The Question is from neet-botany plant-physiology osmosis, 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

NEET Botany Transport in Plants Osmosis MCQs

Select Topic

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

NEET Subjects

Our Bots

Company