8/20/2019 PRACTI~1 (2)

1/28

1

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 A simple experiment to investigate the effect of exercise on breathing in humans

9 Experiments to investigate the role of environmental factors in determining the rate of transpiration

from a leafy shoot

10 A simple experiment to investigate the effect of exercise on heart rate in humans

11 A simple experiment to show how the sensitivity of the skin differs on finger tips, back of hand, wrist

and forearm

12 A practical exercise comparing floral structure in insect-pollinated and wind-pollinated flowers

13 Controlled experiments to demonstrate phototropic and geotropic plant growth responses

14 The use of quadrats to estimate the population size of an organism in two different areas

15 A simple experiment to investigate carbon dioxide production by yeast in different conditions.

8/20/2019 PRACTI~1 (2)

2/28

8/20/2019 PRACTI~1 (2)

3/28

3

2. Controlled experiments to illustrate how enzyme activity can be affected by changesin temperature

1. The effect of amylase on starch digestionAt each temperature selected, from 0°C to 100°C, samples of amylase solution and of starch solution arebrought 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 andadded to individual iodine drops on a white tile. Iodine tests for the presence of starch – iodine goes blackwhen 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 becontrolled using water baths.

The digestive enzyme amylase breaks down starch into the sugar maltose. If the speed at which thestarch 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 andoxygen. The rate of oxygen production can be measured as an indication of enzyme activity. An upturnedboiling tube previously filled with water can be used to collect and measure the volume of oxygenevolved. Or the time take for ‘bubbling’ to stop can be measured. The liver/potato can be exposed todifferent 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 ourbody 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 notliving things, they cannot be killed .

Graph showing the effect of temperature on enzyme reactions

8/20/2019 PRACTI~1 (2)

4/28

8/20/2019 PRACTI~1 (2)

5/28

5

3. Simple experiments on diffusion and osmosis using living and non-livingsystems

a. Cubes of agar jelly placed into solutions of methylene blue or potassium permanganate will absorb thepigment by diffusion. The cubes are left in the pigmented solution for different measured periods of timeand 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 jellyAgar 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 thepermanganate disappears.A Petri dish is prepared which contains a 2 cm deep layer of agar jelly, dyed purple with potassiumpermanganate. Three cubes of different sizes are cut out of the jelly, with side lengths 2 cm, 1 cm and 0.5cm.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 (theshorter 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.

8/20/2019 PRACTI~1 (2)

6/28

6

8/20/2019 PRACTI~1 (2)

7/28

7

c. To demonstrate osmosis, Visking tubing (dialysis tubing) can be tied at one end and filled with 20 percent sucrose solution. The other end is attached to a capillary tube. The level of the sucrose can be notedbefore and after the tubing has been placed in a beaker of water for about 30 minutes.Visking tubing hasmicroscopic holes in it, which let small molecules like water pass through (it is permeable to them) but isnot 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. Thewater molecules can pass through the membrane in either direction, but those on the right are attractedto the sugar molecules. This slows them down and means that they are less free to move – they have lesskinetic 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 casepure water) to the more concentrated solution.

8/20/2019 PRACTI~1 (2)

8/28

8

d. Onion epidermis can be peeled away, cut into squares and mounted on slides in differentconcentrations 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 firmFlaccid - Plant cells in a hypertonic solution, the cell shrinks away from the cell wall and the cell is limp.

8/20/2019 PRACTI~1 (2)

9/28

9

e. Red blood cells in blood obtained from a butcher may be mounted on slides in hypotonic, isotonic andhypertonic 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 intosolutions with different concentrations, the blood cells will gain or lose water by osmosis. This can bedemonstrated 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 examinedunder 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 theleft(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 damagedby the effects of osmosis. The red blood cells placed in water absorb the water by osmosis, swell up andburst, leaving a red solution of haemoglobin in the test tube. When placed in 3% salt solution, the redblood 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 solidparticles from solution.

