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.
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- A sample of an unknown peptide was divided into two aliquots. One aliquot was treated with trypsin and 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—Tyr—Asp—Met—Phe—Ala—Arg Asp—Trp—Asn—Arg Gln—Met—Tyr—Cys—Pro—Ile—Arg Gln—Cys Cyanogen bromide treatment Tyr—Cys—Pro—Ile—Arg—Asn—Tyr—Asp—Met Asp—Trp—Asn—Arg—Gln—Met Phe—Ala—Arg—Gln—CysA 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 dividedinto two aliquots. One aliquot was treated with trypsin; the otherwas treated with cyanogen bromide. Given the following sequences(N-terminal to C-terminal) of the resulting fragments, deduce thesequence of the original peptide.
- A sample of a peptide of unknown sequencewas treated with trypsin; another sample of the same peptide wastreated with chymotrypsin. The sequences (N-terminal to C-terminal)of the smaller peptides produced by trypsin digestion were as follows:A heptapeptide when treated with trypsin produced two peptides. T1 (D, G, Y) and T2 (K, F, V, A). When the heptapeptide was treated with chymotrypsin, three peptides were produced: CT1 (K,,Y, G), CT2 (F,A, V), and CT3 (D). The sequences of these peptides is not known, however. When the peptide was treated with Sanger’s Reagent and hydrolyzed, DNP-K and DNP-A were recovered. What is the amino acid sequence of the heptapeptide?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?
- 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).You are trying to determine the PTMs on your protein of interest, so you set up a mass spectrometry experiment. You get the following data. Starting sequence: PRTEINSKICKASSDER After Trypsin digest, you get the following major peptides and weights from MS: TEINSKICK/1167 Da ASSDER/664 Da a. Which of these peptides is modified? b. What is the modification?The genetic code consists of a series of three-base wordsthat each code for a given amino acid.(a) Using the selections from the genetic code shown below, de-termine the amino acid sequence coded by the following seg-ment of RNA: UCCACAGCCUAUAUGGCAAACUUGAAG AUG= methionine ;CCU= proline; CAU= histidine ;UGG= tryptophan AAG= lysine ; UAU= tyrosine ;GCC= alanine ;UUG= leucine ;CGG= arginine ;UGU= cysteine ;AAC =asparagine ;ACA=threonine ;UCC= serine ;GCA=alanine ;UCA=serine(b) What is the complementary DNA sequence from which this RNA sequence was made? (c) If you were sequencing the DNA fragment in part (b), how many complementary chain pieces would you obtain in the tube containing ddATP?
- Consider the following coding 71 nucleotide DNA template sequence (It does not contain a translational start): 5’- GTTTCCCCTATGCTTCATCACGAGGGCACTGACATGTGTAAACGAAATTCCAACCTGAGCGGCGT GTTGAG-3’ By in vitro translating the mRNA, you determined that the translated peptide is 15 amino acids long. What is the expected peptide sequence in single letter abbreviations?Which of the following amino acid sequences would yield the most optimal oligonucleotide probe? Ala-Met-Ser-Leu-Pro-Trp Gly-Trp-Asp-Met-His-Lys Cys-Val-Trp-Asn-Lys-Ile Arg-Ser-Met-Leu-Gln-AsnDetermine the isoelectric point of the peptide product of the mutated sequence: 5' - AUG UCC AUG AUU CUG GAA AUU ACC UCC AUC AUG AAG CGC UGA CCC AUU AUU AA - 3'