Hlp with hardy weinberg: in a study 48% of mice have heterzygous gene, what is the frequency ofdominant allele

Question

i just need help with this hardy weinberg, full question is: In a study of population of field mice, you find that 48% of the mice have a coat color that indicates that they are heterozygous for a particular gene. What would be the frequency of the dominant allele in this population?

Answer 1

I think the dominant allele is either 60% or 40% and you can't tell which of those it is.

Here's my reasoning.

p^2 + 2 pq + q^2 = 1 That's Hardy-Weinberg
If 48% of the mice are heterozygous, then 2 pq = 0.48

If q is the recessive gene and we're trying to solve for p,
p + q = 1
q = 1 - p for substitution purposes.

Now make substitutions in the Hardy-Weinberg equation
p^2 + 2 pq + q^2 = 1
p^2 + 0.48 + (1-p)(1-p) = 1 Simplify.
p^2 + 0.48 + (1 - 2p + p^2) = 1 Keep simplifying.
2 p^2 - 2p + 0.48 = 0 Divide all by 2.
p^2 - p + 0.24 = 0 Factor.
(p - 0.6)(p - 0.4) = 0
p = 0.6 = 60%
p = 0.4 = 40%

So one allele is 60% and one allele is 40%, but I don't see how you would be able to tell the difference.

Answer 2

p^2+2pq+q^2=1
2pq=.48
q=1-p
2p(1-p)=.48
2p-2p^2=.48
0=p^2-p+.24
p=(1+/-.2)/2
p=.6 or p=.4
40% or 60%