A tobacco plant heterozygous for recessive trait of albinism is selfed and 1200 seeds are obtained. How many seedlings obtained from such seeds will have parent genotype?

Options

A. 900
B. 300
C. 600
D. 1200

Explanation

To solve the question, we need to determine how many seedlings will have the parental genotype when a tobacco plant heterozygous for the recessive trait of albinism is selfed, resulting in 1200 seeds.

### Step-by-Step Solution:

1. **Identify the Genotype of the Parent Plant**:
- The parent plant is heterozygous for the trait of albinism. We can denote the dominant allele (normal color) as "R" and the recessive allele (albinism) as "r". Therefore, the genotype of the parent plant is Rr.

2. **Determine the Possible Gametes**:
- When the heterozygous plant (Rr) self-fertilizes, it can produce two types of gametes: R (dominant) and r (recessive).

3. **Set Up a Punnett Square**:
- To find the genotypes of the offspring, we can set up a Punnett square:
```
R r
------------
R | RR | Rr |
------------
r | Rr | rr |
```
- The possible genotypes from this cross are:
- RR (homozygous dominant)
- Rr (heterozygous)
- rr (homozygous recessive)

4. **Calculate the Ratios**:
- From the Punnett square, we see the following ratios:
- 1 RR : 2 Rr : 1 rr
- This means that out of 4 offspring, 1 will be RR, 2 will be Rr, and 1 will be rr.

5. **Determine the Proportion of Parental Genotype**:
- The parental genotype (Rr) appears in 2 out of 4 offspring, which is 50% of the total offspring.

6. **Calculate the Number of Seedlings with Parental Genotype**:
- If there are 1200 seeds obtained, we can calculate the number of seedlings with the parental genotype (Rr) as follows:
- Number of Rr seedlings = 50% of 1200 = 0.5 * 1200 = 600.

### Final Answer:
- Therefore, the number of seedlings obtained from the seeds that will have the parental genotype (Rr) is **600**.

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