Assume a population is in hardy-weinberg equilibrium for the b gene, which has two alleles, b and b. in this population of 100,000 individuals, if the frequency of the b allele (q) is 4%, how many homozygous bb individuals would you expect to find in the population?

160

Explanation:

If p = frequency of dominant allele and q = frequency of recessive allele according to Hardy-Weinberg equilibrium,

p + q = 1

p² + 2pq + q² = 1

where p² = frequency of dominant homozygous genotype

q² = frequency of recessive homozygous genotype

2pq = frequency of heterozygous genotype

Here,

b = 4% = 0.04

bb = 0.04 * 0.04 = 0.0016

Number of individuals in population = 100000

Number of homozygous bb individuals = 0.0016 * 100000 = 160

i believe the answer is a or 1

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