PLEASE ANSWER ABCD ASAP Draw the peptide, showing every atom, in the protonation state where every atom has zeroConsider the peptide amino terminus-glycine-valine-proline-carboxylic acid terminusA.formal charge.B.Indicate the hybridization of every non-hydrogen atom.C.protonation state at pH 7.D.pproximnately what fraction of amino terminus is in its neutral protonation state at pH7?Carboxylic acid terminus?AtPH 7, roughly what fraction of the entire peptide is in the state shown in (A)?he peptide reacts with acetyl chloride to form one main organic product. Draw the reaction,showing the mechanism using the arrow formalism.What type of reaction is this (max 2 words)?E.F.G.* At pH , how does the net charge of your product from (F) compare to that of the initial peptidein (C)?Instead of reacting the peptide with acetyl chloride, you instead add an enzyme that facilitatesmethylation of the amino terminus with S-adenosylmethionine (SAM) as the methyl donor.Draw monomethylation of the amino terminus by SAM using the arrow formalism (feel free toabbreviate any parts of the peptide not directly involved in the reaction as "peptide").What type of reaction is this (max 2 words)?Which of the following classifiers best describes the reactive carbon involved in this reaction:methyl, primary, secondary, tertiary?You allow the methylation with SAM to continue until the amino terminus is no longer1.J.K.L.nucleophilic. Draw the final product.1. At pH 7, what fraction of your final product is in the state where every atom has zero formalcharge?

Amino acids are biomolecules with an alpha carbon bonded to 4 valancies. They are ; an amine group (N-terminal ) a carboxylic acid group (C-terminal ) a Hydrogen atom a variable sidechain group.A) Structure of the peptide with all atoms having zero formal charge B) Hybridization state of all non H-atoms The non H-atoms inside brown circle are in sp3 hybridization and the non H-atoms inside green circle are in sp2 hybridization C) Predominant protonation structure at pH 7 D) Isoelectric point (pI) of the peptide=4+102=7 i.e. pH 7 is an inflation point (one which is the midpoint of the vertical component of the titration curve of the peptide ). This means almost all the peptides molecules would have been deprotonated at their C-terminal as pH reaches 7 and the deprotonation of N-terminal has only merely started at pH 7. This means at a pH of 7 almost all molecules of the peptide have a deprotonated (negatively charged) C-terminal and a protonated (positively charged) N-terminal. i.e. practically almost all of the peptide molecules have neither the C or the N-terminal as neutral. Now if you want to calculate the fraction; pKa=pH- log(A-HA) For amino terminus; 10=7- logNH2NH3+NH2NH3+=11000Fraction of peptide molecules with neutral amino terminus (i.e. NH2)=molecules with neutral amino terminustotal number of molecules =11000+1=11001 i.e. only one in every 1001 molecules of the peptide have a neutral amino terminus Similarly for carboxyl terminus; 4=7- logCOO-COOHCOO-COOH=1000Fraction of peptide molecules with neutral carboxyl terminus (i.e COOH)=11000+1=11001 i.e. only one in every 1001 molecules of the peptide have a neutral carboxyl terminus