We can use the above information to construct a statistical map giving the relative order of the three genes and the relative distances between those genes.  This statistical map can eventually be related to the physical map of the chromosomes.

In the example given above, parental gametes (the most frequent ones) are ADE and ade.  The least frequent  pair of gametes is AdE & aDe. These are the double crossover gametes.

When I compare the double crossover (DC) gametes to  the parental gametes  I see that the Spock/earless  allele has "switched" chromosomes.

This means that the Spock-earless gene (D)  is in the middle and the map order is:

A----------------D-----------E

Now I calculate the distances from the % of single and double recombinant gametes (phenotypes).  For convenience I will call the gametes from one of the single crossover events Pair X and the gametes of the other single crossover event Pair Y.  When I compare Pair X to the parental types (Pair W) I see that spinner-nonspinner gene (E) has "switched" chromosomes. This represents a crossover between the gene in the middle (D) and the gene on one end (E). The distance between D and E is the % recombination shown by the Pair X plus the frequencies of Double crossovers (since the DC's also represent a recombination event between D and E)

Distance D-E =  [(681 + 716)/20875] + [(1 + 5)/20875]

Distance D-E = 6.72% + .02%  = 6.74% = 6.74 map units

Similarly, I compare SC-Pair Y to  the parental gametes  In this case the alleles for Red eyes/blue eyes (A) have "switched" chromosomes. This represents a crossover between gene A on the end and gene D in the middle. The distance is the single crossover frequency  plus the DC frequency.

Distance A - D =  9.62% + .02%  = 9.64 map units

The Genetic Map is:

In rumbunnies:

Bulb nose (B) is dominant to pug nose (b)

Mossy teeth (M) is dominant to slick teeth (m)

Bad breath (S) is dominant to minty (s)

P₁ Bms/Bms x bMS/bMS

F₁ Bms/bMS (parental gametes are Bms and bMS)

B₁ Bms/bMS x bms/bms

Expected Phenotypes - with three loci we expect 2 x 2 x 2 = 8 phenotypes in a 1:1:1:1:1:1:1:1 ratio. The eight phenotypes will be (B:b) x (M:m) x (S:s)

BMS bulb mossy bad breath

BMs bulb mossy minty

BmS bulb slick bad breath

Bms bulb slick minty

bMS pug mossy bad breath

bMs pug mossy minty

bmS pug slick bad breath

bms pug slick minty

The observed phenotypes are:

That is, we are missing:

BmS bulb slick bad breath 0

bMs pug mossy minty 0

Now, for ease of working I am going to "pair up" the single crossover and parental types. That is, if one genotype had the B allele, the paired partner must have the b allele; if one has the M allele the other has the m allele, and if one has the S allele the other has the s allele,

Comparing the DC gametes to the parental gametes we see that the S locus has switched positions. It must be in the middle and the B and M loci are on the ends. The order of the map is.

B--------------S---------M

Now look at one of the single crossovers  (Pair X) and compare those gametes to the parental gametes.

The difference is that the M locus has crossed over with the middle (S). The distance is the % of  single crossover phenotypes (for Pair X) plus any Double crossover types (zero in this case).

Distance M --- S = (293 + 281)/8676 = 6.6% + 0 map units.

Now look at the the single crossover gametes  (Pair Y) and compare those gametes to the parental gametes.

The difference is that the B locus has crossed over with the middle S locus.

Distance B ----- S = (445 + 443)/8676 = 10.2% = 10.2 map units. The final map is then: