Study Guide for Campbell Biology
Study Guide for Campbell Biology
11th Edition
ISBN: 9780134443775
Author: Lisa A. Urry, Michael L. Cain, Steven A. Wasserman, Peter V. Minorsky, Jane B. Reece, Martha R. Taylor, Michael A. Pollock
Publisher: PEARSON
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Chapter 36, Problem 2IQ
  1. a. A flaccid plant cell has a water potential of –0.6 MPa. Fill in the water potential equation for this cell.

    ψ p = + ψ s = _ ψ = 0.6 MPa

  2. b. The cell is then placed in a beaker of water (Ψ = 0 MPa). Fill in the equation for the cell after it reaches equilibrium in pure water. Explain what happens to the Ψ of this cell.

    ψ p = + ψ s = _ ψ =

  3. c. Explain what would happen to the same cell if it is placed in a solution that has a water potential of –0.8 MPa. Fill in the equation for the cell after it reaches equilibrium in the solution.

    ψ p = + ψ s = _ ψ =

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Students have asked these similar questions
The osmotic potential of the plant cell is -0.1 MPa and it is placed in an 9 mM sorbitol solution (21 °C). Calculate the turgor pressure of the cell when the water potential of the cell has equilibrated to the same as that of the solution.
The water potential of three adjacent plant cells  are as follows: X 0kPa Y  -1000kPa Z -4000kPa In which direction will water move?
Consider a plant cell. The value for solute concentration in a plant cell is -0.12 MPa and the turgor pressure is 0.12 MPa. 1- What is the water potential in this plant cell? 2- If this plant cell were placed in a solution with water potential of -0.1 MPa, what would happen to the cell? (Hint: explain where the water goes and what that does to the cell).
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