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.

   $\begin{array}{l}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}{\text{ψ}}_{\text{p}}=\\ \underset{\_}{+\text{}\text{}\text{}\text{}\text{}{\text{ψ}}_{\text{s}}=}\\ \text{ψ}=-0.6\text{MPa}\end{array}$

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.

   $\begin{array}{l}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}{\text{ψ}}_{\text{p}}=\\ \underset{\_}{+\text{}\text{}\text{}\text{}\text{}{\text{ψ}}_{\text{s}}=}\\ \text{ψ}=\end{array}$

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.

   $\begin{array}{l}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}{\text{ψ}}_{\text{p}}=\\ \underset{\_}{+\text{}\text{}\text{}\text{}\text{}{\text{ψ}}_{\text{s}}=}\\ \text{ψ}=\end{array}$