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School biology notes: Explaining diffusion, osmosis, transport, active transport

experiment to investigate osmosis with potato and sugar solution

Movement of substances in living systems - introduction to diffusion, osmosis, transport and active Transport

See also Surface exchange of substances in animal organisms

 Doc Brown's school biology revision notes: GCSE biology, IGCSE  biology, O level biology,  ~US grades 8, 9 and 10 school science courses or equivalent for ~14-16 year old students of biology

Sub-index for this page

1a. Introduction to diffusion and demonstration experiments

1b. A particle model and factors affecting the rate of diffusion

2. Factors affecting the rate of diffusion and Fick's Law of diffusion calculations

3a. The action of partially permeable cell membranes - selective diffusion

3b. Examples of diffusion in living organisms

4a. Osmosis - examples, explanation and experiments

4b. Some details of examples of osmotic action in individual animal or plant cell types

5. Active transport - explanation and examples

6. A comparison of diffusion, osmosis and active transport

See also on other pages:

Examples of surfaces for the exchange of substances in animal organisms

Enzymes - section on human digestion, metabolism and synthesis

The human circulatory systems



What is diffusion? Why does diffusion happen?   What is osmosis? How does osmosis work?

Why is osmosis so important in plants and animals?  What is active transport? How does active transport work?

Why is active transport needed in plants and animals?

You should appreciate that it is important that dissolved substances must be able to get in and out of a cell through the cell membranes, otherwise the cell could not live or reproduce!


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1a. DIFFUSION what is it? how to demonstrate it!

Experiments to show diffusion (adapted from my states of matter page)

Particles are always moving at random and this causes them to spread throughout in a container if a gas or spread out throughout a solution if dissolved in a solvent.

This is the process of spreading, called diffusion, naturally occurs in gases and liquids, because they have sufficient kinetic energy to move around freely.

Diffusion is almost impossible in solids because the particles cannot move freely from one position to another.

It is this continuous random movement of particles that allows diffusion to take place.

Diffusion is the natural net movement of particles from an area of higher concentration to an area of lower concentration.

Diffusion of a specific material will continue until an equilibrium is reached when the concentration is evenly distributed and a concentration gradient for that material no longer exists.

and

In both experiment you start with a container of a colourless medium (air or water), add a coloured material (gas or soluble solid), make sure the container is sealed to prevent any air disturbance (or gas escaping).

The container is left to stand, preferably at a constant temperature to prevent mixing due to convention. Immediately the coloured particles spread (gases mix, solid dissolves and spreads) due to random natural particle movement, from an area of high concentration to one of low concentration.

The spreading is self-evident and direct experimental evidence for the natural constant random movement of particles (molecules or ions).

After many hours, due to diffusion, the colour is evenly distributed due to the random movement of ALL the particles in the gas or liquid mixture.

As you can see, diffusion readily occurs in liquids or gases and it is faster in gases because of the greater distance between the particles.

Diffusion is almost impossible in solids because of the stronger interparticle bonding forces holding the particles in fixed positions.

 

Another demonstration of diffusion - with a bit of added chemistry!

1. Agar gel cubes are prepared with a little sodium hydroxide solution and a few drops of phenolphthalein indicator added. Phenolphthalein turns pink-red in alkaline solution.

The agar jelly cubes are placed in a beaker of dilute hydrochloric acid.

2. When the beaker is left to stand, the acid will slowly diffuse into the agar jelly cubes and neutralise the alkali ...

hydrochloric acid + sodium hydroxide ===> sodium chloride + water

HCl(aq)  +  NaOH(aq)  ===> NaCl(aq)  +  H2O(l)

... and the agar jelly cubes begin to turn colourless because the indicator phenolphthalein turns colourless in acid.

3. Gradually, all of the alkali is neutralised, and the whole of the agar gel turns colourless as the acid diffuses right to the centre of the cubes.

Notes:

It is also true to say that the alkali can also diffuse out of the cubes and be neutralised in the acid solution. Either way, it doesn't matter, all the particles are on the move constantly at random in all directions.

At the start, the diffusion gradient for the acid is into the agar gel cubes.

