All questions
Question 1
A mutation replaces a residue involved in a salt bridge with a partner carboxylate (–COO−) at pH 7.4. Which new residue is most likely to preserve the electrostatic interaction by remaining positively charged at pH 7.4?
General amino acid: H3N+−CH(R)−COO−
- Glutamine
- Tyrosine
- Lysine (correct answer)
- Glycine
Explanation: This question tests the understanding of amino acid structure and classification, focusing on charged side chains in electrostatic interactions. Basic amino acids like lysine have high pKa amine groups (~10.5), remaining protonated and positively charged at pH 7.4 to form salt bridges with anions. The mutation replaces a residue in a salt bridge with a carboxylate, requiring a new positive charge to preserve the interaction. Lysine (choice C) suits this as its side chain stays +1 at pH 7.4, maintaining the bridge. Glutamine (choice A) fails being polar uncharged, lacking charge for electrostatics, reflecting a misconception equating polarity with ionicity. Verify by comparing side chain pKa to ensure charge state. This underscores salt bridges' role in protein stability.
Question 2
A folded protein is exposed to increasing pH from 7.0 to 11.0, and a surface patch becomes more negatively charged, increasing electrostatic repulsion. Which residue type on the surface is most consistent with gaining negative charge over this pH range due to side-chain deprotonation?
General amino acid: H3N+−CH(R)−COO−
- Lysine (basic amine side chain)
- Tyrosine (phenolic –OH side chain) (correct answer)
- Leucine (aliphatic side chain)
- Arginine (guanidinium side chain)
Explanation: This question tests the understanding of amino acid structure and classification, examining pH-dependent deprotonation and charge effects on protein surfaces. Amino acids with ionizable side chains change charge with pH; tyrosine's phenolic OH (pKa ~10) deprotonates to negative in the 7-11 range, unlike basics that neutralize. The protein surface becomes more negatively charged from pH 7 to 11, increasing repulsion via side-chain deprotonation. Tyrosine (choice B) fits as it shifts from neutral to -1, gaining negative charge and boosting repulsion. Lysine (choice A) fails by neutralizing from +1 to 0, reducing positive repulsion not increasing negative, a misconception about charge gain. Check pKa ranges for deprotonation windows. This illustrates pH-induced unfolding via electrostatics.
Question 3
A peptide containing a single histidine is analyzed by isoelectric focusing in a buffer that is stepped from pH 5.5 to pH 8.5. The residue of interest has an imidazole side chain (pKa near 6). Which conclusion about the histidine side chain is most consistent with its predominant charge state at pH 8.5?
General amino acid: H3N+−CH(R)−COO−
- It is predominantly protonated and positively charged
- It is predominantly deprotonated and neutral (correct answer)
- It is predominantly deprotonated and negatively charged
- It is predominantly zwitterionic with a +1 net charge on the side chain
Explanation: This question tests the understanding of amino acid structure and classification, particularly the ionization behavior of side chains in response to pH changes. Amino acids have side chains with varying pKa values that determine their protonation state and charge at a given pH, with histidine's imidazole group having a pKa around 6, allowing it to switch between charged and neutral forms near physiological pH. In this isoelectric focusing experiment, the histidine side chain is examined at pH 8.5, which is above its pKa, leading to deprotonation. This state is predominantly deprotonated and neutral (choice B) because at pH > pKa, the imidazole loses its proton, resulting in a net zero charge on the side chain. Conversely, choice A fails by assuming it remains protonated and positive, a misconception ignoring the Henderson-Hasselbalch equation where pH >> pKa favors deprotonation. A transferable check is to calculate the fraction deprotonated using 10^(pH - pKa)/(1 + 10^(pH - pKa)). This clarifies how histidine acts as a pH-sensitive switch in proteins.
Question 4
A researcher compares two peptides that differ at one position: one has an R-group of –CH2–OH and the other has –CH2–CH2–S–CH3. In an aqueous buffer at pH 7.4, which conclusion about that position is most consistent with amino acid classification?
