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Osmosis and investigating water potential

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Explainer Video

Tutor: Holly

Summary

Osmosis and investigating water potential

In a nutshell

The transport of water from a higher water potential to a lower water potential is called osmosis. There are various factors, including the water potential gradient and thickness of the membrane, that will impact the rate of osmosis. In this summary you will also learn how to investigate the water potential of plant cells using potato cylinders and solutions of known concentrations.

Osmosis

Definition

Osmosis is the passage of water molecules from a region where it has a higher water potential to a region where it has a lower water potential through a partially permeable membrane.

Note: A partially permeable membrane is one that allows the passage of water molecules and a few other small molecules but not larger molecules.

Water potential

Definition

Water potential is the pressure created by water molecules. It is represented as psi ( $\psi$ ) and it is measured in kilopascals (kPa).

Pure water has a water potential of zero. Water and a solute will have a lower water potential (below zero). The more solute that is added (thus the more concentrated the solution), the lower the water potential. This means that free water molecules travel from a solution with a lower concentration of solute to a solution with a higher concentration of solute. The movement of water molecules continues until the solutions on either side are equally concentrated (isotonic).

Explanation of osmosis

The solution on one side of the membrane has a lower concentration of solute molecules whilst the other side has a higher concentration. The solute and water molecules have kinetic energy and move in a random motion. However, only the water molecules can pass through the partially permeable membrane and they diffuse from the side with high water potential to the side with low water potential. Thus, the water diffuses down the water potential gradient. When the water potentials on either side of the partially permeable membrane are equal, there is a dynamic equilibrium and no further net movement of water.

Rate of osmosis

The rate of osmosis is dependent on the factors below.

Factor	Explanation
Water potential gradient	The higher the water potential gradient, the faster the osmosis will occur.
Thickness of the surface	The thinner the membrane or surface, the faster the rate of osmosis. This is because there is a shorter distance for the ions to travel so they will travel quicker.
Surface area	The surface area of the exchange surface will impact the rate of osmosis. If the surface is large, the rate of osmosis will be greater.

Osmosis in animal cells

Animal cells contain lots of different solutes. Therefore, if they are placed in water, the water will enter the cell as it has a lower water potential. If lots of water enters the cell, it will swell and burst through a process known as cytolysis. To prevent this, the animal cells are found in a liquid with the same water potential as the cells.

Example

Red blood cells are found in plasma, which has the same water potential as the cell. Therefore, there is no net movement of water and the cell doesn’t burst. If the cell is placed in a different solution, the following scenarios may occur.

Water potential of the external solution compared to the cell solution.	Lower	Equal	Higher
Net movement of water	Water leaves the cell.	No net movement of water.	Water enters the cell.
Change observed	The cell shrinks and is known as crenated.	No change in the shape of the cell.	The cell swells and bursts (cytolysis).
Diagram	1. Haemoglobin is more concentrated so the cell looks bigger, 2. Crenated cell, 3. Normal cell, 4. Cellular contents being released, 5. Remains of the cell surface membrane.

Osmosis in plant cells

Plant cells also contain lots of different solutes. These are mainly located in the vacuole. When they are placed in water, water will enter the plant cell as the surrounding solution has a higher water potential than inside the plant cell. This causes the cell to swell and pushes the protoplast against the cell wall. The pressure of the protoplast against the cell wall prevents the entry of more water and the cell is turgid.

Note: The protoplast consists of the cell membrane, nucleus, cytoplasm and vacuole membrane.

If the plant cell is placed in a solution with a lower water potential, the water will leave the cell by osmosis. Eventually, the protoplast will start to peel away from the cell wall. This stage is known as incipient plasmolysis. As more water leaves the cell, the protoplast peels further away from the cell wall. When it is completely detached, the cell is plasmolysed. The responses are summarised below.

