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 use the relationship between water potential (Ψw), osmotic potential (Ψs), and turgor pressure (Ψp). The equation is:

\[ \Psi_w = \Psi_s + \Psi_p \]

Where:
- Ψw = Water potential
- Ψs = Osmotic potential (also known as solute potential)
- Ψp = Turgor pressure (pressure potential)

### Step-by-Step Solution:

1. **Identify the given values:**
- Water potential (Ψw) = -30 bar
- Osmotic potential (Ψs) = -62 bar

2. **Set up the equation:**
Using the formula:
\[ \Psi_w = \Psi_s + \Psi_p \]
We can rearrange it to find Ψp:
\[ \Psi_p = \Psi_w - \Psi_s \]

3. **Substitute the known values into the equation:**
\[ \Psi_p = -30 - (-62) \]

4. **Simplify the equation:**
\[ \Psi_p = -30 + 62 \]
\[ \Psi_p = 32 \]

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

### Final Answer:
The value of the turgor pressure developed in the cell is **32 bar**.

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