/*! 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} Q2. The hypothetical elementary reac... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Q2. (page 398)

The hypothetical elementary reaction 2A鈫�B+Chas a rate constant of 10鈥�6M鈥�1s鈥�1. What is the reaction velocity when the concentration of A is 10mM?

Short Answer

Expert verified

The reaction velocity for the hypothetical elementary reaction 2A鈫�B+C is 10-10Ms-1.

Step by step solution

01

Introduction

The pace of an elementary reaction is proportional to the frequency with which the reacting molecules come in contact with each other at a constant temperature. Thus, the rate constant of an elementary reaction can be used to calculate the reaction velocity.

02

Formula to calculate reaction velocity

For a bimolecular reaction, the reaction velocity is calculated as

v=k[A]2

Where

vis reaction velocity.

Kis reaction constant.

Ais the concentration of A.

03

Calculation

It is given

Concentration of A=10mM (0.01M)

Rate constant=10-6M-1s-1

On substituting the values,

v=k[A]2

=10-6 x (0.010M)2

=10-10Ms-1

Thus, for the hypothetical elementary reaction, the reaction velocity is 10-10Ms-1.

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

List some advantages of phosphorylation/ dephosphorylation cascade systems over simple allosteric regulation.

Determine the type of inhibition of an enzymatic reaction from the following data collected in the presence and absence of the inhibitor.

[S] (mM)

v鈧� (mM min鈦宦�)

v鈧� with I present (mM min鈦宦�)

1

1.3

0.8

2

2.0

1.2

4

2.8

1.7

8

3.6

2.2

12

4.0

2.4

Sphingosine-1-phosphate (SIP) is important for cell survival. The synthesis of SIP from sphingosine and ATP is catalyzed by the enzyme sphingosine kinase. An understanding of the kinetics of the sphingosine kinase reaction may be important in the development of drugs to treat cancer. The velocity of the sphingosine kinase reaction was measured in the presence and absence of threo-sphingosine, a stereoisomer of sphingosine that inhibits the enzyme. The results are shown below.

[Sphingosine]

(饾泹惭)

v鈧� (mg min鈦宦�)

(no inhibitor)

v鈧� (mg min鈦宦�)

(with threo-sphingosine)

2.5

32.3

8.5

3.5

40

11.5

5

50.8

14.6

10

72

25.4

20

87.7

43.9

50

115.4

70.8

Construct a Lineweaver-Burk plot to answer the following questions:

(a) What are the apparent KM and Vmax values in the presence and absence of the inhibitor?

(b) What kind of an inhibitor is threo-sphingosine? Explain.

Enzyme Acatalyzes the reactions S to P and has a KM of 50渭惭and a Vmaxof 100nMs鈥�1. Enzyme Bcatalyzes the reaction S鈫�Q and has a KM of 5mMand a Vmax of 120nMs鈥�1. When 100渭惭 of S is added to a mixture containing equivalent amounts of enzymes A andB , after 1minute which reaction product will be more abundant: P or Q?

Why might an enzyme鈥檚 substrate, transition state, and product all serve as starting points for the design of a competitive inhibitor?
See all solutions

Recommended explanations on Biology Textbooks

Cell Communication

Read Explanation

Biology Experiments

Read Explanation

Ecological Levels

Read Explanation

Reproduction

Read Explanation

Ecology

Read Explanation

Bioenergetics

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.