Hardy-Weinberg equation?

[marshall]

I'm having some trouble with a question about calculating the frequency of alleles.

I understand that :
q= the frequency of recessive alleles (a)
p= the frequency of dominant alleles (A)
2pq= the frequency of Aa

The question asks us to find q2, q, p and 2pq for a population of 16 pigs (four of which are black and homozygous ressive aa, and the rest are white.)

This is all the information i have been given and i am completely confused.

I assume that q2=aa, and therefore q2=4 (because there are four black pigs, right?). But thats as much as i know. Can anyone help me find out what q, p and 2pq are?

This may seem really easy, but i just dont get it. Am i supposed to get a negative answer for p?

[mith]

no, there's no negative answers.... and I don't think you can solve until you know how many of the whites are hetero/ homo

[marshall]

Apparently the population of pigs is at equilibrium. That means that there would be 8 hetro and 4 homo, right?

[biostudent84]

No. If they were in Hardy-Weinberg Equilibrium, there would be an equal number of homogeneous alleles as heterozygous alleles. It is your 1:2:1 ratio.

[mith]

So assuming that there are 4 aa, 4 AA, and 8 Aa,

q^2=4/16= 25%
p^2=4/16= 25%
p= sqrt(p^2) = sqrt(25%)= 50%
q= sqrt(q^2) = sqrt(25%)= 50%

[victor]

Umm.. how about like this:
from the question we got:
there are 16 pigs and 4 of them are black (recessive aa).
it means that : 16 pigs are 100%
4 pigs are 25% (aa 25%)
q2=0.25 -------> q=0.5
p + q = 1 ---------->p + 0.5 = 1 -----> p = 0.5
AA (homozygotus) are p2----> 0.5 quadrat = 0.25 = 25%
2Aa (heterozygotus) are 2pq----> 2 x (0.5) x (0.5) = 0.5 = 50%
so the equition is: AA + 2Aa + aa = 1
25% + 50% +25% = 1