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?

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%

after making a punnet square using bv bv bv bv you will find that the ratio will be 9: 7 or a 128.6% probability.

hope this

the answers are a and c