Chapter 22, Problem 2P

Interpretation Introduction

Interpretation :

standard Gibbs free energy and Gibbs free energy for the gluconeogenesis in the erythrocytes should be calculated.

Introduction :

In gluconeogenesis glucose is produced from the non-carbohydrate substances. This is an anabolic reaction and can be considered as the reverse of the catabolic reactions. For this question, generation of glucose from pyruvate is considered. This reaction sequence can be considered as the reverse of the glycolysis that occurs with the help of set of anabolic enzymes and energy carrier molecules.

Explanation of Solution

All the reactions in glycolysis are reversed in the gluconeogenesis. However, energy of the reactions that occur with hexokinase, phosphofructokinase, and pyruvate kinase enzymes changes depending on the ATP, ADP, Pi, GTP and GDP concentrations. Therefore, ΔG values changes. For other reactions, ΔG values can be obtained by reversing the sign of the respective values indicated in 18.1 table as they are the reverse reactions.

ΔGfor conversion of Pyruvateto PEP

ΔG^ofor the conversion of PEP to pyruvate is $-\text{31}.{\text{7 kJmol}}^{-\text{1}}$ . For the reverse reaction this value should be, $+\text{31}.{\text{7 kJmol}}^{-\text{1}}$ . In the reverse reaction, GTP hydrolysis is also taking place. If the energy of the GTP hydrolysis is $-\text{30}.{\text{5 kJmol}}^{-\text{1}}$ , the total energy for the reverse reaction will be, $\text{31}.{\text{7 kJmol}}^{-\text{1}}-30.5{\text{ kJmol}}^{-\text{1}}=1.2{\text{ kJmol}}^{-\text{1}}$ .

That is the ΔG^ofor the reverse reaction.

  $\begin{array}{l}\Delta G=\Delta G^0+RT\ln \frac{[PEP]{[ADP]}^2[Pi]}{[\text{pyruvate}]{[ATP]}^2}\\ \Delta G=1.2+\text{8}.\text{314 }\times \text{1}{0}^{-\text{3}}\times \text{31}0\ln \frac{[(0.0\text{23 }\times {\text{10}}^{\text{-3}}]{[0.14\times {\text{10}}^{\text{-3}}]}^2[1\times {\text{10}}^{\text{-3}}]}{[\text{0}\text{.051}\times {\text{10}}^{\text{-3}}]{[1.85\times {\text{10}}^{\text{-3}}]}^2}\\ \Delta G=-31.96{\text{ kJmol}}^{\text{-1}}\end{array}$

ΔGfor conversion of Fructose-1,6-bisphosphate to fructose-6-phosphate

F6P = fructose-6-phosphate

F1,6BP = Fructose-1,6-bisphosphate

  $\begin{array}{l}\Delta G=\Delta G^0+RT\ln \frac{[F-6-P][Pi]}{[\text{F-1,6-BP}]}\\ \Delta G=-16.7+\text{8}.\text{314 }\times \text{1}{0}^{-\text{3}}\times \text{31}0\ln \frac{[0.014\times {\text{10}}^{\text{-3}}][1\times {\text{10}}^{\text{-3}}]}{[\text{0}\text{.031}\times {\text{10}}^{\text{-3}}]}\\ \Delta G=-36.6{\text{ kJmol}}^{\text{-1}}\end{array}$

ΔGfor conversion of glucose-6-phosphate to glucose

G6P = glucose-6-phosphate

  $\begin{array}{l}\Delta G=\Delta G^0+RT\ln \frac{[{\text{glucose][P}}_i]}{[\text{G-6-P}]}\\ \Delta G=-13.9+\text{8}.\text{314 }\times \text{1}{0}^{-\text{3}}\times \text{31}0\ln \frac{{[5\times {\text{10}}^{\text{-3}}]}^2[1\times {\text{10}}^{\text{-3}}]}{[\text{0}\text{.083}\times {\text{10}}^{\text{-3}}]}\\ \Delta G=-21.1{\text{ kJmol}}^{\text{-1}}\end{array}$

Therefore, with the other values final ΔG for the entire process,

  $\begin{array}{l}\Delta G=(-31.96-1.1-0.83-0.1+1.29-2.41+0.23-36.6+2.92-21.1){\text{ kJmol}}^{\text{-1}}\\ \Delta G=-89.66{\text{ kJmol}}^{\text{-1}}\end{array}$