De-starching the Leaves of a Plant

Contents
Chapter one – Diffusion
1. Diffusion in gases
2. Diffusion in liquid
3. Diffusion and size
4. Diffusion through a membrane
5. Control of diffusion
6. Two-way diffusion

Chapter Two – Enzymes
1. Effect of amylase on starch
2. Effect of temperature
3. Hydrolysis of starch
4. Saliva on starch
5. Catalase
6. Effect of enzyme concentration
7. Pepsin on egg white
8. Action of lipase
9. Pepsin and PH
10. The Reaction of Starch Phosphorylase
11. Dehydrogenase in Yeast


         Chapter Three – Food Tests
1. The Test for Starch
2. The Test for Glucose
3. The Test for Protein
4. The Test for Fats
5. How Sensitive is the Iodine Test?
6. Testing Food Sample for Starch
7. Comparison of Vitamin C Content
8. How specific is Benedicts’ Reagent?
9. The test for sucrose
10. Testing food sample for the presence of sugar (other than sucrose)
11. Testing food sample for protein

12. Testing food sample for fats (emulsion test)
13. Comparison of energy value of food


              Chapter Four – Osmosis

1. Osmosis
2. Selective permeability
3. Turgor
4. Turgor in plant tissue
5. Turgor in potato tissue
6. Stomatal movements
7. Plasmolysis
8. Surface area and osmosis

 
               Chapter Five – Photosynthesis
     
1. Production of gas by pondweed
2. Testing a leaf for starch
3. The need for light
4. The need for chlorophyll
5. The need for carbon dioxide
6. Collecting the gas from pondweed
7. Gaseous exchange in pondweed
8. The need for mineral elements
9. Condition affecting production of oxygen
10. Effect of light on gas production
11. Effect of carbon dioxide on gas production
12. Effect of temperature on gas production
13. Starch production in darkness
14. Carbohydrate production in iris leaves
15. Need for carbon dioxide, using iris leaves
16. Chromatographic analysis of chlorophyll
       
 

                   Chapter six – Respiration

1. Oxygen uptake
2. Carbon dioxide output
3. Exhaled air (1)
4. Exhaled air (2)
5. Respiration in living organisms
6. Anaerobic respiration
7. Energy release during respiration
8. Changes in mass during germination
9. Measuring the uptake of oxygen
10. Temperature effect on respiration
11. Oxygen uptake in blowfly larva
12. The effect of temperature on fermentation rate
13. Production of combustion
14. Vital capacity
15. Muscle contraction


 
                        Chapter seven – Human Senses
                             
1. Reaction time
2. The blind spot (1) and (2)
3. Inversion of the image
4. The iris diaphragm (1) and (2)
5. Retinal capillaries
6. Binocular vision : eye dominance
7. Binocular vision : double vision
8. Judgement of distance
9. Eye and hand coordination
10. Perception
11. Sensitivity of the skin to touch
12. Recognition of separate stimuli
13. Sensitivity to temperature
14. Location of stimuli



                             Chapter Eight – Transport in Plants

1. Uptake and evaporation in leaves
2. Uptake of water by shoots
3. Rate of transpiration and water uptake
4. Uptake of water by an uprooted plant
5. Condition affecting evaporation
6. Water tension in the stem
7. Pathways for gases in a leaf
8. Evaporation from the leaf surface
9. To collect and identify the product of transportation
10. To trace the path of water through a shoot
11. Measuring the transpiration rate of a potted plant












                                 Chapter Nine – Germination and Tropisms

                                 
1. The need for oxygen
2. Effect of temperature
3. The need for water
4. The role of cotyledons
5. Use of food reserves in germination
6. Geotropism in radicles
7. The region of growth and response in radicles
8. Region of detection and response to one-sided gravity in radicles
9. The effect of one-sided lighting on shoots
10. The effect of one-sided lighting on cress seedlings
11. The region of detection and response to one-sided lighting in coleoptiles
12. The effect of indole- acetic acid on wheat coleoptiles
13. Effect of indole-acetic acid on maize coleoptiles
14. The effect of light on shoots















Chapter one – Diffusion
Experiment 1. Diffusion in gases

You are provided with two glass tubes equal in length and diameter and marked at 2 cm intervals.

(a) Copy the table given below into your notebook and collect a little cold water in a beaker or other container .

(b) Using forceps, pick up a square of red litmus paper, dip it in the water, shake off surplus water and place the paper inside one of the glass tubes. Use the wire or glass rod to manipulate the litmus paper until it is stuck to the glass immediately under the 10 cm mark (see Fig. 2,  p.1.02). Repeat this operation for the remaining marks in both tubes, working from the end of the tube nearer the marks.

(c) Close the 28 cm end of both tubes with the ordinary cork bungs.
The next operation involves ammonia solutions, one of which is very strong and gives off a pungent vapour. It is harmless enough provided you do not deliberately sniff it at close range.

(d) Take the corks with cotton wool plugs to the central dispensing point in the laboratory and use the dropping pipette to place about 20 drops of strong (9N) ammonia solution on the cotton wool in cork A and an equal number of drops of dilute (2N) ammonia solution on the cotton wool in cork B.

(e) Note the time and insert each cork in the appropriate tube at the same time.

(f) In your table, note the time interval required for each square of litmus to turn completely blue and continue recording until the litmus at 28 cm in one of the tubes has turned blue.

(g) Make a graph of distance against time, plotting the values for both tubes on the same graph, with time on the horizontal scale.


Time
started

........... Distance along tube in cm 10 12 14 16 18 20 22 24 26 28
A Strong ammonia solution
Number of minutes from start
B Weak ammonia solution
Number of minutes from start






Experiment 1. Discussion

In order to turn the litmus blue, molecules of ammonia must have travelled from the cotton wool at one end of the tube to the litmus paper at the other end. This process of molecular movement is called DIFFUSION.

I Why is it unlikely that (a) air currents through the tube or (b) convection currents inside the tube could have distributed the ammonia ?

2 What influence on the rate of diffusion does the concentration of the source have in
this case?








Experiment 1. Discussion –answers

1 The corks should effectively prevent any through currents of air, and unless one part of the tube is warmer than another, convection currents should not occur.

2 Ammonia from the more concentrated source will diffuse more rapidly than that from the weak source. This is particularly noticeable in the first 5 minutes. In the final 5 minutes, the rates may be very similar








                                               








Experiment 1. Diffusion in gases – preparation

Outline:  Ammonia diffuses along a glass tube turning litmus paper blue as it does so. Rates of diffusion are compared for strong and weak ammonia solutions.

Prior knowledge : Ammonia turns red litmus blue. Meaning of 'convection'. Meaning of 'molecules'. Precautions in handling ammonia.

Advance preparation and materials:

Tubes. Glass tubes, 20 mm diameter or more, 30 cm long and marked with a spirit marker at 2 cm intervals starting at 10 cm. Allow 2 tubes per group.
To cut the tubes, score a line with a glass cutter round the tube and heat this line with a 25 cm length of 26 swg (0.45 mm) Nichrome wire connected to the 12 V supply
(Fig. I, p.1.03). Flame polish the cut ends.
Pusher. A piece of wire, bicycle spoke, knitting needle or glass rod about 25 cm long for
arranging the litmus squares inside the tube. Allow one per group.
Corks. Use a No.6 borer to make a hole through the cork. Eject the plug of cork from the
borer, cut 10 mm from it and replace the rest of the plug in the hole in the cork leaving a
cavity at the narrow end (Fig. 2). Plug this cavity tightly with cotton wool. Allow 2 of these
corks and 2 ordinary corks of the same size per group. Label the bored corks A and B in equal numbers.

Litmus paper. Cut strips of red litmus paper into squares (6 or 7 per strip). Allow 20 squares per group.
Ammonium hydroxide
(9N) Dilute 0.880 ammonia solution with its own volume of water
(2N) Dilute 11cm3 0.880 ammonia solution with 89 cm3 water. Allow 3 cm3 of each solution per group.
It is advisable to have a single dispensing point for at least the strong solution at the teacher's bench so that it can be supervised.

Apparatus-per group

beaker or small container for water
forceps
container for litmus paper squares
pusher, wire or glass
2 marked tubes, 30 cm long 
2 corks with cotton wool plugs
2 ordinary corks
dropping pipettes ( 4 or 5 at dispensing point)

-per class
1 clock or several seconds timers
NOTE. Using 20 mm diameter tubes and 9N ammonia solution, the litmus at 28 cm should go blue in 20 minutes.





Experiment 2. Diffusion in liquid


You are provided with a boiling tube containing 10% gelatin.

(a) Melt the gelatine in the tube by immersing it almost up to the rim in a beaker or jar of hot water. While waiting for it to melt continue with (b).

(b) Use a spirit marker to label three test-tubes A, B and C and make three marks on tubes A and B as shown in Fig. I.  Add your initials to these two tubes.

(c) Pour the liquid gelatin into tubes A and B up to mark 1 and allow it to set. Setting can be hastened by dipping the tube upright into cold water. Do not allow it to set with the surface oblique.

(d) Pour about 20 mm liquid gelatin into tube C and use a dropping pipette to add one drop of methylene blue solution.* Swirl the tube gently to mix the dye with the gelatin.
Make sure the gelatin in tube A has set and use a clean dropping pipette to transfer enough blue gelatin to fill the space between marks 1 and 2. Insert the pipette well down inside tube A so that blue gelatin does not touch the sides above mark 2. Use the pipette to draw off any air bubbles which form.

(e) Add another 9 drops of methylene blue solution to tube C and use the dropping pipette to transfer this blue gelatin to tube B in the same way as before. Allow the blue gelatin in tubes A and B to set firmly (about 5 minutes in cold water). While waiting, continue with instruction (g).

(f) When the blue gelatin has set, run a little cold water into both tubes to ensure that no liquid gelatin remains. Pour off the water and fill both tubes up to mark 3 with cool but liquid gelatin and cool it quickly. Cork both tubes (Fig. 2)
.
(g) In your notebook make diagrams, similar to Fig. I, of tubes A and B to show the position of the gelatin and the blue dye. Leave space for two more diagrams.

(h) After a week, examine the tubes again and make two more diagrams beside the first two to show the distribution and intensity of the blue colour. Answer discussion questions 1 and 4 while the tubes are still available for examination.

* TAKE CARE. Methylene blue temporarily stains the skin and permanently stains
   clothing. 

Experiment 2. Discussion

1. Did the diffusion of methylene blue take place equally upwards and downwards in the tube ? Give figures to support your answer.

2 In what direction would you expect diffusion to occur if a drop of methylene blue were surrounded on all sides by a large volume of gelatine ?

3 If you did experiment I, comment on the relative speeds of diffusion in air and in gelatine.

4 What difference was there in the rates of diffusion of methylene blue in tubes A and B, bearing in mind that the concentration in tube B was about ten times greater than in tube A ? Give measurements to support your answer .

NOTE. You may not regard gelatine as a true liquid but water would have been unsatisfactory for two main reasons: (a) it is difficult to start with a distinct boundary between two liquids which can mix and (b) convection currents can occur and so distribute the dye. The gelatine prevents the water from flowing so that the dye can
move only by diffusion.



                                                                                             



Experiment 2. Discussion - answers

I Diffusion of methylene blue should take place equally in both directions.

2. One would expect the dye to diffuse equally in all directions.

3 The methylene blue takes a week to diffuse about 30 mm whereas the ammonia diffused 30 cm in 20 minutes. It looks as if gaseous diffusion is a more rapid process than diffusion in liquids. On the other hand the students may attribute the difference to the different nature or concentration of the diffusing substances rather than the medium in which they diffuse.

4 The methylene blue from the more concentrated solution should travel further from the source and produce a deeper colour. The difference in distance, however, is not proportional to the difference in concentration and some students may claim there is no difference.
   


Experiment 2. Diffusion in liquid – preparation

Outline : Methylene blue of different concentrations is allowed to diffuse into clear gelatine for one week.

Prior knowledge : 10% gelatine is 90% water.

Advance preparation and materials

Gelatine:  Make a 10% solution by dissolving the crystals in tap-water and heating to boiling point. Keep the solution moving about or it will burn on the bottom. Stopper the solution while it is hot and thus reduce the chances of bacterial contamination and subsequent liquefaction.
Before the experiment, dispense the liquid gelatine into boiling tubes allowing about 60 cm3 per group.
Methylene blue: Make a 1% solution. Allow 1 cm3 per group.

Apparatus-per group

beaker or jar for warm water
2 corks
beaker or jar for cold water*
1 boiling tube (150 X 25 mm)
3 test-tubes (150 X 19 mm) and rack
spirit marker
container for methylene blue
ruler with millimetres
2 dropping pipettes

NOTE. If time is limited, the first layer of gelatine can be poured and allowed to set before the lesson. In this case, the instructions will begin at (d). Starting from (a) the experiment takes 30-40 minutes to set up.

* Unless the tap water is really cold (15 °C or less), it is advisable to have some ice available to help speed the gelation.

Experiment 3. Diffusion and size

The gelatine block contains cresol red, a pH indicator which is red in alkali and yellow in acid.

(a) Copy the table given below into your notebook.

(b) Place the gelatine block on a tile or Petri dish lid and use a scalpel or razor blade to cut it in half, producing two cubes of 10 mm side (1 cm cube).

(c) Keep one of these cubes intact and cut the other in half.

(d) Repeat this cutting operation as shown in the flow chart on p. 3.02 so that you end up with five blocks of the dimensions shown in the table.

(e) Fill a test-tube to within 10 mm of the, top with dilute hydrochloric acid.

(f) Note the time and, starting with the largest block, drop all the blocks into the acid in the test-tube and close it securely with a rubber bung or cork.

(g) Tilt the tube about to spread the gelatine blocks along its length. Hold the tube horizontally and rotate it so that you can see each block clearly and from all sides in turn.

Try not to warm the tube too much with your hands or the gelatine may dissolve.
Note the time taken for the acid to penetrate to the centre of the block as indicated by the disappearance of the red colour. Record this time in your table.


1 2 3 4 5
Cube size in mm 10 x 10 x 10 10 x 10 x 5 10 x 5 x 5 5 x 5 x 5 5 x 5 x 2.5
Time for acid to penetrate





                  Experiment 3. Discussion
1.  In general terms, what is the relationship between the rate of penetration of acid into the gelatine block and the size of the block?

2.How much more rapidly did acid penetrate to the centre of the smallest block than to the largest block?

3 (a) Was the rate of penetration the same or nearly the same for any of the blocks?
  (b) Suggest a reason to support your observation.

4 Imagine that, on a much smaller scale, the gelatine block represents a cell of a single-celled creature :(a) what substances need to penetrate to all parts of a living cell;
  (b) what substances need to diffuse out of the cell in the opposite direction ?

5 Most single-celled animals are less than 1 mm across and the largest are only about
2 mm long.
By reference to your results in this experiment suggest one reason why single-celled animals are no larger than this ?

6 On the basis of the results of this experiment, the penetration of substances into a large, multi-cellular organism such as a fish should take a very long time. What aspects of structure and organisation in a fish allow it to survive and be active despite such a slow rate of diffusion ?

7 Suggest one or more shapes for a volume of one cubic centimetre of gelatine, which would increase the rate of penetration of substances.





                  .
Experiment 3. Discussion - answers

I The smaller the block, the faster will the acid penetrate to the centre.

2 Probably the acid will take about ten times longer to reach the centre of the 1 cm cube than it will to clear the smallest cube.

3 (a) The answer will depend on the accuracy of cutting the gelatine blocks and the acuity of the observation. Some of the blocks e.g. 2 and 3  (depending on how they are cut) may take about the same time since the minimum distance to the centre is 2.5 mm in each case.
   (b) If the student says that the times are all different he will perhaps justify his answer by saying that the acid has a shorter distance to travel as the blocks get smaller. (This is true for the average distance.)
If the student observes that two or three blocks clear at almost the same time, he may justify this as in 3(a) above.

4 (a) Oxygen and possibly water and dissolved nutrients need to penetrate to all parts of a cell.
   (b) Carbon dioxide and other metabolic waste products must diffuse out of the cell.

5 One reason for the limitation on the size of protozoa might be the rate of penetration of oxygen through the cell surface. The greater the diameter of the cell, the longer will it take oxygen to reach all parts of the cytoplasm. Beyond a certain distance, the rate of penetration of oxygen might be too slow to maintain an adequate rate of activity. The same sort of argument applies to the elimination of waste products but there are likely to be many other reasons for the limitation on size, e.g. mechanical and structural considerations.

6 The question ignores, for the sake of discussion, the fact that an epidermis is not comparable to a cell membrane when considering diffusion.
For absorbing oxygen the fish has gills whose surface area is probably far greater than the external surface of the animal. The blood circulation (a) maintains a steep diffusion gradient at the gill surface and (b) distributes oxygen to all parts of the body.

7 Any shape which increases the surface area and diminishes the distance of penetration would be effective, e.g. a very thin sheet of gelatine.

Experiment 3. Diffusion and size - preparation

Outline A gelatine block containing cresol red is cut into different sized pieces which are immersed in dilute acid. The penetration of the acid can be seen as the indicator changes to yellow.

NOTE. Even if the questions and discussion are too sophisticated for the students, the experiment illustrates very clearly a pattern of diffusion analogous to that which must occur in a cell.
The experiment is not directly concerned with surface area/volume ratios.

Prior knowledge Change of colour in a pH indicator. Preferably some knowledge of diffusion in liquids, e.g. experiment 2. To answer the questions the students will need to have some knowledge of cells, single-celled organisms and their need to absorb oxygen and eliminate carbon dioxide. Fish possess gills and a circulatory system.

Advance preparation and materials

Gelatine. Stir 10 g gelatine crystals into 100 cm3 tap-water and heat to boiling point.       Add 5 cm3 1% cresol red solution and 2 cm3 bench (2M) ammonium hydroxide. Pour the gelatine to a depth of 10 mm in plastic Petri dishes lightly smeared with a little oil, and when cool cut into blocks of 20 x 10 x 10 mm..
Allow 5 cm3 gelatine per group.
Cresol red. Dissolve 0.5 g cresol red in 20 cm3 ethanol, dilute to 50 cm3 with distilled water and filter.
Hydrochloric acid (2M). Dilute 100 cm3 concentrated hydrochloric acid with 400 cm3
tap-water. Allow 30 cm3 per group.
Ammonia (2M). Dilute 11 cm3 0.880 ammonia with 89 cm3 tap-water.

Apparatus-per group
test-tube and bung or cork
tile or Petri dish lid
scalpel or razor blade

-per class
clock

TIME. It takes about 10-15 minutes for the 1 cm cube to clear.


Experiment 4. Diffusion through a membrane

You are provided with a length of cellophane tubing which has been soaked in water to make it flexible.

(a) If the cellophane tubing is not already knotted at one end, tie a knot close to one end of the tubing, leaving the other end open.

(b) Use a dropping pipette to half fill the cellophane tubing with starch solution.

(c) Place the cellophane tubing, with its starch solution, inside a test-tube, fold the open end of the tubing over the rim of the test-tube and secure it with an elastic band as shown in the Figure below.

(d) Wash the test-tube and cellophane tubing under a running tap to remove all traces of starch solution from the outside of the  tubing and then fill the test-tube with dilute iodine solution and leave it in the rack for 10-15 minutes.

(e) After 10-15 minutes, examine the iodine and starch solutions in the test-tube and record any colour changes.                                            .








Experiment 4. Discussion

I What colour change is usually seen when iodine solution is added to starch, or starch to iodine ?

2 After 10-15 minutes, what colour was (a) the starch solution in the cellophane tubing and (b) the iodine solution in the test-tube ?

3 How can you explain the fact that the same colour change had not occurred in both solutions ? (There are two possible explanations.)

4 What result would you expect if the iodine solution was in the cellophane tubing and the starch solution in the test-tube ?

5 What properties must the cellophane tubing possess if these results are to be explained ?

Experiment 4. Discussion - answers

I When iodine and starch solutions are mixed, a blue colour appears.

2 The starch solution in the dialysis (cellophane) tubing should go blue and the iodine solution should not have changed from its original yellow colour.

3 Either (a) the dialysis membrane allows iodine to pass but not starch or (b) the dialysis membrane allows substances to pass in but not out.

4 The answer will depend on which explanation is selected in answer to question 3.
(a) The solution in the test-tube will go blue and the iodine in the dialysis tubing will remain yellow, or (b) the starch will pass into the iodine in the dialysis tubing so producing a blue colour inside while remaining colourless outside, i.e. the same result as the original experiment.

5 Either (a) the tubing allows some substances to pass but not others or (b) the tubing allows substances to pass in only one direction.

Although reversing the solutions is the obvious control to counter explanation 3(b) it is unsatisfactory in practice because the blue starch-iodide complex adheres to the outside of the dialysis tube so that it is still the latter which looks blue. The starch solution outside remains colourless. The blue deposit can, however, be washed off the dialysis tubing to reveal that the iodine inside has not changed colour. The alternative control is to turn the dialysis tubing inside out, a tricky operation with such narrow tubing.


Experiment 4. Diffusion through a membrane - preparation

Outline A piece of dialysis tubing containing starch solution is immersed in dilute iodine solution. Only the starch goes blue.

Prior knowledge Starch and iodine react in solution to give a blue colour.

Advance preparation and materials

Starch solution. Make a 1% starch solution by shaking the starch powder with water and heating to boiling point. Allow 3 cm3 per group.
Iodine solution. Grind 1g iodine and 1g  potassium iodide in a mortar and add distilled water a little at a time. Make the solution up to 100 cm3 with distilled water. Shortly before the experiment dilute 5 cm3 of this stock solution with 200 cm3 tap-water.
Allow 30 cm3 per group.
Dialysis tubing. Cut the 6 mm (½ in) tubing into 15 cm lengths and, before the experiment, soak these in water and tie a knot in one end.. Allow one length per group.
The tubing can be used several times. After the experiment wash out the starch solution and leave the tubing to soak, if necessary, to remove the blue colour. Store the tubing dry or in 30% alcohol.

Apparatus

test-tube and rack
container for iodine solution
dropping pipette
length of dialysis tubing
container for 1% starch solution
elastic band


Experiment 5  Control of diffusion

(a) Use a cork borer to cut cylinders of tissue from a beetroot. (DO NOT cut on to the bench but use a  plastic Petri dish lid or similar). Push the cores of tissue out of the borer with the flat end of a pencil
(Fig. 1 p. 5.02). Place these cylinders of beetroot on a tile or Petri dish lid and use a scalpel or razor blade to cut them into discs about 3 mm thick (Fig 2). You will need 12 discs.

(b) Place the discs in a beaker or jar and wash them in cold water for at least 5 minutes. Either leave them under a running tap or fill and empty the container about five times during the washing process, but in either case continue with (c) and (d).

(c) Label two test-tubes with the figures 30, and 70, and use a syringe or graduated pipette to put  6 cm3 of cold tap-water in each.

(d) Prepare a water bath by half filling a beaker or can with cold water, place it on a tripod and gauze and put a thermometer in the water but DO NOT start heating it yet.

(e) When the beetroot discs have been washed, impale 6 of them on a mounted needle with a space between each one (Fig. 3).

(f) Heat the water bath with a small Bunsen flame till the water reaches 25 oC, remove the burner and allow the temperature to rise to 30°, heating again briefly if necessary.

(g) Note the time and place the mounted needle with the discs in the water bath for exactly 1 minute. At the end of this time, take the discs off the needle and drop them into the tube labelled 30.

(h) Heat the water bath again till the thermometer reads about 67°, remove the flame and allow the temperature to rise to 70°. Impale 6 more discs, immerse them in the water bath for 1 minute and then transfer them to the tube labelled 70. Leave  the discs in the test-tubes for 20 minutes or more.

(i) After 20 minutes or more, shake the tubes, hold them up to the light and compare the colours of the liquids in each.





Experiment 5. Discussion

The colour of beetroot is due to the presence of a red pigment in the cell sap.

(1) What was the colour of the liquid (a) in tube 30 (b) in tube 70?

(2) Does the red pigment from the cell sap diffuse out of the cell at (a) 30oC (b) 70oC?

(3) Which structures in Fig. 4 are composed of living material?

(4) What is the effect of high temperature on living material?

(5) From your answers suggest , in general terms, what controls diffusion of the red pigment.















Experiment 5 Discussion – answers

1 There should be no escape of pigment in tube 30. Tube 70 should have a deep red colouration.

2 The red pigment does not diffuse out of the cell at 30oC but it does so at 70oC.

3 The cytoplasm and nucleus are composed of living material. Cell sap and the cellulose cell wall are not living materials.

4 High temperature kills most living materials (by denaturing their proteins, e.g. enzymes and structures in the cell membrane).

5 It seems likely that a living process in the cytoplasm controls the diffusion of the pigment. Diffusion of pigment is prevented when the cytoplasm and cell membrane are intact but not when they are ‘killed’.
 Students may suggest that the nucleus is the controlling factor since they have learned that the nucleus controls processes in the cell. If this were the case, the intact nucleus would have to influence some process in the cytoplasm, making it impermeable to the red pigment.


Experiment 5. Control of diffusion – preparation

Outline Discs of beetroot are immersed in water at temperatures of 30oC and 70oC to see how much red pigment subsequently escapes.

Prior knowledge Plant cell walls contain cellulose. Cytoplasm contains proteins. Protein is denatured by high temperatures.

Advance preparation and materials

Beetroot. Buy fresh, uncooked beetroot.  500g (about 1lb) is enough for 18 groups if they collect their cylinders from one point. If the material is to be given out, allow about ¼ beetroot per group.
If time is short, the discs can be prepared beforehand, washed overnight and the experiment started from (c).
The tubes can be left for 24 hours while the pigment escapes but longer periods lead to decomposition of the discs and escape of the pigment from all samples.

Apparatus-per group
cork borer No.5 or 6                                                  2 test-tubes and rack
plastic Petri dish lid for cutting beetroot                    Bunsen burner
scalpel or razor blade                                                tripod
beaker or jar for washing discs                                 gauze
spirit marker                                                              mounted needle
beaker or can for water bath                                     means of timing 1 minute
graduated pipette or syringe (10 or 25 cm3)             heat mat (to protect bench)
thermometer (0-100 °C)                                   



Experiment 6. Two-way diffusion in a liquid

(a) Melt the gelatine in the boiling tube by immersing it almost up to the rim in a beaker or jar of hot water. While waiting for the gelatine to melt, continue with (b).

(b) Use a spirit marker to label two test-tubes A and B and draw lines round them at 8, 12 and
13.5 cm from the bottom of the tube (see Fig. I).

(c) Pour liquid gelatine into both tubes up to the first mark and use a dropping pipette or syringe to add 10 drops of 1% cresol red (a pH indicator) to each. Close each tube in turn with your thumb or a rubber bung and invert it several times to mix the indicator with the gelatine.

(d) Allow the coloured gelatine to set firm. This can be hastened by dipping the tubes in cold water. Meanwhile, make three copies of Fig. 1 in your notebook, labelling only the first one.

(e) When the yellow gelatine has set, re-melt the gelatine in the boiling tube if necessary and pour a further layer of cool but liquid gelatine into both tubes up to the second mark and cool quickly.
The next operation uses ammonia solution, which gives off a pungent gas. It is harmless enough
provided you do not sniff it at close range.

(f) When the clear gelatine has set in tube A, add 1 % ammonia solution up to the third mark and insert the bung (Fig. 2). Do the same for tube B but use 10% ammonia solution.

(g) Put your initials on the tubes and leave them on their sides for 3-7 days.

(h) After 3-7 days, examine the tubes and use crayons or shading to indicate the distribution and
colour of the indicator on your two remaining drawings.



















Experiment 6. Discussion

NOTE. Assume that diffusion alone is responsible for the movements of ammonia and
cresol red.
Cresol red is a pH indicator, yellow in acid* and red in alkali. Ammonia is an alkali.

1 What evidence is there that the ammonia has diffused through the gelatine?

2 What evidence is there to show that the cresol red has also diffused?

3 Write a sentence to compare the direction and distance of diffusion in the two substances.

4 How could you explain the fact that ammonia diffuses further in tube B than in tube A?

5 Does the diffusion of ammonia in one direction appear to affect the diffusion of cresol red in the other direction? What evidence is there to support your answer?

* (between pH 1.8-7.2).

Experiment 6. Discussion - answers

1 The cresol red will have changed from yellow to red showing that ammonia has diffused into the gelatine.

2 The cresol red will have coloured the clear gelatine above the first mark showing that it too has
diffused.

3 The ammonia has diffused further than the cresol red and in the opposite direction.

4 The ammonia in tube B is ten times more concentrated than that in tube A. This fact probably accounts for the more rapid diffusion.

5 The cresol red diffuses about the same distance in both tubes. If the diffusion of ammonia in the opposite direction was affecting the diffusion of cresol red, it might be expected that the
cresol red would not have diffused so far in the presence of more concentrated ammonia.

NOTE. The experiment shows that the direction of diffusion depends on the concentration gradient and that two substances can diffuse simultaneously in opposite directions as in e.g. the alveoli of the lungs.
It could be pointed out that the cresol red molecule is considerably larger than the ammonia molecule and that this might contribute to its slower diffusion. However, even though tube A contains 1% cresol red and 1% ammonia, they are not equimolecular concentrations and, also, the differences in solubility might be playing a part.


Experiment 6. Two-way diffusion in a liquid - preparation

Outline Ammonia is allowed to diffuse through clear gelatine into gelatine coloured with cresol red. The indicator diffuses in the opposite direction, into the clear gelatine.

Prior knowledge The significance of 'pH' and 'pH indicator'. Ammonia is an alkali. Precautions
in handling strong ammonia solution.

Advance preparation and materials

Gelatine. Make a 10% solution by dissolving the crystals in tap-water and heating to boiling point. Keep the solution moving about or it will burn on the bottom. Stopper the solution
while it is hot and thus reduce the chances of bacterial contamination and subsequent liquefaction.
Before the experiment, dispense the liquid gelatin into boiling tubes allowing about 60 cm3 per
group.

Cresol red solution. Dissolve 0.5 g cresol red in 20 cm3 ethanol, dilute to 50 cm3 with distilled water and filter. Allow 1 cm3 per group.

Ammonia. Dilute 0.880 ammonia solution with twice its own volume of tap-water to make the 10% solution and dilute this solution with nine times its own volume of water to make the 1% solution. Allow 5 cm3 of each solution per group.
It is advisable to have a single dispensing point for at least the stronger solution at the teacher's bench so that it can be supervised.

Bungs to fit the test-tubes- No. 17 bungs to fit 19 mm test-tubes.

Apparatus

2 test-tubes (19 x 150 mm)
test-tube rack
ruler with millimetres
2 bungs for test-tubes
containers for hot or cold water
boiling tube (25 x 150 mm) for gelatine
container for cresol red
spirit marker
dropping pipette

NOTE. The experiment takes 20-30 minutes to set up.






Chapter Two – Enzymes

Experiment 1. The action of amylase on starch

(a) Prepare a water bath by using a Bunsen burner to heat some water in a beaker on a
tripod and gauze till it boils; then turn down the flame to keep the water just boiling.
While waiting for the water to boil carry on from (b ).

(b) Label four test-tubes 1-4.

(c) Using a syringe or graduated pipette, place 5cm3 2% starch  solution in each tube.

(d) Rinse the syringe or pipette and use it to place 2cm3  amylase solution in each of tubes 2 and 3 and shake the tubes to mix the contents. Leave the tubes for 5 minutes and copy the table below into your notebook.

(e) After 5 minutes, add 3 drops of iodine solution to tubes 1 and 2.

(f) Rinse the syringe or graduated pipette and use it to add 3 cm3 Benedict's solution to tubes 3 and 4 and place both tubes in the water bath for 5 minutes.

(g) Compare the final colours in the tubes and complete the table of results.


Tube Contents Tested with Result Interpretation
1 2% starch solution iodine
2 2% starch solution + amylase iodine
3 2% starch solution + amylase Benedict’s
4 2% starch solution Benedict’s














Experiment 1. Discussion

I What normally happens when iodine solution is added to starch?

2 Tube 2 contained starch solution at the beginning of the experiment; how do you
explain the reaction with iodine at the end of the experiment ?

3 What food substance is Benedict's solution a test for ?

4 Was this food substance present in tubes 3 or 4 at the beginning of the experiment?

5 What evidence do you have to support your answer ?

6 What evidence is there to suggest that this food substance is present in tube 3 at the
end of the experiment ?

7 What chemical change could have taken place in tubes 2 and 3 after adding amylase, which would explain the results in these tubes after applying the iodine test and Benedict's test?

8 What part could amylase have played in this chemical change ?

9 Suggest a control to the experiment which would help to support your answer to 7.


Experiment 1. Discussion - answers

I Iodine added to starch usually produces a blue colour.

2 The failure to produce a blue colour means that starch is absent or in very low concentration.

3 Benedict's solution is a test for a reducing sugar, in this case, glucose.

4 No glucose was present in tubes 3 or 4 at the beginning of the experiment.

5 Tube 4, containing only starch solution, should not give a yellow or red precipitate when tested with Benedict's solution, though it may go green and a little cloudy.

6 Tube 3 gives an orange precipitate with Benedict's solution; indicative of  reducing sugar (glucose).

7 (i) Starch could have been changed to sugar by the action of amylase.
   (ii) Amylase could have been changed to sugar by the action of starch.
   (iii) Starch and amylase could have combined to form sugar.

8 The answer will depend on which of alternatives (i)-(iii) are offered and on the student's knowledge, if any, of the chemical composition of starch and sugar and the role of catalysts and enzymes. In the absence of any knowledge of this kind, (i), (ii) and (iii) seem equally valid interpretations of the experimental results, though if (ii) were correct, one would hardly expect the starch to disappear.

9 Unless the students have some knowledge of the facts outlined above, a control will be almost impossible to design. If they already know something about enzymes they might say that if amylase is an enzyme which converts starch to sugar, boiling the amylase should prevent the reaction from working.



Experiment 1. Diffusion in gases – preparation

Outline Ammonia diffuses along a glass tube turning litmus paper blue as it does so. Rates of diffusion are compared for strong and weak ammonia solutions.

Prior knowledge Ammonia turns red litmus blue. Meaning of 'convection'. Meaning of 'molecules'. Precautions in handling ammonia.

Advance preparation and materials

Tubes. Glass tubes, 20 mm diameter or more, 30 cm long and marked with a spirit marker at 2 cm intervals starting at 10 cm. Allow 2 tubes per group.
To cut the tubes, score a line with a glass cutter round the tube and heat this line with a 25 cm length of 26 swg (0.45 mm) Nichrome wire connected to the 12 V supply
(Fig. I, p.1.03). Flame polish the cut ends.
Pusher. A piece of wire, bicycle spoke, knitting needle or glass rod about 25 cm long for
arranging the litmus squares inside the tube. Allow one per group.
Corks. Use a No.6 borer to make a hole through the cork. Eject the plug of cork from the
borer, cut 10 mm from it and replace the rest of the plug in the hole in the cork leaving a
cavity at the narrow end (Fig. 2). Plug this cavity tightly with cotton wool. Allow 2 of these
corks and 2 ordinary corks of the same size per group. Label the bored corks A and B in equal numbers.

Litmus paper. Cut strips of red litmus paper into squares (6 or 7 per strip). Allow 20 squares per group.
Ammonium hydroxide
(9N) Dilute 0.880 ammonia solution with its own volume of water
(2N) Dilute 11cm3 0.880 ammonia solution with 89 cm3 water. Allow 3 cm3 of each solution per group.
It is advisable to have a single dispensing point for at least the strong solution at the teacher's bench so that it can be supervised.

Apparatus-per group

beaker or small container for water
forceps
container for litmus paper squares
pusher, wire or glass
2 marked tubes, 30 cm long 
2 corks with cotton wool plugs
2 ordinary corks
dropping pipettes ( 4 or 5 at dispensing point)

-per class
1 clock or several seconds timers
NOTE. Using 20 mm diameter tubes and 9N ammonia solution, the litmus at 28 cm should go blue in 20 minutes.




Experiment 2. The effect of temperature on enzyme activity

(a) Label six test-tubes 1-6.

(b) Using a syringe or  graduated pipette, place 5 cm3 1% starch solution in tubes 1-3.

(c) To each of these tubes add also 6 drops of dilute iodine solution.

(d) Wash the syringe or graduated pipette and use it to place 1 cm3 amylase solution in each of tubes 4-6.

(e) Prepare three water baths by half filling 250 cm3 beakers or jars as follows :

 (i)  ice and water, adding ice during the experiment to keep the temperature at
       about 10 °C.
(ii) water from cold tap at room temperature, about 20 °C.
(iii) warm water at about 35 °C by mixing hot and cold water from the tap.

(f) Place tubes 1 and 4 in the cold water, 2 and 5 in the water at room temperature
and 3 and 6 in the warm water. Leave the apparatus for five minutes so that the amylase and starch solution attain the temperature of the water bath. (See Figure on p.2)

(g) Draw up a table in your notebook similar to the one below.

(h) Note the time and then pour the amylase solution from tubes 4, 5 and 6 into the corresponding tube of starch solution, starting with the coldest one. Shake the tube to mix the contents and return it to appropriate water bath.

(i) Watch the tubes for the disappearance of the blue colour and note the time when each solution becomes colourless. Record the time intervals in your notebook.
Experiment 2. Discussion

I What is the significance of the disappearance of the blue colour from the starch/iodide
mixture ?

2 At what temperature did the reaction proceed most rapidly ?

3 Make a general statement connecting the rate at which amylase acts on starch and the temperature at which it is acting. Put in any qualifications you think necessary.

4 Would you expect to find any upper limit of temperature beyond which the reaction
ceases to get any faster ?

Tube Temperature Time for blue colour to disappear Speed of reaction (fast, medium slow)
1 10oC
2 20oC
3 35oC

Experiment 2. Discussion – answers

I The disappearance of the blue colour means that starch is removed from the mixture. Alternatively, it could be argued that the iodine has been removed, or that the starch-iodide complex is separated in some way.

2 35 °C is probably the temperature at which the solution loses its blue colour most rapidly.

3 The rate of reaction is greater at higher temperatures. The qualifications are
(a) assuming that the disappearance of blue colour is correctly attributed to the action             of amylase,
(b) no experiment has been carried out above 35-40 °C so it cannot be predicted that the rate of reaction will continue to increase with rise in temperature.

4 The student may know that enzymes are proteins and so denatured at temperatures above 50 °C ; or more simply that boiling destroys enzymes.






Experiment 2. The effect of temperature on enzyme activity - preparation

Outline  A comparison is made of the time taken for the hydrolysis of starch solution at three different temperatures. The time for hydrolysis is judged by the disappearance of blue colouration from starch-iodide.

Prior knowledge  Starch/iodide reaction.

Advance preparation and materials-per group

1% starch solution *   20 cm3                                 dilute iodine 5 cm3
ice 3-4 cubes                                                       5% amylase solution  3cm3

NOTE. The presence of iodine inhibits the reaction. If too much iodine is added it may prevent the hydrolysis altogether. The instructions are based on the assumption that the iodine solution has been freshly prepared and in accordance with the instructions on p.01.

Apparatus-per group
test-tube rack and 6 test-tubes                         3 X 250 cm3 beakers or jars
6 labels or spirit marker                                    thermometer
syringe or graduated pipette 10 cm3                 dropping pipette

--per class
clock


.



Experiment 3. The hydrolysis of starch with hydrochloric acid

(a) Prepare a water bath by half filling a 250 cm3 beaker with warm water and heating it to boiling point on a tripod and gauze, with a Bunsen burner. When the water boils, reduce the flame to keep the water at boiling point.

(b) Label four test-tubes 1-4.

(c) Copy the table given below into your notebook.

(d) In each tube place 5 cm3 3% starch solution.

(e) Using a syringe or graduated pipette, add 3 cm3 Benedict's solution to the starch solution in
tube 1 and place the tube in the boiling water bath for five minutes.

(f) Rinse the syringe or pipette and use it to add 1 cm3 dilute hydrochloric acid to the starch solution in each of tubes 2, 3 and 4. Note the time and place all three tubes in the water bath.
 (They will be removed at five, ten and fifteen minutes respectively).

(g) Remember to remove tube 1 from the water bath after five minutes if you have not
already done so.

 (h) After five minutes, remove tube 2 from the water bath and cool it under the tap. Neutralize the acid by adding solid sodium bicarbonate, a little at a time, until the addition of one portion produces no fizzing. Place tube in the rack and return to tube 3.

(i) After ten minutes in the water bath, remove tube 3, cool and neutralize the contents as described in (h). Place the tube in the rack.

(j) After fifteen minutes in the water bath, remove tube 4; cool and neutralize as before, and place it in the rack.

(k) With a dropping pipette, remove a sample of the liquid from tube 2 and place 3 drops
on a spotting tile. Rinse the pipette and repeat the procedure for tubes 3 and 4. Add
one drop of dilute iodine to each drop of liquid on the tile.

(l) Rinse the syringe or pipette and use it to place 3 cm3 Benedict's solution in each of
tubes 2, 3 and 4. Return all three tubes to the water bath and heat for five minutes.
After this time replace the tubes in the rack and allow them to cool sufficiently to
handle. Hold the tubes to the light to compare the colours of the solutions and compare
also the colours and relative quantities of any precipitates.

Appearance after testing with Benedict’s
Tube Containing Appearance with iodine Colour of solution Colour of precipitate Relative quantity of precipitate
1 3% starch solution only

2 3% starch solution boiled for 5 min with dilute HCL
3 3% starch solution boiled for 10 min with dilute HCl
4 3% starch solution boiled for 15 min with dilute HCl

Experiment 3. Discussion

1 What was the point of adding sodium bicarbonate to tubes 2, 3 and 4?

2 What food substance is Benedict's solution a test for?

3 At the end of the experiment, what food substance was present in tubes 3 and 4 that
was not there at the beginning?

4 What evidence have you that this substance was not present at the beginning of the
experiment?

5 How do you account for the difference, after testing with Benedict's solution, between
tubes 2, 3 and 4?

6 How do you interpret the results of the iodine test in tubes 2, 3 and 4?

7 What relationship is there between the interpretation of the results with the iodine test and the Benedict's test?

8 The starch molecule consists of a long chain of carbon atoms with hydrogen and oxygen
atoms attached. Sugars, such as glucose, consist of six carbon atoms with hydrogen and oxygen atoms attached.






(*many H atoms omitted)
Assuming that the hydrochloric acid is acting only as a catalyst in the reaction, attempt
an explanation of the chemical change which takes place in tubes 3 and 4.

9 In this experiment, the emphasis is on the conversion of starch to something else using
hydrochloric acid. What control experiment would have to be carried out to show that
hydrochloric acid played a significant part in bringing about this change?
















Experiment 3.  Discussion - answers

1 The hydrogencarbonate neutralizes the hydrochloric acid which would otherwise interfere with Benedict's reagent.

2 Benedict's solution is a test for reducing sugars.

3 Tubes 3 and 4 should have a red precipitate, indicative of a reducing sugar.

4 Tube 1, containing starch solution, should have given little or no colour change with Benedict's
solution.

5 Tube 4 will probably have a more intense red colour or a more dense precipitate than the others, indicating that a greater quantity of reducing sugar has been formed. The liquid in tube 2 may still be blue, indicating unchanged Benedict's solution.

6 The blue colour is progressively less intense or absent altogether in tube 4 indicating that starch is 'disappearing'.

7 In tube 4, at least, starch has disappeared and sugar has appeared. It could be that:
    (i) Hydrochloric acid has changed starch into sugar.
   (ii) Hydrochloric acid has combined with starch to form sugar.
  (iii) Starch has converted hydrochloric acid to sugar.

8 If students are not overwhelmed by the sight of the structural formulae they might notice that
by breaking the starch chain at the -0- linkages, adding H- to one side and -OH to the other, glucose molecules would be produced. The possibility of disaccharides is ignored at this juncture.

9 To show that starch solution is not converted to sugar by simply boiling it, a control should be carried out by boiling 5 cm3 3% starch solution for 10 minutes and then testing with Benedict's
reagent.




Experiment 3 . The hydrolysis of starch with hydrochloric acid - preparation

Outline. The experiment illustrates the conversion of starch to a reducing sugar by the action of hydrochloric acid at boiling point. The longer the starch is exposed to the acid the further hydrolysis proceeds. The experiment is intended to show the contrast with enzymes, which do not need high temperatures and prolonged exposure to reagents and give a quick reaction.

Prior knowledge. Benedict's reaction, starch/iodide reaction.

Advance preparation and materials-per group

3% starch solution, freshly prepared* 25cm3         sodium hydrogencarbonate (bicarbonate)
Benedict's reagent 20 cm3                                         powder about 5 g
dilute hydrochloric acid, 2M or 10%                      iodine solution (dilute)  5 cm3  +
    (bench strength) 5 cm3

* NOTE. Some brands of starch are readily hydrolysed and might give a positive reaction with tube 1. A 3% starch solution should be tested with Benedict’s solution to see if it withstands hydrolysis after 5 minutes in a water bath.

Apparatus -  per group

test-tube rack and 4 test-tubes                  tripod
4 labels or spirit marker                           gauze
graduated pipette or syringe 10 cm3        heat mat   
250 cm3 beaker                                        spatula for adding sodium hydrogencarbonate
Bunsen burner                                         dropping pipette (if not supplied with iodine bottle)

- per class

clock

Time The experiment needs from 30-45 minutes

+ See instructions for making dilute iodine on p. 01




Experiment 4. The action of saliva on starch

Study the flow chart on p. 9.02 for a few minutes to gain an idea of the outline of the experiment.

(a) Prepare a water bath by using a Bunsen burner to heat some water in a beaker on a
tripod and gauze till it boils; then turn the flame down to keep the water just boiling. While waiting for the water to boil, carry on from (b).

(b) Label eight test-tubes 1 - 8 and in tube 1 collect saliva as follows:

    (i) Thoroughly rinse the mouth with water to remove food residues
   (ii) Collect about 50 mm saliva.

(c) Pour half the saliva into tube 2 and place the tube in the boiling water bath for 3 minutes.

(d) Using a graduated pipette or syringe, add 5 cm3 2% starch solution to tubes 3,4 and 7.

(e) Rinse the pipette or syringe and use it to transfer 5 cm3 boiled saliva from tube 2 to tube 3. Shake the tube sideways to mix the contents.

(f) Use the graduated pipette or syringe to transfer 5 cm3 unboiled saliva from tube 1 to tube 4.
Shake the tube to mix the contents.

(g) Leave tubes 3 and 4 to stand for five minutes and copy the table below into your notebook.

(h) After five minutes, pour half the contents of tube 3 (the boiled saliva and starch) into tube 5 and add three drops of iodine solution to tube 5.

(i) To the remaining liquid in tube 3, add about 20 mm Benedict's solution and place
the tube in the boiling water bath for 5 minutes.

(j) Pour half the contents of tube 4 (starch and saliva) into tube 6 and then add three drops of iodine to tube 6.

(k) Test the remaining liquid in tube 4 with Benedict's solution as you did in (i).

(l) Pour half the contents of tube 7 (starch solution) into tube 8 and test the two samples
respectively with iodine as in (h) and Benedict's solution as in (i). Record the results in your table.

Tube Contents Tested with Result Interpretation
3 starch and boiled saliva Benedict’s solution
4 starch and saliva Benedict’s solution
5 starch and boiled saliva iodine
6 starch and saliva iodine
7 starch solution (control) iodine
8 starch solution (control) Benedict’s solution




Experiment 4.  Discussion

1 What substances do iodine and Benedict's solution test for?

2 What change takes place when starch and saliva are mixed, according to the results in tubes 4 and 6?

3 Tubes 3 and 5 probably did not give the same results as tubes 4 and 6. In what way were the contents treated that could account for this difference?

4 (a) Are your results consistent with the hypothesis (theory) that an enzyme in saliva
         has changed starch to sugar?
   (b) Do your results prove that an enzyme in saliva has changed starch to sugar?

5 In what way do the results with tubes 3 and 5 support the enzyme hypothesis?

6 Do your experimental results rule out the possibility that (a) starch converts unboiled saliva to sugar or (b) starch and unboiled saliva combine chemically to form sugar?

7 The starch molecule consists of a long chain of carbon atoms with oxygen and hydrogen
atoms attached. A sugar, such as glucose, has molecules consisting of six carbon atoms
with oxygen and hydrogen atoms attached (see p. 8.02). Using this information, suggest a
way in which sugar could be formed from starch. What part would an enzyme play in
this reaction?



8 If you tried experiment 8, state in what ways the conditions for the reaction between starch and hydrochloric acid differed from those in the starch/saliva reaction.



























Experiment 4.  Discussion - answers

1 Iodine is a test for starch; Benedict's solution is a test for reducing sugars.

2 Tube 4 gives an orange precipitate with Benedict's solution and shows that sugar is present. Tube 6 fails to give a blue colour with iodine and thus shows that starch has gone.

3 In tubes 3 and 5, the only difference was that the saliva had been boiled.

4 (a) The results are consistent with the hypothesis that an enzyme in saliva changes starch to
         sugar, but
   (b) they do not prove it (see question 6).

5 If an enzyme in saliva was responsible for converting starch to sugar, boiling would denature the enzyme and prevent the reaction from taking place The failure of the starch to disappear in tube 5 and of sugar to appear in tube 3 thus supports the enzyme hypothesis.

6 The experimental results by  themselves do not rule out these alternative interpretations, except
that it seems unlikely that starch would disappear if hypothesis (a) were correct.

7 The student might see that by breaking the starch molecule at the -O- linkages and adding
OH to one side and H to the other, glucose molecules would be formed. (The fact that it is
actually maltose that is produced is not considered of vital importance at this stage.)
  By definition, an enzyme would catalyse the reaction in some way.

8 To hydrolyse 3% starch in experiment 8, 15 minutes at 100 °C was needed. With saliva, only
5 minutes at room temperature was needed.



Experiment 4.  The action of saliva on starch - preparation

NOTE  The use of saliva in school experiments is not banned but certain precautions should be taken:
Students should work with only their own saliva
They should wash out their own glassware
The tubes should be sterilized in 1% sodium hypochlorite (sodium chlorate (1)) (See the ASE’s ‘Safeguards in the School Laboratory’ 11e p.95

Outline This is similar to ‘Enzymes’ experiment 1 but includes a control. It is shown that normal saliva will act on starch to produce sugar while boiled saliva will not..

