Given the basic formula (and the assumption that mating is random), it is possible to use the formula to calculate the frequency of the dominant allele "p" and the recessive allele "q". We can then apply this to follow the allele frequencies through several generations.
p2 + 2pq + q2 = 1
If the recessive allele shows a different phenotype i.e. albino (bb), then q2 =
If we know q then p = 1 - q
p = 1 - 0.5
p = 0.5
We can use this to calculate the frequency of the homozygous dominant (p2)and the heterozygous (2pq ).
BB = p2 Bb = 2pq
BB = 0.5 x 0.5 Bb = 2 x (0.5 x 0.5)
BB = 0.25 Bb = 0.5
Once these frequencies are established we can calculate how many of a given population are homozygous dominant, heterozygous or homozygous by multiplying the population by the frequency of that phenotype. Given a population of 1000
BB = 1000 x 0.25 = 250 Bb = 1000 x 0.5 = 500 bb = 0.25 x 1000 = 250
The next step is for each of these genotypes to undergo selection. Not all will survive to become breeding adults and how well they survive is a measure of their fitness. Fitness is describes as being a value, 1 or less. If all juveniles survive to reproduce it is 1. If only 9 out of ten survive then it is 0.9, if 1 in a 100 then 0.01. This value is called fitness and is represented by W. Note that because the homozygous dominant and heterozygote are outwardly the same it is usual for them to have the same W.
To calculate the breeding adults, you simply multiply the juvenile population for that genotype by W. For our calculations we will assume BB (W) and Bb (W) = 1 and bb (W) = 0.1.
Breeding adults
BB = 250 x 1 = 250 Bb = 500 x 1 = 500 bb = 250 x 1 = 25
When these adults reproduce they will be doing so randomly but note that the frequency of the recessive allele has been lowered. We need to re-calculate q (and p) as we did at the start.
p =0.65 q = 0.35
We can calculate the size of the new generation by multiplying the existing population of adults by the value entered for fecundity. (Fecundity simply means breeding ability) In the example shown each adult (on average) will produce two young.
New population of offspring (juveniles) = 2 x (250 + 500 + 25)
= 1550
This population will contain three genotypes BB, Bb and bb. We can calculate the fraction they are of the population as follows.
BB = p2 Bb = 2pq bb = p2
= 0.65 x 0.65 = 2 x (0.65 x 0.35) = 0.35 x 0.35
= 0.42 = 0.46 = 0.12
To find how many of each genotype there are we simply multiply the phenotypic ratio by the population size as follows -
BB = 0.42 x1550 Bb = 0.46 x 1550 bb = 0.12 x 1550
BB = 651 Bb = 713 bb = 185
This now brings us full circle back to selection and calculating the new p and q values in the next generation of breeding adults following selection.
The spreadsheet shown below calculates all of the above into the fifth generation.