The aye-aye exhibits incomplete dominance for eye color. in a population of aye-ayes there are red-eyed individuals exhibiting the dominant phenotype and yellow-eyed individuals exhibiting the recessive phenotype. if there are 66 yellow-eyed individuals in the current population of 459, how many would be expected to exhibit the heterozygous phenotype in the next generation which will consist of 410 total individuals?

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=

q² =  

q² = 0.16

q =

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 =

0.48 =

x = 0.48 × 410

x = 196.8 individuals

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