Q.58 In a random mating population, Y and y are dominant and recessive alleles, respectively.
If the frequency of Y allele in both sperm and egg is 0.70, then the frequency of
Yy heterozygotes (rounded off to two decimal places) is _________.
Hardy-Weinberg equilibrium governs genotype frequencies in random mating populations. With Y allele frequency (p) at 0.70, the heterozygote Yy frequency is 0.42. This calculation applies directly to the given scenario.
Correct Answer
The frequency of Yy heterozygotes is 0.42 (rounded to two decimal places).
Step-by-Step Calculation
In Hardy-Weinberg equilibrium for a random mating population, genotype frequencies derive from allele frequencies where p + q = 1. Here, dominant Y allele frequency p = 0.70 in both sperm and eggs, so recessive y allele frequency q = 1 – 0.70 = 0.30.
Heterozygote frequency equals 2pq: 2 × 0.70 × 0.30 = 0.42. This holds under assumptions of infinite population size, no selection, mutation, migration, or drift.
Why This Formula Applies
Random mating ensures gametes combine independently, yielding YY = p² (0.49), Yy = 2pq (0.42), and yy = q² (0.09), summing to 1. The equal allele frequencies in sperm and eggs confirm panmictic conditions for equilibrium.
Common Mistakes Explained
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Mistaking heterozygote for p² (0.49): Applies only to homozygous dominant YY.
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Using pq without 2 (0.21): Ignores two ways to form Yy (Y from sperm/y from egg, or vice versa).
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Forgetting q = 0.30: Leads to incorrect 2pq values.
