Question

A single plant cell is placed in an open beaker containing deionised water. Very slowly, solutes are added to the water, which is gently stirred. The cell is observed through a microscope. Eventually, the plasma membrane begins to pull away from the cell wall. a) Briefly explain why this occurred. [2 marks] The moment before the plasma membrane began to pull away from the cell wall, the osmotic potential in the bathing solution was –2.00 MPa. The bathing solution was later replaced with deionised water and the cell was given time to reach equilibrium. The plasma membrane was once again pushing against the cell wall. b) What is the pressure potential inside the cell now? Explain your answer.

Answer 1

Answer:

• When the solute is added to the deionized water, the solution becomes hypertonic for the plant cell which causes exosmosis by which the water content inside the cell sap of plant cell gets reduced and the cell membrane gets detached from the cell wall and becomes shrink.
• This process occurs because the concentration gradient of the solute outside the the plant cell gets higher in comparison to the concentration gradient of the solute inside the cell.
• This concentration gradient of the solute pulls the water content out from the plant cell into the exterior solution.
• The new osmotic potential inside the cell will be +2.0 mPa.
• The positive sign shows the the opposite direction of the prior osmotic potential which was -2.0 mPa.
• On addition of deionized water into the surrounding solution of the cell causes the reduction in the concentration gradient of the solute in the solution and this causes the solution becomes hypotonic in comparison to the cell sap concentration gradient of the solute.
• The movement of water inside the plant cell causes the cell membrane to exerts a pressure against the cell wall and the cell wall prevents the cell bursting.

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