biochem Trying to figure out the AA seq to this polypeptide
(biochem) Trying to figure out the AA seq. to this polypeptide. Not sure what DNP-S means... Think the N-terminal starts with S and C-terminal ends with L and that\'s about it.
2. Determine the sequence of the 14-mer peptide given the following data: a. Amino acid composition: (4S, 2L, F, G, l, K, M, T, W, Y) b. Treatment of peptide with 1-fluoro-2,4-dinitro benzene followed by hydrolysis: DNP-S c. Carboxy peptide digestion: L d. Trypsin digestion: (3S, 2L, F I, M, T, W) and (G, K, S, Y) e. Chymotrypsin digestion: (F, I, S) and (G, K, L) and (L, S) and (M, T) and (S, W) and (S, Y) Note: chymotrypsin can cut after leucine and methionine in some conditions, though with less efficiency than the aromatic residues. f. Treatment of (F, I, S) with Sanger reagent followed by hydrolysis: S g. Cyan ogen bromide treatment: (2S, F, G, I, K, L, homoserine, T, Y) and (2S, L, W)Solution
(b) Treatment with 1-Fluoro-2,4-dinitrobenzene (DNFB) is used for identification of the N-terminal amino acid of a given peptide. Upon hydrolysis, the labeled dinitrophenyl (DNP) N-terminal residue is released.
note: DNP-S is dinitrophenyl serine.
Therefore, the N-terminal amino acid is Serine
(c) Carboxypeptide digestion cuts the peptide at the C-terminal end: L
Therefore, the C-terminal amino acid is Leucine
At this point we can write the amino acid sequence as: NH2-S-(G, Y, K, 3S, L, F, I, M, T, W)-L-COOH
(d) Trypsin cuts peptide at R or K
Therefore K is at the 4th position from N-terminal
The sequence can be written as NH2-S-(G, Y)-K-(3S, L, F, I, M, T, W)-L-COOH
(e) Chymotrypsin cuts at aromatic amino acid site (Y, W, F, and sometimes M and L also)
Y is the 2nd position from N-terminal, followed by G-K-L. Also one S is at the 2nd position from the C-terminal end.
The sequence can be written as NH2-S-Y-G-K-L-(F, I, S, M, T, S, W)-S-L-COOH
(f) Sanger reagent binds to N-terminal amino acid.
so among the chymotrypsin cut (F, I, S) residue, the order of amino acid is S-I-F
(g) Cyanogen bromide cleaves at M.
Therefore the final residues are : S-Y-G-K-L-S-I-F-T-homoserine + S-W-S-L
So the amino acid sequence is:
S-Y-G-K-L-S-I-F-T-M-S-W-S-L
