Question

In sheep, coat color is influenced by two genes. Gene A influences pigment production, while gene B produces black or brown pigment. If two heterozygous white sheep resulted in 12 white sheep, 3 black sheep, and 1 brown sheep, which genotype(s) of the white sheep explain this data? a) The white sheep could be A_B_ or aaB_. b) The white sheep could be A_B_ or aabb. c) The white sheep could be A_B_ or A_bb

Answer 1

c) The white sheep could be A_B_ or A_bb.

Step-by-step explanation:

This is a case of dominant epistasis which falls under the category of intergenic interaction (gene interaction) and produces phenotypes in 12:3:1 ratio rather than producing ideal 9:3:3:1 ratio of Mendelian inheritance for dihybrid crosses.

The progeny produced after such dihybrid cross and their phenotypes will be as under:

AABB - 1 white

AABb - 2 white

AAbb - 1 white

AaBB - 2 white

AaBb - 4 white

Aabb - 2 white

aaBB - 1 black

aaBb - 2 black

aabb - 1 brown

In the example as shown above, gene A is masking the expression of gene B and influencing the coat color.

Whenever gene A is present in dominant and heterozygous form, the coat color is white but in the progeny where gene A is homozygous recessive the coat color is either black or brown depending upon whether gene B is present or not. When gene A is absent and gene B is present in dominant and heterozygous form, the coat color is black but when gene B is also absent then the coat color is brown.

So it is clear that the product of gene A will cause the product of gene B to be white always as a result of which the product of gene B will express only in the absence of gene A.