3 Movement into and out of cells

Focus
In the last chapter you were introduced to the main features of animal, plant and bacterial cells and
their functions. You studied examples of a range of specialised cells and how they are adapted to
carry out their roles. You learned about the levels of organisation within the organism and how to
calculate the magnification of microscopic structures. In this chapter, you will find out about how
materials move into and out of cells. These materials include gases like oxygen and carbon dioxide,
water, mineral ions, waste products and nutrients. How does the cell protect itself from gaining
substances that could be toxic, or from losing vital resources? How can a plant keep its shape if
it has no clear means of support? By studying the chapter carefully and following the practical
suggestions, you should be able to answer these questions.

Cells need food materials, which they can respire for The molecules of a gas like oxygen are moving
energy or use to build up their cell structures. They about all the time. So are the molecules of a liquid
also need mineral ions and water, which play a part or a substance like sugar dissolved in water. As
in chemical reactions in the cell. Finally, they need a result of this movement, the molecules spread
to get rid of substances like carbon dioxide, which themselves out evenly to fill all the available
would upset some of the chemical reactions or even space (Figure 3.1).
poison the cell if they built up.
Substances may pass through the cell membrane
either passively by diffusion or actively by some
form of active transport.

Diffusion molecules moving about become evenly distributed

▲ Figure 3.1 Diffusion
FOCUS POINTS
★ What is diffusion? This process is called diffusion. One effect of
★ Where does the energy for diffusion come from? diffusion is that the molecules of a gas, a liquid or
★ How do substances move into and out of the cell a dissolved substance will move from a region where
by diffusion? there are a lot of them (i.e. concentrated) to regions
★ How important is diffusion to living organisms? where there are few of them (i.e. less concentrated),
★ What effects do surface area, temperature, until the concentration everywhere is the same. In
concentration gradient and distance have most organisms substances have to move through
on diffusion?
cell membranes. Some substances move by diffusion.
Figure 3.2(a) is a diagram of a cell with a high
Key definitions concentration of molecules (e.g. oxygen) outside
Diffusion is the net movement of particles from a region
and a low concentration inside. The effect of this
of their higher concentration to a region of their lower difference in concentration is to make the molecules
concentration (i.e. down a concentration gradient), as a diffuse into the cell until the concentration inside
result of their random movement. and outside is the same, as shown in Figure 3.2(b).

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 3 Movement into and out of cells

The importance of diffusion of

gases and solutes
Gases
Most living things need a reliable source of oxygen
for respiration. This moves into the organism by
diffusion down a concentration gradient. Small
animals with a large surface area to volume ratio
(a) greater concentration (b) concentrations equal on both
may get oxygen through their body surface. Larger
outside cell sides of the cell membrane animals need gas exchange organs like lungs or gills,
▲ Figure 3.2 Molecules entering a cell by diffusion which provide a large surface area for gas exchange.
They also need a circulatory system to move the
Whether this will happen or not depends on if the oxygen to all their cells. Carbon dioxide, produced
cell membrane will let the molecules through. Small during aerobic respiration, can be toxic if it builds
molecules like water (H2O), carbon dioxide (CO2) and up. It is removed in the same way, by diffusion.
oxygen (O2) can pass through the cell membrane Photosynthetic plants need carbon dioxide for
quite easily. So, diffusion tends to balance the making their food. This diffuses through the stomata
concentration of these molecules inside and outside in the leaves (see Chapter 8) into the air spaces in
the cell all the time. the mesophyll, before reaching the palisade cells.
When a cell uses oxygen for its aerobic respiration, Oxygen produced during photosynthesis, as well
the concentration of oxygen inside the cell falls as water vapour from the transpiration stream,
and so oxygen molecules diffuse into the cell until diffuses out of the leaf through the stomata. The
the concentration is raised again. During tissue rate of diffusion of water vapour depends on the
respiration, carbon dioxide is produced and so its temperature, humidity and wind speed (see ‘Water
concentration inside the cell increases. Once again uptake’ in Chapter 8). Any oxygen needed for
diffusion takes place, but this time the molecules respiration (some is produced by photosynthesis)
move out of the cell. In this way, diffusion can and carbon dioxide produced (some is used up by
explain how a cell takes in its oxygen and gets rid of photosynthesis) also diffuses through the stomata
its carbon dioxide. of the leaves.

Going further
Nitrogen is the most common gas in the atmosphere is not used by the body tissues, so it builds up. When
(78% of the air is nitrogen). Nitrogen gas also enters the the diver begins to return to the surface of the water,
bloodstream by diffusion, but it is not used by the body. the pressure decreases and the nitrogen can come out
It is an inert (unreactive) gas so normally it causes no of solution, forming bubbles in the blood if the diver
problems. However, divers are at risk. As a diver swims goes back to the surface too quickly. These bubbles can
deeper, the surrounding water pressure increases. This block blood flow and become stuck in joints, resulting
raises the pressure in the diver’s air tank. An increase in a condition called decompression sickness, or ‘the
in nitrogen pressure in the air tank results in more bends’. Unless the diver goes up slowly in planned
nitrogen diffusing into the diver’s tissues, the amount stages, the effect of the nitrogen bubbles can be lethal
increasing the longer the diver stays at depth. Nitrogen and can only be stopped by rapid recompression.

Solutes In the ileum, water-soluble vitamins like vitamin C

Scientists think that some mineral ions in solution, are absorbed into the bloodstream by diffusion.
like nitrates and magnesium, diffuse across the In the kidneys, some solutes, like urea and mineral
tissues of plant roots, but that most are absorbed ions, pass back into the bloodstream by diffusion. At
into the roots by active transport. first, glucose is reabsorbed by diffusion, but active
transport is also involved.

