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

Short Answer

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Resonance forms of protonated guanidino group

Strongly basic isoelectric point of arginine is due to the high basicity of the guanidino group which is due to the resonance stabilisation in its protonated form.

Step by step solution

01

Step-1. Resonance forms of protonated guanidino group:

Protonation of guanidino group gives a resonance-stabilized cation which has all octets filled and positive charge delocalised over three nitrogen atoms. Delocalisation of charge and filled octets brings stability to the system. Though, positive charge on electronegative atom is not stable as in this case, positive charge is at nitrogen atom but nitrogen has filled octet in all resonating structures, so it is stable.

Resonance forms of protonated guanidino group

02

Step-2. Reason for arginine’s strongly basic isoelectric point:

The isoelectric point of arginine is 10.76 which is highest among amino acids. This is primarily due to higher basicity of the guanidino group which is attached to the side chain in arginine. Protonated form of guanidino group is resonance stabilised which increases the basicity of guanidino group and hence, isoelectric point of arginine.

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Most popular questions from this chapter

Show how you would use a Strecker synthesis to make

  1. Leucine (b) Valine (c) Aspartic acid

Answer

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

Peptides often have functional groups other than free amino groups at the N terminus and other than carboxyl groups at the C terminus.

(a) A tetrapeptide is hydrolyzed by heating with 6 M, and the hydrolysate is found to contain Ala, Phe, Val, and Glu. When the hydrolysate is neutralized, the odor of ammonia is detected. Explain where this ammonia might have been incorporated in the original peptide.

(b) The tripeptide thyrotropic hormone releasing factor(TRF) has the full name pyroglutamylhistidylprolinamide. The structure appears here. Explain the functional groups at the N terminus and at the C terminus.

(c)On acidic hydrolysis, an unknown pentapeptide gives glycine, alanine, valine, leucine and isoleucine. No odor of ammonia is detected when the hydrolysate is neutralized. Reaction with phenyl isothiocyanate followed by mild hydrolysis gives nophenylthiohydantoin derivative. Incubation with carboxypeptidase has no effect. Explain these findings.

  1. Show how you would use a Strecker synthesis to make phenylalanine.
  2. Propose a mechanism for each step in the synthesis in part (a).

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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