You can do an investigation with different concentrations of acid to change the diffusion gradient and time the results. You need to keep the recipe of the agar jelly cubes constant and conduct the experiments at the same temperature. You should be able to predict the pattern of results!


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1b. A particle model and factors affecting the rate of diffusion

A particle model of diffusion in gases and liquids:

This picture could represent diffusion of molecules or ions in cell fluids or blood stream or gases in the lungs. Imagine the diffusion gradient from left to right for the green particles added to the blue particles on the left.

The blue particles could we water and the green particles could be a sugar, protein or carbon dioxide molecule. So, for the green particles, net migration is from left to right and will continue, in a sealed container, until all the particles are evenly distributed (as pictured).

BUT, as in living organism, if the green particles are removed or used in some process on the right, then net migration (net diffusion) would continue until there was not enough green particles to create a diffusion gradient from left to right i.e. become evenly very dilute.

 

===> ===>
 

Be able to define diffusion as the movement of particles from an area of high concentration to an area of lower concentration.

You experience the gas diffusion experiment (or the diffusion particle picture above!) if somebody sprays perfume or deodorant into a room (green particles in the diagram above!).

Even without draughts or convection, the odour will eventually enter your nose and be detected by your sense of smell in any area of the room.

Similarly you can smell petrol or diesel fumes throughout garage due to the diffusion of fuel vapour molecules,

You should know that all liquid or dissolved particles have kinetic energy and so in constant random motion in all directions and tend to spread in all directions, BUT, on average, they will tend to migrate from a region of higher concentration to a region of lower concentration.

The two experiments described above illustrate this random spreading, but by the nature of the experiment design you will see initially the spreading on average is upwards because the coloured substance starts off at the bottom of the container where the concentration will be very high.

Note:

(i) The bigger the concentration difference between two adjacent regions, the steeper the diffusion gradient and the faster the rate of diffusion takes in terms of the net transfer of a particular molecule or ions (eg sugar or sodium ions etc.).

(ii) If the system is warmer, at a higher temperature, the particles gain kinetic energy and can on average move faster and so diffusion is faster.

 


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2. Factors affecting the rate of diffusion and Fick's Law of diffusion

The diffusion situation might be exchange of gases in the lungs or movement of molecules and ions through a cell membrane. Three rate of diffusion factors are described and explained in the context of transferring substances through a membrane.

Factors affecting the rate of diffusion of particles through a membrane - you may be talking about diffusion or osmosis and active transport.

Expressed as how diffusion rate is increased - since that's usually what you want in organisms!

(i) The larger the surface area of the membrane, the greater the rate of diffusion.

This factor can also be expressed as increase in surface/volume ratio, so increasing the rate of substance transfer -  there is a bigger chance of a particle passing through a given larger area.

(ii) The steeper the concentration gradient, meaning the greater the difference between the highest concentration and the lowest concentration area on either side of membrane, the greater the rate of diffusion.

So, in a given time, more particles will diffuse from the area of highest concentration to the area of lowest concentration at a greater rate, the greater the concentration difference - more particles available to move to area of lower concentration.

If the concentration was uniform i.e. equal on both sides of the membrane, there would be no net diffusion - no net transfer (ignoring active transport which can operate against a diffusion gradient).

(iii) The shorter the distance the particles have to diffuse - e.g. the thinner the membrane.

The shorter the time needed to transfer particles, the greater the rate of diffusion - think of how thin membranes are!

(iv) Rates of transfer of substance by diffusion will increase with increase in temperature - the particles (molecules or ions) have more kinetic energy and their average speed increases - e.g. particles can move in and out of cells down diffusion gradients more rapidly.

BUT, there will be a limit e.g. in mammals, many enzyme reactions have an optimum temperature of 37oC, and malfunction if overheated!

Therefore, for a healthy organism at constant temperature, its not an important factor.

Fick's Law on the rate of diffusion of particles relating to a membrane

Fick's Law expresses the three diffusion factors (i) to (iii) described above in a ,proportional' mathematical formula:

Rate of diffusion (surface area x concentration difference) (thickness of membrane)

Rate factor (i) x factor (ii) factor (iii)

(i) A bigger surface area of membrane - bigger rate of diffusion.

If you can double or triple the surface area in an organ, you can double or triple the rate of diffusion = rate of transfer of substances.

This assumes a constant diffusion gradient due to constant concentrations, same thickness of membrane (and constant temperature).

In terms of particles, you can argue there is a bigger chance of a particle passing through a given larger area.

(ii) A bigger concentration difference - bigger rate of diffusion

For a given membrane of fixed surface area and thickness, the bigger the difference in concentration between the two sides of the membrane, the steeper the diffusion gradient, the faster the particle diffusion rate.

Suppose in terms of concentrations on either side of a membrane

(a) the concentrations were 0.05 mol/dm3 and 0.10 mol/dm3

(b) the concentrations were 0.025 mol/dm3 and 0.15 mol/dm3

concentration differences:

(a) 0.10 - 0.05 = 0.05;  (b) 0.15 - 0.025 = 0.125

For a given membrane (constant surface area and thickness) the ratio of the rates of diffusion will be 0.125/0.05 = 2.5

\In other words the rate of diffusion in situation (b) is 2.5 times faster than situation (a).

This argument assumes the same thickness of membrane and the same surface area (and constant temperature).

(iii) A thinner membrane - bigger rate of diffusion

Less distance for particles to travel, so less time needed for transfer.

If you can halve the thickness of a membrane you can double the rate of diffusion through it because you are halving the distance and time needed to diffuse through the membrane and for the same concentrations you are doubling the diffusion gradient.

This assumes a constant surface area and a constant the diffusion gradient.

 

Some examples of Fick's Law calculations

Ex. 1. Suppose an exchange surface has an area 20 m2 and a membrane thickness of 0.005 m.

If the concentrations of a substance are 0.2 mol/dm3 and 0.05 mol/dm3 on either side of the membrane, calculate a relative rate of diffusion.

Relative diffusion rate surface are x concentration difference thickness of membrane

Relative diffusion rate = 20 x (0.2 - 0.05) / 0.005 = 600 a.u.

 

Ex. 2.

 

Other examples of diffusion in living organisms are described in detail on

 Surface exchange of substances in animal organisms

 


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3a. The action of cell membranes - selective diffusion

Although a cell membrane holds the cell together it lets substances in and out, but these substances must be dissolved in water in order pass to and fro through the cell membrane by diffusion.

You can think of a partially permeable membrane (*) as having tiny molecular sized holes in it, that only allow certain small, but NOT large, particles through.

Particles can be ions (Na+, Cl-) or molecules (H2O, C6H12O6).

(*) other phrases used are semi-permeable membrane or selectively permeable membrane

However, only small molecules and ions can diffuse through the cell membrane

e.g. relatively small molecules like glucose and oxygen for respiration, waste carbon dioxide from respiration, urea waste, amino acids for protein synthesis and of course water itself, as well as being the solvent.

BUT big molecules cannot get through the cell membrane e.g. starch and proteins.

In the particle model of a cell membrane on the right, the thick black dotted line represents the membrane.

Think of the grey circles as the larger molecules like proteins or starch which cannot pass from left to right through the cell membrane.

Imagine the blue circles are water - they can pass through the membrane in any direction - BUT, the net transfer is from an area of higher concentration to an area of lower concentration - in this case from right to left.

Imagine the green circles are small molecules or ions - they can also pass through the cell membrane in either direction, but the concentration is greater on the left than the right.

Therefore for the green particles the diffusion gradient is from left to right and there is a net movement of the green particles (smaller molecules) from the left higher concentration to the right lower concentration passing through the cell membrane in the process.

Also bare in mind that the larger the surface area of a membrane, the faster the net rate of diffusion of a particular molecule or ion.


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3b. Examples of diffusion in living organisms

The process of respiration.

The thin cell membranes allow the diffusion of small molecules in and out of cells.

Since the capillaries are thin and numerous, the diffusion distance from cells is short, so transfer of nutrients in, and waste products out, is as efficient as possible.

As the cells respire they use up oxygen/glucose, so their concentration falls in the cell. Therefore the external concentrations (e.g. in capillaries) is higher, so more oxygen/glucose will diffuse into the cell.

