Answer:
Explanation:
Since the mice crossed were pure breeding, each parent produced sperms of one kind, both alleles of each parente were equal. Thus, the first (A) and second (B) statements are false.
Assume pure breeding black mice produced BB sperm and pure breeding brown mice produced bb sperm.
The punnett square for the first generation, F₁ is would be:
b b
B Bb Bb
B Bb Bb
This is, all the F₁ generation mice are Bb.
All the 992 in F₁ generation had black coats means black is dominant.
Now when F₁ generation is crossed, the punnet square reveals the second generation, F₂:
B b
B BB Bb
b bB bb
The brown allele that was hidden now appears.
Based on the assumption that black is dominant, BB, Bb and bB are black, which is 3/4 or 75%. This, is 75% of the mice in the F₂ genrertion are black.
The combination bb is brown. Thus, 1/4 or 25% of the mice in the F₂ generation are brown.
When the mice in the first generation were crossed, they yielded 961 black coated mice and 317 brown coated mice. That is:
Therefore, it is proven that an approximately three-to-one ratio of black to brown cated mice in F₂ is accounted fro the black allele being dominant over the brown allele (choice C.)
Also, the brown allele is not independent from the black allele and dissapears in the F₁ generation: it is not independent because it presence will only be shown when the black allele is not present, and it is said that it dissapears because it is not revealed, it is hidden.
