Biology and Health Sciences

Key unit competency

To be able to explain the processes involved in the movement of water molecules and ions into and out of a cell.

Learning objectives

After studying this unit, I should be able to:

• Define diffusion and osmosis.

• Describe the importance of diffusion and osmosis.

• Explain turgor pressure.

• Investigate diffusion and osmosis through experiments.

• Appreciate the importance of turgidity in plant cells.

Introduction

Look at the diagrams below. Can you tell what is happening in flasks A, B, C and D?

In real life situations, what can we compare the observations above to? What does this tell you what this topic is about?

Our bodies are made of cells that carry out several metabolic and physiological processes. In order to carry out these life processes, a cell needs to take in various substances. It also produces certain substances, some of which are waste products which may be toxic and can harm the organism, hence need to be removed from cells.

Other products are useful to cells within the tissue. These useful substances are transferred to cells where they are needed for important metabolic processes like respiration.

Therefore substances are always moving into and out of the cells. The way substances move into and out of the cells depends on certain properties of the substances, for example, size of the molecule and the type of substance. There are three main physiological processes by which substances move in and out of cells. These are diffusion, osmosis and active transport.

3.1  Diffusion of gases and solutes

When a drop of ink is placed into a glass of water, the ink particles spread in the water until all the water is uniformly coloured.

You are also able to smell perfume that other people have worn because the particles of perfume diffuse from them through the air to our organs of smell, the nose.

Activity 3.1:

To investigate diffusion using copper (II) sulphate solution

Requirements

• Crystals of copper (II) sulphate

• 250 ml beaker

• Water

• Glass rod (open ended)

Procedure

1.  Insert the open ended glass rod into the empty beaker.

2.  Drop a crystal of potassium permanganate to the bottom of the beaker through the upper end.

3. With the glass rod still intact, pour water to fill the beaker.

4.  Gently remove the glass rod so as not to tamper with the potassium permanganate crystal.

5.  Make your observations.

6.  Share your findings with the rest of the class.

Study questions

(a) What observation did you make when the glass rod is lifted from the beaker?

(b) Describe the movement of particles of the copper (II) sulphate crystals.

(c) Why does the movement occur?

(d) Other than using the above experiment, describe how you would demonstrate diffusion using perfume.

Diffusion involves movement of particles (ions or molecules) from a region of high concentration to a region of low concentration. This process continues until the particles are uniformly distributed throughout the system or until equilibrium is reached. Diffusion is a product of constant random motion (kinetic energy) of all atoms, molecules, or ions in a solution. The area with higher concentration of the particles has more random motion resulting to the net movement of the particles to the area with lower concentration.

Net movement of particles will always take place whenever there is a difference in concentration of particles between two regions. This difference is known as concentration gradient. Diffusion is important since it enables useful molecules to enter the cell and waste products to be removed.

Factors that affect the rate of diffusion

Discussion corner

1.  What do you think affects the rate of diffusion?

2.  Are the factors that affect the rate of diffusion related?

The rate of diffusion of particles refers to the time taken for the particles to move within an available space (fixed) until they are evenly distributed. Several factors affect the rate of diffusion. They include:

a. Temperature

When the temperature of particles is high, their kinetic energy increases and the particles move faster. Therefore, the higher the temperature, the faster the particles will diffuse while the lower the temperature the lower the rate of diffusion.

b. Concentration gradient

This is the difference in the amount of particles present in two regions. A greater difference in concentration of particles between two regions, results in a steeper concentration gradient which causes diffusion rate to be faster. When the concentration gradient is low, diffusion rate is also slower.

c. Size of molecules

Larger molecules are heavier and will diffuse at a slower rate compared to smaller molecules which are lighter.

d. Diffusion distance

The rate of diffusion depends on the distance that particles have to travel in order to be evenly distributed within the available space. An even distribution of particles is reached faster when the distance involved in diffusion is small compared to longer distances. It takes a longer time for molecules to diffuse across a thick membrane while It takes less time for molecules to diffuse across a thin membrane.

