The hybrid genotypes bB and Bb are identical so the probability of this
occurring is 0.21 + 0.21 = 0.42. Expressed as percentages, the ratios of
genotypes bb, bB and BB in the new generation are 9%, 42% and 49%. Because
B is the dominant factor, 42% + 49% = 91% of the first generation will have
brown eyes. Only an individual with genotype bb will display the observable
characteristics of the b factor, so only 9% of the population will have blue eyes.
The initial distribution of genotypes was 20%, 20% and 60% and in the new
generation the distribution of genotypes is 9%, 42% and 49%. What happens
next? Let’s see what happens if a new generation is obtained from this one by
random mating. The proportion of b-factors is 0.09 + ½ × 0.42 = 0.3, the
proportion of B-factors is ½ × 0.42 + 0.49 = 0.7. These are identical to the
previous transmission probabilities of the factors b and B. The distribution of
genotypes bb, bB and BB in the further generation is therefore the same as for
the previous generation, and in particular the genotype bb which gives blue eyes
does not die out but remains stable at 9% of the population. Successive
proportions of genotypes during a sequence of random matings are therefore
20%, 20%, 60% → 9%, 42%, 49% →... → 9%, 42%, 49%
This is in accordance with the Hardy–Weinberg law: after one generation the
genotype proportions remain constant from generation to generation, and the
transmission probabilities are constant too.
Hardy’s argument
To see that the Hardy–Weinberg law works for any initial population, not just
the 20%, 20% and 60% one that we selected in our example, we can do no
better than refer to Hardy’s own argument which he wrote to the editor of the
American journal Science in 1908.
Hardy begins with the initial distribution of genotypes bb, bB and BB as p, 2r
and q and the transmission probabilities p + r and r + q. In our numerical
example (of 20%, 20%, 60%), p = 0.2, 2r = 0.2 and q = 0.6. The transmission
probabilities of the factors b and B are p + r = 0.2 + 0.1 = 0.3 and r + q = 0.1
- 0.6 = 0.7. What if there were a different initial distribution of the genotypes
bb, bB and BB and we started with, say, 10%, 60% and 30%? How would the
Hardy–Weinberg law work in this case? Here we would have p = 0.1, 2r = 0.6
and q = 0.3 and the transmission probabilities of the factors b and B are