Answer :
Alright, let's clarify and solve this genetics problem step-by-step:
### Problem Statement
- We are given a genotype table:
[tex]\[ \begin{tabular}{|c|c|c|} \hline & \text{R} & \text{R} \\ \hline \text{r} & \text{Rr} & \text{Rr} \\ \hline \text{r} & \text{Rr} & \text{Rr} \\ \hline \end{tabular} \][/tex]
- [tex]\( \text{R} \)[/tex] denotes the dominant allele for smooth seeds.
- [tex]\( \text{r} \)[/tex] denotes the recessive allele for wrinkled seeds.
### Analysis
Based on the genotype table, let's deduce the results for the F1 generation:
1. Possible Genotypes in F1 Generation:
- All alleles in this Punnett square are represented as [tex]\( \text{Rr} \)[/tex].
- Each entry in the F1 generation grid is [tex]\( \text{Rr} \)[/tex], so there are four [tex]\( \text{Rr} \)[/tex] genotypes.
2. Classification by Genotype:
- Homozygous: Genotypes which are [tex]\( \text{RR} \)[/tex] or [tex]\( \text{rr} \)[/tex]. In this case, there are no [tex]\( \text{RR} \)[/tex] or [tex]\( \text{rr} \)[/tex] genotypes. So, the number of homozygous individuals is 0.
- Hybrid: Genotypes which are [tex]\( \text{Rr} \)[/tex]. Here, all four genotypes are [tex]\( \text{Rr} \)[/tex]. Hence, the number of hybrids is 4.
3. Phenotype (Physical Traits):
- Smooth Seeds: Seeds will be smooth if at least one [tex]\( \text{R} \)[/tex] allele is present (dominant trait). Since all individuals have the genotype [tex]\( \text{Rr} \)[/tex], all of them will have smooth seeds. So, the number of individuals with smooth seeds is 4.
- Wrinkled Seeds: Seeds will be wrinkled if the genotype is [tex]\( \text{rr} \)[/tex] (recessive trait). In this scenario, there are no [tex]\( \text{rr} \)[/tex] genotypes. Therefore, the number of individuals with wrinkled seeds is 0.
4. Percentage of Wrinkled Seeds:
- To find the percentage of individuals having wrinkled seeds, we calculate:
[tex]\[ \text{Percentage} = \left( \frac{\text{Number of wrinkled seed individuals}}{\text{Total number of individuals}} \right) \times 100 = \left( \frac{0}{4} \right) \times 100 = 0\% \][/tex]
### Summary
- Number of homozygous individuals: [tex]\(0\)[/tex]
- Number of hybrid individuals: [tex]\(4\)[/tex]
- Number of individuals with smooth seeds: [tex]\(4\)[/tex]
- Number of individuals with wrinkled seeds: [tex]\(0\)[/tex]
- Percentage of individuals with wrinkled seeds: [tex]\(0\%\)[/tex]
### Final Solution
The given table results in:
[tex]\[ (0, 4, 4, 0, 0) \][/tex]
Thus, all individuals in the F1 generation are hybrids with the genotype [tex]\( \text{Rr} \)[/tex], they all display smooth seeds, and none have wrinkled seeds. The percentage of wrinkled seed individuals is 0%.
### Problem Statement
- We are given a genotype table:
[tex]\[ \begin{tabular}{|c|c|c|} \hline & \text{R} & \text{R} \\ \hline \text{r} & \text{Rr} & \text{Rr} \\ \hline \text{r} & \text{Rr} & \text{Rr} \\ \hline \end{tabular} \][/tex]
- [tex]\( \text{R} \)[/tex] denotes the dominant allele for smooth seeds.
- [tex]\( \text{r} \)[/tex] denotes the recessive allele for wrinkled seeds.
### Analysis
Based on the genotype table, let's deduce the results for the F1 generation:
1. Possible Genotypes in F1 Generation:
- All alleles in this Punnett square are represented as [tex]\( \text{Rr} \)[/tex].
- Each entry in the F1 generation grid is [tex]\( \text{Rr} \)[/tex], so there are four [tex]\( \text{Rr} \)[/tex] genotypes.
2. Classification by Genotype:
- Homozygous: Genotypes which are [tex]\( \text{RR} \)[/tex] or [tex]\( \text{rr} \)[/tex]. In this case, there are no [tex]\( \text{RR} \)[/tex] or [tex]\( \text{rr} \)[/tex] genotypes. So, the number of homozygous individuals is 0.
- Hybrid: Genotypes which are [tex]\( \text{Rr} \)[/tex]. Here, all four genotypes are [tex]\( \text{Rr} \)[/tex]. Hence, the number of hybrids is 4.
3. Phenotype (Physical Traits):
- Smooth Seeds: Seeds will be smooth if at least one [tex]\( \text{R} \)[/tex] allele is present (dominant trait). Since all individuals have the genotype [tex]\( \text{Rr} \)[/tex], all of them will have smooth seeds. So, the number of individuals with smooth seeds is 4.
- Wrinkled Seeds: Seeds will be wrinkled if the genotype is [tex]\( \text{rr} \)[/tex] (recessive trait). In this scenario, there are no [tex]\( \text{rr} \)[/tex] genotypes. Therefore, the number of individuals with wrinkled seeds is 0.
4. Percentage of Wrinkled Seeds:
- To find the percentage of individuals having wrinkled seeds, we calculate:
[tex]\[ \text{Percentage} = \left( \frac{\text{Number of wrinkled seed individuals}}{\text{Total number of individuals}} \right) \times 100 = \left( \frac{0}{4} \right) \times 100 = 0\% \][/tex]
### Summary
- Number of homozygous individuals: [tex]\(0\)[/tex]
- Number of hybrid individuals: [tex]\(4\)[/tex]
- Number of individuals with smooth seeds: [tex]\(4\)[/tex]
- Number of individuals with wrinkled seeds: [tex]\(0\)[/tex]
- Percentage of individuals with wrinkled seeds: [tex]\(0\%\)[/tex]
### Final Solution
The given table results in:
[tex]\[ (0, 4, 4, 0, 0) \][/tex]
Thus, all individuals in the F1 generation are hybrids with the genotype [tex]\( \text{Rr} \)[/tex], they all display smooth seeds, and none have wrinkled seeds. The percentage of wrinkled seed individuals is 0%.
