Problem 30 - Biochemistry: A Short Course

Question

The enzyme aldolase catalyzes the following reaction in the glycolytic pathway: Fructose \(1,6-\) bisphosphate \(\rightleftharpoons\) dihydroxyacetone phosphate \(+\) glyceraldehyde 3 -phosphate The \(\Delta G^{\circ \prime}\) for the reaction is \(+23.8 \mathrm{kJ} \mathrm{mol}^{-1}(+5.7 \mathrm{kcal}\) \(\left.\operatorname{mol}^{-1}\right),\) whereas the \(\Delta G^{\circ}\) in the cell is \(-1.3 \mathrm{kJ} \mathrm{mol}^{-1}\) \(\left(-0.3 \mathrm{kcal} \mathrm{mol}^{-1}\right) .\) Calculate the ratio of reactants to products under equilibrium and intracellular conditions. Using your results, explain how the reaction can be endergonic under standard conditions and exergonic under intracellular conditions.

Step-by-Step Solution

Answer

Standard conditions show an unfavorable reaction (high \(K_{eq}\)), but intracellular conditions alter the balance (low \(Q\)), making it favorable, aided by cellular conditions.

Step 1: Understanding the Reaction

The reaction involves the conversion of fructose 1,6-bisphosphate into dihydroxyacetone phosphate and glyceraldehyde 3-phosphate. The standard change in Gibbs free energy \( \Delta G^{\circ \prime} \) is \(+23.8 \, \mathrm{kJ \cdot mol}^{-1}\), whereas the cellular change is \(-1.3 \, \mathrm{kJ \cdot mol}^{-1}\). This suggests that while the reaction is not favorable at standard conditions, it is favorable in the cell.

Step 2: Calculate Equilibrium Ratio Using Standard Conditions

The ratio of products to reactants under standard conditions can be calculated using the formula \( \Delta G^{\circ \prime} = -RT \ln(K_{eq}) \). Given \( \Delta G^{\circ \prime} = +23.8 \, \mathrm{kJ \cdot mol}^{-1}\), we rearrange the formula to find \( K_{eq} = e^{-\Delta G^{\circ \prime}/RT} \). Using \( R = 8.314 \, \mathrm{J \cdot mol}^{-1} \cdot \mathrm{K}^{-1} \) and \( T = 298 \, \mathrm{K} \), calculate \( K_{eq} \).

Step 3: Evaluate Gibbs Free Energy in Cell

For cellular conditions, we use \( \Delta G^{\circ} = -RT \ln(Q) \), where \( Q \) is the reaction quotient. With \( \Delta G^{\circ} = -1.3 \, \mathrm{kJ \cdot mol}^{-1} = -1300 \, \mathrm{J \cdot mol}^{-1}\), rearrange to find \( Q = e^{-\Delta G^{\circ}/RT} \). Calculate \( Q \) using the same \( R \) and \( T \).

Step 4: Interpretation and Comparison

Compare \( K_{eq} \) and \( Q \). If \( K_{eq} \ll Q \), it illustrates that the cellular environment significantly shifts toward a different balance, allowing the reaction to proceed forward exergonically despite being unfavorable under standard conditions, through mechanisms like substrate concentration and other cellular factors influencing \( \Delta G \).

Key Concepts

Aldolase Enzyme

The aldolase enzyme plays a critical role in the glycolytic pathway, transforming sugar molecules to release energy for the cell. Specifically, it catalyzes the splitting of fructose 1,6-bisphosphate into two three-carbon sugars: dihydroxyacetone phosphate and glyceraldehyde 3-phosphate. This step is crucial as it sets the stage for subsequent reactions that extract usable energy.

• Aldolase acts by facilitating the cleavage of carbon bonds within the sugar molecules.
• The enzyme supports the conversion of a six-carbon sugar into two three-carbon sugars, thus continuing the pathway efficiently.

This reaction also shows that enzymes like aldolase can work under varying conditions to maintain essential cellular processes, enabling both glucogenic and ketogenic paths.

Gibbs Free Energy

Gibbs free energy, represented by \(\Delta G\), measures the availability of energy to perform work during a biochemical reaction. In the glycolytic pathway, two values— standard and cellular—are compared to understand reaction spontaneity.

• Under standard conditions, \(\Delta G^{\circ \prime} = +23.8 \, \mathrm{kJ/mol}\) indicates that the reaction is endergonic and non-spontaneous, requiring input of energy.
• However, within the cell, \(\Delta G^{\circ} = -1.3 \, \mathrm{kJ/mol}\) suggests the reaction is exergonic, proceeding naturally because it releases energy.

This dichotomy highlights how cellular conditions such as reactant and product concentrations can drastically alter the energy dynamics of a chemical reaction.

Reaction Equilibrium

Reaction equilibrium occurs when the rates of the forward and reverse reactions are equal, leading to no net change in the concentrations of reactants and products. In this glycolytic step:

• Equilibrium under standard conditions is unfavorable, as reflected by a high positive \(\Delta G^{\circ \prime}\). This means equilibrium lies towards reactants.
• In cellular conditions, equilibrium shifts due to factors like substrate concentration, aligning more favorably towards product formation.

Thus, even inherently endergonic reactions can appear feasible when manipulated by cellular environments, indicating how thermodynamic principles adapt to biological needs.

Enthalpy Change

Enthalpy change, symbolized by \(\Delta H\), represents the heat absorbed or released during a reaction. It provides insight into the energy transfer during the formation and breaking of bonds within the molecules. Although the concept itself is not directly calculated here, understanding \(\Delta H\) helps decode the broader energetics of reactions.

• An endothermic reaction has \(\Delta H > 0\), absorbing heat, which under standard conditions requires more energy, aligning with a positive \(\Delta G\).
• Conversely, an exothermic reaction releases heat \(\Delta H < 0\), helping in making an exergonic cellular condition possible.

Recognizing enthalpy in context underscores how heat exchange interacts with spontaneous reaction potential, fitting the larger picture of dynamic cellular metabolism.