An octapeptide with the amino acids < Ile, Arg, Met, Phe, Pro, Thr, Tyr, Val > w
ID: 61128 • Letter: A
Question
An octapeptide with the amino acids < Ile, Arg, Met, Phe, Pro, Thr, Tyr, Val > was subjected to digestion reactions A, B and C in order to deduce its amino acid sequence. The following results were obtained. The exact position occupied by each of the amino acids (positions 1-8 in the sequence) will be known when the results are analyzed
A. Digestion with Trypsin yielded a [Thr-Arg] dipeptide and a [hexapeptide] with the remaining amino acids.
Question (A1): Explain the reasoning for the fragmentation pattern
Question (A2): What can you deduce regarding the exact positions occupied by Thr and Arg in the sequence?
B. Cyanogen bromide treatment of the octapeptide also yielded a dipeptide [Phe-Pro] and a hexapeptide.
Question (B1): Explain the reasoning for the above fragmentation pattern (3 points) Question (B2): What can you deduce regarding the exact positions occupied by Phe and Pro in the sequence?
C. Chymotrypsin cleaved the octapeptide into two tetrapeptides, one with and the other
Question (C1): Explain the reasoning for the above pattern Question (C2): What can you deduce regarding the exact positions occupied by Val and Ile in the sequence?
Question (D): What is the sequence of the peptide?
Explanation / Answer
Let say the seq be
NH2 1 2 3 4 5 6 7 8 COOH
A. Trypsin digestion
A1. Trypsin cleaves peptide chains mainly at the carboxyl side of the amino acids lysine or arginine, except when either is followed by proline.
NH2 1 2COOH NH2 3 4 5 6 7 8 COOH
A2. Therefore probable sequence will be
NH2 Thr Arg COOH NH2 3 4 5 6 7 8 COOH
B. After CNBr digestion
B1. Cyanogen bromide hydrolyzes peptide bonds at the C-terminus of methionine residues.
B2. And it is given that dipeptide is Phe – Pro means the met is at 6 position and Phe-Pro will be at 7 and 8 position.Therefore seq will be
Phe will be at 7 position because the chymotrysin will not cut the hydrophobic group at C terminal when followed by pro so dipeptide will be phe-pro
NH2 Thr Arg 3 4 5 met Pro Phe COOH
C. After Chymotrypsin digestion
<Thr, Arg, Tyr, Val> and the other <Ile, Met, Phe, Pro>
C1. Chymotrypsin preferentially cleaves peptide amide bonds where the carboxyl side of the amide bond (the P1 position) is a large hydrophobic amino acid (tyrosine, tryptophan, and phenylalanine).
Therefore the cut will be at 5 position,and phe is already at 7 position so tyr will be at 5 position.
C2. Chymotrypsin preferentially cleaves peptide amide bonds where the carboxyl side of the amide bond (the P1 position) is a large hydrophobic amino acid (tyrosine, tryptophan, and phenylalanine). Therefore Tyrosine must be followed by large hydrophobic containing group thet is isoleucine.
Therefore the probable sequence will be
D. NH2 Thr Arg Val Tyr COOH
NH2 Ile met pro phe COOH
D. Therefore the sequence will be
NH2 Thr Arg Val Tyr Ile Met Phe Pro COOH