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

For spontaneous folding (UF → F), we need ΔG < 0.

$$\Delta G = G_F - G_{UF} = (H_F - TS_F) - (H_{UF} - TS_{UF})$$

$$\Delta G = (H_F - H_{UF}) - T(S_F - S_{UF})$$

$$\Delta G = \Delta H - T\Delta S$$

Given: $H_F < H_{UF}$, so $\Delta H = H_F - H_{UF} < 0$ (favorable)

For spontaneous folding: $\Delta G < 0$

$$(H_F - H_{UF}) - T(S_F - S_{UF}) < 0$$

$$T(S_F - S_{UF}) > (H_F - H_{UF})$$

(D) is correct: $(S_F - S_{UF}) > (H_F - H_{UF})/T$

This can be rewritten as:

$$S_F - S_{UF} > \frac{H_F - H_{UF}}{T}$$

Case 1: If $S_{UF} < S_F$ (option A):

• Then $S_F - S_{UF} > 0$
• Both $\Delta H < 0$ and $\Delta S > 0$ favor folding
• $\Delta G < 0$ at all temperatures

Case 2: If $S_{UF} = 0$ (option B):

• Then $S_F - S_{UF} = S_F$
• Condition becomes: $S_F > (H_F - H_{UF})/T$
• This satisfies condition (D)

Case 3: If $S_{UF} = S_F$ (option C):

• Then $S_F - S_{UF} = 0$
• Condition becomes: $0 > (H_F - H_{UF})/T$
• Since $H_F - H_{UF} < 0$, we have $0 > \text{negative}/T$, which is true
• This satisfies condition (D)

All options (A), (B), (C), and (D) are correct.