Analysis of a decapeptide revealed the presence of the follo

Analysis of a decapeptide revealed the presence of the following products: NH_4^+ Asp, Glu, Tyr, Arg, Met, Pro, Lys, Ser, Phe The following facts were observed: a. Neither cartboxylpeptidase A nor B treatment had any effect. b. Trypsin treatment yielded two tetrapeptides and free Lys. c. Clostripain treatment yielded a tetra peptide and a hexapeptide. d. Cyanogen bromide treatment yielded an octapeptide and a dipeplide of sequence NP (using one letter codex). e. Chymotrypsin treatment yielded two tripeptides and a tetrapeptide. The N-terminal chymotiyptic peptide had a net charge of-1 at neutral pH, and a net charge of-3 at pH 12. f. One cycle of the Edman degradation gave the PTH derivative: What is the amino acid sequence of this decapeptide? Use the three letter codes for amino acids in your answer, and if there is more than one possible sequence just enter one. Separate amino acid using spaces.

Solution

1. Carboxypeptidase A will cleave at the C-terminus except for Pro, Lys, and Arg whereas Carboxypeptidase B cleavage occurs at C-terminal Arg and Lys. No action by either indicates either a circular peptide or proline as C-terminus. One cycle of Edman degradation identifies serine as the N-terminal amino acid.

Thus, Ser_ _ _ _ _ _ _ _ _.

2. Cyanogen Bromide cleaves at C-terminus of the Met and release two products that accounts for 10 amino acid residues. Thus, their must be one Met and it must be located near the C-terminus to release the dipeptide NP.

Ser_ _ _ _ _ _ Met Asn Pro

3. Trypsin cleaves at the C-terminus of the positively charged residues. Trypsin cleavage yielded two tetrapeptides and free lysine. Looking at the sequence we have constructed so far, the C-terminus must be in one of the tetrapeptides and so either an Arg or Lys must be at position 6. Position 2 and 3 cannot be either Arg or Lys because the trypsin products do not include a di- or tripeptide. So, there must be an Arg or Lys at position 4. Since, 4 and 6 both contain Arg or Lys, the free Lys must come from position 5. Thus

Ser_ _(Arg or Lys) Lys (Arg or Lys) _ Met Asn Pro

4. Now the peptide contains both Arg and Lys at possibilities of positions 4, 5 and 6 are (Arg Lys Arg), (Arg Lys Lys), and (Lys Lys Arg). Only (Arg Lys Lys) has a possibility to generate free lysine. This would put Arg at position 4, which is consistent with clostripain treatment. Thus,

Ser_ _ Arg Lys Lys _ Met Asn Pro

5. Phe or Tyr must be located at position 7 to ensure that the chymotrypsin cleavage produces a tripeptide form the C-terminal. Also, an aromatic amino acid must be located at position 3. The only amino acid unaccounted for is either Glu or Gln which is located at position 2. thus,

Ser (Glu/Gln) (Phe/Tyr) Arg Lys Lys (Phe/Tyr) Met Asn Pro

6. For the N-terminal chymotryptic peptide Ser (Glu/Gln) (Phe/Tyr), to have a net charge of -1 at neutral pH and -3 at pH =12 , it must contain glutamic acid and tyrosine to account for the charge. Thus

Ser Glu Tyr Arg Lys Lys Phe met Asn Pro is the amino acid sequence for the decapeptide.