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The frequency of the recessive allele would be 0.55, that of the dominant allele would be 0.45, and that of the heterozygous individuals would be 0.495
Population genetics
Recall that for a population in Hardy-Weinberg equilibrium:
p2 + 2pq + q2 = 1
Also, p + q = 1
Where:
p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p2 = percentage of homzygous dominant individuals
q2 = percentage of homzygous recessive individuals
2pq = percentage of heterozygous individuals
In this case, the percentage of homzygous recessive individuals can be calculated as:
Homzygous recessive = those that could not taste PTC = 65/215
q2 = 65/215 = 0.302
q = 0.55
Since p + q = 1
p = 1 - 0.55
= 0.45
The frequency of the heterozygous individuals = 2pq
= 2 x 0.45 x 0.55
= 0.495
More on Hardy-Weinberg equilibrium can be found here: https://brainly.com/question/3406634