You have been given an unknown low weight polypeptide to sequence. Luckily, you have a plentiful supply. You aregiven a limited amount of Edmond's reagent-just enough to perform 1 restricted digestion for every fragment that yougenerate but not for complete digestion of every fragment. You have the following enzymes and reagents available to doyour digestion analysis: Chymotrypsin, V8, HCI (1M), Cyanogen Bromide (CnBr) and 50mM of ammonium bicarbonate.You are only given enough Carboxypeptidase for one fragment analysis. You fortunately have a plentiful supply of ionchromatography columns, eluting buffers, and SDS-PAGE gels to help you separate individual amino acids for analysisand fragment size (length) determination. And you have a really good spec and an HPLC. From the information givenbelow, you will need to construct the sequence of your peptide!You also have the following information:1. Digestion of the intact polypeptide with 1M HCl will liberate all amino acids (you get a solution of free aminoacids)2. Elution via ion chromatography allows for separation of individual amino acids.3. Identification of individual amino acids is via spectrophotometry and HPLC.4. SDS-PAGE facilitates fragment separation; and using a standard allows you to know the size of each fragment.Remember that using SDS-PAGE the shorter fragments move faster!5. Edmond's releases the N-terminal amino acid from a polypeptide6. Carboxypeptidase releases the Carboxyl-terminal amino acid.7. Cyanogen Bromide cleaves on the carbonyl side of Met8. V8 cleaves on the carbonyl side of Glu* or Asp (*only when50mM of ammonium bicarbonate is added).9. Chymotrypsin cleave on the carbonyl side of Phe, Trp, and Tyr.These are your results:I. Using an aliquot of your intact peptide, you completely digest it with 1M HC1. You elute your sample over an ionchromatography column collecting individual amino acids that you then analyze for molar proportions (how manyof each amino acid) and also you then analyze via HPLC/spec for the identity of the amino acids. Your Resultsare as follows:a. Your peptide is 12 amino acids long.b. Your molar ratios are 1:1:1:1:1:1:1:1:1:1:2c. The identity of the amino acids in your peptide are: Ala, Asn, Asp, His, Leu, Lys, Met, Phe, Thr, Trp,Val (at this point you do not know the order or the amount of each)You then perform a restricted Edmond's degradation on an aliquot of intact peptide and it yields an 11 amino acidfragment and Leu.You then perform a Carboxypeptidase digest on another aliquot of your peptide and it yields an 11 amino acidfragment and LysYou take a deep breath.You perform a CnBr digest on another aliquot of your intact peptide and it yields three fragments:a. CnBr Frag 1 is 2 amino acids longb. CnBr Frag 2 is 4 amino acids longc. CnBr Frag 3 is 6 amino acids longYou perform a restricted Edmond's on each of your CnBr fragments:a. CnBr Frag 1 yields a free Thr and 1 residueb. CnBr Frag 2 yields a free Ala and a 3 residue peptidec. CnBr Frag 3 yields a free Leu and a 5 residue peptide.You perform a Chymotrypsin digest on another aliquot of your intact peptide and it yields three fragments:a. Chym Frag 3 is 6 amino acids longII.III.IV.V.VI.VII. b. Chym Frag 2 is 4 amino acids longc. Chym Frag 1 is 2 amino acids longYou perform a restricted Edmond's on each of your Chym fragments:a. Chym Frag 3 yields a free His and a 5 residue peptide.b. Chym Frag 2 yields a free Leu and a 3 residue peptidec. Chym Frag 1 yields an identified Val and 1 residueBreathe againYou perform a V8 digest on another aliquot of your intact peptide and it yields two fragments:a. V8 Frag 1 is 3 amino acids longb. V8 Frag 2 is 9 amino acids longYou perform a restricted Edmond's on each of your V8 fragments:a. V8 Frag 1 yields a free Leu and a dipeptideb. V8 Frag 2 yields a free Phe and an 8 residue peptide.You have now gotten about all the information you need but you decide to perform 3 more digestons.On CnBr Frag 3 you perform a Chymotrypsin digesta. ChymCnBr3 Frag 1 is 4 amino acids long and an Edmond's yields a free Leub. ChymCnBr3 Frag 2 amino acids long and an Edmond's yields a free HisOn Chym Frag 3 you perform a CnBr digesta. CnBrChym3 Frag 1 is 4 amino acids long and an Edmond's yields a free Hisb. CnBrChym3 Frag 2 is 2 amino acids long and an Edmond's yields a free AlaOn V8 Frag 2 you perform a CnBr digesta. A. CnBr V8 2 Frag 1 is 3 amino acids long and an Edmonď's yields a free Pheb. CnBr V82 Frag 2 is 2 amino acids long and an Edmond's yields a free Thrc. CnBr V82 Frag 3 is 4 amino acids long and an Edmond's yields a free AlaCan you determine the sequence of your peptide from the results of your experiment? Write the sequence below:VIIIIX.X.XI.XII.XII.XIV.XV.XVI.1.345 678 910 11 12XVII.If you could do one more digestion what would you use and what would you do it (fragment or intact peptide)would you use. It has to be one of the enzymes you have in stock.

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Proteins are composed of amino acids, which are bound together by peptide linkage. Amino acids…