Practical work and student investigations
1 Tests for glucose and starch, lipid and protein
2 Controlled experiments to illustrate how enzyme activity can be affected by changes in temperature
and pH
3 Simple experiments on diffusion and osmosis using living and non-living systems
4 Controlled experiments to investigate photosynthesis, showing the evolution of oxygen from a water
plant, the production of starch and the requirements of light, carbon dioxide and chlorophyll
5 A simple experiment to determine the energy content of a food sample
6 Controlled experiments to demonstrate the evolution of carbon dioxide and heat from respiring seeds
or other suitable living organisms
7 Simple controlled experiments to investigate the effect of light on net gas exchange from a leaf, using
hydrogen-carbonate indicator
8 Experiments to investigate the role of environmental factors in determining the rate of transpiration
from a leafy shoot
9 A simple experiment to investigate the effect of exercise on breathing in humans
10 A simple experiment to show how the sensitivity of the skin differs on finger tips, back of hand, wrist
and forearm
11 Controlled experiments to demonstrate phototropic and geotropic plant growth responses
12 The use of quadrats to estimate the population size of an organism in two different areas
1
1 Tests for glucose, starch, lipid and protein
The tests that are expected:
• Benedict’s test for glucose
Testing for sugars
Benedict’s solution is blue.
Add Benedict’s solution and heat in a water bath. Solution should turn from blue to green to yellow and
finally red if glucose is present.
If solution remains blue, then glucose is not present.
• iodine test for starch (test not required by specification)
Testing for starch
Iodine is orange/brown
Add iodine solution, if it turns blue black it means starch is present. If it stays yellow/orange means there
is no starch present
• the emulsion test for lipid
Testing for lipid (fat)
Dissolve the lipid in ethanol then add water and shake vigorously, if it goes cloudy white lipid is present.
• biuret reagent for protein(test not required by specification).
Testing for protein
Biuret reagent is pale blue.
Add Biuret reagent to a solution of food, if it turns purple protein is present. If it remains pale blue no
protein is present
These tests could be carried out:
• on prepared samples of pure substances
• on foods that contain the substances
• in a context.
Example
To demonstrate health applications, a fake ‘urine sample’ can be made by colouring water with iodine
solution. A similar solution, with glucose added, could mimic the urine of an untreated diabetic. Students
can use Benedict’s test to identify the diabetic – if glucose is present in urine the patient is diabetic.
The iodine test could be introduced when teaching that plants manufacture starch by photosynthesis, or
when investigating the effect of amylase on starch digestion.
2
Controlled experiments to illustrate how enzyme activity can be affected by changes in
temperature
1. The effect of amylase on starch digestion
At each temperature selected, from 0°C to 100°C, samples of amylase solution and of starch solution are
brought to temperature before being added together. The mixture is then kept at the same temperature.
To measure the rate of reaction, drops of the mixture can be collected at intervals of one minute and
added to individual iodine drops on a white tile. Iodine tests for the presence of starch – iodine goes black
when starch is present. The time taken for the starch to disappear is recorded for each temperature –
indicated when the iodine no longer changes colour (remains orange/brown). The temperature can be
controlled using water baths.
The digestive enzyme amylase breaks down starch into the sugar maltose. If the speed at which the
starch disappears is recorded, this is a measure of the activity of the amylase.
2. The effect of catalase on the breakdown of hydrogen peroxide.
Catalase is found on the surface of potato and liver. Catalase converts hydrogen peroxide into water and
oxygen. The rate of oxygen production can be measured as an indication of enzyme activity. An upturned
boiling tube previously filled with water can be used to collect and measure the volume of oxygen
evolved. Or the time take for ‘bubbling’ to stop can be measured. The liver/potato can be exposed to
different temperature to demonstrate the effect of temperature.
Enzymes are proteins that speed up chemical reactions in our cells.
Enzymes work best at their optimum temperature. This is why homeostasis is important - to keep our
body temperature at a constant 37°C.
As the temperature increases, so does the rate of chemical reaction. This is because heat energy causes
more collisions, with more energy, between the enzyme molecules active sites and other molecules.
However, if the temperature gets too high, the enzyme is denatured and stops working.
A common error in exams is to write that enzymes are killed at high temperatures. Since enzymes are not
living things, they cannot be killed.
