Epistasis, Allelism and Complimentation

In Chapter 4, we have an example of two lines are crossed, and produce a new phenotype. In particular, the two lines are "mutant" phenotypes of White and their hybrid offspring are pigmented, Purple. The complementation is exhibited in the F₁ generation. If the F₂ generation is observed, and there is free recombination (Independent Assortment) we have a modified ratio of 9 Purple : 7 White phenotypic classses.

The cross is AAbb X aaBB which gives AaBb.

When mutant alleles, a or b, are present, there is no effective enzyme at step A or B, respectively. The last product in the chain of reactions thereby becomes the phenotype, either White or Purple. When we cross two individuals that are homozygous for aa in one, and bb in the other, the hybrid has an active allele for both loci, and the full phenotypic expression of Purple results. The genotypes at the two loci compliment one another. Therefore, the loci are different, and the A allele is at a different locus than the B allele. A and B are NOT allelic.

Alternatively, had we been working in the context of multiple alleles for a single locus, we might have had two lines showing mutant (White) phenotypes, and crossed them. The F₁ would have been mutant:

Therefore, the test for allelism using complimentation has the following steps:

• Cross two mutant lines
• If the offspring has a wild phenotype, the mutatnts compliment and the mutants are at different loci
• If the offspring has a mutant phenotype, the mutants fail to compliment, and the mutants are at the same locus

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Last updated06/16/97