Question 1

Thermodynamics of unfolding Protein G: ΔH° (unfolding) = +210.6 kJ/mol. What does the sign and magnitude imply about enthalpic favorability?

Accepted Answer

Introduction: Protein stability reflects a balance of enthalpy (ΔH) and entropy (ΔS). The enthalpy change for unfolding indicates how favorable intramolecular interactions are in the folded state compared to the unfolded state. Given Data / Assumptions: Unfolding reaction: Folded → Unfolded with ΔH° = +210.6 kJ/mol. Sign convention: positive ΔH° means the reaction absorbs heat. Concept / Approach: If unfolding is endothermic (positive ΔH°), then favorable enthalpic interactions (H-bonds, van der Waals contacts, salt bridges) are being broken. Hence, folding is enthalpically preferred; unfolding requires heat input. Step-by-Step Solution: 1) Reaction considered: Folded → Unfolded, ΔH° = +210.6 kJ/mol.2) Positive ΔH° implies energy absorbed to break stabilizing interactions.3) Therefore, enthalpy favors the folded state; unfolding is not enthalpically favored.4) Overall stability depends on ΔG° = ΔH° − T * ΔS°. Verification / Alternative check: Calorimetry typically shows endothermic unfolding transitions for many proteins, consistent with enthalpically stabilized folded states. Why Other Options Are Wrong: “Unfolding enthalpically favored” contradicts positive ΔH°. Statements about entropy cannot be concluded solely from ΔH°. ΔH° certainly conveys enthalpic information. Common Pitfalls: Equating enthalpy with spontaneity; forgetting the role of temperature and ΔS° (hydrophobic effect often gives positive ΔS° for unfolding of hydrophobic surfaces into solvent). Final Answer: Folding is favored enthalpically (unfolding is endothermic)

Question 2

Hydrophobicity and folding: what does the correlation between ΔG of transfer (aqueous → organic) and solvent-accessible surface area (SASA) of amino acid residues tell us?

Accepted Answer

Introduction: Empirical hydrophobicity scales often derive from the free energy change for transferring residues or analogs from water to an organic phase. Correlating these ΔG values with solvent-accessible surface area (SASA) informs how burial of surfaces contributes to folding. Given Data / Assumptions: ΔG(transfer) approximates residue hydrophobicity. SASA decreases upon burial inside the folded core. Peptide backbone and side chains both contribute to solvation energetics. Concept / Approach: During folding, hydrophobic surface area is buried, reducing unfavorable interactions with water and releasing structured water, which contributes to favorable ΔG. The correlation between ΔG(transfer) and area emphasizes surface burial as a central term. Step-by-Step Solution: 1) Measure ΔG for residue analog transfer water → organic.2) Relate ΔG per unit area to loss of SASA upon folding.3) Conclude that reduced exposed area stabilizes the folded state. Verification / Alternative check: Fold stability often scales with hydrophobic core size; mutation studies that increase exposed hydrophobic area typically destabilize proteins. Why Other Options Are Wrong: Limiting to polar residues ignores dominant nonpolar contributions; peptide bond terms can be accounted for; SDS denaturation exposes surfaces (opposite of folding); electrostatics alone cannot explain the broad correlation. Common Pitfalls: Overinterpreting ΔG(transfer) without considering context (backbone, specific interactions, temperature). Final Answer: It reflects the reduction in solvent-accessible area during protein folding

Question 3

Protein packing in globular proteins — Typical coordination number: buried hydrophobic side chains usually fit into a hole formed by how many other side chains?

