Answered

\begin{tabular}{|l|l|l|}
\hline
& R & R \\
\hline
r & Rr & Rr \\
\hline
r & Rr & Rr \\
\hline
\end{tabular}

R = smooth seeds, r = wrinkled seeds

Which of the following statements is true for the F1 generation?

A. The F1 generation will be homozygous.
B. The entire F1 generation will be hybrid.
C. The entire generation will have smooth seeds.
D. The entire generation will have wrinkled seeds.
E. 50% of the generation will have wrinkled seeds.



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%.