Question

The wild-type color of horned beetles is black, although other colors are known. A black horned beetle from a pure-breeding strain is crossed to a pure-breeding green female beetle. All of their F1 progeny are black. These F1 are allowed to mate at random with one another, and 320 F2 beetles are produced. The F2 consists of 239 black, 61 brown, and 20 green. Use these data to explain the genetics of horned beetle color.One hypothesis to explain the results is that:A) two genes are involved with 12:3:1 epistasis, such that A_B_ and A_bb are black, aaB_ is brown, and aabb is greenB) two genes are involved with 12:3:1 epistasis, such that aabb and A_bb are black, aaB_ is brown, and A_B_ is green.C) two genes are involved with 12:3:1 epistasis, such that A_B_ and A_bb are black, a_BB is brown, and AABB is green.D) two genes are involved with 9:6:1 epistasis, such that A_B_ is black, A_bb and aaB_ are brown, and aabb is green.E) two genes are involved with 9:4:3 epistasis, such that A_B_ is black, A_bb is brown, and aaB_ and aabb are green.

Answer 1

A) two genes are involved with 12:3:1 epistasis, such that A_B_ and A_bb are black, aaB_ is brown, and aabb is green .

Explanation:

The F2 proportions approximate 12:3:1 ratios:

• 239/320 ≅ 12/16

• 61/320 ≅ 3/16

• 20/320 = 1/16

These ratios are explained by dominant epistasis: if one gene has a dominant allele, then the expression of another gene will be masked, regardless of what alleles that second gene has.

The pure-breeding crosses were:

P AABB x aabb

F1 AaBb

F1 x F1 crosses

AaBb x AaBb

F2 (from Punnett Square):

9 A_B_ black

3 A_bb black

3 aaB_ brown

1 aabb green

12/16 individuals of the F2 are black, 3/16 are brown and 1/16 is green