The data given in the question 59, for: a. Generation of the Lineweaver–Burk plot. b. Explaining the significance of horizontal and vertical intercept, and slope. c. Identification of the type of inhibition. Substrate (moles) V 0 (micromoles per minute) V 0I (micromoles per minute) 6* 10 -6 20.8 4.2 110 -5 29 5.8 210 -5 45 9 610 -5 67.6 13.6 1.810 -4 87 16.2 Introduction: The activity of an enzyme in a reaction can be reduced by several inhibitors. The inhibitors can be competitive (increase the K m value), uncompetitive (decrease the K m value), or non- competitive (do not alter the K m value), depending upon the mechanism used to inhibit the activity of the enzyme. Some of them bind to the free enzyme, some bind only to enzyme-substrate complex, whereas some inhibitors can bind with both.

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Question

Chapter 6, Problem 60TQ

Summary Introduction

To review:

The data given in the question 59, for:

a. Generation of the Lineweaver–Burk plot.

b. Explaining the significance of horizontal and vertical intercept, and slope.

c. Identification of the type of inhibition.

Substrate (moles) V₀ (micromoles per minute) V_0I (micromoles per minute)

6* 10^-6 20.8 4.2

110^-5 29 5.8

210^-5 45 9

610^-5 67.6 13.6

1.810^-4 87 16.2

Introduction:

The activity of an enzyme in a reaction can be reduced by several inhibitors. The inhibitors can be competitive (increase the K_m value), uncompetitive (decrease the K_m value), or non- competitive (do not alter the K_m value), depending upon the mechanism used to inhibit the activity of the enzyme. Some of them bind to the free enzyme, some bind only to enzyme-substrate complex, whereas some inhibitors can bind with both.

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Answer 2

Question

Chapter 6, Problem 60TQ

Summary Introduction

To review:

The data given in the question 59, for:

a. Generation of the Lineweaver–Burk plot.

b. Explaining the significance of horizontal and vertical intercept, and slope.

c. Identification of the type of inhibition.

Substrate (moles) V₀ (micromoles per minute) V_0I (micromoles per minute)

6* 10^-6 20.8 4.2

110^-5 29 5.8

210^-5 45 9

610^-5 67.6 13.6

1.810^-4 87 16.2

Introduction:

The activity of an enzyme in a reaction can be reduced by several inhibitors. The inhibitors can be competitive (increase the K_m value), uncompetitive (decrease the K_m value), or non- competitive (do not alter the K_m value), depending upon the mechanism used to inhibit the activity of the enzyme. Some of them bind to the free enzyme, some bind only to enzyme-substrate complex, whereas some inhibitors can bind with both.

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Answer 3

Question

Chapter 6, Problem 60TQ

Summary Introduction

To review:

The data given in the question 59, for:

a. Generation of the Lineweaver–Burk plot.

b. Explaining the significance of horizontal and vertical intercept, and slope.

c. Identification of the type of inhibition.

Substrate (moles) V₀ (micromoles per minute) V_0I (micromoles per minute)

6* 10^-6 20.8 4.2

110^-5 29 5.8

210^-5 45 9

610^-5 67.6 13.6

1.810^-4 87 16.2

Introduction:

The activity of an enzyme in a reaction can be reduced by several inhibitors. The inhibitors can be competitive (increase the K_m value), uncompetitive (decrease the K_m value), or non- competitive (do not alter the K_m value), depending upon the mechanism used to inhibit the activity of the enzyme. Some of them bind to the free enzyme, some bind only to enzyme-substrate complex, whereas some inhibitors can bind with both.

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Answer 4

Question

Chapter 6, Problem 60TQ

Summary Introduction

To review:

The data given in the question 59, for:

a. Generation of the Lineweaver–Burk plot.

b. Explaining the significance of horizontal and vertical intercept, and slope.

c. Identification of the type of inhibition.

