Amino Acids, Peptides, and Protein Structure (5D)
Help Questions
MCAT Chemical and Physical Foundations of Biological Systems › Amino Acids, Peptides, and Protein Structure (5D)
A short coiled-coil protein mediates dimerization in a signaling complex. The heptad repeat places hydrophobic residues at positions that pack at the dimer interface, while charged residues are typically solvent-exposed. A single substitution introduces a charged residue at an interface position, and the dimer dissociation constant increases markedly in 150 mM NaCl. Which alteration to the amino acid sequence would most likely disrupt protein function by weakening coiled-coil association?
Replace a solvent-exposed Glu with Asp on the outer surface of the helix
Replace an interface Ile with Glu at a hydrophobic packing position
Replace a solvent-exposed Lys with Arg on the outer surface of the helix
Replace an interface Leu with Val at a hydrophobic packing position
Explanation
This question tests understanding of coiled-coil structure and the importance of hydrophobic packing at dimer interfaces. Coiled-coils feature a heptad repeat pattern where positions 'a' and 'd' typically contain hydrophobic residues that pack at the interface, while positions 'b', 'c', 'e', 'f', and 'g' are often polar or charged and solvent-exposed. Replacing an interface isoleucine with glutamate (choice A) introduces a negatively charged residue into the hydrophobic core, disrupting favorable van der Waals interactions and creating unfavorable desolvation of the charged group. The increased dissociation constant in physiological salt confirms weakened association. Choices B and C involve conservative substitutions of surface residues (Glu→Asp maintains negative charge, Lys→Arg maintains positive charge) that wouldn't significantly affect dimerization. Choice D (Leu→Val) is a conservative hydrophobic substitution that might slightly weaken packing but wouldn't cause marked disruption. To analyze coiled-coil mutations, identify whether the position is at the hydrophobic interface or solvent-exposed surface, then evaluate how the substitution affects the critical hydrophobic packing.
A soluble, globular protein contains a buried ion pair (Glu-Lys) that contributes to tertiary stability. In a physiological context of febrile temperature (40b0C), the wild-type protein remains folded. A mutant replaces the buried glutamate with glutamine (Eb2Q) without changing overall size or expression level. Differential scanning calorimetry shows a lower $T_m$ and increased aggregation propensity. Given the described structure, which interaction is most directly lost in the mutant, leading to reduced stability?
A covalent bond between Glu and Lys that normally links two secondary structure elements
A disulfide bond between Glu and Lys that prevents unfolding at elevated temperature
A hydrophobic interaction between two charged side chains that is stronger than an ion pair
A salt bridge between a negatively charged Glu and a positively charged Lys in the hydrophobic interior
Explanation
This question tests understanding of buried ion pairs and their contribution to protein stability. While hydrophobic residues typically occupy protein interiors, buried ion pairs can form when the energetic cost of desolvation is offset by strong electrostatic attraction. The E→Q mutation removes the negatively charged glutamate carboxylate, preventing salt bridge formation with the positively charged lysine (choice B). This loss of electrostatic interaction destabilizes the protein, lowering Tm and increasing aggregation propensity as hydrophobic regions become exposed. Choice A is incorrect because Glu and Lys don't form covalent bonds. Choice C is false because disulfide bonds require cysteines. Choice D contradicts physical chemistry principles as hydrophobic interactions occur between nonpolar groups, not charged residues. When evaluating buried charged residues, consider that their presence usually indicates functionally important ion pairs or catalytic sites, as the energetic penalty of burying charges is significant unless compensated by strong interactions.
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 Trp at the Asn-Gly site
Replace Gly with Ala at the Asn-Gly site
Replace Gly with Asp at the Asn-Gly site
Replace Gly with Lys 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 62-residue, secreted peptide hormone analog is engineered to resist proteolysis while retaining receptor binding. Circular dichroism indicates it remains predominantly b1-helical at pH 7.4, but its melting temperature ($T_m$) drops by 12b0C after a single substitution within the hydrophobic face of the helix. In serum-like buffer (150 mM NaCl, 37b0C), receptor affinity decreases 8-fold, consistent with partial loss of tertiary packing between two helices. Which alteration to the amino acid sequence would most likely disrupt protein function under these conditions?
Substitute a buried Phe with Tyr within the hydrophobic core, preserving aromatic character
Substitute a solvent-exposed Lys with Arg on the polar face of the helix
Substitute a buried Leu with Asp within the hydrophobic helical interface
Substitute a surface Ser with Thr in a flexible loop distant from the binding site
Explanation
This question tests understanding of how amino acid substitutions affect protein stability and function, particularly focusing on hydrophobic interactions in α-helical structures. The hydrophobic effect drives burial of nonpolar residues away from aqueous solvent, creating stable protein cores that maintain tertiary structure. In this peptide hormone, the 12°C drop in Tm and 8-fold decrease in receptor affinity indicate disruption of hydrophobic packing between helices. Substituting a buried leucine with aspartate (choice B) introduces a charged, polar residue into the hydrophobic interface, severely disrupting the favorable van der Waals interactions and forcing energetically unfavorable exposure of charged groups to the nonpolar environment. The other options involve either conservative substitutions (Lys→Arg maintains positive charge, Phe→Tyr preserves aromaticity) or changes in regions distant from critical interfaces (surface loop substitution). When analyzing protein stability questions, identify whether mutations affect buried hydrophobic regions, surface-exposed areas, or functional sites, then predict how the chemical properties of the substituted amino acids will impact local interactions.
