What proportion of the population is expected to be heterozygous (Aa)?

Answer: 0.42 2pq = 2 x 0.7 x 0.3 = 0.42. Approximately 42% of the population is expected to be heterozygous.

What fraction of newborns is expected to be affected (aa)?

Answer: 0.04 q² = 0.2² = 0.04. About 4% of newborns will be homozygous recessive (affected) under HW equilibrium.

The Hardy-Weinberg principle predicts allele/genotype frequencies when the population is

Answer: Not evolving (large, randomly mating, no selection, no migration, no mutation) HW provides the null hypothesis: stable allele frequencies in an idealized non-evolving population.

AP Biology · Topic 7.5

Hardy-Weinberg Equilibrium Practice

Part of Natural Selection.(EVO-1.E)

Practice questions

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Sample questions

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Sample 1difficulty 1/5
$What proportion of the population is expected to be heterozygous (Aa)?$
- A
  $0.42$
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- B
  $0.09$
- C
  $0.49$
- D
  $0.21$
Why
$2pq = 2 x 0.7 x 0.3 = 0.42. Approximately 42% of the population is expected to be heterozygous.$
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Sample 2difficulty 2/5
$A recessive disease (aa) is lethal before reproduction. The current allele frequencies are p = 0.8, q = 0.2 in a HW population.$
$What fraction of newborns is expected to be affected (aa)?$
- A
  $0.20$
- B
  $0.16$
- C
  $0.04$
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- D
  $0.32$
Why
$q² = 0.2² = 0.04. About 4% of newborns will be homozygous recessive (affected) under HW equilibrium.$
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Sample 3difficulty 2/5
$The Hardy-Weinberg principle predicts allele/genotype frequencies when the population is$
- A
  $Highly inbred (frequent mating between close relatives reduces heterozygosity)$
- B
  $Small and isolated (subject to substantial genetic drift and founder effects)$
- C
  $Not evolving (large, randomly mating, no selection, no migration, no mutation)$
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- D
  $Actively evolving (small population with strong directional selection acting on alleles)$
Why
$HW provides the null hypothesis: stable allele frequencies in an idealized non-evolving population.$
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Sample 4difficulty 2/5
$The Hardy-Weinberg equation is$
- A
  $p + q = 1$
- B
  $p² + 2pq + q² = 1$
- C
  $p \times q = 1$
- D
  $Both A and B$
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Why
$p + q = 1 (allele frequencies); p² + 2pq + q² = 1 (genotype frequencies). Both must hold under HWE.$
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Sample 5difficulty 2/5
$A population of 1000 individuals is at Hardy-Weinberg equilibrium with p² = 0.49 and q² = 0.09.$
$What is the expected number of heterozygotes (Bb)?$
- A
  $210$
- B
  $840$
- C
  $490$
- D
  $420$
  check_circle
Why
$p² + 2pq + q² = 1, so 2pq = 1 - 0.49 - 0.09 = 0.42. With N = 1000, expected Bb = 0.42 x 1000 = 420.$
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