Trypsin cleaves a polypeptide backbone at the C-terminal side of Arg or Lys residues, whereas chymotrypsin cleaves after aromatic residues. A polypeptide was treated with trypsin to generate a series of fragments of the following sequences: Gly-Gly-Ile-Arg Ser-Phe-Leu-Gly Trp-Ala-Ala-Pro-Lys Ala-Glu-Glu-Gly-Leu-Arg And the same polypeptide was treated with chymotrypsin to generate the following fragments: Leu-Gly Ala-Glu-Glu-Gly-Leu-Arg-Trp Ala-Ala-Pro-Lys-Gly-Gly-Ile-Arg-Ser-Phe Assemble the protein sequence.
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A: The sequence are:
Trypsin cleaves a polypeptide backbone at the C-terminal side of Arg or Lys residues, whereas chymotrypsin cleaves after
And the same polypeptide was treated with chymotrypsin to generate the following fragments: Leu-Gly Ala-Glu-Glu-Gly-Leu-Arg-Trp Ala-Ala-Pro-Lys-Gly-Gly-Ile-Arg-Ser-Phe
Assemble the protein sequence.
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- A polypeptide is digested with trypsin, and the resulting segments are sequenced: Val-Gly Ala-Ala-Gly-Leu-Trp-Arg Arg-Asp-Pro-Gly-Lue-Met-Val-Leu-Tyr-Ala-Ala-Asp-Glu-Lys And the following fragments are produced by chymotrypsin fragmentation: Ala-Ala-Gly-Leu-Trp Arg-Arg-Asp-Pro-Gly-Leu- Met-Val-Leu-Tyr Ala-Ala-Asp-Glu-Lys-Val-Gly What is the sequence of the whole original polypeptide? (Recall that trypsin cleaves a polypeptide backbone at the C-terminal side of Arg or Lys residues, whereas chymotrypsin cleaves after aromatic amino acid residues).A polypeptide is digested with trypsin, and the resulting segments are sequenced: Val-Gly Ala-Ala-Gly-Leu-Trp-Arg Arg-Asp-Pro-Gly-Lue-Met-Val-Leu-Tyr-Ala-Ala-Asp-Glu-Lys And the following fragments are produced by chymotrypsin fragmentation: Ala-Ala-Gly-Leu-Trp Arg-Arg-Asp-Pro-Gly-Leu- Met-Val-Leu-Tyr Ala-Ala-Asp-Glu-Lys-Val-Gly What is the sequence of the whole original polypeptide?Assume that the 3 polypeptide strands shown below form a parallel B-sheet. Select amino acids AA1, AA2, and AA3 so that the parallel B-sheet is amphipathic and remains stable. Glu-lle-Asn-AA1-Cys-Val Ser-AA2-GIn-Leu-Lys-Phe Lys-Met-Cys-Leu-AA3-Val O AA1 = Pro, AA2 = Leu, AA3 = lle O AA1 = Val, AA2 = Leu, AA3 = Asn O AA1 = Ala, AA2 = Gly, AA3 = Leu AA1 = Phe, AA2 = Arg, AA3 = Ala O O
- Identify the primary sequence for the polypeptide that yields these fragments upon treatment: His-met-thr-met-ala-trp; Leu-asn-asp-phe; Val-lys obtained from chymotrypsin Leu-asn-asp-phe-his-met; Ala-trp-val-lys; Thr-met obtained from CNBRA sample of a peptide of unknown sequence was treated with trypsin; another sample of the same peptide was treated with chymotrypsin. The sequences (N-terminal to C-terminal) of the smaller peptides produced by trypsin digestion were as follows: Trp-Arg-Thr-Gin Ser-Trp-Arg-His-Trp-Ala-Lys Asp-Val-Ala-Ala-Lys Asn-Ser-Asn-Val-Ile-Arg The sequences of the smaller peptides produced by chymotrypsin digestion were as follows: Arg-His-Trp Arg-Thr-Gin Ala-Lys-Asn-Ser-Asn-Val-Ile-Arg-Trp Asp-Val-Ala-Ala-Lys-Ser-Trp The original peptide sequence was: Asp-Val-Ala-Ala-Lys-Ser-Trp-Ala-Lys-Asn-Ser-Asn-Val-Ile-Arg-Trp-Arg-His-Trp-Arg-Thr-Gin Asp-Val-Ala-Ala-Lys-Asn-Ser-Asn-Val-Ile-Arg-Trp-Arg-Thr-Gin-Ser-Trp-Arg-His-Trp-Ala-Lys Trp-Arg-Thr-Gin-Asn-Ser-Asn-Val-Ile-Arg-Ser-Trp-Arg-His-Trp-Ala-Lys-Asp-Val-Ala-Ala-Lys Arg-His-Trp-Arg-Thr-Gln-Ala-Lys-Asn-Ser-Asn-Val-Ile-Arg-Trp-Asp-Val-Ala-Ala-Lys-Ser-Trp Asp-Val-Ala-Ala-Lys-Ser-Trp-Arg-His-Trp-Ala-Lys-Asn-Ser-Asn-Val-Ile-Arg-Trp-Arg-Thr-Gin…Treatment of a polypeptide by 2-mercaptoethanol yields two polypeptides that have the following amino acid sequences: Ala-Phe-Cys-Met-Tyr-Cys-Leu-Trp-Cys-Asn Val-Cys-Trp-Val-Ile-Phe-Gly-Cys-Lys Chymotrypsin-catalyzed hydrolysis of the intact polypeptide yields polypeptide fragments with the following amino acid compositions: (Ala, Phe) (Asn, Cys2, Met,Tyr) (Cys, Gly, Lys) (Cys2, Leu,Trp2,Val) (Ile, Phe,Val) Indicate the positions of the disulfide bonds in the original polypeptide. I don't understand how the order of amino acids in Step 2 was arranged. Please explain. Thank you.
