Table 1. Calculating frequencies of different genotypes in different loci (simplified example for the 2-locus case).
| Contribution to genotype frequency | |||||||
| Genotype | Recessive homozygote | Heterozygote | |||||
| g | Chromatid | Chromatid | Freq. | Locus | Locus | Locus | Locus |
| 1 | 2 | 1 | 2 | 1 | 2 | ||
| 1 | 0 0 | 0 0 | f 1 | 0 | 0 | 0 | 0 |
| 2 | 0 0 | 0 1 | f 2 | 0 | 0 | 0 | f 2 |
| 3 | 0 0 | 1 0 | f 3 | 0 | 0 | f 3 | 0 |
| 4 | 0 0 | 1 1 | f 4 | 0 | 0 | f 4 | f 4 |
| 5 | 0 1 | 0 0 | f 5 | 0 | 0 | 0 | f 5 |
| 6 | 0 1 | 0 1 | f 6 | 0 | f 6 | 0 | 0 |
| 7 | 0 1 | 1 0 | f 7 | 0 | 0 | f 7 | f 7 |
| 8 | 0 1 | 1 1 | f 8 | 0 | f 8 | f 8 | 0 |
| 9 | 1 0 | 0 0 | f 9 | 0 | 0 | f 9 | 0 |
| 10 | 1 0 | 0 1 | f 10 | 0 | 0 | f 10 | f 10 |
| 11 | 1 0 | 1 0 | f 11 | f 11 | 0 | 0 | 0 |
| 12 | 1 0 | 1 1 | f 12 | f 12 | 0 | 0 | f 12 |
| 13 | 1 1 | 0 0 | f 13 | 0 | 0 | f 13 | f 13 |
| 14 | 1 1 | 0 1 | f 14 | 0 | f 14 | f 14 | 0 |
| 15 | 1 1 | 1 0 | f 15 | f 15 | 0 | 0 | f 15 |
| 16 | 1 1 | 1 1 | f 16 | f 16 | f 16 | 0 | 0 |
| Genotype frequency at different loci: | Paa(1) = Σ(…) | Paa(2) = Σ(…) | PAa(1) = Σ(…) | PAa(2) = Σ(…) | |||
Paa and PAa are frequencies of recessive (mutated) homozygotes and heterozygotes respectively and fg is frequency of a genotype with number g. Each chromatid is represented as a bit string, where 1 denotes a mutated locus and 0 a non-mutated one. A genotype contributes to the frequency of recessive homozygote at locus n if both chromatids have mutation at position n. Similarly, a heterozygote at locus n is contributed by the genotype if there is one and only one mutation in both chromatids at position n. The frequency of mutated alleles is calculated in each locus according to the equation p(n) = Paa (n) + 0.5 PAa (n).