Accepted Answer

Introduction: Protein cores are tightly packed like a three dimensional jigsaw. This question tests understanding of typical packing density in the hydrophobic interior of globular proteins, where nonpolar side chains interlock with neighbors to exclude water and maximize stabilizing contacts. Given Data / Assumptions: Context: a folded, soluble globular protein under physiological conditions. Focus: how many neighboring side chains typically surround one buried hydrophobic side chain. No requirement to identify a specific residue; the topic is general core packing statistics. Concept / Approach: In protein interiors, packing approaches that of organic crystals but with some voids. Each buried side chain makes multiple van der Waals contacts with nearby residues. Empirical analyses of high resolution structures show a typical local coordination number around six, often described as a side chain fitting into a hole made by about five to seven neighbors. Step-by-Step Solution: 1) Recognize that burial is driven by the hydrophobic effect, which favors removing nonpolar surface from water.2) In the core, tight packing maximizes van der Waals complementarity while minimizing void volume.3) Structural surveys report that a buried side chain is usually surrounded by approximately 5–7 other side chains at contact distance.4) Therefore, the most representative choice is 5–7 other amino acids. Verification / Alternative check: Consider common core residues such as Leu, Ile, Val, and Phe. Rotamer libraries and contact maps show each of these typically engages many short range contacts that sum to about half a dozen distinct neighbors, consistent with the 5–7 range rather than very low or very high counts. Why Other Options Are Wrong: 1–3: underestimates real packing; would leave excessive cavities. 9–12 and 13–15: unrealistically high in a sterically crowded core. 2–4: still too low compared to observed coordination numbers. Common Pitfalls: Confusing the number of atomic contacts with the number of distinct neighboring residues. One neighbor can contribute many atom–atom contacts, but the count of distinct side chains remains near six on average. Final Answer: 5–7 other amino acids.

Question 4

Hydrogen bonding in protein cores — How often are unpaired donors and acceptors found in the hydrophobic interior?

Accepted Answer

Introduction: Hydrogen bonding is a key determinant of protein stability. This question examines the prevalence of unsatisfied hydrogen bond donors or acceptors within hydrophobic cores, where burying polar groups without partners imposes a significant energetic penalty. Given Data / Assumptions: A folded soluble protein in aqueous solution. Hydrophobic interior largely excludes water. Polar backbone and side chain groups prefer to form hydrogen bonds if buried. Concept / Approach: When a polar group is buried without a complementary partner, the desolvation penalty is high. Proteins avoid this by forming internal hydrogen bonds (for example, backbone hydrogen bonds in helices and sheets) or by keeping polar and charged groups at the surface. Therefore, unpaired donors or acceptors in the core are uncommon and usually indicate strain, catalytic requirements, or specific design features such as internal salt bridges. Step-by-Step Solution: 1) Evaluate the energetic balance: burying an unsatisfied polar group costs free energy due to loss of favorable interactions with water.2) Folding pathways and natural selection favor structures that satisfy backbone hydrogen bonds via regular secondary structures.3) Side chain polar groups that are buried typically form hydrogen bonds or ion pairs; otherwise they are surface exposed.4) Conclusion: unpaired donors and acceptors in the core are rare. Verification / Alternative check: High resolution structures show most buried backbone NH and CO groups are paired in helices or sheets. When exceptions occur, they are often near functional sites and compensated by local networks or bound ligands to reduce the energetic cost. Why Other Options Are Wrong: Only at helix ends: helix capping exists but does not confine the phenomenon to helix termini. Only at beta turns: polar groups appear throughout the fold, not only in turns. Only on Pro: Pro lacks an amide hydrogen and is not the sole determinant of unsatisfied groups. Frequently: contradicts fundamental energetic considerations. Common Pitfalls: Assuming that the hydrophobic core can tolerate many unsatisfied polar groups. In reality, a few exceptions are carefully compensated by geometry and local interactions. Final Answer: rarely.

Question 5

Thermodynamics relation — Interpreting ΔG° = -RTln(K): effect of a tenfold change in the equilibrium constant

