Two genes R and Y are located very close on the chromosomal linkage map of maize plant. When RRYY and rryy genotypes are hybridized the `F_(2)` segregation will show

Options

A. segregation in the expectected 9:3:3:1 ratio
B. segregation in 3:1 ratio
C. higher number of parental types
D. higher number of the recombinant types

Answer

To solve the question regarding the F2 segregation of two linked genes R and Y in maize, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Genotypes**:
- The parental genotypes are RRYY (homozygous dominant) and rryy (homozygous recessive).
- The F1 generation will be RrYy, which is heterozygous for both traits.

2. **F1 Generation Gametes**:
- The F1 plants (RrYy) can produce gametes. However, since R and Y are linked, the gametes produced will be predominantly parental types (RY and ry) with fewer recombinant types (Ry and rY).

3. **F2 Generation**:
- When F1 (RrYy) is self-fertilized, the expected gametes will be in the ratio of 2:2:1:1 (RY, ry, Ry, rY).
- However, due to the close linkage of genes R and Y, the law of independent assortment does not apply, and crossing over is rare.

4. **Gamete Frequency**:
- The majority of the gametes will be parental types (RY and ry), and only a small proportion will be recombinant types (Ry and rY).

5. **F2 Phenotypic Ratio**:
- As a result, the F2 generation will show a higher number of parental phenotypes compared to recombinant phenotypes. This leads to a skewed ratio favoring the parental types.

6. **Conclusion**:
- Therefore, the F2 segregation will show a higher number of parental types compared to recombinant types.

### Final Answer:
The F2 segregation will show a higher number of parental types.