For this question, we will utilize a population of martians that is in hardy weinberg equilibrium. the dominant martian phenotype is the possession of 2 antennae. in this population, 84 of the martians have 2 antennae, while 16 lack antennae. what is the frequency of heterozygotes in this population?
When a population is in Hardy Weinberg equilibrium the genotypic frequencies are:
freq (AA) = p²
freq (Aa) = 2pq
freq (aa) = q²
p is the frequency of the dominant A allele and q is the frequency of the recessive a allele.
In this population of 100 individuals, 84 martians have the dominant phenotype and 16 have the recessive phenotype.
q² = 0.16
q = 0.4
And p+q=1, so:
p = 1 - q
p = 1-0.4
p = 0.6
The frequency of heterozygotes is:
freq (Aa) = 2pq = 2 × 0.4 × 0.6
freq (Aa) = 0.48
it has one origin of replication, but it proceeds in two directions - d)