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.∆G^o’ for the hydrolysis of glucose-6-phosphate is –13.8 kJ/mol (kilojoule per mole).

Explanation of Solution

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

$$$\text{Glucose}-6-\text{phosphate}\to \text{Glucose}+p_i$

In case the pH (potential of hydrogen) is 7, the change in Gibbs free energy is denoted by ∆G^o’. ∆G^o’ 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 $\left(\Delta G\right)$, standard Gibb’s free energy change $\left(\Delta G{°}^{\text{'}}\right)$, and equilibrium constant K_eq are related as

$\Delta G=\Delta G°+RT\ln K_{\text{eq}}$

At equilibrium, $\Delta G=0$, thus, the above equation will be:

$\begin{array}{l}\Delta G=\Delta G{°}^{\text{'}}+RT\ln K_{\text{eq}}\\ 0=\Delta G{°}^{\text{'}}+RT\ln K_{\text{eq}}\\ \Delta G{°}^{\text{'}}=-RT\ln K_{\text{eq}}\\ \ln K_{\text{eq}}\text{=}\frac{-\Delta G{°}^{\text{'}}}{RT}\end{array}$

Calculate the value of equilibrium constant by substituting the values of $\Delta G{°}^{\text{'}}=-13.8\text{kJ/mol}$, $R=0.008314kJ{K}^{-1}mo{l}^{-1}$, and $T=\text{298}\text{K}$ in the above expression:

$\begin{array}{l}\ln K_{\text{eq}}=\frac{-(-13.8\text{ kJ/mol )}}{0.008314{\text{ kJ K}}^{-\text{1}}{\text{mol}}^{-\text{1}}\times 298\text{ K}}\\ \ln K_{\text{eq}}=\frac{13.8\text{ kJ/mol }}{2.477}\\ \ln K_{\text{eq}}=5.57\\ K_{\text{eq}}=262.43\end{array}$

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

$$$\text{Glucose}-6-\text{phosphate}\to \text{Glucose}+p_i$

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

$\begin{array}{l}{K}_{\text{eq}}=\frac{\left[\text{Glucose}\right]\left[P_i\right]}{\left[\text{Glucose}-6-\text{phosphate}\right]}\\ \Rightarrow \left[P_i\right]=\frac{K_{\text{eq}}\times \left[\text{Glucose}-6-\text{phosphate}\right]}{\left[\text{Glucose}\right]}\end{array}$

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:

$\left[P_i\right]=K_{\text{eq}}\times \left[\text{Glucose}-6-\text{phosphate}\right]$

Substitute $K_{\text{eq}}=262.43$ and $\left[\text{Glucose}-6-\text{phosphate}\right]=4\text{mM}$ in the above expression to get:

$\begin{array}{l}\left[P_i\right]=262.43\times 4\text{mM}\\ \text{=1049}\text{.72}\text{mM}\approx \text{1}\text{.0}\times {\text{10}}^3\text{mM}\end{array}$

Conclusion

Therefore, it can be concluded that the final concentration of inorganic phosphate is $\text{1}\text{.0}\times {\text{10}}^3\text{mM}$.