Graph showing the effect of temperature on enzyme reactions
3
3. Effect of changing pH on enzyme’s rate of reaction.
1 cm3 sodium
carbonate solution
0.5 cm3 sodium
carbonate solution
2 cm3 acetic acid
4 cm3 acetic acid
Place 5 cm3 starch solution in each tube. Note the time and add 1 cm3 amylase solution to each
Then test for starch using iodine – take a sample every 30 seconds and time how long it takes for the
iodine to no longer turn blue/black.
The most rapid breakdown of starch will be in the region of pH 6-7.
When the pH in the mouth was tested, the result was probably about pH 7.
Logically you would expect the pH in the mouth to be the one which favoured a rapid breakdown of
starch. The other pH values seem to slow/inhibit the action of the enzyme – pH can denature the enzyme
so that the enzyme’s active site changes shape and the substrate can no longer bind to it..
4
Simple experiments on diffusion and osmosis using living and non-living systems
a Cubes of agar jelly placed into solutions of methylene blue or potassium permanganate will absorb the
pigment by diffusion. The cubes are left in the pigmented solution for different measured periods of time
and are then sliced open. The distance between the edge of each cube and the edge of the coloured agar
may be used as a measure of the distance the pigment molecules have moved by diffusion.
Demonstration of diffusion in a jelly
Agar jelly has a consistency similar to the cytoplasm of a cell. Like cytoplasm, it has a high water content.
Agar can be used to show how substances diffuse through a cell.
This demonstration uses the reaction between hydrochloric acid and potassium permanganate solution.
When hydrochloric acid comes into contact with potassium permanganate, the purple colour of the
permanganate disappears.
A Petri dish is prepared which contains a 2 cm deep layer of agar jelly, dyed purple with potassium
permanganate. Three cubes of different sizes are cut out of the jelly, with side lengths 2 cm, 1 cm and 0.5
cm.
The cubes are carefully dropped, at the same time, into a beaker of dilute hydrochloric acid (Figure 1.10).
The time is taken for each cube to turn colourless.
The smaller the cube the larger its surface area to volume ratio, so the quicker the rate of diffusion (the
shorter the time for the cube to go colourless).
b A crystal of potassium permanganate can be dropped into a beaker of water and the appearance of the
water noted over time.
5
6
c To demonstrate osmosis, Visking tubing (dialysis tubing) can be tied at one end and filled with 20 per
cent sucrose solution. The other end is attached to a capillary tube. The level of the sucrose can be noted
before and after the tubing has been placed in a beaker of water for about 30 minutes. Visking tubing has
microscopic holes in it, which let small molecules like water pass through (it is permeable to them) but is
not permeable to some larger molecules, such as the sugar sucrose. This is why it is called ‘partially’
permeable.
The sucrose molecules are too big to pass through the holes in the partially permeable membrane. The
water molecules can pass through the membrane in either direction, but those on the right are attracted
to the sugar molecules. This slows them down and means that they are less free to move – they have less
kinetic energy. As a result of this, more water molecules diffuse from left to right than from right to left.
In other words, there is a greater diffusion of water molecules from the more dilute solution (in this case
pure water) to the more concentrated solution.
7
d Onion epidermis can be peeled away, cut into squares and mounted on slides in different
concentrations of sucrose solution. Observation under a microscope will show the effects of osmosis.
Plasmolysis - The state of a plant cell in a hypertonic solution, the cell shrinks.
Turgid - Plant cells in a hypotonic solution, the cell swells against the cell wall and the cell is rigid and firm
Flaccid - Plant cells in a hypertonic solution, the cell shrinks away from the cell wall and the cell is limp.
8
e Red blood cells in blood obtained from a butcher may be mounted on slides in hypotonic, isotonic and
hypertonic saline, and observed under a microscope to show the effects of osmosis.
Blood plasma has a concentration equivalent to a 0.85% salt solution. If fresh blood is placed into
solutions with different concentrations, the blood cells will gain or lose water by osmosis. This can be
demonstrated using sterile animal blood
(available from suppliers of biological materials).
