Chapter 18, Problem 15P

Interpretation Introduction

(a)

Interpretation:

The standard energy change for the phosphoglycerate reaction should be calculated

Concept Introduction:

The energy which allows predicting the reaction spontaneously at a particular temperature is called free energy. The reaction which proceed both forward and backward reaction is called as equilibrium constant K.

Answer to Problem 15P

  $-19.1\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

In phosphoglycerate reaction ATP combine with hydrolysis to obtain following equation.

  ${\text{ATP+H}}_{\text{2}}\text{O}\to {\text{ADP+P}}_{\text{i}}$

  ${\text{ATP+H}}_{\text{2}}\text{O}\to {\text{ADP+P}}_{\text{i}}$

In the first reaction hydrolysis of 1,3-BPG mixes to give

  $1,3-\text{BPG}+{\text{H}}_{\text{2}}\text{O}\to \text{Pyruvate}+P_i;△G=-4\text{9}\text{.6 kJ/mol}$

In the second reaction, it is reverse process of hydrolysis of ATP. In this reaction we use opposite value of free energy.

  ${\text{ADP+P}}_{\text{i}}\to {\text{ATP+H}}_{\text{2}}\text{O}△G=30.\text{5 kJ/mol}$

In the coupled reaction free energy stated as

  $△G^{0\text{'}}{}_{tota{l}_{}}=△G^{o\text{'}}{}_1+△G^{0\text{'}}{}_2$

  $\begin{array}{l}△G^{0\text{'}}{}_{total}=△G^{o\text{'}}{}_1+△G^{0\text{'}}{}_2\\ △G^{0\text{'}}{}_{total}=-49.6\frac{\text{kJ}}{\text{mol}}\text{+30}\text{.5}\frac{\text{kJ}}{\text{mol}}\\ △G^{0\text{'}}{}_{total}=-19.1\frac{\text{kJ}}{\text{mol}}\\ K_{eq}\\ △G^{o\text{'}}=-RT\times \ln K^1{}_{eq}\\ K^1{}_{eq}=e^{\frac{△G^{0\text{'}}}{RT}}\\ K^1{}_{eq}=e^{\frac{-19.1}{8.314\times {10}^{-3}\times 298}}\end{array}$

In phosphoglycereate kinase reaction the standard free energy obtained is $-19.1\frac{\text{kJ}}{\text{mol}}$ .

Interpretation Introduction

(b)

Interpretation:

The equilibrium constant of the reaction should be calculated.

Concept Introduction:

The energy which allows predicting the reaction spontaneously at a particular temperature is called free energy. The reaction which proceed both forward and backward reaction is called as equilibrium constant K

Answer to Problem 15P

  $K^1{}_{eq}=1647.74$

Explanation of Solution

  $K_{eq}$ is the equilibrium constant for the reaction

  $△G^{o\text{'}}=-RT\times \ln K^1{}_{eq}$

Hence $K^1{}_{eq}=e^{\frac{△G^{0\text{'}}}{RT}}$

The temperature of the body is $37{}^{\circ }\text{C}$

Convert Celsius to Kelvin

  $37{}^{\circ }\text{C}=310.5{}^{\circ }\text{C}$

Substitute the equation in following formula.

  $\begin{array}{l}{K}^1{}_{eq}=e^{\frac{-19.1}{8.314\times {10}^{-3}\times 310.15}}\\ K^1{}_{eq}=e^{7.40}\\ K^1{}_{eq}=1647.74\end{array}$

The equilibrium constant for the reaction is $K^1{}_{eq}=1647.74$ .

Interpretation Introduction

(c)

Interpretation:

The ratio of $\frac{\left[\text{ATP}\right]}{\left[\text{ADP}\right]}$ should be calculated.

Concept Introduction:

The energy which allows predicting the reaction spontaneously at a temperature is called free energy. The reaction which proceed both forward and backward reaction is called as equilibrium constant K

Answer to Problem 15P

  $13.73$

Explanation of Solution

The steady state concentration for the reaction is

Concentration ratio for $\left[3-PG\right]\left[1,3-BPG\right]$ in erythrocyte reaction is

  $K^1{}_{eq}=\frac{\left[\text{ATP}\right]\left[\text{3-PG}\right]}{\left[\text{ADP}\right]\left[\text{1,3-BPG}\right]}$

Substitute the value $120\mu M$ and $1\mu M$

  $K^1{}_{eq}=\frac{\left[\text{ATP}\right]\text{×120μM}}{\left[\text{ADP}\right]\text{×1μM}}$

Rearrange the equation to get ratio of ATP AND ADP

  $\begin{array}{l}\frac{\left[\text{ATP}\right]}{\left[\text{ADP}\right]}=\frac{K^1{}_{eq}}{120}\\ \frac{\left[\text{ATP}\right]}{\left[\text{ADP}\right]}=\frac{1647.74}{120}\\ \frac{\left[\text{ATP}\right]}{\left[\text{ADP}\right]}=13.73\end{array}$

The ratio of $\left[ATP\right]$ and $\left[ADP\right]$ in phophoglycereate kinase reaction is $13.73$