Question
Using the Punnett Square below, select the probability of each option. (2 points)Green peasYellow peas
Answers
GPT-4o mini
To determine the probability of offspring characteristics using a Punnett Square, it's crucial to know the genotypes of the parent plants.
Let’s assume:
- "G" represents the dominant allele for green peas.
- "g" represents the recessive allele for yellow peas.
If you have:
1. Two homozygous green peas (GG)
2. One homozygous green pea (GG) and one homozygous yellow pea (gg)
3. One heterozygous green pea (Gg) and one homozygous yellow pea (gg)
Let’s break down the scenarios:
1. **Two Homozygous Green Peas (GG x GG)**:
- All offspring will be GG (100% Green Peas).
2. **One Homozygous Green Pea and One Homozygous Yellow Pea (GG x gg)**:
- All offspring will be Gg (100% Green Peas).
3. **One Heterozygous Green Pea and One Homozygous Yellow Pea (Gg x gg)**:
- The Punnett Square would look like this:
| | g | g |
|--------|------|------|
| G | Gg | Gg |
| g | gg | gg |
- Probability of Green Peas (Gg): 50%
- Probability of Yellow Peas (gg): 50%
Please provide the specific parent genotypes if you need probabilities calculated for a different pairing.
Let’s assume:
- "G" represents the dominant allele for green peas.
- "g" represents the recessive allele for yellow peas.
If you have:
1. Two homozygous green peas (GG)
2. One homozygous green pea (GG) and one homozygous yellow pea (gg)
3. One heterozygous green pea (Gg) and one homozygous yellow pea (gg)
Let’s break down the scenarios:
1. **Two Homozygous Green Peas (GG x GG)**:
- All offspring will be GG (100% Green Peas).
2. **One Homozygous Green Pea and One Homozygous Yellow Pea (GG x gg)**:
- All offspring will be Gg (100% Green Peas).
3. **One Heterozygous Green Pea and One Homozygous Yellow Pea (Gg x gg)**:
- The Punnett Square would look like this:
| | g | g |
|--------|------|------|
| G | Gg | Gg |
| g | gg | gg |
- Probability of Green Peas (Gg): 50%
- Probability of Yellow Peas (gg): 50%
Please provide the specific parent genotypes if you need probabilities calculated for a different pairing.