# Hardy-Weinberg equation?

• Author
Posts
• #861

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?

• #22229

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

• #22230

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

• #22236

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.

• #22248

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%

• #23653