Leaf extract from an infected plant was passed through a filter with a pore size of 0.05 mm diameter. The infectious agent was detected in the filtrate. Which ONE of the following is the likely infectious agent?

Options

A. Bacteria
B. Virus
C. Nematode
D. Fungus

Answer

To determine the likely infectious agent detected in the filtrate after passing a leaf extract from an infected plant through a filter with a pore size of 0.05 mm, we can analyze the options given based on the size of the infectious agents.

### Step-by-Step Solution:

1. **Understanding the Filter Size**: The filter used has a pore size of 0.05 mm (which is equivalent to 50 micrometers). This means that any infectious agent larger than this size will be retained by the filter, while smaller agents will pass through.

2. **Analyzing the Options**:
- **A. Bacteria**: Bacteria typically range in size from about 0.5 micrometers (0.0005 mm) to several micrometers. Since the smallest bacteria are around 0.5 micrometers, they would pass through the filter.
- **B. Virus**: Viruses are much smaller, generally ranging from about 20 nanometers (0.00002 mm) to 500 nanometers (0.0005 mm). Therefore, they can easily pass through the filter.
- **C. Nematode**: Nematodes are larger, usually around 1 mm or more in size. They would be retained by the filter.
- **D. Fungus**: Fungal spores and hyphae are also larger than 0.05 mm and would not pass through the filter.

3. **Conclusion**: Since the infectious agent was detected in the filtrate, it must be smaller than the filter size. The only option that fits this criterion is **B. Virus**, as it is the smallest infectious agent among the options provided.

### Final Answer:
**B. Virus**