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CHAPTER 25 1063 25.69. The following retrosynthetic analysis reveals the three amino acids (tyrosine, serine, and glycine) that are necessary for biosynthesis of the fluorophore: 25.70. If a tripeptide does not react with phenyl isothiocyanate, then it must not have a free N terminus. It must be a cyclic tripeptide. Below are the two possible cyclic tripeptides: 25.71. (a) Trypsin catalyzes the hydrolysis of the peptide bond at the carboxyl side of arginine, giving the following two fragments: Arg + Pro-Pro-Gly-Phe-Ser-Pro-Phe-Arg (b) Chymotrypsin catalyzes the hydrolysis of the peptide bonds at the carboxyl side of phenylalanine, giving the following three fragments: Arg-Pro-Pro-Gly-Phe + Ser-Pro-Phe + Arg 25.72. The R group (highlighted) in the PTH derivative indicates the identity of the N-terminal residue. Since this R group is a benzylic group (CH2Ph), the N- terminal residue must be phenylalanine. 25.73. The first Edman degradation indicates that the N-terminal residue is valine (the R group is isopropyl). The second Edman degradation indicates that the N- terminal residue of the dipeptide is alanine. And finally, the remaining amino acid (glycine) must be at the C- terminus of the tripeptide. In summary, the tripeptide is Val-Ala-Gly, drawn here: 25.74. Only one of the trypsin fragments has a C terminus that is not arginine or lysine. This fragment (which ends with threonine), must be the last fragment in the peptide sequence. The remaining three trypsin fragments can be placed in the proper order by analyzing the chymotrypsin fragments. The correct peptide sequence is: His-Ser-Gln-Gly-Thr-Phe-Thr-Ser-Asp-Tyr-Ser-Lys- Tyr-Leu-Asp-Ser-Arg-Arg-Ala-Gln-Asp-Phe-Val-Gln- Trp-Leu-Met-Asn-Thr There cannot be any disulfide bridges in this peptide, because it has no cysteine residues, and only cysteine residues form disulfide bridges. 25.75. Prior to acetylation, the nitrogen atom of the amino group is sufficiently nucleophilic to attack phenyl isothiocyanate. Acetylation converts the amino group into an amide group, and the lone pair of the nitrogen atom is delocalized via resonance, rendering it much less nucleophilic. www.MyEbookNiche.eCrater.com