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 deprotonates the amide nitrogen, eliminating planarity and preventing cleavage
Low pH converts peptide bonds into disulfide bonds, increasing backbone rigidity
Low pH increases carbonyl electrophilicity via protonation, facilitating nucleophilic attack during hydrolysis
Low pH strengthens amide resonance, making the peptide bond more resistant to hydrolysis
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
Elimination of a salt bridge because glycine is normally positively charged at physiological pH
Increased formation of b1-helices at the loop tip because valine is a strong helix breaker
Global denaturation driven by disruption of all peptide bonds adjacent to the mutated residue
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 small, single-domain protein (90 residues) is studied to model folding in the crowded cytosol. In dilute buffer at 25b0C, it folds cooperatively to a native state dominated by a hydrophobic core. In 6 M urea, the protein becomes unfolded; when urea is rapidly diluted to 0.5 M, folding is slower if 10 mM dithiothreitol (DTT) is present, despite no change in primary sequence. The protein contains two cysteines that are close in the native structure. Based on the vignette, how does the environment influence protein folding?
DTT slows folding by reducing a stabilizing disulfide bond that would otherwise form during refolding
Urea unfolds the protein primarily by forming covalent bonds with peptide backbones
Residual urea accelerates folding by strengthening hydrophobic interactions in the core
DTT slows folding by protonating cysteine thiols, preventing hydrophobic collapse
Explanation
This question tests understanding of protein folding mechanisms and the role of disulfide bonds in stabilizing native structures. DTT is a reducing agent that keeps cysteine residues in their reduced (thiol) form, preventing disulfide bond formation between the two cysteines that are close in the native structure. During normal folding, these cysteines would form a disulfide bond that stabilizes the folded state and accelerates the folding process by reducing conformational entropy. With DTT present, this stabilizing crosslink cannot form, resulting in slower folding kinetics (choice A). Choice B is incorrect because DTT reduces rather than protonates thiols, and protonation wouldn't prevent hydrophobic collapse. Choice C contradicts the known effect of urea as a denaturant that disrupts hydrophobic interactions rather than strengthening them. Choice D is false because urea acts through non-covalent interactions with the protein backbone and side chains. When analyzing folding experiments, consider how chemical additives affect specific interactions: reducing agents prevent disulfide formation, denaturants disrupt hydrophobic and hydrogen bonding interactions, and crowding agents can accelerate folding through excluded volume effects.
A 12-residue peptide is used as a model substrate for proteases in simulated gastric fluid (pH 1.5) versus cytosolic-like buffer (pH 7.4). The peptide contains one Asp and one Lys separated by three residues, and it adopts a compact conformation at neutral pH by intramolecular electrostatic attraction. At pH 1.5, the compactness is lost and the peptide becomes more extended by hydrodynamic radius measurements. Which interaction change most plausibly explains the pH-dependent structural difference?
Deprotonation of Lys at low pH increases negative charge, strengthening attraction to Asp
Hydrophobic interactions are eliminated at low pH because nonpolar residues become ionized
Formation of a new disulfide bond between Asp and Lys at low pH compacts the peptide
Protonation of Asp reduces negative charge, weakening an intramolecular salt bridge with Lys
Explanation
This question tests understanding of pH-dependent ionization and electrostatic interactions in peptides. At pH 7.4, aspartate (pKa ~3.9) carries a negative charge while lysine (pKa ~10.5) is positively charged, allowing attractive electrostatic interactions that compact the peptide. At pH 1.5, which is below the pKa of aspartate, the carboxyl group becomes protonated and neutral, eliminating the negative charge needed for the salt bridge with lysine (choice A). This loss of electrostatic attraction allows the peptide to adopt a more extended conformation. Choice B is incorrect because lysine remains protonated and positively charged at low pH (pH < pKa). Choice C is false because disulfide bonds form between cysteines, not Asp-Lys pairs. Choice D is incorrect because hydrophobic residues don't ionize and their interactions persist across pH ranges. When analyzing pH effects on protein structure, compare the environmental pH to the pKa values of ionizable residues: groups are charged when pH > pKa (for acids) or pH < pKa (for bases).
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?
Creation of a new salt bridge with Ca$^{2+}$ because leucine is positively charged at pH 7.4
Increased Ca$^{2+}$ affinity because hydrophobic side chains preferentially chelate divalent cations
Loss of a negatively charged ligand needed for Ca$^{2+}$ coordination, reducing binding affinity
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.
A soluble enzyme is stabilized by a buried salt bridge between a Lys side chain and an Asp side chain in its tertiary structure. A patient-derived variant substitutes Asp with Asn at this position. Differential scanning calorimetry shows a lower melting temperature ($T_m$) but near-normal secondary structure at room temperature. What is the most likely consequence of the mutation described on protein stability?
Increased stability because Asn forms a stronger ionic interaction with Lys than Asp does
Increased stability because removing Asp increases hydrophobic packing in the protein core
No change in stability because Asp and Asn have the same charge at physiological pH
Decreased stability because the salt bridge is lost and replaced by a weaker polar interaction
Explanation
This question tests understanding of how amino acid substitutions affect protein stability through loss of specific interactions. A salt bridge between lysine (positive) and aspartate (negative) provides significant stabilization energy to the folded structure. Replacing aspartate with asparagine maintains similar size and polarity but removes the negative charge, preventing salt bridge formation. The resulting Lys-Asn interaction is merely a weak polar interaction, providing much less stabilization. The lower melting temperature confirms reduced stability, while maintained secondary structure indicates the overall fold is preserved but less stable. Option A incorrectly suggests stronger interaction with Asn, option C wrongly claims identical charges, and option D misinterprets the structural context of a buried salt bridge. When evaluating mutations affecting charged residues, consider that salt bridges provide 2-5 kcal/mol stabilization, while hydrogen bonds provide only 1-2 kcal/mol.