Here is a compilation of top forty-five experiments on Plant physiology:- 1. Solution, Suspension and Colloids 2. Plant Cellular Organization 3. Cellular Constituents 4. Enzymes 5. Movement of Substances 6. Water Relation 7. Water Loss 8. Absorption, Translocation and Utilisation of Minerals 9. Environmental Effects on Photosynthesis 10. Respiration 11. Growth and Development 12. Plant Movements and Others.
- Experiment on Solution, Suspension and Colloids
- Experiment on Plant Cellular Organization
- Experiment on Cellular Constituents
- Experiment on Enzymes
- Experiment on Movement of Substances
- Experiment on Water Relation
- Experiment on Water Loss
- Experiment on Absorption, Translocation and Utilisation of Minerals
- Experiment on Environmental Effects on Photosynthesis
- Experiment on Respiration
- Experiment on Growth and Development
- Experiment on Plant Movements
- Experiment on Sexual reproduction in Higher Plants
- Experiment on Physiology of Seeds
- Experiment on Transport through Phloem
1. Experiment on Solution, Suspension and Colloids (4 Experiments):
To prepare a percentage solution of sodium chloride.
This is commonly expressed as a gram of solute in 100 ml of a solution. Dissolve 5 g of sodium chloride in water and dilute until the solution attains a final volume of 100 ml. This is 5 per cent solution of sodium chloride.
(If a solute is liquid e.g. alcohol, glycerine then ml of alcohol or glycerine is added to it to make a final volume of 100 ml with water. This type of solution is usually expressed as per cent by volume e.g., 10 ml of glycerine; 90 ml of water, 10% glycerine solution).
To demonstrate the buffer action of salts of organic acids.
Using universal pH paper, determine the pH of IN sodium chloride. Take 10 cc IN of HC1. Mark the pH again. The solution has become acidic.
Note the pH of IN sodium acetate solution. Add 1 ml of IN HC1 to 10 ml of sodium acetate solution and determine its pH. Note very little change in the pH of the new solution. Apparently sodium acetate resists the change in pH when IN HC1 was added. This is called buffer action. In fact sodium acetate gets dissociated.
To determine the buffering capacity of plant sap.
Take a herbaceous twig and crush it to obtain its sap or juice. Obtain an amount of 25 ml of plant juice. Determine the initial pH of the juice using pH indicating papers or pH meter electrodes. Fill the burette to the zero mark with 0.5 N HCl and add 1.0 ml of acid and shake the beaker carefully.
Re-determine pH. Keep on adding 1.0 ml portion of acid until the pH of 20 is attained. Take now 25 ml of juice and add 0.5N KOH from the burette in 1.0 ml increments till 12.0 pH is obtained.
To demonstrate colloidal system.
Suspension: To a solution of barium chloride add a little quantity of dilute sulphuric acid. Precipitate is formed. Shake the precipitate of barium sulphate and allow it to stand. The precipitate settles down. This is suspension.
Hydrophobic solution: Take 1 ml of FeCl3 (30%) solution and pour it in a beaker containing 100 ml distilled water. Boil this solution till its colour becomes brown-red. This is Fe(OH)3 sol which is hydrophobic. Here the dispersion phase is liquid while the dispersed phase comprises particles.
Emulsion: Add a few drops of turpentine oil or paraffin in a few ml of ethyl alcohol. Pour the solution into 100 ml of water. This emulsion of turpentine oil in water is ready. Here both the phases are liquid.
2. Experiment on Plant Cellular Organization (3 Experiments):
To study the structure of a general plant cell.
Take a fine strip of epidermis from the inner scale leaf of an onion bulb and mount it in a drop of weak iodine solution and examine under the microscope (high power).
Study the arrangement and form of cells. Note the cell wall, semi-transparent protoplasm and also the vacuoles. Examine the nucleus lying near the cell wall which becomes distinct after the addition of iodine solution.
To study the plastids.
Chloroplasts. Strip off epidermis of a young leaf of Tradescantia, Hydrilla or a moss plant and observe the chloroplasts filling the cells. Note their form and size.
Chromoplasts. Cut thin sections of carrot root near the outer skin or peel the pericarp of Capsicum (red chillies) fruit. Examine these materials under the microscope and study the shape and size of the chromoplasts.
To demonstrate the occurrence of transfer cells and tracheary element (PAS toluidine Blue O).
1. After localizing polysaccharides employing PAS technique, treat same slides with toluidine Blue O (0.05% in 0.1 M phosphate buffer, pH 6.8).
2. Staining is allowed to continue for 1.5 minute.
3. Pass through xylene and mount in DPX.
Indication: Tracheary elements stain green which contrast sharply with the reddish purple “Transfer cells”.
3. Experiment on Cellular Constituents (4 Experiments):
To demonstrate the presence of starch histochemically.
Prepare a thin section of a potato tuber with the help of a blade or razor. Mount it over a slide in a drop of water. Add a drop of iodine solution (IKI) on it and cover with a glass cover. Examine under the light microscope. The starch grains are stained blue. Old starch grains stain blue black while fresh starch gives red or purple colour.
Estimation of Proteins and Nucleic Acids.
The ovules/seeds (100-1000 mg fresh wt.) are finally homogenized in the chilled pestle and mortar with 95% alcohol and centrifuged at 5000 xg for 15 min. The debris free residue is extracted with 4 ml of 5% Trichloroacetic acid.
(TCA) chilled at 4°C for 15 minutes and is centrifuged and the supernatant containing soluble phosphate is discarded. The residue is then treated with 4 ml of cold absolute alcohol to remove the alcohol soluble phosphate.
The residue is washed 2-3 times with 5 ml ethanol: water (3:1 v/v) for 30 min at 50°C and the supernatant containing lipid- soluble substances is discarded and residue is then extracted with 3 ml of 0.3N NaOH and kept at 30°C for 18 h and is centrifuged.
The residue obtained is washed once again with 1 ml of 0.3 NaOH and the mixture is centrifuged. Final volume is then made to 10 ml with 0.3N NaOH. 2 ml out of this solution is used for protein determination.
The rest (8 ml) of the solution is treated with 4 ml of 15% perchloric acid of pH 1.0, and is kept at 4°C for 40 minutes and is then centrifuged. The supernatant so obtained contains the ribonucleic acid (RNA) fraction.
The residue is suspended in 5 ml of 1N perchloric acid and is kept at 40°C for 30 minutes and is then centrifuged and the supernatant is combined with RNA fraction.
Three successive washings with 1N-perchloric acid are given to the sediments and all the supernatants are combined in RNA fraction and final volume of this fraction is made to 20 ml with IN perchloric acid.