Prior knowledge Starch/iodide reaction, Benedict's reaction.

Advance preparation and materials-per group

2% starch solution * (freshly prepared)
iodine solution 5 cm3
Benedict's solution 15 cm3

Apparatus-per group

test-tube rack and 8 test-tubes        dropping pipette (if not part of iodine bottle)
8 labels or spirit marker                 graduated pipette or syringe (5 or 10cm3)                     
test-tube holder                              beaker (for rinsing pipette)
Bunsen burner



*See note on p. 8.03


Experiment 5. Catalase

Catalase is an enzyme which occurs in the cells of many living organisms. Certain of the
energy-releasing reactions in the cell produce hydrogen peroxide as an end-product. This
compound, which is toxic to the cell, is split to water and oxygen by the action of catalase.
2H2O2 = 2H2O + O2
The investigation below is a fairly critical examination of plant and animal tissues to see if
they contain catalase.

    (a) Label three test-tubes 1-3.
    (b) Pour about 20 mm (depth) hydrogen peroxide into each tube.
    (c) Cut the liver into 3 pieces.
    (d) To tube 1 add a small piece of liver, and to tube 2 add a pinch of dried yeast.
    (e) Insert a glowing splint into tubes 1 and 2, bringing it close to the liquid surface
         or into the upper part of the froth.

1 Describe what you saw happening and the effect on the glowing splint.

2 How do you interpret these observations?

3 Is there any evidence from this experiment so far, to indicate whether the gas is coming
from the hydrogen peroxide or from the solid?

4 Is there any evidence at this stage that an enzyme is involved in the production of gas in this reaction?

    (f) In tube 3 place a few granules of charcoal and observe the reaction.

5 Could charcoal be an enzyme? Explain your answer.

6 Assuming (i) that the gas in (f) is the same as before and (ii) that the charcoal is almost
pure carbon, does the result with charcoal help you to decide on the source of the gas in this and the previous experiments?

   (g) Suppose the hypothesis is advanced that there is an enzyme in the liver and yeast, which
        decomposes hydrogen peroxide to oxygen and water; design and carry out a control
        experiment to test this hypothesis.

7 Record (i) the experiment, (ii) the reasons which led you to conduct it, (iii) the observed
results and (iv) your conclusions.

   (h) Wash out the test-tubes. Design and carry out an experiment to see if the supposed enzyme
         in the plant and animal material can be extracted and still retain its properties. The
        experiment should include a control.

8 Describe briefly your procedure, your results and your conclusions.

9 Assuming that liver and yeast each contain an enzyme which splits hydrogen peroxide,
is there any evidence to show that it is the same enzyme? What would have to be done
to find this out for certain?


Experiment 5 . Discussion - answers

1 Effervescence should be observed in each case but it is more vigorous with yeast than with liver. The glowing splint should relight.

2  Oxygen is being produced.

3 There is no evidence to indicate whether the liquid or solid is giving the gas. If the students think that a solid is unlikely to give off a gas they could be reminded of marble and hydrochloric
acid in which it is the solid producing the carbon dioxide. It seems less likely, however, that yeast and liver would both give off oxygen when treated with hydrogen peroxide, than that hydrogen peroxide should give oxygen when treated with diverse substances.

4 So far, there is no evidence of an enzyme being involved.

5 A gas will come off but not sufficiently rapidly to relight a glowing splint. Charcoal could not
be an enzyme because (a) it is an element and (b) it has been produced by very high temperatures that would destroy enzymes.

6 Charcoal, as an element, could not be giving off oxygen. The gas must be coming from the
hydrogen peroxide.

7 (i) The experiment should involve boiling the tissues and then putting them into hydrogen peroxide.
   (ii) If an enzyme is involved,
   (iii) no gas will be produced.

8 The student should grind the samples with a little sand and distilled water, filter and test the
filtrate with hydrogen peroxide. Oxygen will be evolved with a vigour proportional to that
witnessed when the original substances were tested.
    The student should boil half of each extract and show that it loses its activity.

9 There seems no fundamental reason why yeast and liver should not have different enzymes which catalyse the decomposition of hydrogen peroxide. To be certain on this point, the enzymes would have to be extracted and their chemical composition determined.


Experiment 5. Catalase - preparation

Outline Samples of liver and yeast are dropped into hydrogen peroxide. Oxygen is evolved and the student is asked to extend the experiment to try and decide if an enzyme in the tissues is responsible.

Prior knowledge The existence of inorganic catalysts; enzymes denatured on boiling; oxygen relights a glowing splint.

Advance preparation and materials-per group
20 volume hydrogen peroxide 50 cm3
splint
liver, about 1 cm cube
distilled water 20 cm3
dried yeast about 1 g
clean sand about 1 g
activated charcoal granules, about 1 g

Apparatus-per group
test-tube rack and 4 test-tubes
forceps or seeker for pushing liver into test-tube
4 labels or spirit marker
filter funnel
Bunsen burner
filter paper
test-tube holder
mortar and pestle

NOTE. The experiments and the questions take about one hour.




Experiment 6. The effect of the concentration of enzyme on the rate of reaction

(a) Copy the table given below into your notebook.

(b) Label four test-tubes 1-4.

(c) Using a graduated pipette or syringe, add 5 cm31% urea solution to each tube.

(d) Rinse the pipette or syringe and use it to add 2 cm3 dilute ethanoic acid to each tube.

(e) Add ten drops of BDH universal indicator to each tube, using the dropper incorporated
in the bottle itself or a dropping pipette.

(f) Rinse the graduated pipette and use it to put 3 cm3 boiled urease solution into tube 4.

(g) Now place the following volumes of urease (unboiled) in tubes 1-3:
     (1) 2 cm3, (2) 3 cm3, (3) 5 cm3.

(h) At intervals of 30 seconds shake the tubes, observe the colour of the indicator in each
tube and record the corresponding pH in your table (see the key below).

      BDH Universal Indicator   pH
   
      PINK                                    4 (acid)
      ORANGE                             5
      YELLOW                             6
      GREEN                                7
      BLUE-GREEN                     8 (alkaline)

                                                                     Time intervals (30 seconds)
Tube 5 cm3 urea  solution 2cm3 ethanoic acid,  plus: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 cm3 urease
2 3 cm3 urease
3 4 cm3 urease
4 3 cm3 boiled urease






Experiment 6. Discussion

1 What change in pH took place in tubes 1-3? Is the change from acid to alkali or the
reverse?

2 What evidence is there to suggest that the change was brought about by an enzyme?

3 What was the effect of increasing the amount of enzyme on the rate of reaction? Would
you expect this effect to continue indefinitely with an increasing enzyme concentration
or would you expect a limit to be reached? Explain your answer.

4 The acetic acid was added to make the conditions artificially acid to start with. What substance could urease be producing from urea which could make the conditions alkaline?

The formula for urea is (NH2)2CO or           NH2
                                                                                         CO
                                                                       NH2

5 What part could the enzyme urease be playing in the metabolic process of plants?

6 In general, of what value to a living organism could be the effect of an increased rate of
reaction with increasing enzyme concentration?



Experiment 6. Discussion - answers

1 The changes in tubes 1-3 are from acid to alkali.

2 The fact that there was no discernible change in tube 4 suggests that boiling the urease has stopped its activity. This is what would be expected if urease is an enzyme.

3 Increasing the amount of enzyme accelerates the rate of the reaction which raises the pH of the solutions. The limit might be determined by the availability of the substrate or the accumulation
of the product.

4 The student might see that the NH2-  could fairly easily become NH3 which would make the
solution alkaline.

5 Urease hydrolyses urea to carbon dioxide and ammonia, but its role in, for example, soya beans is not very clear. Urea is produced during the degradation of the amino acid arginine. Presumably the subsequent release of ammonia from the urea makes the nitrogen available for resynthesis of amino acids. The enzyme has been found in many plants and some invertebrate animals but not in vertebrates.
   The students, thinking in terms of animal metabolism will probably try to relate urease activity to excretion. This might be a good opportunity to point out that plants need to conserve rather
than excrete nitrogenous products.

6 By varying the amount of enzyme available, the rate of any particular reaction could be controlled. This could provide a basis for adapting the metabolism of the cell to varying conditions, e.g. if large amounts of a toxic metabolite began to accumulate in a cell, its elimination could be hastened by the production of a greater quantity of the enzyme which
acted on It.

Experiment 6. The effect of the concentration of enzyme on the rate of reaction - preparation

Outline Urease converts urea to ammonia and carbon dioxide. The ammonia neutralizes the ethanoic acid which is placed in the tubes and the pH change is followed with B.D.H. universal
indicator.

Prior knowledge Use of indicators to observe pH changes.

Advance preparation and materials-per group

      1 % urea solution 25 cm3
      5% urease solution 15cm3 +
      M/10 ethanoic acid 10 cm3
      boiled, 5% urease solution 5 cm3
      B.D.H. universal indicator * 5cm3

Apparatus-per group

     test-tube rack and 4 test-tubes
     4 labels or spirit marker
     graduated pipette or syringe
     dropping pipette (if dropping bottles for indicator are not available)
     beaker (for rinsing pipette)

Results

The indicator in the tubes will change from pink to blue within 5-10 minutes.

* Available in 100 cm3 plastic dropping bottles.

+Shake 5 g urease active meal (B.D.H.) with 100 g distilled water in a corked conical flask for several minutes. Allow the suspension to settle for about 10 minutes and decant off the liquid layer. The solution will keep for several days and, if not decanted, will become more active.

Experiment 7. The effect of pepsin on egg white suspension

(a) Label four test-tubes 1-4.

(b) Into each tube place about 5 cm3 (20 mm in test-tube) egg white suspension.

(c) To tubes 2,3 and 4 add three drops of dilute hydrochloric acid.

(d) Using a graduated pipette or syringe place 1 cm3 1% pepsin solution in a clean test- tube and heat it over a small Bunsen flame until the liquid boils. Add the boiled pepsin to the egg-white suspension in tube 4.

(e) Prepare a water bath in a 250 cm3 beaker or jar by mixing hot and cold water from the
tap to attain a temperature of about 40 °C. Have the beaker about half full.

(f) Using a graduated pipette or syringe, add 1 cm3 1% pepsin to tubes 1 and 3 only.

(g) Place all four tubes in the water bath and copy the table below into your notebook.

(h) After five or six minutes remove the four tubes from the water bath and replace them
in the test-tube rack. Compare the appearance of the contents and fill in your table of results.


Tube Contents Results
1 Egg-white suspension & pepsin
2 Egg-white suspension & HCl
3 Egg-white suspension, pepsin & HCl
4 Egg-white suspension, boiled pepsin & HCl




Experiment 7. Discussion

1 Why, do you suppose, does the egg white suspension used in the experiment look cloudy?

2 If the egg-white suspension goes from cloudy to clear, what change must have occurred?

3 In which of the test-tubes are the conditions most like those in the stomach?

4 In general, how is an enzyme affected by boiling?

5 Do the results with tubes 3 and 4 prove that pepsin is an enzyme?

6 Are the results with tubes 3 and 4 consistent with the theory that pepsin is an enzyme that can accelerate the digestion of egg-white?

7 Are the results consistent with the theory that pepsin is an enzyme which, in acid conditions, can digest proteins? .

8 Suppose the hypothesis is advanced that hydrochloric acid is an enzyme but can digest
egg-white only in the presence of unboiled pepsin, what control experiment would help
to eliminate this explanation?




































Experiment 7 . Discussion - answers

1 The cloudy appearance of egg-white suspension is due to solid particles of albumen suspended in water.

2 The clear liquid is now a solution and the solid particles must have dissolved.

3 Tube 3 approximates to the conditions in the stomach having pepsin, hydrochloric acid and a
temperature between 35 °C and 40 °C.

4 Enzymes, being proteins, are denatured above 45-50 °C, i.e. enzyme activity is destroyed by
boiling.

5 The results in tubes 3 and 4 do not prove pepsin to be an enzyme. It could be that hydrochloric
acid hydrolyses egg-white (just as it hydrolysed starch in experiment 8) but will only do so if
unboiled pepsin is present.

6 The results in tubes 3 and 4 are certainly consistent with the hypothesis that pepsin is an enzyme. This hypothesis would account for the fact that the egg white was not dissolved by boiled pepsin.

7 Egg albumen is a protein and the results suggest that pepsin can digest it, but it is not possible to make a generalization which includes all proteins on the basis of an experiment with one of
them.

8 The control would be to repeat the experiment with tube 3 but using boiled hydrochloric acid. If the egg-white suspension does not go clear, it looks as if hydrochloric acid has properties
analogous to an enzyme. (To students with any knowledge of the definition of an enzyme and the structure of proteins and of hydrochloric acid, the hypothesis is patently untenable.)


Experiment 7 . The effect of pepsin on egg-white suspension - preparation

Outline In the presence of hydrochloric acid, pepsin turns a cloudy suspension of egg-white into a clear solution. Controls are conducted with hydrochloric acid by itself, pepsin by itself and boiled pepsin with hydrochloric acid.

Prior knowledge The interpretation of results will be easier if the student knows already that egg-white suspension consists of solid particles of albumen, suspended in water; enzymes are inactivated by boiling, conditions in the stomach are acid.

Advance preparation and materials-per group

1% egg-white suspension * 25 cm            hydrochloric acid
1% pepsin solution 10 cm3                        (bench strength, e.g. 10% or 2M) 5 cm3

Apparatus-per group

test-tube rack and 5 test-tubes        250 cm3 beaker
4 labels or spirit marker                  thermometer
Bunsen burner                                dropping pipette
test-tube holder                               graduated pipette or syringe (1 - 5 cm3)

* See  Reagents for food tests and enzymes p.








Experiment 8 . The action of lipase

There are three or four liquids to be added to each test-tube, in various combinations, so look at the table at the foot of this page to get an idea of the final contents of each tube.

(a) Label three test-tubes 1-3.

(b) Using a graduated pipette or syringe, place 5 cm3 milk in each tube.

(c) Rinse the pipette or syringe and use it to place 7 cm3 dilute (M/I0) sodium carbonate solution in each tube. This solution is to make the mixture alkaline.

(d) Rinse the pipette or syringe and use it to place 1 cm3 3% bile salts solution in tubes 2 and 3 only.

(e) Use a dropping pipette to add phenolphthalein solution to each tube until the contents
are bright pink. About six drops will be sufficient and equal quantities should be added to each tube.
Phenolphthalein is a pH indicator. In alkaline solutions (above pH10) it is pink; in 'acid' solutions (below pH 8.3) it is colourless.

(f) In a spare test-tube, place about 15 mm of 5% lipase solution and, using a test-tube holder, heat the liquid over a small Bunsen flame until it boils for a few seconds.
   Cool the tube under the tap and, using the graduated pipette or syringe, transfer 1 cm3 of the
boiled liquid to tube 2.

(g) With the graduated pipette or syringe, place 1 cm3 unboiled lipase solution in tubes 1 and 3.

(h) Note the time. Shake the tubes to mix the contents, return them to the rack and copy
the table below into your notebook, observing the tubes from time to time.

(i) Note the time required for the contents of each tube to go white and then complete
the table of results.


Action of lipase on milk
Tube All three tubes contain milk, sodium carbonate and phenolphthalein plus: Time taken to change from pink to white
1 lipase only
2 boiled lipase and bile salts
3 lipase and bile salts






Experiment 8. Discussion

1 What food substances are present in milk?

2 If phenolphthalein changes from pink to colourless, what kind of chemical change must have taken place in the tube?

3 Recall (or look up) the final products of digestion of the principal classes of food and
write down which of these products could be formed by the digestion of milk.

4 Which of the final products of digestion of milk could be responsible for the change of
conditions in the test-tube?

5 Which part of the experiment suggests that lipase acts as an enzyme?

6 What chemical change could the lipase be producing which would account for the colour change in the test-tubes?

7 Which part of the experiment indicates that bile salts do not contain an enzyme which affects milk (at least in the way being investigated here)? Explain your reasoning.
8 From the results, assuming that lipase is an enzyme, what part do the bile salts appear to be playing in the reaction (in general terms)?

9 Do the results tell you whether lipase is acting on the fat or the protein in milk? Explain.




























Experiment 8. Discussion - answers

1 Protein (mainly caseinogen), fat (butter fat) and sugar (lactose), are the food substances present in milk.

2 The contents of the tube must have become less alkaline (more acid).

3 Amino acids, fatty acids, glycerol and glucose could be produced by the digestion of milk.

4 Amino acids are amphoteric electrolytes and therefore unlikely to produce the pH changes 
observed in this experiment. The students are unlikely to know this fact, however, and should
consider the production of both amino acids and fatty acids as possible causes of the drop in pH.

5 In tube 2 the lipase is boiled and there is no apparent chemical change when the boiled lipase is added to milk. If lipase is an enzyme, one would expect it to be denatured and rendered inactive above 50 °C

6 The lipase could be converting milk fats into fatty acids and glycerol, or milk protein to amino
acids. Either of these changes might be thought to increase the acidity in the test-tubes.

7 If bile salts had an enzyme-like action on milk, one would expect some change to take place in
tube 2 where the lipase had been inactivated but the bile salts had not.

8 The bile salts appear to make the reaction go faster. This is not a simple pH effect as shown by  tube 2 in which the pH does not change (at least, it is not reduced below pH 8.3). That bile salts
are not essential to the reaction is shown by tube 1 in which the milk fat is hydrolysed in the absence of bile salts.

9 In the absence of information about the relative acidity of fatty acids and amino acids (as explained in 4 above) it is not possible to eliminate the latter. However, the faster reaction in the presence of bile salts does suggest fats rather than proteins as the source of the acid.


Experiment 8. The action of lipase - preparation

Outline Lipase hydrolyses the fat in milk to fatty acids which react with sodium carbonate to
lower the pH of the mixture. This pH change is observed by using phenolphthalein.

Prior knowledge The use of indicators to observe pH changes; the final digestion products of
proteins, carbohydrates and fats.

Advance preparation and materials - per group

milk 20 cm3 phenolphthalein 5 cm3  [1]
3% bile salts 10 cm3  [2] 5% lipase solution 10 cm3 [3]
0.05M sodium carbonate 40 cm3

Apparatus - per group

test-tube rack and 4 test-tubes Bunsen burner
3 labels or spirit marker test-tube holder,
dropping pipette beaker or jar (for rinsing pipette or syringe)
graduated pipette or syringe 10 cm3

- per class
clock

Results Tubes 1 and 3 will probably change from pink to white in about 4 minutes. However, since the efficacy of lipase varies, it is advisable to try out the experiment (for tube 3 only) beforehand and, if the reaction is too slow, reduce the volume of sodium carbonate solution or place the tubes in a water bath at 35°C.

1. 1 g dissolved in 200 cm3 ethanol.
2. Available as sodium tauroglycocholate from Philip Harris.
3. Preferably freshly prepared but will keep for a day or two in a refrigerator. It is easier to make a solution (or suspension) of lipase if it is first made into a paste with a little water.





Experiment 9. The effect of pH on the reaction between pepsin and egg-white

(a) Label five test-tubes 1-5.

(b) Using a graduated pipette or syringe place 5 cm3 of egg-white suspension in each tube.

(c) Using a graduated pipette or syringe, add acid or alkali to the tubes as indicated in the table below.

(d) Prepare a water bath in a 250 cm3 beaker or jar by mixing hot and cold water from the tap to give a temperature of about 40ºC. Have the beaker half full.

(e) Using a graduated pipette or syringe, add 1 cm3 1% pepsin solution to each tube.

(f) Place all five tubes in the water bath.

(g) Copy the table below into your notebook.

(h) After five minutes, return the tubes from the water bath to the test-tube rack and
compare the appearance of the contents.

(i) Compare the pH of each tube by taking a sample with a clean dropping pipette and touching the tip of the pipette on to a piece of pH test paper so that a small drop of liquid runs on to it. Compare the colour produced on the paper, with the standard chart supplied. Repeat this for each tube, rinsing the pipette between samples.

(j) Record your results in the table in your book.


Tube Egg-white suspension and pepsin plus: pH Appearance of contents after 5 minutes
1 2 cm3 sodium carbonate solution
2 0.5 cm3 sodium carbonate solution
3 Nothing
4 1 cm3 hydrochloric acid
5 2 cm3 hydrochloric acid




Experiment 9 .  Discussion

1 In the experiment, at what pH was the hydrolysis of egg-white most effective?

2 This pH is not necessarily the optimum (best) pH for the hydrolysis of egg-white by pepsin. Why not?

3 Can you determine from your results the pH least favourable to the reaction of pepsin
on egg-white?  Explain. 

4 Suggest ways in which the sodium carbonate and hydrochloric acid could influence the reaction between pepsin and egg-white apart from merely altering the pH.

5 Do you think that the pH in your stomach corresponds approximately to the pH in the experiment which resulted in the most rapid hydrolysis of egg-white?  If not, why
not?

6 See if you can find out from a book what the pH in the human stomach really is.

7 Would you expect it to be the same in all people and at all times? Explain.



















































Experiment 9 . Discussion - answers 

1 At pH 2-3 the egg-white suspension will probably clear most rapidly.

2 There may be little to distinguish between the results of pH 2 or 3 as far as rapidity of clearing
is concerned. The experiment would have to be slowed down by, for example, reducing
concentration of pepsin and then the intermediate pH values tested. The most acid atmosphere tested is likely to be the most effective; pH values of less than 2 have not been tried and might prove to be optimal.

3 Since none of the tubes 1-3 is likely to clear in about 20 minutes, it is not possible to determine
the least favourable pH. Perhaps any pH above pH 3 arrests the reaction completely or at least
makes it very slow.

4 The reagents hydrochloric acid and sodium carbonate might have reacted or combined with the egg white or the enzyme and a1tered their properties in a manner similar to that of specific
enzyme inhibitors.

5 Unless students already know the pH in the stomach, they might be reluctant to believe that the acidity is as much as pH 2. They may say that the acid would damage the tissues.

6 The pH of normal, fasting gastric juice is given as pH 1.0 to 1.5.

7 The pH of the stomach might be expected to vary slightly between individuals and to a more marked extent in connection with feeding. Drinking, for example, will dilute the stomach
contents; more hydrochloric acid might be secreted after a protein meal.





Experiment 9. The effect of pH on the reaction between pepsin and egg-white - preparation

Outline The rates at which a cloudy egg-white suspension goes clear are compared at different
pH values.

Prior knowledge Hydrolysis of egg-white by pepsin as in Experiment 10; use of indicators to
estimate acidity and alkalinity; idea of pH as a method of expressing degrees of acidity and
alkalinity.

Advance preparation and materials-per group

1 % egg-white suspension * 30 cm3              pH papers.  Range 2-11     2 strips
0.1M hydrochloric acid    5 cm3                     chart of standard colours (use transparent adhesive                 
0.05M sodium carbonate solution   5 cm3          tape to fix covers from books of pH papers to
1% pepsin solution   10 cm3                              cardboard strips)
                                                                   
Apparatus-per group

test-tube rack and 5 test-tubes                     250 cm3 beaker or jar
5 labels or spirit marker                              thermometer
graduated pipette or syringe  5-10 cm3       dropping pipette



Experiment 10. The Action of Starch Phosphorylase

(a) Cut a piece of potato about 20 mm square and further cut it into small pieces. Place
these in a mortar with a little clean sand and grind them finely with a pestle.

(b) Using a graduated pipette or syringe, add 10 cm3 distilled water to the mixture and continue
grinding for a few seconds.

(c) Filter the extract through a fluted filter paper, into a clean test-tube.

(d) While you are waiting for the filtration, use a dropping pipette to place eight drops of
a 5% solution of glucose phosphate on a spotting tile, in two rows of four drops.

(e) Rinse the dropping pipette and use it to draw up some of the filtered potato extract.
Place one drop in a spare cavity in the tile and test it with a drop of iodine. If it goes blue, the extract will have to be filtered again.

(f) If the sample of filtrate does not give a blue colour when tested with iodine, place one drop of the extract on each drop of glucose phosphate in the top row.

(g) Using a test-tube holder, boil the remainder of the potato extract in the test-tube, over a small Bunsen flame; rinse the dropping pipette and use it to place a drop of the boiled extract on each drop of glucose phosphate in the bottom row. Note the time.

(h) Copy the table below into your notebook.

(i) After 5 minutes place four drops of dilute iodine on the first drop of the mixture in each row and record any colour change. At intervals of 5 minutes, repeat this test with the second, third and fourth drops in both rows.
   From time to time add a drop or two of iodine to the samples already tested because the colours tend to fade.

(j) When the experiment is complete, use a spare cavity in the tile to test a sample of glucose phosphate with iodine.


Drops Colour on adding iodine at 5 min intervals
5 min 10 min 15 min 20 min
Top row Potato extract and glucose phosphate
Bottom Row Boiled potato extract and glucose phosphate
Colour on adding iodine to glucose phosphate alone
Colour on adding iodine to filtered potato extract alone




Experiment 10. Discussion

1 Does the filtered potato extract contain any starch?

2  Does the glucose phosphate contain any starch?

3 How do you interpret the colour changes that occur when iodine is added to the mixture of glucose phosphate and potato extract?

