Chapter 13, Problem 11P

(a)

Summary Introduction

To determine: The equilibrium constant for the reaction for formation of glucose 6-phosphate at 37°C and whether this reaction is a reasonable metabolic step for the catabolism of glucose.

Introduction:

The glycolysis is a series of reactions converting the glucose molecules into two molecules of pyruvate in the presence of oxygen. The pyruvate undergoes series of reactions in the citric acid cycle and converts into acetyl CoA (acetyl coenzyme A).

Explanation of Solution

The reaction for the formation of glucose-6-phosphate by reaction of inorganic phosphate and glucose is as follows:

$\text{Glucose}+\text{Pi}\to \text{Glucose-6-phosphate}+{\text{H}}_{\text{2}}\text{O}\text{|}\text{(}\Delta {\text{G}}^{\prime }°\text{)}=\text{13}\text{.8}\text{kJ}/\text{mol}$

The value of gas constant R is 0.0083 kJ/ mole K. The equilibrium constant at the given temperature 37°C (310 K) is calculated as follows-

$\begin{array}{c}(\Delta G^{\prime }°)=-RT\ln ({K^{\prime }}_{\text{eq}})\\ \ln ({K^{\prime }}_{\text{eq}})=\frac{(\Delta G^{\prime }°)}{-RT}\end{array}$

$\begin{array}{c}\ln ({K^{\prime }}_{\text{eq}})=\frac{(13.8\text{kJ}/\text{mol})}{(-0.0083\text{kJ}/\text{mol}\cdot \text{K})(310\text{K})}\\ =\frac{13.8}{-2.58}\\ =-5.348\simeq -5.35\end{array}$

$\begin{array}{c}({K^{\prime }}_{\text{eq}})=e^{(-5.35)}\\ =0.0047{\text{M}}^{-1}\end{array}$

The value of equilibrium constant of glucose-6-phosphate is determined as follows:

$\begin{array}{c}({K^{\prime }}_{\text{eq}})=\frac{\text{[Glucose-6-phosphate]}}{\text{[Glucose][Pi]}}\\ \text{[Glucose-6-phosphate]}=({K^{\prime }}_{\text{eq}})\text{[Glucose][Pi]}\end{array}$

$\begin{array}{c}[\text{Glucose-6-phosphate}]=[4.75\times {10}^{-3}{\text{M}}^{-1}]\times [4.8\times {10}^{-3}\text{M}]\times [4.8\times {10}^{-3}\text{M}]\\ =10.9\times {10}^{-8}\text{M}\end{array}$

The given reaction will not be favorable based on the levels of glucose in a cell, as it will exceed the concentration of $10.9\times {10}^{-8}\text{M}$.

(b)

Summary Introduction

To determine: Whether the route is physiologically reasonable when the maximum solubility of glucose is less than 1M.

Introduction:

The glycolysis is a series of reactions converting the glucose molecules into two molecules of pyruvate in the presence of oxygen. The pyruvate undergoes series of reactions in the citric acid cycle and converts into acetyl CoA (acetyl coenzyme A).

Explanation of Solution

The value of equilibrium constant is also determined as follows:

$\begin{array}{c}({K^{\prime }}_{\text{eq}})=\frac{\text{[Glucose-6-phosphate]}}{\text{[Glucose][Pi]}}\\ [\text{Glucose]}=\frac{\text{[Glucose-6-phosphate]}}{({K^{\prime }}_{\text{eq}})\text{[Pi]}}\end{array}$

$\begin{array}{c}[\text{Glucose]}=\frac{\text{[Glucose-6-phosphate]}}{({K^{\prime }}_{\text{eq}})\text{[Pi]}}\\ =\frac{[250\times {10}^{-6}\text{M}]}{[4.75\times {10}^{-3}{\text{M}}^{-1}]\times [4.8\times {10}^{-3}\text{M}]}\\ =\frac{250}{22.8{\text{M}}^{-1}}\\ =10.96\simeq 11\text{M}\end{array}$

This route will be unfavorable because the given concentration of glucose exceeds the permissible limit of 1M. The concentration of glucose at given conditions is 11 M and the reaction will not be physiologically feasible because the maximum limit is exceeded by given concentration of glucose.

(c1)

Summary Introduction

To determine: The $({K^{\prime }}_{\text{eq}})$ for the given overall reaction.

Introduction:

The glycolysis is a series of reactions converting the glucose molecules into two molecules of pyruvate in the presence of oxygen. The pyruvate undergoes series of reactions in the citric acid cycle and converts into acetyl CoA (acetyl coenzyme A).

