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Draw a structural formula for Lys-Phe-Ala. Label the \(N\)-terminal amino acid and the C-terminal amino acid. What is the net charge on this tripeptide at \(\mathrm{pH} 6.0\) ?

Short Answer

Expert verified
Answer: The net charge of the Lys-Phe-Ala tripeptide at pH 6.0 is +1. The N-terminal amino acid is Lysine (Lys) and the C-terminal amino acid is Alanine (Ala).

Step by step solution

01

Identify the individual amino acids and their structures

First, identify the individual amino acids and their structures. Lysine (Lys), phenylalanine (Phe), and alanine (Ala) are the three amino acids in the given tripeptide. Their general structure is: Lysine: NH3+ - CH2-CH2-CH2-CH2-NH2 - COO- Phenylalanine: NH3+ - CH2 - C6H5 - COO- Alanine: NH3+ - CH3 - COO-
02

Peptide bond formation in the tripeptide

Now, we form peptide bonds between the amino acids to create the tripeptide. A peptide bond is formed between the carboxyl group of one amino acid (C-terminus) and the amino group of the next amino acid (N-terminus). The order of the amino acids in the tripeptide is Lys-Phe-Ala. Form peptide bonds between Lys-Phe and Phe-Ala to get the tripeptide structure: NH3+ - CH2-CH2-CH2-CH2-NH - CO - CH2 - C6H5 - NH - CO - CH3 - COO-
03

Identifying N- and C-terminal amino acids

The N-terminal amino acid is the one with the free amino group (NH3+), which is at the beginning of the tripeptide. In this case, the N-terminal amino acid is Lysine (Lys). The C-terminal amino acid is the one with the free carboxyl group (COO-), which is at the end of the tripeptide. In this case, the C-terminal amino acid is Alanine (Ala).
04

Calculating the net charge at pH 6.0

We calculate the net charge of the tripeptide at pH 6.0 by analyzing each ionizable group of amino acids and determining if they are protonated or deprotonated at the given pH. The ionizable groups and their pKa values are as follows: Lysine: - α-amino group (NH3+): pKa ~ 9.0 - α-carboxyl group (COO-): pKa ~ 2.2 - ε-amino group (NH2): pKa ~ 10.5 Phenylalanine: - α-amino group (NH3+): pKa ~ 9.0 - α-carboxyl group (COO-): pKa ~ 2.5 Alanine: - α-amino group (NH3+): pKa ~ 9.0 - α-carboxyl group (COO-): pKa ~ 2.3 At pH 6.0: Lysine: - α-amino group: Protonated (NH3+) - α-carboxyl group: Deprotonated (COO-) - ε-amino group: Protonated (NH3+) Phenylalanine: - α-amino group: Protonated (NH3+) - α-carboxyl group: Deprotonated (COO-) Alanine: - α-amino group: Protonated (NH3+) - α-carboxyl group: Deprotonated (COO-) Since in the tripeptide, one N-terminal amino group (Lys) and one C-terminal carboxyl group (Ala) are free, the other ionizable groups involved in peptide bond formation are not considered. The net charge of the tripeptide at pH 6.0 is: Net charge = (+1 Lys-NH3+) + (+1 Lys-ε-NH3+) + (-1 Ala-COO-) = +1 So, the net charge of the Lys-Phe-Ala tripeptide at pH 6.0 is +1.

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Most popular questions from this chapter

Describe the behavior of a mixture of glutamic acid, arginine, and valine on paper electrophoresis at \(\mathrm{pH} 6.0\).

The configuration of the chiral center in \(\alpha\)-amino acids is most commonly specified using the \(\mathrm{D}, \mathrm{L}\) convention. It can also be identified using the \(R, S\) convention (Section 3.3). Does the chiral center in L-serine have the \(R\) or \(S\) configuration?

2,4-Dinitrofluorobenzene, very often known as Sanger's reagent after the English chemist Frederick Sanger who popularized its use, reacts selectively with the \(N\)-terminal amino group of a polypeptide chain. Sanger was awarded the 1958 Nobel Prize in Chemistry for his work in determining the primary structure of bovine insulin. One of the few people to be awarded two Nobel Prizes, he also shared the 1980 award in chemistry with American chemists Paul Berg and Walter Gilbert for the development of chemical and biological analyses of DNA. Following reaction with 2,4 -dinitrofluorobenzene, all amide bonds of the polypeptide chain are hydrolyzed and the amino acid labeled with a 2,4-dinitrophenyl group is separated by either paper or column chromatography and identified. (a) Write a structural formula for the product formed by treatment of the \(N\)-terminal amino group with Sanger's reagent and propose a mechanism for its formation. (b) When bovine insulin is treated with Sanger's reagent followed by hydrolysis of all peptide bonds, two labeled amino acids are detected: glycine and phenylalanine. What conclusions can be drawn from this information about the primary structure of bovine insulin? (c) Compare and contrast the structural information that can be obtained from use of Sanger's reagent with that from use of the Edman degradation.

At what pH would you carry out an electrophoresis to separate the amino acids in each mixture? (a) Ala, His, Lys (b) Glu, Gln, Asp (c) Lys, Leu, Tyr

Distinguish between intermolecular and intramolecular hydrogen bonding between the backbone groups on polypeptide chains. In what type of secondary structure do you find intermolecular hydrogen bonds? In what type do you find intramolecular hydrogen bonding?

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