General amino acid: H3N+−CH(R)−COO−
- The –CH2–OH residue is nonpolar and tends to be buried
- The thioether residue is polar and tends to be solvent-exposed
- The –CH2–OH residue is polar uncharged and can hydrogen bond with water (correct answer)
- Both residues are acidic and will be negatively charged at pH 7.4
Explanation: This question tests the understanding of amino acid structure and classification, comparing polar and nonpolar side chains in aqueous environments. Amino acids are categorized by side chain properties, with polar uncharged like serine (–CH₂–OH) forming hydrogen bonds with water for solvent exposure, while nonpolar like methionine (–CH₂–CH₂–S–CH₃) prefer burial. The peptides differ at a position with these R-groups in pH 7.4 buffer, affecting solvation. Choice C correctly states the –CH₂–OH (serine) is polar uncharged and H-bonds with water, aligning with its classification. Choice B fails by calling the thioether (methionine) polar and exposed, a misconception as methionine is nonpolar despite sulfur. Assess by reviewing standard classifications and hydropathy. This clarifies how side chains dictate protein surface properties.
Question 5
A short peptide is placed in a solution at pH 2.0. Consider a residue with a side chain containing a carboxyl group (–CH2–COOH) in addition to the backbone carboxyl. Which statement is most consistent with the side-chain charge of that residue at pH 2.0?
General amino acid: H3N+−CH(R)−COO−
- The side chain is predominantly deprotonated and carries a −1 charge
- The side chain is predominantly protonated and neutral (correct answer)
- The side chain is predominantly protonated and carries a +1 charge
- The side chain is predominantly zwitterionic with net 0 charge due to internal salt formation
Explanation: This question tests the understanding of amino acid structure and classification, specifically the protonation states of acidic side chains at low pH. Amino acids with carboxyl side chains, like aspartate, have pKa values around 4, meaning they are protonated and neutral below this pH, while deprotonated and charged above it. In this peptide at pH 2.0, the side chain carboxyl (–CH₂–COOH) is below its pKa, favoring protonation. Thus, it is predominantly protonated and neutral (choice B), as the acidic group retains its hydrogen, yielding zero charge. Choice A fails by claiming deprotonation and negative charge, a misconception ignoring that pH < pKa promotes protonation for acids. To verify, use the Henderson-Hasselbalch equation to estimate protonated fraction. This concept extends to predicting peptide behavior in acidic cellular compartments.
Question 6
A membrane protein segment is reconstituted into lipid vesicles. A single residue in the middle of a transmembrane helix is mutated, and the mutant shows reduced insertion efficiency into the hydrophobic bilayer core. Which replacement is most consistent with introducing an energetically unfavorable polar/charged group into the bilayer interior?
General amino acid: H3N+−CH(R)−COO−
- Leucine → Aspartate (correct answer)
- Isoleucine → Valine
- Phenylalanine → Leucine
- Alanine → Glycine
Explanation: This question tests the understanding of amino acid structure and classification, emphasizing the role of side chain polarity in membrane protein insertion. Amino acids are grouped by side chain hydrophobicity, with nonpolar residues favoring the lipid bilayer core, while polar or charged ones incur energetic penalties in hydrophobic environments. Here, a mutation in a transmembrane helix reduces insertion efficiency into lipid vesicles by introducing a polar or charged group into the bilayer interior. Leucine to aspartate (choice A) fits because leucine is nonpolar, but aspartate's carboxyl group is charged and polar, making insertion unfavorable. In contrast, isoleucine to valine (choice B) fails as both are nonpolar aliphatics, not introducing polarity, stemming from a misconception that all substitutions disrupt membranes equally. To check, evaluate side chain hydropathy scores for membrane compatibility. This highlights how charged residues in transmembrane segments can cause misfolding or retention in aqueous phases.
Question 7
Which substitution least disrupts a membrane-spanning alpha helix?
- Val to Ile (correct answer)
- Leu to Glu
- Ala to Ser
- Gly to Pro
Explanation: Replacing one branched hydrophobic side chain with another (Val to Ile) preserves both helix propensity and compatibility with the lipid bilayer, so it perturbs the helix least. A tempting wrong choice is Ala to Ser: both are small, but Ser introduces a polar hydroxyl into the membrane interior, which is unfavorable. By contrast, Leu to Glu adds a charged group and Gly to Pro breaks the helix backbone.
