Numerical Problems/ISC/Class 11
Created by Titas Mallick
Biology Teacher • M.Sc. Botany • B.Ed. • CTET (CBSE) • CISCE Examiner
BioNotesBioNotesCreated by Titas Mallick
Biology Teacher • M.Sc. Botany • B.Ed. • CTET (CBSE) • CISCE Examiner
Online
Advanced Numerical Problems and Solutions on Water Potential, Solute Potential, and Pressure Potential.
Understanding water potential (Ψw) is critical for mastering plant physiology. It determines the direction of water movement, which always flows from a region of higher (less negative) water potential to a region of lower (more negative) water potential.
Formula: Ψw = Ψs + Ψp
Where:
Consider three adjacent plant cells (A, B, and C) forming a tissue pathway.
Determine the exact pathway of net water movement among these three cells and represent it visually.
Step 1: Calculate Water Potential (Ψw) for each cell. Using the formula Ψw = Ψs + Ψp:
Step 2: Compare the water potentials.
Step 3: Determine the direction of flow. Water moves from higher Ψw to lower Ψw.
Water Movement Direction (Higher to Lower Ψw):
A fully flaccid plant cell (Ψp = 0 MPa) has an internal solute potential (Ψs) of -1.5 MPa. It is dropped into a beaker containing a sucrose solution with a water potential (Ψw) of -0.4 MPa. Assuming the cell volume does not change significantly, what will be the pressure potential (Ψp) and solute potential (Ψs) of the cell at equilibrium?
Step 1: Understand the initial state of the cell. Initial Cell Ψw = Ψs + Ψp = -1.5 + 0 = -1.5 MPa. Beaker Ψw = -0.4 MPa. Since Beaker Ψw > Cell Ψw, water will enter the cell (endosmosis).
Step 2: Understand the conditions at equilibrium. At dynamic equilibrium, the water potential of the cell will equal the water potential of the surrounding solution.
Step 3: Determine the Solute Potential at equilibrium. Because the plant cell has a rigid cell wall, a small amount of water influx causes a massive spike in turgor pressure (Ψp) with a negligible change in actual cell volume. Therefore, the internal solute concentration remains practically unchanged.
Step 4: Calculate the final Pressure Potential (Ψp). Using the equilibrium water potential: Ψw = Ψs + Ψp -0.4 = -1.5 + Ψp Ψp = -0.4 + 1.5 = +1.1 MPa
Final Answer: At equilibrium, the cell will be turgid with a Ψp of 1.1 MPa and a Ψs of -1.5 MPa.
Dynamic Equilibrium:
Water exchanges between Beaker and Cell until Ψw is equalized (-0.4 MPa).
Calculate the solute potential (Ψs) of a 0.2 M NaCl solution at 20°C. If a plant cell with Ψw = -0.8 MPa is placed in this solution, will it undergo plasmolysis or become turgid? (Given: Ionization constant (i) for NaCl = 2, Gas constant (R) = 0.00831 liter MPa / mol K)
Step 1: Convert Temperature to Kelvin. T = 20°C + 273 = 293 K
Step 2: Calculate Solute Potential (Ψs) using the formula Ψs = -iCRT.
Ψs = -(2) × (0.2) × (0.00831) × (293) Ψs = -0.974 MPa
Since it's an open beaker, pressure potential (Ψp) = 0, so the water potential (Ψw) of the solution = -0.974 MPa.
Step 3: Determine the fate of the cell.
Water moves from higher (less negative) to lower (more negative) potential. -0.8 MPa > -0.974 MPa. Therefore, water will move out of the cell into the solution (exosmosis). The cell will undergo plasmolysis (become flaccid and eventually plasmolysed).
Question: A plant cell with a solute potential of -1.0 MPa is placed in pure distilled water. What is the pressure potential (Ψp) at equilibrium? The Trap: Students often try to calculate an intermediate value or assume the cell bursts. The Reality: Pure water has a Ψw = 0 MPa. At equilibrium, the cell's Ψw must also be 0 MPa. Since Ψw = Ψs + Ψp → 0 = -1.0 + Ψp → Ψp = 1.0 MPa. The cell becomes fully turgid, and its pressure potential exactly counterbalances its solute potential! Plant cells don't burst due to their rigid cell walls.
Question: If the solute potential in a xylem vessel is -0.1 MPa, and its water potential is -1.5 MPa, what is its pressure potential? The Trap: Students assume Ψp must always be positive because it is in a living plant. The Reality: Xylem tissue is dead and functions under intense tension (transpirational pull), which creates a negative physical pressure. Ψw = Ψs + Ψp → -1.5 = -0.1 + Ψp → Ψp = -1.4 MPa. Negative pressure potentials are normal and necessary in xylem!
The Trap: Believing that a "high concentration" means a "high (positive) solute potential." The Reality: The formula is Ψs = -iCRT. Notice the negative sign! Adding more solute makes the solute potential more negative (lower). Therefore, a highly concentrated solution has a very low (highly negative) water potential, acting like a vacuum to draw water towards it.