Pure tall pea plant with round seeds is crossed with pure dwarf pea plant with wrinkled seeds . The probability of genotype ttRR in `F_2` generation is

Options

A. `1/16`
B. `3/16`
C. `9/16`
D. `6/16`

Answer

To solve the problem, we need to determine the probability of obtaining the genotype \( ttRR \) in the \( F_2 \) generation from the given parental cross.

### Step-by-Step Solution:

1. **Identify the Parental Genotypes:**
- The pure tall pea plant with round seeds has the genotype \( TTRR \) (where \( T \) is for tall and \( R \) is for round).
- The pure dwarf pea plant with wrinkled seeds has the genotype \( ttrr \) (where \( t \) is for dwarf and \( r \) is for wrinkled).

2. **Determine the \( F_1 \) Generation:**
- When we cross \( TTRR \) with \( ttrr \), all offspring in the \( F_1 \) generation will have the genotype \( TtRr \) (tall and round), since both traits are dominant.

3. **Perform a Dihybrid Cross for \( F_2 \) Generation:**
- To find the \( F_2 \) generation, we perform a self-cross of the \( F_1 \) generation:
\[ TtRr \times TtRr \]

4. **Create a Punnett Square:**
- We can create a 4x4 Punnett square for the dihybrid cross:
- The gametes from each parent will be \( TR, Tr, tR, tr \).
- Filling out the Punnett square will give us the following combinations:

| | TR | Tr | tR | tr |
|------|-----|-----|-----|-----|
| **TR** | TTRR | TTRr | TtRR | TtRr |
| **Tr** | TTRr | TTrr | TtRr | Ttrr |
| **tR** | TtRR | TtRr | ttRR | ttRr |
| **tr** | TtRr | Ttrr | ttRr | ttrr |

5. **Count the Genotypes:**
- From the Punnett square, we can count the occurrences of each genotype:
- \( TTRR \): 1
- \( TTRr \): 2
- \( TtRR \): 2
- \( TtRr \): 4
- \( TTrr \): 1
- \( Ttrr \): 2
- \( ttRR \): 1 (this is the genotype we are interested in)
- \( ttRr \): 2
- \( ttrr \): 1

6. **Calculate the Total Combinations:**
- There are a total of 16 combinations in the Punnett square.

7. **Determine the Probability:**
- The probability of obtaining the genotype \( ttRR \) is:
\[ \text{Probability} = \frac{\text{Number of } ttRR \text{ genotypes}}{\text{Total genotypes}} = \frac{1}{16} \]

### Final Answer:
The probability of genotype \( ttRR \) in the \( F_2 \) generation is \( \frac{1}{16} \).