Question 8
Which residue shows the greatest side-chain charge change from pH 6 to 8?
- Histidine (pKa 6.0) (correct answer)
- Lysine (pKa 10.5)
- Cysteine (pKa 8.3)
- Arginine (pKa 12.5)
Explanation: At pH 6, histidine's side chain is half protonated (+0.5 charge); at pH 8 it is almost completely deprotonated (near 0), so it loses about 0.5 charge. Cysteine is the closest competitor because its pKa 8.3 falls in this range, but at pH 8 it is only about one-third deprotonated, giving a change of about 0.33. Lysine and arginine stay fully protonated throughout, so their charges barely change.
Question 9
A protein's 280 nm absorbance comes mainly from which side chains?
- Cys and Met
- Phe and His
- Trp and Tyr (correct answer)
- Ser and Thr
Explanation: Aromatic side chains absorb UV light at 280 nm, and tryptophan and tyrosine have the strong conjugated ring systems that dominate. The most tempting wrong pick is phenylalanine and histidine, but phenylalanine absorbs weakly around 280 and histidine's imidazole ring absorbs much lower, so neither contributes significantly.
Question 10
Raising pH from 7 to 11 disrupts a salt bridge. Which pair is responsible?
- Ser and Glu
- Arg and Glu
- His and Asp
- Lys and Asp (correct answer)
Explanation: A salt bridge needs opposite charges. At pH 7 Lys is protonated and positive while Asp is deprotonated and negative, so they attract. Raising pH to 11 crosses Lys's side-chain pKa near 10.5, so Lys loses its proton and becomes neutral; Asp stays negative and the bridge breaks. Arg and Glu is tempting, but Arg's pKa is about 12.5, so it remains protonated and charged even at pH 11.
Question 11
Which substitution most increases backbone flexibility in a protein loop?
- Ala to Val
- Pro to Gly (correct answer)
- Leu to Ile
- Ser to Thr
Explanation: Proline's rigid ring locks the backbone phi angle, so replacing it with glycine, whose side chain is just a hydrogen, removes that constraint and greatly increases loop flexibility. The tempting wrong choice is Ala to Val, but valine's bulkier branched side chain restricts backbone conformations instead of loosening them.
Question 12
A soluble cytosolic protein is engineered to increase stability by strengthening its hydrophobic core. One candidate substitution introduces an amino acid with R=CH3. The backbone is H3N+−CH(R)−COO− at pH 7.4. Which outcome is most consistent with this residue's classification in a folded, water-soluble protein?
- It is polar and stabilizes the surface via hydrogen bonding
- It is basic and more likely to be solvent exposed
- It is acidic and forms salt bridges with lysine
- It is nonpolar and more likely to be buried in the interior (correct answer)
Explanation: This question tests amino acid classification and its impact on protein stability through core interactions. Side chains classify amino acids as nonpolar (hydrophobic), promoting burial in protein interiors to enhance stability. The substitution introduces R = CH₃, alanine, a nonpolar residue. This nonpolar nature makes it more likely buried, strengthening the hydrophobic core for stability. Suggesting it is basic and solvent-exposed errs by ignoring the methyl group's lack of charge, misclassifying nonpolar as charged. Verify by assessing if the side chain is small and aliphatic without polar atoms. Such classifications guide engineering efforts to optimize protein folding and thermostability.
Question 13
An enzyme active site contains a residue that must act as a nucleophile at pH 7.0 by presenting a deprotonated heteroatom on its side chain. A point mutation replaces this residue with one that is usually protonated (positively charged) at pH 7.0, reducing catalysis. Which substitution is most consistent with this loss of nucleophilicity due to side-chain classification?
- Ser Thr
- Cys Lys (correct answer)
- Asp Glu
- Asn Gln
Explanation: This question tests understanding of amino acid side chain properties and their roles in enzyme catalysis. Nucleophiles are electron-rich species that can attack electrophilic centers, requiring a deprotonated heteroatom (like O, N, or S) at physiological pH. Cysteine has a thiol group (pKa ~8) that can be deprotonated to form a nucleophilic thiolate anion at pH 7.0. The substitution Cys → Lys (choice B) replaces this potential nucleophile with lysine, which has a primary amine side chain (pKa ~10.5) that remains protonated and positively charged at pH 7.0, eliminating nucleophilicity. Options A (Ser → Thr), C (Asp → Glu), and D (Asn → Gln) involve substitutions between similar amino acids that would not dramatically change nucleophilic character. A common misconception is confusing basicity with nucleophilicity; while lysine is basic, its protonated state at pH 7 prevents it from acting as a nucleophile.
