A tobacco plant heterozygous for recessive character is self-pollinated and 1200 seeds are subsequently germinated. How many seedings would have the parental genotype ?

Options

A. 1250
B. 600
C. 300
D. 2250

Explanation

To solve the problem of how many seedlings would have the parental genotype when a tobacco plant heterozygous for a recessive character is self-pollinated, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Genotype of the Parent Plant**:
The tobacco plant is heterozygous for a recessive character, which means its genotype can be represented as Tt (where T is the dominant allele and t is the recessive allele).

2. **Perform a Self-Cross**:
When the heterozygous plant (Tt) is self-pollinated, the cross can be represented as:
\[ Tt \times Tt \]

3. **Determine the Gametes**:
The gametes produced by each parent (Tt) will be:
- T (dominant)
- t (recessive)

4. **Set Up a Punnett Square**:
The Punnett square for the cross Tt x Tt will look like this:

| | T | t |
|-----|-----|-----|
| **T** | TT | Tt |
| **t** | Tt | tt |

5. **Analyze the Offspring Genotypes**:
From the Punnett square, we can see the possible genotypes of the offspring:
- TT (homozygous dominant): 1
- Tt (heterozygous): 2
- tt (homozygous recessive): 1

This gives us a total of 4 offspring with the following ratio:
- 1 TT : 2 Tt : 1 tt

6. **Calculate the Parental Genotype**:
The parental genotype (TT and Tt) consists of 3 out of 4 offspring. Therefore, the fraction of seedlings that have a parental genotype is:
\[ \frac{3}{4} \]

7. **Apply the Ratio to the Total Number of Seeds**:
Given that 1200 seeds are germinated, we calculate the number of seedlings with the parental genotype:
\[ \text{Number of parental genotype seedlings} = \frac{3}{4} \times 1200 = 900 \]

### Final Answer:
Thus, the number of seedlings that would have the parental genotype is **900**.

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