In a population of 100 individuals in Hardy-Weinberg equilibrium with p=0.7, what is the expected frequency of homozygous recessive individuals?

In a population that is in Hardy-Weinberg equilibrium, the expected frequencies of genotypes can be calculated using the allele frequencies. In this scenario, p represents the frequency of the dominant allele (let's call it A) and q represents the frequency of the recessive allele (a). The relationship between p and q is straightforward: since there are only two alleles, p + q = 1.

Given that p = 0.7, we can calculate q as follows:

[ q = 1 - p = 1 - 0.7 = 0.3 ]

The expected frequency of homozygous recessive individuals (genotype aa) is calculated using the formula ( q^2 ).

Here, we find:

[ q^2 = (0.3)^2 = 0.09 ]

Thus, the expected frequency of homozygous recessive individuals in a population of 100 individuals is:

[ 0.09 \times 100 = 9 ]

In summary, the expected frequency of homozygous recessive individuals is 0.09, which reflects the application of the Hardy-Weinberg principle to the genotype frequencies based on the calculated