MCAT Chemical and Physical Foundations of Biological Systems Quiz: 5d Amino Acids Peptides Protein Structure
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5d Amino Acids Peptides Protein StructureQuestion 1 of 20

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?

Lys side chains become deprotonated at high pH, weakening electrostatic attraction to the ligand
The ligand carboxylate becomes protonated at high pH, eliminating its negative charge
High pH strengthens salt bridges by increasing protonation of basic residues
High pH converts peptide bonds into esters, preventing ligand binding
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MCAT Chemical and Physical Foundations of Biological Systems Quiz

MCAT Chemical and Physical Foundations of Biological Systems Quiz: 5d Amino Acids Peptides Protein Structure

Practice 5d Amino Acids Peptides Protein Structure in MCAT Chemical and Physical Foundations of Biological Systems with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on 5d Amino Acids Peptides Protein Structure, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

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?

  1. Lys side chains become deprotonated at high pH, weakening electrostatic attraction to the ligand (correct answer)
  2. The ligand carboxylate becomes protonated at high pH, eliminating its negative charge
  3. High pH strengthens salt bridges by increasing protonation of basic residues
  4. High pH converts peptide bonds into esters, preventing ligand binding
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.

Question 2

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-NH2_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?

  1. Replace Asp at position 5 with Asn (correct answer)
  2. Replace Lys at position 2 with Gln
  3. Replace Leu at position 7 with Ile
  4. Replace Ser at position 6 with Thr
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.

Question 3

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?

  1. Ser provides a nucleophilic hydroxyl that can attack an electrophilic substrate center (correct answer)
  2. Ser forms a disulfide bond with His to stabilize the active site
  3. Ser donates a permanent negative charge to electrostatically repel the substrate
  4. Ser disrupts backbone hydrogen bonding to increase active-site flexibility
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.

Question 4

A cytosolic protein domain binds a small anionic metabolite via a pocket lined with Lys and Arg residues. A variant replaces one Lys with Glu (K;2E) at the binding interface; global folding is unchanged. Which outcome is most likely at physiological ionic strength?

  1. Increased binding due to additional hydrogen bonding from the Glu carboxylate
  2. Decreased binding due to electrostatic repulsion and loss of favorable cation;2anion interactions (correct answer)
  3. No change because Lys and Glu have similar side-chain pKaK_a values
  4. Increased binding because Glu is more hydrophobic than Lys and strengthens the binding pocket core
Explanation: This question tests understanding of electrostatic interactions in protein-ligand binding, particularly the importance of complementary charges. The binding pocket lined with positively charged Lys and Arg residues is optimized for binding anionic (negatively charged) metabolites through favorable electrostatic attraction. Replacing lysine (positive) with glutamate (negative at physiological pH) not only removes a favorable interaction but introduces electrostatic repulsion between the negatively charged Glu and the anionic ligand. This double penalty - loss of attraction plus gain of repulsion - significantly decreases binding affinity. The claim that Glu is more hydrophobic than Lys is incorrect; Glu is polar and charged. The pKa values of Lys (~10.5) and Glu (~4.2) are very different, not similar. When analyzing mutations affecting charged ligand binding, consider both the loss of favorable interactions and the potential introduction of unfavorable ones, and remember that electrostatic effects can be dominant in polar binding sites.

Question 5

A peptide is synthesized with the sequence Ac;2Gly;2Lys;2Asp;2Phe;2NH2_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?

  1. Replace Lys with Arg (correct answer)
  2. Replace Asp with Asn
  3. Replace Phe with Leu
  4. 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.

Question 6

A membrane-associated receptor-binding protein contains a short ;2-hairpin that presents a conserved Gly-Pro motif at the tip of a surface loop. A point mutation replacing glycine with valine (G;2V) 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?

  1. Reduced local backbone flexibility at the loop tip, impairing the loop conformation required for binding (correct answer)
  2. Increased formation of ;1-helices at the loop tip because valine is a strong helix breaker
  3. Elimination of a salt bridge because glycine is normally positively charged at physiological pH
  4. 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.

Question 7

A 40-residue peptide forms an α\alpha-helix in water only when bound to a partner protein. In isolation, it is largely disordered. The helical region is enriched in Leu, Ile, and Ala, with several charged residues at the ends. Which interaction most likely drives helix stabilization upon binding?

  1. Hydrophobic burial of nonpolar side chains at the binding interface, favoring a helical conformation (correct answer)
  2. Covalent crosslinking between peptide and partner backbone atoms
  3. Formation of new peptide bonds between the peptide and partner protein
  4. Replacement of backbone hydrogen bonding with ionic bonds along the helix axis
Explanation: This question probes induced folding in protein-peptide interactions, focusing on hydrophobic stabilization of helices. Binding buries nonpolar residues, driving helix formation in otherwise disordered peptides. The vignette shows helix stabilization upon partner binding with nonpolar enrichment. Hydrophobic burial at the interface favors helical conformation. A distractor like covalent crosslinking misattributes noncovalent driving forces. Approach similar problems by identifying interface properties. Predict folding by assessing entropy gains from hydrophobic effects.

Question 8

A protein is reported to have a collagen-like region with repeating Gly;2X;2Y motifs and forms a triple helix in the extracellular matrix. A point mutation replaces one Gly with Val within the repeating region. What is the most likely structural consequence?

  1. Disrupted triple-helix packing because Gly is required for tight core packing in collagen-like helices (correct answer)
  2. Stabilized triple helix because Val increases backbone hydrogen bonding
  3. No effect because Gly and Val have identical side chains
  4. Formation of a new disulfide bond that replaces the triple helix
Explanation: This question tests knowledge of collagen structure and the critical role of glycine in triple helix formation. Collagen's triple helix structure requires glycine at every third position because the three chains pack so tightly that only glycine's hydrogen side chain can fit in the interior. The Gly-X-Y repeat is essential, with glycine always occupying the first position. Replacing glycine with valine introduces a bulky hydrophobic side chain that creates steric clashes, preventing proper triple helix assembly. This is not about backbone hydrogen bonding changes or disulfide bond formation. Glycine and valine do not have identical side chains - glycine has only a hydrogen while valine has a branched aliphatic group. The key principle is that collagen's structure has evolved to require glycine's minimal size at specific positions. When analyzing mutations in repetitive structural motifs like collagen, consider whether the wild-type residue has unique properties that cannot be substituted without disrupting the structure.

Question 9

A 210-residue cytosolic binding protein was engineered to include a short amphipathic ;1-helix that docks into a hydrophobic groove on a partner protein during osmotic stress signaling. Circular dichroism indicated comparable overall helicity for wild-type and variant proteins, but the variant showed a 15-fold weaker binding affinity (higher KdK_d) in 150 mM NaCl at pH 7.4. The docking helix contains a heptad repeat in which Leu and Ile residues occupy positions that face the groove, while polar residues face solvent. Which alteration to the amino acid sequence would most likely disrupt protein function under these conditions?

  1. Replace a solvent-exposed Glu on the helix with Asp to preserve charge while slightly shortening the side chain
  2. Replace a groove-facing Leu on the helix with Asp to introduce a charged side chain into the hydrophobic interface (correct answer)
  3. Replace a solvent-exposed Ser on the helix with Thr to maintain polarity while increasing side-chain volume
  4. Replace a groove-facing Ile on the helix with Val to retain hydrophobic character with modestly reduced size
Explanation: This question tests understanding of how amino acid substitutions affect protein-protein interactions, specifically in hydrophobic binding interfaces. The amphipathic helix uses hydrophobic residues (Leu, Ile) to dock into a hydrophobic groove, while polar residues face the solvent. Replacing a groove-facing Leu with Asp (option B) introduces a charged, polar residue into the hydrophobic interface, creating unfavorable electrostatic interactions and disrupting the complementary hydrophobic packing. The correct answer recognizes that hydrophobic interfaces require nonpolar residues for stable binding. Option A incorrectly focuses on solvent-exposed residues that don't directly participate in binding, while options C and D maintain appropriate chemical properties for their respective environments. When analyzing protein-protein interactions, identify which residues directly contact the binding partner and ensure mutations maintain compatible chemical properties at the interface.