Three test tubes are set up, containing these solutions:
A 10 cmᶟ of distilled water (hypotonic)
B 10 cmᶟ of 0.85% salt solution (isotonic)
C 10 cmᶟ of 3% salt solution (hypertonic)
1 cmᶟ of blood is added to each tube, and the tubes are shaken. A sample from each tube is examined
under the microscope. The sample from tube A is found to contain no intact cells
The blood cells on the right (tube C) were placed in a 3% salt solution and the normal blood cells on the
left(tube B) were in a 0.85% salt solution.
The cells from tube B look normal, but those from tube C are shrunken, with crinkly edges
It is important that animal cells are surrounded by a solution containing the correct concentration of
dissolved solutes. If the surrounding solution does not have the right concentration, cells can be damaged
by the effects of osmosis. The red blood cells placed in water absorb the water by osmosis, swell up and
burst, leaving a red solution of haemoglobin in the test tube. When placed in 3% salt solution, the red
blood cells lose water by osmosis and shrink.
The three tubes are now placed in a centrifuge and spun around at high speed to separate any solid
particles from solution.
Tube A contains a clear red solution and no solid material at the bottom of the tube = the cells have burst
due to water entering via osmosis and haemoglobin is released into the water
Tubes B and C both contain a colourless liquid and a red precipitate at the bottom = the cells have
remained intact so haemoglobin remains in the cells.
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f Osmosis can be demonstrated by using strips of potato, and this basic experimental method provides a
good opportunity for students to carry out individual whole investigations.
Because of the difficulty of the osmosis concept, it is better to keep this investigation until the latter part
of the course so that students will have had previous experience of carrying out investigations on simpler
topics. Students enjoy the reference to ‘chips’, but should quickly realise that it can be difficult to keep
the size constant – to achieve consistency lengths of potato tissue can be drilled from a potato using a
cork borer. The ‘chips’ are measured by mass or by length and are placed into sucrose solutions of
different concentrations for at least one hour. The percentage change in mass or length is a measure of
the degree of osmosis that has occurred.
A potato tuber is a plant storage organ. It is a convenient tissue to
use to investigate the effects of osmosis on the mass of the
tissue.
A boiling tube is half-filled with tap water and a second with
concentrated (Molar) sucrose solution. A third tube is left empty.
A potato is ‘chipped’ – cut into chips 5cm x 1cm x 1cm, making
these measurements as accurate as possible to ensure the chips
have the same surface area. No skin is left on the potato (it would
provide a waterproof layer).
Each chip is gently blotted to remove excess moisture and
weighed to find the starting mass of each. One chip is placed into each of the three tubes, and after a
fixed time the chips are removed, gently blotted to remove excess moisture and reweighed. The chips
can be felt to see and change in texture and re-measured to see if there is any changed in dimensions
(length/width/breadth).
The change in mass (+/-) is calculated for each chip and the % change is found. (% change in mass takes
into account the difference in starting masses of the chip so they can be compared).
% change = change in mass x 100
Starting mass
Boiling tube with water - water moves into the cells
The water in the tube is more concentrated than the water in the cells, so water molecules will move
from an area of higher concentration of water (in the tube) to an area of lower concentration of water (in
the potato cells).
The cell membrane is semi-permeable so the water molecules can get through and each cell will swell as
the water flows in.
As all the cells swell, the potato chip will increase in mass, and length/width/breadth and become more
rigid.
Boiling tube with concentrated sugar solution - water moves out of the cells
The water in the tube is a lower concentration than the water in the cells, so water molecules will move
from an area of higher concentration of water (in the potato cells) to an area of lower concentration of
water (in the tube).
The cell membrane is semi-permeable so the water molecules can get through and each cell will shrink as
the water flows out.
As all the cells shrink, the potato chip will decrease in mass, and length/width/breadth and become more
flexible/floppy.
Boiling tube with nothing - some water moves out of the cells
Water molecules will move from an area of higher concentration of water (in the potato cells) to an area
of lower concentration of water (in the tube).
The cell membrane is semi-permeable so the water molecules can get through and each cell will shrink as
the water flows out.
As all the cells shrink, the potato chip will decrease in mass, and length/width/breadth and become more
flexible/floppy. The effect will be less than in the tube with concentrated sugar solution.
