Arrange the events for the hydrolysis of amide bonds by chymotrypsin in their correct order 1) attack by Ser 195 to give a tetrahedral intermediate 2) enzyme returns to initial state 3) protonation by His 57 and then cleavage of the C-N bond release the C-terminal fragment 4) protonation by His 57 and then cleavage of the C-O bond release the N-terminal fragment 5) attack by water, leading to formation of a tetrahedral intermediate 6) binding of substrate to properly position the scissile bond for cleavage O 6,2,4, 3, 1, 5 O 6, 1, 3, 5,4,2 O 6, 1,3,5,2,4 O 6, 1, 3,4, 5, 2

6,1,3,5,4,2

Explanation:

Methods for the cleavage of highly stable amide bonds were developed to control their functions. One such method is the use of enzymes, which have active sites and binding pockets for specific amino acids, which is then followed by activation of amide bonds for hydrolysis. Chymotrypsin is one of the families of enzymes which hydrolyses amide bonds. Its catalytic site contains an oxyanion binding hole which consists of: Serine (Ser), Histidine (His) and Aspartic acid (Asp) triad. During catalysis, the triad works in a synergistic manner to break the amide bond. Firstly the side chain of Aspartic acid makes a hydrogen bond with Histidine, making it more nucleophilic. Secondly, His forms a strong hydrogen bond with the hydroxyl group of Ser and abstracts the proton from the hydroxyl group of Ser which in turn attacks the amide bond to form a tetrahedral transition state. This tetrahedral transition state collapses leading to the hydrolysis pf the amide bond by acid-base catalysis.

elements make up everything, and they combine to make compounds.

the answer is c

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