Problem 31 - Biochemistry: A Short Course

Question

On page 262 a reaction, \(\mathrm{A} \rightleftharpoons \mathrm{B},\) with a \(\Delta G^{\circ}=+16.7 \mathrm{kJ} \mathrm{mol}^{-1}\) \(\left(+4.0 \mathrm{kcal} \mathrm{mol}^{-1}\right)\) is shown to have a \(K_{e q}\) of \(1.15 \times 10^{-3}\). The \(K_{\text {eq }}\) is increased to \(2.67 \times 10^{2}\) if the reaction is coupled to ATP hydrolysis under standard conditions. The ATPgenerating system of cells maintains the \([\mathrm{ATP}] /[\mathrm{ADP}]\left[\mathrm{P}_{\mathrm{i}}\right]\) ratio at a high level, typically of the order of \(500 \mathrm{M}^{-1}\). Calculate the ratio of B/A under cellular conditions.

Step-by-Step Solution

Answer

The ratio \([B]/[A]\) under cellular conditions is \( 1.335 \times 10^{5} \).

Step 1: Understand the Given Information

We are given a reaction \( A \rightleftharpoons B \) with a standard free energy change \( \Delta G^{\circ} = +16.7 \text{kJ/mol} \) and an equilibrium constant \( K_{eq} = 1.15 \times 10^{-3} \). The equilibrium constant increases to \( 2.67 \times 10^{2} \) when coupled to ATP hydrolysis. The cellular \( [ATP]/[ADP][P_i] \) ratio is maintained at \( 500 \text{ M}^{-1} \). We need to find the \( [B]/[A] \) ratio under cellular conditions.

Step 2: Use the Relation between Gibbs Free Energy and Keq

The relationship between the Gibbs free energy change and the equilibrium constant is given by:\[ \Delta G^{\circ} = -RT \ln K_{eq} \]where \( R \) is the universal gas constant \( = 8.314 \text{ J/mol K} \) and \( T \) = 298 K (assuming room temperature). This equation helps us relate \( \Delta G^{\circ} \) to \( K_{eq} \).

Step 3: Apply Cellular Conditions

Under cellular conditions, the effective equilibrium constant is modified due to the ATP coupling. We use the modified equilibrium constant \( K'_{eq} \) given by:\[ K'_{eq} = K_{eq} \times 500 \]where \( 500 \text{ M}^{-1} \) is the concentration ratio for \( [ATP]/[ADP][P_i] \).

Step 4: Calculate Effective Equilibrium Constant

Substitute into the formula:\[ K'_{eq} = 2.67 \times 10^{2} \times 500 = 1.335 \times 10^{5} \]This is the effective equilibrium constant under cellular conditions.

Step 5: Calculate [B]/[A] Ratio

The ratio of \([B]/[A]\) is directly the effective equilibrium constant under these cellular conditions. Thus,\[ \frac{[B]}{[A]} = K'_{eq} = 1.335 \times 10^{5} \]

Step 6: Conclusion

Therefore, under cellular conditions, the ratio \([B]/[A]\) is \( 1.335 \times 10^{5} \). This indicates that \( B \) is significantly favored over \( A \) in the cellular environment.

Key Concepts

Gibbs Free Energy

Gibbs Free Energy, often abbreviated as \( \Delta G \), is a crucial concept in biochemistry that helps us understand the spontaneity of chemical reactions. It represents the energy available to do work under constant temperature and pressure conditions. This energy helps indicate whether a reaction will occur spontaneously or requires external energy input.
The equation for Gibbs Free Energy is given by:

• \( \Delta G = \Delta H - T\Delta S \)
• \( \Delta H \) is the change in enthalpy, or total energy.
• \( T \) is the temperature in Kelvin.
• \( \Delta S \) is the change in entropy, or disorder.

A negative \( \Delta G \) indicates a spontaneous reaction, while a positive \( \Delta G \) means the reaction is non-spontaneous under standard conditions. The reaction \( A \rightleftharpoons B \) in the example has a positive \( \Delta G^{\circ} = +16.7 \text{ kJ/mol} \), suggesting it is not spontaneous without additional energy or changes. However, when coupled with another reaction such as ATP hydrolysis, we can manipulate these conditions.

Equilibrium Constant

The equilibrium constant \( K_{eq} \) is a key concept that describes the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction. It is mathematically expressed as:

• For reaction \( A \rightleftharpoons B \): \( K_{eq} = \frac{[B]}{[A]} \)

In terms of biochemistry, \( K_{eq} \) is related to the Gibbs Free Energy with the equation:

• \( \Delta G^{\circ} = -RT \ln K_{eq} \)
• \( R \) is the universal gas constant.
• \( T \) is the temperature in Kelvin (typically 298 K for physiological conditions).

The magnitude of \( K_{eq} \) provides insights into the position of equilibrium. A large \( K_{eq} \) favors products dramatically, while a small \( K_{eq} \) implies more reactants are present at equilibrium. In the given problem, the coupling with ATP hydrolysis alters the \( K_{eq} \) from \( 1.15 \times 10^{-3} \) to a significant \( 2.67 \times 10^2 \), tilting the balance heavily towards products.

ATP Hydrolysis

ATP hydrolysis is a fundamental biochemical reaction that releases energy by converting ATP (adenosine triphosphate) into ADP (adenosine diphosphate) and inorganic phosphate \( P_i \). The reaction is as follows:

• \( \text{ATP} + \text{H}_2\text{O} \rightarrow \text{ADP} + \text{P}_i + \text{energy} \)

This process is exergonic, meaning it releases energy, making it a vital component for coupling with endergonic reactions to drive them forward. The energy released from ATP hydrolysis can be captured effectively by other biochemical reactions to do work, including chemical synthesis, transport, and mechanical work in cells.
For the reaction \( A \rightleftharpoons B \), ATP hydrolysis increases the equilibrium constant significantly because it provides the necessary free energy change to shift the balance towards product formation. The cellular system maintains a high \( [ATP]/[ADP][P_i] \) ratio, optimizing the use of ATP's energy potential and promoting efficient biological processes. This interplay illustrates the interconnectedness of energy needs and supply within cellular systems.