I’d like to screen and them map 2nd mutations that suppress a slow growth phenotype of a 1st mutation.
The suppression phenotype that needs to be mapped is wild type and so I’m thinking of introducing the 1st mutation into the Hawaiian mapping strain by outcrossing.
Can anyone suggest how many times I need to outcross the 1st mutation with Hawaiian?
Thanks,
The right number of backcrosses will depend on whether your initial mutation and your suppressor are linked.
If your suppressor is on a different chromosome, the expected fraction of that chromosome that will still be N2 after backcrossing N times is 2-(N+1). That gives a >99% chance of having Hawaiian DNA in your region of interest after 6 backcrosses. For the chromosome bearing your initial mutation, which you’ll be selecting to retain, the expected proportion that will remain N2 is much larger, and even after 10 backcrosses that chromosome will probably be about 1/3 N2 (and it will still be ~13% after 20 backcrosses; the formula is equation 14.10a in Lynch and Walsh, Genetics and Analysis of Quantitative Traits: fraction N2 after N backcrosses = 2*(1-e(-t/4))/(t/2), where t = N+1 and chromosomes are 50cM). You can speed the process along by genotyping SNPs near your mutation and selecting for Hawaii alleles at those SNPs.
Another thing to consider is linkage to zeel-1 and peel-1 on chromosome I; these are loci that cause an incompatibility between the N2 and Hawaiian strains and make it difficult to convert the left side of chr I from N2 to Hawaiian. See Seidel et al., Science 319:589. This is only an issue if your suppressor is linked to chr I, and even then you can design your crosses to deal with it.
Good luck.
Very helpful. Thanks.