4 Consider two possibilities:
     (i) something in potato extract has converted glucose phosphate to starch.
    (ii) the glucose phosphate has converted something in potato extract to starch.
  (a) In view of the result& with the second row of drops, which of the two possibilities
       is the more likely? Say why.
  (b) Suggest a further control experiment that might help to support one of the two
     alternatives.

5 Are your results consistent with the hypothesis that the potato extract contains an enzyme which converts glucose phosphate to starch?

6 Give a simple chemical explanation of how this change could come about assuming that the phosphate merely makes the glucose more reactive. (Consult the structural formulae
on p. 8.02.)

7 What part could such an enzyme play in the development of a potato tuber?





Experiment 10.  Discussion - answers

1 There should be no blue colour if the potato extract has been properly filtered.

2 Glucose-1-phosphate gives no blue colour with iodine.

3 Colours may range from mauve to blue and suggest the progressive accumulation of starch.

4 (a) Alternative (i) is more likely since boiling the potato extract prevented the formation of   
        starch.
   (b) If the glucose-1-phosphate were boiled and still found to react with potato extract to form
        starch, this would be additional evidence.

5 The results are consistent with this hypothesis.

6 Glucose molecules have been joined together in a long chain, so forming a starch molecule.

7 Soluble sugars, made in the leaves, travel via the phloem to reach the tuber. Enzymes in the tuber convert the sugars to insoluble starch, which is more readily stored.
In certain conditions, starch phosphorylase may catalyse the reverse reaction, i.e. the hydrolysis
of starch.















Experiment 10. The action of starch phosphorylase – Preparation

Outline Potato extract contains an enzyme, starch phosphorylase, which, in vitro at least, converts glucose-1-phosphate to starch. (In living cells, starch phosphorylase catalyses the reverse reaction). The experiment is an illustration of an enzyme catalysing a ‘building up’ reaction in contrast to the other experiments
After 15 minutes with potato extract, glucose-1-phosphate will give a blue colour with iodine showing the formation of starch.

Prior knowledge Starch/iodide reaction; enzymes; the effect of boiling on enzymes; how to fold a ‘fluted’ filter paper.

Advance preparation and materials-per group

potato 1 large one will be sufficient                    dilute iodine+   5 cm3
     for a whole class                                            washed sand   2g
5% solution glucose-1-phosphate* 1 cm3           distilled water l0 cm3

Apparatus-per group

test-tube rack and 1 test-tube           spotting tile
filter funnel                                      dropping pipette
filter paper                                       test-tube holder
mortar and pestle                             Bunsen burner
means of cutting potato safely


-per class

clock

* Glucose-1-phosphate is very expensive and it is better to make up the smallest possible volume in one container and have the students take the drops from this single source under supervision, than to give out 1 cm3 to each group.
A solution of 0.5 g in 10 cm3 distilled water will be enough for 20 experiments. The solution will not keep but the crystals will, if kept in a refrigerator.


Experiment 11 . Dehydrogenase in yeast

During respiration, hydrogen atoms are removed from glucose molecules by enzymes
called dehydrogenases and passed to various chemicals called hydrogen acceptors. As the hydrogen atoms pass from one hydrogen acceptor to another, energy is made available for
chemical reactions in the cell. In this way, substances such as glucose provide energy for vital reactions in living organisms.
In this experiment, a dye called methylene blue acts as an artificial hydrogen acceptor. When this dye is reduced by accepting hydrogen atoms it goes colourless.

(a) Place about 30 mm of yeast suspension in a test-tube and, using a test-tube holder, heat
this suspension over a small Bunsen flame until the liquid boils for about half a minute.
Then cool the tube under the tap.

(b) Label three test-tubes 1-3.

(c) Using a graduated pipette or syringe, place 2 cm3 of the boiled yeast suspension in tube 1.

(d) Using the graduated pipette or syringe, draw up 4 cm3 unboiled yeast suspension and place
 2 cm3 in tube 2 and 2 cm3 in tube 3.

(e) Rinse the pipette or syringe and use it to place 2 cm3 distilled water in tubes 1 and 2.

(f) With the pipette or syringe, place 2 cm3 1 % glucose solution in tube 3.

(g) Prepare a water bath by mixing hot and cold water from the tap to obtain a temperature
between 35 and 45 °C. Place all three tubes in this water bath. Rinse the pipette or syringe.

(h) Copy the table given below into your notebook.

(i) After 5 minutes draw up 6 cm3 methylene blue solution in the pipette or syringe and place
2 cm3 in each tube. Shake all three tubes thoroughly and return them to the water bath, noting the time as you do so. Do not shake the tubes again. 
(j) Watch the tubes to see how long it takes for the blue colour to disappear, leaving the
creamy colour of the yeast. A thin film of blue colour at the surface of the tube may
be ignored but the tubes should not be moved. Record the times in your table.

(k) The experiment may be repeated by simply shaking all the tubes again until the blue
colour returns.


Tube Contents Time for methylene blue to go colourless
1 Boiled yeast
2 Unboiled yeast
3 Unboiled yeast  + 1% glucose




Experiment 11. Discussion

1 Why was distilled water added to tubes 1 and 2?

2 What causes the methylene blue solution to go colourless (according to the introduction
on p. 14.01)?

3 How do you explain the results with tube 1?

4 In which of tubes 2 and 3 was the methylene blue decolourized more rapidly? How can
this result be explained?

5 If the hydrogen atoms for the reduction of methylene blue come from glucose, why should the methylene blue in tube 2 become decolourized at all?

6 What do you think would be the effect of increasing the glucose concentration in tube 3? Explain your answer.

7 How could you extend the experiment to see if enzymes in yeast are capable of reducing
methylene blue?

8 Why, do you think, the colour retuned on shaking the tubes?






















Experiment 11. Discussion - answers

1 The addition of distilled water to tubes 1 and 2 keeps the concentration of yeast and methylene blue the same in all three tubes.

2 The methylene blue accepts hydrogen atoms removed from glucose molecules during respiration. The reduced form of methylene blue is colourless.

3 Boiling will have killed the yeast. Dead yeast is therefore incapable of carrying out one or more stages in the transfer of hydrogen from glucose to methylene blue. (A similar answer may be given in terms of enzymes.)

4 Tube 3 will probably lose its blue colour first. Presumably if the hydrogen atoms for reducing methylene blue come from glucose, additional glucose will mean that more hydrogen atoms are available and decolourization will be more rapid.

5 Respiration will continue in yeast cells, using their own carbohydrate reserves such as glycogen.

6 It might be expected that increasing the glucose concentration would increase the rate of
decolourization up to the point where all the available enzyme or enzymes were being used, or where the concentration of glucose was sufficient to plasmolyse the yeast cells.

7 If enzymes (dehydrogenases) are involved, it should be possible to extract them from yeast by
grinding some dried yeast with sand and distilled water, and filtering. This could be the subject
of further experiment, particularly if little or none of the carbohydrate reserve in yeast comes through in the filtrate.

8  Shaking the tubes introduces more oxygen which re-oxidises the methylene blue



Experiment 11. Dehydrogenase in yeast - preparation

Outline Methylene blue, acting as a hydrogen acceptor, is decolourized during the respiration of
yeast. Addition of small amounts of substrate increases the rate of decolourization.

Prior knowledge An elementary idea of respiration as a process which releases energy during
the breaking down of carbohydrates; yeast is a microscopic living organism.

Advance preparation and materials-per group

20% yeast suspension*                           0.005% methylene blue solution+
    (prepared 1~2 days ahead) 10 cm3     1 % glucose solution
distilled water 10 cm3

Apparatus-per group

test-tube rack and 4 test-tubes      Bunsen burner
3 labels or spirit marker               graduated pipette or syringe 5-10cm3
test-tube holder                            beaker or jar, for water to rinse pipette or syringe

-per class

clock

Result The methylene blue: in tubes 2 and 3 should be decolourized in a few minutes with tube 3 changing first.

* Add 40 g dried yeast and 0.4 g potassium dihydrogen phosphate (KH2PO4) to 200 cm3 distilled water in a tall 600 cm3 (or larger) beaker (a large jam jar will do). Cover the mouth of the container with aluminium foil and bubble air through the yeast suspension for one or two days using an aquarium aerator. Observe the suspension from time to time during the first two hours and control the air flow to prevent the yeast suspension frothing out of the jar.

+Dissolve 0.05 g in 1 litre of distilled water.  Methylene blue stains skin and clothing. Lab coats should be wo
Chapter Three – Food Tests

 Experiment 1. The test for starch

(a) Label four test-tubes 1-4.

                                                         10%  glucose solution into tube 1

                                                         1%     starch solution into tube 2
(b) Put about 20 mm (depth) of
                                                         1% albumen solution into tube 3

                                                          water into tube 4

(c) To each tube, using a dropping pipette, add three drops of iodine solution.  Shake the tube (sideways, not up and down) to mix the contents. Look for any colour changes apart from the yellow colour of iodine itself. Copy the table below and record the results in your notebook.



Substance Colour change after adding iodine
10% glucose solution
1% starch solution
1% albumen solution
water




Experiment 1. Discussion

I The substances selected for testing are examples of three of the principle chemical substances in cells, sugar (glucose), starch, protein (albumen). With which of these substances did iodine react to give a colour change ?

2 Does your result indicate that there is, for example, no sugar and no protein that will
give a colour change with iodine ?

3 What experiments would you have to carry out in order to give a confident answer to
question 2 ?

4 Does the result indicate that starch will always react with iodine solution to give a
colour change?

5 What was the point of the water in tube 4 ?

Experiment 1. Discussion - answers

I Only the starch should give a blue colour. The glucose and albumen solutions may take up the colour of the iodine.

2 The point is being made that the results from a single sample do not justify a sweeping generalization.

3 One would have to test samples of all sugars, all proteins and all fats.

4 The results permit confident prediction only about starch as a 1% aqueous solution.

5 Students often say that the unreactive samples have 'turned yellow'. This control shows that the yellow colour is simply the diluted iodine.
 
Experiment 1. The test for starch - preparation

Outline The experiment establishes that iodine gives a blue colour with starch, but not with glucose or albumen.

Prior knowledge Use of test-tubes and dropping pipettes.

Advance preparation and materials *

     10% glucose solution
                                         
     1% starch solution                 10      cm3 per group
                                 
     1% albumen solution

     iodine solution                        5 cm3 per group

Apparatus-per group


test-tube rack  and 4 test-tubes                       
4 labels or spirit marker                                   
dropping pipette (for iodine)
                 





Experiment 2. The test for glucose

(a) Half fill a beaker with tap water and place it on a tripod and gauze. Heat the
water with a Bunsen burner. While waiting for the water to boil, carry on with instructions (b) to (d).

(b) Label four test-tubes 1-4.

                                                  1%     starch solution into tube 1

                                                  10%  glucose solution into tube 2
(c) Put 20 mm (depth) of     
                                                  1% albumen solution into tube 3

                                                   water into tube 4

(d) To each tube add about 10 mm Benedict's solution.

(e) Place the test-tubes in the beaker of hot water (see Figure on p. 2), and adjust the flame to keep the water just boiling and then copy the table below into your notebook

(f) After about 5 minutes, turn out the flame. Place the four tubes in a test-tube rack and compare the colours. Record the results in the form of a table in your notebook and match the final colours as nearly as possible with crayons.



Solution Colour change on heating with Benedict’s reagent Final colour (crayon)
1 1% starch
2 10% glucose
3 1% albumen
4 water


CLEANING THE TUBES. If a coloured deposit sticks to the inside of the tube even after rinsing with water, it can be removed by pouring in a little dilute hydrochloric acid (HCl).
Rinse the tube with water afterwards.

Experiment 2. Discussion

I What colour changes took place when Benedict's solution was added to each liquid ?

2 The solutions selected for testing are examples of three of the principal chemical substances found in cells:  glucose, starch, protein (albumen). With which of these food materials did Benedict's solution give a decisive change on heating?

3 Apart from the colour, what change took place in the consistency of the Benedict's
solution ?

4 Do your results indicate that any sugar (e.g. sucrose, fructose and maltose) will give the same colour as glucose did when tested with Benedict's solution?

5 Do your results allow you to say that (a) no protein will give a colour change when
heated with Benedict's solution, (b) albumen never reacts with Benedict's solution to
give a colour change ?

6 Can you predict that glucose will always give the same result with Benedict's solution
as it did in your experiment ?

7 Why was water included in the test ?




Experiment 2. Discussion - answers

I No colour change should occur in the cold.

2 The only 'decisive' change, i.e. a red/orange colour, is with the glucose solution. The 1% starch may give a cloudy green colour .

3 Students may have observed the change from a clear solution to a suspension or precipitate.

4 Glucose is an example of a sugar and not 'sugars' in general. Fructose and maltose will react with Benedict’s solution but sucrose will not.

5 Neither of the generalizations are admissible from the results of this single experiment.

6 Such a prediction is not justified from the results of a single experiment.

7 Although failure to give a red precipitate with albumen and starch provides a control, it could be argued that Benedict's solution always gives a red precipitate on boiling and that this change is inhibited by albumen and starch but not sugar. Water acts as a control against this hypothesis.

Experiment 2. The test for glucose- preparation

Outline The Benedict's test is applied to starch, glucose and a protein to establish that it is a specific test for glucose.

Prior knowledge  None



Advance preparation and materials

     10% glucose solution
     1% starch solution *                 5-10 cm3 per group
     1% albumen solution

     Benedict's solution + 30 cm3 per group

Apparatus-per group

test-tube racks and 4 test-tubes                              test-tube brush
Bunsen burner                                                         4 labels or spirit marker
tripod and gauze                                                      250 cm3 beaker
heat mat (to protect bench)

*See p. 01
+ Details of preparation given on p.01. Benedict's solution is preferred to Fehling's      solution since (a) it can be kept for a long time as a single solution, (b) it is less caustic.

Experiment 3. The test for protein

(a) Label four test-tubes 1-4.

                                                              1%.        starch solution into tube 1

                                                              10%       glucose solution into tube 2
(b) Put about 20 mm (depth) of
                                                              1% albumen solution into tube 3

                                                              water into tube 4

(c) Pour into each tube, about 5 mm dilute sodium hydroxide. (CARE *)

(d) Add to this about 5 mm dilute copper sulphate solution. Shake the tube sideways to
mix the contents.

(e) Return the tubes to the rack, leave for a few seconds and record the resulting colours
in your notebook.



Substance Reaction with copper sulphate and sodium hydroxide
1
2
3
4






* CARE. Sodium hydroxide is caustic and dissolves clothing, skin and bench tops. It is
destructive rather than dangerous so if any is spilt on the bench, neutralize it at once with an equal volume of dilute hydrochloric acid and wipe dry. If spilt on clothing do the same but follow with a wash in as much water as possible. If spilt on the skin do not add acid but wash under the tap until the 'soapy' feeling is removed.
Experiment 3. Discussion

I The substances selected for testing are examples of three of the principal types of chemical substances in cells; starch, sugar (glucose), protein (albumen).
 With which of these samples did the reaction give a purple colour ?

2 The substance which gave a purple colour was a single example of its class of substances. Would you expect all other samples of this class  to give the same reaction?

3 What was the point of using test-tube 4 with the water?

NOTE This test is called the ‘biuret’ test. ‘Biuret’ is the name of the compound which gives the purple colour.
Experiment 3. Discussion - answers

I Only the albumen should give a purple colour.

2 The students should not be prepared to make a generalization about all proteins from a single reaction with albumen solution. If albumen is a typical protein, then it is proteins that will normally give the biuret reaction. In fact it is given by all substances having at least two peptide linkages (-CO-NH-) which includes all proteins and most peptides.

3 It could be argued that sodium hydroxide and copper sulphate will produce the characteristic purple colour whether or not proteins are present but the reaction is suppressed in the presence of glucose or starch. The control with test-tube 4 makes this explanation less likely.
 
Experiment 3. The test for proteins - preparation

Outline The Biuret test is applied to starch, glucose and albumen solutions to show that only the protein gives a positive result.

Prior knowledge None, except perhaps that sodium hydroxide has caustic properties.

Advance preparation and materials*

     1% starch solution

     10% glucose solution              5 cm3 per group

     1% albumen solution

     sodium hydroxide 2N or 10%
                                                           10 cm3       per group
     1% copper sulphate solution

(Dilute hydrochloric acid should be available to neutralize any caustic soda spilt.)

Apparatus-per group

test-tube rack and 4 test-tubes
4 labels or spirit marker

*Instructions for preparing the reagents are given on p.04



Experiment 4. The test for fats

ALL APPARATUS MUST BE DRY. ALL FLAMES MUST BE EXTINGUISHED.

(a) Label four test-tubes 1-4.

(b) Into tubes 1 and 2 pour about 20 mm (depth) alcohol (propan-2-ol).

(c) To tube 1 add one drop of vegetable oil, and shake the tube sideways until the oil
dissolves in the alcohol.

(d) In tubes 3 and 4 pour about 20 mm water.

(e) Pour the contents of tube 1 into tube 3 and the contents of tube 2 into tube 4.

(f) Record your results as below:


Result when added to water
1 Oil dissolved in water 3
2 Alcohol alone 4


CLEANING THE TUBES. Keep the oily tubes separate from the others and clean them with hot water and liquid detergent.

Experiment 4. Discussion

I What was the only difference between the contents of tubes 1 and 2 ?

2 What was the visible difference between tubes 3 and 4 after adding the contents of tubes1 and 2 to the water in them?

3 How would you attempt to explain the appearance of the liquid in tube 3 ?

4 What difficulties can you foresee in using this test with samples which contain
both lipids and water ?

5 Water and alcohol mix in all proportions. What precautions must be taken in selecting
the alcohol for use in this experiment and why is this precaution necessary?

6 How would you design an experiment to find out the sensitivity of this test for fats ?
(Describe the method in outline only; details of apparatus etc. are not required).
Experiment 4. Discussion - answers

I Tube 1 contained 1 drop of oil. Tube 2 contained no oil.

2 Tube 3 should contain a cloudy or white liquid and tube 4 a clear liquid.

3 The student may attribute the difference merely to the presence of oil in tube 1 or give a more sophisticated explanation about the fat coming out of solution from the alcohol to form an emulsion in the water.

4 If a sample contains water and lipid, a cloudy emulsion will be formed immediately on extracting with alcohol. Provided that nothing else is causing the cloudiness, this might be acceptable evidence for the presence of lipid.

5 The alcohol needs to be free from water otherwise an emulsion will form on attempting the extraction. This does not necessarily invalidate the test as argued above.

6 A serial dilution, e.g. 1 cm3 oil dissolved in 100 cm3 alcohol, diluted with alcohol to 0.1%, .01 % etc. Equal volumes of the diluted extracts to be poured into equal volumes of water.

Experiment 4. The test for fats – preparation

Outline A little oil is dissolved in alcohol and the resulting solution poured into water to show the formation of an emulsion.

Prior knowledge Some ideas about emulsions (not essential).

Advance preparation and materials

    propan-2-ol     10 cm3 per group
    vegetable oil   1 cm3 per group

Apparatus-per group

    test-tube rack and 4 test-tubes
    4 labels or spirit marker
    dropping pipette (for oil)






Experiment 5. How sensitive is the iodine test?

Make a serial dilution of starch solution as follows:

(a) Place five clean test-tubes in a rack and label them 1-5.

(b) Using a 10 cm3 pipette or syringe, place 10 cm3 1% starch solution in tube 1.

(c) Take 1 cm3 of the solution from tube 1 with the pipette or syringe and place it in tube 2.
(see Figure on p. 5.02)

(d) Add 9 cm3 water with the pipette or syringe to tube 2 and shake sideways to mix the contents.
The 1 cm3 of 1% starch solution has now been diluted ten times to make a 0.1 % solution.

(e) Transfer 1 cm3 of this 0.1% starch solution from tube 2 into tube 3.

(f) Add 9 cm3 water to tube 3 and shake. This will give a .01 % solution.

(g) Transfer 1 cm3 solution from tube 3 to tube 4 and dilute with 9 cm3 water so
 making a .001% solution.

(h) Repeat the operation for tube 5.

(i) Tubes 1-4 will now have 9 cm3 solution in them but tube 5 will have 10 cm3 so remove
1 cm3 of this solution from tube 5 so that the subsequent test is a fair one.

(j) Add one drop of iodine solution to each tube and shake to mix. You can add more
iodine to each tube, if you think it will make any difference, so long as you add an
equal number of drops to each.

(k) Record your results in a table similar to the one below, matching the colours as nearly as possible with crayons.

Strength of solution Colour after adding iodine Colour (crayon)











Experiment 5.  Discussion


I What is the least concentration of starch that iodine solution will detect?

2. If your answer to question 1 is a single percentage, it will need a qualifying statement.

3 Do you think a .0001 % solution contains any starch at all?







Experiment 5. Discussion - answers

I A blue colour will probably be apparent at .01%.

2 If the student says, however, that .01% is the least concentration that iodine will detect, he or she is overlooking the fact that concentrations intermediate between .01 and .001% have not been tested. Iodine may very well give results with .005% starch. 

3 In theory there must come a point in a serial dilution when the sample withdrawn by the pipette may contain no molecules of solute at all. Provided one has a large enough volume of a .0001 % solution, there should be some starch molecules present.

Experiment 5. How sensitive is the iodine test? - preparation

Outline A serial dilution of starch solution is made, starting from 1% and
going down to .00010%
The resulting solutions are tested with iodine to estimate its sensitivity.

Prior knowledge Experiment 1 (not essential). Use of graduated pipette or syringe for measuring small volumes of liquid.

Advance preparation and materials
1 % starch solution 15 cm3 per group
iodine solution 5 cm3 per group

Apparatus-per group
10 cm3 graduated pipette or syringes*
 test-tube rack and 5 test-tubes
dropping pipette (for iodine)
beaker (for water)
5 labels or spirit marker

*A 10 cm3 syringe can be used for adding the water; a 1cm3 syringe can be used for withdrawing the samples




Experiment 6   Testing food for the presence of starch

PRECAUTION. Experiment 3 showed that the starch test is a sensitive one, so if you use a
test-tube for more than one experiment, make sure it is thoroughly cleaned out each time or
else traces of starch from one food sample may remain on the sides and give a positive
result for a sample which, in fact, contains no starch. For the same reason, the mortar and pestle must be washed between each test.

(a) Label six test-tubes 1~6.
Copy the table given below into your notebook.

(b) Crush a 1 cm cube, or equivalent quantity, of the first food sample in a mortar with
about 10 cm3 water (50mm in a test-tube).

 (c) Pour the mixture into a clean test-tube and, using a test-tube holder, heat the mixture
in a small flame of a Bunsen burner till it boils for a few seconds, shaking the tube
gently all the time.

(d) Cool the tube under a running tap.

(e) Add five drops of iodine solution. 

(f) Record your results in the table in your notebook and repeat the experiment with the next food sample.

Food sample Colour change with iodine Interpretation of result
1    Potato
2    Onion
3    Bread
4    Banana
5    Apple
6    Dried milk

















Experiment 6 . Discussion


I If a food sample, extracted and tested as above, gives a blue colour with iodine solution,
can you assume that starch is present in the sample?

2 Does the result indicate that the only food present is starch?

3  Does the result mean that starch will be present in all samples of the same food? 

4 If the food sample, when tested, gives no blue colour, does it mean invariably that no
starch is present?

5 What difficulties are there in trying to answer the question, 'Do apples contain starch?'

If you did experiment 5, answer questions 6 and 7.

6 If, with your sample of potato, you obtained a blue colour that corresponded closely to
your 1% result in experiment 5, why would you not be justified in saying that 1% of
the potato sample was starch? (There are several reasons.)

7 If the blue colour had corresponded to the 0.1 % result and the potato sample had been
a one centimetre cube, what would you be prepared to say about the concentration of
starch in the potato? 
What reservations would you make in giving your answer?


Experiment 6.  Discussion - answers

I Provided that all precautions are taken to exclude starch from other sources, it is reasonable to say that the food sample contains starch. 

2 The result shows that starch is present but there may be other substances also.

3 Other samples may not contain starch, e.g. old peas will contain starch, but young peas may not. Also, different parts of the same vegetable or fruit may have different amounts of starch, e.g. embryo and endosperm of seeds.

4 Failure to obtain a blue colour may mean that starch is not present or is not in sufficient
concentration to give a blue colour with iodine or that it has not been successfully extracted (though the latter seems unlikely).

5 The problem is one of sampling: (a) different varieties of apple may have different food reserves, (b) at various stages of development, starch may or may not be present, (c) there may even be diurnal fluctuations (unlikely with apples but possible with leafy vegetables), (d) there may be too little starch for iodine to detect, (e) the starch may be difficult to extract.

6 (a) The 1 % solution was the strongest tested; the colour may be so intense that the stronger solutions would show no visible difference. Thus potato may contain considerably more than
1 % starch.
(b) The statement would assume that all starch had been extracted from the potato.
(c) Unless the volume of the sample is known, the diluting effect of 10 cm3 water cannot be calculated but the content of 'undiluted' potato would be more than 1 %.

