In the hydrophobic core of a folded protein, there are three alanine and five phenylalanine residues that are buried, and do not interact with water.

Assume:

In solution, waters can take on seven energetically equal states.

Two waters are ordered around each alanine in the unfolded state.

Six waters are ordered around each phenylalanine in the unfolded state.

In the unfolded state, waters are ordered around alanine or phenylalanine residues and can take on only two energetically equal states.

What is the difference in the entropy of the water due to the burying of these residues as the protein folds?

Question

Lauren Hawkins

Biology

9 July, 17:38

In the hydrophobic core of a folded protein, there are three alanine and five phenylalanine residues that are buried, and do not interact with water.

Assume:

In solution, waters can take on seven energetically equal states.

Two waters are ordered around each alanine in the unfolded state.

Six waters are ordered around each phenylalanine in the unfolded state.

In the unfolded state, waters are ordered around alanine or phenylalanine residues and can take on only two energetically equal states.

What is the difference in the entropy of the water due to the burying of these residues as the protein folds?

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Answers (1)

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Genesis Kramer · Answer 1

The difference In The entropy of The western due to The burying of these residues as the protein folds is 291J/K

Explanation:

According to the data we know that

total water molecule = 2 * number of Ala + 6 * number of phenylalanine

replacing we get

2 * 2 + 6 * 4 = 4 + 24 = 28

total water molecule = 28

next step

S folded = Rln728

we replace the data and we get that

8.31 * 28 * ln7 = 452J • K - 1 • mol - 1

S unfolded = Rln228

we replace the data and we get that

8.31 * 28 * ln2 = 161J • K - 1 • mol - 1

∆ S = S folded - S unfolded = 452-161 = 291J / K