Question

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?
a.o.16
b.0.4
c.0.6
d.0.24
d.0.48

Answer 1

Answer:

d.0.48

Explanation:

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.

Therefore:

q²=16/100

q² = 0.16

q=√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