Chymotrypsin Mechanism

Step 1

Substrate binding compresses Asp-102 and His-57 to form a hydrogen bond

Step 2

The hydrogen-bonded imidazole ring of His-57 is now a better proton acceptor (base) and will deprotonate the hydroxyl group of Ser-195. His-57 now has an imidazolium ring

Step 3

The deprotonated Ser-195 is now very nucleophilic and will attack the peptide’s carbonyl carbon atom

a. this forms the 1st tetrahedral intermediate

b. the resulting negatively charged oxygen from the peptide’s carbonyl carbon on the tetrahedral intermediate is stabilized by the oxyanion hole (consisting of the gly-193 and the amide hydrogen of ser-195)

Step 4

The imidazolium ring, now acting as an acid, donates a proton to the amino group of the peptide. The peptide bond is cleaved and the acyl-enzyme intermediate is formed

Step 5

After cleavage of the peptide bond, the amine product (peptide) is released

Step 6

Water enters and donates a proton to the imidazole ring of his-57 (similar to ser-195)

Step 7

The hydroxyl group of the water attacks the carbonyl carbon of the acyl-enzyme intermediate

a. this forms the 2nd tetrahedral intermediate

b. the resulting negatively charged oxygen from the peptide’s carbonyl carbon on the tetrahedral intermediate is stabilized by the oxyanion hole

Step 8

The imidazolium ring of His-57 donates a proton “back to” ser-195 which leads to the release of the carboxylate product