Explanation of Solution

The two given chemical reactions havening their energy change $(\Delta G^{\prime }°)$ are as follows:

$\begin{array}{l}\begin{array}{cc}\text{Glucose}+\text{Pi}\to \text{Glucose}\text{6}-\text{phosphate}+{\text{H}}_{\text{2}}\text{O}& (\Delta G^{\prime }°)=+13.8\text{kJ}/\text{mol}\\ \text{ATP}+{\text{H}}_{\text{2}}\text{O}\to \text{ADP}+\text{Pi}& (\Delta G^{\prime }°)=-30.5\text{kJ}/\text{mol}\end{array}\\ \underset{\_}{}\\ \\ \begin{array}{cc}\text{Glucose}+\text{ATP}\to \text{Glucose-6-phosphate}+\text{ADP}& (\Delta G^{\prime }°)=-16.7\text{kJ}/\text{mol}\end{array}\end{array}$

The value of gas constant R is 0.0083 kJ/ mole K. The equilibrium constant at the given temperature 37°C (310 K) is calculated as follows-

$\begin{array}{c}(\Delta G^{\prime }°)=-RT\ln ({K^{\prime }}_{\text{eq}})\\ \ln ({K^{\prime }}_{\text{eq}})=\frac{(\Delta G^{\prime }°)}{-RT}\end{array}$

$\begin{array}{c}\ln ({K^{\prime }}_{\text{eq}})=\frac{(-16.7\text{kJ}/\text{mol})}{(-0.0083\text{kJ}/\text{mol}\cdot \text{K})(310\text{K})}\\ =\frac{16.7}{2.58}\\ =6.47\end{array}$

$\begin{array}{c}({K^{\prime }}_{\text{eq}})=e^{(6.47)}\\ =645\end{array}$

Conclusion

The $({K^{\prime }}_{\text{eq}})$ for the overall reaction is $\underset{\_}{645}$.

(c2)

Summary Introduction

To determine: The concentration of glucose needed to achieve 250 mM intracellular concentration of glucose-6-phosphate when the concentrations of ATP and ADP are 3.3 mM and 1.32 mM.

Introduction:

The glycolysis is the process of series of reactions converting the glucose molecules into two molecules of pyruvate in the presence of oxygen. The pyruvate undergoes series of reactions in the citric acid cycle and converts into Acetyl CoA (acetyl coenzyme A).

Explanation of Solution

The value of equilibrium constant is determined as follows:

$\begin{array}{c}({K^{\prime }}_{\text{eq}})=\frac{\text{[Glucose-6-phosphate][ADP]}}{\text{[Glucose][ATP]}}\\ [\text{Glucose]}=\frac{\text{[Glucose-6-phosphate][ADP]}}{({K^{\prime }}_{\text{eq}})\text{[ATP]}}\end{array}$

$\begin{array}{c}[\text{Glucose]}=\frac{\text{[Glucose-6-phosphate][ADP]}}{({K^{\prime }}_{\text{eq}})\text{[ATP]}}\\ =\frac{[250\times {10}^{-6}\text{M}][1.32\text{mM}]}{[645]\times [3.38\text{mM}]}\\ =\frac{330\times {10}^{-6}\text{M}}{2181.8}\\ =0.15\times {10}^{-6}\text{M}\\ =\text{1}\text{.5}\times {\text{10}}^{-7}\text{M}\end{array}$

Conclusion

The concentration of glucose needed is $\underset{\_}{1.5\times 10^{-7}M}$.

(c3)

Summary Introduction

To discuss: Whether this coupling process provides feasible route at least in principle for phosphorylation of glucose in the cell.

Introduction:

The glycolysis is the process of series of reactions converting the glucose molecules into two molecules of pyruvate in the presence of oxygen. The pyruvate undergoes series of reactions in the citric acid cycle and converts into Acetyl CoA (acetyl coenzyme A).

Explanation of Solution

This route will be favorable because the given concentration of glucose does not exceed the permissible limit of 1 M. The coupling process will be feasible for given route for phosphorylation of glucose in the cell.

(d)

Summary Introduction

To determine: Whether the route is reasonable given that the coupling requires a common intermediate, the route uses ATP hydrolysis to raise the intracellular concentration of Pi.

Introduction:

The glycolysis is a series of reactions converting the glucose molecules into two molecules of pyruvate in the presence of oxygen. The pyruvate undergoes series of reactions in the citric acid cycle and converts into acetyl CoA (acetyl coenzyme A).

Explanation of Solution

The high levels of phosphates are needed for the precipitation of divalent cations of phosphate salts under normal physiological conditions, and thus, the condition will not be possible. Therefore, the given route is not reasonable for the requirement of common intermediate in coupling.

(e)

Summary Introduction

To determine: The advantages of route of transfer of phosphate group by ATP to glucose by enzyme glucokinase.

Introduction:

The glycolysis is a series of reactions converting the glucose molecules into two molecules of pyruvate in the presence of oxygen. The pyruvate undergoes series of reactions in the citric acid cycle and converts into acetyl CoA (acetyl coenzyme A).

Explanation of Solution

This reaction of transfer of phosphate group to glucose by ATP is catalyzed by glucokinase. The phosphate group is transferred and the potential of ATP is used and generation of high levels of intermediate molecules is not required. The reaction is catalyzed by glucokinase and no intermediates are formed by utilization of potential of ATP.