(e)  Surface area to volume ratio

When the surface area to volume ratio is large, more of the substance diffuses across it than when it is small. This takes place as long as the concentration and temperature of the diffusing molecules remain the same.

Note: A larger surface area to volume ratio does not increase the speed of diffusion of the particles. It simply enables more particles to diffuse across it in a given time.

The figures below show two cubes, calculate the surface area (S.A) of the two cubes).

Also calculate the volume (V) of each cube. Divide the surface area (S.A) with the volume to obtain the surface area to volume ratio (SA/V).

• Which is bigger A or B?

• How will this affect the rate of diffusion?

Activity 3.2: Relating surface area to volume ratio with the size of an organism

Requirements

• Potato

• Razor blade

• Ruler

Procedure

1.  Using the razor blade, cut five cubes of potato each of sides 1cm long. Name this cube X.

2.  Cut another five cubes each of sides 3 cm long. Name this cube Y.

3.  Calculate the surface area to volume ratio of the cubes.

Study questions

(a) Which cube has a smaller surface area to volume ratio?

       • Calculate the surface area (S.A) of the two cubes.

       • Also calculate the volume (V) of each cube.

       • Divide the surface area (S.A) with the volume (V) to obtain the surface area to volume ratio (S.A)/(V) .

        • Which is bigger X or Y? How will this affect the rate of diffusion?

(b) What is the significance of surface area to volume ratio?

Small organisms such as Amoeba have a greater surface area compared to volume than larger organisms, for example human beings. Therefore, diffusion of substances into and out of smaller organisms is faster than in larger organisms. Such small organisms can absorb oxygen and other materials from the environment much more rapidly than large ones. They also can excrete waste products at a faster rate than large organisms. In human beings and other larger organisms, diffusion of substance into their bodies would be slow. Therefore, their bodies have developed a complex system of transport called the blood circulatory system.

Discussion corner

1. Discuss with a classmate the importance of diffusion. Use the following guidelines:

             • Gas exchange

             • Excretion

             • Absorption of materials in both plants and animals

2.  Share your work with the rest of the class.

Importance of diffusion in plants and animals

(a)  Plants absorb water, mineral salts and oxygen from the soil through the root hairs by diffusion.

(b)  Digested food such as glucose and amino acids move from the small intestine into the blood of animals by diffusion. These substances move from the blood to the cells and tissues by diffusion as well.

(c)  Cells and unicellular organisms such as Amoeba get rid of waste substances by diffusion.

(d)  Diffusion is involved in exchange of gases in stomata, skin of frogs and in the lungs of animals.

Self-evaluation Test 3.1

1.  Diffusion is a passive process, explain.

2. State the condition that must be in place for diffusion to take place.

3. The rate of diffusion increases if the

            A.  Temperature of solution decreases

            B.  Concentration gradient decreases

            C.  Viscosity of solution decreases

            D.  All of the above.

4.  Concerning the process of diffusion, at equilibrium ______

          A.  Random movement of molecules continues.

          B.  The concentration of particles is equal throughout the solution.

          C.  Net movement of particles either side is equal.

          D.  The diffusion gradient increases.

3.2 Osmosis

Osmosis is the movement of water molecules from a region of high water potential (dilute solution) to a region of low water potential (concentrated solution) through a partially permeable membrane.

Activity 3.3: To demonstrate osmosis using visking tubing

Requirements 

• 10 cm long visking tubing

• Distilled water

• Concentrated salt solution

• Capillary tube

• Two pieces of strings each measuring 30 cm

• 250 ml beaker

Procedure

1.  Tie one end of the visking tubing using a string.

2.  Open the other end of the visking tubing and half-fill it with the salt solution.

3.  Tightly tie the open end of the visking tubing and allow part of the string to hang.

4.  Fill the beaker with distilled water.

5.  Gently put the visking tubing containing the salt solution into the beaker with distilled water.

6.  Tie the hanging end of the string on the visking tubing onto the capillary tube  and inverse the visking tubing into the distilled water

7. Support the capillary tube using a rhetort stand as shown below.

Note: The capillary tube level of the solution in the visking tubing and the beaker at the beginning of the experiment.

8.  Leave the set-up undisturbed for at least 30 minutes.

9.  Remove the visking tubing from the beaker and make your observations at the end of the experiment.

Study questions

(a)  State three observations you made at the end of the experiment.

(b) Account for your observations in (a) above.

(c) Discuss your findings with your class members.

Since the concentration of solutions is defined in terms of solute concentration and not in terms of water content; water molecules diffuse from less concentrated solution (fewer solutes, more water) to a more concentrated solution (more solute, less water).

Activity 3.4: To demonstrate osmosis in a tuber

Requirements

• Fresh arrow roots, cassava, sweet potatoes or Irish potatoes

• Strong salt solution

• Distilled water

• Scalpel

• Large beaker or small basin

Procedure

1.  Using a scalpel, peel a large Irish potato. You can also use an arrow root, cassava or sweet potato.

2.  Cut off a piece so that it stands at least 6 cm high.

3.  Cut and scoop out a deep hollow portion in its middle and pour strong salt solution halfway up the hollow portion.

4.  Mark the level of the salt solution using a scalpel.

5.  Place the potato in a petri-dish containing distilled water. Let it stand for several hours then note the level of solution in potato.

6.  Repeat the experiment with boiled pieces of Irish potato, sweet potato, cassava or arrow roots.

Study questions

(a) Draw a diagram to represent the results of the experiment.

(b) Is the level of the salt solution still the same at the end of the experiment?

(c) Explain what brings about the change in level of the salt solution.

(d) Compare these results with those when boiled Irish potato is used.

Activity 3.5: To investigate osmosis in a plant tissue

Requirements

• Tubular part of a pumpkin leaf or black jack stem

• Distilled water

• 5% sucrose solution

• Pair of scissors

• Ruler

• Three Petri-dishes

• Two 200 ml beakers

• Labels • Pair of forceps

Procedure

1.  Using a ruler measure 4 cm of the tubular part of the leaf or stem and cut using a pair of scissors.

2.  Repeat the procedure (1) to obtain two other pieces of the leaf or stem.

3.  Make four cuts, each 2 cm long in each of the cut pieces of the leaf or stem.

4.  Put 100 cm3 distilled water into a petri-dish and label water.

5.  Put 100 cm3 of 5% sucrose solution into another petri-dish and label sucrose solution.

6.  Put one of the cut pieces into the petri-dish containing water, another in the petri-dish containing sucrose solution and the last one in an empty Petridish labelled air.

7.  Leave the set-up undisturbed for 25 minutes.

8.  Remove the pieces of the tubular leaf from their respective conditions using a pair of forceps.

 9.   Hold the pieces from distilled water and sucrose solution between your fingers and record your observations in a table below.

                              Table 3.1: Results for osmosis in plant tissues

11. Share your findings with the rest of the class.

Study questions

(a)  Draw the pieces as they appear after the experiment.

(b) Account for any differences that you have observed for the three pieces of leaves.

(c) What was the purpose of putting one piece in the petri-dish exposed to air?

Types of solutions

From your knowledge of chemistry; can you recall what the word solution means? What about solute and solvent?

When a solid is dissolved in water, we get a solution. The solid that is dissolved in this solution is called the solute. The liquid that dissolves the solid is known as the solvent.

                                                 Solute + Solvent = Solution

Concentration of a solution depends on the amount of solute dissolved. A dilute solution has more water molecules compared to solute molecules whereas a concentrated solution has more solute molecules than water molecules.

Suppose a dilute solution is separated from a concentrated solution by a semi-permeable membrane as shown in figure 3.8. Water molecules will move from the dilute solution to the concentrated solution. This is because the dilute solution has more water molecules than the concentrated one. Water molecules are very small and pass easily through the channels or pores of the cell membrane. On the other hand, solute molecules are too large to pass through the pores.

Note:

• The concentration of water in a solution containing mixtures of different solute molecules depends only on the total solute concentration and not the types of solutes.

• If the total concentration of all ions and solutes on both sides of a membrane are the same, there will be no osmosis.

