A recessive lethal gene in chickens causes circulatory failure and death of the embryo at 70 hours.A commercial hatchery finds that a hatching failure due to this gene of greater than 4% is unacceptable.What is the upper limit for the frequency of this allele in the breeding population of fowls that is acceptable to the hatchery managers?

Answer 1

Hardy-Weinberg Law

p^2 + 2pq + q^2 = 1 and p + q = 1

p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population

p^2 = percentage of homozygous dominant individuals
q^2 = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals

so in this problem, q^2=.04, which implies p=1-q=1-.2=.8
and 2pq=2*.8*.2=.32

So the upper limit for those carrying the gene in the breeding population is 32 percent.

Answer 2

Thank you so much. I don't really understand this law at all.