198. A protein completely unfolds in 8 M Urea. The stability of the folded state in water is
calculated to be 8 kcal/mole at 25°𝐶. What is the likely ∆𝐺 of the following reaction in 8 M Urea?
𝐴(𝑛𝑎𝑡𝑖𝑣𝑒) → 𝐴(𝑢𝑛𝑓𝑜𝑙𝑑𝑒𝑑)
Choose the correct answer:
A. Zero
B. Less than zero
C. 82 kcal/mole
D. 8 kcal/mole

 

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Answer: B. Less than zero

Explanation:

The given scenario describes the stability of a protein in 8 M urea, where the native state is stable with a ∆G of 8 kcal/mole at 25°C in water. However, when the protein is exposed to 8 M urea, the urea acts as a denaturant and disrupts the protein’s folded structure. As a result, the protein will be destabilized in the presence of urea.

• ∆G for unfolding in 8 M urea is less than zero because urea destabilizes the native state, making the unfolded state more energetically favorable. In this case, the protein will unfold more readily due to the breaking of non-covalent interactions in its structure caused by urea.

Why the Other Answers Are Incorrect:

• A. Zero: This would imply there is no net change in the stability of the protein. However, urea destabilizes the protein’s folded structure, making the unfolding process energetically favorable.

• C. 8 kcal/mole: This corresponds to the stability of the protein in water. However, in the presence of urea, the protein is less stable, and the unfolding reaction becomes energetically favored, leading to a negative ∆G.

• D. 8 kcal/mole: This would indicate that the unfolding in urea is just as energetically favorable as in water, which is not correct. Urea destabilizes the protein, making the unfolding process easier, and the ∆G should be negative.

Conclusion:

In 8 M urea, the ∆G of the unfolding reaction is negative because urea disrupts the folded protein, making the unfolded state more favorable. Therefore, the correct answer is B. Less than zero.