Chapter 15, Problem 1P

(a)

Summary Introduction

To determine: The reaction quotient for the given reaction, for the PFK-1 reaction under physiological conditions.

Introduction:

Reaction quotient is the ratio of product of concentration of reaction products to the concentration of reactants at a particular time during the course of the reaction. It is calculated to determine the progress of the reaction.

Explanation of Solution

Explanation:

The concentration of fructose- $6$-phosphate is 87 µM.

The concentration of fructose- $1,6$-bisphosphate is 22 µM.

The concentration of ATP is 11400 µM.

The concentration of ADP is 1320 µM.

The reaction quotient of the given reaction is calculated by the formula,

$Q=\frac{\left[\text{fructose-1,6-bisphosphate}\right]\left[\text{ADP}\right]}{\left[\text{fructose-6-phosphate}\right]\left[\text{ATP}\right]}$

Where,

• [fructose-1,6-bisphosphate] is the concentration of fructose-1,6-bisphosphate.
• [fructose-6-bisphosphate] is the concentration of fructose-6-phosphate.
• [ADP] is the concentration of ADP.
• [ADP] is the concentration of ATP.
• Q is the reaction quotient.

Substitute the values of [fructose-1,6-bisphosphate], [fructose-6-bisphosphate], [ADP], and [ATP] in the above equation.

$\begin{array}{c}Q=\frac{22\text{μM}\times 1320\text{μM}}{87\text{μM}\times 11400\text{μM}}\\ =\underset{\_}{2.93\times 10^{-2}}\end{array}$

Conclusion

Conclusion:

The reaction quotient of the given reaction is $\underset{\_}{2.93\times 10^{-2}}$.

(b)

Summary Introduction

To determine: The equilibrium constant of the given reaction.

Introduction:

Equilibrium constant is the ratio of product of concentration of reaction products to the concentration of reactants at equilibrium stage of the reaction. The equilibrium constant is equal to the reaction quotient when the reaction is at equilibrium.

Explanation of Solution

Explanation:

The standard free energy change of the reaction is -14.2 kJ/mol.

The standard free energy change of the given process is calculated by the formula,

$\Delta G^{\circ }=-RT\ln K$

Where,

• ΔG°is the standard free energy change of the reaction.
• R is the gas constant.
• T is the temperature of the reaction.
• K is the equilibrium constant of the reaction.

The equilibrium constant = $\text{In}K{\text{'}}_{eq}=-\Delta G\text{'}°/RT$ (1)

Substitute the values of R, T and K in Equation (1)

$\begin{array}{c} \text{In }K{\text{'}}_{eq}=-(-14.2\text{ kJ/mol})/(8.413\text{J/mol}\text{.K}\times 298\text{K})\\ =-(-14.2\text{ kJ/mol})/(2480\text{J/mol}\text{.K})\\ =-(-14.2\text{ kJ/mol})/(2.48\text{kJ/mol}\text{.K})\\ =5.73\end{array}$

$\begin{array}{c}K{\text{'}}_{eq}=e^{-5.73}\\ =308\end{array}$

Conclusion

Conclusion:

The equilibrium constant of the given reaction is 308. .

(c)

Summary Introduction

To determine: The comparison of reaction quotient and equilibrium constant of the given reaction.

Introduction:

Equilibrium constant is the ratio of product of concentration of reaction products to the concentration of reactants at equilibrium stage of the reaction. The equilibrium constant is equal to the reaction quotient when the reaction is at equilibrium.

Explanation of Solution

Explanation:

The progress of the given reaction is determined by comparing the values of reaction quotient and equilibrium constant for the given reaction. If the reaction quotient is less than the equilibrium constant, then the reaction moves in the backward direction. This means that the concentration of reactants is more than that of the products in the given reaction. If the value of reaction quotient is more than the equilibrium constant, then the reaction moves in the forward direction and the concentration of products is higher than that of the reactants. The reaction is at equilibrium when the reaction quotient is equivalent to the equilibrium constant.

The reaction quotient for the given reaction is much less than the equilibrium constant. Therefore, the given reaction is far away from the equilibrium. Therefore, the concentration of reactants which is fructose-6-phosphate and ATP is higher and the reaction is in the backward direction.

Summary Introduction

To determine: The role of PFK-1 as a regulatory enzyme.

Introduction:

Enzymes are commonly known as the biological catalyst which initiates and speeds up various reactions occurring in the human body. The activity of the given enzyme as catalyst is determined by the rate of the reaction which is increased by the addition of the enzyme.

Explanation of Solution

Explanation:

PKF-1 which is phosphofructokinase-1 is the most important regulatory enzyme which acts as a catalyst for the steps involved in the glycolysis. It speeds up the reaction of fructose- $6$-phosphate and ATP to form fructose- $1,6$-bisphosphate and ADP in the given experiment.

The activity of the given enzyme is very less for the given reaction because the reaction occurring in the cells would not attain equilibrium easily. Therefore, the activity of the PKF- $1$ enzyme as a regulatory enzyme is very slow in the given reaction as compared to glycolysis.