1. Partial hydrolysis of lysozyme yielded an octapeptide that was later identified as the N-terminal segment of the protein. From the following information, determine the sequence of this octapeptide.
a. The following amino acids were identified after complete hydrolysis of the octapeptide:
Arginine (Arg) Glycine (Gly) Phenylalanine (Phe)
Cysteine (Cys) Leucine (Leu) Valine (Val)
Glutamic Acid (Glu) Lysine (Lys)
b. Treatment of the octapeptide with dinitroflurobenzene (Sanger reagent) followed by complete hydrolysis gave lysine (Lys) labeled with two dinitrophenyl (DNP) groups.
c. Treatment of the octapeptide with trypsin gave lysine (Lys), a tripeptide, and a tetrapeptide – the sequence of the tripeptide was found to be Cys-Glu-Leu.
d. Hydrolysis of the tetrapeptide from part c with chymotrypsin gave a two dipeptides:
Dipeptide 1: Contains arginine (Arg) and glycine (Gly)
Dipeptide 2: Contains phenylalanine (Phe) and valine (Val)
Write the sequence of amino acids in the octapeptide and explain how this sequence was determined.
2. Information about the structure of an octapeptide is given below.
a. Hydrolysis of the peptide gives the following amino acids in equal amounts: Ala, Asp, Glu, Met, Lys, Phe, Tyr, and Val.
b. Treatment of the peptide with 2,4-dinitrofluorobenzene followed by hydrolysis yields Asp labeled with a 2,4-DNP group.
c. When the peptide is treated with carboxypeptidase, the concentration of Val increases rapidly.
d. Hydrolysis of the peptide with trypsin gives a tripeptide (A) and a pentapeptide (B).
e. The tripeptide (A) is cleaved by chymotrypsin to give Phe and a dipeptide with a pI of about 5.6.
f. The pentapeptide (B) is cleaved by chymotrypsin to give a dipeptide and a tripeptide. Reaction of the tripeptide with cyanogen bromide gives Tyr.
Write the sequence of amino acids in the octapeptide and explain how this sequence was determined.
Step 1- identify the N terminal amino acid and C terminal amino acid, sangers reagent is used for the labeling of the C terminal peptide .
Step 2- identify the peptides obtained by the treatment with trypsin, cleaves at lys and arg
Step 3- identify the peptide obtained by chymotrypsin, cleaves at bulky aromatic amino acid ( phe, tyr, trp)
Step 4- compare the data obtained by different steps.
1. Partial hydrolysis of lysozyme yielded an octapeptide that was later identified as the N-terminal segment of...
Please be specific about the solution and thank you O points) Deduce the sequence of a pentapeptide that contains the amino acids Ala, Lys, Gly, Ser, and Tyr, from the following experimental data. Edman degradation cleaves Gly from the pentapeptide, and trypsin treatment gives Ala and a tetrapeptide. Treatment of the pentapeptide with chymotrypsin forms a dipeptide and a tripeptide. Partial hydrolysis forms Gly, Ser, and the tripeptide Tyr-Lys-Ala.
4. Complete hydrolysis of a heptapeptide T gave the following amino acids: 2Ảla, Glu, Leu, Lys, Phe, Val Deduce the amino acid sequence of T from the following data: i. Sequential treatment of T with 2,4-dinitrofluorobenzene (DNFB) and incomplete hydrolysis gave DNP-Val followed by DNP-Leu. Hydrolysis of T with carboxypeptidase gave Ala followed by Glu. Partial enzymatic hydrolysis of T gave a dipeptide and a tripeptide V. U with DNFB and hydrolysis gave DNP-Leu followed by DNP-Lys. 2. Complete hydrolysis...
An octapeptide contains the following amino acids: Met, Thr, Cys, Asp, Phe, Arg, Glu, Gln. Carboxypeptidase treatment of the octapeptide forms Asp and a heptapeptide. Treatment of the octapeptide with Chymotrypsin forms forms two tetrapeptides, A and B. Treatment of A with Trypsin yields two dipeptides, C and D. Edman degradation cleaves the following amino acids from each peptide: Cys (octapeptide), Cys (A), Gln (B), Cys (C), Glu(D). Partial hydrolysis of tetrapeptide B forms Gln-Thr in addition to other products....