Accepted Answer

Introduction: The standard free energy change links directly to the equilibrium constant through the fundamental relation ΔG° = -RTln(K). This question checks quick, order of magnitude reasoning for decimal changes in K without a calculator. Given Data / Assumptions: R is the gas constant and T is absolute temperature. Natural logarithm is used in the equation. Goal: estimate the change in ΔG° when K changes by a factor of 10. Concept / Approach: Use the identity ln(10) ≈ 2.303. Substituting into ΔG° = -RTln(K) shows how a tenfold change in K maps to a fixed multiple of RT independent of the actual K value, aside from sign direction. Step-by-Step Solution: 1) Start with ΔG° = -RTln(K).2) For a 10-fold increase, K becomes 10K_old, so ln(K_new) = ln(10) + ln(K_old).3) The change in ΔG° is -RTln(10) ≈ -2.3RT.4) Therefore, ΔG° decreases by about 2.3RT for a 10-fold increase in K. Verification / Alternative check: If K decreases by 10, then ln(K) decreases by 2.3, so ΔG° increases by about 2.3RT. The direction is consistent with favorable equilibria (larger K) corresponding to more negative ΔG°. Why Other Options Are Wrong: 10-fold change in ΔG°: free energy is not proportional to K linearly. Decrease in K decreases ΔG°: sign is wrong; ΔG° becomes less negative. 10-fold decrease increases ΔG° by 10-fold: again mixes linear and logarithmic relations. 2-fold increase gives RT: ln(2) is about 0.693, not 1.0, so the magnitude is off. Common Pitfalls: Mixing log base 10 and natural log. Always convert decimal folds to natural logs with ln(10) ≈ 2.303 for quick estimates. Final Answer: a 10-fold increase in K decreases ΔG° by about 2.3RT.

Question 6

Protein thermal denaturation — Interpreting the midpoint (Tm) of a temperature transition curve

Accepted Answer

Introduction: Thermal unfolding of proteins is often monitored by spectroscopy or calorimetry and summarized by a sigmoidal temperature transition. The midpoint temperature, Tm, has precise thermodynamic meaning that connects populations and free energy. Given Data / Assumptions: Two state unfolding model: Native ⇌ Unfolded. Definition of Tm as the temperature where the transition is halfway complete. Equilibrium constant Keq defined as [Unfolded] / [Native]. Concept / Approach: At the midpoint of a two state transition, populations of the native and unfolded states are equal. Therefore Keq = 1, and ΔG = -RTln(Keq) equals zero. This also means that exactly half of the protein molecules are denatured in a macroscopic sense. Step-by-Step Solution: 1) At Tm, by definition, the unfolding curve passes through 50 percent transition.2) Equal populations imply [Native] = [Unfolded].3) Therefore Keq = [U]/[N] = 1.0.4) Substituting into ΔG = -RTln(Keq) gives ΔG = 0 at Tm. Verification / Alternative check: Differential scanning calorimetry shows peak heat absorption near Tm. Although heat capacity changes can be nonzero across the transition, the free energy criterion at Tm remains ΔG = 0 for a two state system with Keq = 1. Why Other Options Are Wrong: Individual statements (a), (b), and (c) are each correct; the combined answer that includes all of them is most complete. Heat capacity change is zero at Tm: not a general rule; ΔCp can be significant and does not necessarily vanish at Tm. Common Pitfalls: Assuming that half denatured means half of every molecule is unfolded. The interpretation is population based: about half the molecules are native and half unfolded. Final Answer: All of these.

Question 7

Distribution of residues — Which statement about charged and hydrophobic amino acids in folded proteins is most accurate?

Accepted Answer

Introduction: Residue distribution in proteins balances hydrophobic effects, hydrogen bonding, and functional needs. This question asks for the most accurate generalization about where charged and hydrophobic residues are found in folded structures. Given Data / Assumptions: Typical soluble globular proteins in aqueous solution. Charged residues include Asp, Glu, Lys, Arg, and often His. Hydrophobic residues include Leu, Ile, Val, Phe, and others. Concept / Approach: Electrostatic and solvation considerations favor exposure of charged groups to water. However, evolution sometimes buries charged residues where they form salt bridges, bind ligands, tune pKa values, or contribute to catalysis. Hydrophobic residues preferentially occupy the core but are not universally buried because surfaces also require hydrophobic patches for interactions and stability. Step-by-Step Solution: 1) Identify the trend: charged residues usually occur on the surface where they remain solvated.2) Note exceptions: buried ion pairs, active sites, and transport channels can include charged residues.3) Evaluate hydrophobics: most core residues are hydrophobic, yet some hydrophobics appear on the surface within mixed patches or interfaces.4) Choose the statement that captures trends without making absolute claims. Verification / Alternative check: Statistical analyses of protein structures show significant enrichment of charged residues at solvent accessible surfaces, with a smaller fraction buried as functional exceptions. Tyrosine and other aromatics are amphipathic and frequently located near interfaces rather than exclusively in cores. Why Other Options Are Wrong: Never buried: too absolute; buried charged residues do exist. All hydrophobics are buried: many are partly exposed, especially at interaction interfaces. Tyrosine only interior: Tyr is amphipathic and often at boundaries. Glycine always surface: Gly can occur in cores, especially where tight turns or packing flexibility are needed. Common Pitfalls: Treating rules of thumb as strict rules. Protein structure is governed by tendencies with meaningful exceptions that support function and stability. Final Answer: Charged amino acids are seldom buried in the interior of a protein.

