A mapping exercise usually requires three genes with loci on the same chromosome pair. Two genes are set- up on one of the homologues, the third gene on the other homologue. This is accomplished by mating true breeding strains of one individual, bearing two mutations, with a true breeding strain of individuals carrying the remaining mutant. The F1 of such a cross is a wild type individual which if crossed to an individual that was homozygous for all three genes, would produce offspring that would display the parental phenotypes for the most part. Regard the diagram on the right. A stock of corn homozygous for the three recessive genes
[ colorless c, shrunken sh, and waxy wx] is crossed with a stock homozygous for the dominant alleles. The F1 is wild for all three characteristics but, of course, heterozygous... do you know why? When the F1 is backcrossed to the homozygous recessive stock the following different progeny appear in the F2. The first two listings are like the parents in the P1 and are called parental phenotypes. The remaining plants represent different combinations of the parental strain phenotype and are the recombinant classes. To actually run this exercise, flip to the next card.