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g A variation on this theme is to cut potato cubes of different sizes, which have different surface area to
volume ratios. After measuring and recording the masses of the cubes, they are immersed in water. After
one hour, the cubes are blotted dry and their masses measured and recorded again. The percentage
increase in mass for cubes of different surface area to mass ratio can be compared, to explore the
concept of how surface area to volume ratio influences water uptake.
The surface area over which the osmosis occurs can change the speed at which it happens dramatically.
The larger the surface area between the two solutions the faster osmosis will happen. This is due to more
of each solution will be in contact with the membrane at one moment. This
means that a larger amount of the free water molecules in the solution can move across the membrane
at the same time therefore making the process happen much faster.
Eight 1cm³ cubes
One 8cm³ cube
11
Controlled experiments to investigate photosynthesis, showing the evolution of oxygen
from a water plant, the production of starch by leaves and the requirements for light,
carbon dioxide and chlorophyll
a The evolution of oxygen from a water plant can be seen if a water plant (typically elodea or a similar
species) is placed in a beaker of water and covered with a glass funnel that has a water-filled test tube
placed over its opening. After 24 hours, a colourless gas will have displaced water from the test tube. A
test for oxygen is then carried out.
Re-light a glowing
splint
12
b To measure the rate of oxygen production, the stem of a water plant is cut under water, and the plant
kept immersed in water in a beaker or boiling tube. The number of bubbles of gas given off over a
measured time period can be counted. This simple experimental set up can enable students to carry out
individual investigations into the effect of different factors on the rate of bubble production. Suitable
variables include: light intensity (the plant is exposed to a light source and the rate of bubble production
measured at different light intensities by changing the distance between the light source and the water
plant); colour/wavelength of light (coloured filters are placed between the plant and the light source);
and carbon dioxide availability (the plant is immersed in solutions of different concentration of sodium
hydrogen carbonate).
It is vital to have a stabilising period of a minimum 5 minutes before any measurements are taken.
An increase in light intensity should bring about an increase in gas production and, until a limiting factor
comes into operation, the two should be proportional.
Increasing the light intensity makes available a greater amount of energy to drive the photosynthetic
reaction. Since oxygen is a product of photosynthesis, a higher rate of photosynthesis should result in a
higher rate of oxygen production.
Unless the gas produced by Elodea is pure oxygen it would be possible for the plant to double its oxygen
output without doubling the volume of gas if the proportion of oxygen in the gas mixture were increased.
One would need to know that the percentage of oxygen in the bubbles remains constant, before it could
be assumed that doubling the volume of gas meant doubling the production of oxygen.
It is quite likely that the gas production will increase in direct proportion to the light intensity over the
whole range of this experiment. Limiting factors may operate, however, at the higher light intensities.
It is unlikely that gas production would continue to increase indefinitely with increasing light intensity
since carbon dioxide would become limiting or internal factors such as the available chloroplasts or the
diffusion rate of carbon dioxide would impose a limit on the rate of photosynthesis.
As the bench lamp comes closer to the beaker it might be expected that there would be a rise in
temperature of the water in the beaker. Since a rise in temperature speeds up many chemical reactions it
might also accelerate some stages of photosynthesis. (In practice, the temperature is unlikely to rise more
than 1° C.)
13
c Starch production can be investigated by placing a plant in the dark for 24 hours to de-starch the leaves.
A starch test on a leaf from a plant that has been kept in the dark will not give a blue-black colour,
whereas a similar test on a control leaf from a plant kept in the light will give a blue-black colour.
Remove a green leaf from a plant that has been exposed to sunlight for a few hours
Half-fill a beaker with water. Heat the water until it boils. Keep the water at boiling point.
Use the forceps to place the leaf in the boiling water. Boil for 2 minutes.
Boiling the leaf in water:
 Removes the waxy cuticle which prevents entry of iodine/potassium iodide solution.
 Ruptures cell membranes to make starch granules in cytoplasm and chloroplasts accessible to
iodine/potassium iodide solution. Cell membranes are selectively permeable and do not readily
allow the penetration of iodine.
 Denatures enzymes, particularly those which convert starch to glucose e.g. diastase.
 Boiling arrests all chemical reactions, since enzymes which catalyse the reactions are denatured.
Denatured enzymes have altered or destroyed active sites due to heat, pH, ionic concentration
Place the boiled leaf in a boiling tube containing 90% ethanol.
