Epistasis

=Epistasis=

**Epistasis** is an interaction between genes at different loci where the expression of one gene (the //hypostatic gene//) is dependent upon the expression of one or many others (the //epistatic or modifier gene(s)//). Epistasis is distinct from dominance, which is the interaction of different alleles at the same locus to produce a phenotype.

For instance, consider two genes in a flower: the **P** gene, which codes for pigmentation, and the **R** gene, which codes for pigment colour. The dominant allele of the former locus, **P**, codes for pigment, while the recessive allele, **p**, does not. The dominant allele of the latter locus, **R**, codes for red colouration while the recessive allele, **r**, codes for white colouration. If the allele at locus **P** is recessive, **p**, then it does not matter what the allele at locus **R** is: no pigment will be produced, and no colour expressed. This is specifically called **recessive epistasis**: when the epistatic gene, **P**, is homozygous recessive then the phenotype of the hypostatic gene, **R**, is automatically masked.

Often epistasis occurs when multiple gene products are needed as enzymes in a metabolic sequence. For example, consider the simple biochemical pathway below:

Substrate 1--(enzyme **A**)--> Substrate 2 --(enzyme **B**)-> Substrate 3

Suppose enzyme **A** is encoded by gene **A**, and enzyme **B** is encoded by gene **B**. If gene **A** is double-recessive, **aa**, then enzyme **A** will not be synthesised. If substrate 2 is the necessary substrate for enzyme **B**, then regardless of the expression of gene **B,** substrate 3 will not be produced because substrate 2 is not available. This is why gene **B** is in recessive epistasis to gene **A**.

**Epistasis analysis** can be used to determine the order of genes in a biochemical pathway such as the above. Mutations are induced in different genes and the phenotypic effects are observed. Suppose, in the above pathway, substrate 1 has no obvious phenotypic effect, substrate 2 produces a green phenotype and substrate 3 produces a blue one. If the metabolic pathway proceeds normally, then the individual is blue. However, if a mutation in a gene causes the individual to become green then we can confirm that the mutation was in gene **B**, since the metabolic pathway has successfully proceeded at least as far as substrate 2.

__** Epistasis analysis using mutant yeast as an example: **__ W ild-type yeast are capable of synthesising adenine, and form white colonies. Two mutants, both incapable of synthesising adenine, can be produced:
 * //ade4// mutants also produce white colonies;
 * // ade1 // mutants instead accumulate an intermediate that produces red colonies.

Since the red phenotype is produced in a mutant who cannot synthesise adenine, the red intermediate must come before adenine in some biochemical pathway, like so:

--(step 1)-> red intermediate -(step 2)---> adenine

The //ade1// mutation must occur in step 2, because the //ade1// mutant reaches the red phenotype but does not synthesise adenine. This means that the //ade1// gene encodes an enzyme that converts the red intermediate into something else, and when mutated the red intermediate instead accumulates.

The //ade4// mutation, however, could be in either step 1 or step 2. The white colonies might occur because the //ade4// gene encodes an enzyme that is responsible for producing the red intermediate in the first place (hence a mutation in step 1 prevents production of the red intermediate), or because the //ade4// gene encodes an enzyme that is responsible for taking the product of the //ade1// enzyme and converting it to adenine (hence the //ade1// enzyme has already made the phenotype white, but the //ade4// mutation has prevented progression to adenine synthesis). So our two possible scenarios are:

-(//ade4// gene)--> red phenotype //(ade1// gene)> adenine OR ---> red phenotype ---(//ade1// gene)--(//ade4// gene)---> adenine

To establish which scenario is correct, a double-mutant (i.e. one with //ade1// and //ade4// mutations) is required. Such double-mutants produce white colonies. This must mean that the //ade4// gene encodes an enzyme responsible for making the red intermediate (i.e. the first scenario is correct). Because the //ade4// gene is mutated in the double-mutant, the red intermediate is never made and the double-mutant colony is white. The //ade4// gene is epistatic to the //ade1// gene.

You might wonder how we can be sure of this, but consider the reverse scenario. If the //ade4// gene came //after// the //ade1// gene in this pathway, then the colonies would have to be red because the mutation in //ade1// would mean that the red intermediate could not be metabolised, regardless of the mutation in //ade4// down the line (i.e. //ade1// would be epistatic to //ade4//). Because the double-mutants produce white colonies, we know that this cannot be the case.

__**Epistasis and Inheritance: **__

In inheritance, epistasis can cause certain phenotypes that were not present in the parents to emerge in the offspring. This is the likely to be the case where an F2 generation segregates with a phenotypic ratio of 9:4:3 (or some variation; e.g. 9:7 or 9:6:1). Consider the example below.

Two true-breeding strains of summer squash with yellow fruits and a spherical shape are crossed to give an F1 generation that all have green fruit and disc shapes.

// From here we can conclude that all the F1 individuals are heterozygous (since their parents are homozygous) and that green is dominant to yellow and disc-shaped dominant to spherical. //

The F1 individuals are then self-fertilised to give an F2 generation comprising 9 green-fruited and disc-shaped, 6 yellow-fruited and spherical-shaped and 1 yellow-fruited and elongated-shaped.

//The ratio here is 9:6:1 which is similar to 9:4:3, indicating that epistasis is probably occurring between these two genes. The yellow and elongated phenotype is '1', indicating that is probably doubly recessive. //

We can conclude that the dominant phenotype, green and disc-shaped, requires a dominant allele at both loci: i.e. GgDd. In the parents, the genotypes would have been ggDD and GGdd, causing them both to display recessive phenotypes for both traits in spite of carrying a dominant allele for each trait, because each dominant allele at one locus required a dominant allele at the other. In the F1 generation, the offspring all became heterozygous at both loci, GgDd, allowing them to express both dominant phenotypes. In the F2 generation, self-fertilisation produced a mixture of genotypes: where there were dominant alleles at both loci, the dominant phenotype appeared (9 in the ratio); where either one of the loci (but not both) was entirely recessive, the recessive phenotype appeared (6 in the ratio); and where both loci were entirely recessive (ggdd) a completely new phenotype emerged (1 in the ratio).