Question

A scientist was studying a population of mice. In mice, the brown allele (B) has complete dominance over the white allele (b). There were 1000 mice in the population. Of these mice, 840 had brown fur and 160 had white fur. Feel free to use your notes/equations from the review to help you with this problem. What is the frequency of the recessive allele? What is the frequency of the dominant allele?

q= 0.4, p= 0.6
q= 0.6 , p= o.4
q= 0.16, p=.36
q= .36,p= 0.16​

Answer 1

The frequency of the recessive allele (b) can be determined using the Hardy-Weinberg equation:

p^2 + 2pq + q^2 = 1

where p is the frequency of the dominant allele (B) and q is the frequency of the recessive allele (b).

We know that 840 mice had brown fur, which means they were either BB or Bb. Since B is dominant, we know that all the BB mice will have brown fur, and only the heterozygous Bb mice will have brown fur. Therefore, the frequency of the dominant allele (B) can be calculated as follows:

p^2 + 2pq = 840/1000
p^2 + 2pq = 0.84

We also know that 160 mice had white fur, which means they were homozygous recessive (bb). Therefore, we can use this information to calculate the frequency of the recessive allele (b):

q^2 = 160/1000
q^2 = 0.16
q = sqrt(0.16)
q = 0.4

Now that we know the value of q, we can substitute it back into the first equation to solve for p:

p^2 + 2pq = 0.84
p^2 + 2p(0.4) = 0.84
p^2 + 0.8p - 0.84 = 0
(p - 0.6)(p + 1.4) = 0

Since p cannot be negative, the only valid solution is:

p = 0.6

Therefore, the frequency of the recessive allele (b) is q = 0.4, and the frequency of the dominant allele (B) is p = 0.6.

So the correct answer is option (a) q = 0.4, p = 0.6.