It is important to first understand the terms used to describe the solutions of different concentration with respect to cells function.

Water relations in plant cells

Discussion corner

1.  What will happen to an animal cell if placed in: 

2. Compare the structures of an animal cell and a plant cell in the cases  above.

3.  Draw the structure of turgid plant cell and plasmolysed plant cell.

4.  Share your findings with other class members.

Living cells are surrounded by a fluid medium which may be isotonic, hypertonic or hypotonic to the cell contents. If the fluid is isotonic, there will be no net movement of water into or out of the cell. If the external fluid is hypertonic to the cell contents, then, water leaves the cell. If it is hypotonic, then water enters the cell. The movement of water into and out of the cell and the effects that such movements have on the cell may be described as water relations in the cell.

In the next section, we find out how various solutions affect how plant cells behave.

(a) Plant cells in hypotonic solutions

Plant cells have a large vacuole that contains a fluid called cell sap. The sap contains salt and sugar molecules. If a plant cell is surrounded by a hypotonic solution, water molecules move from the surrounding fluid through the cell wall and cell membrane into the vacuole by osmosis.

As it receives water, the vacuole, swells and pushes the cytoplasm and nucleus outwards against the cell wall. This pressure exerted by the cell contents against the cell wall is called turgor pressure. As turgor pressure increases due to intake of more water by osmosis, the cell wall exerts a pressure that is equal to turgor pressure on the protoplasm called wall pressure.

A point will reach when no more water can enter the plant cell. At this point, the wall pressure is equal to turgor pressure but opposite in direction. Because the cell wall is made of rigid cellulose material, it does not stretch very much and the cell does not burst.

However, turgor pressure causes the cell to become stiff or firm. Such a cell is described as being turgid. A plant in which all the cells are turgid always appears firm and erect.

Turgidity in plant cells is important because the stiff cells give support to the soft tissues such as petals and sepals. Turgor pressure enables soft, nonwoody plant stems to remain upright despite the downward force of gravity. If there is inadequate water in the environment, turgor pressure cannot be maintained in a plant.

(b) Plant cells in hypertonic solutions

A plant cell that is surrounded by a hypertonic solution will lose water. Water is lost from the cytoplasm then from the vacuole. The turgor pressure in the cell begins to decrease.  If this continues, the cell membrane and cytoplasm shrink away from the cell wall. The vacuole in turn reduces in size.

The moving of the cell membrane and cytoplasm away from the cell wall is called plasmolysis. The cell is said to be plasmolysed. When a plasmolysed cell is placed in distilled water, it will become turgid again. The cell is said to undergo deplasmolysis. The point at which plasmolysis just begins is called incipient plasmolysis. At this point, the cell protoplasm that is, cell membrane and cytoplasm no longer exerts any pressure against the cell wall. The turgor pressure is zero, and the cell therefore loses its turgidity. It is now a flaccid cell. 

Plants that are deprived of water for several days have their leaf cells plasmolysed. The overall effect is a drooping plant which is said to be wilting. If this persists for long, the plant cells become flaccid. The plant then dies. However, if the plant is well watered before the cells become flaccid, its turgidity is restored.

Note:

Turgidity in plant cells is caused by osmotic flow of water from an area of low solute concentration to outside the cell into the cell vacuole.

The cell vacuole has a higher solute concentration. Healthy plant cells are turgid. Plants rely on turgidity to maintain rigidity and stand upright.

Activity 3.6:  To investigate turgor pressure in plant cells

Requirements

• Raw Irish potatoes

• Boiled Irish potato

• Distilled water

• Sucrose

• Spatula

• Glass rod

• Cork borer

• 10 cm ruler

• Four 200 ml beakers

• Labels

• Scapel

Procedure

Your teacher will help you to prepare these solutions:

1.  Prepare 5% sucrose solution by dissolving 17.1 g of sucrose in 100 ml of water in a beaker. Stir the solution well until all the sucrose dissolves. Label this 5% sucrose.

2.  Prepare 20% sucrose solution by dissolving 68.4g of sucrose in 100 cm3 of water in a beaker.Stir the solution well until all the sucrose dissolves. Label this 20% sucrose.

