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HardyWeinberg equation?Moderator: BioTeam
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HardyWeinberg equation?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? My signature sucks.
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
Living one day at a time;
Enjoying one moment at a time; Accepting hardships as the pathway to peace; ~Niebuhr
No. If they were in HardyWeinberg Equilibrium, there would be an equal number of homogeneous alleles as heterozygous alleles. It is your 1:2:1 ratio.
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% Living one day at a time;
Enjoying one moment at a time; Accepting hardships as the pathway to peace; ~Niebuhr
hardyweinberg equationUmm.. 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
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