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

In a population of Martians in Hardy-Weinberg equilibrium, to calculate the frequency of heterozygotes (2pq) when given the number of individuals expressing the dominant and recessive phenotypes, we determine p and q and then use the formula 2pq. The frequency of heterozygous Martians was calculated to be 0.48.

d is correct

Step-by-step explanation:

The subject in question involves the Hardy-Weinberg principle, which is used to calculate allele frequencies within a population that is not evolving. Given the information that 84 Martians display the dominant phenotype and 16 do not, we're tasked with determining the frequency of heterozygotes. In a population in Hardy-Weinberg equilibrium, the frequencies can be calculated using the formula p² + 2pq + q² = 1, where p represents the frequency of the dominant allele, q represents the frequency of the recessive allele, and 2pq represents the frequency of the heterozygotes.

First, we determine the total number of individuals to be 100 (84 with antennae and 16 without). The frequency of recessive homozygotes (q²) can be calculated as 16/100 = 0.16. Taking the square root of q² gives us q = 0.4. The frequency of the dominant allele p can be found using p = 1 - q, which equals 1 - 0.4 = 0.6.

Now, to find the frequency of heterozygotes (2pq), we multiply 2 by the frequency of the dominant allele (p) and the frequency of the recessive allele (q). This gives us 2pq = 2 * 0.6 * 0.4 = 0.48. Therefore, the frequency of heterozygotes in this Martian population is 0.48.

Answer 2

d.0.48

Step-by-step 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