`DeltaH` for solid to liquid transitions for protein `A` and `B` are `2.73 kcal//mol` and `3.0 kcal//mol` .The two melting points are `0^(@)C` and `30^(@)C` respectively. The entropy changes `DeltaS_(A)` and `DeltaS_(B)` at two transition temperatures are related as

Options

A. `DeltaS_(A) = DeltaS_(B)`
B. `DeltaS_(A) lt DeltaS_(B)`
C. `DeltaS_(A) gt DeltaS_(B)`
D. `DeltaS_(B)= (303 DeltaS_(A))/(273)`

Explanation

To solve the problem, we need to calculate the entropy changes (ΔS) for both proteins A and B during their melting transitions and compare them. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify Given Data
- For protein A:
- ΔH_A = 2.73 kcal/mol
- Melting point (T_A) = 0°C = 273 K

- For protein B:
- ΔH_B = 3.0 kcal/mol
- Melting point (T_B) = 30°C = 303 K

### Step 2: Calculate the Entropy Change for Protein A (ΔS_A)
The formula for entropy change during a phase transition is given by:
\[ \Delta S = \frac{\Delta H}{T} \]

For protein A:
\[ \Delta S_A = \frac{\Delta H_A}{T_A} = \frac{2.73 \text{ kcal/mol}}{273 \text{ K}} \]

Calculating ΔS_A:
\[ \Delta S_A = \frac{2.73}{273} \approx 0.01 \text{ kcal/mol/K} \]

### Step 3: Calculate the Entropy Change for Protein B (ΔS_B)
Using the same formula for protein B:
\[ \Delta S_B = \frac{\Delta H_B}{T_B} = \frac{3.0 \text{ kcal/mol}}{303 \text{ K}} \]

Calculating ΔS_B:
\[ \Delta S_B = \frac{3.0}{303} \approx 0.0099 \text{ kcal/mol/K} \]

### Step 4: Compare the Entropy Changes
Now we compare ΔS_A and ΔS_B:
- ΔS_A ≈ 0.01 kcal/mol/K
- ΔS_B ≈ 0.0099 kcal/mol/K

Since ΔS_A > ΔS_B, we conclude that the entropy change for protein A is greater than that for protein B.

### Final Conclusion
The relationship between the entropy changes is:
\[ \Delta S_A > \Delta S_B \]