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Chapter 24: Q20P (page 1282)

Show the third and fourth steps in the sequencing of oxytocin. Use figure 24-14 as a guide.

Short Answer

Expert verified

The most efficient method for sequencing peptides is Edman degradation. In this process, a peptide is treated with phenyl isothiocyanate that is followed by hydrolysis. The products formed from the degradation process are the peptide chain which is a shortened one and a heterocyclic derivative of the terminal amino acid known as phenylthiohydantoin.

Step by step solution

01

Edman degradation

The most efficient method for sequencing peptides is Edman degradation. In this process, a peptide is treated with phenyl isothiocyanate that is followed by hydrolysis. The products formed from the degradation process are the peptide chain which is a shortened one and a heterocyclic derivative of the terminal amino acid known as phenylthiohydantoin.

02

 First two steps in sequencing of oxytocin

The first two steps in sequencing oxytocin involves Edman degradation which cleaves the terminal amino acid and as a result of it, phenylthiohydantoin derivative is formed. The shortened peptide is readily available for the next steps.

03

 Third and fourth steps in sequencing of oxytocin

The third and fourth step in the sequencing of oxytocin are represented by the figures as shown below.

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Most popular questions from this chapter

Draw the resonance forms of a protonated guanidino group and explain why arginine has such a strongly basic isoelectric point.

Complete hydrolysis of an unknown basic decapeptide gives Gly, Ala, Leu, Ile, Phe, Tyr, Glu, Arg, Lys, and Ser. Terminal residue analysis shows that the N terminus is Ala, and the C terminus is Ile. Incubation of the decapeptide with chymotrypsin gives two tripeptides, A and B, and a tetrapeptide, C. Amino acid analysis shows that peptide A contains Gly, Glu, Tyr, and; peptide B contains Ala, Phe, and Lys; and peptide C contains Leu, Ile, Ser, and Arg.Terminal residue analysis gives the following results.

Incubation of the decapeptide with trypsin gives a dipeptide D, a pentapeptide E, and a tripeptide F. Terminal residue analysis of F shows that the N terminus is Ser and the C terminus is Ile. Propose a structure for the decapeptide and for fragments A through F.

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(a) Explain why an NHS ester is much more reactive than a simple alkyl ester.

(b) Propose a mechanism for the reaction shown.

(c) Propose a mechanism for the reaction of the NHS ester with an amine, R-NH2


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