At the same time, the concentration of the waste product carbon dioxide builds in the cell, and so carbon dioxide will then naturally diffuse out of the cell to the lower concentration region in the capillaries.

For more details on gas exchange and other transfer systems involving diffusion see notes on:

Surface exchange of substances in animal organisms


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4a. OSMOSIS - a special case of diffusion

Introduction

Know and understand that water often moves across boundaries by osmosis - a special case of particle (water) diffusion down a concentration gradient.

Know that osmosis is the diffusion or net bulk movement of water from a dilute to a more concentrated solution through a partially permeable membrane (also described as a partly/semi-permeable membrane) that allows the passage of very small molecules like water (diagram on right).

(Caution: Don't assume other small molecules cannot pass through a membrane, the term osmosis refers to the passage of water, but things are more complicated when dealing with other small molecules and ions passing through membranes).

You can also express osmosis as the movement of water from a higher concentration to a lower concentration - this is effectively from a lower solute concentration to a higher solute concentration.

A partially permeable membrane has extremely small pores or holes that only allow the tiniest of molecules like water through e.g. even relatively small molecules like sucrose will not pass through a partially permeable membrane (see experiment in section 2b. below).

Particle theory: All the particles in a solution are moving at random because of their kinetic energy store, so if a water molecule (blue circle) hits the right spot on the membrane, it can diffuse through. However, a larger molecule (purple circles) like a sugar molecule, will just bounce back off the surface of the membrane.

In osmosis only water gets through a partially permeable membrane and depending on the relative concentration of dissolved substances on either side of the membrane, osmosis can happen in either direction - meaning water can diffuse through the membrane in either direction.

Although the water molecules (and any other particles) are moving around at random, there will be a net transfer of water in one direction at a time through a partially permeable membrane.

The net direction of diffusion of water is from a less concentrated solute solution (more water molecules) to a more concentrated solute solution (less water molecules) i.e. from the higher concentration of water molecules to a lower concentration of water molecules across the membrane.

You can also express this as water moving from a higher concentration to a lower concentration,

OR, from water moving from a higher water potential to lower water potential.

Therefore a more concentrated solution becomes more dilute in the process.

This osmosis diffusion can occur in either direction depending on the relative concentration of the solutes in the cell fluids or tissue fluids and concentrated solutions e.g. of sugars on either side of a partially permeable membrane.

But, whatever, the more concentrated solution will tend to become diluted by water passing through the partially permeable membrane.

The movement of water in and out of cells

The soft cell wall, or outer membrane of an animal cell, acts as a partially permeable membrane.

The water surrounding cells, the tissue fluid, contains the dissolved molecules the cell needs to survive e.g. sugars, amino acids, oxygen, as well as waste carbon dioxide etc.

(a) If the cells are short of water ('partially dehydrated'), the concentration of dissolved substances increases, so water diffuses through the cell membrane into the cells to dilute the cell fluids until equilibrium is established.

The term water potential is used to describe these situations.

You can talk about the water potential gradient across a partially permeable membrane.

Water will diffuse from a high water potential to a low water potential.

In situation (a), the fluid outside the cells has a high water potential than the solute solution in the cells, so water moves into the cells.

(b) Conversely, if the cell solution is too dilute, then water will diffuse out by osmotic action across the semi-permeable membrane of the cell wall.

In situation (b), the solution inside the cells has a high water potential than the fluid outside the cells, so water moves out of the cells.

Therefore the diagram on the right could represent the passage of water (blue circles) in or out of a cell, depending on the relative concentrations of water on either side of the membrane.

 

Osmosis - plant cells and water potential

(i) When you water a plant it increases the water potential of the soil around it.

Therefore the plant cells will draw water in by osmosis until they become turgid - fatter and swollen.

The cell fluids (contents of the cell) will push against the cell wall, known as turgor pressure, and this helps support the plant tissues (therefore the plant as a whole).

(ii) If the soil is very dry, lacking in water, the plant starts to wilt and the water potential of the plant is greater than the surrounding soil.

The result is the plant cells become flaccid and begin to lose water.

The plant doesn't droop (flop) completely and retains much of its shape because the strong cellulose cell wall is relatively inelastic and helps the plant retain its shape.

 

Osmosis - animal cells and water potential

In the case of animal cells, they do not have strong walls and can respond adversely to change in the ambient water pressure.

If animal cells are surrounded by a solution of greater water potential (less concentrated in solutes), they can absorb so much water by osmosis that they burst - which kills the cells - see red blood cell example below.

In extreme cases you can die of over-hydration, but its a complicated effect that reduces the level of salt (= sodium ions) in the blood to dangerous levels.

If cells are surrounded by a too concentrated salt solution, they lose so much water they can shrink and shrivel up and the dehydrated kills the cells - see red blood cell example below.

 

Osmosis is important for the function of many animal organs

It isn't just all about individual cells. e.g. water is absorbed into the bloodstream from the large intestine to form faeces in the appropriate physical state!

 

See also Homeostasis - osmoregulation - ADH - water control  gcse biology revision notes

and Transport and gas exchange in plants, transpiration, absorption of nutrients, leaf and root structure

 


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4b. Some details of examples of osmotic action in individual animal or plant cell types

(i) The effect or pure water and salt (sodium chloride) solution on red blood cells.

effect of salt solution and water on red blood cells swelling up bursting shrinking shrivelling killing them

If cells are placed in pure water (distilled or deionised) OR in a salt solution, the movement of water through the partially permeable cell membrane by osmosis can have some pretty devastating effects - illustrated by the diagram above concerning those rather vital red blood cells, but it can happen to most cells!

On the left: The cells are in a less concentrated solute solution compared to the cytoplasm - or in just water.

If red blood cells are put in pure water, the greater external water potential (more concentrated) of the less dilute solution, means that water will pass into the cell's cytoplasm by osmosis.

The diffusion gradient is into the cells - which have a greater solute concentration than pure water.

The result is the cells swell up, burst open and die.

On the right: The cells are in a more concentrated solution compared to the concentration of solutes in the cytoplasm.

If red blood cells are put in a salt solution, the greater internal water potential of the more dilute solution of the cell's cytoplasm, means that water will pass into the cell by osmosis

The result is the cells shrink and shrivel up and die.

(ii) The formation of a plasmolysed plant cell

plasmolysed plant cell plasmolysis turgid flaccid effects of osmosis

1. Turgid plant cell

When a plant has sufficient water, the water passes into the cells by osmosis and the vacuole fills and swells up.

The vacuole pushes against the cell wall making the cell turgid.

This gives the plant structural support so it doesn't droop/wilt - tall trees are an impressive example of this!

2. Flaccid plant cell

If water passes out of the cells by osmosis, the vacuole shrinks and the plant cell becomes flaccid.

The cytoplasm can begin to move away from the cell wall.

3. Plasmolysed plant cell

If a plant cell loses a lot of water by osmosis, cytoplasm of the cell peels away from the cell wall, leaving gaps between the cell wall and the membrane and making the plant cell shrink and crumple - wilt and droop.

Plasmolysis is the shrinking of the cytoplasm of a plant cell in response to diffusion of water out of the cell and into a high salt concentration solution by osmosis. During plasmolysis, the cell membrane pulls away from the cell wall, but this does not happen in low salt concentration because of the rigid plant cell wall.

Not surprisingly, plasmolysis can happen in very dry conditions, but on watering (rain or us), most paler wilted plants recover to the fully 'green' upright plant.

 


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2b. Simple demonstrations of osmosis

(i) Partially permeable membrane bags osmosis experiment

osmosis experiment with partially permeable membrane bags and sugar solution (semi-permeable membrane)

Two experiments to demonstrate a partially permeable membrane is a two-way system are illustrated above..

All you need are partially permeable membrane bags and a sugar solution of e.g. sucrose and pure water.

(pure water can distilled water or deionised water - its all the same!)

The bags are fixed on to a vertical piece of capillary tubing enabling you to observe the direction of water.

Just use the same beakers, same volumes of solution and do all the experiments in the same temperature.

Experiment A

In experiment A, you fill the bag with the sugar solution and place it in a beaker of pure water.

Observation: You find the bag swells up and the water moves up the tube (picture A1 to A2).

Conclusion: Water has moved from the surrounding pure water into the bag of sugar solution through the partially permeable membrane.

Explanation:

The solution in the bag is more concentrated with solute than the water surrounding the membrane.

The water concentration (potential) is higher in the water around the bag than in the bag, therefore the water must diffuse into the bag through the partially permeable membrane.

The water diffusion gradient is from inside the bag into the surrounding solution, again, as the water tries to dilute the solution, but this time in the opposite direction through the partially permeable membrane.

Experiment B

In experiment B, you fill the bag with pure water and place it in the beaker of sugar solution.

Observation: The bag shrinks and the water moves down the tube (B1 to B2).

Conclusion: Pure water has moved out of the bag through the partially permeable membrane into the surrounding sugar solution.

Explanation:

The solution in the beaker is more concentrated with solute than the water in the bag.

The water concentration (potential) is higher in the bag than the surrounding water, therefore the water must diffuse out the bag through the partially permeable membrane.

The water diffusion gradient is from around the bag into the bag as the water tries to dilute the solution.

Remember: In osmosis, only the water moves through the partially permeable membrane, NOT the sugar molecules!

 


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(II) Potato cylinders osmosis experimental investigation

Set-up and method

You can do a simple experiment to demonstrate osmosis by placing blocks or cylinders of potato into pure water and then a series of sugar solutions (e.g. glucose/sucrose) of increasing in concentration (increasingly higher molarity e.g. from 0.0 to 1.0 mol/dm3) i.e. from pure water, and a dilute to a very concentrated sugar solution.

The dependent variable is the potato mass and the independent variable is the concentration of the sugar solution.

The different concentrations of sugar represent different water potentials - the more concentrated, the lower the water potential.

In the diagram the sugar molecules are purple and the water molecules blue.

All the other variables should be kept constant - so make sure:

the original potato blocks are identical in size and mass - with no peel left on, this ensures the cells have the same surface area exposure to the water or solution,

same temperature - so rate is unaffected,

same time left to change - another rate factor,

same sugar and same volume of liquid,

these are all about a 'fair test' and controlling variables and just changing the concentration of sugar.

You measure and record the initial mass of the potato blocks and place them individually in pure water and the range of sugar solutions in a series of beakers.

Leave the beakers for as longer times as possible and allowing time at the end to make the necessary measurements - best for class if it can be done in a lesson.

Carefully remove the potato blocks from the liquid, dry them with a paper towel and re-weigh them.

Different pupil groups can use one particular concentration and use 3 lots of potato and submit an average to the whole class results.

Results

From the weighings work out the mass gain/loss from each potato block.

You can convert the weighing into % mass gain/loss.

% change in mass = 100 x (final mass - initial mass) / initial mass

You can plot a graph of mass gain/loss (g or %) versus the sugar concentration (mol/dm3), and the graph might not be quite what you expect!

Osmosis is taking place with water and most of the sugar solutions - but not always in the same direction!

from measurements (a) to (e) etc.

Typical results from an osmotic experiment using potato tubes or blocks.

Initially the potato blocks gain mass, then there is little change in mass and then the blocks lose mass.

Explanation

Diagram (a) to (c): With the potato cylinder in pure water, initially the concentration of the sugar in the water is less than the solute concentration in the potato cells of starch.

You can say the water concentration in the external sugar solution is greater than that in the potato cells - pure water has a greater water potential than the cell fluids.

Therefore when osmosis takes place with pure water, or very dilute solutions of sugar, the potato cells absorb water by osmosis giving a percentage mass increase.

The water will diffuse through the partially permeable membranes of the potato cells to try and dilute the internal solute solution of the potato cells - osmosis is happening.

On diagram equal to (a) At a 'medium' concentration of sugar, its concentration matches the concentration of the solutes in the potato cells and there is no net osmosis.

You can say the water concentration in the external sugar solution is the same as in the potato cell - no osmosis - both the potato cells and sugar solution have the same water potential.

Or you can say the solute concentrations are the same.

When there is no change in mass, the two solutions have the same water/solute concentration and the solutions are said to be isotonic - identical water potentials.

On the graph, this corresponds to ~0.3 mol/dm3 sugar concentration and equivalent to (b) on the diagram.

Diagram (d) to (e): At higher external sugar concentrations (above 0.3 mol/dm3 in this case), the osmotic effect reverses direction and water will diffuse through the partially permeable membranes out of the potato cells to try and dilute the external more concentrated sugar solution.

You can say the water concentration in the external sugar solution is less than that in the potato cell, so the water diffusion gradient is out of the potato cells giving a net mass loss for the potato cylinders. In other words, the water potential of the potato cell fluids is greater than the external sugar solution water potential.

Ultimately the greater the concentration, the greater the osmotic effect - a higher concentration of sugar will draw out more water - a greater rate of diffusion and osmosis - the more concentrated the sugar solution, the greater the mass loss of the potato block.

Sources of error

Obviously, all experiment can be repeated e.g. groups of students in the same class, this reduces errors - some groups of pupils might be more careful than others. You can then use mean values in your numerical analysis.

If the potato blocks are not completely dried, your will record a smaller mass of water loss than actually happened.

Especially if the room is warm, evaporation would increase the concentration of the sugar solution, this would increase the mass loss. This error can be eliminated by putting 'lids' over the beaker - paper covers held with an elastic bands will do.

Other similar experiment

You can repeat the experiments using common salt (sodium chloride, NaCl) and you should get a similar pattern of results.

 

Drinks and hydration

Most soft drinks contain water, sugar and ions.

Sports drinks contain sugars to replace the sugar used in energy release during the activity.

They also contain water and ions to replace the water and ions lost during sweating.

The sports drinks are supposed to be isotonic.

Know and understand that if water and ions are not replaced, the ion / water balance of the body is disturbed and the cells do not work as efficiently.


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5. ACTIVE TRANSPORT

You need to explain how this happens having seen that in osmosis only water passes through a partially permeable membrane.

However, cells cannot just rely on diffusion and osmosis for the input (e.g. nutrients) and output (e.g. waste) of substances.

What is 'active transport'

Active transport is the movement of particles across a membrane against a concentration gradient.

So, sometimes, substances are absorbed by cells against a concentration gradient - a net transfer against the normal diffusion gradient action described above in section 1.

This means transfer occurs in the opposite direction to the natural direction of the diffusion gradient.

e.g. active transport enables cells to absorb ions from very dilute solutions.

BUT, this movement of chemicals across a cell membrane against a natural diffusion gradient, requires the use of energy from respiration (via ATP molecules) and the overall process is called active transport.

Remember that absorption by diffusion down the concentration gradient through membranes doesn't require energy from respiration.

Active transport is very complicated system

BUT, 'basically' it uses transport molecules and ions within the membrane structure and the process powered by ATP from respiration.

By this means soluble nutrient molecules/ions can pass through the membrane into cells.

 

Examples of active transport

The gut and digestion

The villi in the small intestine absorb glucose and other nutrients from the gut and transfer them into the bloodstream.

The diagram illustrates the movement of molecules (green spheres) being moved through the membrane of the gut from the gut into the bloodstream, in the opposite direction to the natural diffusion gradient.

(The blue circles represent water molecules - solvent medium.)

The red circles represent the relatively large red blood cells, which are too large to get through the membrane, so staying in the bloodstream, to be joined by nutrient molecules (green circles) and ions via active transport.

Active transport is required to absorb nutrients (green circles) like amino acids, sugars like glucose etc. from the gut when the concentration in the gut is lower than their concentrations in the blood supply, and a healthy body requires these nutrients all the time.

If the concentrations of nutrients (e.g. sugars, amino acids) in the gut is higher than that in the blood stream, then the nutrients will naturally diffuse into the blood stream because of the direction of the concentration gradient (more concentrated ==> less concentrated).

If the concentration gradient flow is in the direction of the blood stream (higher) to the gut (lower), then respiration powered active transport must be used to work against the natural diffusion flow.

So active transport enables the gut to move nutrients like into the blood even though the natural concentration gradient (diffusion gradient) is the wrong way round.

Glucose can be transferred into the blood stream, even if its concentration is higher in the blood stream, and so conveyed to cells for respiration.

