/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Enzymes are kinetically importan... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 21

Enzymes are kinetically important for many of the complex reactions necessary for plant and animal life to exist. However, only a tiny amount of any particular enzyme is required for these complex reactions to occur. Explain.

Short Answer

Expert verified
Enzymes are biological catalysts that increase the rate of chemical reactions by lowering activation energy. Their high specificity ensures they only work on specific reactions, and their active site binds substrates, forming an enzyme-substrate complex. This leads to a conformational change, called the induced fit, which weakens substrate bonds, lowering activation energy. Due to enzymes' high catalytic efficiency and their ability to rapidly bind substrates, lower activation energy, and release products, they can facilitate complex biochemical reactions in small amounts. This is essential for plants and animals to function efficiently while using minimal resources.

Step by step solution

01

Introduction to Enzymes

Enzymes are biological catalysts that speed up chemical reactions by lowering the activation energy required for the reaction to occur. They play a vital role in various processes in living organisms, including cellular respiration, digestion, and DNA replication. One remarkable characteristic of enzymes is their catalytic efficiency, meaning they can speed up reactions by a factor of millions without getting consumed in the process.
02

Enzyme-Substrate Specificity

Enzymes are highly specific for their substrates, which are the molecules they interact with to facilitate a reaction. The enzyme's active site is a unique region that has a specific shape and chemical properties to bind its substrate. This specificity ensures that enzymes can work only on specific reactions, avoiding unwanted side-reactions or interference with other cellular processes.
03

Enzyme Mechanism of Action

The enzyme's active site is responsible for its catalytic activity and substrate binding. When the substrate binds to the active site, it forms an enzyme-substrate complex. This binding causes a slight change in the enzyme's conformation, known as the induced fit, which enables the enzyme to weaken specific chemical bonds within the substrate, thereby lowering the activation energy required for the reaction to proceed.
04

Enzyme Catalytic Efficiency

Enzymes can be extremely efficient catalysts due to their high turnover number, which describes the number of substrate molecules converted into product by a single enzyme molecule in a specific time. This high efficiency is achieved through rapid binding and release of the substrate and products, allowing each enzyme molecule to catalyze multiple reactions, turning over thousands of substrate molecules into product molecules per second.
05

Conclusion: Tiny Amount of Enzymes Required

Given their high catalytic efficiency and substrate specificity, enzymes can facilitate complex biochemical reactions rapidly and effectively, even when present in tiny amounts. This efficiency is due to their ability to bind substrates, lower the activation energy and release products quickly, enabling enzymes to participate in multiple reactions in a short period. This allows plants and animals to maintain the necessary biochemical processes with minimal consumption of their limited resources.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the following initial rate data for the decomposition of compound AB to give A and B: Determine the half-life for the decomposition reaction initially having 1.00\(M \mathrm{AB}\) present.

How does temperature affect k, the rate constant? Explain.

Upon dissolving \(\operatorname{In} \mathrm{Cl}(s)\) in \(\mathrm{HCl}, \operatorname{In}^{+}(a q)\) undergoes a disproportionation reaction according to the following unbalanced equation: $$ \operatorname{In}^{+}(a q) \longrightarrow \operatorname{In}(s)+\operatorname{In}^{3+}(a q) $$ This disproportionation follows first-order kinetics with a half-life of 667 s. What is the concentration of \(\operatorname{In}^{+}(a q)\) after 1.25 \(\mathrm{h}\) if the initial solution of \(\operatorname{In}^{+}(a q)\) was prepared by dis- solving 2.38 \(\mathrm{g} \operatorname{InCl}(s)\) in dilute \(\mathrm{HCl}\) to make \(5.00 \times 10^{2} \mathrm{mL}\) of solution? What mass of In \((s)\) is formed after 1.25 \(\mathrm{h}\) ?

Consider two reaction vessels, one containing \(\mathrm{A}\) and the other containing \(\mathrm{B}\) , with equal concentrations at \(t=0 .\) If both substances decompose by first-order kinetics, where $$ \begin{array}{l}{k_{\mathrm{A}}=4.50 \times 10^{-4} \mathrm{s}^{-1}} \\\ {k_{\mathrm{B}}=3.70 \times 10^{-3} \mathrm{s}^{-1}}\end{array} $$ how much time must pass to reach a condition such that \([\mathrm{A}]=\) 4.00\([\mathrm{B}]\) ?

Rate Laws from Experimental Data: Initial Rates Method. The reaction $$2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)$$ was studied at \(-10^{\circ} \mathrm{C}\). The following results were obtained where $$\text { Rate }=-\frac{\Delta\left[\mathrm{Cl}_{2}\right]}{\Delta t}$$ $$ \begin{array}{ccc} {[\mathrm{NO}]_{0}} & {\left[\mathrm{Cl}_{2}\right]_{0}} & \text { Initial Rate } \\ (\mathrm{mol} / \mathrm{L}) & (\mathrm{mol} / \mathrm{L}) & (\mathrm{mol} / \mathrm{L} \cdot \mathrm{min}) \\ 0.10 & 0.10 & 0.18 \\ 0.10 & 0.20 & 0.36 \\ 0.20 & 0.20 & 1.45 \end{array} $$ a. What is the rate law? b. What is the value of the rate constant?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.