Substrate (moles) V₀ (micromoles per minute) V_0I (micromoles per minute)

6* 10^-6 20.8 4.2

110^-5 29 5.8

210^-5 45 9

610^-5 67.6 13.6

1.810^-4 87 16.2

Introduction:

The activity of an enzyme in a reaction can be reduced by several inhibitors. The inhibitors can be competitive (increase the K_m value), uncompetitive (decrease the K_m value), or non- competitive (do not alter the K_m value), depending upon the mechanism used to inhibit the activity of the enzyme. Some of them bind to the free enzyme, some bind only to enzyme-substrate complex, whereas some inhibitors can bind with both.

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Answer 5

Chapter 6, Problem 60TQ

Summary Introduction

To review:

The data given in the question 59, for:

a. Generation of the Lineweaver–Burk plot.

b. Explaining the significance of horizontal and vertical intercept, and slope.

c. Identification of the type of inhibition.

Substrate (moles) V₀ (micromoles per minute) V_0I (micromoles per minute)

6* 10^-6 20.8 4.2

110^-5 29 5.8

210^-5 45 9

610^-5 67.6 13.6

1.810^-4 87 16.2

Introduction:

The activity of an enzyme in a reaction can be reduced by several inhibitors. The inhibitors can be competitive (increase the K_m value), uncompetitive (decrease the K_m value), or non- competitive (do not alter the K_m value), depending upon the mechanism used to inhibit the activity of the enzyme. Some of them bind to the free enzyme, some bind only to enzyme-substrate complex, whereas some inhibitors can bind with both.

Answer 6

Chapter 6, Problem 60TQ

Summary Introduction

To review:

The data given in the question 59, for:

a. Generation of the Lineweaver–Burk plot.

b. Explaining the significance of horizontal and vertical intercept, and slope.

c. Identification of the type of inhibition.

Substrate (moles) V₀ (micromoles per minute) V_0I (micromoles per minute)

6* 10^-6 20.8 4.2

110^-5 29 5.8

210^-5 45 9

610^-5 67.6 13.6

1.810^-4 87 16.2

Introduction:

The activity of an enzyme in a reaction can be reduced by several inhibitors. The inhibitors can be competitive (increase the K_m value), uncompetitive (decrease the K_m value), or non- competitive (do not alter the K_m value), depending upon the mechanism used to inhibit the activity of the enzyme. Some of them bind to the free enzyme, some bind only to enzyme-substrate complex, whereas some inhibitors can bind with both.

Answer 7

Chapter 6, Problem 60TQ

Summary Introduction

To review:

The data given in the question 59, for:

a. Generation of the Lineweaver–Burk plot.

b. Explaining the significance of horizontal and vertical intercept, and slope.

c. Identification of the type of inhibition.

Substrate (moles) V₀ (micromoles per minute) V_0I (micromoles per minute)

6* 10^-6 20.8 4.2

110^-5 29 5.8

210^-5 45 9

610^-5 67.6 13.6

1.810^-4 87 16.2

Introduction:

The activity of an enzyme in a reaction can be reduced by several inhibitors. The inhibitors can be competitive (increase the K_m value), uncompetitive (decrease the K_m value), or non- competitive (do not alter the K_m value), depending upon the mechanism used to inhibit the activity of the enzyme. Some of them bind to the free enzyme, some bind only to enzyme-substrate complex, whereas some inhibitors can bind with both.

Answer 8

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Answer 11

Ch. 6 - Prob. 1Q Ch. 6 - Prob. 2Q Ch. 6 - Prob. 3Q Ch. 6 - Prob. 4Q Ch. 6 - Prob. 5Q Ch. 6 - Prob. 6Q Ch. 6 - Prob. 7Q Ch. 6 - Prob. 8Q Ch. 6 - Prob. 9Q Ch. 6 - Prob. 1RQ

Solution Summary: The author explains that the Lineweaver–Burk plot is a reciprocal plot, wherein velocity and substrate concentration are taken into consideration.

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Chapter 6 Solutions

Substrate (moles)	V₀ (micromoles per minute)	V_0I (micromoles per minute)
6* 10^-6	20.8	4.2
1*10^-5	29	5.8
2*10^-5	45	9
6*10^-5	67.6	13.6
1.8*10^-4	87	16.2