Tube A contains a clear red solution and no solid material at the bottom of the tube = the cells have burstdue to water entering via osmosis and haemoglobin is released into the waterTubes 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.

8/20/2019 PRACTI~1 (2)

10/28

10

f. Osmosis can be demonstrated by using strips of potato, and this basic experimental method provides agood 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 partof the course so that students will have had previous experience of carrying out investigations on simplertopics. Students enjoy the reference to ‘chips’, but should quickly realise that it can be difficult to keepthe size constant – to achieve consistency lengths of potato tissue can be drilled from a potato using acork 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 touse to investigate the effects of osmosis on the mass of thetissue.A boiling tube is half-filled with tap water and a second withconcentrated (Molar) sucrose solution. A third tube is left empty.A potato is ‘chipped’ – cut into chips 5cm x 1cm x 1cm, makingthese measurements as accurate as possible to ensure the chipshave the same surface area. No skin is left on the potato (it wouldprovide a waterproof layer).Each chip is gently blotted to remove excess moisture andweighed to find the starting mass of each. One chip is placed into each of the three tubes, and after afixed time the chips are removed, gently blotted to remove excess moisture and reweighed. The chipscan 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 takesinto account the difference in starting masses of the chip so they can be compared).

% change = change in mass x 100Starting mass

Boiling tube with water - water moves into the cellsThe water in the tube is more concentrated than the water in the cells, so water molecules will movefrom an area of higher concentration of water (in the tube) to an area of lower concentration of water (inthe potato cells).The cell membrane is semi-permeable so the water molecules can get through and each cell will swell asthe water flows in.As all the cells swell, the potato chip will increase in mass, and length/width/breadth and become morerigid.

Boiling tube with concentrated sugar solution - water moves out of the cellsThe water in the tube is a lower concentration than the water in the cells, so water molecules will movefrom 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 asthe water flows out.As all the cells shrink, the potato chip will decrease in mass, and length/width/breadth and become moreflexible/floppy.

Boiling tube with nothing - some water moves out of the cellsWater molecules will move from an area of higher concentration of water (in the potato cells) to an areaof 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 asthe water flows out.As all the cells shrink, the potato chip will decrease in mass, and length/width/breadth and become moreflexible/floppy. The effect will be less than in the tube with concentrated sugar solution.

8/20/2019 PRACTI~1 (2)

11/28

g. A variation on this theme is tovolume ratios. After measuringone hour, the cubes are blottedincrease in mass for cubes of difconcept of how surface area to

The surface area over which theThe larger the surface area betof each solution will be in contameans that a larger amount of tat the same time therefore maki

Eight 1cm³ cubes

11

cut potato cubes of different sizes, which havend recording the masses of the cubes, they aredry and their masses measured and recorded agferent surface area to mass ratio can be comparolume ratio influences water uptake.

osmosis occurs can change the speed at which ieen the two solutions the faster osmosis will hat with the membrane at one moment. This

he free water molecules in the solution can moving the process happen much faster.

One 8cm³ cube

different surface area to immersed in water. After

ain. The percentage ed, to explore the

t happens dramatically. ppen. This is due to more

e across the membrane

8/20/2019 PRACTI~1 (2)

12/28

12

4. Controlled experiments to investigate photosynthesis, showing the evolution of oxygen from a water plant, the production of starch by leaves and the requirements forlight, carbon dioxide and chlorophyll

a. The evolution of oxygen from a water plant can be seen if a water plant (typically elodea or a similarspecies) is placed in a beaker of water and covered with a glass funnel that has a water-filled test tubeplaced over its opening. After 24 hours, a colourless gas will have displaced water from the test tube. Atest for oxygen is then carried out.

Re-light a glowingsplint

8/20/2019 PRACTI~1 (2)

13/28

13

b. To measure the rate of oxygen production, the stem of a water plant is cut under water, andthe plantkept immersed in water in a beaker or boiling tube. The number of bubbles of gasgiven off over ameasured time period can be counted. This simple experimental set up canenable students to carry outindividual investigations into the effect of different factors onthe rate of bubble production. Suitablevariables include: light intensity (the plant is exposedto a light source and the rate of bubble productionmeasured at different light intensities bychanging the distance between the light source and the waterplant); colour/wavelength oflight (coloured filters are placed between the plant and the light source); and

carbon dioxideavailability (the plant is immersed in solutions of different concentration of sodiumhydrogen 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 factorcomes into operation, the two should be proportional.Increasing the light intensity makes available a greater amount of energy to drive the photosyntheticreaction. Since oxygen is a product of photosynthesis, a higher rate of photosynthesis should result in ahigher rate of oxygen production.Unless the gas produced by Elodea is pure oxygen it would be possible for the plant to double its oxygenoutput 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 couldbe 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 thewhole 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 intensitysince carbon dioxide would become limiting or internal factors such as the available chloroplasts or thediffusion 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 intemperature of the water in the beaker. Since a rise in temperature speeds up many chemical reactions itmight also accelerate some stages of photosynthesis. (In practice, the temperature is unlikely to rise morethan 1° C.)

8/20/2019 PRACTI~1 (2)

14/28

14

c. Starch production can be investigated by placing a plant in the dark for 24 hours to de-starch theleaves. 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 hoursHalf-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 readilyallow 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 concentrationPlace 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 testfor starch, so the leaf needs to be ' decolourized' for changes to be observed. A decolourized leaf ispale 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.

8/20/2019 PRACTI~1 (2)

15/28

15

d. A starch test on a variegated leaf can be used to demonstrate that chlorophyll is needed forphotosynthesis.

• Place a plant with variegated leaves in the dark for at least 24 hours to make sure that there is nostarch in the leaves i.e. ‘de-starching the leaves’.

• Variegated Leaves: have green and white areas – the green areas have chlorophyll but chlorophyll isabsent 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 airwith no carbon dioxide by inserting it into a conical flask containing a small amount of potassium orsodium hydroxide. The plant is left in good light for 24 hours.

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 chlorophyllmolecules 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.

NaOH removesCO₂

8/20/2019 PRACTI~1 (2)

16/28

16

5. 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 themass 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 andignited (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 watertemperature in the boiling tube is re-measured.• The equation used to calculate the energy content of thefood 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, butunless you use a bomb calorimeter this won’t happen.

NB: The energy gets lost to the surroundings as not all energy is direct to heat the water, also there is aneed to heat the glass apparatus first.

Bomb calorimeter

8/20/2019 PRACTI~1 (2)

17/28

6. Controlled experimentsfrom respiring seeds or otVacuum flasks are needed for thwith a bung. To show heat prodthrough the bung into the flask.Students could demonstrate thawater or hydrogen carbonate in

The temperature will increase sienergy the seeds need to germi

Living seeds

17

to demonstrate the evolution of carboner suitable living organisms

is activity. Surface sterilised seeds are put into action the flask needs a cotton wool bung with a

t they produce carbon dioxide by breathing outdicator.

gnificantly in the flask containing the live seeds.ate and the by-product of the use of the energy

thermome

A B

dioxide and heat

flask, which is sealed thermometer going

through a tube into lime

Respiration produces the is heat (energy).

ter

cotton wool

Dead seeds

8/20/2019 PRACTI~1 (2)

18/28

A glass tube runs from inside thor hydrogen carbonate. The car

The lime water associated with lliving organisms should be clear.Carbon dioxide is the gas whichThe results suggest that carbon

The air from non-living organissufficient concentration from livorganisms rather than the air aNB: The experiment provides evThe failure of dead organisms tofrom a process that occurs in livi

Draw gas off fromrespiring seeds

18

flask, through the bung and into an indicator son dioxide produces changes the colour of the i

iving organisms should be milky. The lime water.

turns lime water milky. dioxide is given out by the organisms.

s failed to turn lime water milky. Thus the carboing organisms to turn lime water milky so mustound them

idence for only the organisms in the test-tubes. turn lime water milky shows that the productio

ing organisms but not in dead ones.

Bubble gas through lim – if it goes cloudy thenpresent

lution of either limewater ndicator solution.

receiving air from non-

n dioxide which was in ave come from the living

n of carbon dioxide results

ewater CO₂ is

8/20/2019 PRACTI~1 (2)

19/28

19

7. Simple controlled experiments to investigate the effect of light on net gas exchangefrom a leaf, using hydrogen carbonate indicatorA water plant can be placed in a sealed tube of air-equilibrated hydrogen carbonate solution (red incolour) and placed in the lightor in the dark. The solution willturn purple if kept in the lightand will turn yellow if kept in

the dark.A variation could involve theuse of water snails or, if notavailable, small land insectsplaced on a gauze platformabove the indicator, with andwithout the water plant. Thisvariation allows students tothink about the balancebetween carbon dioxide usedby photosynthesis and carbon

dioxide produced byrespiration.

Tube Colour of indicator Level of CO ₂ in the water CauseA yellow high RespirationB purple low PhotosynthesisC red normal 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 darkfor the same length of time.

8/20/2019 PRACTI~1 (2)

20/28

20

8. A simple experiment to investigate the effect of exercise on breathing in humansThe 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 bemeasured 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 eachbreath beforeexercise

Volume of eachbreath during exercise

8/20/2019 PRACTI~1 (2)

21/28

21

9. Experiments to investigate the role of environmental factors in determining the rateof 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 bagssimulate darkness and can be compared with plants covered withtransparent polythene bags. Hairdryers can simulate wind. The use of potted plants is also acceptable, where the pot and the soil is sealed withpolythene and the mass of the potted plant is measured before and aftera 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 tosunlight, 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 rapiddiffusion. Direct sunlight will also warm up the leaves and increase the rate of evaporation. Inside theplastic bag, the humidity will rise to 100% saturation and so prevent the establishment of a diffusiongradient for water vapour from the leaf to the atmosphere. In moving air, the water vapour is carriedaway from the leaf so maintaining a steep diffusion gradient.Changes in light intensity and diffusion gradients may affect the rate of photosynthesis. Changes inphotosynthetic 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 effectsof temperature and light in these circumstances.On moving from the laboratory to outside there will be changes in light intensity, temperature, airmovement 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 necessarilytranspired 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 affectevaporation far more than any other process in the leaf.

8/20/2019 PRACTI~1 (2)

22/28

10. A simple experiment tStudents should be shown howA short period of exercise can foand down a flight of stairs for fivrequired.)The pulse rate should now be relevel. This experiment can provi

resting heart rate, the effects ofDiscussion can include measureeffects of long-term exercise on

The m

To get ri

Your hea

Your a

Blood is diverted away from

When the m

The heart c

22

investigate the effect of exercise on heto measure and record their pulse. They can mellow, stepping on and off a low stool, or runninge minutes. (Safety note: careful supervision is

corded for five minutes or until it returns to its r de a great deal of interesting data on the variati

exercise and how quickly the heart recovers. of fitness, heart disease, cardiac output and th stroke volume.

scles start to produce more CO2 as they work harder

of the extra CO2 you start to breathe deeper and fast

rt rate increases to pump more oxygen around the bo

terioles widen to stop your blood pressure increasing

inactive organs (stomach / liver) and towards the worvasoconstriction and vasodilation

scles contract they squeeze blood back towards your

ontracts even stronger to pump more blood with each

art rate in humans sure this at rest.

up

est n in

er

y

ing muscles through

heart

beat

8/20/2019 PRACTI~1 (2)

23/28

11. A simple experiment tback of hand, wrist and foStudents should work in pairs. Ahairpin or two pins 5 mm apart.student to lightly touch the finglooking away. The first studentpoint as a stimulus. The second

whether one or two points werrecorded as correct or incorrect.times for each area of the hand.pins 1 cm apart and 2 cm apart.Students can then identify theshould have the most correct redistance. Conclusions can be masensory nerve endings, receptivthickness of skin. This practical aopportunities to discuss data anand anomalous results, and the

results.

Finger-tips will probably recognii.e. the lower limit of this instruachieve a score of 8-10. Other aPossibilities:(a) One point does not touch a ractually indenting the skin will nwhen they are separated by halfpoor discrimination.

(b) Both points touch receptorsreceive branches from sensoryquite plausible. The recognitionadditional burst of impulses initireceptors are sensitive over anfor the back could hardly be exp(c) There are two receptors andcannot be discriminated by the

(a) Only one receptor affected

23

show how the sensitivity of the skin difearm

piece of hard cardboard or cork can be used toThis is then used by one

rtips of another who is an use both points or one

student then has to judge

used and their response . This can be repeated 10

It is then repeated using

ost sensitive area as this sponses with the smaller

de about the number of field size and the

lso provides alysis, experiment design

benefits of grouping class

ze the two points even when they are only abouent. The back of the hand will probably need a

reas will probably be even less sensitive.

eceptor. It seems unlikely that, on the back of thot affect at least one receptor. Moreover, recoga second suggests that it is not shortage of rece

which feed impulses into one nerve fibre. Sincendings covering a region of several square milliof the non-simultaneous double stimulus couldated by a second, stimulus. It has been claimedrea several centimetres in diameter but the 65lained in this way.

two sensory fibres but unless the impulses are srain.

(c) Only one fstimuli separatime

(b) Both receptorsfeed impulses intothesame fibre

fers on finger tips,

ix the two prongs of a

t 2 mm apart, separation of l0 mm to

e hand, a pressure nition of two stimuli

ptors alone which causes

single sensory fibre may etres this explanation is

e explained by the hat single hair follicle m separation claimed

eparated in time, they

ibre but ted in

8/20/2019 PRACTI~1 (2)

24/28

24

12. A practical exercise comparing floral structure in insect-pollinated and wind-pollinated flowersMost areas should have access to suitable specimens. Insect-pollinated flowers can be examined and thevarious structures observed. Wind-pollinated grasses should be available but have fewer structures tosee.

Insect Pollinated Wind Pollinated

large, brightly coloured petals - toattract insects

small petals, often brown or dull green - no needto attract insects

often sweetly scented - to attractinsects

no scent - no need to attract insects

usually contain nectar - to attractinsects

no nectar - no need to attract insects

moderate quantity of pollen - lesswastage than with wind pollination pollen produced in great quantities - becausemost does not reach another flower

pollen often sticky or spiky - to stick toinsects

pollen very light and smooth - so it can be blownin the wind and stops it clumping together

anthers firm and inside flower - tobrush against insects

anthers loosely attached and dangle out - torelease pollen into the wind

stigma inside the flower - so that theinsect brushes against it

stigma hangs outside the flower - to catch thedrifting pollen

stigma has sticky coating - pollensticks to it

stigma feathery or net like - to catch the driftingpollen

Insect pollinated flowers Wind pollinated flowersrose sweet pea ragweed

8/20/2019 PRACTI~1 (2)

25/28

25

13. Controlled experiments to demonstrate phototropic and geotropic plant growthresponsesPlant 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 asshoe boxes. To create unilateral light a small slit can be cut in one side of the box and light can be shoneinto the box. Control seedlings can either have aluminium caps put on their tips or can be kept in a shoebox 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 lightenters.