Question 14
An enzyme active site requires a residue that can accept a proton near physiological pH and often participates in acid–base catalysis. A mutation replaces this residue with one whose side chain is R=CH2-CH2-COO−. Which substitution is most likely to preserve the original residue's basic classification and catalytic role?
- Replace with lysine (R=(CH2)4-NH3+ at pH 7) (correct answer)
- Replace with leucine (R=CH2-CH(CH3)2)
- Replace with aspartate (R=CH2-COO−)
- Replace with serine (R = \text{CH}_2\text{-OH)
Explanation: This question tests understanding of amino acid classifications and their roles in enzyme catalysis, particularly acid-base functions. Amino acids are classified by side chains: basic ones like lysine can accept protons due to their amine groups. The original residue is basic for proton acceptance in catalysis, but mutation to R = CH₂-CH₂-COO⁻ (glutamate, acidic) disrupts this. Replacing with lysine (R = (CH₂)₄-NH₃⁺) preserves the basic classification and catalytic role via its protonatable amine. A distractor suggesting leucine fails as it is nonpolar, lacking charge for acid-base catalysis, confusing hydrophobicity with basicity. Check by confirming side chain pKa allows proton acceptance near pH 7. This clarifies how basic residues facilitate catalysis in active sites.
Question 15
A peptide (10 residues) is analyzed by isoelectric focusing. At pH 3.0 the peptide migrates toward the cathode, but at pH 10.0 it migrates toward the anode. The investigator suspects a single side chain with a pKa near physiological pH contributes substantially to the charge change. Which residue is most consistent with this behavior based on its side-chain functional group?
- Histidine (imidazole side chain) (correct answer)
- Leucine (aliphatic side chain)
- Aspartate (carboxylate side chain)
- Lysine (primary amine side chain)
Explanation: This question tests understanding of amino acid ionization states and their pH-dependent behavior. Amino acids have characteristic pKa values for their ionizable groups: carboxyl groups (~2-4), amino groups (~9-10), and certain side chains. At pH 3.0, the peptide migrates toward the cathode (negative electrode), indicating a net positive charge, while at pH 10.0 it migrates toward the anode (positive electrode), indicating a net negative charge. Histidine (choice A) has an imidazole side chain with a pKa around 6.0, meaning it transitions from protonated (positive) to deprotonated (neutral) near physiological pH. Leucine (B) has no ionizable side chain, aspartate (C) has a carboxylate with pKa ~3.8 that would already be deprotonated at pH 3, and lysine (D) has a primary amine with pKa ~10.5 that would remain protonated even at pH 10. The dramatic charge change between pH 3 and 10 is best explained by histidine's unique pKa near the middle of this range.
Question 16
A membrane-embedded helix is being designed for stable insertion into a lipid bilayer. A position facing the lipid tails is assigned an amino acid with side chain R=CH2-CH2-S-CH3. Which classification best matches this residue and its compatibility with the bilayer core?
- Basic, favorable for the bilayer center due to positive charge
- Polar uncharged, favorable for the bilayer center due to H-bonding
- Acidic, favorable for the bilayer center due to negative charge
- Nonpolar, favorable for interaction with lipid tails (correct answer)
Explanation: This question tests amino acid classification and compatibility with membrane environments like lipid bilayers. Amino acids with nonpolar side chains are hydrophobic, favoring interactions with lipid tails in bilayer cores. The residue has R = CH₂-CH₂-S-CH₃, methionine, classified as nonpolar despite the sulfur. This makes it favorable for the bilayer center via hydrophobic packing with lipids. Claiming it is polar uncharged for H-bonding in the core fails, as the side chain is mostly aliphatic, misconstruing weak sulfur polarity as strong. Confirm by noting low polarity and preference for non-aqueous environments. This principle applies to designing transmembrane proteins for stable insertion.
Question 17
A researcher compares two residues for placement on the solvent-exposed surface of a globular protein at pH 7.4: (1) R=CH3 and (2) R=CH2-COO−. Which choice is most consistent with maximizing surface compatibility with water?
- Choose (1), because nonpolar groups hydrogen-bond strongly to water
- Choose (2), because an acidic carboxylate is polar/charged and water-compatible (correct answer)
- Choose (1), because hydrophobic residues preferentially face solvent
- Choose (2), because charged residues preferentially partition into lipid bilayers
Explanation: This question tests amino acid suitability for solvent-exposed positions. Charged or polar residues are water-compatible, unlike nonpolar. Comparing CH₃ (nonpolar) and CH₂-COO⁻ (acidic, charged), the latter maximizes water interactions. Choosing the acidic carboxylate leverages its polarity and charge for solvation. Selecting nonpolar errs, as it avoids water, confusing hydrophobicity with compatibility. Evaluate hydration potential by side chain groups. Surface residues influence solubility and folding.
Question 18
A peptide is synthesized with one residue whose side chain is R=H (the backbone is H3N+−CH(R)−COO− at neutral pH). Which classification is most consistent with this residue and its typical role in folded proteins?
- Polar uncharged; always found only on the protein surface
- Acidic; often forms salt bridges due to its carboxylate side chain
- Basic; often binds DNA due to a protonated amine side chain
- Nonpolar; often tolerated in both surface and interior due to minimal side chain (correct answer)
Explanation: This question tests classification of glycine and its protein roles. With R = H, glycine is nonpolar due to minimal, non-polar side chain. It's tolerated in both surface and interior positions owing to flexibility. This small size allows versatile placement without steric issues. Claiming acidic misidentifies H as carboxylate, ignoring actual structure. Confirm by noting lack of functional groups in R. Glycine's nonpolarity aids tight packing in proteins.
Question 19
A transmembrane helix is predicted to span the lipid bilayer. A single residue in the middle of the helix is mutated, and membrane insertion efficiency drops sharply. Which replacement is most likely to cause this effect by introducing an ionizable side chain into the hydrophobic membrane interior?
- Leu Ile
- Ala Val
- Phe Met
- Leu Asp (correct answer)
Explanation: This question tests understanding of amino acid hydrophobicity and membrane protein structure. Transmembrane helices require hydrophobic residues to interact favorably with the lipid bilayer's hydrophobic core. Introducing an ionizable, charged residue into this environment creates an energetic penalty that disrupts membrane insertion. The substitution Leu → Asp (choice D) replaces hydrophobic leucine with aspartate, which has a carboxyl side chain that is ionized (negatively charged) at physiological pH. Options A (Leu → Ile), B (Ala → Val), and C (Phe → Met) all maintain hydrophobic character, as they substitute one nonpolar residue for another. The principle that charged residues are energetically unfavorable in membrane interiors explains why transmembrane domains are enriched in hydrophobic amino acids. This incompatibility between charged groups and lipid environments is fundamental to membrane protein topology.
Question 20
A soluble cytosolic protein is engineered to include a new surface-exposed residue to increase aqueous solubility without adding a formal charge at pH 7.4. Which amino acid is most likely to achieve this by providing a polar, uncharged side chain capable of hydrogen bonding?
- Valine
- Glutamate
- Serine (correct answer)
- Lysine
Explanation: This question tests understanding of amino acid classification based on polarity and charge state. To increase aqueous solubility without adding formal charge at pH 7.4, a polar but uncharged amino acid is needed. Serine (choice C) has a hydroxyl (-OH) side chain that is polar and capable of hydrogen bonding with water molecules, but remains uncharged at physiological pH. Valine (A) is nonpolar and would not increase solubility, glutamate (B) carries a negative charge at pH 7.4 (violating the constraint), and lysine (D) carries a positive charge at pH 7.4. The distinction between polar/uncharged amino acids (Ser, Thr, Asn, Gln, Tyr) and charged amino acids (Asp, Glu, Lys, Arg, His) is crucial for protein engineering. A helpful mnemonic is that hydroxyl and amide groups provide polarity through hydrogen bonding without ionization at neutral pH.