]} \], where \([I]\) is constant.
07

Confirm Michaelis-Menten Form

The rate equation resembles the Michaelis-Menten equation but includes an extra term due to inhibition: \[ V_{max} = k_2[E]_0 \] and \[ K_M' = K_M(1 + \frac{[I]}{K_i}) \], showing that competitive inhibition effectively increases the \(K_M\) value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enzyme Kinetics
Enzyme kinetics is the study of how enzymes bind substrates and convert them into products. At the heart of this study lies the interaction between enzymes and substrates, which occurs through specific binding sites on the enzyme surface. Each enzyme has an optimal way of binding, leading to different rates of reactions known as reaction kinetics.
The concept is crucial since it provides insight into enzyme efficiency and enzyme mechanisms, which are paramount for developing drugs and understanding metabolic pathways. The reaction rate is influenced by several factors, such as enzyme concentration, substrate concentration, and the presence of inhibitors or activators. Inhibitors, like the one observed in competitive inhibition, play a vital role in regulating enzyme activity by affecting how an enzyme interacts with its substrate.
Michaelis-Menten Equation
The Michaelis-Menten equation offers a framework to describe the rate of enzyme-catalyzed reactions. It is a mathematical representation that details how reaction velocity relates to substrate concentration. The equation takes the form:\[ v = \frac{V_{max}[S]}{K_M + [S]} \]Where:- \( v \) is the reaction velocity,- \( V_{max} \) is the maximum reaction velocity,- \( [S] \) is the substrate concentration, and- \( K_M \) is the Michaelis constant, reflecting the substrate concentration at which the reaction velocity is half of \( V_{max} \).The Michaelis-Menten equation provides a clear understanding of how enzymes reach their maximum rate of reaction and is an essential tool for characterizing enzyme kinetics in the presence or absence of inhibitors.
Steady-State Approximation
The steady-state approximation is a simplifying assumption used in the analysis of enzyme kinetics. It assumes that the formation and breakdown of the enzyme-substrate complex (ES) reach a balance, leading to a constant concentration over time.This means, during the reaction, the rate at which the substrate binds to the enzyme to form \([ES]\) is equal to the rate at which \([ES]\) converts back to enzyme and product. This approximation is vital because it allows us to derive simplified equations like the Michaelis-Menten equation. In the context of competitive inhibition, using the steady-state approximation helps to simplify the equations and leads to the derivation of enhanced treatments to understand enzyme behavior under less than ideal conditions.
Enzyme-Inhibitor Complex
An enzyme-inhibitor complex is formed when an inhibitor binds to an enzyme. In competitive inhibition, this mechanism prevents substrate molecules from binding to the enzyme. Consequently, no product is formed as long as the inhibitor occupies the active site of the enzyme.The interaction can be represented as:- Enzyme (E) + Inhibitor (I) ⇌ Enzyme-Inhibitor Complex (EI)These complexes are significant because they directly impact the rate of an enzyme-mediated reaction. In competitive inhibition, the presence of an inhibitor increases the apparent \(K_M\) without affecting \(V_{max}\), making it symbolically similar to enzyme-substrate interactions but with nuanced differences due to the involvement of inhibitors.Understanding enzyme-inhibitor complex dynamics aids in drug design, allowing scientists to develop medications that can specifically target and regulate certain enzyme functions within biological systems.

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Most popular questions from this chapter

In the text, the steady-state rate law was derived with the assumption that the reaction is irreversible and/or only the initial velocity was determined. Derive the steady-state rate law for the reversible enzyme reaction: $$ \mathrm{E}+\mathrm{S} \rightleftharpoons \mathrm{X} \rightleftharpoons \mathrm{E}+\mathrm{P} $$ Show that the rate law can be put into the form $$ v=\frac{\left.V_{\mathrm{S}} / K_{\mathrm{S}} \mid \mathrm{S}\right]-V_{\mathrm{p}} / K_{\mathrm{p}}[\mathrm{P}]}{1+[\mathrm{S}] / K_{\mathrm{S}}+[\mathrm{P}] / K_{\mathrm{p}}} $$ where \(V_{\mathrm{S}}\) and \(V_{\mathrm{P}}\) are the maximum velocities for the forward and reverse reactions, and \(K_{\mathrm{s}}\) and \(K_{\mathrm{p}}\) are the Michaelis constants for the forward and reverse reactions. When equilibrium is reached, \(v=0\). Calculate the ratio of the equilibrium concentrations of \(S\) and \(P,[P] /[S]\), in terms of the four steady-state parameters. This relationship is called the Haldane relationship and is a method for determining the equilibrium constant of the overall reaction.

