Sex Linkage and Complementation

Before reading this, review the basic idea of the complementation test.

We can now add the possibility that one or both genes (depending on how many there actually are) are sex-linked. Below, are some observations for the complementation test taking into account possible sex linkage.

Despite the fact that sex-linked genes are considered, the basic idea remains the same: if you cross two mutants, both recessive to wild type, and observe wild type progeny (as we see here this does not necessarily require that ALL progeny are wild type,, only SOME progeny will be wild type) then the mutations complement. The interpretation of this? The mutations are in DIFFERENT genes.

1) First possibility - mutations in the same gene

We will designate the mutant as a and the wild type as a⁺ (remember, lower case means the mutant allele 'a' is recessive to wild type). If the mutation is in the same gene, both strains will be aa genotype (or X^aY if it is X-linked).

a) Gene is autosomal:

• Cross would be aa x aa. All progeny are aa (recessive phenotype).

b) Gene is X-linked:

• Cross would be X^aY x X^aX^a. Progeny are X^aX^a (females) or X^aY (males). All would have the recessive phenotype.

Therefore, if both mutations are in the same gene, there is no complementation and so all progeny will have the mutant phenotype, regardless of whether the gene is autosomal or X-linked. We do not cover Y-linkage here but you should be able to figure it out.

2) Second possibility - mutations are in 2 different genes

We will designate the two mutant alleles as a and b since we have to diagram the genetics for 2 genes. If the mutations happen to be in different genes then the first strain will have the genotype aa b⁺b⁺ (since we assume it has wild type alleles at the gene that is mutated in the second strain) and the second strain will have the genotype a⁺a⁺ bb. You should keep in mind that, if either aa or bb causes the same mutant phenotype, an individual only has to be homozygous recessive for a or b to show this phenotype.

a) Both gene are autosomal:

• Cross would be a⁺a⁺ bb x aa b⁺b⁺. All progeny are a⁺a b⁺b so all would have the wild type phenotype.

Notice, however, that we have assumed in this cross that both genes are autosomal. What would happen if this is not true? There are two possibilities in this case. The first is that one gene is X-linked, the other autosomal. The second is that both are X-linked.

b) One autosomal, one X-linked:

• Cross is X^aY b⁺b⁺ x X^a+X^a+ bb. Female progeny are X^a+X^a b⁺b. (All wild type.) Male progeny are X^a+Y b⁺b. (All wild type.)

c) Both are X-linked:

• Cross is ab⁺Y x a⁺a⁺ bb. Female progeny are a⁺a b⁺b (all wild type). Male progeny are a⁺bY (all mutant phenotype).

Therefore, if the mutations are in different genes, either all progeny will be wild type phenotype or all females will be wild type and all males mutant phenotype. In any case, if wild type progeny occurs then there has been complementation and you can infer that the mutations are in different genes.

Point to consider: Have we covered every possibility here? Think carefully about the crosses and any assumptions we have made.