In a certain population of fruit flies, 640 out of 1000 have red eyes, while the remainder have sepia eyes. The sepia eye trait is recessive to red eyes. If the population is in H-W equilibrium, how many individuals would you expect to be homozygous for red eye color?

Based on the given question, the total number of fruit flies is 1000, of which red eye flies are 640. Thus, the number of sepia eye flies will be 1000-640 = 360.

On the basis of the Hardy-Weinberg equation, p+q = 1, in which p is the rate of recurrence of one allele, that is, dominant and q is the rate of recurrence of the other allele, that is, recessive.

q^2 for the population is 360/1000 = 0.36, q = 0.6.

Therefore, p = 1-q = 1-0.6 = 0.4.

The frequency of homozygous dominant allele is denoted by p^2 = 0.16.

Hence, the number of homozygous dominant for red eye color would be 16% of 1000 which will be equal to 160.

Phoca Phoca · Answer 2 · 2020-03-26T05:54:11+01:00

Answer:

160 individuals or 16% of population

Explanation:

If the population is in Hardy-Weinberg equilibrium then the population will follow the mathematical expression of the HW equilibrium which are:

1. Genotypic frequencies- P²+q²+2pq=1

2. Allele frequency- p+q=1

In the given question, it is stated that the sepia eye is recessive while red eyes are dominant.

Now, population of red eyes=640

the population of sepia eyes will be- 1000- 640= 360

therefore, q² will be - 360/1000= 0.36

calculating q- = 0.6

q= 0.6

According to p+q=1

p will be= 1- 0.6=0.4

p=0.4

Now to find homozygous of red eye or p² = 0.4 x 0.4

p²= 0.16

This means 16% of total population is homozygous for red color

or 160 individual of 1000 are homozygous for red color.