Question

If a population has only two alleles for a gene (B, b) with equal frequencies and it is in Hardy–Weinberg equilibrium, what are the expected frequencies of genotypes BB, Bb, and bb?

Answer 1

Answer:

According to the Hardy-Weinberg equilibrium equation, heterozygotes are represented by the 2nd term. Therefore, the number of heterozygous individuals (Aa) is equal to 2 pq, which is equivalent to 2 × 0.19 × 0.81 = 0.31 or 31%

Explanation:

Homozygous recessive individuals (aa) are represented by the term q 2 in the equilibrium HWecuation which is equivalent to 0.81 × 0.81 = 0.66 or 66%

Answer 2

As per the Hardy-Weinberg equilibrium equation the heterozygotes can be shown by the 2nd term. Hence, the number of the heterozygous individuals (A a) is equal to two p q, which is equivalent  

• to 2 × 0.19 × 0.81 = 0.31 or 31%.

• The H W equation which is  = to 0.81 × 0.81 = 0.66 or 66 %.

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