Q36.2-1CC. - Campbell Biology

Question

If a plant cell immersed in distilled water has a Ψ_S of -0.7 MPa and a Ψ of 0 Mpa, what is the cell’s Ψp? If you put it in an open beaker of solution that has a Ψ of -0.4 Mpa, what would be its Ψp at equilibrium?

 

Step-by-Step Solution

Answer

The pressure potential (Ψp) of the cell is 0.7MPa. 

 

At equilibrium, the pressure potential would be 0.3 MPa.

Step 1: Water Potential

The potential energy of water per unit volume compared to pure water in reference conditions is known as water potential. The equation for water potential is $\mathrm{\psi }={\mathrm{\psi }}_{\mathrm{p}}+{\mathrm{\psi }}_{\mathrm{s}}$ , where Ψs is the water potential, ΨS is the solute potential, and ΨP is the pressure potential. However, when ${\mathrm{\psi }}_{\mathrm{p}}= {\mathrm{\psi }}_{\mathrm{s}}$, Ψ is zero.

Step 2: Pressure potential of the cell in the distilled water

When the cell is placed in distilled water, the solute potential (Ψs) is -0.7MPa, and water potential (Ψ) is zero, then the pressure potential of the cell is calculated as:

 $\begin{array}{l}{\mathrm{\psi }}_{\mathrm{p}}=\mathrm{\psi }-{\mathrm{\psi }}_{\mathrm{s}}\\ {\mathrm{\psi }}_{\mathrm{p}}=0-(-0.7)\\ {\mathrm{\psi }}_{\mathrm{p}}= 0.7\end{array}$

The pressure potential of the cell is 0.7 MPa.

Step 3: Pressure potential of the cell at equilibrium

When the cell is placed in an open beaker, the water potential is -0.4 MPa. 

It is given that, Ψs is -0.7 MPa, so the pressure potential at equilibrium would be calculated as:

$\begin{array}{l}{\mathrm{\psi }}_{\mathrm{p}}=\mathrm{\psi }-{\mathrm{\psi }}_{\mathrm{s}}\\ {\mathrm{\psi }}_{\mathrm{p}}=(-0.4)-(-0.7)\\ {\mathrm{\psi }}_{\mathrm{p}}=0.3\end{array}$ 

The pressure potential at equilibrium is 0.3 MPa.