Complete hydrolysis of an unknown basic decapeptide gives Gly, Ala, Leu, Ile, Phe, Tyr, Glu, Arg, Lys, and Ser. Terminal residue analysis shows that the N terminus is Ala, and the C terminus is Ile. 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, Ile, Ser, and Arg.Terminal residue analysis gives the following results.

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 Ile. Propose a structure for the decapeptide and for fragments A through F.

Trypsin cleaves the peptide bonds in which basic amino acids (lysine and arginine) contribute the carboxyl group. Trypsin cleaves the peptide bond between the carboxyl group of arginine or the carboxyl group of the lysine and the amino group of the adjacent amino acid. Chymotrypsin cleaves those peptide bonds in which aromatic amino acids (tyrosine, phenylalanine, and tryptophan) contribute the carboxyl group.

Complete hydrolysis of an unknown basic decapeptide gives Gly, Ala, Leu, Ile, Phe, Tyr, Glu, Arg, Lys, and Ser. Terminal residue analysis shows that the N terminus is Ala, and the C terminus is Ile. Means, N-terminal and C-terminal are fixed.

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, Ile, Ser, and Arg. Peptide A analysis gave ammonia in addition to the amino acids, so the Glu in the analysis must have been Glu in the original peptide. From the residue analysis, the N-terminus and C-terminus of fragments A, B, C are fixed. But the position of Ser and Leu is not justified. Thus, for that we need to know fragment F as then only from decapeptide structure we can determine fragment C. For now, let fragment C will be, Arg-Ser-Leu-Ile. Thus, we have,

Fragment A = Glu-Gln-Tyr (middle)

Fragment B = Ala-Lys-Phe (N-terminus)

Fragment C = Arg-Ser-Leu-Ile (C-terminus)

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 Ile. Thus, fragment F is clear, fragment F is Ser-Leu-Ile. Placing fragments D and E below combination of fragments A, B and C we have the following structure of decapeptide and that will also confirm fragments C, D and E.

End groups: N-terminus Ala--------------------------------Ile C-terminus

Chymotrypsin fragments:

A Glu-Gly-Tyr (middle)

B Ala-Lys-Phe

(N-terminus)

C Arg-Ser-Leu-Ile

(C-terminus)

Ala-Lys-Phe-Glu-Gly-Tyr-Arg-Ser-Leu-Ile

Trypsin fragments:

D Ala-Lys

E Phe-Glu-Gly-Tyr-Arg

F Ser-Leu-Ile

On combination of fragments and aligning them, one below the other, we have the following structure of decapeptide,

Ala-Lys-Phe-Gln-Gly-Tyr-Arg-Ser-Leu-Ile

Thereby, confirming the fragments C, D and E.