posted by Nancy .
1. A population which meets the hardy- weinberg requirements:
b. is small and usually isolated
c. has allele frequency in equilibrium
d. changes genotypic distribution from generation to generation
e. always has 75% A and 25% a allele frequencies
2. In a hypothetical population, the frequency of the dominant A allele is 80% and the frequency of the recessive allele a is 20%. that percentage of the population would you expect to be heterozygous(Aa) in genotype?? can you please show work or explain how you did it.
A population that meets Hardy-Weinberg requirements is in genetic equilibrium, and neither its allele frequencies nor its genotype frequencies changes from generation to generation. If allele and genotypes frequencies are not changing from generation to generation, can the population be evolving? What is the requirement for H-W concerning population size?
In H-W equilibrium, the 3 genotype frequencies are
p^2 + 2pq + q^2 = 1
where p is the frequency of the dominant allele (A, which is 0.8) and q is the frequency of the recessive alleled (a, which is 0.2).
(0.8)^2 + 2(0.8)(0.2) + (0.2)^2 = 1
0.64 + 0.32 + 0.04 = 1
F(AA) = 0.64
F(Aa) = 0.32
F(aa) = 0.04
Now just multiply the above 3 frequencies by the total number of individuals in the population to determine how many AA, Aa, and aa individuals there should be.