the key to this problem is recalculating p. new p=1/3 and new q=2/3.
In short version,
hetero dominant(pq) is has 1p and 1q. homozygous recessive has 2 q's.
new allele frequency p=1(50)/(50+100)=1/3
new allele frequency q=[1(50)+2(25)]/(50+100)=2/3
so p^2=1/9
In longer version:
To get to the new p and q, we have to "count" number of p and q alleles in each categories. its easier to think if you think in terms of numbers than in percents. (think there are 100 rabbits total. out of 100, there are 25 homozygous dominant rabbits, 50 hetero dominant, 25 homozygous recessive.)
now that the homozygous dominant rabbits are eliminated, we have 50 heterozygous dominant rabbits and 25 dominant rabbits.
we know that
homozygous dominant (pp) has two p alleles,
heterozygous dominant (pq) has 1 p allele and 1 q allele
homozygous recessive (qq) has two q alleles,
writing this mathmatically,
#p allele = 2*(# homozygous dominant) + 1*(#heterozygous dominant)
#q allele= 2*(# homozygous recessive) + 1*(#heterozygous dominant)
allele frequency is simply #/total #. So,
allele frequency of p= #p/(#q+#p)
allele frequency of q =#q/(#q+#p)
applying the above equation to our situation (where, #homozygous dominant now is zero, #hetero dominant=50, #homozygous recessive=25)
#p=1*50=50
#q=2*25+1*50=100
allele frequency of p=50/(100+50)=1/3
allele frequency of q=100/(100+50)=2/3
so p=1/3 and q=2/3.
now according to hardy weinberg, our new homozygous dominant will be p^2, which is (1/3)^2 or 1/9. This is 0.11.