Determination of Ribonucleic Acid (RNA):
RNA is estimated by the method of Markham (1955) with orcinol as a colour developing reagent.
Orcinol reagent is freshly prepared by mixing:
(i) 40 ml of conc. HCl (A.R.)
(ii) 1 ml of 10% ferric chloride and
(iii) 10 ml of 0.1 % orcinol solution
For the colour development 2 ml of orcinol reagent is added to 1 ml of the sample and the contents are kept for exactly 8 minutes in a boiling water bath.
The green coloured precipitate so obtained is dissolved in n-butanol and the volume is made to 5 ml.
The transparent green colour solution is read at 550 nm in Spectronic-20 colorimeter using IN perchloric acid as a blank. The amount of RNA is expressed in terms of µg ribose.
Estimation of Protein:
Protein is estimated according to the method of Lowry (1951).
For this, the reagents are prepared as follows:
Reagent A: 2% Na2CO3 in 0.1 N NaOH.
Reagent B: 0.5% CuSO4. 5H2O in 1% sodium potassium tartarate.
Reagent C: Prepared by mixing 50 ml of reagent A with 1 ml of reagent B. The reagent is freshly prepared every time.
Reagent D: Folin Ciocalteau reagent diluted with water in 1:1 ratio.
0.5 ml of protein extract is taken in a test tube and 5 ml of solution is added to it. After 10 minutes, 0.5 ml solution D is added rapidly with immediate mixing. After 30 minutes the colour is read at 570 nm. Protein is calculated by using bovine serum albumin as standard. Only soluble proteins are estimated by this method.
Estimation of Free Amino Acids:
Free amino acids can be determined by the method given below:
The seeds are well ground with 70% ethanol in pestle and mortar and the extract is centrifuged at 5000 xg for 20 min.
The residue is washed thrice with 70% ethanol and centrifuged each time and supernatants are combined and are evaporated to dryness on a water bath. The residue is dissolved in 25% ethanol and volume is made to 5 ml.
The qualitative separation of amino acids is done by two dimensional paper chromatography using What-man No. 1 chromatographic paper.
Spotting is done on these papers. After air drying the spots, sheets are run in the solvent system of phenol: water (4:1 v/v) in ammonical atmosphere in a descending order and is allowed to run 3 cm above the other end of the paper.
The paper is taken out, dried at room temperature and is run in second dimensions in second solvent system of butanol: acetic acid: water (4:1:5; v/v). After complete run, the paper is taken out, dried at room temperature for 12 h.
The chromatograms are sprayed with 0.2% ninhydrin made in butanol. The colour is developed by keeping the chromatograms in oven at 60°C for 30 min. The ninhydrin positive spots of amino acids are identified by comparing the Rf values with those of known amino acids.
The amino acids are quantitatively estimated by the method of Yemm and Cocking (1955). The individual spots of the amino acids are cut into small pieces and are eluted with 5 ml of 70% ethanol for 1 hour.
The optical density of the elutant is recorded at 570 nm using 70% ethanol as blank. The quantity of amino acids is worked out taking glutamic acid as the standard.
To localize polysaccharides in a tissue.
Place the tissue sections in 0.5% aqueous HIO4 for 10-15 minutes at room temperature. Wash in running water for 10 minutes. Place in Schiff’s reagent for 15-20 minutes.
Schiff’s reagent: Dissolve 0.5 g of basic fuchsin and 0.5 g of sodium met bisulphite in 100 ml of 0.15N HCl.
Shake the mixture for 2-3 hr. Add 300 mg fresh decolourizing charocal and shake for 5 minutes. Filter and keep the colourless solution in a coloured bottle in a refrigerator.
Rinse the sections in water and transfer into 2% sodium bisulphite for 1 -2 minutes.
Wash again in running water for 5-10 minutes.
Dehydrate and mount in DPX.
Intense purplish-red colour shows the presence of insoluble polysaccharides.
To estimate starch and phenols.
Estimation of Total Phenols:
Total phenols are estimated by the method of Swain and Hills (1959).
Dissolve 35 g of anhydrous sodium carbonate in 100 ml distilled water by heating on a water bath at 70° to 80°C. Cool overnight and filter through glass wool.
Preparation of Extract:
1 g of ovule/seeds is collected and ground in a pestle and mortar with 0.3 N HCl in methanol. Keep the material for 24 h and then filter. Evaporate the methanol from filtrate and dilute it to 5 ml with distilled water.
0. 2.ml of extract is diluted to 5 ml with distilled water and shaken. 0.5 ml of folin Dennis reagent is added and is shaken thoroughly.
Add 1 ml of sodium carbonate solution after exactly 3 min and shake the extract well and leave for 1 h. Record O.D. at 630 nm. The quantity of phenols is expressed in terms of the absorbance at 630 nm per mg fresh weight.
Estimation of Starch:
Reagents for extractions:
1. 80% ethanol
2. Perchloric acid, 9.2N (dilute 793 ml of 70% HC104 to 1 litre).
3. Perchloric acid, 4.6N (dilute 397 ml of 70% HC104 to 1 litre).
Starch is estimated by taking wheat seeds or leaves. These are finally ground in pestle and mortar with ethanol while constantly stirring with a glass rod.
To the residue 1 ml of hot distilled H2O is added and kept for 1/2 hour at 70°C and 1.3 ml of 52% perchloric acid is added with constant stirring.
Supernatant is collected and final volume is made up to 5 ml. 0.5 of above supernatant is added to 5 ml of anthrone reagent and heated for 12 minutes at 100°C in a boiling water bath. The absorbance is read at 630 nm after cooling to room temperature.
The quantity of starch is calculated from the standard curve of glucose and is expressed in terms of glucose.
4. Experiment on Enzymes (2 Experiments):
To demonstrate the activity of enzymes.
(i) Prepare saturated solution of benzidine (5 ml). Cut a thin section of fast growing tissues e.g. root or germinating seeds or take pollen grains. Put the material in phosphate buffer at pH 7.0 and transfer to the above incubation mixture [5 ml benzidine saturated solution + H202 (1 %) 5 ml + ammonium chloride (5%) 1 ml]. Incubate at room temperature for 2-3 min. Note the appearance of dark blue colouration due to the presence of peroxidase.
(ii) Take a starch paste and immerse some materials like germinating pollen or actively growing roots in it. Notice the transformation of starch into sugars. It indicates the occurrence of amylase in the plant tissue studied.
(iii) Take Pinus roxburghii pollen grains and germinate them in distilled water or 15% sucrose solution. Test for the occurrence of starch grains after 15-20 hours. It will be observed that pollen grains lack starch while those germinated in sucrose solution contain abundant starch as observed after staining with IKI2 diluted solution. Presence of starch-phosphorylase is evident.
Demonstrate the activity of amylase.
Take about 50 g of potato tuber and grind in 100 ml distilled water. Filter the homogenate through muslin cloth and obtain the filtrate. Let the liquid stand for 15 min and centrifuge (2400 rpm) to remove sedimented starch. Repeat centrifugation 2—3 times and remove the sedimentated starch.
The latter is added to 100 ml of boiling water to procure starch solution. Now take germinating wheat or barley seeds and extract amylase solution with phosphate citrate buffer (4.0 ml pH 5.6). Decant and centrifuge at 2000 rpm for 5 min. Take about 0.5 ml of this solution in four test tubes.
Keep two test tubes in boiling water for 5—10 min while the other two are kept at room temperature. Add 0.5 ml of starch solution to each of the tubes. After a while test for starch using KI test. The tubes which were kept in boiling water contained starch while the others did not.
5. Experiment on Movement of Substances (2 Experiments):
To demonstrate the process of Endosmosis and Exosmosis using raisins and fresh grapes, respectively.
Place few raisins with their stalks intact in a dish containing water overnight. The raisins become fully swollen due to the entry of water inside. Since the concentration of water outside the raisins is higher than the concentration of water inside the cells, the movement of water occurs (higher concentration) into the cells of raisins (lower concentration). The outer membrane of raisin acts as a semipermeable. It is said to be endosmosis.
To demonstrate exosmosis place few fresh grapes with their stalks intact in concentrated sugar or salt solution and keep as such overnight.
The grapes shrink due to the loss of outward movement of water into the outside solution from inside the grapes where its concentration is higher than its concentration in the outside concentrated solution. As in this osmotic phenomenon there is loss of water from the cells, therefore, it is known as “exosmosis”. Due to loss of water from the cells they are plasmolysed and due to loss of turgor of cells, there is shrinking of grapes.
Raisins and grapes should be with their stalks intact. The sugar or salt solution must be a solution of higher concentration than the solution present inside the grapes.
Effect of temperature on cell membrane permeability.
Take a fresh piece of red sugar beet and cut a few thin slices of equal sizes. Rinse these pieces in running water repeatedly. Take a beaker containing distilled water and put these pieces in it. Take a test tube and heat around 200 ml of distilled water to 70°C.
Immerse the slice in this water. Repeat the process at different temperatures e.g. 65°, 60°, 55°, 50°, 45°, 40°. Use fresh slice each time and a separate test tube containing water at specific temperature. In one of the test tubes distilled water is added and is kept at room temperature. This is used as a control.
Leave the test tubes for an hour or so. Shake them periodically. Compare the relative amount of red pigment which diffused from each slice. Note the temperature at which the cytoplasm is killed and allows the diffusion of red pigment (anthocyanins) from the cell.
6. Experiment on Water Relation (2 Experiments):
To demonstrate root pressure, and its measurement.
Take a potted plant of Petunia or balsam or any Other herbaceous Plant.
Water the pot thoroughly and cut across the stem (Ps) a few centimetres above the soil surface (cut with in water in order to avoid entrance of air bubbles in xylem vessels). Put a drop of oil at the surface of water of the glass tube (GT) to prevent evaporation.
Make all the connections air tight with wax and note the initial level of water in the glass tube. To measure actual pressure, attach to the cut stem, one end of manometer tube instead of glass tube and make the joints air right. The manometer tube and the rubber joint (RT) contain water. The second end of manometer below the bulb is filled with mercury.
After a few hours, notice the rise in the level of water in the glass tube or rise in the second end of manometer tube, if manometer is used for pressure measurement water will ooze out of the stump due to root pressure and this will exert pressure on the water causing the rise in water level or mercury level. (Fig. 7-8). This increase in water level in the glass tube or mercury in the manometer is due to the following reasons:
The water absorbed osmotically (i|/) by the root hairs from the soil accumulates in the tissues of the cortex and this accumulation of water causes turgidity of the cortex cells. Under this condition their walls which are composed of cellulose, exert pressure on the fluid contents and force out the water towards the xylem vessels, and the cortical cells become flaccid.
They again absorb outer water and become turgid. This process continues causing an intermittent pumping action in the cortex of the roots and this process naturally gives rise to considerable pressure causing movement of water under force.
This diffusion of water into the xylem vessels through the unthicken portions and the accumulation of the exuded water under force in the glass tube exhibits the rise in the level of water or mercury and this rise indicates the root pressure.
Select a succulent plant. Stem should be cut under water to avoid the entrance of air bubbles in xylem vessels. Rubber tubing should be fixed carefully and all connections should be made air-tight with wax.
To demonstrate that xylem is the main path of movement of sap in the plant (Ringing experiment).
Remove a ring of bark about an inch in length outside the wood with the help of a blade from the stem of the twig without injuring xylem. This removed girdle of bark contains all the external tissues including phloem.
The wood (xylem) elements are removed by careful manipulation in the other twig, a few cm in length without injuring the phloem and cortex. the ringing operations are done while the twigs are under water. The twigs are then kept in water separately for a few hours.
In case of twig in which the xylem is not removed, there is no change in the condition and the twig remains alive and does not show any sign of wilting the twig in which the xylem was removed, shows wilting within a few hours, due to the failure of water to rise up the stem and leaves as there is transpiration from the aerial parts.
But in case of twig where xylem was intact there is upward movement of water to make the loss of water from the aerial portions by transpiration. The twig should be cut under water. Do ringing the utmost care.
7. Experiment on Water Loss (4 Experiments):
To demonstrate the transpiration.
Take a well-watered potted plant and cover external soil surface of pot with oil paper or rubber sheet to avoid loss of water by direct evaporation. Then after weighing the pot along with its contents, place it under bell jar on a glass slab and seal the edges of bell jar with Vaseline and keep it in sunlight for sometimes.
It will be observed that after a few hours the bell jar becomes misty and its inner wall contains drops of water which may flow down the sides of the bell jar. These water drops have come from the water vapours lost by the aerial parts of the plant during transpiration and consequently condensed on the inner surface of the bell jar.
The pot is again weighed and it is found that there is a decrease in weight, which is also due to the loss of water from the aerial portions of plant during transpiration.
The amount of water lost can be measured by placing a weighed quantity of calcium chloride in a crucible. The amount of water transpired is absorbed by anhydrous calcium chloride and there is increase in the weight of crucible which indicates the amount of water transpired.
A green well-watered potted plant should be taken for this experiment. Pot with its soil surface should be covered properly with water proof cloth or paper.
Bell jar should be made air tight by smearing its edges with Vaseline properly.
To demonstrate stomatal transpiration.
Cover the cut ends of the petioles of leaves or Ficus with wax so as to prevent evaporation. Then treat the leaves as follow:
Smear with Vaseline the leaves as follow: Leaf A—on both surfaces, B—on lower side, leaf C— on upper surface and D—is untreated i.e. normal. Weigh each one of them and leave them hanging by a thread on a stand in an open place (Fig. 8-8). After a day it will be observed that each leaf exhibits a different behaviour.
Leaf A—remains as such, B—some what dries out, C—more dries out and D— most dried out. These leaves dry in order of D to B i.e. D leaf dried first of all, then C, and B and A remains in its original green and fresh condition.
When weighed again it is also found that leaves from A to D lose weight, progressively—number A leaf loses the least amount of weight, leaf B—a little more weight-leaf C—loses much more weight than the first two leaves and leaf D loses weight the most.
The dry and wilting or loss of weight of leaves C and D is due to the reason that the stomatal transpiration is going on at a greater rate in these leaves due to more number of stomata on the lower surface of the leaves. Leaf D is affected the most because here both stomatal and cuticular transpiration is going on.
But in case of A where lower surface having more stomata is blocked the closing of stomata with Vaseline is caused and main loss of water (stomatal transpiration) is prevented. So this leaf remains in its original condition.
But in case of leaf B in which the upper surface having a few number of stomata is exposed is affected a little (mainly cuticular transpiration, as lower stomatal side being covered with Vaseline). It may be concluded that—
(i) Main loss of water in most of the plants is due to stomata (stomatal transpiration),
(ii) Transpiration depends on the number and condition of the stomata,
(iii) Transpiration occurs at a more rapid rate on the lower surface than on the upper surface of leaves.
Cut ends of petioles should be smeared with wax or Vaseline to prevent evaporation. Leaves should be thoroughly smeared with Vaseline. Dorsiventral leaves should be selected for this experiment.
To demonstrate that the amount of water lost by the plant during transpiration is approximately equal to the amount of water absorbed by the roots of the plant.
Take a young herbaceous plant or a seedling and a large glass tube, provided with a graduated side tube, rubber cork, water, oil, wax or Vaseline, weighing balance and weights. Fill the bottle with water. Some of the water automatically passes into the side-tube. Water in the bottle and side-tube attains same level.
A few drops of oil are put above the water level in the side tube to check evaporation. A small, rooted herbaceous plant is fixed in the mouth of the bottle through a hole in the cork in such a way that roots are completely dipped in water. The connection is made air tight by applying little Vaseline around the hole and joints.
Wrap black cloth around the bottle in order to exclude light from the roots. The whole apparatus is weighed and the level of water in the side tube is noted before and after the experiment. Place the apparatus in sunlight. After some time the level of water in the side tube falls down due to absorption by roots. The final level of water is noted.
The difference in the water level is equal to the water absorbed by the plant. The apparatus is now weighed again. The difference between the initial and the final weight of the apparatus is equal to the amount of water lost during transpiration. This decrease in weight of the apparatus is approximately equal to the decrease in level of water in the side tube (1 c.c. = 1 g) during the period of transpiration.
Thus, it is observed that the water absorbed will be approximately equal to water transpired if calculated on the basis that 1 cc of water is equal to 1 g by weight.
The experiment proves that the amount of water lost by the plant during transpiration is approximately equal to the water absorbed by the plant during a certain period. There occurs loss of weight due to transpiration.
The side tube must be fixed in such a way that it remains at a higher level than the bottle, otherwise the experiment cannot be performed accurately. The apparatus should be air tight. Seedling should immediately be inserted into the bottle after taking out frorn pot. The roots of the plant should be well washed to remove all the soil particles. Oil drop should be put on the water column in the side tube to avoid evaporation.
Demonstration of opening and closing of stomata.
Take a leaf of Amaryllis or some plant having large stomata. Strip off a small piece of epidermis from the lower surface and mount on a slide in a drop of water. With the aid of ocular micrometre measure the width of the widely opened stomata.
Add a few drops of 1 M sucrose solution on one side of the cover slip and with the help of filter paper from the opposite side, draw it inside the cover slip. Observe under the microscope and notice a gradual closure of the stomata.
Add a few drops of water on one side of the cover slip and proceed as described above. Repeat the procedure several times with a view to replacing sucrose solution. Observe the stomata under the microscope. Notice reopening of the stomata. Record the time required for the reopening of the stomata.
The stomata can also be closed with the help of diluted solution of ABA or some phenols. One of the convenient substitutes for the phenol is salicylic acid present in Aspro or Anacin tablets.
8. Experiment on Absorption, Translocation and Utilisation of Minerals (3 Experiments):
Demonstrate the presence of phosphate in plant tissues.
Dip some leaves in warm alcohol for some time. Wash in distilled water and cut thin transverse sections. Prepare ammonium molybdate solution by dissolving l g of this substance in 15 ml of nitric acid. Mount the section in one to two drops of this solution. Examine under a microscope after half an hour. Notice yellow drops or crystals of ammonium phosphomolybdate in the leaf.
Determination of total nitrogen in plant tissues.
Weigh about 100 mg of dry plant material which has been powdered and put in a small Kjeldahl flask. Add 2 ml of salicylic-sulphuric acid mixture in it. Thoroughly mix the whole mixture and allow it to stand for 30 min.
Then add 0.3 g sodium thiosulphate and heat till fumes start appearing. Cool at this stage. Add catalyst and 0.5 ml sulphuric acid and digest on low and high flame till colourless solution is formed. Let the material be digested for 30 min. Then run the blank without plant material.
Add small pieces of glass rods or beads in the Kjeldahl flask and then connect with the collecting flask at the receiving end to a distillation. Collect the fumes and distillate in 5 ml 2% boric acid contained in 50 ml conical flask. Add 40% NaOH to the flask through the funnel carefully.
Let the alkaline solution settle down at the bottom of the flask. Again add a few pieces of glass beads to the flask and set the flask with a condenser as shown in figure (Fig. 10.10). Light the burner or put the flame and mix the contents in the flask by rotating the same. The condenser must be immersed in the boric acid before beginning with the distillation.
As a few drops of distilled material have appeared in the condenser, the condenser is raised. The distillation may be continued in order to secure sufficient distillate. Stop distillation and add 2 drops of indicator to the distillate and titrate with N/28 HC1 to a faint pink colour. The percentage of nitrogen is computed as follows:
The percentage of nitrogen is computed as follows:
N2% = [(Va – Vb) N × a × 0.014007 × 100]/ Wt. of plant material
Va = volume of actual titration of HCl, Vb = volume of blank titration of HCl
Na = normality of HCl, and 0.014007 = milliequivalent of N2.
