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AP Biology Practice (Interactive)

AP — Biology Practice

10 auto-graded practice problems. Select an answer, submit, and review the explanation.


Worked Examples

Example 1: Hardy-Weinberg Equilibrium

Problem: In a population of 1000 individuals, 160 show the recessive phenotype. Calculate the frequency of the dominant allele.

Solution: Step 1: Let q2q^2 = frequency of homozygous recessive = 160/1000=0.16160/1000 = 0.16

Step 2: q=0.16=0.4q = \sqrt{0.16} = 0.4 (frequency of recessive allele)

Step 3: p=1−q=1−0.4=0.6p = 1 - q = 1 - 0.4 = 0.6 (frequency of dominant allele)

Step 4: Verify: p2+2pq+q2=0.36+0.48+0.16=1.0p^2 + 2pq + q^2 = 0.36 + 0.48 + 0.16 = 1.0 ✓

Key insight: The Hardy-Weinberg equation p2+2pq+q2=1p^2 + 2pq + q^2 = 1 relates allele frequencies to genotype frequencies. It assumes no selection, mutation, migration, or genetic drift.


Example 2: ATP Calculation

Problem: How many ATP molecules are produced from one glucose molecule through complete aerobic respiration?

Solution: Step 1: Glycolysis → 2 ATP (net) + 2 NADH Step 2: Pyruvate oxidation → 2 NADH (per glucose, 2 pyruvates) Step 3: Krebs cycle → 2 ATP + 6 NADH + 2 FADH₂ Step 4: Total NADH = 2 + 2 + 6 = 10 NADH → ~25 ATP (2.5 ATP per NADH) Step 5: Total FADH₂ = 2 → ~3 ATP (1.5 ATP per FADH₂) Step 6: Grand total: 2 + 25 + 3 = ~30 ATP (modern estimate)

Key insight: Older textbooks say 36-38 ATP, but modern estimates are ~30 ATP due to the cost of transporting NADH into mitochondria.


Example 3: Chi-Square Test

Problem: A genetic cross produces 312 round seeds and 88 wrinkled seeds. Is this consistent with a 3:1 ratio?

Solution: Step 1: Null hypothesis: data fit 3:1 ratio Step 2: Expected: round = 34×400=300\frac{3}{4} \times 400 = 300, wrinkled = 14×400=100\frac{1}{4} \times 400 = 100

Step 3: Calculate χ2\chi^2:

χ2=(312−300)2300+(88−100)2100=144300+144100=0.48+1.44=1.92\chi^2 = \frac{(312-300)^2}{300} + \frac{(88-100)^2}{100} = \frac{144}{300} + \frac{144}{100} = 0.48 + 1.44 = 1.92

Step 4: Degrees of freedom = 2−1=12 - 1 = 1

Step 5: Critical value at p=0.05p = 0.05 is 3.843.84

Step 6: 1.92<3.841.92 < 3.84, so we fail to reject the null hypothesis. The data are consistent with 3:1.

Key insight: A small χ2\chi^2 means observed values are close to expected. A large χ2\chi^2 means they differ significantly.


Chemistry of Life

Intuition: Biology starts with chemistry, the molecules of life (water, proteins, nucleic acids, lipids, carbohydrates) follow chemical rules, but their organisation creates something greater than the sum of parts.

Common Mistakes: Confusing DNA and RNA bases; mixing up enzyme inhibition types; forgetting that enzymes lower activation energy but don’t change ΔG.


Cell Structure and Function


Cellular Energetics


Heredity

Cross-References

  • Chemistry of Life: Water, macromolecules, and enzyme fundamentals.
  • Cell Structure: Cell organelles, membrane transport, and surface-area-to-volume ratio.
  • Heredity: Mendelian genetics, Punnett squares, and chi-square tests.