The frequency of the homozygotic recessive individuals in the population will increase by inbreeding, this is particularly true when the gene frequency is low.
Example: In a population the gene frequency of a
recessive gene is 0.01 corresponding to a frequency of homozygote
recessive individuals at 0.0001 or
1 in 10,000. If full sib mating forms the new generation, only,
the inbreeding coefficient is 25% in all individuals in the next generation.
In this population the frequency of homozygote recessive individuals can be
calculated as follows: It is proportional to the number of heterozygotes
individuals in the parents' generation (2*0,01*0,99 = 0,0198) and their chance that segregation of the recessive occurs
(1/16) by
full sib mating (Figure 4.1), this segregation should equal one of the genes, the disease gene.
Either the grandfather or the grandmother can carry the disease gene. Thus, the
joint probability should be multiplied by two.
Therefore the joint probability of breeding homozygote recessive individuals is 2*0,0198/16 = 0,0025, which
means that it has increased 25 times compared to the outbred base
population. (In these calculations the case where other combinations of herozygosity and homozygosity
of the grandparents are not taken into account, so the
result deviate slightly when
compared to the general formulas shown below). If the formulas given below are
used in our example, the result is q2 + pqF = 0.0001 + 0.99*0.01*0.25 =
0.0026.
Correspondingly, the joint number of homozygote dominants will be increased by the same proportion as the recessive at the expense of the heterozygotic individuals. During inbreeding the proportions would change in comparison to the expected frequency under Hardy Weinberg equilibrium with the following expectations, exp:
---------
Genotype AA Aa aa
Frequency, exp p2 2pq q2
+pqF -2pqF +pqF
By means of the expected frequencies the degree of inbreeding can be calculated in sub populations;
which gives F = (H0 - Hn)/H0 ,
where H0 and Hn are the genotype frequencies for heterozygotes in
respectively the base population and generation n of the subdivided
populations.
Applied on the Albumin example of dogs, section 2.4, the average inbreeding within
dog breeds is F = (0.490 - 0.330)/0.490 = 0.33.