Properties of Biological Macromolecules

Properties of Biological Macromolecules is topic 1.4 of AP Biology, inside Chemistry of Life. This page works through three real practice questions on it, with the full reasoning behind each credited answer.

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71 questionsConcept Explanation55% with a figure
Worked examples

Three real properties of biological macromolecules questions

From the practice pool, not the mock papers — each with the reasoning that produces the answer.

71 practice questions sit on this single topic — enough that it is worth working through by topic rather than meeting it scattered through a unit review.

55% of them come with a figure. Reading the graph or diagram correctly is most of the work here before any content knowledge applies.

Concept Explanation · with figure

A protein contains a charged amino acid at position 82. A mutation replaces it with a nonpolar amino acid. Which outcome is most likely if position 82 normally forms an ionic interaction with another side chain?

  1. AThe tertiary structure is less stable because a side-chain interaction is lost.correct
  2. BThe mutation creates a new phospholipid bilayer around the protein.
  3. CThe peptide backbone is completely hydrolyzed at position 82.
  4. DThe DNA sequence remains unchanged because proteins cannot be affected by mutations.
Why A is correct

Tertiary structure depends partly on interactions among amino acid side chains. Replacing a charged side chain with a nonpolar one can remove an ionic interaction and reduce stability.

The evidence in the prompt matters before the topic label: "A protein contains a charged amino acid at position 82. A mutation replaces it with a nonpolar amino acid. Which...". The supported answer is "tertiary structure is less stable because a side-chain interaction is lost" because Tertiary structure depends partly on interactions among amino acid side chains. Distractors such as mutation creates a new phospholipid bilayer around the protein or peptide backbone is completely hydrolyzed at position 82 either ignore the stated pattern or name a related concept without matching the evidence.

Argumentation

An enzyme is exposed to a solution with a pH far from its optimum. The enzyme's amino acid sequence is unchanged, but its reaction rate decreases sharply. Which explanation is most consistent with the observation?

  1. AWater dissolves ions mainly by forming covalent bonds with each ion.
  2. BThe enzyme gained a new peptide bond at the active site.
  3. CThe enzyme's DNA sequence was immediately altered by the pH change.
  4. DChanges in ionization disrupted tertiary structure near the active site.correct
Why D is correct

A large pH change can alter charges on amino acid side chains, disrupting interactions that maintain tertiary structure. If the active site's shape changes, enzyme activity can decrease even when primary structure is unchanged.

The stem asks students to explain a specific mechanism, not just recall a unit term: "An enzyme is exposed to a solution with a pH far from its optimum. The enzyme's amino acid sequence is unchanged,...". The correct choice is supported by A large pH change can alter charges on amino acid side chains, disrupting interactions that maintain tertiary structure. The tempting alternatives such as Water dissolves ions mainly by forming covalent bonds with each ion or enzyme gained a new peptide bond at the active site fail because they do not produce the same outcome described in the stem.

Concept Explanation

A mutation replaces a cysteine residue that normally forms a disulfide bridge in a secreted protein. Which outcome is most likely?

  1. AProtein will be translated from DNA instead of mRNA
  2. BThe protein will gain a phospholipid bilayer around each amino acid.
  3. CWater dissolves ions mainly by forming covalent bonds with each ion.
  4. DA lost disulfide bridge can destabilize the folded proteincorrect
Why D is correct

Disulfide bridges are covalent interactions between cysteine side chains that can stabilize tertiary structure. Losing one can reduce folding stability without preventing all peptide-bond formation.

The stem asks students to explain a specific mechanism, not just recall a unit term: "A mutation replaces a cysteine residue that normally forms a disulfide bridge in a secreted protein. Which outcome...". The correct choice is supported by Disulfide bridges are covalent interactions between cysteine side chains that can stabilize tertiary structure. The tempting alternatives such as protein will be translated directly from DNA instead of mRNA or protein will gain a phospholipid bilayer around each amino acid fail because they do not produce the same outcome described in the stem.

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Last reviewed 2026-08-28. Topic and unit names follow the College Board course framework. Question counts describe the PrepScore practice bank, not the exam.
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