Question

The pK of the side chain of amino acid X in a polypeptide is normally in the range of 9-10 and carries a positive charge when protonated. Suppose you have a soluble globular protein and you find there is an X that has a pK of 6.5. What is the most likely reason for such a large drop in pK? Circle the correct answer.

a) X is on the surface of the protein where it ion pairs with the carboxylate group of another amino acid side chain.

b) X forms an ion pair with a buffer anion.

c) X is ion paired with a carboxylate group of another amino acid in the hydrophobic interior of the protein.

d) X is buried inside the protein surrounded by hydrophobic groups.

e) X is hydrated thus preventing ion pairing with other groups.

Can you please include an explanation? (The key states the answer is D)

Answer 1

The hydrophobic amino acids surrounding the amino acid X destabilize the protonated form thereby lowering the pka value. The hydrophobic core is non polar which repels the polar charged molecules. Therefore it will destabilize the protonated form of amino acid X

pka of 9-10 means the side of amino acid X will deprotonate at pH 9-10. Below pH9-10 amino acid X's side chain will remain protonated and would be repel by the hydrophobic core. The drop of pka to 6.5 would cause the amino acid X's side chain to deprotonate at low pH( pH6.5), therefore it will not be repelled by the non polar amino acid residues in the core.