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AP Biology — Practice Questions flashcards

6 of 60 questions free to study here · 9 topics · the whole deck is built into Willys

Multiple-choice practice questions, one best answer each, on all eight AP Biology units, with water potential, chi-square, Hardy–Weinberg and surface-area-to-volume calculations.

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Practice deck · part of AP Biology · written against: College Board AP Biology Course and Exam Description (eight units). Updated 2026-10-10.

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  1. In a mutant enzyme, a nonpolar amino acid buried in the hydrophobic core is replaced by an amino acid with a charged R group. Which outcome is most likely?

    A) The fold changes, which can alter the shape of the active site
    B) The enzyme becomes more stable inside the lipid bilayer
    C) No peptide bond can form at that position in the chain
    D) The primary structure is unchanged, so activity is unaffected
    A) The fold changes, which can alter the shape of the active site

    A charged R group is not stable in a water-excluding core, so the tertiary structure shifts and the active site can lose its fit for the substrate. The primary structure has in fact changed — a substitution is a change in sequence — and the backbone still forms a normal peptide bond.
    Chemistry of Life › Macromolecules
  2. An enzyme's rate is measured with and without compound Q. At low substrate concentration Q halves the rate, but at very high substrate concentration the rates are equal. What best explains this?

    A) Q makes the reaction's free-energy change less negative
    B) Q competes with the substrate for the active site
    C) Q binds an allosteric site and changes the enzyme's shape
    D) Q denatures the enzyme by breaking its hydrogen bonds
    B) Q competes with the substrate for the active site

    A competitive inhibitor can be outcompeted when substrate is abundant, so maximum rate is restored. Allosteric (noncompetitive) inhibition and denaturation would still lower the rate at high substrate.
    Cellular Energetics › Enzymes
  3. A drug blocks phosphodiesterase, the enzyme that breaks down cAMP. How would this affect a liver cell's response to epinephrine, which signals through cAMP?

    A) The response is unchanged because the receptor is unaffected
    B) The response lasts longer because cAMP stays elevated
    C) The response is blocked because epinephrine cannot bind
    D) The response is weaker because less cAMP is produced
    B) The response lasts longer because cAMP stays elevated

    cAMP is the second messenger; removing the enzyme that breaks it down keeps the signal on after epinephrine leaves. The drug acts downstream of the receptor, so binding and cAMP production are unaffected.
    Cell Communication
  4. In a species with 2n = 8, one homologous pair fails to separate during meiosis I, and meiosis II is normal. What chromosome numbers do the four gametes have?

    A) Two have 5 and two have 3
    B) All four have the normal 4
    C) One has 5, one has 3 and two have 4
    D) Two have 8 and two have 0
    A) Two have 5 and two have 3

    When homologs fail to separate in meiosis I, one cell gets both and the other gets neither, so every gamete is abnormal. One gamete with 5, one with 3 and two normal is the pattern for nondisjunction in meiosis II.
    Heredity › Meiosis
  5. A double-stranded DNA sample is 30 percent adenine. What percentage of its bases are guanine?

    A) 40 percent
    B) 70 percent
    C) 30 percent
    D) 20 percent
    D) 20 percent

    Adenine pairs with thymine, so A + T = 60 percent, leaving 40 percent for G + C, which are equal: 20 percent guanine. 40 percent is the combined G + C value.
    Gene Expression › DNA Structure and Replication
  6. In a food chain, secondary consumers receive about 30 kilocalories per square meter per year. Using the 10 percent rule, roughly how much energy did the producers capture?

    A) 300 kcal per square meter per year
    B) 30,000 kcal per square meter per year
    C) 3,000 kcal per square meter per year
    D) 90 kcal per square meter per year
    C) 3,000 kcal per square meter per year

    Working backward, primary consumers have about 300 and producers about 3,000 kilocalories. 300 stops one level short, at the primary consumers.
    Ecology › Energy and Cycles

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About these cards

Review status: Second-reader reviewed, with corrections applied.

Checked against:

  • OpenStax Biology 2e, consulted as a reference; no text copied.
  • Checked against Campbell Biology, 12th ed. (reference only; no text copied).
  • Topic coverage follows the published AP Biology course units in the Course and Exam Description effective Fall 2025 (the AP name is used nominatively; no exam or course material reproduced).

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