Movement of substances in living systems - introduction to
diffusion, osmosis, transport and active Transport
IGCSE AQA GCSE Biology Edexcel
GCSE Biology OCR GCSE Gateway Science Biology OCR GCSE 21st Century
Science Biology 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. Some of these revision notes on diffusion in and out
of cells, explaining osmosis transport of substances like oxygen,
glucose, nutrients, carbon dioxide and waste products are suitable for UK KS3 Science-Biology
(~US grades 6-8)
See also Surface
exchange of substances in
animal organisms
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
7.
Learning
objectives Diffusion, Osmosis, Transport 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
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1a.
DIFFUSION what is it? how to demonstrate it!
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!
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)
Rate
(surface area) x (concentration difference - gradient)
÷ (thickness of membrane - diffusion gradient
distance
(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
µm2 and a membrane thickness of 0.005
µm.
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
Diffusion and
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.
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
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
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!
(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.
The water moves by diffusion from a higher concentration
to a lower concentration through the cell membrane.
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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7. 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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