Heterozygous round and yellow seeded pea plants were selfed and total 800 seeds are collected. What is the total number of seeds with first dominant and second recessive traits ?

Options

A. 950
B. 300
C. 200
D. 150

Answer

To solve the problem of determining the total number of seeds with the first dominant (round) and second recessive (green) traits from the selfing of heterozygous round and yellow seeded pea plants, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Genotypes**:
- The heterozygous round seeded pea plant has the genotype RrYy, where:
- R = Round (dominant)
- r = Wrinkled (recessive)
- Y = Yellow (dominant)
- y = Green (recessive)

2. **Determine the Dihybrid Cross**:
- When two heterozygous plants (RrYy x RrYy) are selfed, we perform a dihybrid cross. The phenotypic ratio for a dihybrid cross is typically 9:3:3:1.

3. **Calculate the Phenotypic Ratio**:
- From the dihybrid cross, the phenotypic ratio can be broken down as follows:
- 9 Round Yellow (R_Y_)
- 3 Round Green (R_yy)
- 3 Wrinkled Yellow (rrY_)
- 1 Wrinkled Green (rryy)

4. **Identify the Required Traits**:
- We are interested in the seeds with round shape (dominant trait) and green color (recessive trait), which corresponds to the phenotype R_yy. According to the phenotypic ratio, this is represented by the "3" in the ratio.

5. **Calculate the Total Seeds with R_yy**:
- The total number of seeds collected is 800. To find the number of seeds with the round and green traits, we use the ratio:
\[ \text{Number of R_yy seeds} = \frac{3}{16} \times \text{Total Seeds} \]
\[ \text{Number of R_yy seeds} = \frac{3}{16} \times 800 = 150 \]

6. **Final Answer**:
- Therefore, the total number of seeds with the first dominant (round) and second recessive (green) traits is **150**.