TABLE 3.
Base substitution mutations | ||||
---|---|---|---|---|
Strain | No. mutations | % genome ≥7× coverage | Genome Size (bp) adjusted | Mutation rate (per base per gen ×10−9) |
Mut2 | 6 | 41 | 9,898,136 | 3.8 |
Mut3 | 9 | 69 | 16,657,838 | 3.4 |
Mut4 | 13 | 84 | 20,279,107 | 4.0 |
Average | 3.7 | |||
Single-nucleotide indel mutations in 5- to 13-nt HP tracts | ||||
Strain |
No. mutations |
No. HP tracts >7× coverage |
Mutation rate (per HP tract/generation × 10−7) |
|
Mut2 | 6 | 57,502 | 6.5 | |
Mut3 | 15 | 99,714 | 9.4 | |
Mut4 | 27 | 122,816 | 14 | |
Average | 10 | |||
Single-nucleotide indel mutations in 8- to 13-nt HP tracts | ||||
Strain |
No. mutations |
No. HP tracts >7× coverage |
Mutation rate (per HP tract/generation × 10−7) |
|
Mut2 | 4 | 2,820 | 89 | |
Mut3 | 10 | 7,054 | 89 | |
Mut4 | 19 | 8,696 | 140 | |
Average | 110 |
The base substitution mutation rate was determined by calculating the percentage of the genome in which at least sevenfold DNA sequencing coverage to unique regions was obtained. This was done because our statistical analysis did not have sufficient power to reliably detect heterozygous mutations in regions with lower coverage. This information was used to calculate the mutation rate on the basis of the following formula: (number of mutations)/(160 generations)/(adjusted genome size), with the diploid S. cerevisiae genome size determined as 24,141,794 bp (http://www.yeastgenome.org/). To obtain indel mutation rates, we first determined the number of HP tracts of a given length in unique regions of the genome which had ≥ sevenfold sequence coverage. We then used the following equation to calculate mutation rate: (number of indels)/(160 generations)/(number of HP tracts with ≥ sevenfold coverage).