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What is the general two-step mechanism by which most enzymes work?

Short Answer

Expert verified
Most enzymes work in two steps: first by binding to the substrate and forming an enzyme-substrate complex, and second by converting the substrate into the product and releasing it.

Step by step solution

01

Understanding Enzyme Function

Recognize that enzymes are biological catalysts that speed up chemical reactions in cells. They do this without being consumed or altered in the process. Most enzymes work by lowering the activation energy of reactions, which makes the reactions occur more easily and rapidly.
02

Identifying the Two Steps of Enzyme Action

Learn the two steps that most enzymes use to catalyze reactions: (1) Binding to the substrate: The enzyme provides an active site, a specific place where the reactant molecules, known as substrates, bind. (2) Converting substrate to product: The enzyme provides an environment that facilitates the chemical reaction, transforming the substrates into the final product(s). Once the product is formed, it is released from the active site, and the enzyme can bind to a new substrate.
03

Recognizing the Induced Fit Model

Understand the concept of the 'induced fit model'. This model explains how enzymes change their shape slightly to accommodate the shape of the substrates. Once the substrates are bound to the active site, the enzyme changes conformation to better fit the substrates, which facilitates the catalysis process.

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Key Concepts

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

Biological Catalysts
Enzymes are remarkable molecules that play a critical role in facilitating chemical reactions within biological systems. They act as biological catalysts, meaning they are not only naturally occurring but they also accelerate reactions without being consumed or changed permanently. This efficiency is particularly vital for life, as it allows complex biochemical processes to occur at temperatures and conditions that are compatible with living organisms. Enzymes are highly specific, meaning each type of enzyme catalyzes a particular reaction, ensuring order and regulation within the cellular environment.

What sets enzymes apart from non-biological catalysts is their unparalleled selectivity and efficiency. They often allow reactions to proceed millions of times faster than they would without such catalysis. Moreover, because enzymes are not irreversibly altered during the reaction, they can repeatedly engage in the reaction cycles, acting upon multiple substrate molecules and producing vast amounts of product over time.
Activation Energy
Every chemical reaction requires a certain amount of energy to get started; this is known as activation energy. In biochemical contexts, the activation energy is the threshold that must be crossed for reactants to convert into products. Enzymes dramatically lower this threshold, which is crucial for the timely execution of metabolic processes.

They achieve this feat by stabilizing the transition state of the reaction, which is a high-energy, unstable state that reactants must pass through on the way to becoming products. By lowering the activation energy, enzymes do not change the overall energy of the reactants or the products; instead, they facilitate a path that requires less energy input. This principle is a cornerstone of enzymatic function and a key to understanding how life can carry out complex reactions quickly and efficiently at mild temperatures.
Substrate Binding
The interaction between an enzyme and its substrate is akin to a lock-and-key mechanism where the enzyme's active site matches the shape of the substrate. Substrate binding is the initial step in enzymatic catalysis, where the substrate—the molecule that will undergo the reaction—attaches to the enzyme's specifically designed active site.

Enzymes have evolved to have highly specific active sites that can recognize and bind to their particular substrate(s) among the myriad of molecules found within a cell. This specificity is the result of intricate three-dimensional structures and the presence of amino acids within the active site that interact with the substrate through various types of bonding and interactions (hydrogen bonds, ionic bonds, van der Waals forces etc.). Properly binding the substrate is fundamental because it aligns the reactant molecules in the precise orientation necessary for the chemical reaction to occur efficiently.
Product Formation
After the substrate is securely bound, the enzyme performs its catalytic function, facilitating the transformation of the substrate into the product or products. This conversion is the essence of enzymatic action—the result of all previous steps in the enzyme mechanism. The environment within the active site of the enzyme is meticulously structured to promote the conversion—lowering activation energy, stabilizing the transition state, and providing necessary co-factors or co-enzymes when required.

Product formation may involve the making or breaking of covalent bonds, electron transfers, or other chemical changes. Once the product is formed, it no longer fits perfectly in the active site, which results in the release of the product from the enzyme. This release is essential as it clears the active site for a new substrate molecule to bind, allowing the enzyme to continue the catalytic cycle.
Induced Fit Model
The induced fit model of enzyme action expands upon the lock-and-key concept, providing a more dynamic perspective on substrate interaction. This model proposes that while the active site of the enzyme is complementary to the substrate, it is not a perfect fit from the start. Instead, binding of the substrate induces a conformational change in the enzyme, enhancing the fit between the enzyme and the substrate.

