Amino Acids, Peptides, and Protein Structure (5D)
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MCAT Chemical and Physical Foundations of Biological Systems › Amino Acids, Peptides, and Protein Structure (5D)
A 55-residue intracellular protein binds a small anionic metabolite. The binding site includes a Lys side chain that forms a key electrostatic interaction with the ligands carboxylate. At pH 10.5 the protein retains its folded structure by CD but shows markedly reduced binding. Which environmental interpretation best explains the reduced binding at high pH?
The ligand carboxylate becomes protonated at high pH, eliminating its negative charge
High pH converts peptide bonds into esters, preventing ligand binding
Lys side chains become deprotonated at high pH, weakening electrostatic attraction to the ligand
High pH strengthens salt bridges by increasing protonation of basic residues
Explanation
This question tests protein-ligand interactions, emphasizing pH effects on electrostatic binding via ionizable residues. Lysine's positive charge at physiological pH forms salt bridges with anionic ligands; deprotonation at high pH neutralizes this. The vignette shows reduced binding at pH 10.5 despite retained fold. Lys deprotonation at high pH weakens electrostatic attraction to the ligand. A distractor suggesting ligand protonation at high pH confuses acidification with basification. For related queries, examine pKa and charge states. A method is to predict binding changes by tracking ionization shifts in key residues.
A 20-residue peptide is synthesized with the sequence Ac-Ala-Lys-Gly-Asp-Ser-Leu-Val-Phe-Gly-Lys-Leu-Ala-Asp-Gly-Ser-Val-Leu-Lys-NH$_2$. The peptide is designed to be net neutral near physiological pH due to N-terminal acetylation and C-terminal amidation. At pH 7.4, which alteration to the amino acid sequence would most likely increase the peptides net positive charge?
Replace Ser at position 6 with Thr
Replace Lys at position 2 with Gln
Replace Asp at position 5 with Asn
Replace Leu at position 7 with Ile
Explanation
This question assesses peptide charge properties, specifically how substitutions alter net charge at physiological pH. Amino acids contribute charge based on side-chain pKa; replacing acidic Asp with neutral Asn removes a negative charge. The vignette describes a net neutral peptide design, querying charge-increasing alterations. Replacing Asp with Asn eliminates the negative charge, increasing net positive. A distractor like Lys to Gln removes a positive, decreasing net positive instead. For analogous questions, calculate net charges pre- and post-mutation. Predict effects by comparing pKa and charge contributions of residues.
A cytosolic enzyme active site contains Asp, His, and Ser residues arranged in close proximity, and mutagenesis shows that replacing Ser with Ala abolishes catalysis while substrate binding remains measurable. The enzyme functions at pH 7.4 in aqueous solution. Given the described active site, which role is most consistent with Ser in catalysis?
Ser provides a nucleophilic hydroxyl that can attack an electrophilic substrate center
Ser donates a permanent negative charge to electrostatically repel the substrate
Ser disrupts backbone hydrogen bonding to increase active-site flexibility
Ser forms a disulfide bond with His to stabilize the active site
Explanation
This question assesses enzyme active-site function, particularly the catalytic roles of residues like serine in nucleophilic attacks. Serine provides a hydroxyl group that, when activated, acts as a nucleophile in mechanisms like those in serine proteases. The vignette highlights Ser's necessity for catalysis but not binding, with Asp and His nearby suggesting a triad. Ser's nucleophilic hydroxyl attacks electrophilic substrate centers, enabling catalysis. A distractor proposing disulfide formation with His ignores Ser's lack of thiol and misassigns its role. To solve similar problems, identify residue functions based on side-chain chemistry. Predict mutation effects by evaluating loss of specific catalytic capabilities like nucleophilicity.
A peptide is synthesized with the sequence Acb2Glyb2Lysb2Aspb2Pheb2NH$_2$ and studied at pH 7.4. The N-terminus is acetylated and the C-terminus is amidated, eliminating terminal charges. Which modification would most likely increase the propensity for an intramolecular salt bridge within this peptide at pH 7.4?
Replace Lys with Arg
Replace Asp with Asn
Replace Phe with Leu
Replace Gly with Ala
Explanation
This question tests understanding of salt bridge formation and amino acid ionization states at physiological pH. At pH 7.4, lysine (pKa ~10.5) is positively charged while aspartate (pKa ~3.9) is negatively charged, creating potential for electrostatic attraction. Replacing lysine with arginine maintains the positive charge (Arg pKa ~12.5) while potentially strengthening the interaction due to arginine's more distributed charge and greater propensity for salt bridge formation. The other modifications either remove charges (Asp→Asn eliminates negative charge) or don't affect charged residues (Phe→Leu, Gly→Ala are both uncharged). Students often forget to consider pKa values when predicting ionization states at specific pH values. When analyzing potential for electrostatic interactions, always verify that the residues involved are actually charged at the pH of interest, and remember that Arg generally forms stronger salt bridges than Lys due to its resonance-stabilized guanidinium group.
