Chapter 16, Problem 30P

Interpretation Introduction

Interpretation:

The equilibrium concentrations of fructose-1,6-bisphosphate, dihydroxyacetone (DHAP), and glyceraldehyde-3-phosphate (G3P) under given conditions should be determined.

Concept introduction:

By the action of aldolase enzyme, fructose-1,6-bisphopshate is converted into DHAP and G3P. Aldolase uses zinc ion as its cofactor. This is the fourth step in the glycolysis process, and the last step of the preparatory phase.

Answer to Problem 30P

Equilibrium concentration of fructose-1,6-bisphosphate = $\text{7}\text{.8 }\times {\text{10}}^{-4}\text{ M}$

Equilibrium concentration of DHAP = $\text{2}\text{.2}\times {10}^{-4}\text{M}$

Equilibrium concentration of G3P = $\text{2}\text{.2}\times {10}^{-4}\text{M}$

Explanation of Solution

Given information:

1 mM (0.001M) of fructose-1,6-bisphosphate is incubated with aldolase under standard conditions.

The reaction is:

$\text{Fructose-1,6-bisphosphate}\stackrel{\text{aldolase}}{\rightleftharpoons }\text{DHAP+G3P ; }\Delta {\text{G}}^{\circ }=23.9\text{kJ/mol = 23}\text{.9 }\times {\text{ 10}}^3\text{J/mol}$

Creating the ICE table as:

$\begin{array}{l}\text{Reaction - Fructose-1,6-bisphosphate}\stackrel{\text{aldolase}}{\rightleftharpoons }\text{DHAP+G3P ; }\Delta {\text{G}}^{\circ }=23.9\text{kJ/mol = 23}\text{.9 }\times {\text{ 10}}^3\text{J/mol}\\ \text{Initial concentration- 0}\text{.001M 0M 0M}\\ \text{Difference in conc - -xM +xM +xM}\\ \text{Final Concentration - (0}\text{.001-x)M xM xM}\end{array}$

Calculation of equilibrium constant, ${\text{K}}_{\text{eq}}$ :

$\begin{array}{l}\Delta {\text{G}}^{\circ }=-{\text{RTlnK}}_{\text{eq}}\\ \text{23}\text{.9 }\times {\text{ 10}}^3\text{J/mol}=-8.314\text{J/mol K }\times \text{ 298K}\times {\text{ln K}}_{\text{eq}}\\ {\text{K}}_{\text{eq }}\text{= 6}\text{.5}\times {\text{10}}^{-5}\end{array}$

Now,

$\begin{array}{l}\text{ Keq = }\frac{\text{concentration of products}}{\text{concentration of substrates}}\\ \text{Keq = }\frac{\text{[DHAP]}\times \text{[G3P]}}{\text{[fructose-1,6-bisphosphate]}}\\ \text{6}\text{.5}\times {\text{10}}^{-5}=\frac{\text{x }\times \text{ x}}{0.001-\text{x}}\\ 6.5\times {10}^{-5}(0.001-{\text{x) = x}}^2\\ 6.5\times {10}^{-8}-\text{6}\text{.5}\times {\text{10x = x}}^2\\ {\text{x}}^2+\text{6}\text{.5}\times {\text{10}}^{-5}\text{x}-6.5\times {10}^{-8}=0\end{array}$

Solving for x,

$\begin{array}{l}\text{x = }\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \text{x = }\frac{-(6.5\times {10}^{-5})\pm \sqrt{{(6.5\times {10}^{-5})}^2-4\times 1\times (-6.5\times {10}^{-8})}}{2\times 1}\\ \text{x = 2}\text{.2}\times {10}^{-4}\text{M}\end{array}$

So, equilibrium concentrations of DHAP and G3P = $\text{2}\text{.2}\times {10}^{-4}\text{M}$

Equilibrium concentration of fructose-1,6-bisphosphate = $\text{(0}\text{.001-2}\text{.2}\times {\text{10}}^{-4}\text{) M = 7}\text{.8 }\times {\text{10}}^{-4}\text{ M}$

Conclusion

Equilibrium concentration of fructose-1,6-bisphosphate = $\text{7}\text{.8 }\times {\text{10}}^{-4}\text{ M}$

Equilibrium concentration of DHAP = $\text{2}\text{.2}\times {10}^{-4}\text{M}$

Equilibrium concentration of G3P = $\text{2}\text{.2}\times {10}^{-4}\text{M}$