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MCAT Protein Structure, Folding and Denaturation Practice Questions

By StudyStore · Published 2026-10-20 · 8 original questions · about 12 min

Eight free MCAT-style practice questions on peptide bonds, alpha helices, beta sheets, disulfides, the hydrophobic effect, prions and denaturation, with full explanations.

Quick concept primer

Protein structure questions come in two flavors: name the level of structure, or predict what happens when a force that holds a level together is removed. Keep a short ledger. Primary structure is the covalent amino acid sequence. Secondary structure is local backbone hydrogen bonding (alpha helix, beta sheet). Tertiary structure is the full fold of one chain, stabilized by the hydrophobic effect, hydrogen bonds, ionic interactions, disulfide bonds and van der Waals contacts. Quaternary structure is the arrangement of separate chains.

The peptide bond has partial double bond character from resonance, so the six atoms around it lie in a plane and rotation is restricted; the freedom lives in the phi and psi angles at each alpha carbon. Folding is driven mainly by the hydrophobic effect: burying nonpolar side chains frees ordered water, raising the entropy of the surroundings.

Traps: urea, heat and detergents disrupt noncovalent forces but leave peptide bonds intact, while strong acid hydrolysis breaks them; and reducing agents break disulfides, which urea alone does not.

Free notes for this topic: Protein Structure: Primary to Quaternary, and What Denaturation Destroys
Free tools: Amino acid quiz

8 practice questions

Try each question before opening the explanation. Click "Show answer and explanation" to check yourself. All questions are original and written for study practice.

Question 1

Resonance in the peptide bond has which direct consequence for protein structure?

  1. The amide nitrogen becomes strongly basic and is usually protonated at pH 7
  2. The carbonyl carbon, carbonyl oxygen, amide nitrogen, amide hydrogen and the two neighboring alpha carbons lie in one plane, and rotation about the C-N bond is restricted
  3. The peptide bond is much easier to hydrolyze than an ordinary amide
  4. The C-N bond rotates freely, allowing a large variety of backbone conformations
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Answer: B. The carbonyl carbon, carbonyl oxygen, amide nitrogen, amide hydrogen and the two neighboring alpha carbons lie in one plane, and rotation about the C-N bond is restricted

Delocalization of the nitrogen lone pair into the carbonyl gives the C-N bond partial double bond character, which holds the amide group planar and strongly favors the trans arrangement. Backbone flexibility comes only from rotation about the N-C(alpha) and C(alpha)-C bonds (phi and psi). Choice C is the opposite of the truth, since resonance makes the bond more stable, and the amide nitrogen is not basic because its lone pair is shared with the carbonyl.

Question 2

Which residue is LEAST likely to be found in the middle of an alpha helix?

  1. Alanine
  2. Leucine
  3. Proline
  4. Glutamate
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Answer: C. Proline

In an alpha helix the carbonyl oxygen of residue i accepts a hydrogen bond from the N-H of residue i+4. Proline's nitrogen is part of a ring and has no amide hydrogen to donate, and the ring locks phi at a value that does not fit a helix, so proline introduces a kink. Alanine, leucine and glutamate are all common helix residues because their side chains do not interfere with the backbone.

Question 3

A researcher unfolds a small enzyme containing four disulfide bonds with 8 M urea and a reducing agent. She then removes both by slow dialysis in air, and 90 percent of the original activity returns. Which conclusion is best supported?

  1. Chaperone proteins must have contaminated the sample, since folding cannot occur without them
  2. Denaturation broke peptide bonds, which then reformed spontaneously
  3. Disulfide bonds are not part of the final native structure
  4. The amino acid sequence contains enough information to specify the native fold under the right conditions
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Answer: D. The amino acid sequence contains enough information to specify the native fold under the right conditions

The recovery of activity from a fully unfolded, reduced chain shows that the information needed to reach the native fold is encoded in the primary sequence. The disulfides re-form during air oxidation, so choice C is wrong. Urea and reducing agents do not break peptide bonds (choice B), and nothing in the experiment points to chaperones, although many proteins in cells do rely on them to avoid aggregation.

Question 4

Which treatment would change a protein's primary structure?