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

8/20/2019 PRACTI~1 (2)

26/28

26

A few beans are soaked in water overnight then placed on wet cotton wool for a few days until the firstroot of each seed (the radicle) grows to a length of about 2cm. Wet cotton wool is attached to the corkdisc of two clinostats. Three or four of the germinating beanseeds are pinned onto each of the discs with their radiclespointing outwards. Covers are placed on them to keep theair moist. The clinostats are turned on their sides. One is leftswitched on, the other is switched off to act as the control(to allow you to compare your results) Both clinostats areleft set up for a few days. The cotton wool is kept damp.After this time the radicles of the beans on the controlclinostat will have grown downwards, whereas those onthe beans on the rotating clinostat will be growing straightout horizontally.

Explanation: Gravity would have been 'pulling' consistentlyat 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.

8/20/2019 PRACTI~1 (2)

27/28

27

14. The use of quadrats to estimate the population size of an organism in two differentareasQuadrats can be used to sample part of each area. Calculation will be needed to work out the estimatedpopulation 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 quadratsmultiplied by 10.It isimportant 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 andask 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 theestimation was.

A quadrat is a square (of either metal,wood, or plastic) used in ecology andgeography to isolate a sample, usuallyabout 1m 2 or 0.25m 2. The quadrat is

suitable for samplingplants, slow-moving animals (such as millipedesand insects), and some aquaticorganisms.When an ecologist wants to knowhow many organisms there are in aparticular habitat, it would not befeasible to count them all. Instead, heor she would be forced to count asmaller representative part of thepopulation, called a sample. Sampling

of plants or animals that do not movemuch (such as snails), can be doneusing a sampling square called aquadrat. A suitable size of a quadratdepends on the size of the organismsbeing sampled. For example, to countplants growing on a school field, onecould use a quadrat with sides 0.5 or1 metre in length.It is important that sampling in an area is carried out at random, to avoid bias. For example, if one weresampling 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.

8/20/2019 PRACTI~1 (2)

28/28

15. A simple experiment tconditionsThis allows students to see thatStudents add yeast to glucose sputting a drop of oil (cooking oilthe side arm of the test tube anpipette is placed under water to

the easiest condition to investig

The lime water associated withclear. Tube A will show signs ofchange.Lime water going milky is evide

down of carbohydrates to carbothat respiration is taking place.If a simple chemical reaction beproduced in tube B as well as inby temperatures as low as 100 °Air should be bubbled through libubbles) as in the experiment tsame time interval as in the expThe experiment was designed texpel the dissolved gases, includthe start of the experiment.

The glucose is the substrate fordown to alcohol and carbon dio

3-way tap

yeast andglucose A

lime water

capillarytube

investigate carbon dioxide production

carbon dioxide is released during anaerobic resplution in a side-arm test tube. Anaerobic conditiwill do) onto the yeast and glucose mixture. A ra glass pipette is inserted at the other end of t

allow the bubbles of carbon dioxide gas to be c

ate. Glucose concentration and pH could also be

ube A should go milky. Lime water connected toermentation, bubbles of gas and frothing. Tube

ce of the presence of carbon dioxide. Respiratio

n dioxide. The production of carbon dioxide thu

ween the yeast and glucose was producing carbtube A, assuming that the chemicals in yeast haC.

me water, using a pump or filter pump, at the sdiscount the idea that CO₂ in the air is causing t

eriment the lime water remains clear, the argumshow that it was due to anaerobic respiration b

ing oxygen. The layer of oil prevented air re-diss

respiration. The yeast obtains energy from the glide.

dead yeast andglucose B

y yeast in different

iration (fermentation). ons are achieved by

bber tube is attached to e rubber tube. The

unted. Temperature is

investigated.

tube B should remain B should show little or no

n involves the breaking

supports the contention on dioxide it should be

not been decomposed

me rate (counting the he change. If after the

ent is refuted. y boiling the water to

olving in the solution at

lucose by breaking it

water bath

thin layer of oil

tap open

tap closed