7 Since the 0.1 % colour is in the middle of the range of standards produced by Experiment 5, comparison is justified. One cubic centimetre of potato in 10 cm3 water is diluted just over x10 so that a 0.1 % result on the colour comparison means a 1% concentration in the actual potato.
The reservations are: (i) assuming almost total extraction of starch, (ii) accuracy of colour comparison.



Experiment 6.  Testing food for the presence of starch - preparation

Outline Food samples are crushed in water and the extract tested with iodine.

Prior knowledge Experiment I (or knowledge of starch-iodide reaction). Heating test-tubes in
Bunsen flames.

Advance preparation and materials
Food samples as indicated on p. 6.01, cut into pieces of about 1 cm3. If the teacher uses the samples suggested, there will be a range of results from no starch (apple and onion), a little starch (banana), to much starch (potato).

If seeds are to be used, they should be soaked the day before.

iodine solution 5 cm3 per group.

Apparatus-per group
test-tube rack and 6 test-tubes
dropping pipette (for iodine)
6 labels or spirit marker
6 watch-glasses (for samples)
test-tube holder
Bunsen burner
test-tube brush
Beaker for water
 small mortar and pestle

-per lab.
one or two balances reading to nearest gram if samples are to be weighed

Experiment 7. A comparison of the vitamin C content of fruit juices

Vitamin C is ascorbic acid: one of its chemical properties is that it is a 'reducing agent',
ie. removes oxygen from or adds hydrogen to other chemicals.
DCPIP is dichlorophenol-indophenol, a blue dye which is decolourized by ascorbic acid
on account of its reducing properties.

(a) Copy the table below into your notebook. (b) Label six test-tubes 1-6.

(c) Use a 2 cm3 syringe to put 1 cm3 DCPIP solution into each tube.

(d) Use a fresh syringe to draw up 2 cm3 of 0.1 % ascorbic acid solution (ie. pure vitamin C solution). Draw the solution into the pipette until the bottom of the plunger is against the 2.0 cm3 mark on the barrel of the syringe (Fig. 1 p.7.02).

(e) Squeeze the plunger of the syringe carefully so that you add one drop at a time of
ascorbic acid solution to the DCPIP in tube 1.
Shake the tube and go on adding the ascorbic acid from the syringe until the blue dye just changes to a colourless liquid *. (If you empty the syringe before the DCPIP goes colourless, refill it and add 2 cm3 to your final volume.)

(f) Note the position of the plunger in the syringe and subtract the reading from 2.0 (or
simply count the divisions from the top mark on the barrel to the bottom of the plunger).
This will tell you the volume of ascorbic acid solution you have added (Fig. 2).

(g) Write this volume into the appropriate space in the table (not forgetting to add 2 cm3
for each time you refilled the syringe.)

(h) Wash the syringe and repeat the experiment with one of the fruit juices, using the
DCPIP in tube 2.

(i) Repeat the experiment with an acidified solution or sodium sulphite, which is not a
fruit juice or a vitamin, but is a reducing agent.

*Note. The acid in the fruit juice may turn the DCPIP from blue to red but it is the point at which the dye becomes colourless which counts. Coloured fruit juices, however, will not give a colourless solution. The end point is when the blue (or red) colour disappears and the original colour returns.

Tube Liquid Volume needed to decolourize DCPIP
1 0.1% ascorbic acid
2 Fresh lemon juice
3 Fresh orange juice
4 Canned or bottled orange juice
5 Fresh grapefruit juice
6 Fresh grape juice
7 Orange squash
8 Orange juice that has stood in an open beaker for ………days


Experiment 7. Discussion

I If the volume of fruit juice needed to decolourize DCPIP is greater than the volume of 0.1 % ascorbic acid, does this mean that the juice contains more or less vitamin C than
ascorbic acid solution?

2 On this basis, which of your samples contained the most vitamin C?

3 Did the canned fruit. juice differ greatly from the fresh juice?

4 Suppose that 0.8 cm3 orange juice decolourized 1 cm3 DCPIP but it took 1.6 cm3grapefruit
juice to decolourize the same volume of DCPIP; does this mean that oranges contain more vitamin C than grapefruit?

5 What might take place at a canning factory which could (a) reduce, (b) increase the
vitamin C content of fruit juice?

6 Did the fruit juice which had been exposed to air for several days have significantly less
vitamin C than the fresh juice?

7 Sodium sulphite is not vitamin C and yet it decolourized DCPIP - Why did it do this?
(See introductory notes on p. 7.01 and instruction (i).)

8 Since chemicals other than vitamin C will decolourize DCPIP, does this mean that using the dye as a test for vitamin C is unsatisfactory?

9 If canned or bottled juice has sulphur dioxide or sodium sulphite added as a preservative,
will this affect the reliability of your results with DCPIP?

 
















Experiment 7. Discussion - answers

I The greater the volume of juice needed to decolourize DCPIP the less ascorbic acid it contains.

2 The answer will vary according to the source and brand of the various juices but usually fresh orange and lemon juices contain more concentrated vitamin C than the others. Canned grapefruit juice may be substantially weaker and orange squash very weak in vitamin C compared with the others. Grape juice may contain so little that the student has to give up after squirting several
syringes full of juice into the DCPIP.

3 Canned juice is frequently weaker than fresh juice.

4 The result shows that the orange juice in the experiment has a higher vitamin C content than the grapefruit but a generalization about all oranges and all grapefruit is not justified. There is probably a great deal of variation with species, season and age of fruit.

5 (a) Heating the juice might destroy some of its vitamin C content.
(b) Heating the juice might concentrate the vitamin. Ascorbic acid might be added artificially to increase its concentration.

6 There is likely to be a significant difference in the vitamin C content of fruit juices exposed to
the air for several days.

7 Sodium sulphite is a reducing agent and so decolourizes DCPIP.

8 Since chemicals other than vitamin C will affect DCPIP, the test is unsatisfactory unless it can
be assumed that reducing agents other than ascorbic acid are unlikely to be present.

9 If sulphur dioxide or sodium sulphite is present, presumably less of the juice will be needed to decolourize the DCPIP. Whether this will make the results substantially unreliable will depend
on the relative concentrations of sulphite and ascorbic acid in the juice.
Experiment 7. A comparison of the vitamin C content of fruit juices -preparation

Outline A comparison is made of the volumes of different juices needed to decolourize DCPIP solution. The more vitamin C there is in a juice, the less of it will be needed to decolourize
the dye.

Prior knowledge None.

Advance preparation and materials
0.1 % dichlorophenol-indophenol (DCPIP) 10-15 cm3 per group
0.1 % ascorbic acid (freshly prepared) and fruit juices as listed on p. 7.01.  2 cm3 per group. If the effect of exposure to air is to be examined, about 100 cm3 of orange juice should be prepared
and left in a beaker for several days in advance.
Grape juice may be obtained from health food stores or by crushing grapes. 10 cm3 per group
Orange squash allow 5-10 cm3 per group
Acidified sodium sulphite solution; mIx equal volumes of 10% sodium sulphite solution and
10% (bench strength) hydrochloric acid: allow 5-10 cm3 per group
NOTE. The effect of boiling fruit juices or leaving them exposed to air seems very unpredictable. In some cases it seems to have no effect at all.

Apparatus-per group
two 2 cm3 disposal syringes (one exclusively for DCPIP)
test-tube rack and 6 test-tubes
6 labels or spirit marker





Experiment 8. How specific is Benedict's reagent?

(a)  Half fill a beaker with tap water and place it on a tripod and gauze. Heat the water with a Bunsen burner. While waiting for the water to boil carry on with instructions (b) - (d).

(b) Label four test-tubes 1-4.

(c) Pour about 20mm (depth) of the following liquids into each tube:

    1. 10% glucose solution
    2. 10% sucrose solution
    3. 10% maltose solution
    4. 10% fructose solution

(d) To each tube add about 10 mm Benedict's solution.

(e) Place the test-tubes in the beaker of hot water and adjust the flame to keep the water just boiling.

(f) After 5 minutes, turn out the flame and compare the colours. Record the results in your table and match the colours as nearly as possible with crayons.

Sugar Final colour after heating with Benedict’s solution Colour change
1 10% glucose
2 10% sucrose
3 10% maltose
4 10% fructose

CLEANING THE TUBES. If, after rinsing the tubes, a reddish deposit still adheres to the inside, it can be removed with a little dilute hydrochloric acid followed by rinsing with water


Experiment 8. Discussion

1 Glucose, sucrose, maltose and fructose are all sugars but they differ from each other in
chemical composition. Did they all react with Benedict's solution to give the same
result?

2 If a food sample contains a sugar, will it give a visible reaction when boiled with Benedict's solution?

3 If a food sample contained glucose, would it give a colour change when heated with
Benedict's reagent?

4 If a food sample contained sucrose, would it give a colour change on heating with Benedict's reagent?


Experiment 8  Discussion - answers

1 Glucose, fructose and maltose should give a red precipitate, but sucrose should not.

2 If a food sample contains a sugar, it will give a result with Benedict's solution only if the sugar is a reducing sugar, e.g. glucose.

3 A food sample containing glucose should normally give a red precipitate on heating with Benedict's solution.

4 A food sample containing sucrose and no other sugar would not normally react with Benedict's
solution.

NOTE: It is left to the teacher to decide how involved he or she wishes to become with reducing and non-reducing sugars. He or she may decide to settle for the Benedict's reaction as a test for sugars, at one end of the scale or, with the top flight of students, as a reaction with aldehyde groups, which are present in certain sugars.



Experiment 8. How specific is Benedict's reagent? - preparation

Outline Glucose, sucrose, maltose, and fructose are tested with Benedict’s solution.
Sucrose fails to give a red precipitate.

Prior knowledge Reaction of Benedict’s solution with glucose (not essential).

Advance preparation and materials

      10%  glucose
      10%  sucrose                 
      10%  maltose
      10%  fructose
      Benedict’s solution  30 cm3 per group

Apparatus-per group

test-tube rack and 4 test-tubes    250 cm3 beaker
test-tube holder                           tripod and gauze
Bunsen burner                             heat-resistant mat (to protect bench)
4 labels or spirit marker


Experiment 9   The test for sucrose

(a) Prepare a water bath by half filling a beaker with water and placing it on a tripod and
gauze. Heat the water with a Bunsen flame until it boils and then turn down the flame
so that the water just continues to boil.

(b) While waiting for the water to boil, label three test-tubes 1 - 3 near the rim.

(c) To tube 1 add about 20 mm (depth) 10% sucrose solution followed by two drops of dilute
hydrochloric acid. Now place tube 1 in the water bath for about two minutes.

(d) In tube 2 place about 20 mm of 10% sucrose solution.

(e) In tube 3 place about 20 mm of water and two drops only of dilute hydrochloric acid.

(f) After tube 1 has been in the water bath for 2 minutes, add about 10 mm Benedict's
solution to all three tubes.

(g) Place all three tubes in the water bath for one or two minutes.

In your notebook record the results as follows:

Treatment Colour change
1 10% sucrose boiled with HCl and tested with Benedict’s solution
2 10% sucrose tested with Benedict’s solution
3 Benedict’s solution with hydrochloric acid

CLEANING THE TUBES. If a red deposit remains on the inside of the tubes after rinsing
them out, it can be removed with a little dilute hydrochloric acid followed by rinsing with water.



Experiment 9  Discussion

1 In tube 2, did the sucrose give a colour change on heating with Benedict's solution?

2 In tube 3, did Benedict's solution give any reaction when heated with acid?

3 How was the sucrose solution treated before it would react with Benedict's solution?

4 What was the point of heating Benedict's solution with dilute acid in tube 3?

5 If a food sample contained only sucrose, how would you demonstrate that a sugar was
present?

6 If a food sample contained only glucose, what would you observe if you boiled it with dilute acid and then tested it with Benedict's solution?

7 If the only sugar present in a sample of food was sucrose, you would, have to carry out
two successive tests to prove that this particular sugar was present. What are these two
tests and why are they both necessary?


Experiment 9. Discussion - answers

1 There should be no change on heating sucrose with Benedict's solution.

2 In tube 3 there should be no change on heating Benedict's solution with the acid.

3 Before the sucrose would react with the Benedict's solution it had to be heated with dilute hydrochloric acid.

4 Heating Benedict's solution with acid was a control to avoid the criticism that in tube 1 the final precipitate was due not to the hydrolysed sucrose but to the acid itself.

5 A food sample boiled with dilute acid would subsequently give a red precipitate with Benedict's solution showing a sugar (but not specifically sucrose) to be present.

6 A food sample containing glucose would give a red precipitate with Benedict's solution whether or not it was previously boiled with acid.

7 The first test would be to boil the sample with Benedict's solution. A negative result shows that no reducing sugar is present. The second test is to hydrolyse the sample with acid before heating with Benedict's reagent and show that a red precipitate is formed. If the first test is not carried
out there is nothing to indicate that the precipitate is not the result of a reducing sugar (i.e. not sucrose).





Experiment 9. The test for sucrose - preparation

Outline Sucrose is hydrolysed with dilute hydrochloric acid and then tested with Benedict's solution.

Prior knowledge Benedict's reaction (not essential).

Advance preparation and materials

10% sucrose solution                20 cm3 per group
Benedict's solution                    20 cm3 per group
dilute hydrochloric acid (2N)     5 cm3 per group

Apparatus-per group

250 cm3 beaker or tin can (for water bath)        tripod
test-tube rack and 3 test-tubes                           gauze
Bunsen burner                                                   heat-resistant mat                                             
dropping pipette (for the acid)                          3 labels or spirit marker



Experiment 10. Testing food samples for the presence of sugar (other than sucrose)

 (a) Prepare a water bath by half filling a beaker with water (hot water from the tap will
save time) and heating it on a tripod and gauze over a Bunsen burner. When the water boils, turn the flame down so that boiling point is just maintained.

(b) While waiting for the water to boil, draw up in your notebook a table like the one
below so that the results can be filled in as you obtain them.

(c) Label six test-tubes 1-6. If you use sticky labels, place these near the tops of the tubes so that they do not float off in the water bath.

(d) Crush about 1 cm cube, or equivalent volume, of the food in a mortar with about
5 cm3 (20 mm in a test-tube) of water. (If the sample is a liquid, simply pour 20 mm of it
into a test-tube.)

(e) Pour about 20 mm of the crushed mixture into a test-tube and add about 10 mm Benedict's solution.

(f) Place the test-tube in the water bath for 5-10 minutes and prepare the next food sample.

(g) Repeat the test with the remaining food samples and compare the colours produced in the test-tubes.

Enter the results in the table in your notebook.

Food sample Colour change on heating with  Benedict’s solution Interpretation
1  Onion
2  Milk (or dried milk)
3  Rice
4  Raisin (or sultana)
5  Potato
6  Banana

CLEANING THE TEST-TUBES. If, after rinsing, a film of red cuprous oxide adheres to the
inside of the test-tubes, it can be removed with dilute hydrochloric acid followed by rinsing with water.








Experiment 10 Discussion.

1 If a food sample, extracted and tested as in experiment 9, gives a yellow or red precipitate
on heating with Benedict's solution, can you assume that it therefore contains sugar of
some kind?

2  Does a yellow or red precipitate indicate that only a sugar is present and not starch or
protein?

If you did experiment 7 and/or experiment 8 answer questions 3 and 4.

3 Do your results enable you to say whether sucrose is present or absent in any of these
food samples?

4 If one of the samples gave no reaction on heating with Benedict's solution, what further
tests could you do to see if sucrose was present?












                                           Using the water bath 
Experiment 10. Discussion - answers

1 A red precipitate in this context is indicative of a sugar (reducing sugar).

2 The test gives no information about proteins and starch which could be present in addition to sugar.

3 Sucrose could be present in any of the samples but the results here will not indicate either its presence or absence.

4 To detect sucrose, the food sample would have to be boiled with dilute hydrochloric acid and then tested with Benedict's reagent. A red precipitate would indicate that sucrose had been present but only if no other sugar had been detected in the previous experiments.


Experiment 10 Testing food samples for the presence of sugar (other than sucrose) - preparation

Outline Samples of food are crushed in water and tested with Benedict's solution

Prior knowledge The Benedict's reaction,
Advance preparation and materials

     Food samples as in the table on p. 9.01
     Benedict's solution 50 cm3 per group
     (If seeds are to be used, they should be soaked for 24 hours beforehand):

Apparatus-per group

     test-tube rack and 6 test-tubes                      tripod
     small mortar and pestle                                gauze
     test-tube brush                                              heat resistant mat
     6 watch-glasses for samples                         6 labels or spirit marker
     250 cm3 beaker or tin can for water bath

    -per lab.
    One or two balances, weighing to 1g if samples are to be weighed.




Experiment 11. Testing food samples for protein

(a) In your notebook draw up a table like the one below.

(b) Label six test-tubes 1-6.

(c) If the food is a solid, crush a small piece (about 5 mm cube) in a mortar with about
10 cm3 water. (If the food is a liquid, simply pour about 20 mm of it into a test-tube.)

(d) Pour about 20 mm of the crushed mixture into a test-tube.

(e) Add about 5 mm of sodium hydroxide solution. (CARE *)

(f) Add about 5 mm dilute copper sulphate solution and shake the tube gently sideways
to mix the contents. Return the tube to the rack and wait for a few seconds.

(g) Record the results in the table in your notebook.

Food Result on adding sodium hydroxide and copper sulphate Interpretation
Milk
Egg white
Onion
Apple
Beans
White meat


* CARE. Sodium hydroxide is caustic and dissolves clothing, skin and bench tops. It is
destructive, so if any is spilt on the bench, neutralize it at once with an equal volume of dilute hydrochloric acid and wipe dry. If spilt on clothing, do the same but follow with a wash in as much water as possible. If spilt on the skin, do not add acid but wash under the tap until the 'soapy' feeling is removed.


Experiment 11 Discussion

1 In which of your samples do you think protein is fairly abundant?

2 In which samples do you think protein is absent or in very low concentration?

3 If one of your samples gives no purple colour with the biuret test, does this result mean
that the sample contains no protein?

4 If a food sample gives a purple colour with the biuret test does this mean that it contains
only protein?


Experiment 11. Discussion - answers

1 Meat and egg white should give a strong colour.

2 Milk and beans should give an indication of protein. Apple and onion should show little or no
protein reaction.

3 Possibilities:
   (a) the sample contains no protein.
   (b) the sample does not contain enough protein to react.
   (c) it contains a protein which has not been extracted by the crushing in water.

4 The sample could contain other classes of food in addition to protein.



Experiment 11 Testing food samples for protein – Preparation

Outline Food samples are crushed in water and the biuret test applied.

Prior knowledge The biuret reaction (not essential). The caustic nature of sodium hydroxide.

Advance preparation and materials

   Food samples as on p. 10.01
   Beans, peas, etc. should be placed to soak the day before the experiment.
   The white meat can be chicken or fish; preserved dogfish meat will do.
   Boil the egg white with tap water to make a suspension; 10 cm3 egg white per 100 cm3 water   
   is sufficient. Allow 5 cm3 per group.
 
   2M or 10% sodium hydroxide solution 
   1 % copper sulphate solution

Apparatus-per group

small mortar and pestle                     Bunsen burner
test-tube rack and 6 test-tubes           6 labels or spirit marker
test-tube holder                                  6 watch-glasses (for food samples)





Experiment 12. Testing food samples for fats (the emulsion test)

ALL APPARATUS MUST BE DRY. ALL FLAMES MUST BE EXTINGUISHED.

(a) Label four test-tubes 1-4.

(b) Cut a small sample of food (not more than 5 mm cube).

(c) Crush the sample in a dry mortar and pestle* with about 10 cm3 alcohol (propan-2-ol).

(d) Filter the crushed mixture into a dry test-tube.*

(e) While you are waiting for the first liquid to filter, copy the table given below into your 
 notebook.
While the second and third samples are filtering, the next food sample can be prepared.

(f) Pour the filtrate into a test-tube containing about 1 cm (depth) water.

(g) Examine the liquid in the test-tube and record in your table whether it is clear, cloudy or slightly cloudy.

Food Appearance of filtrate when added to water Interpretation
Nuts
Cheese
Potato
Meat

* The mortar, pestle and filter funnel must be rinsed and carefully dried between each experiment



Experiment 12. Discussion

1 Which of the food samples do you consider contain fat (or oil)?

2 If the filtrate was cloudy before adding it to water, does this mean that your results are
going to be unreliable?

3 What could be the cause of a cloudy filtrate?

4 If a clear filtrate, when added to water, produces a cloudy liquid, does this indicate that the food sample contained fat (or oil)?

5 If a cloudy filtrate, when added to water, remains the same or goes less cloudy, does it
mean that the food sample contained no fat (or oil)?

6 Would it be sufficient to add a crushed food sample directly to water and look for the
formation of a cloudy liquid?


. Experiment 12. Discussion - answers

1 Nuts and cheese will probably give emulsions, indicating the presence of fats.

2 If the filtrate is cloudy before adding to water:
   (a) it may still be possible to see if it goes more cloudy when poured into water.
   (b) the production of a cloudy extract after filtration may in itself be an indication that fats are 
   present, unless something else is causing the cloudiness.

3 A cloudy filtrate could be produced by
   (a) the presence of water in the food sample, the alcohol or the apparatus,
   (b) some other matter suspended in the alcohol, fine enough to pass through the filter.

4 An emulsion produced by a clear filtrate poured into water is a fairly reliable indication of the presence of fats.

5 Failure to produce an emulsion means either that fat is not present or it has not been extracted effectively.

6 Unless the fat is extracted as a solution in alcohol (or other suitable solvent), it is unlikely that an emulsion will be formed. Cloudiness due to oil droplets will be difficult to distinguish from cloudiness due to food particles.





Experiment 12. Testing food samples for fats - preparation

Outline Samples of food are crushed in alcohol and the filtrate poured into water to see if an
emulsion is formed.

Prior knowledge Experiment 4 (Food tests) or some knowledge of emulsions (not essential).

Advance preparation and materials

food samples (e.g. list on p. 11.01)
(if seeds are used, they should not be soaked)
propan-2-ol 40-50 cm3 per group

Apparatus-per group

test-tube rack and 8 test-tubes                                    small mortar and pestle
4 watch glasses (for samples)                                    4 filter papers
scalpel or something to cut food samples                  filter funnel
4 labels or spirit marker                                             4 watch glasses (for samples)







Experiment 13. Comparison of energy value of food

READ ALL THE INSTRUCTIONS BEFORE STARTING.

(a) Place a little sugar in the bottom of a dry test-tube.

(b) Using a test-tube holder, heat the bottom of the tube strongly in a Bunsen flame.

(c) Look for steam coming off or water condensing on the cooler upper part of the test-tube.

(d) When smoke starts to come out of the tube, try to set it alight with a lighted splint. Keep heating the tube and go on trying to light the smoke until no more appears.

(e) DO NOT PUT THE HOT TEST-TUBE ON THE BENCH OR IN THE RACK but place it on a heat mat until cool.

(f) Draw up the table below in your notebook.

(g) Repeat this test with different food samples and record your results.


Food How much steam or condensed water? How much smoke? Did it burn vigorously, briefly or not at all?
Sugar
Apple
Cheese
Potato
Nuts
Dried potato









Experiment 13 . Discussion

1 Is it likely that what happens to food in the test-tube is similar to what happens in the body? Give reasons.

2 In this experiment, in what forms is the energy released from the food?

3 In the body, in what forms is energy released?

4 From the experiment, what kind of result would you regard as showing a great deal of energy present in a food sample?

5 On this basis, which food samples contained most energy?

6 Does it follow that the foods listed in the answer to 5 will supply most energy to the body? Explain.

7 When a food sample, on heating, fails to produce any inflammable gas, does this signify that it contains little energy? Explain.

8 Suppose a food sample is rich in energy and does give an inflammable gas on heating but
the food also contains a great deal of water. What problems can you see in conducting and interpreting this experiment?

9 What is the point of eating food which contains very little energy?






Experiment 13. Discussion - answers

1 It seems most unlikely that anything resembling the destructive distillation of food takes place
in the body except in general terms such as the decomposition of food and release of energy.

2  Energy is released as heat and light (flame).

3 In the body, energy is released as heat and movement (manifestly).

4 A large flame lasting for a long time would indicate abundant energy.

5 Cheese and nuts, from these examples, will probably appear to contain most energy as judged
by their flame.

6 It does not follow that these foods will necessarily provide most energy in the body but they certainly seem to have most energy available.

7 Failure to produce any inflammable gas may mean little available energy, too low a temperature to volatilize the constituents or too much steam.

8 An energy-rich food containing much water may produce so much steam that the inflammable
gases cannot be ignited.

9 Food with little energy content is usually eaten for its flavour, its vitamin content or roughage value.





Experiment 13. Comparison of energy value of food - preparation

NOTE It is not worth embarking on this experiment (a) unless you have a supply of expendable, cheap, soda-glass test-tubes, because they are difficult or impossible to clean afterwards, and (b) unless it can be staged at the end of a morning or afternoon session because the lab is filled with the smell of burning food. It could be done as a demonstration to small groups.

Outline Samples of different food are destructively distilled by heating in a soda-glass test-tube.
The water is driven off as steam, some of which condenses on the sides of the tube. The volatile gases are ignited and the size and duration of the flames give an indication of the energy value.

