Chapter 27, Problem 10P

Interpretation Introduction

Interpretation:

The equilibrium constant for cellular respiration and carbon dioxide fixation under the same conditions should be calculated.

Concept introduction:

ATP will work phosphorylation, breaking a reaction down into 2 energetically favored steps connected by a phosphorylated (phosphate-bearing) intermediate. This strategy is employed in several metabolic pathways within the cell, providing some way for the energy discharged by changing nucleotide to ADP to drive different reactions forward. ATP reaction is coupled to a synthesis reaction. However, nucleotide reaction may also be coupled to different categories of cellular reactions, like the form changes of proteins that transport different molecules into or out of the cell.

Answer to Problem 10P

The equilibrium constant can then be calculated.

$\begin{array}{l}{K}_{eq}=e^{(970/RT)}\\ =e^{391}\\ ={10}^{169}\end{array}$

The equilibrium constant can then be calculated.

$\begin{array}{l}{K}_{eq}=e^{(340/RT)}\\ =e^{173}\\ ={10}^{75}\end{array}$

Explanation of Solution

Suppose that cellular respiration goes from glucose to${\text{CO}}_{\text{2}}$and${\text{H}}_{\text{2}}\text{O}$since the question talks about CO₂fixation. Total oxidation of glucose produces two ATP to pyruvate and two NADH, which accounts for six ATP, for a total of eight ATP. When two pyruvate molecules are converted to 2 acetyl-CoA via the citric acid cycle, 2 GTP which accounts for 2 ATP, 2${\text{FADH}}_{\text{2}}$which account for 4 ATP, and 8 NADH which accounts for 24 ATP are generated. A total of 38 ATP are regenerated and at 50kJ/mol, this would create 1,900 kJ per mole glucose. The metabolic cost is 12 NADH and 18 ATP according to the stoichiometry of the metabolic pathway of${\text{CO}}_{\text{2}}$fixation. The equation is as given below:

$\begin{array}{l}12\text{NADH}+12{\text{H}}^{+}+18\text{ATP}+6{\text{CO}}_{\text{2}}+12{\text{H}}_{\text{2}}\text{O}\to \\ {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}+12{\text{NADP}}^{+}+18\text{ADP}+18{\text{P}}_i\end{array}$

This is the same as$12\times 4+18=48+18=6\text{6 ATP}$, and this would form 3,300 kJ per mole per 6$C{O}_2$at 50 kJ/mol.

We cannot determine the equilibrium constants without knowing the$△G^{°\text{'}}$for both reactions. So,$K_{eq}$has to be calculated given$△G^{°\text{'}}$

$=-RTIn{K}_{eq}$. We can assume that the energy we determine above would correspond to$△G$values for ATP hydrolysis or ATP production. We can also assume that the following$△G^{°\text{'}}$corresponds to the oxidation to carbon dioxide and water. The reaction is as follows:

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}+6{\text{O}}_{\text{2}}\to 6{\text{CO}}_2+6{\text{H}}_{\text{2}}\text{O}\\ △G^{°\text{'}}=-278\text{0 kJ/mol}\end{array}$

The above reaction is cellular respiration with no ATP synthesis rejoining.

Because$△G$is a state function, it will follow the following rule:

$△G^{°\text{'}}{}_{\text{glycolysis}}=△G^{°\text{'}}{}_{\text{glucoseoxidation}}+△G^{°\text{'}}{}_{\text{ATPhydrolysis}}$

We can determine$\Delta G°{\text{'}}_{\text{glycolysis}}$

$\begin{array}{l}\Delta G°{\text{'}}_{\text{glcolysis}}=+1,900-2870\\ =-970\text{ kJ/mol}\end{array}$

The equilibrium constant can then be calculated.

$\begin{array}{l}{K}_{eq}=e^{(970/RT)}\\ =e^{391}\\ ={10}^{169}\end{array}$

The equilibrium constant can then be calculated.

$\begin{array}{l}{K}_{eq}=e^{(340/RT)}\\ =e^{173}\\ ={10}^{75}\end{array}$

Positive equilibrium constants for both reactions indicate that both are favorable under these conditions. The value of$\Delta G$for both reactions is negative, therefore, both are spontaneous.

Conclusion

The metabolic cost is 12 NADH and 18 ATP according to the stoichiometry of the metabolic pathway of${\mathrm{CO}}_2$fixation. The equation is as given below:

$\begin{array}{l}12\text{NADH}+12{\text{H}}^{+}+18\text{ATP}+6{\text{CO}}_2+12{\text{H}}_{\text{2}}\text{O}\to \\ {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}+1{\text{2NADP}}^{+}+18\text{ADP}+18{\text{P}}_i\end{array}$

This is the same as$12\times 4+18=48+18=6\text{6 }\frac{1}{2}\text{ATP}$, and this would form 3,300 kJ per mole per 6${\text{CO}}_{\text{2}}$at 50 kJ/mol.

We cannot determine the equilibrium constants without knowing the$△G^{°\text{'}}$for both reactions. So,$K_{eq}$has to be calculated given$△G^{°\text{'}}$

$=-RTIn{K}_{eq}$. We can assume that the energy we determine above would correspond to$△G$values for ATP hydrolysis or ATP production.