2. The most abundant pairing of phenotypes must be the parental types (AB and ab) and the least abundant pair must be the Recombinant types (Ab and aB), because crossing over is RARE
3. The parental types show you which chromosomes the allelese were on in the parent. So, in this double heterozygous parent, one homologous chromosome had 'AB' and the other had 'ab'. So, during gamete formation, when NO CROSSING OVER occurred these gametes were produced. This is the most frequent event, so these gametes were most common - resulting in lots of offspring expressing these traits.
4. The other gametes, 'aB' and 'Ab', are the recombinant types. They were produced when crossing over DID occur.
5. Well, because the frequency of crossing over is directly related to the distance between genes, we can use the frequency of crossing as a direct measure of the distance between the genes. So, what was the frequency of crossing over? Well, in this case, it was 32/95 offspring. That = 33.4%, so we say that the genes are 33.4 map units (or centiMorgans).
Study Question:
Consider these results from the following testcross:
AaBb x aabb
F1 Phenotypes:
AB 15
Ab 22
aB 29
ab 13
Conduct a chi-square tst of independence. Using the p = 0.05 level (and critical X2 of 3.84), make a conclusion about whether the genes are linked or assort independently.
If the genes are linked, map their positions in the heterozygous parent, showing which alleles are on which homolog, and determining the distance between loci.