Question

The sap of a plant cell has an osmotic potential of -10 bars and there is a wall pressure of 2 bars, when this cell is placed with an osmotic potential of -3 bars, the force causing water to enter the cell is:

• A. -8 bar
• B. -7 bar
• C. -5 bar
• D. -3 bar

Hint:

To quickly find the answer using the DPD method:
- \(\text{DPD of cell} = 10 - 2 = 8 \text{ bars}\).
- \(\text{DPD of solution} = 3 \text{ bars}\).
- \(\text{Net driving force} = 8 - 3 = 5 \text{ bars}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Water movement in and out of plant cells is governed by the water potential gradient (\(\Delta \Psi_w\)) or Diffusion Pressure Deficit (DPD) gradient.
Water always moves from a region of higher water potential (less negative) to a region of lower water potential (more negative).
Alternatively, water moves from a region of lower DPD to a region of higher DPD.

Step 2: Key Formula or Approach:
The water potential of a cell (\(\Psi_{\text{cell}}\)) is given by: \[ \Psi_{\text{cell}} = \Psi_s + \Psi_p \] where \(\Psi_s\) is the osmotic potential and \(\Psi_p\) is the turgor pressure (equal in magnitude to wall pressure, WP).
The driving force causing water entry is: \[ \text{Driving Force} = \Psi_{\text{external}} - \Psi_{\text{cell}} \] Using the DPD method: \[ \text{DPD} = \text{Osmotic Pressure (OP)} - \text{Turgor Pressure (TP)} \] \[ \text{Driving Force} = \text{DPD}_{\text{cell}} - \text{DPD}_{\text{external}} \]

Step 3: Detailed Explanation:
Given for the plant cell:
Osmotic potential of cell sap, \(\Psi_s = -10 \text{ bars}\)
Wall pressure (equal to turgor pressure), \(\Psi_p = 2 \text{ bars}\)
Calculate the water potential of the cell: \[ \Psi_{\text{cell}} = -10 \text{ bars} + 2 \text{ bars} = -8 \text{ bars} \] Given for the external solution:
Osmotic potential of external solution, \(\Psi_{\text{external}} = -3 \text{ bars}\)
Now, calculate the net driving force causing water to enter the cell: \[ \text{Driving Force} = \Psi_{\text{external}} - \Psi_{\text{cell}} = -3 \text{ bars} - (-8 \text{ bars}) = +5 \text{ bars} \] In terms of tension/deficit conventions, the force causing water entry is represented as -5 bar.

Step 4: Final Answer:
The driving force is -5 bar, corresponding to option (C).