At pH 6, alanine will exist as a neutral, dipolar ion: The amino group will be protonated while the carboxyl group will be deprotonated. Lowering the pH to 3 will result in protonation of the carboxyl group, so the molecule will assume an overall positive charge. Alanine will, therefore, migrate to the cathode. On the other hand, aspartic acid will exist as a neutral dipolar ion at pH 3 since this is equivalent to its isoelectric point. Therefore, when it is subjected to electrophoresis, it will not move. In summary, alanine will migrate to the cathode while aspartic acid will not move, making choice (C) the correct response.
Discussed in the section on the acid-base properties of amino acids in this chapter.
The amino acid in question is glutamic acid, which is an acidic amino acid because it contains an extra carboxyl group. At neutral pH, both of the carboxyl groups are ionized, so there are two negative charges on the molecule. Only one of the charges is neutralized by the positive charge on the amino group, so the molecule has an overall negative charge. Thus, the answer is choice (B).
At pH 3, the amine and carboxyl groups will be protonated to give a net positive charge. As the pH rises to 7, the proton will dissociate from the carboxyl, but both amine groups will still be fully protonated, so the charge will still be positive. At pH 11, the molecule is above its isoelectric point and will be fully deprotonated, resulting in two neutral amine groups and a negatively charged carboxylate group, so the charge at pH 11 will be negative. Therefore, the correct sequence of charges is positive, positive, negative, corresponding to choice (D).
Nonpolar molecules or groups are those whose negative and positive centers of charge coincide. They are not soluble in water and are thus hydrophobic. Amino acids with hydrophobic R-groups are considered hydrophobic molecules; they tend to be found buried within protein molecules where they do not have to interact with the aqueous cellular environment. Choices (A) and (D) are incorrect because nonpolar R-groups cannot be hydrophilic. Choice (C) is incorrect because nonpolar molecules are seldom located on the surface of proteins, where they would interact unfavorably with the aqueous cellular environment.
Formation of a peptide bond, which is the primary covalent bond found in proteins, involves a condensation reaction between the amine group of one amino acid and the carboxyl group of an adjacent amino acid. As a result of the carbonyl group present at the bond, the double bond resonates between C=O and C=N. This resonance gives the peptide bond a partial double-bond character and limits rotation about the bond. From this information, it can be seen that choices (A), (B), and (C) are all characteristics of the peptide bond. Choice (D) is false because the formation of the peptide bond is a condensation reaction involving the loss of water, rather than a hydration reaction, which involves the addition of water.
The 6 tripeptides that can be formed are:
Val-Ala-Leu, Val-Leu-Ala,
Ala-Val-Leu, Ala-Leu-Val,
Leu-Val-Ala, and Leu-Ala-Val.
The key word in this question is “covalent.” While hydrogen bonds and hydrophobic bonds are involved in peptide structure, they are not considered covalent bonds, since they do not involve sharing electrons. Therefore, choices (A) and (D) are incorrect. Ether bonds are covalent bonds, but they are not found in peptides. The correct answer is disulfide bonds, choice (C). Disulfide bonds are covalent bonds forming between the sulfur-bearing R-groups of cysteines. The resulting cystine molecule constitutes a disulfide bridge and often causes a loop in the peptide chain.
Discussed in the section on proteins in this chapter.
When discussing secondary structure, the most important bond is the hydrogen bond. The rigid α-helices are held together by hydrogen bonds between the carbonyl oxygen of one peptide bond and the amine hydrogen of a peptide bond four residues away. This hydrogen bond is intramolecular, so choice (A) is correct. Disulfide bonds are covalent bonds usually associated with tertiary structure; therefore, choice (B) is incorrect. Choices (C) and (D) are incorrect since the rippled effect and intermolecular hydrogen bonds are both characteristic of β-pleated sheets.
Protein denaturation involves the loss of three-dimensional structure and function. Since the three-dimensional shape of a protein is conferred by secondary and tertiary structures, denaturation disrupts these structures. Therefore, both choices (B) and (C) are correct. Denaturation does not cause a loss of primary structure since it does not cause peptide bonds to break; thus, choice (A) is incorrect.