Question 10

A small cytosolic protein (140 residues) folds into a ;2-sandwich stabilized by a hydrophobic core rich in Val, Leu, and Phe. Differential scanning calorimetry shows a melting temperature (TmT_m) of 56b0C in 10 mM phosphate buffer at pH 7.4. When the NaCl concentration is increased from 20 mM to 500 mM, TmT_m increases to 61b0C with minimal change in secondary structure content. Based on the vignette, how does the environment influence protein folding?

  1. Higher ionic strength screens repulsion among surface charges, favoring the folded state without directly strengthening the hydrophobic core (correct answer)
  2. Higher ionic strength disrupts all hydrogen bonds in ;2-sheets, forcing the protein to rely on hydrophobic interactions and raising TmT_m
  3. Higher ionic strength protonates acidic residues at pH 7.4, increasing salt bridges and thereby raising TmT_m
  4. Higher ionic strength weakens the hydrophobic effect by making water less structured, which increases TmT_m by reducing unfolding entropy
Explanation: This question explores how ionic strength affects protein stability through electrostatic screening. Higher salt concentrations shield electrostatic repulsions between like-charged surface residues, reducing the energetic penalty of the folded state without directly affecting the hydrophobic core. The β-sandwich structure relies primarily on hydrophobic core packing for stability, while surface charge repulsions can destabilize the folded form. The correct answer recognizes that ionic screening stabilizes proteins by reducing unfavorable electrostatic interactions at the surface. Option B incorrectly claims salt disrupts hydrogen bonds in β-sheets, while option D misunderstands the hydrophobic effect's temperature dependence. When analyzing salt effects on protein stability, consider that ionic strength primarily affects surface electrostatics rather than core hydrophobic interactions or hydrogen bonding networks.

Question 11

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 (E;2Q) without changing overall size or expression level. Differential scanning calorimetry shows a lower TmT_m and increased aggregation propensity. Given the described structure, which interaction is most directly lost in the mutant, leading to reduced stability?

  1. A covalent bond between Glu and Lys that normally links two secondary structure elements
  2. A salt bridge between a negatively charged Glu and a positively charged Lys in the hydrophobic interior (correct answer)
  3. A disulfide bond between Glu and Lys that prevents unfolding at elevated temperature
  4. A hydrophobic interaction between two charged side chains that is stronger than an ion pair
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.

Question 12

A 12-residue peptide hormone is stored in secretory granules at pH 5.5 and released into blood at pH 7.4. The peptide contains one Asp, one Lys, and one His, and adopts a compact conformation in granules but becomes more extended after secretion. Assume typical side-chain pKaK_a values: Asp ;c4.0, His ;c6.0, Lys ;c10.5. Based on the vignette, which interaction is most likely present in granules but diminished at physiological pH, contributing to the conformational change?

  1. A salt bridge between protonated His and deprotonated Asp (correct answer)
  2. A salt bridge between deprotonated Lys and protonated Asp
  3. A disulfide bond between His and Lys side chains
  4. Backbone hydrogen bonding that requires Asp to be protonated
Explanation: This question tests understanding of pH-dependent ionization states and their effects on protein interactions. At pH 5.5 (in granules), histidine (pKa ~6.0) would be predominantly protonated and positively charged, while aspartate (pKa ~4.0) would be deprotonated and negatively charged, allowing formation of a salt bridge. At pH 7.4 (in blood), histidine becomes predominantly deprotonated and neutral, disrupting this electrostatic interaction and causing the conformational change. Lysine remains positively charged at both pH values (pKa ~10.5), and option C incorrectly suggests disulfide bonds between residues that cannot form them. Option D misunderstands that aspartate would be deprotonated at both pH values. To solve pH-dependent interaction problems, compare the pH to each residue's pKa: when pH < pKa, the residue is protonated; when pH > pKa, it's deprotonated.

Question 13

A bacterial DNA-binding protein contains an α\alpha-helix that inserts into the major groove. The helix presents several polar side chains to form sequence-specific contacts, while the helix backbone is stabilized by internal hydrogen bonding. A mutation replaces one helix residue from Glu to Gln at a DNA-contacting position, and binding weakens modestly. What is the most likely explanation?

  1. Loss of negative charge reduces electrostatic complementarity while preserving hydrogen-bonding capacity (correct answer)
  2. Backbone hydrogen bonds in the helix are eliminated because Gln cannot form peptide bonds
  3. A new disulfide bond forms between Gln and DNA, preventing binding
  4. Gln is nonpolar and disrupts the hydrophobic core of the helix, fully unfolding the protein
Explanation: This question assesses protein-DNA interactions, emphasizing side-chain roles in sequence-specific binding. Glu provides negative charge for electrostatics and H-bonding; Gln maintains H-bonding but neutralizes charge. The vignette describes weakened binding upon Glu to Gln in a DNA-contacting helix. Loss of charge reduces electrostatic complementarity while preserving H-bonding. A distractor suggesting disulfide formation ignores Gln's non-thiol nature. For comparable questions, differentiate charge and H-bond contributions. Predict outcomes by comparing ionic and polar properties.

Question 14

A protein domain binds DNA through a helix that inserts into the major groove. The interface is enriched in Lys, Arg, and Asn residues. A mutation replaces one Arg with Leu (R;2L) at the interface. Which change is most likely in binding affinity under physiological conditions?

  1. Increased affinity because Leu forms strong ionic interactions with the DNA phosphate backbone
  2. Decreased affinity because a positively charged side chain that stabilizes DNA binding is removed (correct answer)
  3. No change because Arg and Leu have similar hydrogen-bonding patterns
  4. Increased affinity because Leu increases helix dipole moment and strengthens base pairing
Explanation: This question tests understanding of protein-DNA interactions and the role of electrostatic complementarity. DNA has a negatively charged sugar-phosphate backbone due to phosphate groups. Proteins typically bind DNA using positively charged residues (Lys, Arg) that form favorable electrostatic interactions with these phosphates. Arginine is particularly important for DNA binding due to its ability to form multiple hydrogen bonds while maintaining positive charge. Replacing Arg with Leu removes the positive charge and hydrogen bonding capability, eliminating favorable interactions with DNA phosphates. Leucine is hydrophobic and cannot form ionic interactions. Arg and Leu have completely different properties - Arg is charged and polar while Leu is hydrophobic. Leucine doesn't affect helix dipoles or base pairing. The key principle is that protein-DNA interfaces are typically enriched in basic residues that complement DNA's negative charge. When analyzing mutations at protein-DNA interfaces, consider how changes in charge complementarity affect binding affinity.

Question 15

A membrane-associated signaling protein contains a short cytosolic helix with multiple Lys residues that interact with negatively charged phospholipid headgroups. A mutation replaces two Lys with Gln (K;2Q) while preserving helix length. Which outcome is most likely for membrane association at physiological pH?