A 60-residue peptide hormone is stored in secretory granules at pH 5.2 and released into blood at pH 7.4. The peptide contains two Cys residues that form an intramolecular disulfide bond in the oxidizing lumen of the granule. In vitro, the reduced (disulfide-free) peptide shows markedly lower thermal stability and increased susceptibility to proteolysis, despite similar amino acid composition. Given the described structure, which interaction stabilizes the protein most directly in the oxidized form?
c0c0 stacking between cysteine side chains that stabilizes tertiary structure in aqueous solution
A covalent Sb3Sb3 bond between cysteine side chains that constrains the peptides conformational entropy
Ionic pairing between cysteine side chains and nearby acidic residues that increases at higher pH
Hydrogen bonding between cysteine thiols and backbone carbonyls that locks in an b1-helix
Explanation
This question tests knowledge of disulfide bond formation and its role in protein stability. Disulfide bonds form between cysteine residues through oxidation of their thiol groups, creating covalent S-S linkages that dramatically constrain conformational flexibility. The oxidizing environment of secretory granules promotes disulfide formation, while the reduced form lacks this covalent constraint, explaining the decreased stability and increased protease susceptibility. The correct answer identifies the covalent nature of disulfide bonds and their entropy-reducing effect. Option B incorrectly describes hydrogen bonding between thiols and carbonyls, while option C wrongly attributes ionic character to cysteine. When analyzing protein stability, recognize that disulfide bonds provide the strongest structural constraints among common post-translational modifications, particularly important for extracellular proteins.
A globular enzyme contains a short b2-turn that positions a catalytic Ser side chain. The turn is stabilized by a Pro at position i+1 and a Gly at position i+2 (relative to the turn start). A point mutation replaces Gly(i+2) with Val. Enzyme activity decreases substantially, while mass spectrometry confirms the protein is full-length and properly translated. What is the most likely consequence of the mutation described?
Increased flexibility of the turn due to Vals additional rotatable bonds, improving catalytic positioning
Unchanged local structure because Gly and Val are both nonpolar and therefore interchangeable in turns
Disrupted turn geometry because Val is bulkier and less conformationally permissive than Gly, misaligning the catalytic Ser
Enhanced stability of the turn because Val forms an additional backbone hydrogen bond not possible for Gly
Explanation
This question examines how specific amino acids enable tight turns in protein structures. Glycine lacks a side chain (R = H), providing exceptional conformational flexibility needed for sharp turns, while valine's branched side chain creates steric clashes in the restricted geometry of β-turns. The Pro-Gly combination is particularly common in turns because Pro restricts the backbone to turn-promoting angles while Gly provides the flexibility to complete the turn. The correct answer recognizes that Val's bulk prevents adoption of the φ,ψ angles required for the turn, misaligning the catalytic serine. Option A incorrectly attributes flexibility to Val, while option D wrongly suggests Val can form additional backbone hydrogen bonds. When analyzing turn structures, remember that Gly and Pro have unique conformational properties that often make them irreplaceable in tight turns and loops.
A peptide therapeutic is formulated for subcutaneous injection. During accelerated stability testing, deamidation is observed at an Asn-Gly sequence, and the product shows reduced potency. The formulation team considers substituting the glycine to reduce local backbone flexibility and limit access of water to the susceptible amide. Which alteration to the amino acid sequence would most likely reduce deamidation while minimally perturbing overall peptide size and polarity?
Replace Gly with Lys at the Asn-Gly site
Replace Gly with Ala at the Asn-Gly site
Replace Gly with Trp at the Asn-Gly site
Replace Gly with Asp at the Asn-Gly site
Explanation
This question tests understanding of deamidation mechanisms and rational peptide engineering to improve stability. Asn-Gly sequences are particularly susceptible to deamidation because glycine's flexibility allows the backbone to adopt conformations that facilitate nucleophilic attack by the following peptide bond nitrogen on asparagine's side chain amide. Replacing glycine with alanine (choice A) introduces a small methyl side chain that restricts backbone flexibility while maintaining similar size and polarity, reducing deamidation rates without significantly altering peptide properties. Choices B and C introduce charged residues (Asp negative, Lys positive) that would dramatically change local polarity and potentially affect biological activity. Choice D introduces tryptophan, a large aromatic residue that would significantly alter peptide size and hydrophobicity. When designing stability-enhancing mutations, choose substitutions that address the mechanistic cause (here, excessive flexibility) while minimizing changes to other critical properties like charge, size, and hydrophobicity.
A laboratory studies peptide bond stability under different chemical environments relevant to digestion and cellular homeostasis. In aqueous solution, the peptide bond has partial double-bond character due to resonance, restricting rotation and favoring a planar amide. Under strongly acidic conditions, proteolysis rates increase for many proteins even without changes in enzyme concentration. Based on the vignette, how does the environment most likely influence peptide bond cleavage susceptibility?
Low pH strengthens amide resonance, making the peptide bond more resistant to hydrolysis
Low pH increases carbonyl electrophilicity via protonation, facilitating nucleophilic attack during hydrolysis
Low pH deprotonates the amide nitrogen, eliminating planarity and preventing cleavage
Low pH converts peptide bonds into disulfide bonds, increasing backbone rigidity
Explanation
This question tests understanding of peptide bond chemistry and how pH affects hydrolysis susceptibility. The peptide bond exhibits partial double-bond character due to resonance between the C=O and C-N bonds, creating a planar, rigid structure. Under strongly acidic conditions, protonation of the carbonyl oxygen increases the electrophilicity of the carbonyl carbon, making it more susceptible to nucleophilic attack by water during hydrolysis (choice A). This acid-catalyzed mechanism explains increased proteolysis rates at low pH even without enzymes. Choice B is incorrect because protonation weakens, not strengthens, resonance stabilization. Choice C is false because peptide bonds cannot convert to disulfide bonds, which form only between cysteine residues. Choice D is incorrect because the amide nitrogen is already protonated under normal conditions and further protonation at low pH would actually disrupt planarity. When analyzing pH effects on peptide stability, consider that extreme pH conditions can either facilitate hydrolysis (acid/base catalysis) or denature proteins by disrupting electrostatic interactions.
A membrane-associated receptor-binding protein contains a short b2-hairpin that presents a conserved Gly-Pro motif at the tip of a surface loop. A point mutation replacing glycine with valine (Gb2V) is introduced at the loop tip. Binding assays show a large decrease in affinity, while size-exclusion chromatography indicates the protein remains monomeric and similar in overall size. What is the most likely consequence of the mutation described?
Reduced local backbone flexibility at the loop tip, impairing the loop conformation required for binding
Global denaturation driven by disruption of all peptide bonds adjacent to the mutated residue
Increased formation of b1-helices at the loop tip because valine is a strong helix breaker
Elimination of a salt bridge because glycine is normally positively charged at physiological pH
Explanation
This question tests understanding of how amino acid properties affect local protein structure, particularly in flexible loop regions. Glycine lacks a side chain beyond hydrogen, providing exceptional backbone flexibility that allows tight turns and unusual conformations often found at loop tips. The Gly-Pro motif is particularly important for creating sharp turns in β-hairpins. Replacing glycine with valine introduces a branched, bulky side chain that restricts backbone rotation through steric clashes, preventing the loop from adopting its binding-competent conformation (choice A). Choice B is incorrect because valine is actually a β-sheet favoring residue, not a helix breaker, and wouldn't promote helix formation at a loop tip. Choice C is false because glycine is not ionizable and cannot form salt bridges. Choice D is incorrect because single amino acid substitutions don't break peptide bonds or cause global denaturation in stable proteins. To analyze loop mutations, consider that glycine and proline have unique conformational properties: glycine provides maximum flexibility while proline restricts rotation, and substituting either can dramatically affect local structure.
A calcium-binding protein contains an EF-hand-like loop in which a glutamate side chain coordinates Ca$^{2+}$ through a carboxylate oxygen. In a mutant, this glutamate is replaced by leucine (Eb2L). In vitro, the mutant shows substantially reduced Ca$^{2+}$ affinity and altered thermal stability, while overall secondary structure content remains similar by circular dichroism. What is the most likely consequence of the mutation described?
Loss of a negatively charged ligand needed for Ca$^{2+}$ coordination, reducing binding affinity
Increased Ca$^{2+}$ affinity because hydrophobic side chains preferentially chelate divalent cations
Creation of a new salt bridge with Ca$^{2+}$ because leucine is positively charged at pH 7.4
Gain of a stronger ionb4dipole interaction because leucine is more polar than glutamate
Explanation
This question tests understanding of metal coordination in proteins and how amino acid properties determine binding capability. EF-hand motifs coordinate calcium through oxygen atoms from acidic residues (Asp, Glu) and backbone carbonyls, with the negative charges stabilizing the divalent cation. Glutamate's carboxylate side chain provides negatively charged oxygen atoms ideal for Ca²⁺ coordination. The E→L mutation replaces this charged, oxygen-containing side chain with leucine's nonpolar, aliphatic group that cannot coordinate metal ions (choice A). This loss of a critical ligand reduces calcium affinity and alters protein stability. Choice B is incorrect because leucine is hydrophobic, not polar. Choice C is false because leucine is nonpolar and uncharged at any physiological pH. Choice D contradicts coordination chemistry principles as hydrophobic residues cannot chelate metal ions, which require electron-donating atoms like oxygen, nitrogen, or sulfur. When analyzing metal-binding sites, identify residues with lone pair electrons (Asp, Glu, His, Cys) that can serve as ligands, and consider how mutations affect the coordination geometry and charge complementarity.