Place the boiling tube in hot water and boil for 10 minutes or until the leaf decolourizes.
Boiling the leaf in ethanol:
 Removes chlorophyll which is a green pigment and so masks the colour change of the iodine test
for starch, so the leaf needs to be ' decolourized' for changes to be observed. A decolourized leaf is
pale yellow or green. Ethanol is an organic solvent and so extracts chlorophyll from the leaf.
Gently remove the leaf and wash with a fine trickle of cold tap water.
Spread the leaf evenly on a white tile.
Add a few drops of iodine/potassium iodide solution to the leaf and note any observations.
The iodine solution will turn from brown to blue - black if starch is present.
14
d A starch test on a variegated leaf can be used to demonstrate that chlorophyll is needed for
photosynthesis.

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





Place a plant with variegated leaves in the dark for at least 24 hours to make sure that there is no
starch in the leaves i.e. ‘de-starching the leaves’.
Variegated Leaves: have green and white areas – the green areas have chlorophyll but chlorophyll is
absent in the white areas.
Leave the plant in the light with de-starched leaves in good light for 6 hours at room temperature
(20˚C).
Then test the leaf for the presence of starch using iodine.
Compare the stain pattern with the colour map.
The green chlorophyll rich parts are blue-black with starch.
The white parts without chlorophyll are yellow-brown without starch.
Since starch is only produced if chlorophyll is present, then chlorophyll is needed for photosynthesis
e To show that carbon dioxide is needed for photosynthesis, a leaf on a plant may be surrounded by air
with no carbon dioxide by inserting it into a conical flask containing a small amount of potassium or
sodium hydroxide. The plant is left in good light for 24 hours.
NaOH removes
CO₂
The test leaf and a control leaf from the plant are then tested for starch.
To test leaves for starch:
• drop into very hot/boiling water for one minute (to destroy the cell membranes so that chlorophyll
molecules can pass through)
• drop into hot ethanol (to remove/dissolve the green chlorophyll)
• place leaf into water (to rehydrate and soften the leaf so that it can be spread out)
• drop iodine solution onto the leaf (test for starch) – blue-black colour will show the presence of starch.
15
A simple experiment to determine the energy content of a food sample
Fat-containing foods such as dried crisps work very well.
• A known mass of the food sample is weighed and the
mass noted.
• A boiling tube is prepared, containing a known volume of
water. The water temperature is recorded.
• The food sample is put in a crucible or burning spoon and
ignited (for example in the flame of a Bunsen burner).
• The food sample is quickly placed under the boiling tube.
As soon as the food sample has completely burnt the water
temperature in the boiling tube is re-measured.
• The equation used to calculate the energy content of the
food is:
Energy content of food sample (joules per gram) =
mass of water heated (g) x temperature rise (oC) x 4.2
mass of food sample (g)
To obtain an accurate result, all the energy in the food sample needs to be transferred to the water, but
unless you use a bomb calorimeter this won’t happen.
Bomb calorimeter
NB: The energy gets lost to the surroundings as not all energy is direct to heat the water, also there is a
need to heat the glass apparatus first.
16
Simple controlled experiments to investigate the effect of light on net gas exchange
from a leaf, using hydrogen carbonate indicator
A water plant can be placed in a sealed tube of air-equilibrated hydrogen carbonate solution (red in
colour) and placed in the light
or in the dark. The solution will
turn purple if kept in the light
and will turn yellow if kept in
the dark.
A variation could involve the
use of water snails or, if not
available, small land insects
placed on a gauze platform
above the indicator, with and
without the water plant. This
variation allows students to
think about the balance
between carbon dioxide used
by photosynthesis and carbon
dioxide produced by
respiration.
Tube
A
B
C
Colour of indicator
yellow
purple
red
Level of CO₂ in the water
high
low
normal
Cause
Respiration
Photosynthesis
Respiration and Photosynthesis
A useful demonstration uses four tubes containing hydrogen carbonate solution: one with water plant
only, one with animals only, one with both water plant and animals and one with no living organisms.
One set of the tubes is exposed to light and left for 12 to 24 hours and another set is placed in the dark
for the same length of time.