  1. Stronger association because Gln is more positively charged than Lys
  2. Weaker association because neutralizing positive charges reduces electrostatic attraction to anionic lipids (correct answer)
  3. No change because membrane binding is dominated solely by backbone hydrogen bonding
  4. Stronger association because Gln increases hydrophobicity and inserts into the bilayer core
Explanation: This question tests understanding of electrostatic interactions in membrane binding and how charge neutralization affects protein-lipid interactions. Many membrane-associated proteins use positively charged residues (Lys, Arg) to interact with negatively charged phospholipid headgroups through electrostatic attraction. Replacing lysine (positively charged at pH 7.4) with glutamine (polar but uncharged) removes these favorable electrostatic interactions, weakening membrane association. Glutamine is not more positively charged than lysine - it's neutral. While Gln is slightly more hydrophobic than Lys, it's not hydrophobic enough to insert into the bilayer core. Membrane binding involves multiple interaction types, not solely backbone hydrogen bonding. The key principle is that electrostatic complementarity between cationic protein surfaces and anionic lipid headgroups is a major driver of membrane association. When analyzing mutations affecting membrane binding, consider how changes in surface charge distribution affect electrostatic interactions with lipids.

Question 16

An intracellular enzyme has a catalytic Ser in a tight turn region. A variant replaces this Ser with Thr (S;2T). The overall fold remains unchanged, but the catalytic rate decreases modestly. Which rationale best accounts for the effect based on structure;2function considerations?

  1. Thr is less polar than Ser and cannot participate in hydrogen bonding
  2. Thr introduces a methyl group that can sterically hinder optimal positioning of the nucleophilic hydroxyl (correct answer)
  3. Thr is negatively charged at physiological pH, altering electrostatics in the active site
  4. Thr forces formation of a disulfide bond, rigidifying the turn and preventing catalysis
Explanation: This question tests understanding of structure-function relationships in enzyme active sites, particularly how subtle amino acid differences affect catalysis. While serine and threonine both have hydroxyl groups capable of nucleophilic attack, threonine's additional methyl group creates steric hindrance that can affect the precise positioning required for optimal catalysis. In tight turn regions where space is limited, even small steric clashes can disrupt the catalytic geometry. The hydroxyl groups of both residues can participate in hydrogen bonding, and neither is charged at physiological pH. Threonine cannot form disulfide bonds as it lacks sulfur. The key concept is that catalytic efficiency often depends on precise geometric arrangements, and even conservative substitutions can have functional consequences. When evaluating mutations in enzyme active sites, consider not just the chemical similarity of residues but also their size differences and how these might affect the three-dimensional arrangement of catalytic groups.

Question 17

A 55-residue protein folds into a compact structure with one buried Trp that exhibits strong fluorescence in the folded state. Upon heating, fluorescence decreases as Trp becomes solvent-exposed. A point mutation replaces a buried Leu with Asp (L;2D). Which prediction is most consistent with protein stability principles?

  1. Higher stability because introducing a charged residue strengthens hydrophobic packing
  2. Lower stability because burying a charged residue is energetically unfavorable and can promote unfolding (correct answer)
  3. No change because Leu and Asp have similar size and polarity
  4. Higher stability because Asp forms disulfide bonds with Trp, rigidifying the core
Explanation: This question tests understanding of the hydrophobic effect and the energetic cost of burying polar groups in protein cores. Leucine is a hydrophobic residue well-suited for burial in the protein interior, while aspartate is charged at physiological pH (pKa ~3.9). Burying a charged residue in the low-dielectric hydrophobic core is highly unfavorable energetically, as there are no water molecules or other polar groups to stabilize the charge. This destabilization promotes unfolding, reducing the melting temperature. The notion that Asp could form disulfide bonds with Trp is incorrect - only cysteine can form disulfide bonds. Leu and Asp do not have similar size and polarity; Leu is hydrophobic while Asp is polar and charged. When evaluating mutations in protein cores, remember that hydrophobic residues are strongly preferred and that burying charges without compensation is one of the most destabilizing changes possible.

Question 18

A helical segment in a soluble protein displays an amphipathic pattern: hydrophobic residues occur every 3;24 positions, consistent with one face forming a hydrophobic interface with another helix. A mutation replaces a surface-exposed Glu on the polar face with Leu (E;2L). Which outcome is most likely?

