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

Question

A plant cell is placed in a solution with a water potential of -0.8 MPa. The cell's initial solute potential is -1.2 MPa and its pressure potential is 0.4 MPa.  After equilibration, which of the following statements about the cell's state is MOST accurate?

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, water movement.
 **Prerequisites**: Understanding of water potential ($\Psi$), solute potential ($\Psi_s$), and pressure potential ($\Psi_p$).
## SOLUTION PATHWAY
**Given Information**:
  Water potential of solution ($\Psi_{solution}$) = -0.8 MPa
  Initial solute potential of cell ($\Psi_{s,initial}$) = -1.2 MPa
  Initial pressure potential of cell ($\Psi_{p,initial}$) = 0.4 MPa
  Important conditions: At equilibrium, the water potential of the cell will equal the water potential of the solution.
**Method & Steps**:
 1. Calculate the initial water potential of the cell:
  Reasoning: Water potential is the sum of solute potential and pressure potential.  $\Psi = \Psi_s + \Psi_p$
  Formula/Rule: $\Psi = \Psi_s + \Psi_p$
  Working: $\Psi_{initial} = -1.2 MPa + 0.4 MPa = -0.8 MPa$
 2. Determine the direction of water movement:
  Reasoning: Water moves from a region of higher water potential to a region of lower water potential.
  Working: Initially, the cell's water potential (-0.8 MPa) is equal to the solution's water potential (-0.8 MPa).  Therefore, there is no net water movement.
 3. Analyze the cell's state after equilibration:
  Reasoning: Since the solution's water potential is less negative than the cell's initial solute potential, water will leave the cell to reach equilibrium.  This will lead to a decrease in pressure potential.
  Result: The cell will lose water, leading to plasmolysis (separation of the cell membrane from the cell wall). The pressure potential will decrease to zero.
  Verification: At equilibrium, $\Psi_{cell} = \Psi_{solution} = -0.8 MPa$. Since $\Psi_p$ becomes zero at plasmolysis, $\Psi_s$ will be -0.8 MPa.
 ## OPTION ANALYSIS
A :x: The cell will not be turgid because water is leaving the cell. The pressure potential will not be greater than 0.4 MPa.
B :x: The cell will not be simply flaccid; it will undergo plasmolysis because the external solution has a lower water potential.
C ✓: Correct. Plasmolysis occurs when the cell loses water, causing the protoplast to shrink and detach from the cell wall, resulting in a pressure potential of zero.
D :x: Incipient plasmolysis occurs when the cell just begins to lose turgor, but the pressure potential is still slightly above zero.  In this case, the water potential difference is significant enough to cause full plasmolysis.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Confusing solute and water potential: Remember that water potential is the sum of solute and pressure potential.
  2.Ignoring the direction of water movement: Water always moves from higher to lower water potential.
 **Quick Tips**:
  1.Use the formula $\Psi = \Psi_s + \Psi_p$ to solve problems related to water potential.
  2.Remember that plasmolysis is a result of water loss, leading to a decrease in pressure potential.

Mr. NARESH KUMAR K · Answer

The cell will be turgid, with a pressure potential greater than 0.4 MPa.

Mr. NARESH KUMAR K · Answer

The cell will be flaccid, with a pressure potential of zero.

Mr. NARESH KUMAR K · Answer

The correct Option is "The cell will be plasmolyzed, with a pressure potential of zero."

Mr. NARESH KUMAR K · Answer

The cell will experience incipient plasmolysis, with a pressure potential slightly above zero.

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