Let's illustrate this principle by considering a mapping cross to determine the distance between two genes (a 'two-point' mapping cross). The two genes we'll map are autosomal genes, which, when mutated, produce vestigial wings (this gene is therefore called vestigial (vg)) and a black body (this gene is calledblack (b)). We'll begin by crossing a true-breeding wild-type fly with a true-breeding vestigial/black fly.

Wild type

X

Vestigial/black

Wild type

This produces a generation of flies that are wild type but heterozygous. The heterozygous condition of these flies is important. One chromosome in each cell has both wild-type alleles, and the other has both mutant alleles. When meiosis occurs in these flies, any crossover between the two genes will generate chromosomes with one wild-type and one mutant allele. To see whether a crossover has occurred, all we have to do is perform a testcross with these flies. That way, if a non-crossover wild-type chromosome is contributed, the resulting offspring will be wild type. If a non-crossover mutant chromosome is contributed, the resulting offspring will have vestigial wings and black bodies. These two phenotypes are the same as the original parents, and are therefore called the parentals.If a crossover has occurred, the resulting offspring will have one wild-type trait and one mutant trait. These flies are referred to as recombinants. Therefore, simply by assessing the phenotypes of the F2 flies, we can tell what percentage of the chromosomes experienced a crossover.

F1 heterozygote

X

Vestigial/black

Wild type

415

vestigial/black

405

vestigial

92

black

88

Out of the 1000 offspring in the F2 generation, 180 (92 + 88) are recombinants (the result of a crossover event). The distance between the vestigial and black genes is therefore:

180/1000 = 0.18 = 18% = 18 map units

 

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