Question

In the metabolism of glucose, the first step is the conversion of glucose to glucose 6 -phosphate: glucose \(+\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow\) glucose 6 -phosphate \(+\mathrm{H}_{2} \mathrm{O}\) $$ \Delta G^{\circ}=13.4 \mathrm{~kJ} / \mathrm{mol} $$ Because \(\Delta G^{\circ}\) is positive, this reaction does not favor the formation of products. Show how this reaction can be made to proceed by coupling it with the hydrolysis of ATP. Write an equation for the coupled reaction, and estimate the equilibrium constant for the coupled process.

Short answer

The coupled reaction is exergonic with \( \Delta G^{\circ} = -17.1 \text{ kJ/mol} \) and \( K_{eq} \approx 984.6 \).

Step-by-step solution

Understanding the Reaction

We begin with the conversion of glucose to glucose 6-phosphate with a positive \( \Delta G^{\circ} \) of 13.4 kJ/mol, indicating that the reaction is not spontaneous under standard conditions.

ATP Hydrolysis

The hydrolysis of ATP to ADP and inorganic phosphate is a highly exergonic reaction with \( \Delta G^{\circ} = -30.5\, \text{kJ/mol} \). This helps in providing the necessary energy for endergonic reactions.

Coupling the Reactions

By coupling the conversion of glucose with the hydrolysis of ATP, we can sum the reactions: \[\text{glucose} + \text{ATP} \rightarrow \text{glucose 6-phosphate} + \text{ADP}\]The overall \( \Delta G^{\circ} \) becomes \( 13.4 \text{ kJ/mol} - 30.5 \text{ kJ/mol} = -17.1 \text{ kJ/mol} \).

Calculate the Equilibrium Constant

The equilibrium constant \( K_{eq} \) for the coupled reaction can be estimated using the relationship:\[\Delta G^{\circ} = -RT \ln K_{eq}\]Rearranging, we find \( K_{eq} = e^{-\Delta G^{\circ}/RT} \). Using \( \Delta G^{\circ} = -17.1 \text{ kJ/mol} \) and \( R = 8.314 \times 10^{-3} \text{ kJ/mol K} \), at \( 298 \text{ K} \), calculate \( K_{eq} \).

Substitute Values

Substituting the values, we get:\[ K_{eq} = e^{17.1 / (8.314 \times 10^{-3} \times 298)} \approx e^{6.89} \approx 984.6 \]

Final Conclusion

The equilibrium constant \( K_{eq} \) is much greater than 1, indicating the coupled reaction strongly favors the formation of products.

Key concepts

ATP hydrolysis

ATP hydrolysis is a key process that provides energy in cells. It involves the breakdown of Adenosine Triphosphate (ATP) into Adenosine Diphosphate (ADP) and an inorganic phosphate. This process releases energy due to the high-energy phosphate bond being broken. The change in Gibbs free energy (\( \Delta G^{\circ} \)) for ATP hydrolysis is negative, specifically \(-30.5 \, \text{kJ/mol}\). This negative value indicates that this reaction easily occurs and releases energy.

In cellular activities, ATP hydrolysis transfers energy where it is needed. This libera(ftted energy can drive many other biological processes, especially those that require the input of energy to proceed. Think of ATP as the cell's energy currency, facilitating processes that would not naturally occur. Understanding ATP hydrolysis is crucial for grasping how our bodies maintain their metabolic functions.

Gibbs free energy

Gibbs free energy is an important concept in chemical thermodynamics. It helps predict whether a reaction will occur spontaneously. The symbol \( \Delta G^{\circ} \) represents the standard change in Gibbs free energy for a reaction:

• \( \Delta G^{\circ} < 0 \) indicates the reaction is exergonic and likely spontaneous.
• \( \Delta G^{\circ} > 0 \) means the reaction is endergonic and requires an input of energy to proceed.
• \( \Delta G^{\circ} = 0 \) suggests the system is at equilibrium and the reaction does not progress in either direction.

In the metabolism of glucose, the conversion step to glucose 6-phosphate has a positive \( \Delta G^{\circ} \) of 13.4 kJ/mol, meaning it requires energy input to occur. This is why coupling with ATP hydrolysis, which has a \( \Delta G^{\circ} \) of \(-30.5 \, \text{kJ/mol}\), allows the overall process to proceed spontaneously. By combining the two processes, we effectively manage energy across reactions.

Equilibrium constant

The equilibrium constant, denoted as \( K_{eq} \), is a measure of a reaction's tendency to favor products or reactants once it reaches equilibrium. For a chemical reaction at a given temperature, \( K_{eq} \) can be directly related to Gibbs free energy through the equation:
\[ \Delta G^{\circ} = -RT \ln K_{eq} \]

Where \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. When '\( K_{eq} \)' is much greater than 1, it indicates a strong product formation at equilibrium.

In our coupled glucose phosphorylation reaction, the \( K_{eq} \) value is approximately \( 984.6 \). This high value shows a significant shift toward product formation, indicating that the coupled reaction is favorable under standard conditions.

Coupled reactions

Coupled reactions are a fascinating mechanism used by cells to ensure that energy-consuming (endergonic) reactions happen.This involves pairing an endergonic reaction, which requires energy, to an exergonic reaction, which releases energy, such as ATP hydrolysis.

In our example, the glucose conversion to glucose 6-phosphate has \( \Delta G^{\circ} > 0 \), needing energy to proceed. By coupling with ATP hydrolysis (\( \Delta G^{\circ} = -30.5 \text{ kJ/mol} \)), the resulting equation
\[ \text{glucose} + \text{ATP} \rightarrow \text{glucose 6-phosphate} + \text{ADP} \] ends up with \( \Delta G^{\circ} = -17.1 \text{ kJ/mol} \), indicating the process is spontaneous.

Cells routinely use this coupling technique, ensuring that essential biochemical processes occur as needed, even when individual reactions are not energetically favorable.