If the frequency of genotype AA is p^2, where p is the frequency of allele A, then the frequency of genotype AA would be p^2.
Hardy-Weinberg problems typically involve calculating allele frequencies and genotype frequencies in a population under certain assumptions. For example, you may be asked to determine the frequency of individuals with a specific genotype, or to calculate the frequency of a particular allele in a population.
Heterozygotes. If AA X AA, or AA X AA, is all the mating allowed, then Aa will lose frequency in the population.
The Hardy-Weinberg Equilibrium equation: p2 + 2pq + q2 = 1 p is frequency of dominant allele A q is frequency of recessive allele a p + q always equals 1 pp or p2 is probability of AA occurring qq or q2 is probability of AA occurring 2pq is probability of Aa occurring (pq is probability of Aa, qp is probability of aA, so 2pq is probability of all heterozygotes Aa) These add up to 1 because they represent all possibilities. The frequency of the homozygous recessive genotype
No, a parent with AS and AA genotype cannot give birth to a child with AC genotype. The parent can only pass on either the A allele or the S allele to their child, resulting in genotypes of either AA or AS.
50% AA and 50% Aa
If the parents both have the genotype Aa, their children could have the genotypes AA, Aa, or aa. The possible phenotypes for their children would be individuals with type A blood (AA or Aa genotype) or type O blood (aa genotype).
No because AA and SS create the genotype AS :)
The AA genotype typically produces the phenotype associated with the dominant allele A. This means that the dominant trait will be expressed in the individual with this genotype.
The offspring's genotype will be AA. Both parents are homozygous dominant, AA, having only dominant alleles to pass on to their offspring. So each parent can pass on only the dominant allele (A) to its offspring. So the offspring will also be homozygous dominant, AA.
The number 1,000,000 is incorrect and should be 100,000. Reduce all three numbers to get 100, 75 and 25. Out of 100 agouti rabbits the offspring will be 75% agouti and 25% will be self (white). The bucks and does all look agouti (in this case a very pale gray) but carry the recessive gene for self (white). The alleles for both are Aa. When you cross these Aa X Aa, you will get one block in the Punnet Square to be AA, two will be Aa and one will be aa. So the ratio is 1:2:1. AA and Aa are agouti and aa is white. See link below and under rabbit genetics they will show the work to get this answer:
the equation you would use would be Aa=2pq
The allele frequency in a population determines the genotype frequency. Allele frequency refers to how often a particular version of a gene appears in a population, while genotype frequency is the proportion of individuals with a specific genetic makeup. Changes in allele frequency can lead to changes in genotype frequency within a population over time.