Question 8

Folding thermodynamics — Which factor is most unfavorable during protein folding?

Accepted Answer

Introduction: Protein folding reflects a competition between favorable and unfavorable contributions to free energy. This question asks which factor opposes folding the most, highlighting the entropic cost of ordering a flexible polypeptide chain. Given Data / Assumptions: Folding is considered under aqueous, near physiological conditions. Forces include hydrophobic effect, van der Waals packing, electrostatics, and covalent crosslinks. Conformational entropy refers to the loss of chain configurational freedom upon folding. Concept / Approach: The free energy of folding can be sketched as ΔG_fold = ΔH - T*ΔS. Hydrophobic interactions, van der Waals contacts, electrostatic pairs, and disulfide bonds generally contribute favorable enthalpy or solvent entropy. In contrast, restricting the chain from a vast ensemble of unfolded conformations to a single native state yields a large negative entropy change for the polypeptide, which is unfavorable (increases ΔG). Step-by-Step Solution: 1) Consider the unfolded state: many accessible conformations provide high entropy.2) Folding restricts backbone torsions and side chain rotamers, lowering chain entropy.3) Favorable contributions must compensate this penalty: hydrophobic effect increases solvent entropy by releasing ordered water, while packing and electrostatics contribute favorable enthalpy.4) Therefore, the most unfavorable term is the loss of conformational entropy of the chain. Verification / Alternative check: Calorimetric measurements and computational models consistently require large favorable terms to offset the chain entropy cost, which explains marginal stability and temperature sensitivity of many proteins. Why Other Options Are Wrong: Hydrophobic interactions: major favorable driver of folding. Van der Waals interactions: favorable packing in the core. Electrostatic interactions: often favorable through salt bridges and dipole interactions, though context dependent. Disulfide bonds: covalent crosslinks stabilize the folded state. Common Pitfalls: Confusing solvent entropy with chain entropy. Hydrophobic collapse increases solvent entropy even while chain entropy decreases. Final Answer: Conformational entropy.

Question 9

Noncovalent forces — When do attractive van der Waals interactions occur?

Accepted Answer

Introduction: Van der Waals interactions include dispersion, induction, and orientation components. This question probes the generality of the attractive dispersion term that underlies packing in protein interiors and molecular crystals. Given Data / Assumptions: Atoms or molecules are neutral overall. Distances are within the range where electron cloud fluctuations can interact but are not so close as to cause Pauli repulsion. Temperature and phase can vary. Concept / Approach: London dispersion forces arise from instantaneous correlated fluctuations in electron density, leading to transient dipoles that attract each other. Because all atoms and molecules exhibit electron fluctuations, the attractive component is universal. The interaction strength depends on polarizability and distance and is balanced by short range repulsion at very close approach. Step-by-Step Solution: 1) Recognize that dispersion does not require permanent polarity or a specific phase.2) Any two nearby atoms experience an attractive dispersion term that scales approximately with 1/r^6 at intermediate separations.3) Optimal packing places atoms near the minimum of the Lennard–Jones potential, maximizing dispersion while avoiding repulsion.4) Therefore, the statement that any pair of nearby atoms experiences attractive van der Waals forces is correct. Verification / Alternative check: Molecular dynamics force fields include a universal dispersion term for all atom pairs, modulated by atom types and distance dependent parameters, reflecting this generality. Why Other Options Are Wrong: Apolar molecules in liquid only: too narrow; van der Waals act in all phases. Polar molecules in solid only: again too narrow. Only if other forces are less favorable: dispersion acts regardless of other interactions. Only between identical atoms: dispersion occurs between any atom types. Common Pitfalls: Assuming polarity is required for attraction. Dispersion is present even for noble gases and saturated hydrocarbons, explaining liquefaction and crystal formation. Final Answer: any pair of nearby atoms.

