Question

In water, the enthalpy of a protein in its folded state (\(H_f\)) is lower than that in its unfolded state (\(H_{UF}\)). The entropies of the folded and unfolded states are \(S_f\) and \(S_{UF}\), respectively. The condition(s) under which this protein spontaneously folds in water at a temperature \(T\), is(are)

• A. \( S_f > S_{UF} \)
• B. \( S_f = 0 \)
• C. \( S_f > S_{UF} \, \text{and} \, (H_f - H_{UF}) / T \)
• D. \( S_f > S_{UF} \, \text{and} \, (H_f - H_{UF}) / T \)

Hint:

For spontaneous protein folding, the enthalpy must decrease (\( H_f < H_{UF} \)) and the entropy must increase (\( S_f > S_{UF} \)).

Answer 1

The Correct Option is A, B, C, D

Step 1: Understanding protein folding.
For a protein to spontaneously fold, the Gibbs free energy \( \Delta G \) must be negative. \( \Delta G = \Delta H - T\Delta S \), where \( \Delta H \) is the enthalpy change and \( \Delta S \) is the entropy change. For spontaneous folding at a given temperature \( T \), we need \( H_f < H_{UF} \) (so that \( \Delta H \) is negative) and \( S_f > S_{UF} \) (so that \( \Delta S \) is positive).

Step 2: Analyzing the options.
- Option (A) \( S_f > S_{UF} \) is necessary for spontaneity, but not sufficient on its own.
- Option (B) \( S_f = 0 \) is incorrect as it would not favor folding.
- Option (C) is correct as it satisfies the condition for spontaneous folding, where \( S_f > S_{UF} \) and the term \( (H_f - H_{UF}) / T \) ensures the system is energetically favorable.

Step 3: Conclusion.
The correct answer is (C).