Biochemistry
Biochemistry
6th Edition
ISBN: 9781337359573
Author: Reginald H. Garrett; Charles M. Grisham
Publisher: Cengage Learning US
Question
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Chapter 20, Problem 9P
Interpretation Introduction

(a)

To calculate:

The change of standard free energy for the given reaction.

Introduction:

Succinate dehydrogenase is also known as SQR succinate Q reductase is seen in the cells of bacteria as a complex enzyme. It is also found in the eukaryotes in the membrane of inner mitochondria. Thus enzyme takes part in the electron transport chain and citric acid cycle.

Expert Solution
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Explanation of Solution

The electron acceptor is O2 and the electron donor is succinate which is determined from the half reactions:

  Fumarate+2H++2eSuccinate+(E0'=+0.031V)

  12O2+2H++2eH2O,(E0'=+0.816V)

The free change of energy for the reaction is calculated by determining the change of the standard potential:

  ΔE0'=acceptorΔE0'donorΔE0'

  =+0.816V(0.031V)=+0.785V

  ΔG0'=nFΔE0'n=2F=96,485kJ/Vmol

  =2(96.485kJ/V.mol)(0.785V)=151.5kJ/mol

Interpretation Introduction

(b)

To calculate:

The equilibrium constant of the reaction.

Introduction:

Succinate dehydrogenase is also known as SQR succinate Q reductase is seen in the cells of bacteria as a complex enzyme. It is also found in the eukaryotes in the membrane of inner mitochondria. Thus enzyme takes part in the electron transport chain and citric acid cycle.

Expert Solution
Check Mark

Explanation of Solution

  ΔG0'=RTlnKeqKeq=e ΔG RT

  =e (151.5) (8.314× 10 3 )(298)=3.60×1026

Interpretation Introduction

(c)

To determine:

The release of the actual free energy accompanying the coenzyme Q reductase of NADH is equal to releasing amount under the normal condition

Introduction:

Succinate dehydrogenase is also known as SQR succinate Q reductase is seen in the cells of bacteria as a complex enzyme. It is also found in the eukaryotes in the membrane of inner mitochondria. Thus enzyme takes part in the electron transport chain and citric acid cycle.

Expert Solution
Check Mark

Explanation of Solution

From the given we have,

  G=151.3kJ/mol for 2 ATP

So with 70% efficiency,

  151.3kJ/mol×0.70=106.1kJ/mol

For 1 ATP,

  53.1kJ/mol

With the equation,

  ΔG=ΔG0+RTln(ATP)(ADP)(Pi)

  (ATP)(ADP)=(Pi)×eΔGΔG 0'RT

  =(1mM)×e (53.0530.5) (8.314× 10 3 )(298)=8.89

The ratio of ATP and ADP maximum for the phosphorylation oxidative occurring at Pi =1 mM is 8.89.

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