3.  Put 100 cm3 of distilled water in the other empty beaker and label it distilled water.

4.  Peel three Irish potatoes; insert the cork borer through one end of the potato and press hard so it comes out through the other end of the potato to produce a cylinder. Cut off the rough end of the potato cylinder and measure 6 mm, cut off the remaining part.

5.  Repeat procedure (4) above so as to obtain at least six potato cylinders.

6.  Put 2 potato cylinders in each of the three solutions in the beakers.

7. Put the boiled potato tissue in the fourth beaker containing distilled water.

8. Leave the set-ups uninterrupted for 45 minutes.

9.  Prepare a table like the one shown below to record your observations.

10.  Remove the cylinders from the different solutions and measure their final length using the 10 cm ruler. Record your results in the table you have prepared.

11. Compare your findings with those of other class members.

Study questions

(a) Work out the change in length of the potato cylinders in:

           • Distilled water with raw potato tissue.

           • 5% sucrose solution.   

           • 20% sucrose solution.

           • Distilled water with boiled potato tissue.

(b) Account for the change in length in (a) above.

(c) Why was it necessary to include the set up with boiled potato tissue in the experiment?

Role of osmosis in plants and animal cells

Discussion corner

1. Discuss the following with a classmate.

         • How are plants supported by turgor pressure?

         • What is the importance of osmosis in plant and animal cells?

2. Share your findings with other members of your class.

1.  Uptake of water by roots  Plants absorb water from the soil inner cells in the root.

                                     Fig. 3.11: Movement of water from cell to cell in tissues

2.  Opening and closing of stomata 

When guard cells gain water by osmosis from the surrounding cells, they become turgid. This changes their shape, making the stomata to open. As a result:

by osmosis. The cell sap in root hair cells usually has a higher solute concentration than water in the soil. Therefore, root hair cells are able to take up water from the soil by osmosis.

3.   Movement of water from cell to cell in tissues 

When a cell takes up water by osmosis, its solute concentration becomes lower (it becomes more dilute) compared to the cell adjacent to it. As a result, water moves from it to the adjacent cell whose solute concentration is higher (more concentrated). In this way, water moves from the root hair cells to the

                   • Gaseous exchange takes place through the open stomata.

                   • Water vapour also exits through the open stomata.

When guard cells lose water through osmosis to surrounding cells, their size and shape changes. This makes the stomata to close.

4. Feeding in insectivorous plants 

    Insectivorous plants such as Pitcher plant prey on insects.

They trap insects when there is a sudden change in their turgor pressure when disturbed by the insect.

Self-evaluation Test 3.2

1.  Which one of the following describes osmosis.

2.  What will happen to an animal cell when it is placed in an isotonic solution?

3. What kind of a membrane is partially permeable?

4. A learner set-up an experiment shown below to investigate osmosis.

5.  An animal cell bursts when placed in water but a plant cell does not. Explain

6. Using your knowledge of plant and animal cell structure. Explain why plant cells have regular shapes while animal cells have irregular shapes.  

Unit summary

• Passive movement of substances into and out of cells involves diffusion and osmosis. These processes involve kinetic energy of the molecules involved.

• Diffusion is the movement of particles from an area of higher concentration to one of lower concentration along a concentration gradient. These particles move by their natural kinetic energy.

• A concentration gradient is the difference in solute concentration between two points separated by a distance.

• The end result of diffusion is a uniform distribution of the particles (molecules or ions).

• Factors that influence diffusion include: temperature, concentration gradient, surface area and distance.

• Osmosis involves movement of water molecules from a region of their higher concentration to a region of their lower concentration across a semi-permeable membrane.

• Osmotic pressure is a force generated to prevent movement of water molecules from a solution across a partially permeable membrane.

• Plant and animal cells behave differently when placed in an isotonic solution, hypertonic and hypotonic solution.

• Osmosis enables uptake of water into the plant through the roots, turgidity in plant cells and opening and closing of stomata.

• Turgidity in plant cells provides support to plant tissues.