Question

An incompletely dominant gene controls the color of chickens so that BB produces black, Bb produces a slate-gray color called blue, and bb produces splashed white. A second gene controls comb shape, with the dominant gene R producing a rose comb and r producing a single comb. If a pure-breeding black chicken with a rose comb is mated to a splashed white chicken with a single comb in the F2 generation, what fraction of the offspring will be blue with a single comb?

A) 1/16.
B) 3/16.
C) 9/16.
D) 12/16.

Answer 1

The probability of an F2 offspring being blue with a single comb, after performing a dihybrid cross of BbRr (blue with rose comb) chickens, is 1/8 or 2/16. However, this option is not listed in the provided choices, which could indicate a trick question or an error in the possible answers.

Step-by-step explanation:

The question involves the concept of incomplete dominance and independent assortment in genetics. Given the pure-breeding black chicken with a rose comb (BBRR) is mated to a splashed white chicken with a single comb (bbrr), the F1 generation will be all BbRr (blue with rose comb), as per Mendel's principles of inheritance. For the F2 generation, we perform a dihybrid cross by mating two BbRr chickens.

The fraction of the offspring that will have blue feathers with a single comb from the F2 generation can be calculated using a Punnett square or the product rule for probabilities. We first find the ratio for the feather color (incomplete dominance): Bb (blue) is of 1:2:1 for BB (black) : Bb (blue) : bb (white). For the comb shape (complete dominance): Rr (rose comb) is of 3:1 for RR or Rr (rose) : rr (single).

Using the product rule: Probability of blue (Bb) is 1/2, and probability of single comb (rr) is 1/4. Thus, the probability of an F2 offspring being blue with a single comb is (1/2) × (1/4) = 1/8 or 2/16, which is not present in the options provided. Therefore, it might be a trick question, or there might be an error in the provided choices.