This procedure for placing autosomal genes on an existing map is similar to that presented for the X chromosome, but requires an extra step because the phenotype of the new mutation cannot usually be scored in the G2 (the second generation). This example is essentially a restatement of Marty Chalfie's TIGS diatribe: M. Chalfie, (1997) Is the Traditional Way of Teaching Three-Factor Mapping Sufficient? Trends Genet. 13, 94-95.

Consider the following generic example: Genes A and B are located at 20 and 28 cM on the 2nd chromosome of Drosophila. A new mutation c, maps close to the A and B loci. Since A and B have been extensively characterized, the ab/ab homozygous mutant is readily available.

The crosses are performed as follows: (relative map order is unknown)

Parentals: a b / a b X c / c

G1: a b + / + + c X a b / a b

G2: Identify recombinants for the a-b interval (A- B+ [aaBb] and A+ B- [Aabb] phenotypes) and test them for the C- [cc] phenotype. This must be done with another cross: i.e. the G2 a-b recombinants X c/c homozygous mutants. This will allow you to determine the proportion of a-b recombinants that carry c. From this you can determine the relative order of a b and c and the map position of c.

For example: Suppose there are 22 a+/ab and 16 + b/ab G2 recombinant progeny.

Of the 22 a + /ab G2 flies, 7 gave some c/c progeny in G3.

Of the 16 + b/ab G2 flies, 9 gave some c/c progeny in G3.

Consider what these numbers mean. For the 22 a+/ab G2 flies: 7 carried the c allele. These 7 must have arisen from a crossover between a and c and were thus a c + /a + b. The other 15 came from came from gametes with a crossover between b and c marking them as a + + /a + b. This suggests that C is in the middle, and that C is closer to A than to B because there are fewer A-C than C-B crossovers.

For the 16 + b/ab G2 flies: 9 came from a crossover between b and c making them genotype + c b/a + b, and 7 arose from gametes with crossovers between a and c making them genotype + + b / a + b. This suggests again that C is in the middle and that C is closer to A than it is to be because there were fewer A-C than C-B recombinants.

These data place C between A and B, and also indicate that C maps closer to A than to B. Restating the results, of the 38 a-b recombinants, 14 had a crossover between and a and c, and 24 had a crossover between c and b. (You should draw the three possible gene orders to convince yourself of this.)

The map distance is thus 20cM + (14/38) 8 cM = 22.9 cM or 28 cM - (24/38) 8 cM = 22.9 cM.

Study questions:

1) What are the problems with using a triply homozygous tester stock?

2) Why are only a-b recombinants from the G2 analyzed? Why not look at the parental types as well?

3) What about double cross-over progeny?

4) What could you say about the relative position of c if none of the G2 a+/ab gave c/c progeny in the G3, and if all of the G2 +b/ab progeny gave some c/c progeny in the G3?