How will you estimate nitrate in a given sample of plant.
Estimation of Nitrate :
Fruit and seeds can be used for the estimation of nitrate which is determined by the method of Wodley (1960). The technique is first standardized for the materials used :
(a) For standard curve, a stock solution of 5 mM KN03 is made and further dilutions are done accordingly. Those are ranging from 100 a mole to 2000 i mole.
(b) 20% acetic acid solution containing 0.2 ppm Cu+2 (20 (a g/100 ml of copper sulphate)
(c) Power mixture :
(i) 100 g of barium sulphate
(ii) 10 g of Magnesium sulphate dehydrate
(iii) 2 g of finely powered zinc
(iv) 75 g of citric acid
(v) 4 g of sulphanilic acid
(vi) 2 g 1-naphthylamine
The above coarse (mixtures) salts are ground separately and then mixed together to a fine mixture and stored in a brown coloured bottle since light affects 1-Naphthylamine. This reagent is stable for many months in a bottle painted black.
An anueous extract of plant material is made taking 100 mg of dried seeds and fruit wall at different stages of development. The samples are dried at 80°C for 48 hours.
For extraction 5 ml of distilled water is added to 100 mg of the sample. The aqueous extract is kept in a shaker at 50°C for 40 minutes and then boiled on boiling water bath for exactly 5 minutes. After boiling, the extract is cooled immediately.
To one ml of this extract 9 ml of 20% acetic acid containing 0.2 ppm copper sulphate and one gram of powdered mixture are added. The sample is shaken for 15 seconds and again shaken 5 minutes later. After 5 minutes the sample is shaken for the third time.
The extract is centrifuged at 1000 * g for 3 minutes and supernatant is filtered through miracloth. Light absorbance is measured at 520 nm against a standard solution and the amount of nitrate is calculated from standard curve and expressed in terms of mmole/g dry weight.
9. Experiment on Environmental Effects on Photosynthesis (3 Experiments):
To demonstrate that oxygen is liberated during photosynthesis.
Fill a large beaker with water and place some freshly collected bunch of aquatic plant (e.g. Hydrilla) in the beaker. Cover them with an inverted funnel as seen in Fig. 14-3. Cut the plant at the base and the cut ends tied with the help of a thread. Cut ends are kept facing towards the tube of a funnel. On the upper end of the funnel a test tube full of water is inverted. The whole apparatus is exposed to bright sun.
After some time, a chain of gas bubbles is seen to emerge from the cut ends of Hydrilla. These bubbles ascent through the water, collect at the upper end of the inverted test tube displacing water. This gas, on testing (by taking glowing splinter to it or by introduction of pyrogallate solution of potash which absorbs this gas and level again rises) is found to be oxygen.
The same experiment can also be set to show that
a. CO2 is essential for photosynthesis.
b. Light is essential for photosynthesis.
c. Chlorophyll is essential for photosynthesis.
In order to check the leakage of gas at different places do not injure the plant except at one place where it is cut. Experiment should be kept in bright light.
(a) Effect of sodium bicarbonate on photosynthesis:
When sodium bicarbonate is added to water of the beaker more evolution of oxygen bubbles is seen. Sodium bicarbonate is hydrolysed to form sodium carbonate and carbonic acid as below :
2NaHC03 + H2 O Na2 C03 + H2 C03
The carbonic acid is unstable and decomposes into water and carbon dioxide as below :
H2C03 H20 + C02
With an increase in the percentage of C02 rate of photosynthesis enhances and hence more 02 is evolved.
(b) Effect of boiling water on photosynthesis:
If boiling water is added in the beaker no evolution of oxygen bubbles is seen even though the green Hydrilla plant is getting light and water. Boiling the water depletes the C02 content in it and hence the plant does not get the C02 required for photosynthesis. Higher temperature of water retards the activity of protoplast.
(c) Effect of common salt (NaCI) on photosynthesis:
Addition of common salt to the water of the beaker results in no photosynthesis and hence no evolution of oxygen bubbles is seen. Common salt increases the osmotic pressure of water and there is
plasmolysis due to exosmosis.
Cell sap is gradually deprived of the water contents and they becomes flaccid. All the physiological activities are affected. The absence of water in the cells affects photosynthesis. Plasmolysis also affects the permeability of protoplasm and diffusion of C02 cannot take place.
(d) Effect of chloroform on photosynthesis:
When chloroform (or any other injurious substance) is added to the water there is no evolution of oxygen bubbles indicating that photosynthesis is not taking place. Chloroform is an injurious chemical which gradually kills the living protoplasm of the cells and hence no vital activity is possible.
(e) Effect of KOH solution on photosynthesis:
If KOH is added in the water no evolution of oxygen gas is seen. The reason being that when KOH is added, it absorbs C02 gas dissolved in water. Therefore water becomes free of C02. In the absence of C02 the green plants cannot undertake photosynthesis. In the absence of photosynthetic activity, no evolution of oxygen is possible.
(f) Effect of keeping terrestrial twig in the beaker:
If a terrestrial twig (e.g. Tecoma) is placed in a beaker, no evolution of oxygen gas is seen because of lack of photosynthesis. The terrestrial twig experiences entirely different atmosphere within the water. Stomata remain closed and no entrance of C02 is possible. This results in the failure of photosynthetic activity, and no evolution of oxygen is seen.
(g) If the apparatus is placed in red and blue light separately:
If the apparatus (Fig. 14-4) is placed in red light there is more vigorous evolution of oxygen bubbles. Red light is most effective in the process of photosynthesis because a large amount of energy is absorbed from it by the chlorophyll. Thus, the rate of photosynthesis is increased in the red light and hence more rapid evolution of oxygen.
Apparatus placed in blue light shows less evolution of oxygen bubbles as compared to the above experiment. Clearly the rate of photosynthesis is more in the red wavelength as compared to the blue wavelength. In the blue wavelength plant can absorb less amount of energy from it by chlorophyll and thus less evolution of oxygen bubbles.
(h) Effect of covering the beaker by a black paper:
If the beaker is covered with a black paper, no evolution of 02 bubbles occurs. Plant is precluded from getting light and hence no photolysis of water occurs. Hence no evolution of oxygen gas is seen.
Leaves used in photosynthetic experiments are depleted of starch or are de-starched simply by keeping the plant in the dark over-night. All the starch present in the leaf is converted into sugar and taken away from the leaf in the dark. Hence it becomes starch free. Leaf or the pot, having a plant containing de-starched leaves is taken out of dark just before the commencement of the experiment.
