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.
The formula to calculate heterozygosity in a population is H = 2pq(1-F) where p and q are allele frequencies and F is the inbreeding coefficient. Given allele frequencies of 0.6 and 0.4, and an inbreeding coefficient of 0.40, the heterozygosity would be H = 2 * 0.6 * 0.4 * (1-0.40) = 0.288.
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.
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.
Heterozygotes. If AA X AA, or AA X AA, is all the mating allowed, then Aa will lose frequency in the population.
To calculate the frequency of alleles in a population, you would count the number of each specific allele present and divide by the total number of alleles in the population. This can be represented as a fraction or decimal. For example, if there are 200 alleles for a specific gene in a population and 50 of them are the "A" allele, the frequency of the "A" allele would be 50/200 = 0.25 or 25%.
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.
Yes, a child's genotype can be different from that of both parents due to genetic variations and recombination during the formation of gametes, which can result in unique combinations of genes in the child. Additionally, spontaneous mutations can also lead to differences in the child's genotype compared to the parents.
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 equation you would use would be Aa=2pq