Activity 11Transport of Water and Other Materials In and Out of the Plants

One of the basic necessities of life is the ability to transport materials that are needed by the cells to perform their function. Plants accomplish this task through their vascular tissues namely the xylem for water transport, and phloem for sugar and other mineral transport. There are two types of transport at cellular level: passive transport, the movement along a concentration gradient and active transport, the movement of particles across a concentration gradient which requires ATP expenditure. Equilibrium must be maintained, thus the cell membranes play their role of regulating the transport processes undertaken by the cell. Objectives: To determine the mechanisms undertaken by plant to transport the materials that it needs To demonstrate the processes by which materials are transported and transpired To identify the effects of different factors in the rate of diffusion To demonstrate how different factors affect the integrity of membranes To demonstrate how plants transport water through the stems To identify different factors that affect the rate of transpiration in plants To differentiate guttation from transpiration I. Diffusion of Selected Plant PigmentsA. IntroductionDiffusion is a type of passive transport; it is the random movement of particles in a solution from regions with higher concentration to regions with lower concentration. Diffusion is very important to plants. Carbon dioxide required for photosynthesis enters the plant through diffusion in the in intercellular spaces, oxygen is liberated from plants through diffusion, and roots absorb water and minerals through diffusion.B. Methodology1. Put 1 g of atsuete seeds each into four test tubes.2. Do the ff. in each test tube: Test tube 1 put 10 ml of distilled water Test tube 2 put 10 ml distilled water and place in boiling water bath Test tube 3 put 10 ml vegetable oil Test tube 4 put 10 ml heated vegetable oil3. After 30 minutes, shake the test tubes and compare the color intensities with +, ++, +++.C. ResultsSubstanceColor Intensity

Test tube 1+

Test tube 2+++

Test tube 3++

Test tube 4++++

D. DiscussionQuestions:1. What is the effect of concentration on the rate of diffusion? When concentration is higher, the rate of diffusion is faster. The greater the difference in concentration between two areas, the greater the rate of diffusion.2. What is the effect of size of diffusing particles on the rate of diffusion? When particles are larger, there is a slower rate of diffusion. Consequently, smaller particles diffuse faster.3. Which plant pigment traveled the farthest? Explain your answer. The pigment in test tube 4 traveled the farthest. When vegetable oil was heated, temperature increases the rate of molecular movement (kinetic energy), making them travel farther and faster.II. OsmosisA. IntroductionOsmosis is the diffusion of water across a selectively permeable membrane driven by a difference in concentration on the two sides of the membrane.B. Methodology1. Strip off a thin section of the lower epidermis of the bangka-bankaan2. Prepare a wet mount section. 3. Examine the cell, which is turgid, under LPO.4. Without removing the slide, draw off the water by using a small piece of tissue.5. Replace the water with 5% salt solution.6. Examine the plasmolyzed cell.C. Results

D. Discussion1. What is the direction of diffusion of water in a turgid cell? In a plasmolyzed cell? When cells(as in the experiment) are placed in water, water diffuses into the cell causing the protoplast to swell (turgor pressure) and presses against the cell wall which also presses with an equal force. When turgor pressure and wall pressure of the cell are equal, the cell is in a turgid state. The cell and the solution are now in equilibrium, no net water movement occurs( water molecules do move but equally in both direction. When the cell is placed in salt solution, water diffuses out of the cell. As it continues to lose water, the protoplast shrinks and no longer presses the wall. The cell has been plasmolyzed.2. Is the cell wall still intact in a plasmolyzed cell? The cell wall is still intact, only the protoplasm shrinks.III. Factors affecting integrity of cell membraneA. IntroductionCell membranes are important to proper cell functioning. It consists of a lipid bilayer having a dula nature in that they are both hydrophilic (water loving) and hydrophobic (water-fearing). The inner and outer surfaces are hydrophilic and interact with aqueous solutions while the core is hydrophobic which does not permit entry of water and polar molecules.Because of this dual nature the cell membrane serves many functions : it acts a selectively permeable membrane that allows entry/exit of materials, and it houses proteins and carbohydrates necessary for cell to cell and extracellular communication.Cell membranes are also sensitive to environmental conditions like temperature , pH and organic solvents. Since cells work under narrow range of conditions, we can hypothesize that higher temperatures and extreme low temperatures will cause severe damage to cell membranes. The same goes for organic solvents and pH.B. Methodology1. Make 7 separate sections of peel of red apple.2. Place three sections in three test tubes (A, B,C) filled with 10 ml distilled water. Put tube A under room temperature, B inside a refrigerator, anc C in a water bath with temperature 60oC. 3. After 30 minutes, remove the sections and prepare a wet mount. Examine under microscope and compare color intensities. 4. Place the other sections in four slides (D, E,F,G). Add 50% choloroform to D, 50% acetone to E, 0.1M NaOH to F, and 0.1M HCl to G.5. Examine the cells immediately, after 15 minutes and after 30 minutes.C. ResultsTest tube Color intensityTest tubeColor intensity