All the subsequent tests that are to be described need the demonstration of starch in the leaf due to photosynthesis. Starch is stained blue with I2. If iodine solution is put on a leaf as such nothing will
happen since starch is present within the living cells and masked by the chlorophyll. To perform “Iodine test” or “Starch-test” in leaf-to-be-tested is killed and its chlorophyll is extracted. This is done by boiling the leaf in alcohol.
The leaf-to-be-tested is killed by putting it in a beaker containing boiling water. When water turns yellow, it indicates that leaf has been killed. This leaf is than transferred to alcohol (50-70%) heated on water bath (50°- 60°C). Alcohol extracts chlorophyll from leaf and turns green. Then leaf is put in a Petri-dish having iodine solution. If leaf contains starch, it turns blue, otherwise it is not affected.
The solution of iodine is prepared as below :
= lg = 2g = 30 c
= lg = 2g = 30 c
First dissolve the potassium iodide in water and then add iodine. This iodine solution is used in starch test which is also called iodine test.
To demonstrate that light is essential for photosynthesis.
Take a wide-leaved potted plant (Colocasia; Tropaeolum) and keep in the dark for about 24-48 hours to deplete the leaves from starch. A leaf which is still attached to the plant is inserted between the lid and the open end of the box.
The spring-like handle keeps the lid tightly fixed. A detatched leaf is also used but its petiole is allowed to dip in water. The whole apparatus is exposed to light and then chlorophyll is extracted and subsequently tested for starch.
Only the portion of the leaf exposed to light when tested with iodine becomes blue. The pattern cut in the lid is transmitted to the leaf.
The starch is formed in the green plants through photosynthesis for which C02, water, light and chlorophyll are essential. As exposed parts of the leaf get all the basic materials for starch formation, the latter is formed and accumulated in it while the part of the plant which is shaded is not able to undergo photosynthesis because of lack of light.
To demonstrate the measurement of photosynthesis by dry weight method :
Take a plant which has been kept in the dark overnight and select a symmetrical leaf early in the morning and from one half of this leaf cut off a few discs with Ganong’s leaf punch. Heat the leaf discs in an oven at 110°C till the weight is constant and note that weight. Now place the plant in the sun. In the evening, remove the same number of discs of leaves from the remaining half.
Treat them similarly and determine the dry weight. An increase in dry weight in case of the discs obtained in the evening will be observed. This is due to the accumulation of carbohydrates (starch) formed during this interval. An increase in the dry weight of the plant is due to an accumulation of carbohydrates formed during the process of photosynthesis.
10. Experiment on Respiration (3 Experiments):
To demonstrate the process of ‘anaerobic’ respiration.
A dish half-filled with mercury is taken. Fill a hard glass test tube with mercury and invert it over a dish full of mercury (Fig. 18-7). Introduce a few (6-8) seeds soaked in water whose test have been removed into the test tube, with the help of forceps from below. The seeds will float up to the top of the test tube due to lesser density than that of mercury. Hold the tube vertical on the mercury of the dish by a stand.
After a few hours of setting up of the experiment, the mercury level falls in the upper end of the inverted test tube due to accumulation of carbon dioxide gas liberated from the respiring seeds. Now introduce a few pieces of caustic potash in the tube through the open end.
These will absorb this gas and hence the mercury column will again rise. Therefore, this gas is C02 since KOH can only absorb C02. Due to the vacuum created by its absorption, the mercury from the dish has again risen.
It is a fact that in all the higher plants respiration goes on for a while even in the absence of oxygen and C02 is liberated. This is known as anaerobic respiration. In the present experiment the soaked (germinating) seeds are cut off from all the external supply of oxygen, i.e., they are kept in a vacuum, respire anaerobically and produce C02 which can be represented by the following equation:
C6 H,2 06 2 C2 H5 OH + 2 kcals
Because free oxygen is not available in this process, carbohydrates undergo incomplete oxidation (partial oxidation) and this produces ethyl alcohol. The production of C02 and energy in anaerobic respiration is due to the shift in the molecular arrangements in the substances.
Therefore, it is also intermolecular respiration. Thus in the present experiment the C02 produced by anaerobic respiration of the soaked seeds, the mercury level falls down. As soon as KOH is introduced, the C02 is absorbed and mercury level again rises.
Even in the absence of free oxygen respiration can go on using intermolecular oxygen present in the material itself and liberating C02 (anaerobic respiration).
Seeds should be soaked (germinating) and their testa should be removed carefully before introducing into the tube. Test tube filled with mercury should be inverted carefully not to leave any space at the top of the tube.
To study the R. Q. of different respiratory substrates by Ganong’s respirometer.
Ganong’s respirometer consists of a graduated glass tube and a levelling tube. The two tubes are connected at their basal ends with a rubber tubing. The graduated tube has a bulb at the upper end. The open end of the bulb has a glass stopper fitted into its neck.
The stopper and the neck of the bulb are provided with holes which can be brought in line with one another to facilitate communication with external air or if desired, to cut off the air inside the apparatus from that of the outer, thereby making the apparatus a closed chamber. The apparatus is perfectly fixed with mercury, since C02 cannot get dissolved in it.
A saturated solution of common salt may also be taken instead of mercury since C02 is only slightly soluble in it. The volume of the apparatus is 100 c.c. but due to the projection at the base of the bulb, it is actually slightly more i.e., 102 c.c. or so.
The plant material, the R. Q. of which is to be determined, is kept inside the bulb. The level of saline solution is brought to 100 c.c. mark. Initially the holes of the stopper and the neck of bulb are kept in line. As the experiment is started the stopper is turned and the holes of the neck are closed.
(a) When the respiratory material is carbohydrate:
Ganong’s respirometer is partly filled with saline water. Some germinating pea seeds are placed in bulb. The stopper of the bulb is then twisted so that the two holes come opposite to each other and outside air can enter the bulb.
The levelling tube is then so adjusted that the levels of saline water in the graduated and levelling tube are paralleled. The stopper is twisted, two holes are separated and bulb is closed. Respiration proceeds in this closed chamber. The level of the saline water in the graduated tube is noted.
The second respirometer is also fixed in a similar way but its bulb is kept empty. This acts as a control.
The level of saline water in the graduated tube remains stand still. When KOH solution is added into the saline water, the level in the graduated tube increases. The increase is noted after equalling the levels of graduated and levelling tube by adjusting the latter. In the control the level remains always the same.
