Biochemistry: Concepts and Connections (2nd Edition)
Biochemistry: Concepts and Connections (2nd Edition)
2nd Edition
ISBN: 9780134641621
Author: Dean R. Appling, Spencer J. Anthony-Cahill, Christopher K. Mathews
Publisher: PEARSON
Question
Chapter 12, Problem 1P
Interpretation Introduction

Interpretation:

Whether the phosphofructokinase reaction in muscle is more or less exergonic than under standard conditions should be identified along with its amount.

Concept introduction: The formula used:

ΔG=ΔG0+RT ln[products][reactants]

Where,

ΔG = calculated Gibbs free energy (change)

ΔGo = standard Gibbs free energy (change)

R = universal gas constant

T = temperature

[ products] = concentration of products

[reactants] concentration of reactants

Expert Solution & Answer
Check Mark

Explanation of Solution

The phosphofructokinase reaction is as follows:

Fructose6phosphate+ATPFructose-1,6-biphosphate+ADP

Thus,

Fructose6phosphate+ATP are reactants and Fructose-1,6-biphosphate+ADP are products.

The value of standard Gibbs free energy of the reaction is -14.2 kJ mol-1 .

The expression for ΔG will be as follows:

ΔG=ΔG0+RT ln[F-1,6-bisP][ADP][F-6-P][ATP]

Putting the values,

ΔG=-14200 Jmol-1+(8.315 Jmol-1K-1)(298K) ln[0.01 M][0.0005 M][0.001 M][0.005 M]

ΔG=14200 Jmol1+(8.315 Jmol1K1)(298K)ln(1)

ΔG=14200 Jmol1+0=14200 Jmol1

It is determined that the phosphofructokinase reaction in muscle is neither less or nor more exergonic instead it is equally energetic in reactions occur under standard conditions as above calculated Gibbs free energy is equal to standard Gibbs free energy.

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