The K m value for the inhibited and uninhibited reactions using a Michaelis–Menten plotis to be determined. The data for the enzyme-catalyzed hydrolysis reaction in the presence and absence of inhibitor is given as follows: Substrate (moles) V 0 (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 Introduction: The K m value is defined as the substrate concentration at which half of the maximum velocity (1/2 V max ) is achieved in the reaction. 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. The inhibitors act by decreasing the rate of a catalyzed reaction by interacting with free enzyme, enzyme–substrate complex, or both.

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Answer 1

Question

Chapter 6, Problem 59TQ

Summary Introduction

To review:

The K_m value for the inhibited and uninhibited reactions using a Michaelis–Menten plotis to be determined. The data for the enzyme-catalyzed hydrolysis reaction in the presence and absence of inhibitor is given as follows:

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

Introduction:

The K_m value is defined as the substrate concentration at which half of the maximum velocity (1/2 V_max) is achieved in the reaction. 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. The inhibitors act by decreasing the rate of a catalyzed reaction by interacting with free enzyme, enzyme–substrate complex, or both.

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The two half reactions for beginning and end of the electron transport chain are given below in standard form. Calculate & for the overall process. Using the Nernst equation (AG° = -n Fo, F= 96.485 kJ/volt mol), calculate AG°. Explain the need for a stepwise process in the electron transport chain. NAD* + H+ + 2 e- = NADH ½ 0г + 2H+ + 2е- = H20 = -0.32v E = +0.82V

Answer 2

Question

Chapter 6, Problem 59TQ

Summary Introduction

To review:

The K_m value for the inhibited and uninhibited reactions using a Michaelis–Menten plotis to be determined. The data for the enzyme-catalyzed hydrolysis reaction in the presence and absence of inhibitor is given as follows:

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

Introduction:

The K_m value is defined as the substrate concentration at which half of the maximum velocity (1/2 V_max) is achieved in the reaction. 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. The inhibitors act by decreasing the rate of a catalyzed reaction by interacting with free enzyme, enzyme–substrate complex, or both.

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Answer 3

Question

Chapter 6, Problem 59TQ

Summary Introduction

To review:

The K_m value for the inhibited and uninhibited reactions using a Michaelis–Menten plotis to be determined. The data for the enzyme-catalyzed hydrolysis reaction in the presence and absence of inhibitor is given as follows:

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

Introduction:

The K_m value is defined as the substrate concentration at which half of the maximum velocity (1/2 V_max) is achieved in the reaction. 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. The inhibitors act by decreasing the rate of a catalyzed reaction by interacting with free enzyme, enzyme–substrate complex, or both.

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Answer 4

Question

Chapter 6, Problem 59TQ

Summary Introduction

To review:

The K_m value for the inhibited and uninhibited reactions using a Michaelis–Menten plotis to be determined. The data for the enzyme-catalyzed hydrolysis reaction in the presence and absence of inhibitor is given as follows:

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

Introduction:

The K_m value is defined as the substrate concentration at which half of the maximum velocity (1/2 V_max) is achieved in the reaction. 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. The inhibitors act by decreasing the rate of a catalyzed reaction by interacting with free enzyme, enzyme–substrate complex, or both.

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Answer 5

Chapter 6, Problem 59TQ

Summary Introduction

To review:

The K_m value for the inhibited and uninhibited reactions using a Michaelis–Menten plotis to be determined. The data for the enzyme-catalyzed hydrolysis reaction in the presence and absence of inhibitor is given as follows:

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

Introduction:

The K_m value is defined as the substrate concentration at which half of the maximum velocity (1/2 V_max) is achieved in the reaction. 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. The inhibitors act by decreasing the rate of a catalyzed reaction by interacting with free enzyme, enzyme–substrate complex, or both.

Answer 6

Chapter 6, Problem 59TQ

Summary Introduction

To review:

The K_m value for the inhibited and uninhibited reactions using a Michaelis–Menten plotis to be determined. The data for the enzyme-catalyzed hydrolysis reaction in the presence and absence of inhibitor is given as follows:

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

Introduction:

The K_m value is defined as the substrate concentration at which half of the maximum velocity (1/2 V_max) is achieved in the reaction. 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. The inhibitors act by decreasing the rate of a catalyzed reaction by interacting with free enzyme, enzyme–substrate complex, or both.

Answer 7

Chapter 6, Problem 59TQ

Summary Introduction

To review:

The K_m value for the inhibited and uninhibited reactions using a Michaelis–Menten plotis to be determined. The data for the enzyme-catalyzed hydrolysis reaction in the presence and absence of inhibitor is given as follows:

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

Introduction:

The K_m value is defined as the substrate concentration at which half of the maximum velocity (1/2 V_max) is achieved in the reaction. 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. The inhibitors act by decreasing the rate of a catalyzed reaction by interacting with free enzyme, enzyme–substrate complex, or both.

Answer 8

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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

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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