Chapter 27, Problem 1P

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant for the reaction should be calculated.

Concept Introduction:

A metabolic pathway may be a connected series of chemical reactions occurring at intervals a cell. The reactants, products, and intermediates of a catalyst reaction are called metabolites that are changed by a sequence of chemical reactions catalyzed by enzymes.

Answer to Problem 1P

The equilibrium constant for the reaction is 360,333

Explanation of Solution

The equilibrium constant can be calculated as follows:

  $\begin{array}{l}△{\text{G}}^{\text{°}}{\text{=-RT In K}}_{\begin{array}{l}\text{eq}\\ \end{array}}\\ {\text{K}}_{eq}{\text{=e}}^{\frac{-△G°}{RT}}\\ {\text{ = e}}^{\frac{-(31.7kJ/mol)}{(8.314\times {10}^{-3}kJ/mol-K)(298K)}}\\ \text{ = 360,333}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The factor that must exceed in the reaction to proceed in the reverse direction should be explained.

Concept Introduction:

For the reverse reaction, the expression for $\Delta G$ related to $\Delta G^{°}$and concentration of all reactants and products are as follows:

  $\Delta G=\Delta G^{°}+RT\ln \frac{\text{[ATP][pyruvate]}}{\text{[ADP][PEP]}}$

Answer to Problem 1P

The reaction will proceed in the reverse direction when G > 0. The natural log term must be >1.

Explanation of Solution

The reaction will proceed in the reverse direction when G > 0. The natural log term must be >1.

  $\begin{array}{l}\Delta G=\Delta G^{°}+RT\ln \frac{\text{[ATP][pyruvate]}}{\text{[ADP][PEP]}}\\ \Delta G{°}^{\text{'}}+RT\ln \frac{\text{[ATP][pyruvate]}}{\text{[ADP][PEP]}}>0\\ \ln \frac{\text{[ATP][pyruvate]}}{\text{[ADP][PEP]}}>\frac{-\Delta G°\text{'}}{RT}\\ \frac{\text{[ATP][pyruvate]}}{\text{[ADP][PEP]}}>e^{\frac{-\Delta G°\text{'}}{RT}}\end{array}$

We already know that [ATP] = [ADP]. By using this reaction and the value obtained in part a gives:

  $\begin{array}{l}\frac{\text{[pyruvate]}}{\text{[PEP]}}>360,333\\ [pyruvate]>360,333[PEP]\end{array}$

(c)

Interpretation Introduction

Interpretation:

The value of K_eqshould be explained.

Concept Introduction:

The same magnitude relation of merchandise and reactants are going to be reached. K_eq is independent of pressure. Basically, changes in pressure are like changes in concentration for gases. K_eq can rely upon the relative energies of reactants and merchandise.

Answer to Problem 1P

The value of K_eq= $\text{0}\text{.724}$

Explanation of Solution

With the help of the formula described in part a, K_eqcan be calculated.

  $\begin{array}{l}{K}_{eq}=e^{\frac{-\Delta G°}{RT}}\\ \text{ =}{e}^{\frac{-(0.8kJ)}{(8.314\times {10}^{-3}kJ/mol•K)(298K)}}\\ \text{ =0}\text{.724}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The ratio of [PEP] / [pyruvate] should be explained.

Concept Introduction:

For the reaction, the expression for the equilibrium constant is as follows:

  ${\text{K}}_{eq}=\frac{[PEP][ADP]GDP][P_i]}{[\text{pyruvate][ATP][GTP]}}$

Answer to Problem 1P

The ratio of [PEP] / [pyruvate] is equal to 724

Explanation of Solution

Told that [ATP] = [ADP], [GTP] = [GDP], and [P_i] = 1 1 mM, we can figure out the ratio of [PEP] to [pyruvate].

  $\begin{array}{l}{\text{ K}}_{eq}=0.724\\ \frac{[PEP][ADP]GDP][P_i]}{[\text{pyruvate][ATP][GTP]}}=0.724\\ \frac{[\text{PEP][}\text{.001]}}{[\text{pyruvate]}}=0.724\\ \frac{[\text{PEP]}}{[\text{pyruvate]}}=724\end{array}$

(e)

Interpretation Introduction

Interpretation:

The reason for the substrate cycle to be strongly favored in both directions under physiological conditions should be explained.

Concept Introduction:

The same magnitude relation of merchandise and reactants are going to be reached. K_eq is independent of pressure. Basically, changes in pressure are like changes in concentration for gases. K_eq can rely upon the relative energies of reactants and merchandise.

For the reaction, the expression for the equilibrium constant is as follows:

  ${\text{K}}_{eq}=\frac{[PEP][ADP]GDP][P_i]}{[\text{pyruvate][ATP][GTP]}}$

Answer to Problem 1P

The reaction will be favorable in both directions under the following conditions:

  $\text{724>}\frac{[PEP]}{[pyruvate]}>\frac{1}{360,000}$

Explanation of Solution

With the help of the observation from part b and part d of this question, the reaction will be favorable in both directions under the following conditions.

  $\text{724>}\frac{[PEP]}{[pyruvate]}>\frac{1}{360,000}$