Biochemistry, The Molecular Basis of Life, 6th Edition
Biochemistry, The Molecular Basis of Life, 6th Edition
6th Edition
ISBN: 9780190259204
Author: Trudy McKee, James R. McKee
Publisher: Oxford University Press
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Chapter 4, Problem 26RQ
Summary Introduction

To review:

The concentration of phosphate at equilibrium is to be calculated.

Introduction:

Glucose-6-phosphatase is the enzyme thatcarries out the hydrolysis of glucose-6-phosphate. The end product of the hydrolysis is a phosphate group and a free glucose molecule. This hydrolysis reaction is an instrumental step in the production of glucose through gluconeogenesis.∆Go’ for the hydrolysis of glucose-6-phosphate is –13.8 kJ/mol (kilojoule per mole).

Expert Solution & Answer
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Explanation of Solution

The hydrolysis reaction of glucose-6-phosphate is shown below:

Glucose6phosphateGlucose+pi

In case the pH (potential of hydrogen) is 7, the change in Gibbs free energy is denoted by ∆Go’. ∆Go’ for the hydrolysis of glucose-6-phosphate is –13.8 kJ/mol (kilojoule per mole) and the concentration of glucose-6-phosphate is4mM (millimolar).

Change in Gibb’s free energy (ΔG), standard Gibb’s free energy change (ΔG°'), and equilibrium constant Keq are related as

ΔG=ΔG°+RTlnKeq

At equilibrium, ΔG=0, thus, the above equation will be:

ΔG=ΔG°'+RTlnKeq0=ΔG°'+RTlnKeqΔG°'=RTlnKeqlnKeq=ΔG°'RT

Calculate the value of equilibrium constant by substituting the values of ΔG°'=13.8kJ/mol, R=0.008314 kJ K1mol1, and T=298K in the above expression:

lnKeq=(13.8 kJ/mol )0.008314 kJ K1mol1×298 KlnKeq=13.8 kJ/mol 2.477lnKeq=5.57Keq=262.43

The hydrolysis reaction of glucose-6-phosphate is as follows:

Glucose6phosphateGlucose+pi

The equilibrium constant is related to equilibrium concentrations of the reactants and products as

Keq=[Glucose] [Pi][Glucose6phosphate][Pi]=Keq×[Glucose6phosphate][Glucose]

Here, the concentration of glucose is ignored because glucose is exported out of the cell as soon as it is produced through the hydrolysis reaction. Thus, the above expression becomes:

[Pi]=Keq×[Glucose6phosphate]

Substitute Keq=262.43 and [Glucose6phosphate]=4mM in the above expression to get:

[Pi]=262.43×4mM=1049.72mM1.0×103mM

Conclusion

Therefore, it can be concluded that the final concentration of inorganic phosphate is 1.0×103mM.

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