A peptide was cleaved into two smaller peptides with cyanogen bromide (CNBr) and into two different peptides by trypsin (Tryp). Their sequences were as follows: CNB1 1: Gly-Thr-Lys-Ala-Glu CNB 2: Ser-Met Tryp 1: Ser-Met-Gly-Thr-Lys Tryp 2: Ala-Glu What was the sequence of the original peptide?
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- 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-MetA 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: 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…
- You have digested a small protein with two different proteases, each protease acting on the protein in separate reactions. These are the peptide fragments generated: Trypsin: IPVK: Ile-Pro-Val-Lys ALEL: Ala-Leu-Glu-Leu HRPGDR: His-Arg-Pro-Gly-Asp-Arg FGADAEDGAMNK: Phe-Gly-Ala-Asp-Ala-Glu-Asp-Gly-Ala-Met-Asn-Lys YLEFISECIIQVLQSK: Tyr-Leu-Glu-Phe-Ile-Ser-Glu-Cys-Ile-Ile-Gln-Val-Leu-Gln-Ser-Lys Chymotrypsin: (Reaction conditions were used to maximize cutting to the C-terminal size of F, W, and Y.) LEF: Leu-Glu-Phe IPVKY: Ile-Pro-Val-Lys-Tyr GADAEDGAMNKALEL: Gly-Ala-Asp-Ala-Glu-Asp-Gly-Ala-Met-Asn-Lys-Ala-Leu-Glu-Leu ISECIIQVLQSKHRPGDRF: Ile-Ser-Glu-Cys-Ile-Ile-Gln-Val-Leu-Gln-Ser-Lys-His-Arg-Pro-Gly-Asp-Arg-Phe I am not sure what these questions are asking.A 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
- 1. A certain polypeptide was treated with trypsin and yielded the following Fragments: Leu-Glu Gly-Tyr-Asn-Arg Gln-Ala-Phe-Val-Lys The same polypeptide was treated with chymotrypsin and yielded the following fragments: Gln-Ala-Phe Asn-Arg-Leu-Glu Val-Lys-Gly-Tyr What is the amino acid sequence of this polypeptide? Instructions Make use of the table below to determine the sequence of the mystery protein.4) A polypeptide is subjected to the following digestion procedures and the fragments were sequenced. What is the sequence of the original peptide? Cyanogen bromide Trypsin digestion Asp-lle-Lys-Gln-Met Lys-Phe-Ala-Met Tyr-Arg-Gly-Met Gln-Met-Lys Gly-Met-Asp-lle-Lys Phe-Ala-Met-Lys and Lys Tyr.Arg Sequence of original peptide:Determine the sequence of a polypeptide treated with trypsin and chimotripsine. Below are the fragments generated with each treatment. Determine the original sequence for both fragmentations (reduerde that they must be equal in the order of amino acids) Quimotripsina 1. Leu-His-Lys-Gln-Ala-Asn-Gln-Ser-Gly-Gly-Gly-Pro-Ser 1. Gln-Gln-Ala-Gln-His-Leu-Arg-Ala-Cys-Gln-Gln-Trp 2. Arg-lle-Pro-Lys-Cys-Arg-Lys-Phe Trypsin 1. Arg 2. Ala-Cys-Gln-GIn-Trp-Leu-His-Lys 3. Cys-Arg 4. Gln-Ala-Asn-Gln-Ser-Gly-Gly-Gly- Pro-Ser 5. lle-Pro-Lys 6. Light 7. Phe-Gin-Gln-Ala-Gln-His-Leu-Arg
- Translate the following DNA sequence into amino acids 5'ATAGTACCGCAAATTTATCGCT3 O met-ala-phe-lys-stop O met-ala-phe-lys- met-tyr-his-gly-val-stop-met-gly O met-ala-ser-gly-thr-stop O tyr-his gly-val-stop-met-ly O ala-phe-lys stopA solution of peptide of unknown sequence was divided into 2 samples. One sample was treated with trypsin and the other one with pepsin. The fragments produced are given below: TRYPSIN A. Pro-Gly-Met-Phe-Leu-Arg B. Gln-Ile-Pro-Lys C. Ala-Gly-Trp-Lys PEPSIN A. Phe-Leu-Arg-Ala-Gly В. Pro-Gly-Met С. Тrр-Lys-Gln-Пе-Pro-Lys Deduce the original polypeptide chain.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