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 Diffusion

Rates of diffusion
Worked example
Molecules and ions in liquids and gases move
around randomly using kinetic energy (energy The diagram in Figure 3.5 shows a section of the small
from movement). The speed with which a substance intestine, with villi. These increase the surface area to make
diffuses through a cell wall or cell membrane will the diffusion of digested food molecules more efficient.
depend on many conditions, including:
» the surface area across which the diffusion
is happening
» the temperature X
» the difference between its concentration inside
and outside the cell
» the distance it diffuses.
Y
Surface area
If 100 molecules diffuse through 1 mm2 of a W
membrane in one minute, then an area of 2 mm2
V
should allow twice as many molecules through in the
same time. So, the rate of diffusion into a cell will
depend on the cell’s surface area. A larger surface
area will result in faster diffusion. Cells which are ▲ Figure 3.5 Section through the small intestine to
involved in rapid absorption, like those in the show villi
kidney or the intestine, often have their exposed
Tasks
surface membrane formed into hundreds of tiny 1 Use a piece of cotton or string.
projections called microvilli (see Figure 3.3). These Hold one end of the cotton against point X. Now spread
increase the absorbing surface to make diffusion the cotton along the surface of the villi, weaving
faster. downwards and upwards until you reach point Y. Use a
pen to mark this point on the cotton.
microvilli ‘free’ (absorbing) surface Now hold the cotton against a ruler and measure its
length. Record this length as the length between X and Y.
It may take two or more attempts to follow the surface
of the villi from X to Y. It works best to trap the end of the
cotton at point X, then lie the cotton a short distance, for
example, until the first bend and trap it with a finger from
your other hand. Then lie the cotton along the next section
and keep repeating the procedure until you get to point Y.
2 Repeat the process, measuring V to W, which is the length
the surface would be without the presence of villi. Record
this length.
▲ Figure 3.3 Microvilli 3 Calculate the percentage increase in length by comparing
X to Y (the surface with villi) with V to W (the surface
The shape of a cell will also affect the surface area. without villi).
For example, the cell in Figure 3.4(a) has a greater To calculate the percentage increase in length
surface area than the cell in Figure 3.4(b), even new length(X to Y)
% increase = × 100
though they both have the same volume. original length(V to W)

The percentage increase shows how important the villi

are in the small intestine for increasing the surface area
(a) (b) for absorption of digested food molecules.
Bear in mind that this increase does not take into account
▲ Figure 3.4 Surface area. The cells both have the same
the presence of microvilli!
volume but the cell in (a) has a much larger surface area

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 3 Movement into and out of cells

Temperature
An increase in temperature gives molecules
and ions more kinetic energy. This allows them
to move faster, so the process of diffusion
speeds up. molecules will move from
the densely packed area
Concentration gradient
▲ Figure 3.6 Concentration gradient
The greater the difference in the concentration
of a substance on either side of a membrane, the Distance
faster it will diffuse. The difference is called a
concentration gradient (Figure 3.6). If a substance Cell membranes are all about the same thickness
on one side of a membrane is steadily removed, (approximately 0.007 µm), but plant cell walls vary
the concentration gradient stays the same. When in their thickness and permeability (how easily
oxygen molecules enter a red blood cell they materials pass through them). Usually, the thicker the
combine with a chemical (haemoglobin), which wall, the slower the rate of diffusion. When oxygen
takes them out of solution. So, the concentration diffuses from the alveoli of the lungs into red blood
of free oxygen molecules inside the cell is kept cells, it travels through the cell membranes of the
very low and the concentration gradient for alveoli, the blood capillaries and the red blood cells,
oxygen stays the same. as well as the cytoplasm of each cell. This increased
distance slows down the diffusion rate.

Practical work
For safe experiments/demonstrations which 2 Diffusion and temperature
are related to this chapter, please refer to the l Set up two beakers with equal volumes of hot
Biology Practical Skills Workbook that is also water and iced water.
part of this series. l Add a few grains of potassium
permanganate to each beaker and observe
Safety
how quickly the dissolved dye spreads
l Eye protection must be worn. through the water in each beaker. An
l Note your teacher’s advice for using a knife. alternative is to use tea bags.
l Take care using methylene blue or potassium
permanganate solution – they will stain skin Safety
and clothing. l Eye protection must be worn.
l Take care with hot water handling. l Take care, concentrated ammonia solution
is corrosive and irritant and should only
Experiments on diffusion
be used in a fume cupboard. Wear disposable
1 Diffusion and surface area gloves.
l Use a block of starch agar or gelatine at least 3 Diffusion and concentration gradients and
3 cm thick. Using a ruler and a sharp knife, distance (teacher demonstration only)
measure and cut four cubes from the jelly with
l Use a wide glass tube that is at least 30 cm
sides of 3.0 cm, 2.0 cm, 1.0 cm and 0.5 cm.
long and corked at one end. Using a glass
l Place the cubes into a beaker of methylene
rod or wire, push squares of moist red litmus
blue dye or potassium permanganate solution.
paper into the tube, so that they stick to the
l After 15 minutes, remove the cubes with
side and are evenly spaced out, as shown
forceps and place them on to a white tile.
in Figure 3.7. (It is a good idea to mark 2 cm
l Cut each of the cubes in half and measure the
intervals along the outside of the tube, starting
depth to which the dye has diffused.

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 Diffusion

at 10 cm from one end, with a permanent Which of the ‘animals’ would be able to
marker or white correction fluid before get enough oxygen by diffusion through
inserting the litmus paper.) their surface to keep them alive? Explain
l Close the open end of the tube with a cork your answer.
carrying a plug of cotton wool saturated 3 Try cutting different shapes, for example,
with a strong solution of ammonia. Start cutting a block 3.0 cm long, 1.0 cm wide and
a stopwatch. 0.5 cm deep. Research what type of animal
l Observe and record the time when each square this would represent and how this type of
of litmus starts to turn blue. Use this information animal obtains its oxygen.
to calculate the rate at which the alkaline 4 Explain the results you observed in
ammonia vapour diffuses along the tube. experiment 2.
l Repeat the experiment using a dilute solution 5 A 10% solution of copper sulfate is separated
of ammonia. by a partially permeable membrane from a
l Plot both sets of results on a graph, labelling 5% solution of copper sulfate.
each plot line. Will water diffuse from the 10% solution to the
5% solution or from the 5% solution to the
Cotton wool soaked with
ammonia solution 10% solution? Explain your answer.
6 If a fresh beetroot is cut up, the pieces
Wet litmus paper washed in water and then left for an hour in
a beaker of water, little or no red pigment
escapes from the cells into the water. If the
beetroot is boiled first, the pigment does
escape into the water.
Using your knowledge of the properties of a
living cell membrane, explain the difference
in results.
▲ Figure 3.7 Experiment to measure the rate of diffusion 7 In experiment 3, which ammonia solution
of ammonia in air diffused faster? Can you explain why?
8 Study the graph you produced for this
Practical work questions experiment. What happened to the rate of
1 Calculate the surface area and volume of each diffusion as the ammonia travelled further
cube used in experiment 1 and construct a along the tube? Can you explain why?
table of your data. Remember to state the units
in the heading for each column.
2 Imagine that the cubes in experiment 1 are
animals, with the jelly representing living
cells and the dye representing oxygen.

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 3 Movement into and out of cells

Going further
Artificial partially permeable membranes are made fail, the patient’s blood composition has to be controlled
from cellulose acetate in sheets or tubes. These are by a dialysis machine. In the same way, the accident
used for a process called dialysis for patients suffering victim can be kept alive on a dialysis machine until their
from kidney failure. The pore size can be altered blood pressure is returned to normal.
during manufacture so that large molecules cannot get
Simply, a dialysis machine consists of a long cellulose
through at all.
tube coiled up in a water-bath. The patient’s blood is
led from a vein in the arm and pumped through the
The dialysis machine (‘artificial kidney’) cellulose (dialysis) tubing (Figures 3.8 and 3.9). The tiny
Kidney failure can be the result of an accident involving pores in the dialysis tubing allow small molecules, like
a drop in blood pressure or of a disease of the kidneys. those of salts, glucose and urea, to leak out into the
In the first example, recovery is usually natural and water-bath. Blood cells and protein molecules are too
quick, but if it takes longer than 2 weeks, the patient can large to get through the pores (see experiment 5 on
die because of a potassium imbalance in the blood. This page 50). This stage is like the filtration process in the
causes heart failure. In the case of kidney disease, the glomerulus (see Chapter 13).
patient can stay alive with only one kidney, but if both
solution
changed

blood
flow

tank of
bubble trap water, salts
and glucose
pump

substances dialysis
diffuse out tubing
of blood

▲ Figure 3.8 The principle of the kidney dialysis machine

To prevent a loss of glucose and important mineral ions

from the blood, the water-bath contains a solution of
mineral ions and sugar of the correct composition, so that
only the substances above this concentration can diffuse
out of the blood into the bathing solution. In this way, urea,
uric acid and excess mineral ions are removed.
The bathing solution is also kept at body temperature and
is regularly changed as the unwanted blood solutes build
up in it. The blood is then returned to the patient’s arm vein.
A patient with total kidney failure spends 2 or 3 nights
each week connected to the machine (Figure 3.9).
With this treatment and a carefully controlled diet, the
patient can lead quite a normal life. A kidney transplant ▲ Figure 3.9 Kidney dialysis machine. The patient’s blood
is a better solution though, because then the patient is pumped to the dialyser, which removes urea and
does not need to use a dialysis machine. excess mineral ions

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 Osmosis

sugars and amino acids from the leaves to where

Test yourself they are used or stored (see Chapter 8).
1 Look at Figure 9.21 on page 161. The symbol O2 is Water is important in the process of excretion
an oxygen molecule. in animals. It is a powerful solvent for excretory
Explain why oxygen is entering the cells drawn on materials, like nitrogenous molecules (e.g. urea), as
the left but leaving the cells on the right. well as mineral ions, used hormones and drugs. The
2 Look at Figure 11.5 on page 185. It shows one of the
water has a diluting effect, so excretory materials
small air pockets (an alveolus) that form the lung.
a Suggest a reason why the oxygen and carbon are less toxic (poisonous).
dioxide are diffusing in opposite directions.
b What might happen to the rate of diffusion if the
blood flow were to speed up? Going further
The physical and chemical properties of water are
different from those of most other liquids. These
Osmosis make it uniquely effective in helping living activities.
For example, water has a high capacity for heat (high
thermal capacity). This means that it can absorb a
lot of heat without its temperature rising to levels
FOCUS POINTS that damage the proteins in the cytoplasm. However,
★ What is osmosis? because water freezes at 0 °C most cells are
★ What is the role of water in living organisms? damaged if their temperature falls below this and ice
★ How does water move into and out of cells crystals form in the cytoplasm. (Despite this, rapid
by osmosis? freezing of cells in liquid nitrogen at below –196 °C
★ What role does water have in supporting plants? does not harm them.)
★ How would you investigate osmosis?
★ What is the effect of immersing plant cells in
solutions of different concentrations? Diffusion of water
If a dilute solution is separated from a concentrated
★ What do the terms turgid, turgor pressure, solution by a partially permeable membrane, water
plasmolysis, flaccid and water diffuses across the membrane from the dilute to the
potential mean? concentrated solution. This is called osmosis and is
shown in Figure 3.10.

Roles of water partially permeable

membrane
Most cells contain about 75% water and will die if level rises
their water content falls much below this. Water is a
good solvent and many substances move about the level falls
cells in a watery solution.
Water molecules take part in many vital chemical concentrated dilute
reactions. For example, in green plants, water solution solution

combines with carbon dioxide to form sugar (see

Chapter 6). In animals, water helps to break down
▲ Figure 3.10 Osmosis. Water will diffuse from the dilute
and dissolve food molecules (see ‘Chemical digestion’ solution to the concentrated solution through the partially
in Chapter 7). Blood is made up of cells and a liquid permeable membrane. As a result, the liquid level will
called plasma. This plasma is 92% water and is a rise on the left and fall on the right
way of transporting many dissolved substances, like
carbon dioxide, urea, digested food and hormones. A partially permeable membrane is permeable but
Blood cells are carried around the body in the plasma. allows water to pass through more rapidly than
Water is also a way of transporting materials in dissolved substances.
plants. Water passes up the plant from the roots Since a dilute solution effectively contains more
to the leaves in xylem vessels and carries dissolved water molecules than a concentrated solution, there
mineral ions. Phloem vessels transport dissolved is a concentration gradient, which encourages the

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 3 Movement into and out of cells

passage of water from the dilute solution to the cells are surrounded by a solution which is more
concentrated solution. concentrated than the cytoplasm, water will pass
In living cells, the cell membrane is partially out of the cell by osmosis and the cell will shrink.
permeable and the cytoplasm and vacuole (in plant Too much uptake or loss of water by osmosis may
cells) contain dissolved substances. As a result, damage cells. For this reason, it is very important
water tends to diffuse into cells by osmosis if they that the cells in an animal’s body are surrounded
are surrounded by a weak solution (e.g. fresh water). by a liquid which has the same concentration as
If the cells are surrounded by a stronger solution the liquid inside the cells. In vertebrates, the
(e.g. sea water), the cells may lose water by osmosis. brain monitors the concentration of the blood and
These effects are described more fully later. the kidneys adjust it, as described in Chapter 13.
By keeping the blood concentration within narrow
Animal cells limits, the concentration of the tissue fluid
The cell in Figure 3.11 is shown surrounded by remains more or less constant (see ‘Homeostasis’ in
pure water. Nothing is dissolved in the water; it Chapter 14). So, cells are not swollen by taking in
has 100% concentration of water molecules. So, too much water or dehydrated by losing too much.
the concentration of free water molecules outside
the cell is greater than the concentration of water Plant cells
molecules inside. As a result, water will diffuse The cytoplasm of a plant cell and the cell sap in its
into the cell by osmosis. The membrane allows vacuole contain mineral ions, sugars and proteins.
water to go in or out. So, in our example, water This reduces the concentration of free water
can move into or out of the cell. The cell membrane molecules inside the cell. The cell wall is freely
is partially permeable to most of the substances permeable to water and dissolved substances, but
dissolved in the cytoplasm. So, although the the cell membrane of the cytoplasm is partially
concentration of these substances inside may be permeable. If a plant cell is surrounded by water or
high, they cannot diffuse freely out of the cell. The a solution more dilute than its contents, water will
water molecules move into and out of the cell, but pass into the vacuole by osmosis. The vacuole will
because there are more of them on the outside, expand and press outwards on the cytoplasm and
they will move in faster than they move out. The cell wall. The cell wall of a mature plant cell does
liquid outside the cell does not have to be 100% not stretch, so the inflow of water is limited by the
pure water. If the concentration of water outside inelastic cell wall, as shown in Figure 3.12.
is higher than that inside, water will diffuse in
2
by osmosis. cytoplasm

3 3
2 2
1
3
water 3 cell wall

2
(a) There is a higher (b) The extra water makes the
concentration of free water cell swell up. 1 since there is effectively a lower
molecules outside the cell concentration of water in the cell sap
than inside, so water diffuses 2 water diffuses into the vacuole
into the cell. 3 and makes it push out against the cell wall

▲ Figure 3.11 Osmosis in an animal cell ▲ Figure 3.12 Osmosis in a plant cell

Water entering the cell will make it swell up and, This has a similar effect to blowing up a soft bicycle
unless the extra water is removed in some way, tyre. The tyre is like the firm cell wall, the floppy
the cell will burst. In the opposite situation, if the inner tube is like the cytoplasm and the air inside

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 Osmosis

is like the vacuole. If enough air is pumped in, it

pushes the inner tube against the tyre and makes
the tyre hard.
When plant cells have absorbed a maximum
amount of water by osmosis they become very
rigid, due to the pressure of water pressing
outwards on the cell wall. As a result, the stems
and leaves are supported. If the cells lose water,
there is no longer any water pressure pressing
outwards against the cell walls. So, the stems and
leaves are not supported any more. At this point,
(a) plant wilting (b) plant recovered after watering
the plant becomes limp and wilts (droops) (see
Figure 3.13). ▲ Figure 3.13 Wilting

Practical work
For safe experiments/demonstrations which l Watch the level of liquid in the capillary tubing
are related to this chapter, please refer to the over the next 10–15 minutes.
Biology Practical Skills Workbook that is also
part of this series.

Safety
capillary tube
l Eye protection must be worn.
l Take care handling glass capillary tube, follow
your teacher’s guidance to avoid breakage.

Experiments on osmosis
Some of the experiments use ‘Visking’ dialysis
tubing. It is made from cellulose and is partially
permeable, allowing water molecules to diffuse
through freely, but limiting the passage of some
dissolved substances.
4 Osmosis and water flow
l Take a 20 cm length of dialysis tubing that has first level

been soaked in water and tie a knot tightly at

one end.
l Place 3 cm3 of a strong sugar solution in the elastic band
tubing using a plastic syringe and add a small
amount of coloured dye. water
l Fit the tubing over the end of a length of
capillary tubing and hold it in place with an Dialysis tube
containing
elastic band. Push the capillary tubing into the sugar solution
dialysis tubing until the sugar solution enters (with red dye)
the capillary.
l Now clamp the capillary tubing so that the
dialysis tubing is totally covered by the water ▲ Figure 3.14 Demonstration of osmosis
in the beaker, as shown in Figure 3.14.

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 3 Movement into and out of cells

Result Result
The level of liquid in the capillary tube rises. The starch inside the dialysis tubing goes blue
but the iodine outside stays yellow or brown.
Interpretation
Water must be passing into the sugar solution Interpretation
from the beaker. This is what you would expect The blue colour is normal for the reaction that
when a concentrated solution is separated from takes place between starch and iodine. This is
water by a partially permeable membrane. used as a test for starch (see Chapter 4).
The results show that iodine molecules have
A process like this may be involved in moving
passed through the dialysis tubing into the
water from the roots to the stem of a plant.
starch, but the starch molecules have not
5 Partial permeability moved out of the tubing into the iodine. The
l Take a 15 cm length of dialysis tubing that has dialysis tubing is partially permeable because
been soaked in water. Tie a knot tightly at one of its pore size. Starch molecules are very large
end. and probably cannot get through the pores.
l Use a dropping pipette to partly fill the tubing Iodine molecules are much smaller and can
with 1% starch solution. get through.
l Put the tubing in a test tube and hold it in place
Note: This experiment shows that movement
with an elastic band, as shown in Figure 3.15. of water is not necessarily involved, and the
l Rinse the tubing and test tube under the tap
pore size of the membrane makes it truly
to remove all traces of starch solution from the partially permeable with respect to iodine
outside of the dialysis tube. and starch.
l Fill the test tube with water and add a
few drops of iodine solution to colour the 6 The effects of water and sugar solution on
water yellow. potato tissue
l Leave for 10–15 minutes. l Push a No.4 or No.5 cork borer into a large
l After this time, observe any changes in the potato.
solution in the test tube. Caution: Do not hold the potato in your hand;
use a board as in Figure 3.16(a).
l Push the potato tissue out of the cork borer
elastic band
using a pencil, as in Figure 3.16(b). Prepare
a number of potato cylinders in this way and
choose the two longest. (They should be at
dilute iodine least 50 mm long.) Cut these two accurately
to the same length, e.g. 50, 60 or 70 mm.
Measure carefully.
l Label two test tubes A and B and place a
potato cylinder in each. Cover the potato
dialysis tubing containing
starch solution tissue in tube A with water; cover the tissue in
B with a 20% sugar solution.
l Leave the tubes for 24 hours.
l After this time, remove the cylinder from
tube A and measure its length. Notice also
whether it is firm or flabby. Repeat this for
▲ Figure 3.15 Experiment to demonstrate the effect of a the potato in tube B but rinse it in water
partially permeable membrane before measuring it.

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 Osmosis

measurement, but if most of the class record an

increase in length, then experimental error is
unlikely to be the cause.
7 The effects of varying the concentration of
sucrose solution on potato tissue
l Push a No.4 or No.5 cork borer into a
large potato.
Caution: Do not hold the potato in your hand;
use a board as in Figure 3.16(a).
l Push the potato tissue out of the cork borer
(a) place the potato on a board using a pencil, as in Figure 3.16(b). Prepare
six potato cylinders in this way and cut them
all to the same length. (They should be at least
50 mm long.) Measure them carefully.
l Label six test tubes with the concentration of
sucrose solution in them (e.g. 0.0 mol dm–3,
0.2 mol dm–3, 0.4 mol dm–3, 0.6 mol dm–3,
0.8 mol dm–3 and 1.0 mol dm–3) and place them
in a test-tube rack.
l Add the same volume of the correct sucrose
(b) push the potato cylinder out with a pencil solution to each test tube.
l Weigh a cylinder of potato, record its mass and
▲ Figure 3.16 Obtaining cylinders of potato tissue
place it in the first test tube. Repeat until all the
Result test tubes have been set up.
l Leave the tubes for at least 30 minutes.
The cylinder from tube A should have increased
l After this time, remove the potato cylinder from
in length and feel firm. The cylinder from tube B
the first tube, surface dry the potato and re-
should have decreased in length and feel flabby.
weigh it. Notice also whether it is firm or flabby.
Interpretation Repeat this for the other potato cylinders.
The cells of the potato in tube A have absorbed l Calculate the change in mass and the
water by osmosis, causing an increase in the percentage change in mass for each cylinder.
length of the potato cylinder.
(change in mass)
In tube B, the sugar solution is more concentrated Percentage change in mass = × 100
(mass at start)
than the cell sap of the potato cells, so these cells
have lost water by osmosis. As a result, the potato l Plot the results on a graph with sucrose
cylinder has become flabby and shorter. concentration on the horizontal axis and
An alternative to measuring the potato cores percentage change in mass on the vertical axis.
is to weigh them before and after the 24 hours’
Note: there will be negative as well as positive
immersion in water or sugar solution. The
percentage changes in mass, so your graph axes
core in tube A should gain mass and that in
will have to allow for this.
tube B should lose mass. It is important to
blot the cores dry with a paper towel before Result
weighing them. The cylinders in the weaker sucrose solutions
Whichever method is used, the changes may will have gained mass and feel firm. One of the
be quite small. So it is a good idea to collect cylinders may have shown no change in mass.
the results of the whole class. An increase in The cylinders in the more concentrated sucrose
length of 1 or 2 mm might be due to an error in solutions will have lost mass and feel limp.

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 3 Movement into and out of cells

Interpretation used, when would you expect the net flow of

If the potato cells are in a solution that has a water from the beaker into the dialysis tubing
lower concentration than the solution in the to stop?
cell vacuoles, water will move into the cells 11 For experiment 5, explain how the iodine got
by osmosis. The potato will increase in mass into the dialysis tubing.
because of the extra water it has gained. The 12 For experiment 5, suggest what would
cells swell up and this makes the potato feel firm. happen if you did not rinse the dialysis tubing
thoroughly before placing it in the test tube.
If the potato cells are in a solution that has a 13 For experiment 6, explain why the potato
higher concentration than the solution in the cylinder in test tube A increased in length.
cell vacuoles, water will move out of the cells 14 For experiment 6, suggest two safety
by osmosis. The potato will decrease in mass precautions you need to take when carrying
because it has lost water. The cell vacuoles are out this experiment.
no longer full of fluid and this makes the potato 15 Use data from your graph in experiment 7 to
feel limp. describe and explain the effect of changing
concentration of sucrose on potato tissue.
Practical work questions
If you do not have your own data, use the
9 In experiment 4 (Figure 3.14), what do you information in the table below to plot a graph
think would happen in these cases? first.
a A much stronger sugar solution was placed
in the cellulose tube. concentration of sucrose percentage change in
b The beaker contained a weak sugar solution/mol dm–3 mass
solution instead of water. 0.0 +3.3
c The sugar solution was in the beaker and
0.2 –2.5
the water was in the cellulose tube?
0.4 –8.3
10 In experiment 4, the column of liquid
accumulating in the capillary tube applies a 0.6 –10.0
steadily increasing pressure on the solution in 0.8 –10.8
the dialysis tubing. If a very long capillary is 1.0 –12.5

The partially permeable membrane does not act

Key definitions
like a sieve here. The sugar molecules can diffuse
Osmosis is the net movement of water molecules from
a region of higher water potential (dilute solution) to a
from right to left. However, they are bigger and
region of lower water potential (concentrated solution) surrounded by a cloud of water molecules. So they
through a partially permeable membrane. diffuse more slowly than the water, as shown in
Figure 3.18.
The cell membrane behaves like a partially
How osmosis works permeable membrane. The partial permeability may
When a substance like sugar dissolves in water, depend on pores in the cell membrane. However,
the sugar molecules attract some of the water the processes involved are much more complicated
molecules and stop them moving freely. This than in an artificial membrane. They depend on the
effectively reduces the concentration of water structure of the membrane and on living processes
molecules. In Figure 3.17 the sugar molecules on in the cytoplasm. The cell membrane contains fats
the right have held on to half the water molecules. and proteins. Anything that denatures proteins,
There are more free water molecules on the left for example, heat, also destroys the structure
of the membrane than on the right, so water will and the partially permeable properties of a cell
diffuse more rapidly from left to right across the membrane. If this happens, the cell will die
membrane than from right to left.

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 Osmosis

because essential substances will diffuse out of the different water potentials are in contact with each
cell and harmful chemicals will diffuse in. other, a water potential gradient is made. Water
will move from a cell with a higher water potential
water molecule partially permeable sugar molecule
membrane (a more dilute solution) to a cell with a lower water
potential (a more concentrated solution). This
explains one way in which water moves from root
hair cells through to the xylem of a plant root (see
Figure 8.11 on page 134).

The importance of water potential

and osmosis in plants
▲ Figure 3.17 The concentration gradient for water. There A plant cell with the vacuole pushing out on the
are more free water molecules on the left, so more will cell wall is turgid (it is swollen because the cell
diffuse from left to right than in the other direction. has taken up water) and the vacuole is exerting
Sugar molecules will diffuse more slowly from right
turgor pressure on the inelastic cell wall. Turgor
to left
pressure is the force inside the cell which pushes
partially permeable sugar molecules pass
membrane through pores more slowly
outwards, pushing the cell membrane against the
cell wall.
If all the cells in a leaf and stem are turgid, the
stem will be firm and upright. The leaves are held
out straight. If the vacuoles lose water for any
reason, the cells will lose their turgor (a process
fewer water molecules
called plasmolysis) and become flaccid. (See
go in this direction experiment 9 ‘Plasmolysis’ on page 56.) Plasmolysis
is the process of losing water from the cell. When a
cell loses water and is between the states of being
turgid and plasmolysed it is flaccid. If a plant has
flaccid cells, the leaves will be limp and the stem
will droop. A plant that loses too much water is
more water molecules hydrated sugar wilting (see Figure 3.13).
go in this direction molecule
Root hair cells are touching water trapped
high concentration of low concentration of between soil particles. When the water potential
free water molecules free water molecules
of the cell sap is lower than the water potential
▲ Figure 3.18 The diffusion theory of osmosis of the soil water, the water will enter the cells by
osmosis. This gives the plant the water it needs.
Water potential (This process is described in more detail in ‘Water
The water potential of a solution is a measure of uptake’ in Chapter 8.)
whether it is likely to lose or gain water molecules When a farmer spreads chemical fertiliser on
from another solution. A dilute solution has a the soil, the fertiliser dissolves in the soil water.
high proportion of free water molecules. So, it Too much fertiliser can lower the water potential
has a higher water potential than a concentrated of the soil water. This can draw water out of the
solution. Water will flow from the dilute to the plant root hair cells by osmosis. The plants can
concentrated solution (from a high potential to a wilt and die.
low potential). Pure water has the highest possible Irrigation of crops (the supply of controlled
water potential because water molecules will flow amounts of water to plants as they need it) can
from it to any other aqueous solution, even if have a similar effect. Irrigation which provides just
it is very dilute. When cells containing sap with enough water for the plant can lead to a build-up

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 3 Movement into and out of cells

of salts in the soil. The salts will eventually cause the same water potential as the cell contents.
the soil water to have a lower water potential This prevents any net flow of water into or out
than the plant root cells. Crops can then no longer of the cells. If the bathing fluid has a higher
be grown on the land, because they wilt and die water potential (a weaker concentration) than the
through water loss by osmosis. Much agricultural cells, water will move into the cells by osmosis,
land in hot countries has become unusable because causing them to swell up. As animal cells
of the side-effects of irrigation (Figure 3.19). have no cell wall and the membrane has little
strength, water would continue to enter. The
cells will eventually burst. Single-celled animals
like Amoeba (see Figure 1.35 on page 21) living
in fresh water obviously have a problem. They
avoid bursting by having a contractile vacuole.
This collects the water as it enters the cell and
regularly releases it through the cell membrane,
keeping the water content of the cell under
control. When surgeons carry out operations on
a patient’s internal organs, they sometimes need
to rinse a wound. Pure water cannot be used as
this would enter any cells it met and cause them
to burst. A saline solution (salt solution), with
▲ Figure 3.19 An irrigation furrow
the same water potential as tissue fluid, has to
be used.
Some countries apply salt to roads in the winter to During physical activity, the body may sweat
stop the formation of ice (Figure 3.20). However, to keep a steady temperature. If liquids are not
vehicle wheels splash the salt on to plants at drunk to make up for water loss through sweating,
the side of the road. The build-up of salts in the the body can become dehydrated. Loss of water
roadside soil can kill plants living there, due to from the blood results in the plasma becoming
water loss from the roots by osmosis. more concentrated (its water potential decreases).
Water is then drawn out of the red blood cells
by osmosis. The cells become plasmolysed. Their
surface area is reduced, so they are less effective
in carrying oxygen (see Figure 3.21).

▲ Figure 3.20 Salt gritter at work to prevent ice formation

on a road

The importance of water potential

and osmosis in animal cells
and tissues
▲ Figure 3.21 Plasmolysed red blood cells
It is important that the fluid which bathes cells
in animals, like tissue fluid or blood plasma, has

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 Osmosis

People doing sport sometimes use sports drinks

(Figure 3.22) which are isotonic (they have
the same water potential as body fluids). The
drinks contain water, salts and glucose. They
are designed to replace lost water and salts,
and provide energy, without creating osmotic
problems to body cells. However, use of these
drinks when not exercising vigorously can lead
to weight gain in the same way as the prolonged
use of any sugar-rich drink.

Figure 3.22 People may use isotonic sports drinks      

▲

Test yourself
3 Explain why the long-term use of irrigation in 5 When soil becomes saturated with water, it fills
farming can result in making the soil unsuitable for up the air spaces between the soil particles.
growing crops. Suggest why root hair cells may die in water-
4 Explain why it is more damaging for animal cells to logged soil.
be immersed in water than plant cells.

Practical work
For safe experiments/demonstrations which l After this time, remove the dialysis tubing
are related to this chapter, please refer to the from the water and note any changes in how
Biology Practical Skills Workbook that is also it looks or feels.
part of this series.
Result
Safety The tubing will become firm, swollen by the
l Eye protection must be worn. solution inside.
Further experiments on osmosis Interpretation
8 Osmosis and turgor The dialysis tubing is partially permeable
and the solution inside has fewer free water
l Take a 20 cm length of dialysis tubing that
molecules than outside. So, water has diffused
has been soaked in water and tie a knot in and increased the volume and the pressure
tightly at one end. of the solution inside.
l Place 3 cm3 of a strong sugar solution
in the tubing using a plastic syringe This is a simple model of what is thought to
(Figure 3.23(a)) and then knot the open end happen to a plant cell when it becomes turgid.
of the tube (Figure 3.23(b)). The partly filled The sugar solution is like the cell sap and the
tube should be quite floppy (Figure 3.23(c)). dialysis tubing is like the cell membrane and
l Place the tubing in a test tube of water for cell wall together.
30–45 minutes.

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 3 Movement into and out of cells

l Place the epidermis on a slide with a drop

of water and cover with a cover-slip (see
Figure 2.12(b)).
l Put the slide on a microscope stage and find
a small group of cells.
l Place a 30% solution of sugar at one edge
of the cover-slip with a pipette. Move the
solution under the cover-slip by placing a
piece of blotting paper on the opposite side.
l Study the cells you identified under the
microscope and watch for any changes in
their appearance.

Result
(a) place 3cm3 sugar solution in the dialysis tube The red cell sap will appear to get darker and
shrink, pulling the cytoplasm away from the cell
wall and leaving clear spaces. (It is not possible
to see the cytoplasm, but its presence can be
(b) knot tightly, assumed because the red cell sap seems to
after expelling have a distinct outer boundary where it has
the air bubbles separated from the cell wall.) Figure 3.24 shows
the turgid and plasmolysed cells.

(c) the partly filled tube should

be flexible enough to bend
water

dialysis tube
containing
sugar solution

(a) Turgid cells (×100). The cells are in a strip of epidermis

from a rhubarb stalk. The cytoplasm is pressed
▲ Figure 3.23 Experiment to model turgor in a plant cell against the inside of the cell wall by the vacuole.
▲ Figure 3.24 Demonstration of plasmolysis in
9 Plasmolysis rhubarb cells
l Peel a small piece of epidermis (the outer
layer of cells) from a red area of a rhubarb
stalk (see Figure 2.12(c) on page 31).

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 Osmosis

10 The effects of varying the concentration of

sucrose solution on potato tissue
Refer back to experiment 7.
Interpretation
If the cells of the potato have absorbed water
by osmosis, there will be an increase in the
mass of the potato cylinder. This happens
when the external solution has a higher water
potential than the water potential inside the
potato cells. (The sucrose solution is less
concentrated than the contents of the potato
cells.) Water molecules move into each
(b) Plasmolysed cells (×100). The same cells as they
cell through the cell membrane. The water
appear after treatment with sugar solution. The
vacuole has lost water by osmosis, shrunk and pulled molecules move from a higher water potential
the cytoplasm away from the cell wall. to a lower water potential. The cells become
▲ Figure 3.24 Demonstration of plasmolysis in rhubarb turgid, so the cylinder feels firm.
cells (continued) If the cells of the potato have lost water by
Interpretation osmosis, there will be a decrease in mass of
The interpretation in terms of osmosis is given in the potato cylinder. This happens when the
Figure 3.25. The cells are plasmolysed. external solution has a lower water potential
than the water potential inside the potato cells.
1
(The sucrose solution is more concentrated
3
3 than the contents of the potato cells.) Water
2
molecules move out of each cell through the cell
1 2
membrane. The water molecules move from a
1
2 higher water potential to a lower water potential.
2
3 The cells become plasmolysed or flaccid, so the
cylinder feels flabby.
1
1 the solution outside the cell is more
Practical work questions
concentrated than the cell sap
2 water diffuses out of the vacuole 16 a Which part of a plant cell do the parts of
3 the vacuole shrinks, pulling the cytoplasm the model represent?
away from the cell wall, leaving the cell flaccid
i) dialysis tube
▲ Figure 3.25 Plasmolysis ii) the contents of the dialysis tube.
The plasmolysis can be reversed by drawing b Explain how the process you have
water under the cover-slip in the same way that observed would be useful in a plant.
you drew the sugar solution under. It may need 17 a Explain how a cell becomes plasmolysed.
two or three lots of water to move out all the b How could a plasmolysed cell be returned
sugar. If you watch a group of cells, you should to full turgor?
see their vacuoles expanding to fill the cells 18 Study your graph from experiment 7. Can
once again. you predict the sucrose concentration which
would be the equivalent to the concentration
Rhubarb is used for this experiment because of the cell sap in the potato cells?
the coloured cell sap shows up. If rhubarb is not 19 Would you expect to get the same results if
available, the epidermis from a red onion scale the potato cylinders had been boiled before
can be used. the investigation? Explain your answer.

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 3 Movement into and out of cells

Active transport
FOCUS POINTS Key definitions
★ What is active transport? Active transport is the movement of particles
through a cell membrane from a region of lower
★ Why is active transport needed for moving concentration to a region of higher concentration
(i.e. against a concentration gradient) using energy
molecules or ions across membranes?
from respiration.
★ How are protein carriers involved in
active transport?

The importance of active active transport needs a supply of energy from

respiration. Figure 3.26 shows a possible model to
transport explain active transport.
If diffusion was the only way a cell could take The protein carrier molecules shown in
in substances, it would have no control over Figure 3.26 are protein molecules. As the name
what went in or out. Anything that was more suggests, their job is to transport substances
concentrated outside would diffuse into the cell across the membrane during active transport, as
even if it was harmful. Substances which the shown in (b).
cell needed would diffuse out as soon as their Epithelial cells in the villi of the small
concentration inside the cell increased above intestine have the job of absorbing glucose
their concentration outside it. However, the against a concentration gradient. The cells
cell membrane has a lot of control over the contain large numbers of mitochondria in which
substances which enter and leave the cell. respiration takes place. The chemical energy
In some cases, substances are taken into or released is converted into kinetic energy for
removed from the cell against the concentration the movement of the glucose molecules. The
gradient. For example, sodium ions may same type of process happens in the cells of
continue to pass out of a cell even though the the kidney tubules for the reabsorption of
concentration outside is greater than inside. glucose molecules into the bloodstream against
The cells lining the small intestine take up their concentration gradient.
glucose against a concentration gradient. The Plants need to absorb mineral salts from the
processes by which substances are moved against soil, but these salts are in very dilute solution.
a concentration gradient are not fully understood. Active transport allows the root hair cells of
The processes may vary for different substances, plant roots to take up salts from this dilute
but they are described as active transport. solution against the concentration gradient.
Anything which interferes with respiration, Again, chemical energy from respiration is
like a shortage of oxygen or glucose, stops converted into kinetic energy for movement of
active transport happening. This shows that the salts.

