A peptide toxin produced by a red spider was sequenced to study itspotential medical use. Sequencing produced the following results:• Peptide composition is phe x4, arg x1, val x1, glu x1• Treatment of the intact peptide with chymotrypsin gave 2 aminoacids, a dipeptide and a tripeptide.• Streptococcal protease treatment cleaved the peptide to a tripeptideand a tetrapeptide of composition arg1phe2val1.• When the peptide toxin was treated with trypsin, a hexapeptide andphe were obtained.Phe-arg-val-glu-phe-phe-phePhe-glu-phe-phe-val-arg-phePhe-phe-glu-phe-val-arg-pheArg-phe-val-phe-glu-phe-pheVal-phe-phe-glu-phe-arg-phe

Peptides are macro molecules made up of Aminoacids joined by peptide bond. Dipeptide is made up of…

Answered: A peptide toxin produced by a red…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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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—Cys
The sequence of a 29 aa long peptide can be determined from the following data: Treatment of the peptide with dansyl chloride reveals that the amino-terminal is Val. Trypsin digestion, separation of peptides, and Edmann technique give the sequences for peptide fragments as follows: T-1 V-G-A-H-A-G-E-Y-G-A-E-A-T-E T-2 A-A-W-G-KT-3 V-L-S-P-A-K T-4 T-N-V-K
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 cleaved into peptides by treatment with trypsin and cyanogen bromide, and then the peptides are purified and sequenced. The sequences of the peptides are shown below. Trypsin peptides Cyanogen bromide peptides Based on sequences of the overlapping peptides generated by treatment with trypsin and cyanogen bromide (shown above), which of the following peptides represents the N-terminus of the original polypeptide? A. T-1 B. T-4 C. C-2 D. C-4 T-1 FENYAT-2 ELIMVPKT-3 NFEEGSKT-4 ITGLAIHQKT-5 GSMDPVALMTLR C-1 DPVALMC-2 VPKGSMC-3 ITGLAIHQKELIMC-4 TLRNFEEGSKFENYA
A peptide was analyzed with the following results. Whatis the primary sequence of the peptide? Briefly state the conclusions of each experimental result and indicate your logic. a)Mass spectrometry was consistent with a peptide of 7 amino acids containing W,C,D,M,N,K, and I. b)One cycle of Edman degradation of the peptide yielded a single product, the PTH derivative of I c)Cleavage of the peptide with CNBr yielded two products. One contained 5 amino acids, and the other 2. One cycle of Edman degradation yielded the PTH derivatives of Iand N, respectively. d)Treatment of the peptide with trypsin yielded 2 products. One of them had significant light absorbance at 280 nm, and Edman degradation yielded the PTH derivative of I. The other trypsin-generated product contained 4 amino acids, and Edman degradation yielded the PTH derivative of Asp.
VII. Analysis of a peptide antibiotic purified from a strain of Bacillus brevii resulted in the following significant information: Complete hydrolysis of the peptide yielded equimolar amounts of Leu, Orn, Phe, Pro, and Val Molecular weight of the peptide was estimated to be aroung 1,200 Da The peptide failed to undergo hydrolysis when treated with carboxypeptidase Partial hydrolysis and then chromatographic separation of products yielded the following di- and tripeptides: Leu-Phe, Phe-Pro, Orn-Leu, Val-Orn, Val-Orn-Leu, Phe-Pro-Val, Pro-Val-Orn Treament with dinitrofluorobenzene (DNFB) followed by complete hydrolysis yielded only free amino acids and the DNFB-derivatized ornithine at its side chain Using this set of information available to you, deduce the amino acid sequence of this peptide antibiotic. Write your final peptide sequence only using 3-letter abbreviations.
"Regarding resolution of mass spectrometers and various ways mass of peptides can be represented, which is or are true? " "A larger peptides will have a smaller difference between the monoisotopic and average masses, compared to smaller peptides, because the contribution of 13C is greater. " Higher resolution allows investigators to distinguish between monoisotopic masses and other masses of otherwise identical peptides. "Average mass size is always smaller than monoisotopic mass, because 13C and 15N contribute to monoisotopic mass. " "Monoisotopic mass is calculated as if there are no heavy isotopes within the peptide, and so is always smaller than the average observed mass."
The sequence of kassinin, a tachykinin dodecapeptide from the African frog Kassina senegalensis, was determined. A single round of Edman degradation identifies Asp as the N-terminus. A second sample of the peptide is treated with chymotrypsin. Two fragments are released with the following amino acid compositions: fragment I (Gly, Leu, Met, Val) and fragment II (Asp2,Gln, Lys, Phe, Pro, Ser, Val). Next, a third sample of the peptide is treated with trypsin, which results in two fragments with the following amino acid compositions: fragment III (Asp, Pro, Lys, Ser) and fragment IV (Asp, Gln, Gly, Leu, Met, Phe, Val2). Treatment of another sample with elastase yields a single Gly residue and three fragments—fragment V (Leu, Met), fragment VI (Asp,Lys, Pro, Ser, Val), and fragment VII, which was sequenced: Asp–Gln–Phe–Val. A fourth sample was treated with CNBr, but the dodecapeptide was not cleaved.
A peptide has the following amino acid composition:2 Met, 2 Phe, 2 Glu, 1 Arg, 1 Lys, 1 Val, 1 Leu, 1 Gly, 1 SerReaction of the intact peptide with Sanger’s Reagent followed by acid hydrolysis creates a derivative of Met. A specific cleavage of the intact peptide using chymotrypsin produces fragments with the following sequences:Fragment A: Glu-Gly-Lys-Phe Fragment B: Met-Ser-Leu-Arg Fragment C: Met-Val-Glu-Phe What is the polypeptide sequence? (You may use the three-letter symbol for each amino acid in the sequence).
Chymotrypsin digestion, separation of peptides, and Edmann technique give the sequences for peptide fragments as follows: C-1 G-A-E-A-T-E C-2 G-K-V-G-A-H-A-G-E-Y C-3 V-L-S-P-A-K-T-N-V-K-A-A-W What is the sequence for the peptide? NOTE: Label the fragments when reconstructing the peptide chain e.g. for trypsin T-1, T-2 etc...
What method could be used to make an immobilised library of 6 different peptides, all differing by one amino acid? E.g. AMFMDE AAFMDE AMAMDE AMFADE AMFMAE AMFMDA
Sir Bong wanted to determine the a amino acide sequence og BONG. The summary of the analyses performed to the peptide is attached. 1. The three letter designation for the N terminal amino acid residue of CUTE is? 2. The three letter designation for the N terminal amino acid residue of POGI is? 3.The three letter designation for the N terminal amino acid residue of BONG is? 4. The ONE letter designation for the N terminal amino acid residue of CUTE is? 5. The ONE letter designation for the N terminal amino acid residue of POGI is? 6. The ONE letter designation for the N terminal amino acid residue of BONG is? 7.The ONE letter designations for the fragments produced upon treatment of POGI with chymotrypsin are? 8. The ONE letter designations for the fragments produced upon treatment of BONG with thermolysin are?

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