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:

$\text{ΔG=}{\text{ΔG}}^{\text{0}}\text{+RT ln}\frac{\text{[products]}}{\text{[reactants]}}$

Where,

ΔG = calculated Gibbs free energy (change)

ΔG^o = standard Gibbs free energy (change)

R = universal gas constant

T = temperature

[ products] = concentration of products

[reactants] concentration of reactants

Explanation of Solution

The phosphofructokinase reaction is as follows:

$\text{Fructose}-6-\text{phosphate+ATP}\to \text{Fructose-1,6-biphosphate+ADP}$

Thus,

$\text{Fructose}-6-\text{phosphate+ATP}$ are reactants and $\text{Fructose-1,6-biphosphate+ADP}$ are products.

The value of standard Gibbs free energy of the reaction is $\text{-14}\text{.2}{\text{ kJ mol}}^{\text{-1}}$ .

The expression for $\text{ΔG}$ will be as follows:

$\text{ΔG}\text{=}{\text{ΔG}}^{\text{0}}\text{+RT ln}\frac{\text{[F-1,6-bisP][ADP]}}{\text{[F-6-P][ATP]}}$

Putting the values,

$\text{ΔG}\text{=}\text{-14200}{\text{ Jmol}}^{\text{-1}}\text{+(8}\text{.315}{\text{ Jmol}}^{\text{-1}}{\text{K}}^{\text{-1}}\text{)(298}\text{K) ln}\frac{\text{[0}\text{.01 M][0}\text{.0005 M]}}{\text{[0}\text{.001 M][0}\text{.005 M]}}$

$\text{ΔG}=-\text{14200}{\text{ Jmol}}^{-1}+(8.315{\text{ Jmol}}^{{}^{-1}}{\text{K}}^{-1})(298\text{K)}\ln (1)$

$\text{ΔG}=-\text{14200 }{\text{Jmol}}^{-1}+0=-\text{14200}{\text{ Jmol}}^{-1}$

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.