Problem 4 - Biochemistry: Concepts and Connections

Question

The oxidation of glucose to \(\mathrm{CO}_{2}\) and water is a major source of energy in aerobic organisms. It is a reaction favored mainly by a large negative enthalpy change. $$ \begin{array}{cr} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{~s})+6 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow & 6 \mathrm{CO}_{2}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-2816 \mathrm{~kJ} / \mathrm{mol} & \Delta S^{\circ}=+181 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \end{array} $$ (a) At \(37^{\circ} \mathrm{C}\), what is the value for \(\Delta G^{\circ}\) ? (b) In the overall reaction of aerobic metabolism of glucose, 32 moles of ATP are produced from ADP for every mole of glucose oxidized. Calculate the standard state free energy change for the overall reaction when glucose oxidation is coupled to the formation of ATP at \(37{ }^{\circ} \mathrm{C}\). (c) What is the efficiency of the process in terms of the percentage of the available free energy change cantured in ATP?

Step-by-Step Solution

Answer

(a) \(-2872.54 \text{ kJ/mol}\); (b) \(-89.77 \text{ kJ/mol ATP}\); (c) 33.97\% efficiency.

Step 1: Convert Temperature to Kelvin

We need to convert the temperature from Celsius to Kelvin to use it in the Gibbs Free Energy equation. The formula for this conversion is: \[ T(K) = T(°C) + 273.15 \] For this problem, we have: \[ T = 37 + 273.15 = 310.15 \text{ K} \]

Step 2: Calculate Gibbs Free Energy Change (9G0)

Use the following formula to calculate the Gibbs Free Energy Change:\[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \]Substitute the given values (note that \(\Delta S^{\circ}\) needs to be in \( \mathrm{kJ/mol \cdot K} \):\[ \Delta S^{\circ} = \frac{181 \text{ J/mol} \cdot \text{K}}{1000} = 0.181 \text{ kJ/mol} \cdot \text{K} \]\[ \Delta G^{\circ} = -2816 - (310.15 \times 0.181) \]\[ \Delta G^{\circ} = -2816 - 56.53615 \]\[ \Delta G^{\circ} = -2872.54 \text{ kJ/mol} \]

Step 3: Gibbs Free Energy Change per Mol of ATP Formed

Given that 32 moles of ATP are produced per mole of glucose oxidized, calculate the energy per mole of ATP:\[ \text{Energy per mol of ATP} = \frac{-2872.54 \text{ kJ/mol}}{32} \]\[ \text{Energy per mol of ATP} = -89.77 \text{ kJ/mol} \]

Step 4: Efficiency Calculation

Gibbs Free Energy for ATP hydrolysis is typically around -30.5 kJ/mol. Calculate the total energy captured by ATP:\[ \text{Total energy in ATP} = 32 \times 30.5 = 976 \text{ kJ/mol} \]Calculate the efficiency of glucose oxidation:\[ \text{Efficiency} = \left( \frac{976}{2872.54} \right) \times 100 \%\text{Efficiency} = 33.97\% \]

Key Concepts

Aerobic Metabolism

In aerobic metabolism, glucose and oxygen undergo several reactions to produce carbon dioxide and water, alongside a significant release of energy. This energy is converted to ATP, which serves as a cellular energy currency.
Glucose is broken down in the presence of oxygen, which is why this process is called "aerobic", meaning it requires air or oxygen.
As a highly efficient energy conversion process, aerobic metabolism supports high-energy needs of living organisms through a complex series of reactions known as cellular respiration.

Gibbs Free Energy

Gibbs Free Energy, often denoted as 01G01, helps predict the direction of chemical reactions and whether they will occur spontaneously. A negative 01G indicates that a reaction can proceed without external energy input.
In the context of aerobic metabolism, 01G is critical as it measures the available energy from glucose oxidation that can be harnessed for ATP synthesis.

• 01G is calculated using 01H (enthalpy change) and 01S (entropy change), alongside the absolute temperature.
• For glucose oxidation, the large negative enthalpy change contributes to a significantly negative 01G, showing the reaction is highly favorable and spontaneous.

ATP Synthesis

ATP synthesis is a crucial outcome of glucose oxidation. Here, with each molecule of glucose metabolized in aerobic conditions, approximately 32 moles of ATP are produced for cellular energy needs.
The energy released during glucose oxidation is temporarily stored in the high-energy phosphate bonds of ATP.

• The process of converting ADP (adenosine diphosphate) into ATP ensures cells have direct access to energy for vital functions.
• This involves using the energy derived from the large 01G of glucose oxidation to drive ATP synthesis.

Efficient ATP production from glucose oxidation outlines the efficiency and importance of aerobic pathways in energy conservation.

Thermodynamics

Thermodynamics underlies the study of energy changes and transformations in biological processes, such as glucose oxidation.

Principles of Thermodynamics in Biological Systems

These principles help explain how energy is transferred within cells and demonstrates how biological systems remain efficient even under high energy demands:

• First Law: Energy cannot be created or destroyed, only converted from one form to another. This law explains energy conservation during ATP production.
• Second Law: Total entropy, or energy dispersion in a system, tends to increase. Here, the entropy change (01S) in reactions like glucose oxidation helps explain the efficiency of energy transformations.
• Third Law: As temperature approaches absolute zero, entropy reaches a constant minimum. While this law is more theoretical, it frames the broader understanding of metabolic reactions and their heat dynamics.