In a certain population of rabbits, the allele first brown fur is dominant over the allele for white fur. If 60 out of 100 rabbits have white fur, what is the allele frequency for the recessive allele?

The Hardy-Weinberg equation is:

[homozygous dominant]2 + 2[heterozygous] + [homozygous recessive]2 = 1

Well, Your answer will be is 0.01 Because, The Hardy-Weinburg equation is p + q = 1, or p^2 + 2pq + q^2 = 1, where p is dominant and q is recessive. If 10 out of 100 rabbits have white fur, 10% of the rabbits have white fur. Therefore, 90% of the rabbits have brown fur, which can be substituted into the first equation to become 0.9 + 0.1 = 1. Now that we know what p and q equal, we can solve the rest of the equation.

0.9^2 = 0.81

0.9 * 0.1 * 2 = 0.18

0.1^2 = 0.01

So, The best answer will be is 0.01 Good Luck!

From ~Itsbrazts~.