An octapeptide contains the following amino acids: Met, Thr, Cys, Asp, Phe, Arg, Glu, Gln. Carboxypeptidase treatment of the octapeptide forms Asp and a heptapeptide. Treatment of the octapeptide with Chymotrypsin forms forms two tetrapeptides, A and B. Treatment of A with Trypsin yields two dipeptides, C and D. Edman degradation cleaves the following amino acids from each peptide: Cys (octapeptide), Cys (A), Gln (B), Cys (C), Glu(D). Partial hydrolysis of tetrapeptide B forms Gln-Thr in addition to other products....
Complete hydrolysis of an unknown basic decapeptide gives Gly, Ala, Leu, Ile, Phe, Tyr, Glu, Arg, Lys, and Ser. Problem 24-44 ConstantsI Periodic Table Complete hydrolysis of an unknown basic decapeptide gives Gly, Ala, Leu, lle, Phe, Tyr, Glu, Arg, Lys, and Ser. Terminal residue analysis shows that the N terminus is Ala and the C terminus is lle. Incubation of the decapeptide with chymotrypsin gives two tripeptides, A and B, and a tetrapeptide, C. Amino acid analysis shows that...
Styles A decapeptide has the following amina acid composition: Arg. Asp, Gly, Leu, Lys, Met, Phe, Ser. Trp, and Val Reacting the native peptide with FDNB and then hydrolyzing released 2.4- dinitrophenylvaline. Brief incubation of the native peptide with carboxypeptidase yielded free Leu. Incubation with cyanogen bromide yielded two fragments: a tetrapeptide with composition Met, Phe, Ser, and Val, and a hexapeptide. The hexapeptide yielded 2.4- dinitrophenylglycine. Proteolytic cleavage by trypsin of the native peptide gave free Leu, a tripeptide,...
3. Amino acid analysis of a peptide seven residues long gave: Asp, Leu, Lys, Met, Phe, Tyr The following facts were observed: a. Trypsin treatment had no apparent effect b. Reaction with phenylisothiocyanate reagent yielded PTH-Phe c. Treatment with chymotrypsin yielded several products, including a dipeptide and a tetrapeptide. The amino acid composition of the tetrapeptide was Leu, Lys and Met. d. Cyanogen bromide treatment yielded a dipeptide, a tetrapeptide, and free lysine. What is the...
What fragments will be obtained by a trypsin hydrolysis of the following octapeptide? Ala-Val-Trp-Lys-Phe-Gly-Arg-Met A) Ala-Val-Trp-Lys-Phe and Gly-Arg-Met 3) Ala-Val-Trp-Lys-Phe-Gly and Arg-Met - Ala-Val-Trp-Lys and Phe-Gly-Arg and Met ) Ala-Val-Trp-Lys and Phe and Gly-Arg and Met ) Ala-Val-Trp and Lys-Phe-Gly and Arg-Met Bradykinin is a nonapeptide, Arg-Pro-Pro-Gly-Phe-Ser-Pro-Phe-Arg. In addition to one mole of Arg, what peptides are present after hydrolysis of bradykinin with chymotrypsin? A) Arg-Pro-Pro and Gly-Phe and Ser-Pro-Phe B) Pro-Pro-Gly and Phe-Ser-Pro-Phe-Arg C) Arg-Pro-Pro-Gly-Phe and Ser-Pro-Phe ?) Arg-Pro-Pro-Gly-Phe-Ser...
D 8.5-10.5. 10.5 -12.5. 12.5. If you were trying to separate histidine with values of I 80 (carboxyl group), 6.00 (R group 9.20 (amino group) and lysine with pk of 2.18 (carboxyl group), 10.53 group). 8.95 (amino group) on a cation ion-exchange column chromatography, which retains positively charged analyte molecules. What would be your preferred pll of solution? 1.5. 4.0 A. B C. D. E. 80. 11.0. None of the above. Determine the amino acid sequence of the following oligopeptide...
Deduce the sequence of a heptapeptide that contains the amino acids Met, Gln, Phe, Leu, Cys, Val, and Arg, from the following experimental data. Edman degradation cleaves Cys from the heptapeptide, and carboxypeptidase forms Gln and a hexapeptide. Treatment of the heptapeptide with chymotrypsin forms a hexapeptide and a single amino acid. Treatment of the heptapeptide with trypsin forms a pentapeptide and a dipeptide. Partial hydrolysis forms Gln, Cys, Phe, and the tripeptides Leu-Met-Val and Met-Val-Arg. Be sure to answer...