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 of glucose conversion to glucose-6-phosphate and ATP hydrolysis proceeds spontaneously with a \(\Delta G^{\circ}\) of -17.1 kJ/mol. The equilibrium constant for the reaction, \(K_{eq}\), can be found using the relation \(\Delta G^{\circ} = -RTlnK_{eq}\) and substituting relevant values.

Step-by-step solution

Understanding ATP hydrolysis

The hydrolysis of ATP (adenosine triphosphate) into ADP (adenosine diphosphate) and inorganic phosphate is a spontaneous reaction that releases energy. The reaction is given by the equation: ATP \(\rightarrow\) ADP + \(Pi\) with \(\Delta G^{\circ} = -30.5 \mathrm{kJ/mol}\).

Coupling reactions

A coupled reaction involves mixing a spontaneous and non-spontaneous reaction to make the overall process spontaneous. The conversion of glucose to glucose-6-phosphate is coupled with ATP hydrolysis. The equation for the coupled reaction is: glucose + ATP \(\rightarrow\) glucose 6-phosphate + ADP + \(H_{2}O\)

Calculating the \(\Delta G^{\circ}\) for the coupled process

The \(\Delta G^{\circ}\) for the coupled process equals the sum of the \(\Delta G^{\circ}\) values of the individual processes. For this coupled process, \(\Delta G^{\circ}_{coupled} = \Delta G^{\circ}_{glucose \rightarrow glucose-6-phosphate} + \Delta G^{\circ}_{ATP \rightarrow ADP + Pi} = 13.4 \mathrm{kJ/mol} + (-30.5 \mathrm{kJ/mol}) = -17.1 \mathrm{kJ/mol}\). Because \(\Delta G^{\circ}_{coupled}\) is negative, the coupled reaction is spontaneous and will proceed.

Find the equilibrium constant

From the equation \(\Delta G^{\circ} = -RTlnK_{eq}\), we solve for \(K_{eq}\) to find the equilibrium constant. Input \(\Delta G^{\circ}_{coupled} = -17.1 \mathrm{kJ/mol}\), \(R = 8.314 \mathrm{J/K \cdot mol}\), and \(T = 298 K\) (standard temperature). Calculating, we get \(K_{eq} = e^{(\Delta G^{\circ}_{coupled}/RT)}\). Note that the units of \(\Delta G^{\circ}\) should be converted to J/mol before performing the calculation.

Key concepts

Glucose Metabolism

Glucose metabolism is a vital process in living organisms as it provides the energy necessary for cellular activities. It all starts with glucose, a simple sugar found in foods, being transformed into different forms to release energy. The first step in this metabolic journey is the conversion of glucose into glucose 6-phosphate. This reaction involves glucose reacting with inorganic phosphate (H₃PO₄) and results in glucose 6-phosphate and water (H₂O). This change is crucial because it helps trap glucose inside cells, allowing it to continue in the pathway of glycolysis.

However, this reaction alone is not favorable because it has a positive Gibbs free energy change (ΔG°) of 13.4 kJ/mol, meaning the formation of glucose 6-phosphate is not naturally preferred. To overcome this hurdle, cells use a fascinating strategy known as coupled reactions to make the process energetically feasible. This coupling is essential for the efficiency of glucose metabolism and overall energy production in cells.

Coupled Reactions

Coupled reactions are an amazing biochemical strategy used by cells to drive non-spontaneous reactions by pairing them with spontaneous ones. This is what happens in the case of the glucose to glucose 6-phosphate conversion.

By itself, this transformation requires energy as it is not spontaneous. However, when coupled with the hydrolysis of ATP (adenosine triphosphate) into ADP (adenosine diphosphate) and inorganic phosphate (Pi), the overall process releases energy, making it energetically favorable. This is because ATP hydrolysis is a spontaneous reaction with a negative ΔG° of -30.5 kJ/mol.

The chemistry behind this coupling is simple. The energy released from ATP hydrolysis is employed to "pay for" the energy cost of the glucose phosphorylation reaction. By summing the ΔG° values of both reactions, the overall coupled process results in a ΔG° of -17.1 kJ/mol. This negative value indicates a spontaneous and favorable reaction, enabling the pathway to proceed with efficiency.

Equilibrium Constant

Every chemical reaction aims to reach a state of equilibrium. The equilibrium constant ( Keq ) provides a measure of the tendency of a reaction to proceed to completion. When it comes to our coupled reaction in glucose metabolism, we can calculate Keq using the equation: ΔG° = -RT ln Keq .

Here, ΔG° is the Gibbs free energy change for the reaction, R is the universal gas constant (8.314 J/K·mol), and T is the temperature in Kelvin, typically 298 K for standard conditions. For our coupled reaction, ΔG° is -17.1 kJ/mol, which we convert to -17100 J/mol for consistency with R's units.

Solving this gives Keq = e^{(17100 / (8.314 × 298))} . This calculation results in a large Keq , suggesting that at equilibrium, the concentration of products is much higher than that of the reactants, affirming the reaction's spontaneity under biological conditions. Understanding Keq helps appreciate how efficiently energy transformations occur in living organisms, especially during processes like glucose metabolism.