the genotype frequencies of a poplulation are determined to be .6 AA , .0 Aa ( there are no heterozygote) and .4 aa. Its allele frequencies are p = .6 and q .4. is this poplulation at genectic wqulilbrium? why or why not?
2007-02-12 10:34:13 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Biology
No, because there are no hetrozygotes
For the Hardy-Weinberg equilibrium to be maintained, five conditions must be met:
- Very Large population size (irrelevent)
- Isolation from other populations (irrelevent)
- No Mutation (irrelevent)
- Random Mating (irrelevent )
- No natural selection, which means the genotypes are equal in survival, and in this case, this is where is voilates the rules of the genetic equalibrium.
2007-02-12 12:40:16 · answer #1 · answered by -Eugenious- 3 · 0⤊ 0⤋
No. It doesn't match the equation because there is no 2 pq. The heterozygotes should be 2*frequency of A*frequency of a. Since neither A nor a are zero, then 2pq should not be zero.
2007-02-12 10:37:55 · answer #2 · answered by ecolink 7 · 0⤊ 0⤋