Problem 6 - Campbell Biology

Question

A plant cell with a \(\Psi_{\mathrm{s}}\) of -0.65 MIPa maintains a constant volume when bathed in a solution that has a \(\Psi_{\mathrm{s}}\) of \(-0.30 \mathrm{MPa}\) and is in an open container. The cell has a (A) \(\Psi_{\mathrm{P}}\) of \(+0.65 \mathrm{MPa}\). (B) \(\Psi\) of \(-0.65 \mathrm{MPa}\). (C) \(\Psi_{\mathrm{p}}\) of \(+0.35 \mathrm{MPa}\). (D) \(\Psi_{\mathrm{p}}\) of \(0 \mathrm{MPa}\).

Step-by-Step Solution

Answer

The cell has a ψ_p of +0.35 MPa.

Step 1 - Understand ψ (Psi)

Water potential (ψ) is a measure of the potential energy in water as well as the difference in potential between a given water sample and pure water. It is composed of solute potential (ψ_s) and pressure potential (ψ_p). The formula is: ψ = ψ_s + ψ_p.

Step 2 - Analyze given values

The solute potential (ψ_s) of the cell is -0.65 MPa. The solute potential (ψ_s) of the surrounding solution is -0.30 MPa.

Step 3 - Condition for constant volume

For the plant cell to maintain a constant volume when bathed in the solution, the water potential (ψ) of the cell and the surrounding solution must be equal. This implies that ψ_inner (cell) = ψ_outer (solution).

Step 4 - Calculate ψ (Water Potential) for Surrounding Solution

For an open container, the pressure potential (ψ_p) of the surrounding solution is 0 MPa. Thus, ψ_outer = ψ_s_outer + ψ_p_outer = -0.30 MPa + 0 MPa = -0.30 MPa.

Step 5 - Determine ψ (Water Potential) for the Cell

Given that the volume of the cell remains constant, the water potential of the cell must equal the water potential of the surrounding solution: ψ_inner = ψ_outer = -0.30 MPa.

Step 6 - Use the known values to find ψ_p (Pressure Potential) of the Cell

We know that ψ_inner = ψ_s_inner + ψ_p_inner. Substituting the known values, we get -0.30 MPa = -0.65 MPa + ψ_p_inner. Solving for ψ_p_inner, we find ψ_p_inner = -0.30 MPa + 0.65 MPa = +0.35 MPa.

Key Concepts

Solute Potential (ψ_s)

Solute potential, also known as osmotic potential, is a key component in determining the overall water potential in plant cells. It is represented by \(\text{ψ_s}\) and typically has a negative value because solutes lower the potential energy of water. The more solute particles present in the solution, the more negative the \(\text{ψ_s}\).

In our example, the solute potential of the plant cell was \(-0.65 \text{MPa}\), while that of the surrounding solution was \(-0.30 \text{MPa}\). Because the cell contains more solutes compared to the surrounding solution, it has a lower \(\text{ψ_s}\).

Having a detailed understanding of solute potential helps in recognizing how water will move through the cell membranes, balancing concentrations and affecting the cell’s volume and turgor pressure.

Pressure Potential (ψ_p)

Pressure potential, or \(\text{ψ_p}\), represents the physical pressure on a solution. In plant cells, this is also known as turgor pressure and is crucial for maintaining structural integrity and optimal physiological functions. Unlike solute potential, pressure potential can be positive, negative, or zero, depending on the surrounding conditions.

For a plant cell maintaining constant volume, the pressure potential offsets the solute potential. In the exercise, we calculated that the cell's pressure potential was \(+0.35 \text{MPa}\), which balanced its more negative solute potential. When the cell is placed in a solution with lower solutes, it compensates by increasing internal pressure, ensuring that water does not continually enter or leave the cell.

This demonstrates the cell's ability to regulate internal conditions to maintain turgor pressure, which is essential for growth and stability.

Water Potential (ψ)

Water potential (\text{ψ}) is the central concept in plant water relations, combining both solute potential (\text{ψ_s}) and pressure potential (\text{ψ_p}). It dictates the direction in which water will move, from areas of higher water potential to lower water potential. This movement is crucial for processes like nutrient distribution and waste removal within plants.

The formula for water potential is \[ \text{ψ} = \text{ψ_s} + \text{ψ_p} \]

In the given problem, the plant cell maintained its volume by ensuring its internal water potential (\text{ψ_inner}) matched the water potential of the surrounding solution (\text{ψ_outer}). Since the solution was \(\text{ψ} = -0.30 \text{MPa}\) and open to the atmosphere (\text{ψ_p_outer} = 0 \text{MPa}), the cell adjusted its internal pressure potential to +0.35 \text{MPa}, balancing with its solute potential of -0.65 \text{MPa}.

Understanding water potential is crucial for students learning about plant physiology because it explains the principles behind water uptake, nutrient transport, and overall plant health.