Question

In a genetic cross having recessive epistasis, F${}_2$ phenotypic ratio would be

• A. 9:6:1
• B. 15:1
• C. 9:3:4
• D. 12:3:1

Answer 1

The correct option is C 9:3:4

Epistasis is the interaction between genes. Epistasis takes place when the action of one gene is modified by one or several other genes, which are sometimes called modifier genes. The gene whose phenotype is expressed is said to be epistatic, while the phenotype altered or suppressed is said to be hypostatic.
The normal F${}_2$ dihybrid phenotypic ratio is 9:3:3:1 - epistasis results in deviations from this ratio. For recessive epistatis, the phenotypic ratio is 9:3:4 (here, the homozygous recessive allele for a gene {the epistatic gene} masks the expression of the dominant allele for another gene (the hypostatic gene)).
Eg., Gene A makes one pigment, which is then converted to a final pigment by gene B. Thus, genotype aa (4/16) makes no pigment. A-bb makes pigment 1 (3/16) and A-B- (9/16) makes the final pigment.
The 9:3:4 ratio appears for the mouse coat colors albino, black, and agouti in a dihybrid cross with the mouse genes C (color) and A (Agouti).