Can you answer this Hardy-Weinberg problem?

Question

This was on a test for one of my classes and it's driving me crazy:

In a population, 1 in 25,000 people suffer from cystic fibrosis, a recessive disorder. Calculate the frequency of the dominant p allele, the frequency of the recessive q allele, and the frequency of heterozygotes.

I'm just curious what you get for answers, I already (for the most part) know how to do these problems, I just don't think the answers I got were right. The equation for allelic frequency is p+q=1, and the equation for genotypic frequency is p^2+2pq+q^2=1, in case you forgot. Aaaaaanyhoo, thanks!

Answer 1

you know the frequency of the recessive-recessive is 0.00004 = q^2

therefore q = .00632455532... = frequency of recessive allele

p + q = 1 therefore 1- .00632455532... = .993675444...=freq dom.

p^2 = .987390889 = homozygous dominate
2pq = .01256911 = heterozygotes
q^2 = 0.00004 (from above) = homozygous recessive.

Answer 2

Allele frequencies: p = .994 q = .006.
Frequency of heterozygotes = 2pq = 2(.994)(.006) = 0.011928
The frequency of heterozygotes is about 1/84 in the population.

Answer 3

I arise with the subsequent figures: p (dominant allele) = ninety 9.37% q (recessive, CF allele) = 0.sixty 3% heterozygotes = a million.26% I doulbed checked 2pq, and that i evaluate dixie made a mistake.

Answer 4

I come up with the following figures:

p (dominant allele) = 99.37%
q (recessive, CF allele) = 0.63%
heterozygotes = 1.26%

I doulbed checked 2pq, and I think dixie made a mistake.

Answer 5

So if 1:25000 has CF, that means 1:25000 = q^2. Therefore q^2=0.00004, so q=0.0063. 1-q=p 1-0.0063=0.9937.p^2=.987 2pq=0.00008.