Draw the structure of the predominant form of

(a)Isoleucine at pH 11 (b) Proline at pH 2

(c)Arginine at pH 7 (d) Glutamic acid at pH 7

A mixture of alanine, lysine, and aspartic acid at 1). pH 6 ; 2). pH 11; 3). pH 2

Short Answer

Expert verified

(a)

Isoleucine at pH 11

(b)

Proline at pH 2

(c)

Arginine at pH 7

(d)

Glutamic acid at pH 7

(e)

(i)

Alanine at pH 6

(ii)

Lysine at pH 11

(iii)

Aspartic acid at pH 2

Step by step solution

01

Amino acids

Amino acids are defined as compounds containing at least one amino group and one carboxylic group. The pH at which a particular amino acid does not migrate under the influence of an electric field is known as the isoelectric point.

02

Isoelectric point

Amino acids exist in their neutral form at the isoelectric point. The amine group exists as -NH3+, and the carboxyl groups exist as -COO-. However, when the pH rises above the isoelectric point, the amine group becomes -NH2, and the carboxyl groups remain the same as -COO-. Also, when the pH falls below the isoelectric point, the amine group becomes -NH3+, and the carboxyl groups become -COOH.

(a)

The isoelectric point of isoleucine is 6.02. Therefore, the predominant form of isoleucine at pH=11can be represented as shown below:

Isoleucine at pH 11

(b)

The isoelectric point of proline is 6.30. Therefore, the predominant form of proline at pH=2can be represented as shown below:

Proline at pH 2

(c)

The isoelectric point of arginine is 10.8. Therefore, the predominant form of arginine at pH=7can be represented as shown below:

Arginine at pH 7

(d)

The isoelectric point of glutamic acid is 3.2. Therefore, the predominant form of glutamic acid at pH=7can be represented as shown below:

Glutamic acid at pH 7

(e)

The isoelectric point of alanine is 6.1, that of lysine is 9.8, and that of aspartic acid is 3.0. Therefore, their predominant form at pH=6 can be represented as shown below:

Alanine at pH 6

(ii)

The isoelectric point of alanine is 6.1, that of lysine is 9.8, and that of aspartic acid is 3.0. Therefore, their predominant form at pH=11 can be represented as shown below:

Lysine at pH 11

(iii)

The isoelectric point of alanine is 6.1, that of lysine is 9.8, and that of aspartic acid is 3.0. Therefore, their predominant form at pH=2 can be represented as shown below:

Aspartic acid at pH 2

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Most popular questions from this chapter

Most naturally occurring amino acids have chiral centers (the asymmetric α carbon atoms) that are named (S) by the Cahn-Ingold-Prelog convention (Section 5-3). The common naturally occurring form of cysteine has a chiral center that is named (R), however.

(a) What is the relationship between (R)-Cysteine and (S)-alanine? Do they have the opposite three-dimensional configuration (as the names might suggest) or the same configuration?

(b) (S)-Alanine is an L-amino acid (Figure 24-2). Is (R)-cysteine a D-amino acid or an L-amino acid?

A molecular weight determination has shown that an unknown peptide is a pentapeptide, and an amino acid analysis shows that it contains the following residues: one Gly, two Ala, one Met, one Phe. Treatment of the original pentapeptide with carboxypeptidase gives alanine as the first free amino acid released. Sequential treatment of the pentapeptide with phenyl isothiocyanate followed by mild hydrolysis gives the following derivatives:

Propose a structure for the unknown pentapeptide.

Draw three-dimensional representation of the following amino acids:

(a) L-alanine

(b) L-Leucine

(c) D-serine

(d) D-glutamine

The Sanger method for N-terminus determination is a less common alternative to the Edman degradation. In the Sanger method, the peptide is treated with the Sanger reagent, 2,4-dinitrofluorobenzene, and then hydrolyzed by reaction with 6 M aqueous HCl. The N-terminal amino acid is recovered as its 2,4-dinitrophenyl derivative and identified.

(a)Propose a mechanism for the reaction of the N terminus of the peptide with 2,4-dinitrofluorobenzene.

(b) Explain why the Edman degradation is usually preferred over the Sanger method.

Show where trypsin and chymotrypsin would cleave the following peptide.

Tyr-Ile-Gln-Arg-Leu-Gly-Phe-Lys-Asn-Trp-Phe-Gly-Ala-Lys-Gly-Gln-Gln.NH2

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