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Part A 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 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 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. O 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 O 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 O 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 Submit My Answers Give Up Incorrect; Try Again; 5 attempts remaining

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Answer #1

The answer is second one. 12:3:1 epistatis, A_B_ and A_bb: Black,; aaB_ : Brown and aabb : Green

For explanation find the below image

The observed values are almost fit to 12:3:1 ratio.
It is an example for dominant epistatic gene interaction.
In Dominant eptistatic interaction one gene expressess her character even in the presence of dominant allele. But the second gene can express her character only in the absence of dominant allele of first gene.

早 al lar, 구! Klaul Aasb 15 → laalb 12 1: )

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