- 9 albino rabbits (represented by the alleles bb) and
- 91 brown rabbits (49 homozygous [BB] and 42 heterozygous [Bb]).
The gene pool contains 140 B alleles [49 + 49 + 42] (70%) and 60 b alleles [9 + 9 + 42]
(30%) – which have gene frequencies of 0.7 and 0.3, respectively.
Solution
If we assume that alleles sort independently and segregate randomly as sperm and eggs
form, and that mating and fertilization are also random, the probability that an offspring
will receive a particular allele from the gene pool is identical to the frequency of that allele
in the population:
- BB: 0.7 x 0.7 = 0.49
- Bb: 0.7 x 0.3 = 0.21
- bB: 0.3 x 0.7 = 0.21
- bb: 0.3 x 0.3 = 0.09
If we calculate the frequency of genotypes among the offspring, they are identical to the
genotype frequencies of the parents. There are 9% bb albino rabbits and 91% BB and Bb
brown rabbits. Allele frequency remains constant as well. The population is stable – at a
Hardy-Weinberg genetic equilibrium.
A useful equation generalizes the calculations we’ve just completed. Variables include
- p= the frequency of one allele (we’ll use alleleBhere) and
- q= the frequency of the second allele (bin this example).
We will use only two alleles (sop+qmust equal 1), but similar equations can be written
for more alleles.
Allele frequency equals the chance of any particular gamete receiving that allele. There-
fore, when egg and sperm combine, the probability of any genotype is the product of the
probabilities of the alleles in that genotype. So:
Probability of genotypeBB=pXp=p^2 and
Probability of genotypeBb= (pXq) + (qXp) =2pqand
Probability of genotypebb=qXq=q^2
We have included all possible genotypes, so the probabilities must add to 1.0. In our example
0.49 + 2(0.21) + 0.9 = 1. Our equation becomes: