Let's analyze the problem step-by-step:
1. Initial Cross: PP x pp
- Genotype of Parent 1: PP (homozygous dominant, purple flowers)
- Genotype of Parent 2: pp (homozygous recessive, white flowers)
When we cross PP with pp, all offspring will inherit one allele from each parent:
- Parent 1 (PP) can only contribute a P allele.
- Parent 2 (pp) can only contribute a p allele.
So, all offspring will have the genotype Pp (heterozygous):
- Pp
Since the P allele is dominant, these offspring will all have purple flowers.
Therefore, the probability that the offspring from this cross will have purple flowers is:
- [tex]\( 1.0 \)[/tex]
2. Second Cross: Pp x pp
- One offspring from the first cross has the genotype Pp (heterozygous, purple flowers).
- Cross this offspring with a plant that is pp (homozygous recessive, white flowers).
Constructing a Punnett square for Pp x pp:
```
| | p | p |
|---|---|---|
| P | Pp| Pp|
| p | pp| pp|
```
The possible genotypes for the next generation (second cross) are:
- Pp (heterozygous, purple flowers)
- pp (homozygous recessive, white flowers)
Counting occurrences:
- Pp: 2 (purple flowers)
- pp: 2 (white flowers)
Therefore, there are no PP genotypes in the second cross.
The probability of having a homozygous dominant genotype (PP) in this cross is:
- [tex]\( 0.0 \)[/tex]
In summary:
- The probability that the offspring from the first cross (PP x pp) will have purple flowers is [tex]\( 1.0 \)[/tex].
- The probability of an offspring from the second cross (Pp x pp) being homozygous dominant for purple flowers (PP) is [tex]\( 0.0 \)[/tex].
Let's fill in the drop-down menus:
1. The probability that the offspring will have purple flowers is [tex]\( \boxed{1.0} \)[/tex].
2. The probability of homozygous dominant genotype for the flowers will be [tex]\( \boxed{0.0} \)[/tex].
