Out of a population of 800 individuals in `F_(2)` generation of a cross between yellow round and green wrinkled Pea Plants, what would
be number of yellow and wrinkled seeds

Options

A. 800
B. 400
C. 200
D. 150

Answer

To solve the problem of determining the number of yellow and wrinkled seeds in the F2 generation of a cross between yellow round and green wrinkled pea plants, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Cross**:
- We are dealing with a dihybrid cross involving two traits: seed color (yellow vs. green) and seed shape (round vs. wrinkled).
- Yellow (Y) is dominant over green (y), and round (R) is dominant over wrinkled (r).

2. **Determine the F2 Generation Ratio**:
- According to Mendelian genetics, the phenotypic ratio for a dihybrid cross in the F2 generation is typically 9:3:3:1.
- This ratio represents:
- 9 for yellow round (YYRR, YYRr, YyRR, YyRr)
- 3 for yellow wrinkled (YYrr, Yyrr)
- 3 for green round (yyRR, yyRr)
- 1 for green wrinkled (yyrr)

3. **Identify the Relevant Phenotype**:
- We are interested in the yellow and wrinkled seeds, which corresponds to the phenotype represented by the ratio of 3 in the 9:3:3:1 ratio.

4. **Calculate the Total Population**:
- The total population given is 800 individuals.

5. **Calculate the Number of Yellow and Wrinkled Seeds**:
- To find the number of yellow and wrinkled seeds, we use the ratio of yellow wrinkled seeds (3 out of 16 total phenotypes).
- The calculation is as follows:
\[ \text{Number of yellow wrinkled seeds} = \left(\frac{3}{16}\right) \times 800 \]
- Performing the calculation:
\[ \text{Number of yellow wrinkled seeds} = \frac{3 \times 800}{16} = \frac{2400}{16} = 150 \]

6. **Conclusion**:
- The number of yellow and wrinkled seeds in the F2 generation is **150**.

### Final Answer:
The number of yellow and wrinkled seeds is **150**.

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