Chapter 17, Problem 26P

Interpretation Introduction

(a)

Interpretation:

The reason as to why the oxidation reaction of malate by ${\text{NAD}}^{+}$ to produce oxaloacetate proceeds further as written under physiological conditions is to be stated.

Concept introduction:

The cycle which represents the set of chemical reactions that are utilized by all the aerobic or living organisms which helps them in releasing the stored amount of energy by the oxidation process of acetyl-CoA is known as citric acid cycle. This acetyl-CoA is obtained from the carbohydrates, protein and fats in the form of ATP.

Answer to Problem 26P

The oxidation reaction of malate by ${\text{NAD}}^{+}$ to produce oxaloacetate proceeds further as written under physiological conditions because this endergonic reaction gets pulled by exergonic reaction.

Explanation of Solution

The eighth step of citric acid cycle includes the oxidation by ${\text{NAD}}^{+}$. In this step, the oxidation of malate into oxaloacetate takes place with the help of ${\text{NAD}}^{+}$ as shown below.

 

The value of $\Delta G°\text{'}$ for this reaction is $\text{29}\text{kJ}{\text{mol}}^{-1}$. The oxidation reaction of malate by ${\text{NAD}}^{+}$ is an endergonic reaction. This oxidation reaction takes place under standard physiological conditions because the endergonic reactions may be pushed or pulled to move forward. In the oxidation reaction of malate by ${\text{NAD}}^{+}$, the endergonic reaction gets pulled which means that the formed products, oxaloacetate and $\text{NADH}$ are easily cleaned up with the help of an exergonic reactions. Due to this condition, the concentration of products remains low and the reaction is allowed to move further.

Interpretation Introduction

(b)

Interpretation:

The lowest $\left[\text{malate}\right]/\left[\text{oxaloacetate}\right]$ ratio at which oxaloacetate can be formed from malate is to be stated.

Concept introduction:

The set of complex compounds which are responsible for migration of the electrons from the electron donors to the acceptors of the electrons with help of the redox reactions is known as electron transport chain. This electron transport chain also helps in the coupling of transferred electrons with protons transfer inside and outside the cell membrane.

Answer to Problem 26P

The lowest $\left[\text{malate}\right]/\left[\text{oxaloacetate}\right]$ ratio at which oxaloacetate can be formed from malate is $1.57\times {10}^4$.

Explanation of Solution

The oxidation reaction of malate by ${\text{NAD}}^{+}$ to produce oxaloacetate and $\text{NADH}$ is given below.

 

The given ratio of $\left[{\text{NAD}}^{+}\right]/\left[\text{NADH}\right]$ for the oxidation reaction of malate by ${\text{NAD}}^{+}$ to produce oxaloacetate and $\text{NADH}$ is $8$.

The value of $\Delta G°\text{'}$ for this reaction is $\text{29}\text{kJ}{\text{mol}}^{-1}$.

The given temperature is $\text{25°C}$.

The conversion of degrees Celsius into Kelvin is done below.

$\text{0}°\text{C}=\text{273}\text{K}$

So, the temperature becomes $\text{25}+\text{273}\text{K}=\text{298}\text{K}$.

The equilibrium constant for this oxidation reaction is given below.

$K_{\text{eq}}^{\text{'}}=\frac{\left[\text{oxaloaceate}\right]\left[\text{NADH}\right]}{\left[\text{malate}\right]\left[{\text{NAD}}^{+}\right]}$

The expression for calculating the equilibrium constant at $\text{25°C}$ is given below.

$K_{\text{eq}}^{\text{'}}=e^{-\Delta G°\text{'}/RT}$  (1)

Where,

• $R$ is the universal constant. The universal gas constant is $8.315\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}{\text{K}}^{-1}$.
• $T$ is the temperature.
• $\Delta G°\text{'}$ is the standard free energy.

Substitute the values of $R$, $T$ and $\Delta G°\text{'}$ in equation (1).

$\begin{array}{c}{K}_{\text{eq}}^{\text{'}}=e^{-\text{29}\text{kJ}{\text{mol}}^{-1}/8.315\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}{\text{K}}^{-1}\times 298\text{K}}\\ =e^{-\text{29}\text{kJ}{\text{mol}}^{-1}/2.47\text{kJ}{\text{mol}}^{-1}}\\ =7.96\times {10}^{-6}\end{array}$

The expression to calculate the ratio, $\left[\text{malate}\right]/\left[\text{oxaloacetate}\right]$ is given below.

$\frac{1}{K_{\text{eq}}^{\text{'}}}=\frac{\left[\text{malate}\right]\left[{\text{NAD}}^{+}\right]}{\left[\text{oxaloacetate}\right]\left[\text{NADH}\right]}$

Substitute the value of $K_{\text{eq}}^{\text{'}}$ and value of ratio of $\left[{\text{NAD}}^{+}\right]/\left[\text{NADH}\right]$ in the above equation.

$\begin{array}{l}\frac{1}{7.96\times {10}^{-6}}=\frac{\left[\text{malate}\right]}{\left[\text{oxaloacetate}\right]}\times 8\\ 0.1256\times {10}^6=\frac{\left[\text{malate}\right]}{\left[\text{oxaloacetate}\right]}\times 8\\ \frac{0.1256\times {10}^6}{8}=\frac{\left[\text{malate}\right]}{\left[\text{oxaloacetate}\right]}\\ \frac{\left[\text{malate}\right]}{\left[\text{oxaloacetate}\right]}=1.57\times {10}^4\end{array}$

Therefore, the value for the ratio of $\left[\text{malate}\right]/\left[\text{oxaloacetate}\right]$ is $1.57\times {10}^4$.