Answers recombination

From Wikionchus

• Phenotypically "wild type" animals are heterozygous for both mutations

• In a fraction of meioses cross-over events took place that separated the two mutants. This happened when the heterozygous animal, which was a male here, made its sperm. Strictly speaking, this type of cross only examines recombination rates in male spermatogenesis, which might be different than what happens in the hermaphrodite.

• Determination of the recombination frequency in a F1-backcross:

We are looking at 1 potentially recombined gamete per animal, thus we can write:

recombinant gametes	4+6=10
non-recombinant gametes	77+73=150
rec. fraction r	10/160=0.063

• If always exactly one recombination happens per meiosis you might think r should be 100%. It is not, because only 2 out of 4 gametes will receive a recombined chromatid. (Cross-over takes place between the two "neighbouring" non-sister-chromatids in Meiosis I, the two "outward" chromatids are unaffected!)

• Recombination fractions in most organisms are not additive. This is because there are double cross-overs: Two loci on different chromosome ends can seemingly be un-recombined in some animals (e.g. both wild type in our example) , when two (or 4,6...) recombinations took place between the loci. Double cross-over is the more likely the further apart two loci are on a chromosome. Recombination fractions can be approximately made additive by transforming them to centiMorgan as a map unit.

• Additional question: In C. elegans no chromosome is bigger than 50cM. Why could that be?

• Remark: In daily practice you will usually do an F2-mapping by genotyping progeny from a selfed F1. Also our linkage map was constructed that way. I chose another example here because calculation of recombination frequencies in a F2 population is a bit advanced...thats what computers are for!