Answer:
The correct answers are: The entire F1 generation will be hybrid.
The entire generation will have smooth seeds.
Explanation:
If the crossing individuals are both homozygote, then the F1 generation will be 100% heterozygote and will have smooth seeds.
The Punnet square is:
R R
r Rr Rr
r Rr Rr
F1 Genotype: 100% Heterozygote
Phenotype: 100% Smooth, which is the dominant trait.
- The F1 generation will be homozygous. False. As the crossing individuals are both homozygote, the whole F1 generation must be heterozygote.
- The entire F1 generation will be hybrid. True, all the F1 individuals will be heterozygote.
- The entire generation will have smooth seeds: True, as the smooth trait is dominant over the wrinkled trait and all the F1 individuals are heterozygotes, they all express the smooth trait.
- The entire generation will have wrinkled seeds: False because the wrinkled trait is the recessive one, so unless there is a homozygote individual for the trait, it can not be expressed.
- 50% of the generation will have wrinkled seeds: False. This would be the case in which a smooth heterozygote individual is crossed to a wrinkled homozygote individual. In the present example, both parental lines are homozygote.