The pea seeds are starchy seeds and thus they have carbohydrates as their respiratory substrates. The respiration with carbohydrates take place according to the following equation :
C6 H12 06 + 6 02 + 12 H2 O 6 C02 + 12 H2 O + 686 kcals
The equation shows that in such respiration the volume of oxygen absorbed is equal to the volume of C02 liberated. Hence there can be no appreciable change in the volume of the air in the respirometer bulb. That is why the level of saline water remains standstill.
As the KOH solution is added, the C02 is absorbed by it and thus there is rise of the level. By deducting the second level from the first one, the volume of CO? liberated can be calculated. As it is clear from the above discussion, the volume of C02 absorbed is equal to the volume of C02 liberated, the volume of oxygen is also obtained.
As the R.Q. is the ratio of the volume of C02 given out to the volume of 02 absorbed in respiration, it can be calculated from the results thus obtained.
R.Q. = Volume of C02 evolved/Volume of C02 absorbed
In the present case both the volumes are equal and, therefore, R.Q. = 1.
(b) When the respiratory substrates are fats (castor oil seeds):
There is an initial rise in saline water level of the graduated tube and it is noted. Then the KOH solution is added. There is further rise in level. This is again noted after levelling the tube. Suppose the first rise is designated by V,, it is excess of 02 consumed. The further rise of volume after the addition of KOH is denoted by V2. Therefore the V2 represents, volume of C02 liberated, and the total volume of02usedis V, + V2.
R.Q. = V2/V, + V2 = C02/02
The castor oil seeds are fatty seeds, and fats are the respiratory substrates. The fats are poorer in oxygen content and require more of oxygen in their oxidation. Therefore, volume of C02 produced is less than volume of 02 absorbed. This is why there is an initial increase in the level of saline water in the graduated tube. This is clear from the following example:
2C5i H98 06 + 145 02 ->102 C02 + 98 H20
As the KOH solution is added C02 is absorbed and, therefore, there is further rise in the level. As the volumes of 02 and C02 are known, the value of R.Q. can be calculated. It is always less than one in case of fats. It is clear from the above cited example:
R.Q. = C02 / 02 = 102 / 145 = 0.7
(c) When the respiratory substrate is an organic acid (e.g., Opuntia stem):
There is an initial decrease in the level of saline water in the graduated tube. When KOH solution is added, there is an increase in the level of saline water in the graduated tube. Let the fall of the level of the saline solution be designated by V! which represents excess of C02 liberated than 02 used. After adding KOH the volume goes up to V2 which represents the total C02 liberated, therefore,
R.Q. =C02/02 = V2/V2-Vl
In case of succulents, in light, the organic acids are the respiratory substrates. As the organic acids are richer in oxygen content than carbohydrates, they require less of 02 for their oxidation, hence the volume of C02 liberated exceeds the volume of 02 absorbed. This can be visualised by the following example:
2(COOH)2 + 02 4 C02 + 2 H20
As, less of oxygen is absorbed and more of C02 is liberated there is an initial decrease in the level of saline water in the graduated tube. When C02 is absorbed by KOH solution, there is a iise in the level. Let the decrease in the level from the initial be equal to Vi and increase in the level from V! be equal to V2. The volume of C02 liberated will be V2 and volume of 02 absorbed will be V2 – Vj. Thus, knowing the volume of C02 liberated and volume of 02 absorbed, R.Q. can be calculated. In case of organic acids R.Q. is always more than one. This can be exhibited by the above example.
R.Q. = C02 / 02 = 4 /1 = 4
To demonstrate the production of C02 during respiration.
A green potted plant is placed over a glass sheet and under a bell jar which is covered on the outside by a black cloth to avoid light and to prevent photosynthesis. With the help of tubes, the bell
jar is connected on both the sides with the bottles filled with lime water.
One bottle is connected to a bulb or a tube filled with soda lime which is in contact with the outer air. The other one is connected to an aspirator. All the joints are made airtight by applying grease around them and the aspirator is worked to draw current of air into the apparatus through the open end of the tube.
It will be observed that lime water of the bottle which is connected with the soda lime bulb does not turn milky while that of the other bottle turns milky.
Plants respire by taking in oxygen and giving off C02. When the aspirator is put in action, air is drawn through the soda lime bottle and then through the lime water in the bottle. When the air passes over the soda lime its C02 is absorbed, with the result the air passing through the first bottle does not turn its lime water milky.
As the air passes through the bell jar its 02 is used by the plant during respiration which takes place according to the following equation:
C6 H,2 Ofi + 6 02 + 6 H2 O 6 C02 + 12 H2 O + 686 Kcals
The above equation shows that C02 is evolved during respiration. When the air passes through the second bottle, the lime water turns milky due to the following reaction which takes place between C02 and lime water.
Ca(0H)2 + C02 CaC03 + H20
The calcium carbonate, which is white in colour, is precipitated and thus lime water appears to be milky. It is concluded that C02 is evolved during respiration.
All the joints should be perfectly air light. Bell jar should be covered with black cloth.
11. Experiment on Growth and Development (3 Experiments):
To demonstrate the regions of growth in a root.
Take germinating seeds of pea or a bean, with straight radicles having a length of about 2 cm. With the help of fine nib and India ink mark out the apical 1.5 cm of radicle. The length may bedivided into 10-15 points. The points may be separated with an interval of 1 mm each.
Take a beaker half full of water and suspend the seedlings through the help of a threat so that the radicles point towards the lower side. Cover the beaker with a black paper or cloth to prevent the exposure of roots to light. After three to four days remove the seedlings gently and measure the distances between the adjacent marks.
It will be observed that the first two to three marks are closely situated while the rest (i.e., 27-8) of the points separate widely. The distance between the rest of the marks will decrease from tip of the radicle. The increase in the distance between the consecutive marks shows the level of growth.
Apical meristem is present towards the tip and this region has less growth. Further maximum growth takes place in the region of cell elongation of the newly formed cells. This is the region or zone of cell elongation. This zone is followed by a region of cell maturation or differentiation. Least growth takes place in this zone.
To measure the rate of growth of a potted plant.
Take an auxanometer and a seedling grown in a pot. Tie a thread to the growing tip of stem and pass it over the pulley which is attached to a long needle. This needle can move over the arch which is graduated when there is movement in the pulley. The free end of the thread is tied to a small weight to keep the thread in a stretched position.
As the stem tip elongates or grows, the weight will start moving downward and accordingly the thread moves the pulley with its needle on the arch. It is possible to observe the measurements of growth on the graduated arch. It may be added that a pulley with 4″ diameter and 20″ long needle will amplify the growth nearly ten times.
To demonstrate the influence of light on growth.
