Problem 1 - Biochemistry: Concepts and Connections

Question

The process of a protein folding from an inactive unfolded structure to the active folded structure can be represented by the following equation: $$ \text { unfolded protein } \rightleftharpoons \text { folded protein } $$ (c) At what temperature would the ratio of unfolded protein to folded protein be \(1: 5 ?\) The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the folding of the protein lysozyme are: $$ \begin{aligned} \Delta H^{\circ} &=-280 \mathrm{~kJ} / \mathrm{mol} \\ \Delta S^{\circ} &=-790 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \end{aligned} $$ (a) Calculate the value of \(\Delta G^{\circ}\) for the folding of lysozyme at \(25^{\circ} \mathrm{C}\). (b) At what temperature would you expect the unfolding of lysozyme to become favorable?

Step-by-Step Solution

Answer

The temperature is approximately \(36.6^{\circ}C\).

Step 1: Understand the Given Information

We have values for the enthalpy change \(\Delta H^{\circ}\) and entropy change \(\Delta S^{\circ}\) for the folding of a protein. We need to find the temperature where the ratio of unfolded to folded protein is 1:5.

Step 2: Use the Definition of the Equilibrium Constant

The equilibrium constant \(K\) for the unfolded to folded protein reaction is given by the ratio of concentrations: \(K = \frac{[folded]}{[unfolded]} = 5\) as given in the problem.

Step 3: Relate \(K\) to \(\Delta G^{\circ}\)

The relation between the Gibbs free energy change \(\Delta G^{\circ}\) and the equilibrium constant \(K\) is given by the equation: \[\Delta G^{\circ} = -RT \ln K\] where \(R = 8.314 \text{ J/mol}\cdot\text{K}\).

Step 4: Express \(\Delta G^{\circ}\) in Terms of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\)

The Gibbs free energy change \(\Delta G^{\circ}\) is also given by: \[\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\]

Step 5: Combine Equations and Solve for \(T\)

Set the expressions for \(\Delta G^{\circ}\) equal: \[ -RT \ln K = \Delta H^{\circ} - T \Delta S^{\circ}\] Substitute \(\ln K = \ln 5\), \(\Delta H^{\circ} = -280,000 \text{ J/mol}\), and \(\Delta S^{\circ} = -790 \text{ J/mol}\cdot\text{K}\) into the equation, and solve for \(T\):\[ -8.314 \cdot T \cdot \ln 5 = -280,000 - T(-790) \] Simplify and solve for \(T\): \[T = \frac{280,000}{790 - 8.314 \ln 5} \approx 309.75 \text{ K}\]

Step 6: Convert Temperature to Celsius

The final step is to convert this temperature from Kelvin to Celsius: \(T^{\circ}C = T - 273.15\). Therefore, \(T^{\circ}C = 309.75 - 273.15 = 36.6^{\circ}C\).

Key Concepts

Gibbs Free Energy

Gibbs free energy, represented as \( \Delta G^{\circ} \), is a thermodynamic potential used to predict whether a process will occur spontaneously. It combines enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)) to determine the driving force behind chemical reactions and physical transformations, like protein folding.
In the context of protein folding, a negative \( \Delta G \) value signifies that the process is favorable and can occur without any external energy inputs. This means the folded structure of a protein is more stable than its unfolded counterpart.

• Formula: \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \)
• \( \Delta G \) gives insight into the balance between the enthalpic and entropic contributions.
• In our exercise, calculating \( \Delta G \) at a specific temperature helps us predict protein folding behavior.

This balance is crucial in understanding how temperature influences protein stability.

Equilibrium Constant

The equilibrium constant, denoted as \( K \), describes the ratio of product concentrations to reactant concentrations at equilibrium. In protein folding, it indicates the ratio of folded to unfolded proteins when the process reaches equilibrium.
For our given scenario, the equilibrium constant is \( K = 5 \), meaning there are five times more folded proteins compared to unfolded ones. This simple relationship allows us to connect the equilibrium condition with Gibbs free energy through the equation:

• \( \Delta G^{\circ} = -RT \ln K \)
• Where \( R \) is the gas constant \( 8.314 \, \text{J/mol} \cdot \text{K} \)
• The natural logarithm of \( K \) signifies the multiplicative factor that \( \Delta G \) exerts.

Knowing \( K \) informs us about the stability of the protein’s structure at a given temperature.

Enthalpy

Enthalpy change, represented as \( \Delta H^{\circ} \), is the heat released or absorbed during a process under constant pressure. It informs us about the energy bindings in reactions, such as protein folding.
The provided \( \Delta H^{\circ} \) value of \(-280 \, \text{kJ/mol}\) for lysozyme indicates that the folding process releases energy, making it exothermic:

• Negative \( \Delta H \) = energy release = generally favorable folding.
• Exothermic reactions tend to occur spontaneously, particularly when accompanied by favorable entropy changes.

In our exercise, enthalpy supports calculating \( \Delta G \) and understanding why folded proteins are thermodynamically favorable at lower temperatures.

Entropy

Entropy, denoted as \( \Delta S^{\circ} \), measures the disorder or randomness in a system. A protein's folding often leads to a decrease in entropy because it transitions from a more disordered unfolded state to a more ordered folded structure.
Given by \( \Delta S^{\circ} = -790 \, \text{J/mol} \cdot \text{K} \), a negative entropy change typically disfavors protein folding due to the loss of randomness:

• Negative change means the system becomes more ordered.
• The entropy effect pushes against the folding, contributing to \( \Delta G \).
• To counteract this, the temperature \( T \) must be optimized to favor folding.

Thus, while enthalpy favors folding by releasing energy, entropy needs to be balanced against temperature to ensure folding remains a spontaneous process.