Many biological reactions are very sensitive to \(\mathrm{pH}\). This can readily be incorporated into the rate laws because protolytic reactions can be assumed to be much faster than other rates in most cases. For example, in enzyme mechanisms the ionization states of a few key protein side chains are often critical. Suppose that two ionizable groups on the enzyme are critical for catalytic activity and that one of them needs to be protonated and the other deprotonated. The protolytic reactions can be written as $$ \mathrm{EH}_{2} \rightleftharpoons \mathrm{EH}+\mathrm{H}^{\prime} \rightleftharpoons \mathrm{E}+2 \mathrm{H}^{\prime} $$ If only the species \(\mathrm{EH}\) is catalytically active and the protolytic reactions are much more rapid than the other steps in the reaction, all of the rate constants that multiply the free enzyme concentration in the rate law have to be multiplied by the fraction of enzyme present as EH. A. Calculate the fraction of free enzyme present as EH at a given pH. Your answer should contain the concentration of \(\mathrm{H}^{+}\)and the ionization constants of the two side chains, \(\left.K_{\mathrm{E} 1}=[\mathrm{E}]\left[\mathrm{H}^{+}\right] / \mathrm{EH}\right]\) and \(\left.K_{\mathrm{E} 2}=[\mathrm{EH}]\left[\mathrm{H}^{+}\right] / \mathrm{EH}_{2}\right]\). B. Assume that ES in the Michaelis-Menten mechanism (Eq. 5-2) also exists in three protonation states, \(\mathrm{ESH}_{2}\), ESH, and ES, with only ESH being catalytically active. Calculate the fraction of the enzyme-substrate complex present as EHS. Designate the ionization constants as \(K_{\mathrm{ESI}}\) and \(K_{\mathrm{ES} 2}\). C. Use the results of parts \(\mathrm{A}\) and \(\mathrm{B}\) to derive equations for the \(\mathrm{pH}\) dependence of \(V_{\mathrm{m}^{\prime}} K_{\mathrm{M}}\), and \(V_{\mathrm{m} /} K_{\mathrm{M}}\). Measurement of the \(\mathrm{pH}\) dependence of the steady. state parameters permits determination of the ionization constants, and sometimes identification of the amino acid side chains.

Consider the binding of a protein, P, to a DNA segment (gene regulation). Assume that only one binding site for P exists on the DNA and that the concentration of DNA binding sites is much less than the concentration of \(P\). The reaction mechanism for binding can be represented as $$ \mathrm{P}+\mathrm{DNA} \rightleftharpoons \mathrm{P}-\mathrm{DNA} \rightarrow \mathrm{P}-\mathrm{DNA}^{\prime} $$ where the second step represents a conformational change in the protein. Calculate the rate law for the appearance of P-DNA' under the following conditions. A. The first step in the mechanism equilibrates rapidly relative to the rate of the overall reaction and [P-DNA] \ll<[DNA]. B. The intermediate, P-DNA, is in a steady state. C. The first step in the mechanism equilibrates rapidly relative to the rate of the overall reaction and the concentrations of DNA and P-DNA are comparable. Express the rate law in terms of the total concentration of DNA and P-DNA, that is, [DNA] + [P-DNA]. D. The following initial rates were measured with an initial DNA concentration of \(1 \mu \mathrm{M}\). \begin{tabular}{rc} {\([\mathrm{P}](\mu \mathrm{M})\)} & \(10^{4} \operatorname{Rate}(\mathrm{M} / \mathrm{s})\) \\ \hline 100 & \(8.33\) \\ 50 & \(7.14\) \\ 20 & \(5.00\) \\ 10 & \(3.33\) \\ \hline \end{tabular} Which of the rate laws is consistent with the data?

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