You wish to determine the sequence of a 7 residue peptide An
Solution
There are 6 amino acids given: Asp, Glu, Phe, Pro, Trp and Val.
\'Trp\' is present in the end. Edman\'s degradation gives the N\' terminal amino acid.
Three peptides are produced on cleavage with chymotrypsin. So, 2 sites of chymotrypsin are there. Chymotrypsin cleaves at the ends of aromatic amino acids (one is \'Phe\' and the other is \'Trp\'). In total, there are 7 amino acids in the protein. At N-terminal Edman\' degradation has already revealed \'Trp\'.
So, the sequence becomes:
N\'-Trp (1st cut) ---------- (4 amino acids)-----------------------------Phe--(2nd cut)--(//)-C\'
// signifies any other amino acid.
Now, acid hydrolysis reveals three fragments of sizes of 1, 3 and 3 amino acids. Acid hydrolysis will occur at C-terminal ends of acid.
N\'-Trp--Val-- Glu--(1st cut)--Asp--(2nd cut)--Pro--Phe--(//)-C\'
When cleaved with Trypsin, one peptide of 7 amino acids was obtained. Trypsin cleaves at C-terminal of \'Arg\' and \'Lys\'. Please note that the in the initial information given, basic amino acids are not there in this protein. And a single fragment is revealed from trypsin hydrolysis, so you can conclude that at the C-terminal end, the 7th amino acid of this protein is Lysine or Arginine.
So, the final protein sequence becomes:
N\'-Trp--Val-- Glu---Asp--Pro--Phe--(Lys/Arg)-C\'
