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In the hybridization of 2 genes (4 different alleles, 2 of each pair) how does epistasis affect the proportion of phenotypical forms in the F2 generation?
In the dihybridism without epistasis double heterozygous parental individuals cross and in F2 4 phenotypical forms appear. The amount is 9 double dominant to 3 dominant for the first pair, recessive for the second to 3 recessive for the first pair, dominant for the second to 1 double recessive (9:3:3:1).
Considering so as to the epistatic gene is the second pair and that the recessive genotype of the hypostatic gene means lacking of the characteristic, in the F2 generation of the dominant epistasis the following phenotypical forms would emerge: 13 dominant for the second pair or recessive for the first that is the characteristic does not manifest, 3 dominant for the first pair, recessive for the second That is the characteristic manifests. Phenotypical proportion would be 13:3. In recessive epistasis in F2 the phenotypical forms that would emerge are: 9 double dominant (the characteristic manifests), 7 recessive for the first pair or recessive for the second that is the characteristic does not manifest. hence the phenotypical proportion would be 9:7.
These instances show how epistasis changes phenotypical forms and proportions, from the normal 9:3:3:1 in F2 to 13:3 in the dominant epistasis or to 9:7 in recessive epistasis (note that some forms have even disappeared) and if the recessive genotype of the hypostatic gene is active, not simply meaning that the dominant allele does not manifest, the number of phenotypical forms in F2 changes.
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