How do you do the following with reference to the Hardy-Weinberg model. A) Indicate the conditons under....?

....which allelic frequencies (p and q) remains constant from one generation to the next.


B) calculate, showing all work, the frequencies of the alleles and the frequencies of the genotypes in a population of


100,000 rabbits, of which 25, 000 are white and 75, 000 are agouti. (In rabbits the white color is due to a recessive allele, w, and the agouti is due to a dominant all, W)


C) If the homozygous dominant condition were to become lethal, what would happen to the allelic and genotypic frequencies in the rabbit population after two generations?





If SOMEONE CAN HELP ME, I WILL REALLY APPRECIATE ALL YOUR HELP. THANK YOU.

How do you do the following with reference to the Hardy-Weinberg model. A) Indicate the conditons under....?
A) The Hardy-Weinberg equillibrium assumes that:


1. There is random mating (no inbreeding, assortative mating, or genetic drift).


2. There is no selection in favour of either allele (the species is not evolving).


3. No mutation is occuring (no new alleles).


4. There is no migration.





B) The alleles are:


W - dominant, agouti


w - recessive, white





The Hardy Weinberg equillibrium equations are:





Allele Frequencies


f(W) = p


f(w) = q


p + q = 1





Genotype Frequencies


f(WW) = p^2


f(Ww) = 2pq


f(ww) = q^2


p^2 + 2pq + q^2 = 1





We know that the 25,000 rabbits showing the recessive phenotype (white) MUST have the homozygous recesssive genotype (ww). The frequency of this genotype is:





f(ww) = 25,000 / 100,000 = 0.25


q^2 = 0.25


q = 0.5





p + q = 1


p = 1 - q


p = 1 - 0.5


p = 0.5





The allele frequencies are:


f(W) = p = 0.5


f(w) = q = 0.5





The genotype frequencies are:


f(WW) = p^2 = 0.5^2 = 0.25 (or 25%, or 25,000 individuals)


f(Ww) = 2pq = 2x0.5x0.5 = 0.5 (or 50%, or 50,000 individuals)


f(ww) = q^2 = 0.5^2 = 0.25 (or 25%, or 25,000 individuals)





To check the mathematics is correct check that


p^2 + 2pq + q^2 = 1


0.25 + 0.5 + 0.25 = 1





Also, note that f(WW) + f(Ww) = 0.75 = the proportion of rabbits observed with the agouti phenotype.





C) If the homozygous dominant condition were to become lethal then there would be selection in favour of the recessive allele, and the population would no longer be in Hardy-Weinberg equillibrium.





Generation 1





Surviving parent genotypes:


f(Ww) = 50,000 individuals = 2/3 = 0.6667


f(ww) = 25,000 individuals = 1/3 = 0.3333





Parental allele frequencies:


f(W) = p = 50,000 / (2 x 75,000) = 1/3 = 0.3333


f(w) = q = (50,000 + (2 x 25,000)) / (2 x 75,000) = 2/3 = 0.6667





F1 offspring:


f(WW) = p^2 = (1/3)^2 = 1/9


f(Ww) = 2pq = 2 x 1/3 x 2/3 = 4/9


f(ww) = q^2 = (2/3)^2 = 4/9


[note that p^2 + 2pq + q^2 = 1]





Surviving F1 offspring genotypes:


f(Ww) = (4/9) / (8/9) = 1/2 = 0.5


f(ww) = (4/9) / (8/9) = 1/2 = 0.5





F1 allele frequencies:


f(W) = p = 0.5/2 = 0.25


f(w) = q = 0.5/2 + 0.5 = 0.75





Generation 2





F2 offspring:


f(WW) = p^2 = (0.25)^2 = 0.0625


f(Ww) = 2pq = 2 x 0.25 x 0.75 = 0.3750


f(ww) = q^2 = (0.75)^2 = 0.5625


[note that p^2 + 2pq + q^2 = 1]





Surviving F2 offspring genotypes:


f(Ww) = 0.3750 / (1-0.0625) = 0.4


f(ww) = 0.5625 / (1-0.0625) = 0.6





F2 allele frequencies:


f(W) = p = 0.4/2 = 0.2


f(w) = q = 0.4/2 + 0.6 = 0.8





You could then draw a table, or graph the results:





Allele frequencies:


F0: f(W) = 1/3 = 0.3333, f(w) = 0.6667


F1: f(W) = 1/4 = 0.25, f(w) = 0.75


F2: f(W) = 1/5 = 0.2, f(w) = 0.8





The W allele is selected against, so its frequency declines over each successive generation. However, the rate at which it declines will reduce with each successive generation, as a higher proportion of the W alleles occur in the non-lethal heterozygous genotype (Ww).





Genotype frequencies:


F0: f(Ww) = 0.6667, f(ww) = 0.3333


F1: f(Ww) = 0.5, f(ww) = 0.5


F2: f(Ww) = 0.4, f(ww) = 0.6





As the frequency of the W allele declines, so does the frequency of the heterozygous (Ww) genotype.





Eventually we would expect the frequency of the w allele to become so high that it is likely to become fixed [ f(w) = 1 ] by chance.





Hope that helps.