This conformational adjustment can be crucial for catalytic efficiency as it may promote a more favorable alignment of catalytic residues in the active site or assist in destabilizing the substrate, making it more reactive. The induced fit model reflects the flexible nature of enzyme structures and helps to explain how enzymes can exhibit such high specificity and efficiency, even for substrates with similar structures. It underscores the dynamic interaction between an enzyme and its substrate, a dance that is central to the remarkable catalytic capabilities of enzymes.

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Most popular questions from this chapter

Explain the meaning of each term within the Arrhenius equation: activa- tion energy, frequency factor, and exponential factor. Use these terms and the Arrhenius equation to explain why small changes in temperature can result in large changes in reaction rates.

Explain the difference between the average rate of reaction and the in- stantancous rate of reaction.

The tabulated data shown here were collected for the first-order reaction: $$ \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{O}(g) $$ Use an Arrhenius plot to determine the activation barrier and frequency factor for the reaction. $$ \begin{array}{|c|c|} \hline \text { Temperature (K) } & \text { Rate Constant (s }^{-1} \text {) } \\\ \hline 800 & 3.24 \times 10^{-5} \\ \hline 900 & 0.00214 \\ \hline 1000 & 0.0614 \\ \hline 1100 & 0.955 \\ \hline \end{array} $$

The first-order integrated rate law for a reaction \(\mathrm{A} \longrightarrow\) products is derived from the rate law using calculus. \begin{equation} \begin{aligned} \text { Rate } &=k[\mathrm{A}] \quad \text { (first-order rate law) } \\ \text { Rate } &=\frac{d[\mathrm{A}]}{d t} \\\ \frac{d[\mathrm{A}]}{d t} &=-k[\mathrm{A}] \end{aligned} \end{equation} The equation just given is a first-order, separable differential equa- tion that can be solved by separating the variables and integrating: \begin{equation} \begin{aligned} \frac{d[\mathrm{A}]}{[\mathrm{A}]} &=-k d t \\\ \int_{[\mathrm{A}]_{0}}^{[\mathrm{A]}} \frac{d[\mathrm{A}]}{[\mathrm{A}]} &=-\int_{0}^{t} k d t \end{aligned} \end{equation} In the integral just given, \([\mathrm{A}]_{0}\) is the initial concentration of \(\mathrm{A} . \mathrm{We}\) then evaluate the integral: \begin{equation} \begin{aligned}[\ln [\mathrm{A}]]_{[\mathrm{A}]_{0}}^{[\mathrm{Al}} &=-k[t]_{0}^{t} \\ \ln [\mathrm{A}]-\ln [\mathrm{A}]_{0} &=-k t \end{aligned} \end{equation} \begin{equation} \ln [\mathrm{A}]=-k t+\ln [\mathrm{A}]_{0}(\text { integrated rate law }) \end{equation} \begin{equation} \begin{array}{l}{\text { a. Use a procedure similar to the one just shown to derive an inte- }} \\ {\text { grated rate law for a reaction } A \longrightarrow \text { products, which is one-half- }} \\ {\text { order in the concentration of } A \text { (that is, Rate }=k[A]^{1 / 2} )}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { b. Use the result from part a to derive an expression for the half-life }} \\ {\text { of a one-half-order reaction. }}\end{array} \end{equation}

The desorption of a single molecular layer of \(n\) -butane from a single crystal of aluminum oxide is found to be first order with a rate constant of 0.128\(/ \mathrm{s}\) at 150 \(\mathrm{K}\) . \begin{equation} \begin{array}{l}{\text { a. What is the haff-life of the desorption reaction? }} \\ {\text { b. If the surface is initially completely covered with } n \text { -butane at }} \\ {150 \mathrm{K}, \text { how long will it take for } 25 \% \text { of the molecules to desorb? For }} \\ {50 \% \text { to desorb? }}\\\\{\text { c. If the surface is initially completely covered, what fraction will remain }} \\ {\text { covered after } 10 \text { s? After } 20 \mathrm{s?}}\end{array} \end{equation}

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