  1. Heating it in 6 M HCl at 110 degrees C for 24 hours
  2. Dissolving it in 8 M urea
  3. Heating it briefly to 80 degrees C
  4. Adding the detergent SDS
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Answer: A. Heating it in 6 M HCl at 110 degrees C for 24 hours

Primary structure is the sequence of residues joined by covalent peptide bonds. Hot concentrated acid hydrolyzes those bonds and breaks the chain into free amino acids. Urea, brief heating and SDS all unfold the protein by disrupting hydrogen bonds and hydrophobic interactions, which alters higher-order structure only.

Question 5

Why is the folding of a globular water-soluble protein spontaneous even though the protein chain itself becomes more ordered?

  1. The peptide bonds are formed during folding, releasing energy
  2. Burying nonpolar side chains releases ordered water molecules, so the entropy of the surroundings increases enough to make the overall free energy change negative
  3. Folding is highly endothermic, which drives the reaction forward
  4. The entropy of the protein increases on folding
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Answer: B. Burying nonpolar side chains releases ordered water molecules, so the entropy of the surroundings increases enough to make the overall free energy change negative

The hydrophobic effect is an entropy-driven phenomenon. Water forms an ordered cage around exposed nonpolar groups; collapsing these groups into the core frees the water, and the entropy gain of the solvent outweighs the entropy loss of the chain. Choice D has the sign wrong for the chain, choice C misreads the thermodynamics (folding is typically modestly exothermic from hydrogen bonds and van der Waals contacts), and choice A confuses folding with translation.

Question 6

A purified protein runs as a single band at about 160 kDa on a native gel. On SDS-PAGE with a reducing agent it gives a single band at 40 kDa. On SDS-PAGE without a reducing agent it gives a single band at 80 kDa. Which model fits these data?

  1. A tetramer of identical 40 kDa subunits in which pairs of subunits are joined by interchain disulfide bonds
  2. A dimer of two different 80 kDa subunits
  3. Four 40 kDa subunits associated through noncovalent interactions alone
  4. A single polypeptide of 160 kDa that is cleaved by SDS
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Answer: A. A tetramer of identical 40 kDa subunits in which pairs of subunits are joined by interchain disulfide bonds

SDS breaks noncovalent contacts, so any band that persists without a reducing agent reflects covalent links. The 80 kDa band means the subunits are paired by disulfides; adding the reducing agent separates them into the 40 kDa units, and the native 160 kDa mass is four of them. If noncovalent interactions alone held them (choice C), the non-reducing SDS gel would also show 40 kDa.

Question 7

Which statement best describes how a prion causes disease?

  1. The prion destroys the host's chaperone genes
  2. A misfolded, beta sheet-rich conformer of a normal host protein induces other copies of that protein to adopt the same abnormal fold
  3. A mutation in the prion's own RNA changes the amino acid sequence of host proteins
  4. A viral genome integrates into host DNA and directs production of toxic proteins
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Answer: B. A misfolded, beta sheet-rich conformer of a normal host protein induces other copies of that protein to adopt the same abnormal fold

Prion disease is a disease of conformation, not sequence. The infectious form of the prion protein has the same primary structure as the normal form but a different, aggregation-prone fold that templates the conversion of normal protein molecules. No nucleic acid is required, which is why prions resist treatments that destroy DNA and RNA.

Question 8

Which statement about beta sheets is correct?

  1. Hydrogen bonds form only within a single strand
  2. A beta sheet can only contain strands that run in the same direction
  3. Hydrogen bonds form between side chain groups on adjacent strands
  4. Hydrogen bonds form between backbone N-H and C=O groups on adjacent strands, and side chains point alternately above and below the sheet
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Answer: D. Hydrogen bonds form between backbone N-H and C=O groups on adjacent strands, and side chains point alternately above and below the sheet

Beta sheet stability comes from backbone hydrogen bonds between neighboring strands, which can run parallel or antiparallel. Because the backbone is nearly fully extended and pleated, consecutive side chains point to opposite faces. In contrast, an alpha helix is stabilized by hydrogen bonds within a single chain, which is why choice A describes a helix, not a sheet.

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Written by StudyStore from standard scientific facts as a study aid. These are not official exam questions and are not affiliated with or endorsed by any testing organization. Always check values against the reference provided with your exam or course.