C2005/F2401 How to measure an RF with diploids

Suppose you want to measure the RF between gene Alpha (alleles A & a) and gene Beta (alleles B and b). The obvious thing to do is to cross AABB X aabb (parents) --> AaBb F1 (or cross AAbb X aaBB), and then to look at the products of meiosis from the F1 -- how many types, and in what proportions? If you can look directly at the meiotic products (haploid) and determine if they are A or a, B or b, you can tally up the number of recombinant products, the # of parentals, and calculate the RF. But if the products of meiosis are gametes, you can't look at them directly. So you have to go about this in a different way.

First you cross the parents to get an F1 as above. Then you cross the F1 heterozygote to a doubly homozygous recessive (do a testcross) and look at the F2 diploid offspring. Because of the way you set this up, you can tell which gamete the F1 AaBb parent contributed to each F2 offspring, and you can ignore crossing over in the aabb parent. So you can effectively count gametes from the hetero. F1 parent. For example, if an F2 offspring has the double dominant phenotype, than an AB gamete met an ab. If you started with AABB X aabb parents your double dominant F2 = parental = comes from a parental F1 gamete; if you started with AAbb X aaBB your double dominant = recombinant.

Example: Parents = albino tailless mouse X colored with normal tail (all inbred)

F1 = colored with tail
F2 = result of test cross (F1 X albino, no tail)
        72 albino, no tail
        11 albino, normal tail
        8 colored, no tail
        68 colored, normal tail

The phenotype of the F1 tells you colored and with tail are dominant (determined by alleles A and B respectively).
So you should cross the F1 to albino, no tail. When you do so, you get the F2. Because of the way you set this up (which is the standard procedure) the parental phenotypes in F2 = same as those of original parents; all F2 with these phenotypes came from parental gametes of F1.

How get RF?
    = (recombinants/total) X 100
    = (11 + 8/ total) X 100
    = 19/159 X100 = 12% = 12 map units

Note that everything above is written in terms of phenotype, not genotype. That's because you deliberately set it up so genotype could be deduced from phenotype. To be sure you have it straight, go back and put the genotypes in. It helps to draw out the chromosomes as

as well as just AB/AB X ab/ab or AABB X aabb.