genotype) = (total number of Aa genotypes)/N
We do not distinguish between the heterozygotes Aa and aA and treat them as being equivalent genotypes. Therefore, we can express allele frequencies in terms of genotype frequencies by adding together the frequencies of A‐containing and a‐containing genotypes:
(2.2)
(2.3)
Each homozygote contains two alleles of the same type, while each heterozygote contains one allele of each type so the heterozygote genotypes are each weighted by half.
With the variables defined, we can then follow allele frequencies across one generation of reproduction. The first step is to calculate the probability that parents of any two particular genotypes will mate. Since mating is assumed to be random, the chance that two genotypes will mate is just the product of their individual frequencies. As shown in Figure 2.7, random mating can be thought of as being like gas atoms in a balloon. As with gas atoms, each genotype or gamete bumps into others at random, with the probability of a collision (or mating or union) being the product of the frequencies of the two objects colliding. To calculate the probabilities of mating among the three different genotypes, we can make a table to organize the resulting mating frequencies. This table will predict the mating frequencies among genotypes in the initial generation, which we will call generation t.
Figure 2.7 A schematic representation of random mating as a cloud of gas where the frequency of A's is 14/24 and the frequency of a's 10/24. Any given A has a frequency of 12/20 and will encounter another A with probability of 14/24 or an a with the probability of 10/24. This makes the frequency of an A‐A collision (14/24)2 and an A‐a or a‐A collision 2(14/24)(10/24), just as the probability of two independent events is the product of their individual probabilities. The population of A's and a's is assumed to be large enough so that taking one out of the cloud will make almost no change in the overall frequency of its type.
A parental mating frequency table (generation t) is shown below.
| Moms | Frequency | Dads | ||||
| AA | Aa | aa | ||||
| X | Y | Z | ||||
| AA | X | X 2 | XY | XZ | ||
| Aa | Y | XY | Y 2 | YZ | ||
| Aa | Z | ZX | ZY | Z 2 | ||
The table expresses parental mating frequencies in the currency of genotype frequencies. For example, we expect matings between AA moms and Aa dads to occur with a frequency of XY.
Next, we need to determine the frequency of each genotype in the offspring of any given parental mating pair. This will require that we predict the offspring genotypes resulting from each possible parental mating. We can do this easily with a Punnett square. We will use the frequencies of each parental mating (above) together with the frequencies of the offspring genotypes. Summed for all possible parental matings, this gives the frequency of offspring genotypes one generation later, or in generation t + 1. A table will help organize all the frequencies, like the offspring frequency table (generation t + 1) shown below.
| Parental mating | Total frequency | Offspring genotype frequencies | ||
| AA | Aa | aa | ||
| AA × AA | X 2 | X 2 | 0 | 0 |
| AA × Aa | 2XY | XY | XY | 0 |
| AA × aa | 2XZ | 0 | 2XZ | 0 |
| Aa × Aa | Y 2 | Y2/4 | (2Y2)/4 | Y2/4 |
| Aa × aa | 2YZ | 0 | YZ | YZ |
| aa × aa | Z 2 | 0 | 0 | Z 2 |
In this table, the total frequency is just the frequency of each parental mating pair taken from the parental mating frequency table. We now need to partition this total frequency of each parental mating into the frequencies of the three progeny genotypes produced. Let's look at an example. Parents with AA and Aa genotypes will produce progeny with two genotypes: half AA and half Aa (you can use a Punnett square to show this is true). Therefore, the AA × Aa parental matings, which have a total frequency of 2XY under random mating, are expected to produce (½)2XY = XY of each of AA and Aa progeny. The same logic applies to all of the other parental matings. Notice that each row in the offspring genotype frequency table sums to the
