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

Short Answer

Expert verified
Most enzymes work through a two-step mechanism: first, the substrate binds to the enzyme’s active site, and then the enzyme catalyzes the conversion of the substrate into the product.

Step by step solution

01

Binding

The first step in the general two-step mechanism by which most enzymes work is the binding step. In this step, the substrate—the molecule upon which the enzyme acts—binds to the active site of the enzyme. This binding typically involves non-covalent interactions such as hydrogen bonds, ionic bonds, and Van der Waals forces.
02

Catalysis

After binding, the next step is catalysis. During this step, the enzyme facilitates a chemical reaction that converts the substrate into the product(s). The enzyme provides an environment in which the reaction can occur more easily, often by stabilizing the transition state, reducing the activation energy required for the reaction to proceed.

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Key Concepts

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

Enzyme-Substrate Binding
Understanding how an enzyme interacts with its substrate is essential for comprehending enzyme function. The enzyme-substrate binding is the initial step in enzymatic catalysis, where a specific substrate slots into the enzyme's active site with precision akin to a key entering a lock. This is known as the 'lock and key' model. However, the more accurate 'induced fit' model describes the active site as adapting to fit the substrate upon binding.

The interactions that hold the enzyme and substrate together are gentle and reversible, with hydrogen bonds, ionic bonds, and Van der Waals forces playing a pivotal role. The strength and specificity of these interactions are crucial because they ensure that enzymes act only on their intended substrates and not on other molecules.
Catalysis
Enzymatic catalysis is like a master craftsman shaping a delicate work of art. An enzyme speeds up a chemical reaction without being consumed in the process. After the substrate binds to the enzyme, catalysis begins. The enzyme then performs its magic, ably converting the substrate into the desired product.

To make this process more tangible, visualize the enzyme as a highly specialized machine on an assembly line in a factory, tweaking and twisting the substrate until it's been transformed. This remarkable capacity to accelerate reactions is fundamental to life, allowing biological processes to occur at the speed necessary for cells to function.
Activation Energy
Imagine a high wall that separates reactants from becoming products. This wall is the 'activation energy', an energy barrier that must be overcome for a reaction to occur. Enzymes serve as a hidden passageway through this wall, significantly lowering the height of the energy barrier.

In other terms, activation energy is like the initial push needed to start a boulder rolling downhill. Enzymes reduce the force needed to initiate this movement, thus streamlining the path from reactant to product. By lowering the activation energy, enzymes accelerate the rate at which reactions occur, which is instrumental to the efficiency of biological systems.
Transition State
The transition state is a fleeting moment, the pinnacle of a reaction's pathway where the substrate is neither reactant nor product. It's the most unstable state—like a hiker balanced precariously on the mountaintop before descending. Enzymes are adept at stabilizing this transitional character, thereby reducing the amount of energy needed to reach it.

Enzymes achieve this by providing a unique environment, akin to a perfectly molded cradle, that supports and stabilizes the transition state. This crucial assistance helps to dramatically lower the activation energy required and is a key factor in the remarkable catalytic power of enzymes.

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Most popular questions from this chapter

This reaction is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\) $$ \mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{3}(g)+\mathrm{NO}_{2}(g) $$ The rate constant for the reaction at a certain temperature is \(0.053 / \mathrm{s}\) a. Calculate the rate of the reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.055 \mathrm{M}\). b. What would the rate of the reaction be at the concentration indicated in part a if the reaction were second order? Zero order? (Assume the same numerical value for the rate con- stant with the appropriate units.)

What is a catalyst? How does a catalyst increase the rate of a chemical reaction?

The activation energy of a reaction is \(56.8 \mathrm{~kJ} / \mathrm{mol}\), and the frequency factor is \(1.5 \times 10^{11} / \mathrm{s}\). Calculate the rate constant of the reaction at \(25^{\circ} \mathrm{C}\).

Ethyl chloride vapor decomposes by the first-order reaction: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{HCl} $$ The activation energy is \(249 \mathrm{~kJ} / \mathrm{mol}\), and the frequency factor is \(1.6 \times 10^{14} \mathrm{~s}^{-1} .\) Find the value of the rate constant at \(710 \mathrm{~K}\) What fraction of the ethyl chloride decomposes in 15 minutes at this temperature? Find the temperature at which the rate of the reaction would be twice as fast.

Consider a simple reaction in which reactant A forms products: \(\mathrm{A} \longrightarrow\) products What is the rate law if the reaction is zero order with respect to A? First order? Second order? For each case, explain how a doubling of the concentration of A would affect the rate of reaction.

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