17
Experiments to investigate the role of environmental factors in determining the rate of
transpiration from a leafy shoot
A bubble potometer can be used to illustrate the effects of light, wind,
temperature and air humidity. Plants covered with dark polythene bags
simulate darkness and can be compared with plants covered with
transparent polythene bags. Hairdryers can simulate wind. The use of
potted plants is also acceptable, where the pot and the soil is sealed with
polythene and the mass of the potted plant is measured before and after
a period of exposure to the environmental factor.
Effect of different factors on transpiration rate:
An increase in rate will result from transferring the shoot from darkness to light, moving from shade to
sunlight, from still to moving air and removing the plastic bag.
A decrease will occur when the shoot is enclosed in a plastic bag and on any reduction of light intensity.
The stomata probably respond to increased light intensity by opening more widely so allowing more rapid
diffusion. Direct sunlight will also warm up the leaves and increase the rate of evaporation. Inside the
plastic bag, the humidity will rise to 100% saturation and so prevent the establishment of a diffusion
gradient for water vapour from the leaf to the atmosphere. In moving air, the water vapour is carried
away from the leaf so maintaining a steep diffusion gradient.
Changes in light intensity and diffusion gradients may affect the rate of photosynthesis. Changes in
photosynthetic rate are, however, unlikely to produce such large changes in water uptake.
The most dramatic increase in transpiration will be seen when the shoot is moved from shade to sunlight.
The rate may increase to 10 times that in the shade of the laboratory. It is difficult to separate the effects
of temperature and light in these circumstances.
On moving from the laboratory to outside there will be changes in light intensity, temperature, air
movement and perhaps humidity.
The inside of the plastic bag may become misted with condensed water. The significance is:
(a) that the water must have come from the plant
(b) that the air in the bag must be saturated with water vapour.
An increase in light intensity will increase the rate of photosynthesis if other factors are not limiting.
An increased rate of photosynthesis will need more water, which may be reflected in the rate of uptake.
The distinction between uptake and loss is made because not all the water taken up is necessarily
transpired or, alternatively, evaporation may be taking place faster than uptake in some circumstances.
It is assumed that the bulk of the water taken up is evaporated and that the changing conditions affect
evaporation far more than any other process in the leaf.
18
A simple experiment to investigate the effect of exercise on breathing in humans
The breathing rate can be measured at rest and after a period of exercise by counting the number of
inhalations per minute.
Exercise also influences the rate of breathing by increasing the volume of each breath, this can be
measured by taking the volume of one exhalation before and after exercise.
This can be done by breathing through a tube into a plastic container filled with water. The volume of
displaced water can be measured. The breathing rate at rest and after exercise can be calculated as
number of breaths per minute x volume of each breath.
Volume of each
breath before
exercise
Volume of each
breath during exercise
19
A simple experiment to show how the sensitivity of the skin differs on finger tips, back
of hand, wrist and forearm
Students should work in pairs. A piece of hard cardboard or cork can be used to fix the two prongs of a
hairpin or two pins 5 mm apart. This is then used by one
student to lightly touch the fingertips of another who is
looking away. The first student can use both points or one
point as a stimulus. The second student then has to judge
whether one or two points were used and their response
recorded as correct or incorrect. This can be repeated 10
times for each area of the hand. It is then repeated using
pins 1 cm apart and 2 cm apart.
Students can then identify the most sensitive area as this
should have the most correct responses with the smaller
distance. Conclusions can be made about the number of
sensory nerve endings, receptive field size and the
thickness of skin. This practical also provides
opportunities to discuss data analysis, experiment design
and anomalous results, and the benefits of grouping class
results.
Finger-tips will probably recognize the two points even when they are only about 2 mm apart,
i.e. the lower limit of this instrument. The back of the hand will probably need a separation of l0 mm to
achieve a score of 8-10. Other areas will probably be even less sensitive.
Possibilities:
(a) One point does not touch a receptor. It seems unlikely that, on the back of the hand, a pressure
actually indenting the skin will not affect at least one receptor. Moreover, recognition of two stimuli
when they are separated by half a second suggests that it is not shortage of receptors alone which causes
poor discrimination.
(b) Both points touch receptors which feed impulses into one nerve fibre. Since a single sensory fibre may
receive branches from sensory endings covering a region of several square millimetres this explanation is
quite plausible. The recognition of the non-simultaneous double stimulus could be explained by the
additional burst of impulses initiated by a second, stimulus. It has been claimed that single hair follicle
receptors are sensitive over an area several centimetres in diameter but the 65 mm separation claimed
for the back could hardly be explained in this way.
(c) There are two receptors and two sensory fibres but unless the impulses are separated in time, they
cannot be discriminated by the brain.
(a) Only one receptor
affected
(b) Both receptors
feed impulses into the
same fibre
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(c) Only one fibre
but stimuli separated
in time
Controlled experiments to demonstrate phototropic and geotropic plant growth
responses
Plant material such as wheat, maize, oat or cress seedlings can be used to demonstrate phototropism.
Petri dishes containing moist cotton wool and the plant material can be put into light-proof boxes such as
shoe boxes. To create unilateral light a small slit can be cut in one side of the box and light can be shone
into the box. Control seedlings can either have aluminium caps put on their tips or can be kept in a shoe
box without a slit for light. A clinostat needs to be used to demonstrate geotropism.
To test the response to or effect of light on a plant shoot.
A growing herbaceous stem is placed in a box that has a small opening on one side only, where light
enters.
Auxins are plant hormones that make some parts of a plant stem grow faster than others. The result is
that the plant stem bends towards the light.
You may have noticed that a houseplant grows towards the window and turns its leaves towards the
light. It does this because light coming from the window side of the plant destroys the auxin in that side
of the stem. So growth on that side slows down.
On the shaded side of the plant there is more auxin. So growth on this side speeds up. The result is that
the shoots and leaves are turned towards the light for photosynthesis.
Use of a clinostat to show geotropism in seedlings
21
A few beans are soaked in water overnight then placed on wet cotton wool for a few days until the first
root of each seed (the radicle) grows to a length of about 2cm. Wet cotton wool is attached to the cork
disc of two clinostats. Three or four of the germinating bean
seeds are pinned onto each of the discs with their radicles
pointing outwards. Covers are placed on them to keep the
air moist. The clinostats are turned on their sides. One is left
switched on, the other is switched off to act as the control
(to allow you to compare your results) Both clinostats are
left set up for a few days. The cotton wool is kept damp.
After this time the radicles of the beans on the control
clinostat will have grown downwards, whereas those on
the beans on the rotating clinostat will be growing straight
out horizontally.
Explanation: Gravity would have been 'pulling' consistently
at 90° to the horizontal radicles in the stationary box.
In the clinostat the gravitational force also acts at 90° but on all sides of the radicle in turn.
22
The use of quadrats to estimate the population size of an organism in two different
areas
Quadrats can be used to sample part of each area. Calculation will be needed to work out the estimated
population size. For example, if 10 quadrats have been used and the total area amounts to 100 quadrats,
the estimated population size will be the number of organisms counted in the 10 sample quadrats
multiplied by 10.
It is important to place the sample quadrats randomly – this avoids bias and improves the reliability of the
data.
An interesting way to practise the technique is to throw plastic beads on the floor of the classroom and
ask students to guess how many beads there are. The quadrat sampling procedure can be used in front of
students to get an estimate. The beads can then be collected and counted.
The actual number can be compared to the estimated number and used to see how accurate the
estimation was.
A quadrat is a square (of either metal,
wood, or plastic) used in ecology and
geography to isolate a sample, usually
about 1m2 or 0.25m2. The quadrat is
suitable for sampling plants, slowmoving animals (such as millipedes
and insects), and some aquatic
organisms.
When an ecologist wants to know
how many organisms there are in a
particular habitat, it would not be
feasible to count them all. Instead, he
or she would be forced to count a
smaller representative part of the
population, called a sample. Sampling
of plants or animals that do not move
much (such as snails), can be done
using a sampling square called a
quadrat. A suitable size of a quadrat
depends on the size of the organisms
being sampled. For example, to count
plants growing on a school field, one
could use a quadrat with sides 0.5 or
1 metre in length.
It is important that sampling in an area is carried out at random, to avoid bias. For example, if one were
sampling from a school field, but for convenience only placed quadrats next to a path, this might not give
a sample that was representative of the whole field. It would be an unrepresentative, or biased, sample.
One way one can sample randomly is to place the quadrats at coordinates on a numbered grid.
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