Water potential of the external solution compared to the cell solution.	Higher	Equal	Lower
Net movement of water	Water enters the cell.	No net movement of water.	Water leaves the cell.
Change observed	The cell swells and becomes turgid.	No change in the shape of the cell.	The cell shrinks and becomes plasmolysed.
Diagram	1. Cell wall, 2. Protoplast, 3. Nucleus, 4. Protoplast peeling away from the cell wall, 5. Protoplast detached.

Investigating water potential

You can investigate the water potential of plant cells using potato cylinders and solutions of known concentrations.

Creating solutions

You must first create solutions of known concentrations using one of the two methods described below.

Biology; Exchange and transport; KS5 Year 12; Osmosis and investigating water potential — 1. Transfer $5\ cm^3$ and then mix, A. $10\ cm^3$ of $2\ M$ sucrose solution, B. $5\ cm^3$ of distilled water.

Serial dilution method

1.	Place five test tubes in a test tube rack.
2.	Measure and transfer $10cm^3$ of $2M$ sucrose solution to the first test tube.
3.	Measure and transfer $5cm^3$ of distilled water to the rest of the test tubes.
4.	Using a pipette, take $5cm^3$ of the solution from the first tube and transfer it to the second tube (which only contains distilled water).
5.	Mix the solutions together. *Note:* The second test tube now contains *$5cm^3$* of $2M$ sucrose solution and $5cm^3$ of distilled water. The concentration of the solution in this tube is $1M$ .
6.	Repeat this process for the rest of the tubes.
7.	You will end up with five tubes with different sucrose concentrations listed below. $2M, 1M, 0.5M,0.25M,0.125M$

Scale factor method

The scale factor method can also be used to create solutions of known concentrations. Below are the general steps and the steps for this specific example.

Example

You want to make $25cm^3$  of $0.5M$  sucrose solution.

	General	Example
1.	Get a sucrose solution of a known concentration.	A sucrose solution of $1M$ .
2.	Find the scale factor using the formula below. $Scale\space factor = \dfrac{concentration\space of\space that\space solution}{ concentration\space of\space the\space solution\space you\space want\space to\space make}$	$Scale\space factor = \dfrac{1M}{0.5M} = 2$
3.	The solution you want to make is the scale factor times less concentrated than the one you have.	The solution you want to make is two times less concentrated than the one you have.
4.	To make the solution weaker, you use less of it.	So, to make it two times less concentrated, you use two times less of it. So you would use $\dfrac{25cm^3}{2}=12.5cm^3$ .
5.	Transfer the calculated volume into a test tube.	Transfer $12.5cm^3$ * of the sucrose solution into the test tube.*
6.	Top up with distilled water to reach the total volume you want to make.	You wanted $25cm^3$ of the solution so you need $25-12.5=12.5cm^3$ * of distilled water.*

Finding the water potential

You can now use the solutions you have made to investigate the water potential of the potato cells.

1.	Using a cork borer, cut $15$ cylinders out of a potato. These should have a diameter of around $1\ cm.$
2.	Form five groups of three potato cylinders.
3.	Using a mass balance, measure the mass of each group of potato cylinders.
4.	Place a group of potato cylinders into each of the five test tubes of sucrose solution that you have made.
5.	Leave the potato cylinders in the tubes for half an hour.
6.	Remove the cylinders from the tube and dry with absorbent paper towel.
7.	Measure the mass of each group again.
8.	Calculate the percentage change in potato mass for each group of potato cylinders.
9.	Plot a graph with the percentage change in mass on the $y-$ axis and the concentration of sucrose solution on the $x-$ axis.

In the solution with higher water potential (lower concentrations of sucrose), the cylinders will gain water. This will be shown as a greater percentage increase in mass. As the sucrose concentration increases, the water potential decreases. Therefore, there will be more water loss from the potato cylinders and they will have a decrease in mass.

When the percentage change in mass is zero, the water potential of the potato cylinders and the water potential of the sucrose solution is the same. You can find the concentration of the sucrose solution by reading the graph. Then, you can use a textbook or the internet to find the water potential of that concentration of sucrose solution. This will then be the same for the potato cylinders.

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