Hardy-Weinberg equation?

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    • #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

      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

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