A single, 18-amino acid peptide chain is to be assigned from the following experimental information below, your job is to provide the full triple lettercode sequence (if possible) and show your logic behind your answer.
(i) Amino acid analysis of the native peptide: Cx6, Rx2, Ex1, Yx1, Px1, Vx1, Ax1, Mx1, Wx1, Lx1, Kx1, and Tx1.
(ii) Dinitrofluorobenzene (FDNB) analysis indicated the presence of C on reaction of the native peptide.
(iii) The native peptide is subjected to DTT and Pepsin Digestion of the native peptide produced 4 peptide fragments that were then exposed to Edman degradation analysis, they provided the following sequence information:
Fragment 1: YVPRMAC Fragment 2: CC Fragment 3: LCKETC Fragment 4: WRC
(iv) Following disulfide bond reduction, Cyanogen Bromide digestion of the original intact peptide provided 2 peptide, one 6mer + homoserine lactone and one 11mer.
(v) The resultin 11mer was HPLC isolated and exposed to TCEP, this peptide was subjected to Trypsin digestion followed by amino acid analysis. The Trypsin digested peptide fragments: A, B, andC, were assigned the following amino acid composition:
Fragment A: Cx2, Lx1, Kx1 Fragment B: Tx1, Ex1, Cx1 Fragment C: Cx1, Wx1, Ax1, Rx1
(i) This will serve as the amino acid reference for the sequence analysis. The amino acids (AAs) that are key to the peptide's reaction to different chemicals and enzymes should be noted.
(ii) FDNB analysis determines the AA in the N terminal of the peptide:
C _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
(iii) DTT is used for reduction of disulfide bonds to recover the linear primary structure. Pepsin digestion cleaves the N side of the amino acids L, F, W, and Y. Thus, these amino acids should be the starting AA of each fragment, except for the established first AA, which is C. Thus Fragment 2 is first in the sequence.
C C _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
(iv) Cyanogen bromide (CNBr) cleaves the C-side of Met (M). Formation of homoserine lactone is just the side reaction of met with the reagent. There is only one M in the sequence, giving only one point of cleavage. Thus, there are only 2 fragments. Homoserine lactone is M, which makes it the 7th amino acid.Referring to the fragments in (iii), the sequence would be: