As per the Hardy-Weinberg rule, the variable p regularly addresses the recurrence of a specific allele, typically a dominant one.
For instance, expect that p addresses the recurrence of the dominant allele, Y, for yellow pea pods.
The variable q addresses the recurrence of the recessive allele, y, for green pea pods.
In the event that p and q are the lone two potential alleles for this trademark, then, at that point the amount of the frequencies should amount to 1, or 100%.
We can likewise compose this as p + q = 1.If the recurrence of the Y allele in the populace is 0.6, then, at that point we realize that the recurrence of the y allele is 0.4.
From the Hardy-Weinberg standard and the known allele frequencies, we can likewise deduce the frequencies of the genotypes.
Since every individual conveys two alleles for each quality (Y or y), we can anticipate the frequencies of these genotypes with a chi square.
In the event that two alleles are drawn indiscriminately from the genetic stock, we can decide the likelihood of every genotype. In the model, our three genotype prospects are:
pp (YY), delivering yellow peas; pq (Yy), likewise yellow; or qq (yy), creating green peas. The recurrence of homozygous pp people is p2; the recurrence of heterozygous pq people is 2pq; and the recurrence of homozygous qq people is q2.
In the event that p and q are the lone two potential alleles for a given attribute in the populace, these genotypes frequencies will whole to one: p2 + 2pq + q2 = 1.
n our model, the potential genotypes are homozygous dominant (YY), heterozygous (Yy), and homozygous recessive (yy).
In the event that we can just notice the aggregates in the populace, we know just the recessive aggregate (yy).
For instance, in a nursery of 100 pea plants, 86 may have yellow peas and 16 have green peas. We don’t have a clue the number of are homozygous dominant (Yy) or heterozygous (Yy), ergo we do realize that 16 of them are homozygous recessive (yy).
Consequently, by knowing the recessive aggregate and, in this way, the recurrence of that genotype (16 out of 100 people or 0.16), we can compute the number of different genotypes.
Assuming q2 addresses the recurrence of homozygous recessive plants, q2 = 0.16. In this way, q = 0.4. Because p + q = 1, then, at that point 1 – 0.4 = p, and we realize that p = 0.6.
The recurrence of homozygous dominant plants (p2) is (0.6)2 = 0.36. Out of 100 people, there are 36 homozygous dominants (YY) plants. The recurrence of heterozygous plants (2pq) is 2(0.6)(0.4) = 0.48. Thusly, 48 out of 100 plants are heterozygous yellow (Yy).