In a population of gerbils ,long hair is completely dominant over short hair .if 18percent of the population has short hair , calculate the percentage of the population that is expected to be heterozygous

49%

Explanation:

* long hair = P

* shirt hair = q using the Hardy-Weinberg equation for genotypic frequencies

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

*first use the equation to solve for the known value 18% hh

.18 = q^2

Square root .18 = q

So q = .42

* plug this allelic frequency in to p+q = 1

Which will give you the frequency of dominant alleles

1 - .45 = p So P = .58

Then just plug the P and q values found into the portion of the equation for heterozygous offspring.

2(.58)(.42)

Which is .49 or 49%

The entire population should be 33% homozygous dominant 49% heterozygous and 18% homozygous recessive.

49 percent

Explanation:

I got this question correct on the quiz.

looks like youre on your own with this one buddy

the answer to your question is b