Other examples where active transport is essential in animals

Cells in the kidney reabsorb sodium ions from urine - sodium ions are needed for many biochemical processes in the body.

Fish in seawater have cells in the gills than can transfer salt back into the more salty sea water.

Similarly, crocodiles have salt glands in their tongue that can transfer excess salt from their bodies back into the water.

These are two good animal examples of adaptations to their environment.

The thyroid glands have cells that can concentrate iodine, against the diffusion gradient, to make the important hormone thyroxine.

 

For more on the gut and other examples see Surfaces for the exchange of substances in animal organisms

 

Active transport in plants

Active transport is used in the absorption of nitrates and other ions by plant roots.

For details see Transport and gas exchange in plants, transpiration, absorption of nutrients etc.


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6. A comparison of diffusion, osmosis and active transport

 

Mode of transfer Substances being moved Situation conditions Requirements Energy required?
Diffusion Any soluble substance - dissolved gas, liquid or solid From a higher to lower concentration Transfer of substances down a concentration gradient No energy needed
Osmosis Water From a dilute solution to a more concentrated solution Water passing through a partially permeable membrane No energy needed
Active transport Any dissolved substance

(g, l or s)

Transferring a substance from a lower concentration to a higher concentration Moving substances against a natural diffusion concentration gradient through a partially permeable membrane Energy needed from ATP  - respiration

Phrasing: In the context of diffusion - from a higher to a lower concentration means transfer from ...

(i) a more concentrated solution to a more dilute solution,

or (ii) a higher/greater level/concentration of a gas to a lower/smaller level/concentration


See also on other pages:

Examples of surfaces for the exchange of substances in animal organisms

Enzymes - section on human digestion, metabolism and synthesis

The human circulatory system


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Typical learning objectives  Diffusion, Osmosis, Transport and Active Transport

  • Appreciate that we need to understand how biological and environmental systems operate when they are working well in order to be able to intervene when things go wrong.

  • Appreciate that modern developments in biomedical and technological research allow us to do so.

  • Know and understand that the cells, tissues and organs in plants and animals are adapted to take up and get rid of dissolved substances.

  • Know that different conditions can affect the rate of transfer.

  • Sometimes energy is needed for transfer to take place - active transport.

  • You should be able to use your skills, knowledge and understanding to:

    • evaluate the development and use of artificial aids to breathing, including the use of artificial ventilators,

    • evaluate the claims of manufacturers about sports drinks,

    • analyse and evaluate the conditions that affect water loss in plants.

  • Know and understand that differences in the concentrations of the solutions inside and outside a cell cause water to move into or out of the cell by osmosis.

    • The soft cell wall, or outer membrane of an animal cell, acts as a partially permeable membrane.

    • The water surrounding cells, the tissue fluid, contains the dissolved molecules the cell needs to survive eg sugars, amino acids, oxygen, as well as waste carbon dioxide etc.

    • If the cells are short of water ('partially dehydrated'), the concentration of dissolved substances increases, so water diffuses through the cell membrane into the cells to dilute the cell fluids until equilibrium is established. Conversely, if the cell solution is too dilute, then water will diffuse out from osmotic action across the semi-permeable membrane of the cell wall.

  • Know and understand that substances are sometimes absorbed against a concentration gradient.

    • This means transfer occurs in the opposite direction to the natural direction of diffusion and osmosis.

    • Know that this requires the use of energy from respiration and this process is called active transport.

    • Know that active transport enables cells to absorb ions from very dilute solutions.

    • Active transport is required to absorb nutrients like amino acids, sugars like glucose etc. from the gut when the concentration in the gut is lower than their concentrations in the blood supply, and a healthy body requires these nutrients all the time.

    • If the concentrations of nutrients in the gut is higher than that in the blood stream, then the nutrients will naturally diffuse into the blood stream because of the direction of the concentration gradient (more concentrated ==> less concentrated).

    • If the concentration gradient flow is in the direction of blood stream (higher) to gut (lower), then respiration powered active transport must be used to work against the natural diffusion flow.

    • Remember that absorption by diffusion down the concentration gradient through membranes doesn't require energy from respiration


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