  1. Increased solubility because Leu forms additional hydrogen bonds with water
  2. Decreased solubility or increased aggregation because adding hydrophobic surface area can promote nonspecific association (correct answer)
  3. No effect because Glu and Leu are both uncharged at pH 7.4
  4. Increased stability because Leu forms a salt bridge with nearby Lys more strongly than Glu
Explanation: This question tests understanding of amphipathic helices and the consequences of disrupting their hydrophobic/hydrophilic balance. Amphipathic helices have distinct hydrophobic and hydrophilic faces, allowing them to interact with both polar and nonpolar environments. Replacing a charged, hydrophilic glutamate with hydrophobic leucine on the polar face disrupts this amphipathic character. The additional hydrophobic surface area can promote nonspecific protein-protein interactions and aggregation, particularly in aqueous solution where hydrophobic residues seek to minimize water contact. Leucine cannot form hydrogen bonds with water more effectively than glutamate - the opposite is true. At pH 7.4, Glu is charged (negative) while Leu is always uncharged. Leucine cannot form salt bridges as it lacks charged groups. When analyzing mutations in amphipathic structures, consider how changes affect the balance of hydrophobic and hydrophilic surfaces and the potential for unwanted interactions.

Question 19

A short peptide in aqueous buffer forms a stable antiparallel ββ-hairpin stabilized by backbone hydrogen bonds and a hydrophobic cluster (Val, Ile, Phe) at the turn region. A single substitution is introduced at the turn. Which alteration would most likely disrupt the ββ-hairpin by decreasing turn flexibility and introducing steric strain?

  1. Gly ;2 Pro (correct answer)
  2. Val ;2 Leu
  3. Ile ;2 Val
  4. Phe ;2 Tyr
Explanation: This question tests knowledge of amino acid effects on secondary structure, particularly in β-turns. Glycine is the most flexible amino acid due to its lack of a side chain, making it ideal for tight turns in protein structures. Proline, conversely, is the most conformationally restricted amino acid due to its cyclic structure connecting the side chain to the backbone nitrogen. This rigidity makes proline a poor choice for positions requiring flexibility and can introduce steric clashes in tight turns. The other substitutions (Val→Leu, Ile→Val, Phe→Tyr) involve amino acids with similar conformational preferences and would have minimal impact on turn structure. The key concept is that β-hairpin turns require backbone flexibility to achieve the necessary reversal in chain direction. When analyzing mutations affecting turns or loops, prioritize considering how the substitution affects backbone flexibility, with Gly providing maximum flexibility and Pro providing maximum constraint.

Question 20

A 12-residue peptide is designed to form an αα-helix. To test helix propensity, a researcher substitutes one residue at position 6. Which substitution would most likely increase helix stability in aqueous solution by favoring helix formation and reducing conformational entropy of the unfolded state?

  1. Replace Ala with Pro
  2. Replace Ala with Gly
  3. Replace Ala with Leu (correct answer)
  4. Replace Ala with Asp at the helix center to increase charge repulsion
Explanation: This question tests understanding of amino acid helix propensities and the factors affecting secondary structure stability. Leucine is known as a strong helix former due to its hydrophobic nature and favorable conformational preferences. It has restricted conformational freedom in the unfolded state compared to smaller residues, which reduces the entropic cost of helix formation. Proline is a helix breaker that would destabilize the structure. Glycine increases conformational flexibility, actually destabilizing helices by increasing the entropy of the unfolded state. While Asp can participate in helix-stabilizing interactions at terminal positions, introducing it at the helix center would create unfavorable charge burial or helix dipole interactions. The key principle is that amino acids differ in their intrinsic helix-forming tendencies based on both conformational preferences and side chain properties. When designing or analyzing helical peptides, choose residues with high helix propensity (Ala, Leu, Glu, Met) and avoid helix breakers (Pro, Gly) in central positions.