- Treatment of a polypeptide by 2-mercaptoethanol yields two polypeptides that have the following amino acid sequences: Ala-Phe-Cys-Met-Tyr-Cys-Leu-Trp-Cys-Asn Val-Cys-Trp-Val-Ile-Phe-Gly-Cys-Lys Chymotrypsin-catalyzed hydrolysis of the intact polypeptide yields polypeptide fragments with the following amino acid compositions: (Ala, Phe) (Asn, Cys2, Met,Tyr) (Cys, Gly, Lys) (Cys2, Leu,Trp2,Val) (Ile, Phe,Val) Indicate the positions of the disulfide bonds in the original polypeptide.Consider the peptide Asp-Lys-Phe-Glu-Asn-Tyr-Gln-Val-Cys. In a single beaker, you treat this peptide with 2 proteases. One protease cleaves at the N-terminus of aromatic R groups and the other cleaves at the C-terminus of polar, non-ionizable R groups. Following the enzymatic digestion, you want to separate your peptide fragments so that you can identify them. You choose to separate the fragments using an anion exchange column. Beginning at pH=6 you apply your peptide fragments to the column and you gradually decrease the pH of the column stopping the separation when the pH of the column equals 4. Omitting chemical structures, write the amino acid sequence of the peptide fragments that are produced from this digest. Write the order that these fragments will elute from the column (if at all). (Relevant pKa values are: 2.1, 3.8, 4.3, 8.3, 9.6, 10.1, and 10.5)A sample of an unknown peptide was divided into two aliquots. One aliquot was treated with trypsin, and the other with cyanogen bromide. Given the following sequences of the resulting fragments, deduce the sequence of the original peptide. Trypsin treatment: Asn-Thr-Trp-Met-Ile-Lys Gly-Tyr-Met-Gln-Phe Val-Leu-Gly-Met-Ser-Arg Cyanogen Bromide treatment: Gln-Phe Ile-Lys-Gly-Tyr-Met Ser-Arg-Asn-Thr-Trp-Met
- A sample of an unknown peptide was divided into two aliquots. One aliquot was treated with trypsin; the other was treated with cyanogen bromide. Given the following sequences (N- terminal to C-terminal) of the resulting fragments, deduce the sequence of the original peptide. Trypsin treatment Asn-Thr-Trp-Met-Ile-Lys Gly-Tyr-Met-Gln–Phe Val-Leu-GlyMet-Ser-Arg Cyanogen bromide treatment Gln–Phe Val-Leu-Gly-Met Ile-Lys-Gly-Tyr-Met Ser-Arg-Asn-Thr-Trp-MetA sample of a peptide of unknown sequence was treated with trypsin; another sample of the same peptide was treated with chymotrypsin. The sequences (N-terminal to C-terminal) of the smaller peptides produced by trypsin digestion were as follows: Ala Ser Glu-Met-AspLys Cys-His Ile His-Arg Thr-Trp Ala Ile-Phe-Asn-Arg Trp Cys–Cys–Gln The sequences of the smaller peptides produced by chymotrypsin digestion were as follows: Glu-Met-Asp Lys-Trp Asn-ArgAla Ser Cys-His-Ile-His-Arg-Thr-Trp Ala Ile-Phe Cys-Cys-Gin The original peptide sequence was:A sample of an unknown peptide was divided into two aliquots. One aliquot was treated with trypsin; the other was treated with cyanogen bromide. Given the following sequences (N-terminal to C- terminal) of the resulting fragments, deduce the sequence of the original peptide. Trypsin treatment Asn-Thr-Trp-Met-lle-Lys Gly-Tyr-Met-Gln-Phe Val-Leu-Gly-Met-Ser-Arg Cyanogen bromide treatment Gln-Phe Val-Leu-Gly-Met lle-Lys-Gly-Tyr-Met Ser-Arg-Asn-Thr-Trp-Met