Question

Full color in domestic cats is dominant over dilute color. Of 600 cats observed, 200 have dilute color.

1. If this population is in Hardy-Weinberg equilibrium, what is the frequency of the dilute allele?
2. How many of the 400 cats with full color are likely to be heterozygous?

Answer 1

a. According to hardy weinberg equation,

p2 +2pq + q2 = 1

Where ,

p2 = frequency of homozygous dominant individual (AA) (here, full color of cat)

q2 = frequency of homozygous recessive individual (aa) (here, dilute color of cat)

2pq = frequency of heterozygous individual

Now., we know that

p + q = 1

Where, p = frequency of dominant allele (A) (i.e., full color)

q = frequency of recessive allele (a) (i.e., dilute color)

In the above question, it is given that out of 600 total cat , we have 200 dilute cat.

Now, we know that

gene frequency = no of homozygous recessive (aa)/total no of progeny

q2 = 200/600

q2 = 0.33

Therefore, q = 0.57

therefore frequency of dilute allele is 0.57.

B) we found in above part of question that q = 0.57

Since we need to know the heterozygous allele frequency. We will need the individual dominant allele frequency as heterozygous is represented by 2pq.

Finding q, We know that, p + q = 1

So, p = 1- q

i.e., 1- 0.57 = 0.43

Therfore, p = 0.43 ( dominant allele frequency)

Now heterozygous frequency is calculated as 2pq which will be 2× 0.43× 0.57 = 0.49

now ​​​​​​, no of members = frequency × total number

0.49 × 600 = 294

So, out of 400 full color, 294 are in heterozygous condition.