hardy-weinberg equilibrium problem

[niharika]

here i am giving two types of solutions.plz tell me which is correct and why!

A randomly mating population of your favourite diploid animal has the dominant allele |B| and the recessive allele "b" at the frequencies 0.65 and 0.35. A sudden cataclysmic event causes all of the homozygous recessive individuals to die. You are very sad but take comfort in the fact that your favourite animals are resilient creatures that simply carry on with life and get busy mating. What was the percentage of homozygous recessive individuals before the cataclysm, and what is the percentage of homozygous recessive individuals born in the first generation conceived after the cataclysm?

type1:Before the cataclysm, the frequency of homozygous recessive individuals would be given by the term q^2 in the H-W equation. Since q = 0.35, q^2 = 0.1225 or 12.25%

 

If you wiped out all of those homozygous recessive individuals, the remaining population would consist just the dominant phenotyped individuals in the original population. PRIOR to the cataclysm, the proportions of the original population of each of the two dominant genotypes was: p^2 = 0.4225, or 42.25% BB and 2pq = 0.455 or 45.5% Bb individuals. After the cataclysm, the proportion of the remaining population with each genotype would be:

 

BB = 42.25 / (42.25 + 45.5) X 100 = 48.15%

Bb = 45.5 / (42.25 + 45.5) = 51.85%

 

So, p = frequency of B = 2X48.15 + 51.85 / 200 = 0.741

and q = frequency of b = 51.85 / 200 = 0.259

 

The H-W equilibrium will be established immediately, so the frequencies in the next generation will be:

 

p^2 = 0.741^2 = 0.549 = 54.9% BB

2 pq = 2(0.742)(0.259) = 0.384 = 38.4% Bb and

q^2 = 0.259^2 = 0.067 = 6.7% bb

 

type2:before cataclysm:

frequency of recessive allele =.35 => q=.35,genotypic frequency=(.35)^2=.1225

percentage =.1225x100=12.25%

genotypic frequency of BB=.65x.65=.4225

genotypic frequency of Bb=1-(.4225+.1225)=.455

after cataclysm:

crosses possible are BBxBb,BBxBB,BbxBb

genotypic frequencies of above crosses are:.4225x.455,.4225x.4225,.455x.455 respectively

=.1922375, .0.17850625, 0.207025

among these crosses only one/four of third cross are recessive homozygotes=.207025x1/4

=0.05175625

then percentage =0.05175625/total x100=0.05175625/(.1922375+.0.17850625+ 0.207025) x100

=0.05175625/0.57776875 x100=0.089579525x100=8.9579525%