Prior knowledge None.

Advance preparation and materials.

 Food samples as indicated on p. 14.01, ½ cm3 is sufficient. 

Apparatus-per group

6 soda-glass test-tubes          Bunsen burner
test-tube holder                     wooden splints
heat mat                                 6 watch glasses for food samples



Chapter Four – Osmosis
Experiment 1. Osmosis

(a) Use a dropping pipette to fill the cellophane tube and connector with syrup solution
 (Fig. I).

(b) Push the capillary tube into the connector far enough to bring the syrup to a level      somewhere in the lower third of the capillary (Fig. 2).

(c) If there are short columns of liquid and air bubbles trapped in the upper part of the capillary, squeeze the cellophane tubing to force the syrup to the top of the capillary and then let it return slowly.

(d) Clamp the capillary tube vertically in a stand and then lower it to immerse the cellophane tubing in a beaker or jar of water until it is completely covered but not touching the bottom of the vessel (Fig. 3). Leave for a minute so that the syrup can acquire the temperature of the water .

(e) Move the rubber band on the capillary to mark the level of the liquid.
(f) Watch the level of the liquid in the capillary. If it falls rapidly, examine the cellophane tube through the side of the beaker to see if syrup is escaping.
Measure any change in level after 10 minutes.





                                                                                                   



Experiment 1. Discussion

I Assuming that there were no leaks, what happened to the level of liquid in the capillary ?

2 What does this result indicate about the volume of liquid in the cellophane tubing and capillary?

3 What explanation can you offer for this change in volume ?



4 What assumptions have you made about the cellophane membrane and the movement of sugar molecules?

5 This experiment has been conducted without a control. What control do you think should be carried out to exclude any alternative explanations of the results ?


Experiment 1. Discussion - answers

I The syrup should rise up the capillary tube at about 15 mm per minute.

2 The volume of liquid in the cellophane tube and capillary must be increasing.

3 If one rules out temperature effects, the most likely explanation is that water is passing through the cellophane tubing into the syrup.

4 It has been assumed that sugar molecules cannot pass through the membrane or that they do so much more slowly than water molecules

5 The only alternative explanation would involve fairly drastic temperature changes. The control in this case would be to set up the same apparatus with syrup or water both inside and outside the dialysis tube.
Experiment 1. Osmosis - preparation

Outline A dialysis tube containing syrup is immersed in water and connected to a capillary tube. The syrup level is seen to rise.

Prior knowledge Syrup is mainly sucrose. Meaning of 'molecule'. Substances diffuse from regions of high concentration to regions of low concentration.

Advance preparation and materials

Osmometers. Cut 2 cm lengths of 6-7 mm diameter soda glass tubing and flame polish      the ends.
Cut 15 cm lengths of 6 mm (¼ in) dialysis tubing, soak in water for a minute or two and    knot one end securely. Push the glass tubing into the open end of the dialysis tubing and allow to dry. Cut 3 cm lengths of red rubber tubing 8 mm outside diameter, 5 mm bore, and fit these over the end 10 mm of the glass tubing.
The osmometers may be stored in 30% alcohol to keep them moist and flexible, or they 
   can be stored dry though they may develop cracks in this way. In either case, soak the
   osmometers in water for a few minutes before the experiment and allow I per group.
Capillary tubes. Cut 40 cm lengths of I mm bore, 6-7 mm diameter, capillary tubing and
   flame polish the ends. Allow I per group.
Markers. Cut rings about 2 mm wide from the red rubber tubing and fit one on each
   capillary.
Syrup. Dilute golden syrup with its own volume of water and stir to obtain a uniform
   consistency. Allow 5 cm3 per group. Alternatively, dissolve sucrose in its own weight of
   warm water. Colour the syrup solution with a little food colouring (e.g. cochineal
   substitute).

Apparatus-per group


clamp, boss-head and stand
tall beaker or jar, e.g. 150 g coffee jar
osmometer
container for syrup solution













dropping pipette
ruler to measure rise in level
capillary tube and marker










Experiment 2. Selective permeability

(a) Label four test-tubes A to D.

(b) Use a syringe or graduated pipette to place 3 cm3 starch solution and 3 cm3 glucose solution in tube A.
Close the mouth of the tube with your thumb and invert it several times to mix the solutions.

(c) Securely knot one end of a length of cellophane tubing if this has not been done already and use the syringe or pipette to place 3 cm3 of the starch glucose mixture in it (Fig. 1, p.2.02).

(d) Place the cellophane tubing in tube B, fold the open end over the rim and secure it with an elastic band (Fig. 2, p.2.02).

(e) Wash away all traces of starch and glucose from the outside of the cellophane tube by filling and emptying tube B with water several times. Finally, fill the tube with water, note the time and  leave the tube in the rack for 15 minutes or more.

(f) During the 15 minutes, copy the table given below into your notebook and prepare a water bath by half filling a beaker or can with water and heating it to boiling point on a tripod and gauze with a Bunsen burner. When the water boils, reduce the Bunsen flame to keep it just at boiling point.

(g) After 15 minutes, pour some of the water from tube B into tube C to a depth of about 20 mm, add an approximately equal volume of Benedict's solution and place tube C in the water bath for 5 minutes (Fig. 3, p.2.02).

(h) Pour another 20 mm water from tube B into tube D and add 5 drops of iodine solution.

(i) Record in your table the results of the iodine and Benedict's test.


Liquid from tube B tested
with... Colour Interpretation
Benedict’s solution
Iodine solution

Experiment 2. Discussion

I What is the significance of the disappearance of the blue colour from the starch/iodide
mixture ?

2 At what temperature did the reaction proceed most rapidly ?

3 Make a general statement connecting the rate at which amylase acts on starch and the temperature at which it is acting. Put in any qualifications you think necessary.

4 Would you expect to find any upper limit of temperature beyond which the reaction
ceases to get any faster ?







Experiment 2. Discussion – answers

I The disappearance of the blue colour means that starch is removed from the mixture. Alternatively, it could be argued that the iodine has been removed, or that the starch-iodide complex is separated in some way.

2 35 °C is probably the temperature at which the solution loses its blue colour most rapidly.

3 The rate of reaction is greater at higher temperatures. The qualifications are
(a) assuming that the disappearance of blue colour is correctly attributed to the action             of amylase,
(b) no experiment has been carried out above 35-40 °C so it cannot be predicted that the rate of reaction will continue to increase with rise in temperature.

4 The student may know that enzymes are proteins and so denatured at temperatures above 50 °C ; or more simply that boiling destroys enzymes.



Experiment 2. Selective permeability - preparation

Outline A dialysis tube containing a mixture of sugar and starch solutions is left in water for 15 minutes. The dialysate is tested with iodine and Benedict's solutions.
 (See ‘Diffusion’ Experiment 4 for a simpler experiment)

Prior knowledge Benedict's solution is a test for reducing sugars, and iodine is a test for starch.
In order to answer the questions, it helps if the students need to know something of diffusion and osmosis, the relative molecular size or solubility of starch and glucose.

Advance preparation and materials

Glucose solution, Prepare a 30% solution of glucose in tap-water. Allow 5 cm3 per   
group
Starch solution. Prepare a 3% solution by mixing the starch with cold tap-water and                                             heating to boiling point. Allow 5 cm3 per group.
Iodine solution. Grind I g iodine and 1 g potassium iodide in a mortar while adding successive portions of distilled water to make 100 cm3 solution. Allow 1 cm3 per group,
Benedict's solution, Dissolve 17 g sodium citrate crystals and 10 g sodium carbonate crystals in 80 cm3 warm distilled water. Dissolve separately 2 g copper sulphate crystals in 20 cm3 distilled water and add this to the first solution with constant stirring.
Allow 5 cm3 per group.
Dialysis tubing, Cut 15 cm lengths of 6 mm (¼in) dialysis tubing and soak in water before the experiment. The tubing can be used again after washing and may be stored dry or in 30% alcohol to keep it supple.

Apparatus-per group

4 test-tubes and rack                                                     elastic band
4 labels or spirit marker                                                 Bunsen burner
syringe or graduated pipette (5 or 10 cm3)                    tripod and gauze
containers for starch and iodine solutions                     dropping pipette for iodine
15 cm length of dialysis tubing                                      beaker or can for water bath

NOTE. Although results can be obtained in 15 minutes or less, the experiment can be left for 7 days,

Experiment 3. Turgor

You are provided with a length of cellophane tubing which has been soaked in water and knotted at one end.

(a) Use a syringe or graduated pipette to place 3 cm3 syrup solution in the cellophane tubing (Fig. I).

(b) Tightly knot the open end of the dialysis tubing to leave only about 20 mm tubing free above the knot, taking care to exclude air bubbles from the tubing (Fig. 2).

(c) Wash the tubing under the cold tap to remove all traces of syrup solution from the outside. The partly filled tube should be flexible enough to bend through 90° (Fig. 3, p.3.02)
.
(d) Place the cellophane tubing in a test-tube and fill the test-tube with water (Fig. 4, p.3.02).

(e) Label the test-tube with your initials and leave it in the rack for 30-45 minutes.

(f) After 30-45 minutes remove the cellophane tubing filled with syrup and note how it differs from its starting condition.















Experiment 4. Discussion

I What shape were the strips before they were placed in the dishes ?

2 Describe what happened to the strips of stalk when they were placed (a) in the water and (b) in the salt solution, either at first or after you swapped them over .

3 Study Figs. 2 and 3, which show the microscopic structure of the dandelion stalk as seen in transverse section. From these diagrams and your knowledge of cell structure, offer an explanation for the behaviour of the strips of stalk when placed in (a) water and (b) salt solution, assuming that all the cell walls are equally permeable and all the cell sap has the same concentration.

4 How could the forces which produce the movement in dish A help to make the stalk firm and upright in the living plant?






















Experiment 3. Discussion - answers

I The cellophane tube which, at the beginning of the experiment, was only partly filled with syrup and limp enough to bend through 90° is now full of solution and turgid.

2 Both the volume and the pressure of the syrup solution must have increased to distend the cellophane tubing in this way.

3 Osmosis has taken place. Water has passed through the selectively permeable cellophane tubing into the syrup increasing first its volume and finally its outward pressure on the tubing.

4 If the cellophane tubing were strong enough it might have reached a certain degree of distension and then stopped; the diffusion pressure of water inwards being equaled by its outward hydrostatic pressure and the inextensible cell wall..
If the tubing were weak, it could burst as a result of the increasing volume and pressure of the syrup solution. In fact, the tubes will lose their turgor after several hours, possibly as a result of the outward diffusion of syrup.

5 The cellophane tubing with syrup would be expected to distend and press outwards on the glass tube, fitting it exactly. This is analogous in some respects to the cytoplasmic lining within a plant's cell wall.


Experiment 3. Turgor - preparation

Outline A length of dialysis tubing is partially filled with strong sugar or syrup solution, knotted at both ends and left in water for 45 minutes during which time it becomes turgid.

Prior knowledge This experiment should preferably be preceded by Experiment 1 and Experiment 4 of ‘Diffusion’ so that the differential permeability of dialysis tubing is appreciated.

Advance preparation and materials

   Syrup solution. Mix golden syrup with its own volume of water and stir to obtain a      uniform consistency. Alternatively, dissolve sucrose in its own weight of warm water.
 Allow 5 cm3 per group.
   Dialysis tubing. Cut the 6 mm (¼ in) tubing into 20 cm lengths, soak these in water for a few minutes and securely knot one end. Keep the tubing in water till required for the experiment.
The dialysis tubing can be used again. After the experiment, cut the knot from one end and wash out the syrup solution. Store the lengths dry or in 30% alcohol; they can be used for Experiments 1 and 2.

Apparatus - per group

test-tube and rack                                           syringe or graduated pipette (5cm3)
container for syrup solution                            20 cm length of dialysis tubing


Experiment 4. Turgor in a dandelion stalk

(a) Label two watch-glasses or Petri dishes A and B. In A place some tap-water and in B some salt solution.

(b) Use a razor blade or scalpel to cut a piece of dandelion stalk about 40 mm long and then spit it lengthwise (on a Petri dish lid) into strips about l or 2 mm wide (Fig. I).

(c) Place one of these strips in the water in dish A and the other in the salt solution in dish B
.
(d) Watch what happens to each strip in the next few seconds. If any curling takes place notice whether the epidermis side of the strip is on the inside or the outside of the curve.

(e) When no further change takes place, swap the strips over so that the one which was in the water is now in the salt solution and vice versa.

















Experiment 4. Discussion

I What shape were the strips before they were placed in the dishes ?

2 Describe what happened to the strips of stalk when they were placed (a) in the water and (b) in the salt solution, either at first or after you swapped them over .

3 Study Figs. 2 and 3, which show the microscopic structure of the dandelion stalk as seen in transverse section. From these diagrams and your knowledge of cell structure, offer an explanation for the behaviour of the strips of stalk when placed in (a) water and (b) salt solution, assuming that all the cell walls are equally permeable and all the cell sap has the same concentration.

4 How could the forces which produce the movement in dish A help to make the stalk firm and upright in the living plant?


















Experiment 4. Discussion - answers

I The strips may be straight or slightly curved with the epidermis on the inside circumference.

2 (a) In the water the strips, which will be curved already, will curl up more with the epidermis on the inside circumference.
   (b) In the salt solution the strips will straighten out.

3 In the water, the cells of the stalk have their vacuoles separated from water by the selectively permeable cell membrane. The cells will thus absorb water by osmosis to become fully turgid. Because of their shape, the cells of the pith and cortex can expand while the cells of the epidermis and outer cortical layer, with their thick cellulose walls, cannot change their shape. The elongation of the cortex will thus cause the strip of tissue to curl up with the cortex and pith on the outer circumference.

This  explanation assumes that the cell structure revealed in a transverse section is equally applicable to a longitudinal section.

 In the salt solution, water is withdrawn from the cells by osmosis. The cells of the pith and cortex will 'collapse' more than the epidermal cells so that the strip straightens out.
The students may also explain their observations in terms of the thick cellulose walls of the collenchyma impeding the movement of water and solutes.

4 When water is available and the cells approach full turgor, the cortex and pith will tend to increase their volume while the epidermis remains the same. These opposing forces confer a rigidity on the stem. (A good analogy is a helical spring compressed inside a rubber tube.)


Experiment 4. Turgor in a dandelion stalk - preparation

Outline The inflorescence stalk of a dandelion is split lengthways and the strips placed in water or salt solution. In water the strips curl up; in salt solution they straighten out. The results are related to osmosis and the cell structure of the stalk.

Prior knowledge The principles of osmosis applied to plant cells. Knowledge of plant cell structure.

Advance preparation and materials

Dandelion flowers. The flowering shoots are abundant usually in April and May. Select young, turgid stalks preferably not more than 24 hours before the experiment and leave them with the cut ends in water. Allow 1 stalk per group.

Salt solution. Prepare a 10% solution of sodium chloride. Allow 20 cm3 per group.

Apparatus- per group

2 watch glasses or Petri dishes                                     razor blade or scalpel
spirit marker (or label dishes A and B beforehand)       container for salt solution
Plastic Petri dish lid for cutting on

Experiment 5. Turgor in potato tissue

(a) Label three test-tubes A, B and C and add your initials.

(b) Use a syringe or graduated pipette to put 20 cm3 water in tube A and 10 cm3 in
tube B.

(c) Similarly, put 20 cm3 17% sucrose solution in tube C and 10 cm3 in tube B (Fig. 1, p.5.02).

(d) Working on a dissecting board or folded newspaper, cut the ends off a large potato and use a No.4 or 5 cork borer to obtain 4 or 5 cylinders of potato tissue as long as possible (Fig. 2).

(e) Push the cylinders of potato out of the cork borer with the flat end of a pencil (Fig. 3) and select the three longest.

(f) Cut all three to the same length, e.g. 50, 60 or 70 mm, trimming the ends at 90° at the same time Put one cylinder in each test-tube.

(g) Leave the potato tissue for 24 hours.

(h) After 24 hours, use forceps or a mounted needle to remove the potato cylinder from test-tube B. Rinse it in a beaker of water and measure its length in mm. Repeat this operation for the potato cores in tubes A and C noting at the same time whether the tissue is firm or flabby.
Record your results in your notebook.


First length...........mm A Water B 8.5% Sucrose C 17% sucrose
Length after immersion in














Experiment 5. Discussion

I From your knowledge of osmosis in plant cells, what do you suppose has happened to the cells of the potato tissue (a) in 17% sucrose, (b) in water?

2 How would these changes in the cells account for any changes in size of the potato cylinder?

3 What do you think limits the change in length of the potato cylinder which was immersed in water?

4 How could the results of this experiment be used to explain the extension growth of plant shoots and roots ?

5 How could the experiment be modified to find out the osmotic conce













Experiment 5. Discussion – answers

I (a) The sucrose solution is more concentrated than the cell sap; the cytoplasm of the potato cells is selectively permeable, so water passes out of the potato cells into the sucrose, reducing the volume (by plasmolysing) of the cells.
  (b) By the same principles, water enters the vacuoles of the potato cells making them turgid.

2 The plasmolysis of the cells in 17% sucrose solution leads to a reduction in the volume of cell sap and the cells as a whole so that the potato cylinder becomes shorter. In the water, the uptake of water by the potato cells causes the vacuoles to enlarge, increasing the cell size and, consequently, the length of the cylinder as a whole.

3 The final increase in length will be limited by the lack of plasticity in the cell walls which will not allow the vacuoles to increase beyond a certain point. The relative increase in length will depend on how turgid the tissue is to start with.

4 If the cell walls in the region of extension of a shoot or root were fairly plastic, an uptake of water by osmosis in the cells could cause a permanent increase in size. The cells or their arrangement in the tissue would need to favour increase in length rather than width.

5 The student might suggest that, by using more varied sucrose concentrations, one might be found which did not produce extension or shrinkage and this would thus correspond to the concentration of the cell sap.
More perceptive students might realize that this will be true only for tissues which are in a state of 'incipient plasmolysis' since fully turgid cells cannot take up water even if they are immersed in a hypotonic solution.

NOTE. (a) Although the sucrose concentrations are 0.5 and 0.25 molar, they are expressed as percentages in the students' instructions to avoid having to explain molarity if this
concept is not to be introduced.
   (b) The changes in length will be quite small, plus or minus 2 or 3 mm only. It is best to
collect all the results from the class to show that they are all in 'the same direction'
and thus unlikely to have been caused by errors of measurement.




Experiment 5. Turgor in potato tissue - preparation

Outline Potato cylinders are immersed in water, 0.25 M and 0.5 M sucrose. The change in length and turgor is noted after 24 hours.

Prior knowledge It is assumed that the students know how to apply the principles of osmosis to plant cells, e.g. Experiment 4.

Advance preparation and materials

0.5 molar sucrose. Dissolve 17 g sucrose in 100 cm3 tap-water. Allow 20 cm3 per group.

Potatoes. Select the largest available, not 'new' or fully turgid, and 10 cm or more long if possible. With care, 10 cores, 60 mm long can be obtained from a 200 g potato.

Apparatus-per group


3 test-tubes and rack                                                   scalpel or razor blade
3 labels or spirit marker                                               cork borer, No.4, 5 or 6
syringe or graduated pipette (10 or 25 cm3 )                         ruler with millimetres
dissecting board or newspaper to protect bench         forceps or mounted needle                 
       


NOTE. If the experiment is to be left for more than 24 hours, the test-tubes should be placed in a refrigerator or the potato will ferment and decompose.






Experiment 6. stomatal movements

 (a) Tear the leaf with a twisting and pulling action so that some of the lower epidermis is exposed (Fig. 1).

(b) Place a small piece of the leaf with an exposed fringe of epidermis on a microscope slide, lower surface uppermost, and trim off the thick part of the leaf with a scalpel (Fig. 2).

(c) Add 2 or 3 drops of water to cover the epidermis and carefully lower a cover slip over it, excluding all air bubbles (Fig. 3).

(d) Examine the slide under the microscope using the x10 objective. Move the slide about until you find a group of stomata which are open (Fig. 4). Clip the slide firmly to the stage and leave
it there for the rest of the experiment.

(e) With a dropping pipette, place 2 drops of sodium chloride solution on the slide so that it is in
contact with one edge of the cover slip.

(f) Examine the stomata and be ready to observe them again immediately after the next operation.

(g) Place a square of blotting paper on the slide on the opposite side of the cover slip to the salt solution and move it so that it just touches the cover slip. As soon as the
blotting paper starts to draw the solution through, watch the stomata closely.

(h) If nothing seems to happen, draw through 2 more drops of sodium chloride solution.

(i) When the change is complete, try and reverse it by drawing through water instead of salt solution.
The procedure may have to be repeated 2 or 3 times in order to wash away all the salt solution.

















Experiment 6. Discussion

I If the volume of fruit juice needed to decolourize DCPIP is greater than the volume of 0.1 % ascorbic acid, does this mean that the juice contains more or less vitamin C than
ascorbic acid solution?

2 On this basis, which of your samples contained the most vitamin C?

3 Did the canned fruit. juice differ greatly from the fresh juice?

4 Suppose that 0.8 cm3 orange juice decolourized 1 cm3 DCPIP but it took 1.6 cm3grapefruit
juice to decolourize the same volume of DCPIP; does this mean that oranges contain more vitamin C than grapefruit?

5 What might take place at a canning factory which could (a) reduce, (b) increase the
vitamin C content of fruit juice?

6 Did the fruit juice which had been exposed to air for several days have significantly less
vitamin C than the fresh juice?

7 Sodium sulphite is not vitamin C and yet it decolourized DCPIP - Why did it do this?
(See introductory notes on p. 7.01 and instruction (i).)

8 Since chemicals other than vitamin C will decolourize DCPIP, does this mean that using the dye as a test for vitamin C is unsatisfactory?

9 If canned or bottled juice has sulphur dioxide or sodium sulphite added as a preservative,
will this affect the reliability of your results with DCPIP?

 










Experiment 6. Discussion - answers

I The greater the volume of juice needed to decolourize DCPIP the less ascorbic acid it contains.

2 The answer will vary according to the source and brand of the various juices but usually fresh orange and lemon juices contain more concentrated vitamin C than the others. Canned grapefruit juice may be substantially weaker and orange squash very weak in vitamin C compared with the others. Grape juice may contain so little that the student has to give up after squirting several
syringes full of juice into the DCPIP.

3 Canned juice is frequently weaker than fresh juice.

4 The result shows that the orange juice in the experiment has a higher vitamin C content than the grapefruit but a generalization about all oranges and all grapefruit is not justified. There is probably a great deal of variation with species, season and age of fruit.

5 (a) Heating the juice might destroy some of its vitamin C content.
(b) Heating the juice might concentrate the vitamin. Ascorbic acid might be added artificially to increase its concentration.

6 There is likely to be a significant difference in the vitamin C content of fruit juices exposed to
the air for several days.

7 Sodium sulphite is a reducing agent and so decolourizes DCPIP.

8 Since chemicals other than vitamin C will affect DCPIP, the test is unsatisfactory unless it can
be assumed that reducing agents other than ascorbic acid are unlikely to be present.

9 If sulphur dioxide or sodium sulphite is present, presumably less of the juice will be needed to decolourize the DCPIP. Whether this will make the results substantially unreliable will depend
on the relative concentrations of sulphite and ascorbic acid in the juice.

Experiment 6. Stomatal movements - preparation

Outline The lower epidermis from a rhubarb or broad bean leaf is examined under the microscope (x 100). Salt solution is used to plasmolyse the guard cells and cause the stomata to close.

Prior knowledge Appearance of guard cells and stoma. Use of a microscope. Plant cell structure. Principles of osmosis and meaning of 'turgor'. Photosynthesis in a leaf.

Advance preparation and materials

Plant material. Broad bean and rhubarb leaves have a lower epidermis which easily peels off and bears large guard cells. Broad beans will grow in pots, needing 6-10 weeks to reach a
suitable stage of growth. If the leaves are collected long before the experiment or on a sunny afternoon, the stomata may be closed. They can usually be induced to open again by covering the leaves with po1ythene bags and exposing to light. In the case of rhubarb, the leaves are cut and brought into the laboratory, covered with a polythene bag tied round the petiole which is placed in a jar of water. With potted bean plants, the whole shoot may be covered with a large polythene bag tied round the stem. In both cases, exposure to the light of a bench lamp for 30-45 minutes will probably cause most of the stomata to open.

Blotting paper. Cut strips of blotting paper about 20mm wide and cut these into squares. Allow
6 per group.

Sodium chloride. Prepare a 10% solution in tap-water. Allow 5 cm3 per group.

Apparatus-per group

microscope
microscope slide
cover slip
dropping pipette
container for water
container for sodium chloride
scalpel or razor blade
mounted needle or fine forceps


Experiment 7. Plasmolysis

(a) Use a scalpel or razor blade to make a shallow transverse cut in the red epidermis of the piece of rhubarb stalk.

(b) With a pair of fine forceps lift up a strip of the epidermis at one side of the cut. Lift only the
epidermis and not the underlying cortex. Having freed a narrow band of epidermis, pull it off with the forceps (Fig. 1) and press it flat on a slide with the outermost surface upwards.

