The frequency of the heterozygous genotype (i just took the test)-apex
Homozygous recessive genotype
p (represents dominant) and q (represents recessive) although the results will be the same no matter what letter is assigned to which allele.
To keep the information straight a person can also substitute the letters that are normal "tags" for the allele.
For example: T for tall and t for short. T+t=1 T(squared)+2Tt+t(squared)=1 The equation can be solved if there is an adequate statical number for all the homozygous recessive individuals in a population.
Hey, the frequency of the homozygous recessive genotype. Apex
the proportion of homozygotes for the alternative allele
The frequency of the homozygous dominant genotype
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
The Hardy-Weinberg equation is as follows: p2 + 2pq + q2 = 1 p & q represent the frequencies for each allele.
If you have the alleles A and a, then the The hardy-weinberg equation is: A2 + 2Aa + a2 = 1 where a and A represent allele frequencies. So A2 would be the genotype frequency for AA. 2Aa is the genotype frequency for Aa. And a2 is the genotype frequency for aa. Plug in whatever information you have into the equation and you can probably come up with an answer. Save
the frequently of the heterozygous dominant genotype
allele frequencies in a population will remain constant unless one or more factors cause those frequencies to change
The frequency of the homozygous recessive genotype.
p2 + 2pq + q2 = 1 and p + q = 1p = frequency of the dominant allele in the populationq = frequency of the recessive allele in the populationp2 = percentage of homozygous dominant individualsq2 = percentage of homozygous recessive individuals2pq = percentage of heterozygous individuals
The frequency of the homozygous recessive genotype.
The Q is the recessive trait and the P is the dominant trait. Always find Q first when solving Hardy Weinberg equations.
The frequency of the homozygous dominant genotype.
The frequency of the homozygous dominant genotype.
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
It depends on what you make p equal to. P is usually the frequency of the dominant allele, which makes q the frequency of the recessive allele, but they can be switched. As long as p is one frequency and q is the other, the formula will work. So if you have the dominant allele frequency (A) =.6 then the recessive allele frequency (a) =.4, because p+q=1 When you plug the frequencies into the hardy-weinberg equation p^2 +2(p)(q) + (q)^2 = 1 then you have (0.6)^2 + 2(0.4)(0.6) + (0.4)^2 = 1 (0.6)^2 = 0.36 which is the frequency of dominant homozygotes 2(0.4)(0.6)=0.48 which is the frequency of heterozygotes (0.)^2 = 0.16 which is the frequency of recessive homozygotes If you have a population of 100 people, these frequencies would mean that: 36 people would be AA 48 people would be Aa 16 people would be aa Which would mean that 36+48=84 people would exhibit the dominant trait and 16 people would show the recessive trait.
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
The incidence of an allele within the population compared to all other alleles. The Hardy-Weinberg equation is useful to calculate this- p2 + 2pq + q2 = 1 where p is incidence of the dominant allele and q is the incidence of the recessive allele where there are only two versions. If blue eyes is recessive and 0.36, then q2 = 0.36 so therefore q= 0.6 so the frequency of the blue eye allele is 60%.
p represents the square root of the frequency of the homozygous genotype AA.
To determine how allele frequency changes