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 Active transport

substance

carrier protein
molecule

OUTSIDE

energy cell
from membrane
respiration

INSIDE

(a) substance combines with (b) carrier transports substance across (c) substance released
carrier protein molecule membrane using energy from respiration into cell

▲ Figure 3.26 A theoretical model to explain active transport

Going further Test yourself

6 Suggest why a cell stops taking in
substances by active transport which has
Controlled diffusion been exposed to
Although for any one substance the rate of diffusion a high temperature
through a cell membrane depends partly on the b respiratory poison.
concentration gradient, the rate is often faster or 7 State which parts of a cell are
slower than expected. Water diffuses more slowly responsible for
and amino acids diffuse more rapidly through a a making proteins to act as carrier
membrane than expected. In some cases, scientists molecules in the cell membrane
think this happens because the ions or molecules b releasing energy for active transport
can pass through the membrane through special across the cell membrane
pores. There may not be many of these pores, or c controlling cell activities such as
they may be open or closed in different conditions. active transport
In other cases, the movement of a substance can d storing mineral ions which have passed
be speeded up by an enzyme working in the cell through the cell membrane.
membrane. So, ‘simple passive’ diffusion, even of
water molecules, may not be so simple or so passive
where cell membranes are involved.
When a molecule gets inside a cell, there are
many structures and processes which may move it
from where it enters to where it is needed. Simple
diffusion is unlikely to be the only way that this
movement happens.

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 3 Movement into and out of cells

Revision checklist
After studying Chapter 3 you should know and ✔ Water enters cells bathed in dilute solutions but
understand the following: leaves cells bathed in concentrated solutions
✔ Diffusion is the random movement of molecules of because of osmosis.
liquid, gas or dissolved solid. ✔ Osmosis maintains the pressure of water in plant
✔ The molecules of a substance diffuse from a region cells to support the plant.
where they are very concentrated to a region ✔ Active transport involves the movement of
where they are less concentrated. substances against their concentration gradient.
✔ Kinetic energy of molecules and ions results in ✔ Active transport requires energy.
their diffusion.
✔ Substances may enter cells through the cell ✔ Osmosis involves the net movement of water
membrane by simple diffusion or active transport. molecules from a region of higher water
✔ The rate of diffusion is affected by surface area, potential to a region of lower water potential
temperature, concentration gradient and distance. through a partially permeable membrane.
✔ Water has several roles as a solvent in organisms. ✔ The meanings of the terms turgid, turgor
✔ Cell membranes are partially permeable, and pressure, plasmolysis and flaccid.
cytoplasm and cell sap contain many substances ✔ The importance of water potential and
in solution. osmosis in the uptake and loss of water
✔ Osmosis is the diffusion of water through a by organisms.
partially permeable membrane, from a dilute ✔ Active transport is important as it allows
solution of salt or sugar to a concentrated solution, movement of substances across membranes
because the concentrated solution contains fewer against a concentration gradient.
free water molecules. ✔ The role of protein carriers in
cell membranes.

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 Exam-style questions

Exam-style questions
1 Diffusion, osmosis and active transport 5 Some plant cells were placed on a microscope
are processes involved in the movement of slide and covered with sugar solution, which
substances in a plant. was more concentrated than the sugar inside
Complete the table by the cells.
a defining what each term means a Describe what changes would happen in
b giving one example of a substance moved each of the following cell parts:
by the process in the plant. [9] i) cell wall [1]
ii) cytoplasm [1]
name of definition example of a substance moved iii) sap vacuole. [1]
process by the process in the plant
b With reference to water potential gradient,
diffusion explain how these changes occur. [2]
osmosis c i) State the differences between
active diffusion and active transport. [2]
transport ii) Give one example of each process in
living organisms. [2]
2 When a plant leaf is in daylight, its cells make 6 The diagram shows a cylinder of potato tuber,
sugar from carbon dioxide and water. The sugar with a weight attached, in a container of
is turned into starch straight away and stored water. A second, identical cylinder of potato
in plastids. Sugar is soluble in water; starch is was set up in the same way but placed in a
insoluble. With reference to osmosis, suggest why container of concentrated sugar solution. Both
it is an advantage for the plant to convert the cylinders were left for 3 hours.
sugar to starch. [3] a Describe and explain in terms of water
potential what would happen to
3 When doing experiments with animal i) the potato in water [3]
tissues they are usually bathed in Ringer’s ii) the potato in the concentrated
solution, which has a concentration like sugar solution.[3]
the concentration of blood or the fluid that b State the name of the process involved in
surrounds cells. With reference to osmosis, causing any changes to the appearance of
explain why this is necessary. [2] the cylinders. [1]
4 Explain why a dissolved substance reduces c Root hair cells are involved in taking up
the number of ‘free’ water molecules in a water and mineral ions from the surrounding
solution.[2] soil. State how the processes of taking up
these substances are different. [2]

Water Support Weight Concentrated

sugar solution

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 3 Movement into and out of cells

7 The graph shows the absorption of phosphate a State which plant cells absorb the phosphate
ions by the roots of a beech plant when kept in ions.[1]
an atmosphere of air or nitrogen. A represents b Describe the absorption of phosphate
the concentration of phosphate in external ions by the beech plant
solution. i) in an atmosphere of air [2]
10 ii) in an atmosphere of nitrogen. [2]
air c Suggest what process is involved in the
9
absorption of phosphate ions. Explain
absorption of phosphate/arbitrary units

8 your answer.[3]
7

1 nitrogen
A
0
0 5 10 15 20 25 30
hours

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