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.
In the Hardy-Weinberg equation, p2 represents the frequency of the homozygous dominant genotype in a population for a specific gene. It represents the proportion of individuals in the population that have two copies of the dominant allele.
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.
For allele frequencies, we can start by determining the number of alleles in the population. Since each rabbit carries two alleles, there are 200,000 total alleles (100,000 rabbits * 2 alleles each). Next, we calculate the frequency of the white allele: 25,000 white rabbits * 2 alleles each / 200,000 total alleles = 0.25. The frequency of the agouti allele would be 0.75 (1 - 0.25). For genotype frequencies, we can use the Hardy-Weinberg equation. The genotype frequencies would be: white-white (0.25 * 0.25 = 0.0625), white-agouti (2 * 0.25 * 0.75 = 0.375), and agouti-agouti (0.75 * 0.75 = 0.5625).
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.