Answer:
196.8
Explanation:
In a population of Aye-Ayes there are red-eyed individuals exhibiting the dominant phenotype -----i.e RR
yellow-eyed individuals exhibiting the recessive phenotype ----- i.e rr
in the next generation, we will have the following offspring
R R
r Rr Rr
r Rr Rr
All the offspring are heterozygous in nature.
Now, the question goes further by saying, if there are 66 yellow-eyed individuals in the current population of 459, i.e (q² =66)
How many would be expected to exhibit the heterozygous phenotype in the next generation which will consist of 410 total individuals.
In solving Hardy-Weinberg question, we use the following equation below.
p + q = 1
p² + 2pq + q² = 1
where;
p = the frequency of the dominant allele
q = the frequency of the recessive allele
p² = the frequency of individuals with homozygous dominant phenotype
2pq = the frequency of individuals with heterozygous phenotype
q² = frequency of individuals with the homozygous recessive phenotype
Now, If In the next generation, the total population = 410
and q² (yellow-eyed individuals that are recessive in nature) = 66
then; q²= [tex]\frac{individuals with recessive phenotype}{total population}[/tex]
q² = [tex]\frac{66}{410}[/tex]
q² = 0.16
q = [tex]\sqrt{0.16}[/tex]
q = 0.4
Since q = 0.4, we can easily get p by using the formula above:
p + q = 1
p = 1 - q
p = 1 - 0.4
p = 0.6
∴ Since we've known our p and q , we can easily determine how many would be expected to exhibit the heterozygous phenotype in the next generation.
2pq = 2 ( 0.6 × 0.4)
= 2 ( 0.24)
= 0.48
2pq = [tex]\frac{individuals with heterozygous phenotype}{total population}[/tex]
0.48 = [tex]\frac{x}{410}[/tex]
x = 0.48 × 410
x = 196.8 individuals
