Question

Complete hydrolysis of an unknown basic decapeptide gives Gly, Ala, Leu, Ile, Phe, Tyr, Glu, Arg, Lys, and Ser.
Problem 24-44 ConstantsI Periodic Table Complete hydrolysis of an unknown basic decapeptide gives Gly, Ala, Leu, lle, Phe, Tyr, Glu, Arg, Lys, and Ser. Terminal residue analysis shows that the N terminus is Ala and the C terminus is lle. Incubation of the decapeptide with chymotrypsin gives two tripeptides, A and B, and a tetrapeptide, C. Amino acid analysis shows that peptide A contains Gly, Glu, Tyr, and peptide B contains Ala, Phe, and Lys; and peptide C contains Leu, lle, Ser, and Arg. Terminal residue analysis gives the following results. N terminus C terminus A GIn B Ala C Arg Tyr Phe lle Incubation of the decapeptide with trypsin gives a dipeptide D, a pentapeptide E, and a tripeptide F. Terminal residue analysis of F shows that the N terminus is Ser, and the C terminus is lle.

Answer 1

N = Ala d'Ile Deca = 10 Trypsin dc' Lys or Arg Chimon – phe on Trp on TGV Atbs 3 Peplies es Telva Peptide A GIN -Gly-Tyr | 3 Ala - L98 - Phe e Arg - Ser-Lev-Ile. N'Ala-Lys-phe } Glo-Gly-Typ S Arg_ser-Lev-Dec Trypsin. F User-Lev-dee so, Frasment a Arg-ser-Lev-Ile E Phe-Glu-Gly-Sur-AVS A GW-90-Tyy