Take a germination jar and put some broad bean seeds in it after soaking and leave them in the dark. Place a similar jar with seeds in the light. Observe differences in the two sets of seedlings after 2 to 3 weeks. Dark grown seedlings grow faster than those in the light but are weak, having long internodes, small undeveloped leaves which are ‘etiolated’. Seedlings are tall but sickly looking and do not survive long.
12. Experiment on Plant Movements (3 Experiments):
To demonstrate the process of phototropism.
Process of phototropism can be demonstrated by several experiments. These are described below :
1. Sow a few seeds of mustard or balsam in saw dust or vermicule in small Petri dishes and moisten it. Keep the petri dishes in the room near a window. As soon as the seedlings emerge their aerial parts will bend towards the light coming from the window.
2. Take a potted plant of balsam or mustard and place it near a window. Notice after a few days that the stem tip bends towards the light. Also observe that leaves lie at right angle to light.
3. Take a phototropic chamber made of wood and having hole towards one side. Light can pass through this hole. The box is painted black from inside. Place a small pot having young seedling or a Petri dish with some seedlings and place them in the box.
The box is illuminated from one side. Observe bending of the stem towards the hole after a few days. Leaves are bent at right angle to light. If young mustard or rapeseed seedlings are placed in a beaker containing water, after a couple of days it is observed that shoot tip bends towards the light while the roots are negatively phototropic.
To demonstrate geotropism through Clinostat.
Clinostat is made up of a disc attached to the tip of a horizontal rod which can be rotated with the help of a clockwork. Fix a small potted plant on the disc of the clinostat firmly in a horizontal position. Begin the clockwork with 2-6 revolutions per hour. Place the second potted plant horizontally on the table. Observe after two to three days that the pot placed on the table exhibits curvature in the stem and roots.
Notice that stem bends upward while the root bends downward under the force of gravity. In the plant fixed on the working clinostat horizontally there is no bending made out since the force of gravity acted on all the sides of the stem and the root. Clearly roots and stem respond to the geotropism when the gravity is applied unilaterally. Further the stem is negatively geotropic and the root is positively geotropic.
In order to find out the precise regions of graviperception in the stem and roots take young actively growing seedling of maize or gram. Mark out 1 cm distance from the root tip into 10 equal parts. A similar procedure may be adopted for the stem.
Now remove the root apex in one of the seedlings and shoot apex in another seedling. The third seedling may be kept as such. Place all the three seedlings horizontally in a way that they are able to bend both ways. After 2-3 days it will be
seen that both stem and root apices have curved upward. Now notice that stem is negatively geotropic while the root is positively geotropic. Notice maximal elongation in the region of apex. Where the tip of root is removed only stem apex will bend while in case of seedling with stem apex removed only root tip will bend upward.
In case of contact seedling both the parts will bend. Clearly the stimulus for gravity is perceived in the regions of the root and stem apex. The maximal perception was in the region of elongation.
To demonstrate the process of hydrotropism.
Take a small clay funnel with several holes in it and cover it outwardly with filter paper. Put this on top of a wide mouthed bottle in a slightly raised position. Fill the bottle with water. Put some seedlings in the clay funnel. As soon as radicles emerge out they bend towards water in the bottle. They are thus positively hydrotropic.
This experiment can also be done with a dish having perforated base. Put sawdust in the dish and soak the seeds with water. Keep the dish in a tilted position. Notice the emergence of roots after a few days and these move inside the perforated dish. They do so towards the water. To begin with the young roots are positively geotropic and later they becomes positively hydrotropic.
13. Experiment on Sexual reproduction in Higher Plants (3 Experiments):
To demonstrate germination of pollen on stigma.
Take a freshly opened flower of Portulaca dusted with pollen and dissect out the stigma hair in a drop of aceto-carmine. Notice germinating pollen grains in the stigma hair.
To demonstrate pollen germination in virto.
K Take 3% sucrose solution on a slide and sprinkle few pollen grains of Amaryllis or Impatiens balsamina. Examine under the microscope. Observe growing pollen tubes.
To examine the parts of ovule.
Take a few ovules of any orchid and immerse in a drop of aceto-carmine. Notice nucleus, integuments, micropyle and the embryo sac inside. Orchid ovules have transparent integuments and are thus easy to mount and useful for the study of different parts.
14. Experiment on Physiology of Seeds (3 Experiments):
Effect of water soaking of seeds on germination.
Take 10 petridishes and line them with soaked bloating paper on the two dishes. Now obtain 50 seeds of pea and soak them in water for 5, 10 and 20 hours. Germinate 10 seeds of each treatment (i.e. after soaking for 5, 10 and 20 hour) and seeds germinated directly as controls.
Notice the number and amount of seedling growth after 1, 2, 3, 4, 5, 6———- 10 days of sowing. Record the effect of soaking for different periods in water and compare them with the unsoaked seeds.
Effect of soaking of seeds in salt solution of different molarity on germination.
Select pea seeds and soak them in different conc. of NaCl (0.05 M to 0.50 M) for 1 to 10 days. Count the number of seeds after each treatment and tabulate the results.
Demonstrate the effect of different chemicals on seed germination.
Take fresh seeds of pea or lettuce and germinate them in coumarin (0.5 ppm), actinomycin-D (5 ppm), chloramphenicol (50 ppm), 2, 4-dinitrophenol (13 ppm) and GA (5 ppm). Keep the soaked seeds in these above chemicals for 7-8 days. Record percentage germination and inhibition. Notice the effect on root number, length and internode and emergence of radicles.
15. Experiment on Transport through Phloem (3 Experiments):
To demonstrate the translocation of carbohydrates from the leaves.
Take a potted plant and place it in the sun. Remove some leaves after some time and test for starch. When immersed in iodine solution, the leaves will turn blue in colour. Now allow the plant to remain in the dark overnight.
Again remove some leaves and test for starch. Very little or no starch is seen. This shows that very little starch is left in the leaf or most of the starch photosynthesized in the leaf has been translocated to other parts of the plant.
To demonstrate that phloem is the channel of translocation of organic material.
Take a developing cucurbita fruit and insert an hypodermic needle gently into its stalk. Near the surface an exudation of organic material rich sap will be visible. Cut a thin section and notice that organic food material is coming out of the phloem. Exudation also suggests that organic solutes are exudating in the phloem through mass flow.
To demonstrate path of translocation of organic material.
Take a branch of citrus or some other tree where fruits and leaves are present. Select three fruits as shown in Fig. 26- 1. Mark them as 1, 2 and 3. Remove a ring of bark as shown. Notice the differences in the sizes of fruits after several days. It will be inferred that phloem is the channel of the translocation of organic material.