If water potential of a cell is - 30 bar and osmotic potential is - 62 bar then what is the value of the TP developed in the cell.

Options

A. 102 bar
B. `-92` bar
C. `50` bar
D. 32 bar

Answer

To solve the problem, we need to find the value of the turgor pressure (TP) developed in the cell using the water potential (Ψ) and osmotic potential (Ψs).

### Step-by-Step Solution:

1. **Understand the formula for water potential**:
The water potential (Ψ) of a cell can be expressed using the formula:
\[ Ψ = Ψs + TP \]
where:
- Ψ = Water potential
- Ψs = Osmotic potential (also known as solute potential)
- TP = Turgor pressure

2. **Identify the given values**:
From the question, we have:
- Water potential (Ψ) = -30 bar
- Osmotic potential (Ψs) = -62 bar

3. **Substitute the known values into the formula**:
We can rearrange the formula to solve for turgor pressure (TP):
\[ TP = Ψ - Ψs \]
Now substituting the known values:
\[ TP = -30 \, \text{bar} - (-62 \, \text{bar}) \]

4. **Simplify the equation**:
This simplifies to:
\[ TP = -30 + 62 \]
\[ TP = 32 \, \text{bar} \]

5. **Conclusion**:
The value of the turgor pressure (TP) developed in the cell is **32 bar**.