The Question is from neet-botany plant-physiology , and it was created by the author Mr. NARESH KUMAR K

Question

A plant cell with a water potential of -0.65 MPa is placed in a solution with a water potential of -0.30 MPa.  Assuming the cell wall is rigid and fully permeable, and the cell membrane is selectively permeable, what will be the pressure potential of the cell at equilibrium?

Mr. NARESH KUMAR K · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
 **Primary Concept**: Water potential and its components in plant cells.
 **Related Topics**: Osmosis, turgor pressure, plasmolysis.
 **Prerequisites**: Understanding of water potential ($\Psi$), solute potential ($\Psi_s$), and pressure potential ($\Psi_p$).
## SOLUTION PATHWAY
**Given Information**:
  Water potential of the cell ($\Psi_{cell}$) = -0.65 MPa
  Water potential of the solution ($\Psi_{solution}$) = -0.30 MPa
  Cell wall is rigid and fully permeable.
  Cell membrane is selectively permeable.
**Method & Steps**:
 1. At equilibrium, the water potential of the cell will equal the water potential of the solution.
  Reasoning: Water moves from a region of higher water potential to a region of lower water potential until equilibrium is reached.
  Formula/Rule: $\Psi_{cell} = \Psi_{solution}$ at equilibrium.
 2. Water potential is the sum of solute potential ($\Psi_s$) and pressure potential ($\Psi_p$).
  Working:  $\Psi_{cell} = \Psi_{s,cell} + \Psi_{p,cell}$  Since the solute potential remains relatively constant during this process, the change in water potential is due to the pressure potential.
  Key Point:  The cell wall's rigidity prevents further expansion of the cell volume beyond a certain point.  The pressure potential changes to equalize the water potential.
 3. Determine the pressure potential at equilibrium.
  Since $\Psi_{cell} = \Psi_{solution}$ at equilibrium:
      $\Psi_{s,cell} + \Psi_{p,cell} = -0.30 MPa$
      $\Psi_{p,cell} = -0.30 MPa - \Psi_{s,cell}$ 
      We know that the initial $\Psi_{cell} = \Psi_{s,cell} + \Psi_{p,initial} = -0.65 MPa$.  The solute potential remains constant during the process. Therefore, the change in water potential is equal to the change in pressure potential.
      The change in water potential is $-0.30 MPa - (-0.65 MPa) = +0.35 MPa$. This change is entirely reflected in the pressure potential.
      Therefore, the final pressure potential = $\Psi_{p,cell} = +0.35 MPa$ 
  Result: +0.35 MPa
  Verification: The water potential of the cell will be -0.30 MPa at equilibrium, which matches the solution's water potential.
 ## OPTION ANALYSIS
[A] -0.95 MPa :x: Incorrect. This would imply a net water loss from the cell, which is not the case since the solution has a higher water potential.
[B] -0.35 MPa :x: Incorrect. This would imply that the cell is still losing water, which is not true at equilibrium.
[C] +0.35 MPa ✓: Correct. This indicates that the cell has gained water and reached equilibrium with the solution.
[D] +0.95 MPa :x: Incorrect. This value is too high and unrealistic given the initial conditions.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Ignoring the rigid cell wall:  A flexible cell would have a different outcome.
  2.Confusing solute and pressure potential:  These are distinct components of water potential.
 **Quick Tips**:
  1.Remember that water moves from high to low water potential.
  2.At equilibrium, the water potential of the cell and solution are equal.

Mr. NARESH KUMAR K · Answer

-0.95 MPa

Mr. NARESH KUMAR K · Answer

-0.35 MPa

Mr. NARESH KUMAR K · Answer

The correct Option is "+0.35 MPa"

Mr. NARESH KUMAR K · Answer

+0.95 MPa

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