A photoblastic seed is gives following treatment. In which case germination of seeds will be maximum ?
(R = red light of 660 m`mu` and Fr= far red light of 730 m`mu`)

Options

A. Seeds + Fr
B. Seeds + R + Fr
C. Seeds + R + Fr + R
D. Seeds + R + Fr + R + Fr

Explanation

To determine in which treatment the germination of photoblastic seeds will be maximum, we need to analyze the effects of red light (R) and far-red light (FR) on seed germination.

### Step-by-Step Solution:

1. **Understanding Photoblastic Seeds**:
- Photoblastic seeds are seeds that require light for germination. They respond differently to various wavelengths of light, particularly red (660 nm) and far-red (730 nm) light.

2. **Effects of Light on Germination**:
- Red light (R) promotes germination.
- Far-red light (FR) inhibits germination. The last light exposure is crucial in determining the outcome of germination.

3. **Analyzing Each Treatment**:
- **Option 1: Seeds + FR**:
- Since FR inhibits germination, this option will result in no germination.
- **Option 2: Seeds + R + FR**:
- R promotes germination, but since the last light exposure is FR, it will inhibit germination. Thus, this option will also result in no germination.
- **Option 3: Seeds + R + FR + R**:
- The first R promotes germination, then FR inhibits it, but the last R promotes germination again. Since the last exposure is R, this option will lead to germination.
- **Option 4: Seeds + R + FR + R + FR**:
- Similar to option 3, but ends with FR. The last exposure being FR will inhibit germination, resulting in no germination.

4. **Conclusion**:
- The treatment that will yield maximum germination is **Option 3: Seeds + R + FR + R**, as the last light exposure is red light, which promotes germination.

### Final Answer:
**Maximum germination will occur in Option 3: Seeds + R + FR + R.**