Question 10

Driving forces of folding — Which interaction is generally the most favorable contributor to protein folding?

Accepted Answer

Introduction: Several interactions stabilize the native state of proteins. This question targets the principal favorable driver that pushes nonpolar residues into the core and shapes the overall fold in aqueous environments. Given Data / Assumptions: Folding occurs in water at near physiological conditions. Interactions considered include hydrophobic effect, hydrogen bonding, van der Waals, and others. Focus is on overall free energy contribution rather than local specificity. Concept / Approach: The hydrophobic effect is largely entropic from the solvent perspective. Aggregating nonpolar surfaces reduces the structured water shell, releasing water molecules to the bulk and increasing solvent entropy, which strongly favors folding. Other forces such as hydrogen bonds and van der Waals contacts refine the fold and add stability but are often compensatory because similar interactions exist in the unfolded state with water. Step-by-Step Solution: 1) Consider the unfolded state: nonpolar side chains are hydrated, imposing ordered water shells.2) Upon folding, nonpolar surfaces bury against each other, decreasing total nonpolar surface exposed to water.3) This releases ordered water and increases solvent entropy, lowering free energy significantly.4) Hydrogen bonds and van der Waals interactions contribute, but their net gain can be modest compared to the hydrophobic effect because the unfolded chain also forms hydrogen bonds with water. Verification / Alternative check: Mutational studies show that substituting hydrophobic core residues with polar or smaller residues often destabilizes the fold more than disrupting individual hydrogen bonds, highlighting the dominant role of hydrophobic burial. Why Other Options Are Wrong: Conformational entropy: unfavorable for folding. Van der Waals: favorable but typically secondary to hydrophobic driving force. Hydrogen bonds: important for specificity and structure, but many are exchanged with water in the unfolded state. Disulfide bonds: stabilizing but not universal; many proteins fold without them. Common Pitfalls: Equating hydrogen bonding with the main folding driver. Hydrogen bonds are essential for secondary structure but the hydrophobic effect largely powers the global collapse. Final Answer: Hydrophobic interactions.

Question 11

Protein denaturation example — In a hard boiled egg, which structural level of ovalbumin is least affected by denaturation?

Accepted Answer

Introduction: Cooking an egg is a classic demonstration of protein denaturation. Heat disrupts weak interactions, leading to loss of native structure and aggregation. This question asks which level of protein structure remains least affected under such conditions for the major egg white protein, ovalbumin. Given Data / Assumptions: Ovalbumin is heated to the point of irreversible denaturation in a hard boiled egg. No proteolysis or peptide bond hydrolysis is implied by normal cooking times. Denaturation refers to loss of native secondary, tertiary, and quaternary organization. Concept / Approach: Primary structure is the linear amino acid sequence linked by peptide bonds. Heat denaturation primarily disrupts noncovalent forces (hydrogen bonds, hydrophobic interactions, ionic pairs) and can break some disulfide arrangements but generally does not cleave peptide bonds. Therefore, the amino acid sequence remains largely intact, while higher order structures are altered. Step-by-Step Solution: 1) Recognize that secondary and tertiary structures depend on noncovalent interactions sensitive to heat.2) As temperature rises, unfolding exposes hydrophobic regions, promoting aggregation and opacity.3) Primary covalent bonds require harsher chemical conditions or prolonged extremes to break; typical cooking does not hydrolyze peptide bonds.4) Quaternary changes can occur if the native protein is oligomeric or forms aggregates upon denaturation. Verification / Alternative check: Electrophoresis after cooking shows bands corresponding to intact polypeptide mass unless deliberate hydrolysis is performed. Functional assays lose activity, confirming higher order structural loss despite preserved sequence. Why Other Options Are Wrong: Secondary and tertiary: these are the first to change during denaturation. Quaternary: inter subunit arrangements or aggregation are affected by heat. All equally affected: different levels respond differently; covalent backbones are most resistant. Common Pitfalls: Equating denaturation with chemical degradation. Denaturation is not the same as hydrolysis; it is a loss of native conformation without necessarily breaking the peptide chain. Final Answer: The primary structure of ovalbumin.

Protein Stability Questions