(c) Use the scalpel or razor blade to cut about 10 mm of this strip from the thinnest and reddest
portion (Fig. 2) and, using a dropping pipette, cover this with 3 drops of water.

(d) Use the forceps to lower a cover slip carefully on to the water drops, avoiding trapping air bubbles (Fig. 3), and examine the epidermis under the microscope using the x10 objective.

(e) Move the slide about to find a group of clearly defined cells near the edge, with red cell sap, and make a drawing in your notebook to show one of these cells. Draw the cell at least 50 mm long, representing the outline accurately and shading the area filled with cell sap. Clip the slide
securely to the microscope stage and leave it in this position for the rest of the experiment.

(f) Use the pipette to place 2 drops of sucrose solution on the left-hand side of the slide, just touching the edge of the cover slip.

(g) Draw all this solution under the cover slip by applying a strip of blotting paper to the right- hand edge of the cover slip. Try not to move the slide, the cover slip or the epidermis.

(h) Examine the cells again and watch for about 2 minutes. If nothing happens, draw through some more sucrose solution.

(i) When a significant change has occurred in the cells, draw the same cell as before to show the
cell wall and the cell sap. The cell is plasmolysed.

(j) Use the pipette to place 3 drops of water on the left hand' side of the slide and draw it through under the cover slip as before. Do this twice to flush out all the sucrose solution.

(k) Study the cells again for about 2 minutes repeating operation (j) if nothing happens in this time.




Experiment 7. Discussion

1 When the rhubarb cells were exposed to sucrose solution what change did you observe in
(a) the shape of the vacuole and (b) the colour of the cell sap?

2 What change, if any, took place in the shape of the cell?

3 Bearing in mind the fact that liquids cannot be compressed, what must have happened to the cell sap to account for (a) the change in volume and (b) the change in colour?

4 After exposure to the sucrose solution, what do you suppose occupied the space between the
vacuole and the cell wall in the plasmolysis cells?

5 Why did the cell sap not mix with the liquid in this space?

6 Which part of the cell must be 'selectively-permeable' in order to explain these results?

7 How do the results of this experiment lead to the conclusion that the cell wall is permeable not
only to water but also to dissolved sucrose?

8 What effect would it have on the tissues of the whole plant if all the cells were plasmolysed?

9 (a) What changes took place in the cells when the sucrose was replaced by water? (b) How can you explain these changes in terms of osmosis?

10 How would this change, if it applied to all the cells, affect the tissues of the plant as a whole?





















Experiment 7: Discussion - answers

1 (a) The vacuole should diminish and appear to separate from the cell wall in places.
(b) The cell sap may become noticeably deeper in colour when compared with unplasmolysed
cells.

2 There should be no perceptible change in the shape of the cell.

3 The cell sap must have lost part of its contents, presumably water. This would account for its diminution in volume and, since the red pigment would become more concentrated, the intensification of its colour.

4 The space between the vacuole and cell wall must be occupied by the sucrose solution (and, of
course, the cell membrane and cytoplasm bounding the cell sap).

5 The cell membrane, tonoplast and cytoplasm separate the solution and the cell sap.

6 The cell membrane, the cytoplasm, the tonoplast or some combination of these must be selectively permeable.

7 If the cell wall were selectively permeable, the loss of water to the sucrose solution would distort the cell wall and no space would appear between it and the vacuole.

8 If all the cells in a plant were plasmolysed, the unsupported structures would be limp and wilting and the plant would eventually die.

9 (a) When the sucrose solution was replaced by water, the cells recovered from the plasmolysed
 condition and the vacuole expanded to fill the cell once again.
(b) The cell sap now contains a stronger solution than the surrounding fluid and absorbs water - through the selectively permeable cytoplasm.

10 A plant with its cells turgid would have firm tissues, an erect stem and expanded leaves.


Experiment 7:  Plasmolysis – preparation

Outline Red epidermis from rhubarb petiole is mounted in water, examined under the microscope (x 100) and irrigated with sucrose solution to plasmolysed the cells.

Prior knowledge Use of the microscope, plant cell structure, the principles of osmosis, the meaning of 'epidermis' and 'cortex'.

Advance preparation

Rhubarb petioles. Use freshly cut rhubarb with bright red epidermis. Allow about 50 mm stalk.
per group.

Sucrose solution. Prepare a molar solution by dissolving 34 g sucrose in 100 cm3 tap-water.
Allow 5 cm3 per group.

Blotting paper. Cut strips of blotting paper or filter paper approximately 20 x 60 mm. Allow
4 strips per group.

Apparatus-per group

scalpel or razor blade
fine forceps
microscope slide
cover slip
microscope with x 100 magnification
dropping pipette 
container for sucrose solution
container for water
4 strips of blotting paper


Experiment 8: Osmosis and surface area

(a) Use a scalpel or kitchen knife to cut eight cubes of potato with sides 1 cm long (i.e. each cube has a volume of 1 cm3). DO NOT cut on to the surface of the bench but use a tile or
dissecting board.

(b) Similarly, cut one cube of potato with sides of 2 cm (i.e. the cube has a volume of
2 x 2 x 2 = 8 cm).

(c) Weigh the eight 1 cm pieces together and note their total mass.

(d) Weigh the 8 cm3 piece and note its mass.

(e) Place all the cubes in a beaker of water and note the time.

(f) In your notebook draw a table like the one below and enter the two masses in the
appropriate column.

(g) After about one hour, remove the potato cubes from the water, dry them on a piece of
blotting  paper and weigh them. (As before, weigh the eight 1 cm3 pieces together.)
Record the new masses in your table.

(h) If possible, return the cubes of potato to the water and repeat the weighings on the
following day.

(i) After your final weighing, work out the total increase in mass for the eight small cubes
and the increase in weight of the single, larger cube. Record these increases in your table.

(j) Calculate the percentage increase in mass in each case, as follows
     
percentage increase in mass =
                                               
                 
(k) Enter this figure in your table.

(l) If the differences in increase in mass are smaIl, it may be advisable to pool the results from
the whole class and work out the average increase.


Eight 1 cm3 cubes One 8 cm3 cube
1st mass
2nd mass
Increase in mass
Percentage increase in mass







Experiment 8: - Discussion

1 At the start of the experiment, by how much did the total mass of the eight 1cm3 cubes
differ from the mass of the 8cm3 cube?

2 What is the most likely reason for this difference?

3 Would you expect there to be much difference in the total volume of the eight 1 cm3 cubes and the volume of the single 8 cm3 cube?

4 (a) What was the total surface area of the eight 1 cm3 cubes?
   (b) What was the surface area of the single 8 cm3 cube?

5 Why, do you suppose, the mass of the cubes increased after one hour's immersion in water?

6 What difference was there in the percentage increase in mass between the eight 1 cm3 cubes
and the single 8 cm3 cube?

7 What is the most likely explanation of this difference?










Experiment 8: Discussion - answers

1 An 8 cm3cube weighs about 9 or 10 g. The total mass of the eight 1 cm3 cubes might differ from this by up to 1 g. Because the percentage increases in mass are compared, this difference
is not very important.

2 Any difference can be attributed to the inaccuracy involved in cutting the cubes to size.

3 Theoretically there is no difference between the volume of eight 1 cm3 cubes and one 8 cm3 cube. In practice there might be a small difference for the reason given above.

4  (a) The total surface area of the eight 1 cm3 cubes is 48 cm2.
    (b) The surface area of the single 8 cm3 cube is 24 cm2.

5 The increase in mass of the cubes is most likely to be a result of the potato cells taking in water
by osmosis.

6 After an hour, the single 8 cm3 cube will have increased its mass by about 1-5 per cent. The mass of the eight 1 cm3 cubes will have increased by about 3-8 per cent. After 24 hours the
increases might be as much as 8 and 15 per cent respectively.
These increases will depend on the initial state of turgidity in the potato cells.
 
7 Since the eight small cubes present an absorbing surface which is twice as large as that of the - large cube, we would expect them to take up more water in a given time.


Experiment 8: Osmosis and surface area – preparation

Outline The difference in uptake of water between eight 1 cm3 cubes of potato and one 8 cm3
cube is measured.

Prior knowledge Living cells can absorb water by osmosis.

Advance preparation and materials

Potatoes. A 100 g potato should provide enough cubes for 5 experiments.

Apparatus--per group
kitchen knife or scalpel
cutting board (dissecting board or tile)
ruler (with cm)
beaker or jar to hold about 100 cm3 water
paper towel or blotting paper
balance accurate to 10 mg (2 or 3 per class if possible)

*This experiment is based on one described by D. R. B. Barrett in Journal of Biological Education 18 (4) 273.



           Chapter Five – Photosynthesis

Experiments on Photosynthesis – Introduction

The following experiments may be introduced as an example of hypothesis testing.
The hypothesis is that plants combine carbon dioxide and water to make carbohydrates (glucose, sucrose or starch) with the aid of energy from sunlight and with chlorophyll to absorb the energy.
 Accordingly, the ‘IF.....THEN’ test can be applied;  e.g. ‘IF photosynthesis uses the sun’s energy for making starch .....THEN, depriving the plant of sunlight will prevent starch formation (Experiment 3).
The same test can be applied to chlorophyll and carbon dioxide. It cannot be applied to water because depriving the plant of water will kill it before photosynthesis is impaired.
The  boxes indicate the points at which the hypothesis can be tested.





   6CO2         +       6H2O                                                         C6H12O6           +        6O2                                                                                                                                     




The instructions are prescriptive, i.e. they set out the exact sequence of operations that the students should follow. Although this is not ideal, it does ensure that the experiment will produce useful results which can be discussed positively.
 Similarly, precise guidance is provided for preparing the practical work. If these guidelines are followed carefully, there is a very good chance that all the students will achieve meaningful results.
 The expected results of the experiments are not stated. Instead, under the heading of ‘Discussion’, the students are asked a series of questions which guides them towards an interpretation of their results, encourages them to consider alternative interpretations, and directs their attention to inadequacies in the experimental design.
 It is hoped that, in this way, they will be persuaded that interpretations are tentative and that, with further evidence, they may be changed. This is scientifically more acceptable than claiming that the results prove the hypothesis. 
Destarching plants
If the presence of starch is to be taken as evidence that photosynthesis has taken place, it is important to ensure that, at the beginning of the experiment, the leaf is free from starch. This can be done by placing the plant in darkness for 2-3 days, during which time, any starch present will have been converted to sucrose and translocated from the leaves.  On plants growing outside or too large to be enclosed, individual leaves can be covered with aluminium foil, though it must be borne in mind that this could also impair
gaseous exchange. Experiment 3 presents rather a circular argument because it purports to show that starch is not made in darkness which is assumed in the process of destarching in experiments 3-5.
 Bear in mind that students are likely to confuse destarching with decolourizing a leaf.
Light source
It is worth investing in a 4ft (120 cm) or 6ft (180 cm)  fluorescent lamp. This will produce results in a 50-60 min. period at a time when direct sunlight is lacking

It is essential that good laboratory practice is observed as set out in ‘Safeguards in the School Laboratory’  11th edition 2006 published by the ASE (Association for Science Education)


Experiment 1. Production of gas by pond-weed

(a) Fill a beaker or glass jar with tap-water and add about 5 cm3 (20 mm in a test- tube) saturated sodium hydrogen carbonate solution.

(b) Select a pond-weed shoot between 5 and 10 cm long
.
(c) Take a small paper clip, prize it slightly open and slide it over the pond-weed shoot a
 short distance behind the growing point. This is to hold the weed down in the water.

(d) Place the pond-weed in the jar so that it floats vertically with the cut end of the stem      uppermost but still entirely below the water.

(e) Switch on the bench lamp and bring it close to the jar.

(f) After a minute or two, bubbles should appear from the cut end of the stem. If they do  not, obtain a fresh piece of plant material and try again.

(g) When the bubbles are appearing with regularity, switch the lamp off and observe any changes in the production of bubbles.

(h) Switch the lamp on and place it about 25 cm from the pond-weed. Try to count the number of bubbles appearing in a minute. Now move the lamp to about 10 cm from the plant and again try to count the number of bubbles.







Experiment 1. Discussion

I Assuming that there were no leaks, what happened to the level of liquid in the capillary ?

2 What does this result indicate about the volume of liquid in the cellophane tubing and capillary?

3 What explanation can you offer for this change in volume ?

4 What assumptions have you made about the cellophane membrane and the movement of sugar molecules?

5 This experiment has been conducted without a control. What control do you think should be carried out to exclude any alternative explanations of the results ?


Experiment 1. Discussion - answers

I Assuming some continuity between leaves and stem, the leaves could be producing the gas.

2 An increase in light intensity appears to increase the rate of production of gas.

3 Although twice as many bubbles appear per minute, the bubbles might be smaller so that the total volume of gas produced has not doubled.

4 (a) There is no experimental evidence to indicate the composition of the gas in the bubbles.

   (b) Oxygen, nitrogen, hydrogen and carbon dioxide are present in free or combined form in air or water. All or any of these might be present in the bubbles.

5 The sodium hydrogen carbonate provides carbon dioxide for photosynthesis.

Experiment 1. Production of gas by pond-weed - preparation

Outline Shoots of Elodea are held upside down in a beaker of water. Bubbles of gas appear from the stems when the plants are illuminated.

Prior knowledge The gases present in air and water. Sodium hydrogencarbonate is a source of carbon dioxide.

Advance preparation and materials-per group

Elodea shoots about 80 mm long             
 5 cm3 10% sodium hydrogencarbonate solution

The Elodea should be collected before the experiment, the shoots cut and placed in a large container of fresh tap-water. Just before the experiment the shoots are illuminated strongly with the fluorescent tube (or sunlight) so that the students can select those which are bubbling most strongly. Branched shoots with abundant leaves are likely to prove most satisfactory and the pond-weed will still function after several days in the laboratory.
Ceratophyllum will work but less effectively than Elodea.
Potamogeton crispus has worked well but other species have not been tried.

Apparatus-per group

beaker (250 cm3 or larger) or jam jar                                   bench lamp (60 watts)
test-tube (for collecting hydrogencarbonate solution)          paper clip

-per class
clock

NOTE. Some razor blades should be available for cutting the stems cleanly if they fail to bubble satisfactorily but their use must be supervised.

Experiment 2. Testing a leaf for starch

(a) Half fill a 250 cm3 beaker or tin can with water (hot water if available) and place it on a tripod over a Bunsen burner. Heat the water till it boils and then turn down the Bunsen flame sufficiently to keep the water at boiling point.

(b) Hold the leaf in forceps and plunge it into the boiling water for 5 seconds. This will kill the cells, arrest all chemical reactions and make the leaf permeable to alcohol and iodine solution later on.

(c) With the forceps, push the leaf carefully to the bottom of a test-tube and cover it with methylated spirit.

(d) TURN OUT THE BUNSEN BURNER.

(e) Place the test-tube in the hot water and leave it for 5 minutes. The alcohol will boil and dissolve out the chlorophyll in the leaf (See Figure on p.2.02).

(f) Use a test-tube holder to remove the test-tube from the water bath and tip the green alcoholic solution into the receptacle for waste alcohol but take care not to tip the leaf out as well.
If the leaf is white or very pale green, go on to (g).
If there is still a good deal of chlorophyll left in the leaf, boil it for a further 5 minutes with a fresh supply of alcohol, using the hot water bath. If it is necessary to relight the Bunsen to heat the water to boiling point, remove the test-tube and do not replace it until the Bunsen flame is extinguished.

(g) Fill the test-tube with cold water and the leaf will probably float to the top.
     I. Tough leaves (e.g. Tradescantia). Hold the leaf stalk with forceps and dip it into the
     hot water in the water bath for 2-3 seconds. Spread it flat on a tile or Petri dish lid with
     the aid of a little cold water.
    2. Soft leaves (e.g. Busy Lizzie). Use forceps to place the leaf on a tile or back of a       
    Petri dish lid and, holding the leaf stalk firmly against the tile or lid, let a fine trickle of   
    water from the cold tap run over it to wash away the alcohol.

(h) If necessary, use the forceps to spread the leaf quite flat on the tile or lid. Using a dropping pipette cover the leaf with iodine solution for one minute.

(i) Take the leaf to the sink and holding it on the tile or lid, wash away the iodine solution with a fine trickle of cold water .


   Experiment 2. Discussion

I What is the significance of the disappearance of the blue colour from the starch/iodide
mixture ?

2 At what temperature did the reaction proceed most rapidly ?

3 Make a general statement connecting the rate at which amylase acts on starch and the temperature at which it is acting. Put in any qualifications you think necessary.

4 Would you expect to find any upper limit of temperature beyond which the reaction
ceases to get any faster ?









Experiment 2. Discussion – answers

I The disappearance of the blue colour means that starch is removed from the mixture. Alternatively, it could be argued that the iodine has been removed, or that the starch-iodide complex is separated in some way.

2 35 °C is probably the temperature at which the solution loses its blue colour most rapidly.

3 The rate of reaction is greater at higher temperatures. The qualifications are
(a) assuming that the disappearance of blue colour is correctly attributed to the action             of amylase,
(b) no experiment has been carried out above 35-40 °C so it cannot be predicted that the rate of reaction will continue to increase with rise in temperature.

4 The student may know that enzymes are proteins and so denatured at temperatures above 50 °C ; or more simply that boiling destroys enzymes.



Experiment 2. The effect of temperature on enzyme activity - preparation

Outline  A comparison is made of the time taken for the hydrolysis of starch solution at three different temperatures. The time for hydrolysis is judged by the disappearance of blue colouration from starch-iodide.

Prior knowledge  Starch/iodide reaction.

Advance preparation and materials-per group

1% starch solution *   20 cm3                                 dilute iodine 5 cm3
ice 3-4 cubes                                                       5% amylase solution  3cm3

NOTE. The presence of iodine inhibits the reaction. If too much iodine is added it may prevent the hydrolysis altogether. The instructions are based on the assumption that the iodine solution has been freshly prepared and in accordance with the instructions on p.01.

Apparatus-per group
test-tube rack and 6 test-tubes                         3 X 250 cm3 beakers or jars
6 labels or spirit marker                                    thermometer
syringe or graduated pipette 10 cm3                 dropping pipette

--per class
clock


Experiment 3. The need for light in photosynthesis

You are provided with a potted plant which has been kept in darkness for at least 48 hours so that any starch in its leaves has been converted to sugar and removed.

(a) Water the plant if the soil appears to be dry

(b) DO NOT REMOVE ANY LEAVES but select one at the top of the shoot, preferably one held out from the stem nearly horizontally.

(c) Round this leaf wrap a strip of aluminum foil. Press it close to the surface so that this part of the leaf cannot receive any light at all. Put your initials on the foil with a spirit marker.

(d) Place the plant in direct sunlight or under the light source so that the leaf is directly under the fluorescent tube. When the whole class has prepared the plants, the tube can be lowered to within a centimeter or two of the top leaves.

(e) After a period of time (40-50 minutes is the minimum), detach the leaf with the foil and test it for starch as described in Experiment 2.


Experiment 3. Discussion

.I Describe the appearance of the leaf after it had been tested for starch.

2 (a) What is the significance of the colours and their distribution in the leaf ?
   (b) How could this be explained in terms of photosynthesis and light ?
   (c) Suggest at least one other way in which the results could be explained.

3 (a) Apart from cutting off the light from part of the leaf, what other effect might the
        aluminium foil have had on the leaf which could influence the results ?
   (b) What control experiments could be conducted which would check whether these
        effects had influenced the results ?

 4 In this experiment, the plant was not tested at the beginning to ensure that its leaves were free from starch. Why was this precaution omitted?
       
5 Explain how the method used for destarching (not decolourizing) the plant tends to assume the results of this experiment.?

   .


Experiment 3. Discussion - answers

I The leaf should be some shade of blue except where the foil has excluded light. This area will
be yellow from the iodine stain.

2 (a) The significance of the blue colour is that it indicates the presence of starch. The
        distribution of blue colour suggests that starch has been formed only in those parts of the leaf
        which received light.
   (b) Some stages of carbohydrate production in photosynthesis need a supply of energy from
         light. If this energy is lacking, starch (at least) cannot be produced.
   (c) The leaf could be producing sugar everywhere but light is needed to change sugar to starch.
        Alternatively, in darkness the starch might be removed as fast as it is. formed, i.e. light could
        inhibit the conversion of starch to sugar .

3 (a) The aluminium foil will also interfere with the free passage of carbon dioxide into the leaf.
        A shortage of carbon dioxide could reduce photosynthesis as could the absence of light.
   (b) A strip of transparent tape could be applied to the leaf. This will interfere with the passage
        of gases but not light. If the leaf produces starch in the area covered by the tape we can
       assume that the aluminium foil did not prevent gaseous exchange.

4 If the area of leaf under the foil remains unaffected by iodine, this is evidence that the leaf was
free from starch at the beginning of the experiment.

5 One assumes that during 3 days of darkness, starch is removed from the leaves and is not re-
formed.

Experiment 3. The need for light in photosynthesis  - preparation

Outline An aluminium foil strip is wrapped round a leaf of a destarched potted plant. After exposure to light, the leaf is tested for starch.

Prior knowledge Starch is a product of photosynthesis. During darkness starch is removed from the leaves so that they are; starch-free at the beginning of the experiment.
The iodine test for starch. Method of testing a leaf for starch. (This information can be
obtained by following the instructions from  Experiment 2.

Advance preparation and materials
(a) Potted plants. Impatiens (busy Lizzie), Tradescantia, Pelargonium, French bean will all give good results after about 4-6 hours exposure to light. For results within a 
     double lesson, Impatiens is recommended. Allow one plant per group.
(b) Destarching. Water the plants and leave them in a dark cupboard for 3 days
     (72 hours) if possible. Most of the plants will be starch-free after this time.
(c) Starch test on leaf. (See Teachers' Notes, p.2.04.)

Apparatus-per group
1 strip of aluminium foil about 10 X 60 mm
apparatus for starch test as listed in Teachers' Notes p.2.04.
spirit marker

-per class
one or more fluorescent tubes

NOTE. To obtain results with only 45-60 minutes exposure to light, the fluorescent tube must be nearly touching the leaves.

Experiment 4. The need for chlorophyll

You are provided with a potted plant having variegated leaves. This plant has been in darkness for 2 days so that starch has been removed from the leaves.

(a) Detach a leaf from near the top of the plant and test it for starch as described in Experiment 2 on p.2.01 If any blue colour appears, the plant is unsuitable for the experiment.

(b) Water the plant if the soil looks dry.

(c) Place the plant in a situation where the leaf can receive sunlight or arrange the plant
 with the leaf nearly touching a fluorescent tube and leave it for at least 45 minutes.

(d) Detach a leaf from near the top of the plant and make a drawing in your notebook to show the outline and the areas which contain chlorophyll.

(e) Test this leaf for starch as described in Experiment 2,  p.2.01.

(f) Alongside your first drawing of the leaf, make a second drawing after testing the leaf with iodine, to show its outline and the distribution of the blue colour due to starch and the brown colour due simply to staining with iodine.


Experiment 4. Discussion

I What was the relationship between the distribution of chlorophyll in your leaf and the distribution of starch revealed by iodine ?

2 How can you explain this distribution of starch in terms of photosynthesis ?

3 What alternative explanations could there be for any correspondence between the distribution of chlorophyll and the distribution of starch ?

4 Why was it necessary to draw the leaf before testing it for starch ?

5 Why was there no need for a separate control experiment ?

6 How do you know that there was no starch present in the leaf at the beginning of the experiment ?
Are you satisfied with this evidence ?


Experiment 4. Discussion - answers

I There should be an exact correspondence between the previous distribution of chlorophyll and the distribution of starch in the leaf as revealed by the iodine test.

2 Starch has been produced only in the green areas presumably because chlorophyll is necessary for photosynthesis.

3 (a) Perhaps all parts of the leaf produce sugar by photosynthesis but this is converted
         to starch only in the green parts.
   (b) The white parts of the leaf might lack not only chlorophyll but various enzymes
         which are normally essential for photosynthesis.
   (c) The white parts of the leaf might be non-living.

4 When the leaf is fully decolourized it will not be possible to remember exactly where the green parts were in order to compare the pattern of starch distribution with the pattern of chlorophyll distribution.

5 The control in this case is the presence of chlorophyll in the leaf so that a variegated leaf acts as both experiment and control.

6 A similar leaf from the same plant will have given a negative result with the starch test at the beginning of the experiment. This does not prove, however, that the leaf used for the experiment was similarly starch-free. One way of overcoming this difficulty is to cut off the top half of the leaf from a destarched plant and show that it contains no starch. The bottom half of the leaf, still attached to the plant, is exposed to light for an hour or so.

Experiment 4. The need for chlorophyll - preparation

Outline A destarched variegated potted plant is exposed to light and one of its leaves tested for starch.

Prior knowledge Starch is a product of photosynthesis. During darkness starch is removed from the leaves so that they are starch-free at the beginning of the experiment. The iodine test for starch. Method of testing a leaf for starch. (This information can be obtained by following the instructions for Experiment 2, p.2.01).

Advance preparation and materials
(a) Potted plant. Variegated Tradescantia and variegated Impatiens (busy Lizzie) give adequate results in 45-60 minutes but can also be left for a week.
Allow one plant per group.

