Respuesta :
The frequency of allele G (in 1960) is p=0.8
If population is in HW equilibrium than p+q=1, which means that the frequency of the allele q (in 1960) is 0.2.
The frequencies of the genotypes are:
p2=0.64 dominant homozygous
2pq=0.32 heterozygous= 32%
q2=0.04 recessive heterozygous
Answer:
Option B, [tex]32[/tex]%
Explanation:
Given,
Gray (G) is dominant to white (g)
As per Hardy Weinberg's equation, the frequency of dominant allele is represented by "p" and recessive allele is represented by "q"
In 1960, the frequency of allele "G" is [tex]0.8[/tex]
So it means, [tex]p=0.8[/tex]
As per Hardy Weinberg's I equilibrium equation-
[tex]p+q=1\\[/tex]
Substituting the value of "p" in above equation, we get -
[tex]q= 1-9\\q=1-0.8\\q=0.2[/tex]
Then, the frequency of population of moth with genotype "GG" is [tex]p^2[/tex]
[tex]p^2= 0.8^2\\p^2= 0.64[/tex]
the frequency of population of moth with genotype "gg" is [tex]q^2[/tex]
[tex]q^2= 0.2^2\\q^2= 0.04[/tex]
and the frequency of population of moth with heterozygous population with genotype "Gg" is [tex]2pq[/tex]
As per Hardy Weinberg's II equilibrium equation-
[tex]p^2+q^2+2pq=1\\[/tex]
Substituting the available values in above equation, we get -
[tex]0.64+0.04+2pq=1\\2pq=1-0.04-0.64\\2pq=.32[/tex]
Thus, [tex]32[/tex]% moth population was heterozygous in 1960
Hence, option B is correct