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

humans, along with other animals and plants, have linear chromosomes that are arranged in pairs within the nucleus of the cell. the only human cells that do not contain pairs of chromosomes are reproductive cells, or gametes, which carry just one copy of each chromosome.

all answers true and false

explanation: