Problem 6 - Campbell Biology

Question

A plant cell with a \(\Psi_{\text { s }}\) of \(-0.65\) MPa maintains a constant volume when bathed in a solution that has a \(\Psi_{5}\) of \(-0.30\) MPa and is in an open container. The cell has a \begin{equation} \begin{array}{l}{\text { (A) } \Psi \text { ot }+0.65 \mathrm{MPa} \text { . }} \\ {\text { (B) } \Psi \text { of }-0.65 \mathrm{MPa} \text { . }}\\\\{\text { (C) } \Psi_{P} \text { of }+0.35 \mathrm{MPa} \text { . }} \\\ {\text { (D) } \Psi_{\mathrm{P}} \text { of } 0 \mathrm{MPa.}}\end{array} \end{equation}

Step-by-Step Solution

Answer

Option (C) \(\Psi_p = 0.35\) MPa.

Step 1 - Understand the Problem

The problem involves understanding water potential (\(\Psi\)) components in a plant cell. Given are the solute potential (\(\Psi_s\)) and the water potential of the surrounding solution when the cell maintains a constant volume.

Step 2 - Recall Water Potential Equation

Water potential (\(\Psi\)) is given by the equation: \(\Psi = \Psi_s + \Psi_p\), where \(\Psi_s\) is the solute potential and \(\Psi_p\) is the pressure potential.

Step 3 - Define Given Values

Identify the given values: \(\Psi_s\) of the plant cell = \(-0.65\) MPa, \(\Psi_s\) of the surrounding solution = \(-0.30\) MPa. The volume remains constant, implying equilibrium.

Step 4 - Determine \(\Psi\) in Open Container

In an open container, the water potential (\(\Psi\)) of the solution must be equal to the water potential (\(\Psi\)) of the plant cell to maintain equilibrium. This gives \(\Psi_{\text{cell}} = \Psi_{\text{solution}}\). Therefore, \(\Psi_{\text{cell}} = -0.30\) MPa.

Step 5 - Calculate the Pressure Potential (\(\Psi_p\))

Using the equation \(\Psi = \Psi_s + \Psi_p\) and substituting the known values, solve for \(\Psi_p\): \(-0.30 = -0.65 + \Psi_p\). Adding \(0.65\) to both sides: \(\Psi_p = 0.35\) MPa.

Step 6 - Select the Correct Option

Based on the calculated \(\Psi_p\), the correct answer is option (C) \(\Psi_p\) of \(0.35\) MPa.

Key Concepts

Solute Potential

Solute potential, often represented as \( \Psi_{s} \), is a crucial part of understanding water potential in plant cells. It reflects the effect of solutes on the overall water potential. Essentially, the more solute present in a solution, the more negative the solute potential becomes.

In the exercise given, the solute potential of the plant cell is \( -0.65 \) MPa. This indicates that the solutes inside the cell create a negative pressure, which impacts the cell's overall water movement. When the solute potential is negative, it means the solution has fewer free water molecules available to move, lowering the water potential.

For instance, when a plant cell is placed in a surrounding solution with a solute potential of \( -0.30 \) MPa, the difference in solute concentrations between the cell and the surrounding solution helps determine the movement of water. The comparison of these potentials helps establish the conditions for equilibrium.

Pressure Potential

Pressure potential, represented by \( \Psi_{p} \), is another key component of water potential. It is the physical pressure exerted by the cell wall or membrane as water enters or leaves the cell.

In a turgid plant cell, where the vacuole is full of water and pressing against the cell wall, the pressure potential is positive. This pressure is crucial for maintaining the structure and rigidity of the plant cells. In the exercise, after calculating, we find that the pressure potential is \( 0.35 \) MPa. Such a positive value indicates that the cell is exerting pressure to counterbalance the inward pull of water due to the negative solute potential.

In summary, pressure potential acts as a balancing force against the negative solute potential. The interaction between these two forces helps maintain the water potential equilibrium in the plant cell, ensuring that it doesn't shrivel or burst.

Equilibrium in Plant Cells

Equilibrium in plant cells refers to the state where the internal and external water potentials are equal, resulting in no net movement of water. This balance is vital for plant cell stability.

In the given exercise, the plant cell is submerged in a surrounding solution, and it reaches a point where it maintains a constant volume. This happens when the water potential inside the cell equals the water potential of the surrounding solution (\( \Psi_{\text{cell}} = \Psi_{\text{solution}} \)).

We calculated the water potentials in the plant cell and its surrounding solution. The water potential of the cell is \( -0.30 \) MPa, equal to that of the bathing solution. This equalizes the water movement in and out of the cell, leading to a state of equilibrium.

Understanding equilibrium helps in various agricultural practices and environmental adaptations, as it determines how plants manage water uptake, crucial for their survival in different environments. An equilibrium state helps in maintaining the cell's turgidity, vital for the plant's structural integrity and function.