MATHEMATICS IN GENETICS - An Invitation to Biomathematics - page 111

To prove part 2, we use a Punnett square approach to generalize from an
individual mating to any random mating occurring in the population.
This can be done by treating a randomly chosen diploid organism as
being the result of a random pairing of two alleles, each drawn from the
parental pool containing A and a alleles at proportions p and q,
respectively. Because diploid organisms inherit one allele from each of
their parents, and because we assumed random mating and equal
fitness of the genotypes, this treatment is well justified. Thus, the
probability of homozygous dominant P(AA) will be the probability
of randomly drawing A twice from the allelic pool, and we obtain
P(AA)

p 2 . Notice that the assumption we made for an
infinitely large population allows us to assume that the frequencies of
the allele do not change from draw to draw, regardless of the fact that
our selection is without replacement. In the same way, the probability
for a heterozygous Aa organism will be the probability for drawing an
A followed by an a, calculated as pq, and so on. The Punnett square,
Table 3-1, helps us visualize this. In addition to the genotypes resulting
from the cross (see Figure 3-6), we have now also included, in
parentheses, the proportions of the A and a alleles and the proportions
of the resulting genotypes.

¼ (p) (p) ¼

From here, we see that the proportion of the genotypes in F 1 will be:

p 2

P ð AA Þ¼

ð aa Þ¼

q 2

(3-4)

P

P ð Aa Þ¼ pq þ qp ¼ 2pq :

We have calculated P(Aa) as the sum of the proportions of Aa and aA,
both of which represent the same genotype. To finish the proof of claim
2, we need to show that the proportions of the A and a alleles in the F 1
remain the same as in the parental generation. The argument is the same
as in our example and the proof of claim 1 above, but applied this time
to the genotype distribution given by Eq. (3-4). If there are N 1
individuals in the F 1 , of them N 1 p 2 will have the AA genotype; N 1 q 2 will
have aa; and 2N 1 pq will have the Aa genotype. Thus, the proportion of
the A allele in the F 1 will be:

2N 1 p 2

2N 1 pq
2N 1

þ 2N 1 p ð p þ q Þ

¼ ¼ p ð p þ q Þ¼ p ; since p þ q ¼ 1 :

2N 1

A (p) a (q)

p 2 ) A (p) AA (p p ¼ Aa (pq)

q 2 ) a (q) aA (qp) aa (q q ¼

TABLE 3-1.
Punnett square applied to genotype frequencies.