population density are habitats, food, mates, and ... Decreases.
Answer:
D. 1 BB : 2 Bb : 1 bb
Explanation:
This question involves a single gene coding for fur color in guinea pig. Black fur (B) is dominant over white fur (b). This means that, as stated in the question, if a black fur parent (BB) is crossed with a white fur parent (bb), a 100% heterozygous offspring (Bb) with black fur will result.
If two heterozygous guinea pigs are crossed i.e. Bb × Bb, the following gametes will be produced by each heterozygous parent:
Bb = B and b
Using these gametes in a punnet square (see attached image), offsprings with the following genotypic ratio will be produced:
1 BB : 2 Bb : 1 bb
BB and Bb = black fur guinea pigs
bb = white fur guinea pigs
This would be probably true if the assumption that all possible genotypic variations would be equally distributed (so we would have 25% HH, 25% hh and 2x 25 Hh). If this distribution would be true and Huntingtons disease really was a single gene dominant trait diesase, then yes, we could expect such a distribution in the population.