(b) Destarching. See  ‘Introduction’).

(c) Starch test. See Experiment 2, p

Apparatus
fluorescent tube
apparatus and materials for starch test; see p.)

NOTE. In a double lesson, adequate results can be obtained in 45-60 minutes if the leaves are nearly touching the fluorescent tube, but in this case the test to see if the leaves are starch-free to begin with should be conducted concurrently with the exposure of the rest of the plants to light in the hope that they will turn out to be destarched. Exposures of 4 hours or more give very good results.


Experiment 5. The need for carbon dioxide

You are provided with two potted plants which have been in darkness for 48 hours so that any starch present in their leaves has been removed.

(a) Label the pots A and B.

(b) Detach a leaf from near the top of each plant and test each one as described in Experiment 2, p,  to ensure that no starch is present. If either leaf gives a blue colour the plant should not be used for the experiment.

(c) Water the plant if the soil seems dry.

(d) In a small container place about 20 g soda lime. In a similar container pour about
10 cm3 (40 mm in a test-tube) saturated sodium hydrogencarbonate solution. Place the container of soda-Iime on the soil in pot A and the sodium bicarbonate container in pot B (See Figure on page 5.02).

(e) Cover each plant with a transparent plastic bag, after checking it for holes. Secure the bag round the pots with elastic bands and mark the pots with your initials.

(f) Place both plants where they can receive daylight or artificial light for several hours

(g) Label two test-tubes A and B.

(h) Copy the table on p.5.02 into your notebook

(i) After several hours of illumination remove the plastic bags from the plants and detach a leaf from near the top of each plant. Place the leaf in the correct test-tube to identify it.

(j) Test each leaf for starch as described in Experiment 2, p.2.01.

(k) Record your results in the table

NOTE. Soda lime absorbs carbon dioxide; sodium hydrogen carbonate solution decomposes slowly to give carbon dioxide.

Experiment 5. Discussion

I From your knowledge of osmosis in plant cells, what do you think has happened to the cells of the potato tissue (a) in 17% sucrose, (b) in water?

2 How would these changes in the cells account for any changes in size of the potato cylinder?

3 What do you think limits the change in length of the potato cylinder which was immersed in water?

4 How could the results of this experiment be used to explain the extension growth of plant shoots and roots ?

5 How could the experiment be modified to find out the osmotic concentration of cell sap ?














Experiment 5. Discussion – answers

I (a) The sucrose solution is more concentrated than the cell sap; the cytoplasm of the potato cells is selectively permeable, so water passes out of the potato cells into the sucrose, reducing the volume (by plasmolysing) of the cells.
  (b) By the same principles, water enters the vacuoles of the potato cells making them turgid.

2 The plasmolysis of the cells in 17% sucrose solution leads to a reduction in the volume of cell sap and the cells as a whole so that the potato cylinder becomes shorter. In the water, the uptake of water by the potato cells causes the vacuoles to enlarge, increasing the cell size and, consequently, the length of the cylinder as a whole.

3 The final increase in length will be limited by the lack of plasticity in the cell walls which will not allow the vacuoles to increase beyond a certain point. The relative increase in length will depend on how turgid the tissue is to start with.

4 If the cell walls in the region of extension of a shoot or root were fairly plastic, an uptake of water by osmosis in the cells could cause a permanent increase in size. The cells or their arrangement in the tissue would need to favour increase in length rather than width.

5 The student might suggest that, by using more varied sucrose concentrations, one might be found which did not produce extension or shrinkage and this would thus correspond to the concentration of the cell sap.
More perceptive students might realise that this will be true only for tissues which are in a state of 'incipient plasmolysis' since fully turgid cells cannot take up water even if they are immersed in a hypotonic solution.

NOTE. (a) Although the sucrose concentrations are 0.5 and 0.25 molar, they are expressed as percentages in the students' instructions to avoid having to explain molarity if this
concept is not to be introduced.
   (b) The changes in length will be quite small, plus or minus 2 or 3 mm only. It is best to
collect all the results from the class to show that they are all in 'the same direction'
and thus unlikely to have been caused by errors of measurement.




Experiment 5. The need for carbon dioxide - preparation

Outline The shoots of destarched potted plants are enclosed in plastic bags in which carbon dioxide is removed by soda-lime or supplemented by sodium hydrogencarbonate. The plants are exposed to light for several hours and the leaves tested for starch.

Prior knowledge Starch is a product of photosynthesis. During darkness starch is removed from the leaves so that they are starch-free at the beginning of the experiment. The iodine test for starch. Method of testing a leaf for starch. (This information can be obtained by following the instructions for Experiment 2, p.2.01).

Advance preparation and materials

(a) Potted plants. Impatiens (busy Lizzie) seems to be the most suitable for this experiment.
Allow 2 plants per group.

(b) Destarching. See ‘Introduction’..

(c) Starch test. See Experiment 2, p.2.04).

soda lime, allow 20g per group
sodium bicarbonate, allow 10 cm3 10% solution per group

Apparatus-per group
fluorescent tubes, 2 sets if possible
2 plastic bags, about 400 X 250 mm according to size of plant
2 elastic bands to fit round flower pot
2 containers for soda-lime or sodium hydrogencarbonate solution, e.g. 50 mm plastic Petri dishes or 75 X 25 mm specimen tubes
apparatus for the starch test; (see p.2.04).
spirit marker

NOTE. The experiment does not give reliable results during a period of 50-60 minutes but can be left for 1-7 days. Provided the plants are illuminated for several hours (e.g. overnight) before the class needs to test them, the results should be satisfactory.

Experiment 6. Collecting the gas evolved by pond-weed

You are provided with a glass jar having a lid through which a test-tube can pass, a rubber band and a cork with a wide hole in it.

(a) Pass the test-tube up through the hole in the lid and secure it with the elastic band.

(b) Write your initials on the jar and fill it nearly to the top with tap-water. Add about
5 cm3 sodium hydrogencarbonate solution (about 20 mm in a test-tube).

(c) Collect about 10 pieces of pond-weed up to 10 cm long. Arrange the shoots parallel to each other and trim the ends of the stems with a razor blade.

(d) Fill the test-tube with water to the top of the cork. Push the cut ends of the pond-weed stems through the hole in the cork so that they are held firmly but without crushing them (Figure 1, p.6.02).

(e) Hold the test-tube horizontally with the cork over the edge of the jar (Figure 2, p.6.02) and then turn it upside down so that the pond weed enters the jar, the lid fits over the opening and the test-tube is held by the elastic band (Figure 3, p.6.02).

(f) Little or no air should enter during this operation but if, for some reason, more than about 10 mm air gets into the tube, the experiment should be set up again. Make sure that the cork is under the water in the jar.

(g) Repeat the whole operation from (a) to set up an identical experiment but cover the jar with aluminium foil or some opaque material to exclude all light. Place both jars in a position where they can receive maximum daylight or artificial light.

TESTING THE GAS. READ ALL THIS SECTION BEFORE PROCEEDING.

(h) When either of the test-tubes is more than half full of gas, remove the elastic band and lid, lift the tube and weed out of the jar and turn the tube the right way up. Remove the cork and pond-weed and quickly close the mouth of the tube with your thumb (Figure 4, p.6.02).

(i) Light a wooden splint and when it is burning well, blow it out so that the end continues to glow red.

(j) Remove your thumb from the test-tube and at the same time insert the glowing end of the splint into the tube. (By closing the tube again, blowing out the splint and re-inserting it, you may be able to repeat the test several times.)


Experiment 6. Discussion

I What happened to the glowing splint when it was placed in the test-tube?

2 What gas usually causes this reaction ?

3 The test does not prove that only this gas is present in the test-tube. What other gases might be present?

4 What evidence do you have which suggests that the production of this gas depends on light reaching the plant ?

5 What evidence is there to suggest that the gas in the test-tube came from the pond-weed and not directly from the water in the jar ?

6 How could you eliminate this last possibility ?











Experiment 6. Discussion - answers

I The glowing splint should burst into flames when placed in the test-tube.

2 This reaction is caused characteristically by oxygen.

3 Nitrogen and carbon dioxide might also be present. Hydrogen is not likely to be present since it would have caused explosive combustion in the test-tube.

4 The plants in the jar from which light was excluded will have produced little or no gas.

5 The only evidence against the water as a direct source of the gas is the failure of the control to produce any gas but since this water was not exposed to light, it is not a fair comparison. If the students have done experiment 1 they will have evidence that the gas can be seen escaping from the cut stems. of the plant.

6 A further control should be set up with a jar containing sodium hydrogen carbonate solution as in the experiment but without the pond-weed. If, on illumination, little or no gas collects it is unlikely that the gas in the experiment comes from the water independently of the plant.


Experiment 6. Collecting the gas from pond-weed - preparation

Outline Several cut shoots of Elodea are illuminated in a jar of water. The gas from the cut stems is collected in an inverted test-tube supported by the lid of the jar. The gas is tested with a glowing splint.

Prior knowledge Oxygen relights a glowing splint.

Advance preparation and materials-per group
10 shoots of Elodea 5 cm3 10% sodium bicarbonate solution
The pond-weed should be collected before the experiment and placed in fresh tap-water where it will remain healthy for several days.
Jam jars (300-350 cm3) with screw-on or clip-on lids should be collected and a hole cut in the centre of the lid with a pair of sharp-pointed scissors. The hole should be large enough to admit a test-tube but too small to allow the lip to pass through.
Corks to fit the test-tube should be bored with a 10 mm hole (No.5 cork borer).

Apparatus-per group
2 jam jars (250-300cm3) with lids. The lids to have a central hole
sheet of aluminium foil (or similar opaque material), about 100 X 250 mm, to cover
one jar
2 test-tubes)
2 elastic bands                                               
2 corks with holes bored
splint and access to flame (later)

-per class
4 or 5 spirit markers for students to identify their jars
fluorescent tube as light source (one 4 ft tube will illuminate 15 jars)

NOTE. If results are needed by the following day the fluorescent light should be placed on its side and the jars arranged close to the tube. If a week is to elapse before testing the gas, the jars can be left on a window sill in daylight and exposed to the fluorescent light as necessary before the next lesson. The gas production does tend to fall off as time progresses and it is better to accumulate a tube full of gas as soon as possible even if it has to stand for several days.
The experiment can be varied to compare hydrogencarbonate solution with tap-water  which has been boiled to expel carbon dioxide.







Experiment 7 Gaseous exchange in leaves

The experiment depends on the use of hydrogencarbonate indicator, a pH indicator, containing the dyes cresol red and thymol blue in a solution of sodium hydrogencarbonate. This pH indicator is in equilibrium with atmospheric carbon dioxide, i.e. its orange colour when you receive it indicates the acidity of the atmosphere due to carbon dioxide. Carbon dioxide is an acid gas.
Increase in acidity (fall in pH) turns the indicator yellow while decrease in acidity (rise in pH) turns it first red and eventually purple.

(a) Wash three test-tubes in tap-water. Rinse them with distilled water and finally rinse them
with the bicarbonate indicator itself.

(b) Label the tubes 1-3.

(c) Use a graduated pipette or syringe to place 2 cm3 hydrogencarbonate indicator in each tube.

(d) Roll the leaf longitudinally, its upper surface outwards and slide it into tube 1 so that it is held against the wall of the test-tube and does not touch the indicator solution (Fig. 1). Do
the same with tube 2.

(e) Close each tube with a rubber bung.

(f) Cover tube 1 with aluminium foil to exclude light and place the three tubes a few centimetres
away from a bench lamp (or in direct sunlight if possible). Switch on the lamp and leave the
apparatus for 40 minutes.

(g) Copy the table given below into your notebook.

(h) At the end of 40 minutes, hold all three test-tubes against a white background to compare
the colours of the indicator solutions and record these colours in your table. 

Tube Conditions Colour of indicator Change in pH
1 Leaf in darkness
2 Leaf in light
3 No leaf
 






Experiment 7. Discussion

I If the volume of fruit juice needed to decolourize DCPIP is greater than the volume of 0.1 % ascorbic acid, does this mean that the juice contains more or less vitamin C than
ascorbic acid solution?

2 On this basis, which of your samples contained the most vitamin C?

3 Did the canned fruit. juice differ greatly from the fresh juice?

4 Suppose that 0.8 cm3 orange juice decolourized 1 cm3 DCPIP but it took 1.6 cm3grapefruit
juice to decolourize the same volume of DCPIP; does this mean that oranges contain more vitamin C than grapefruit?

5 What might take place at a canning factory which could (a) reduce, (b) increase the
vitamin C content of fruit juice?

6 Did the fruit juice which had been exposed to air for several days have significantly less
vitamin C than the fresh juice?

7 Sodium sulphite is not vitamin C and yet it decolourized DCPIP - Why did it do this?
(See introductory notes on p. 7.01 and instruction (i).)

8 Since chemicals other than vitamin C will decolourize DCPIP, does this mean that using the dye as a test for vitamin C is unsatisfactory?

9 If canned or bottled juice has sulphur dioxide or sodium sulphite added as a preservative,
will this affect the reliability of your results with DCPIP?

 
















Experiment 7. Discussion - answers

I The greater the volume of juice needed to decolourize DCPIP the less ascorbic acid it contains.

2 The answer will vary according to the source and brand of the various juices but usually fresh orange and lemon juices contain more concentrated vitamin C than the others. Canned grapefruit juice may be substantially weaker and orange squash very weak in vitamin C compared with the others. Grape juice may contain so little that the student has to give up after squirting several
syringes full of juice into the DCPIP.

3 Canned juice is frequently weaker than fresh juice.

4 The result shows that the orange juice in the experiment has a higher vitamin C content than the grapefruit but a generalization about all oranges and all grapefruit is not justified. There is probably a great deal of variation with species, season and age of fruit.

5 (a) Heating the juice might destroy some of its vitamin C content.
(b) Heating the juice might concentrate the vitamin. Ascorbic acid might be added artificially to increase its concentration.

6 There is likely to be a significant difference in the vitamin C content of fruit juices exposed to
the air for several days.

7 Sodium sulphite is a reducing agent and so decolourizes DCPIP.

8 Since chemicals other than vitamin C will affect DCPIP, the test is unsatisfactory unless it can
be assumed that reducing agents other than ascorbic acid are unlikely to be present.

9 If sulphur dioxide or sodium sulphite is present, presumably less of the juice will be needed to decolourize the DCPIP. Whether this will make the results substantially unreliable will depend
on the relative concentrations of sulphite and ascorbic acid in the juice.

Experiment 7. Gaseous exchange in leaves - preparation

Outline Detached leaves are placed in closed test-tubes and illuminated or darkened. The rise of carbon dioxide concentration resulting from respiration, or its fall as a result of photosynthesis,
changes the colour of the pH indicator in the bottom of the test-tube.

Prior knowledge Air contains oxygen, carbon dioxide and nitrogen but only carbon dioxide
is an acid gas. What a pH indicator does. What 'equilibrium' means in the context of this
experiment.

Advance preparation and materials

Leaves or leaf portions from deadnettle, iris, rose, plantain, dandelion, dock, forsythia have
given satisfactory results in 30-45 minutes. Allow two leaves per group.

Hydrogencarbonate indicator. Dissolve 0.2 thymol blue and 0.1 g cresol red powders in 20 cm3 ethanol. Dissolve 0.84 g  sodium hydrogencarbonate  (analytical quality) in 900 cm3 distilled water. Add the alcoholic solution to the hydrogencarbonate solution and make the volume up to 1 litre with distilled water.
Shortly before use, dilute the appropriate amount of this solution 10 times, i.e. add 9 times its
own volume of distilled water.
To bring the solution into equilibrium with atmospheric air; bubble air from outside the
laboratory through the diluted indicator using a filter pump or aquarium pump. After about
10 minutes, the dye should be red.
Allow 10 cm3 per group,

Apparatus-per group

Note that the glassware and bungs must be clean. Any trace of acid or alkali will affect the indicator.

3 test-tubes and rubber bungs
1 bench lamp (in the absence of sunlight)
1 test-tube rack
1 piece of aluminium foil, about 120 x 140 mm
1 graduated pipette or syringe, 5 or 10 cm3
 3 labels or a spirit marker
1 pair forceps

NOTE Experiment 8 is identical to Experiment 7 except that an aquatic plant is used. It may be expedient to have half the class doing Experiment 7 and half doing Experiment 8




Experiment 8. The need for mineral elements

(a) Label four test-tubes as follows: +, Ca, N, --.
(b) Fill each tube to within about 20 mm of the top, with the appropriate water culture.
     (+) Solution containing all mineral elements thought to be needed for healthy growth
     (Ca) Solution containing all mineral elements as in + except for calcium
     (N) Solution as in + but lacking nitrate
     (--) Distilled water, i.e. no mineral elements at all
(c) Select four seedlings which appear to be at the same stage of development.
(d) If there is a wide difference between the development of the root systems, reduce all systems
to the same number and approximate length of root as in the least developed.
(e) Leaving the shoot and roots free, roll a strip of cotton wool round the grain to hold the seedling lightly but firmly in place in the mouth of the: test-tube (Fig. 1). Place a seedling in each test-tube so that the root is well covered by the culture solution.
(f) Mark your initials and the date on the rack or container provided and place the four tubes in a position where they can receive daylight or artificial illumination. Leave the seedlings to
grow for two weeks.
(g) During this period of time the levels of the solutions will fall in the test-tubes. They need to be inspected every two days and the level restored if necessary. Carefully remove the cotton wool and top up the test-tubes with DISTILLED WATER from a wash-bottle.
(h) After two weeks transfer the tubes to a rack so that the seedlings can be compared side by side.
(i) Draw up a table similar to the one below.
(j) Study the whole group of seedlings and note in your table  any which show abnormalities of
leaf colour or shape, e.g. dead areas, discoloured patches, pale green colour.
(k) Remove the seedling from the full culture (+), unwind and discard the cotton wool and cut off the leaves as shown in Fig. 2. By placing the leaves end to end along a ruler, measure and
record their total length.
(l) Cut the root system just below the grain (Fig. 2) and use forceps to separate the main roots, working from the top. Place these end to end along a ruler to measure their total length. Ignore the lateral roots for this purpose (unless you have plenty of time and patience).
(m) When you have made the measurements, place the roots and leaves in the container labelled
'+' so that their dry. weight can be found later.
(n) Repeat the measurements for each of the seedlings in turn, placing the leaves and roots in
the appropriate container afterwards.
(o) Plot histograms (Fig. 3) of the root and shoot lengths for each seedling.

Culture solution + Ca N --
Leaf colour
Total leaf length
Total root length
Dry weight*
* Whole class






Experiment 9 - Discussion

1 At the start of the experiment, by how much did the total mass of the eight 1cm3 cubes
differ from the mass of the 8cm3 cube?

2 What is the most likely reason for this difference?

3 Would you expect there to be much difference in the total volume of the eight 1 cm3 cubes and the volume of the single 8 cm3 cube?

4 (a) What was the total surface area of the eight 1 cm3 cubes?
   (b) What was the surface area of the single 8 cm3 cube?

5 Why, do you suppose, the mass of the cubes increased after one hour's immersion in water?

6 What difference was there in the percentage increase in mass between the eight 1 cm3 cubes
and the single 8 cm3 cube?

7 What is the most likely explanation of this difference?











Experiment 9. Discussion - answers

1.Theoretically, the best growth should be in the full culture and the worst in distilled water.
Failure to achieve these results may be attributed to, e.g. genetic variability in the seedlings, disease, impurities in the solutions, damage to seedlings in setting up the experiment or
topping-up the tubes.
  2.Calcium pectate, by its contribution to the middle lamella, affects the adhesion of cells. In the
absence of calcium, the growing points of shoots and roots may become disorganized. Calcium  deficiency sometimes causes a harmfully high level of magnesium absorption, which may explain why some seedlings in the calcium-deficient solution grow less well than those in distilled water. Nitrates are essential as the only source of nitrogen for making amino acids and proteins.

3.If the lack of a particular element resulted in poor root growth, the development of the shoot
would be retarded through inadequate water and mineral supply from a deficient rooting system,
irrespective of any specific effect on the shoot of the lack of that element. Similarly, poor development of the leaves could limit root growth as a result of inadequate food supply to the roots from the leaves.

4.Large seeds such as runner bean have a considerable supply of nutriment in their cotyledons or endosperm. It takes a week or two to exhaust this supply so that the effect of the deficient cultures is delayed.

5.The rates of water uptake and salt uptake are not necessarily the same. If during the course of transpiration the plant removes more water from the test-tube than it removes salts, topping up
with culture solution may increase the concentration of salts too much.

6. The proportion of minerals in the solution may be different from that in the soil. The ions of the elements selected may not be the same as those in the soil. Mineral particles are lacking. The experiment is confined to only the early stages of growth; the plant may have different
requirements as it matures.

7.The carbon needed for making carbohydrates comes from carbon dioxide in the air.


Experiment 9 The need for mineral elements - preparation

Outline Wheat seedlings are grown in water cultures some of which lack essential elements
Measurements are made on the seedlings to find the relative effect of these elements.

Prior knowledge The connection between the element and its salt, i.e. nitrate contains nitrogen. Some idea of the role of mineral elements in metabolism.

Apparatus - per group

test-tube rack and 4 test-tubes                           wash bottle of distilled water for 'topping up'
4 labels or spirit marker                                     graph paper
1 can or box to hold 4 tubes                               labelled receptacle for collecting seedlings from
4 strips cotton wool about 20 x 100 mm              each solution for determining dry weight

Advance preparation and materials

Wheat seedlings. Seven days before the experiment, soak the wheat for 24 hours. On day 1 pour oft the water but leave the wheat covered in the container. On day 2 select those grains which show signs of germinating and roll them in moist blotting paper or newsprint with the seedlings about 20 mm from the top of the roll and the embryo directed downwards. Space the fruits about15 mm apart so that the roots do not get entangled. Place the rolls upright in a plastic bag and allow the wheat to germinate in darkness. Four seedlings are needed per group but germinate twice as many to allow for failures. If several rolls are prepared, the students can select their own seedlings from them.

Water cultures. Either purchase the Sach's water culture tablets or prepare the solutions as in Table 1 (p.9.05), using chemicals of the highest purity available (analytical reagents) and freshly distilled or deionized water. Store the stock solutions in stoppered borosilicate flasks and mix and dilute them when required as described in Table 2. The three solutions selected have been found to give the most consistent and reliable results with wheat,



Experiment 8. Discussion - answers

1 (a) The vacuole should diminish and appear to separate from the cell wall in places.
(b) The cell sap may become noticeably deeper in colour when compared with unplasmolysed
cells.

2 There should be no perceptible change in the shape of the cell.

3 The cell sap must have lost part of its contents, presumably water. This would account for its diminution in volume and, since the red pigment would become more concentrated, the intensification of its colour.

4 The space between the vacuole and cell wall must be occupied by the sucrose solution (and, of
course, the cell membrane and cytoplasm bounding the cell sap).

5 The cell membrane, tonoplast and cytoplasm separate the solution and the cell sap.

6 The cell membrane, the cytoplasm, the tonoplast or some combination of these must be selectively permeable.

7 If the cell wall were selectively permeable, the loss of water to the sucrose solution would distort the cell wall and no space would appear between it and the vacuole.

8 If all the cells in a plant were plasmolysed, the unsupported structures would be limp and wilting and the plant would eventually die.

9 (a) When the sucrose solution was replaced by water, the cells recovered from the plasmolysed
 condition and the vacuole expanded to fill the cell once again.
(b) The cell sap now contains a stronger solution than the surrounding fluid and absorbs water - through the selectively permeable cytoplasm.

10 A plant with its cells turgid would have firm tissues, an erect stem and expanded leaves.






Experiment 10b. The effect of carbon dioxide concentration on gas production 

(a) Set up the apparatus as described on pp. 10.01 & 10.02 Select a vigorously bubbling piece of pond-weed and insert it in the microburette.

(b) Place the bench lamp 15 cm from the beaker and do not move it during the course of the
experiment.

(c) Copy the table given below into your notebook.

(d) At an appropriate moment, note the time and draw any gas which has accumulated in the bulb, up into the tube. Leave the pond-weed to continue bubbling for an exact number of minutes.

(e) At the end of this time, draw the collected gas into the graduated tube, measure the length of
the gas column and record this in your table.

(f) Now add 5 cm3 sodium hydrogencarbonate solution to the jar from a graduated pipette or syringe and force the air from the tube into the bulb.

(g) At an appropriate moment, note the time and draw up the collected gas. Allow the pond-weed to bubble for the same length of time as before and at the end of this time, draw up the accumulated gas from the bulb and measure its length in the graduated tube.

(h) Add another 5 cm3 sodium hydrogencarbonate solution and measure the gas produced in the
time you have selected.

(i) Continue the experiment by adding successive 5 cm3 volumes of hydrogencarbonate solution
until there is no change in the rate of gas production.

(j) Plot a graph of the volume of gas produced against the concentration of hydrogencarbonate.

Volume of hydrogencarbonate in cm3 Volume of gas in mm Time Volume per minute
0
5
10
15
20
25
30
35
40

NOTE If bubbling ceases during the experiment it can often be restarted by cutting a small piece from the stem without having to obtain a fresh shoot and start the experiment all over again