Question

Chymotrypsin selectively cleaves a peptide at the carboxyl side of the amino acids having an aryl sidechain. The tripeptide(s) formed on hydrolysis of the peptide, \(\mathrm{Val\text{-}Phe\text{-}Leu\text{-}Met\text{-}Tyr\text{-}Pro\text{-}Gly\text{-}Trp\text{-}Cys}\), with chymotrypsin is(are)

• A. Leu-Met-Tyr
• B. Phe-Leu-Met
• C. Pro-Gly-Trp
• D. Tyr-Pro-Gly

Hint:

Cut the chain right after every Phe, Tyr and Trp, then read off whatever three-residue piece falls between two consecutive cuts.

Answer 1

The Correct Option is A, C

An easier way to see this is to mark a cut symbol directly after every aromatic residue in the sequence, then just read off the pieces between the cuts.

The sequence is $\mathrm{Val\text{-}Phe\text{-}Leu\text{-}Met\text{-}Tyr\text{-}Pro\text{-}Gly\text{-}Trp\text{-}Cys}$. The three aromatic (aryl-sidechain) amino acids are phenylalanine, tyrosine and tryptophan, so a cut goes right after each one:

\[ \mathrm{Val\text{-}Phe}\ |\ \mathrm{Leu\text{-}Met\text{-}Tyr}\ |\ \mathrm{Pro\text{-}Gly\text{-}Trp}\ |\ \mathrm{Cys} \]

Reading the four pieces left to right: $\mathrm{Val\text{-}Phe}$ (a dipeptide), $\mathrm{Leu\text{-}Met\text{-}Tyr}$ (a tripeptide), $\mathrm{Pro\text{-}Gly\text{-}Trp}$ (a tripeptide) and $\mathrm{Cys}$ alone (a single residue).

Now check each answer option against this list directly:

1. Leu-Met-Tyr: matches the second piece exactly. Correct.
2. Phe-Leu-Met: this string straddles the first cut point (it takes Phe from the first piece and Leu-Met from the second), so it can never appear as a single fragment. Incorrect.
3. Pro-Gly-Trp: matches the third piece exactly. Correct.
4. Tyr-Pro-Gly: this straddles the second cut point (Tyr from the second piece, Pro-Gly from the third), so it is likewise never a real fragment. Incorrect.

So the tripeptides actually formed are Leu-Met-Tyr and Pro-Gly-Trp.