A(room temp.)+++D(chloroform)++

B(refrigerator)++E(acetone)+

C(60o C)+F(NaOH)++

G(HCl)+

D. Discussion1. Where does the pigment come from? What do you call this pigment? These pigments are called anthocyanins. They are stored in vacuoles.2. In order to reach the outside of a cell, through what membranes must the pigment diffuse? The anthocyanin pigment must diffuse out of the tonoplast (or the vacuolar membrane) and the cell membrane itself.EFFECTS OF TEMPERATURE3. What temperature stressed and damaged the membranes the most? Explain. The highest temperature (60o C) stressed and damaged the membranes the most as shown by the great change in color of specimen after being heated. Specimen A has shown the greatest color intensity, its color close to the original, showing that in room temperature there is little to no disruption of membrane. B follows which shows that when subject to cold temperature, cell membrane are disrupted but not as much as when exposed to heat. When you apply heat, you apply energy which makes molecules vibrate rapidly. Lipids become more fluid, proteins untangle and the membrane breaks apart. The cell membrane is damaged and pigments will leak out of their compartments which explains the apparent loss of color.EFFECTS OF ORGANIC SOLVENT4. Which solvent stressed the membranes more? Explain. The color intensity of E which is treated with acetone is lighter than D which is treated with chloroform. This means that acetone is the stronger organic solvent and has done more damage to the cell membrane.EFFECTS OF pH Because cells are usually exposed in normal pH levels, exposure to very basic pH levels (from NaOH) and very acidic pH levels (from HCl) causes cell membrane to break and pigments to leak out.

IV. ImbibitionA. Introduction Imbibition is the disruption of water accompanied by swelling.

B. Methodology1. Individually weigh 2 pieces of wood and rubber, and 2 sets of 10g of corn seeds.2. In the first beaker, immerse one piece of wood, rubber, and 10g of corn seed in water. In the second beaker immerse one piece of wood, rubber, and 10g of corn seed in kerosene.3. After 90 minutes, take out the materials and get their final weight. C. ResultsMediumImbibitantInitial weightFinal Weight% of change in weight

WaterRubber0.5g0.5g0%

Wood0.8g1g25%

Seeds10g10.35g3.5%

KeroseneRubber0.5g0.8g60%

Wood0.8g1.25g56.25%

seeds10g10.75g7.5%

D. Discussion1. Which of the materials tested serve as good imbibitant of water? Kerosene? Both wood and seeds imbibed water but wood is the better imbibitant. All materials absorbed kerosene but wood and rubber are better imbibitants of water.2. Describe the nature of the chemical constituents of water? Kerosene? Water (H2O) is a polar substance (with partial positive and partial negative charge) while kerosene is anon-polar.3. Are living cells necessary in order for imbibitions to take place? No. wood and rubber imbibed kerosene.4. Are there any living cells in the veneer? Yes.5. In the seeds, are living cells involved in imbibitions? Are dead cells involved? The seed is composed of living cells which undergo imbibitions. Dead cells are also involved; the see coat/testa is composed of dead cells which imbibe water and break apart.6. In what way/s is the swelling effect of imibition important to seed germination? Imbibition is the initial step in seed germination. It causes the seed to swell and results in breaking of the testa.V. Movement of water through the stemA. IntroductionAs water is absorbed by roots, it moves through the stem through the xylem. The xylem transports water to the whole of the plant body. This experiment will trace the movement of water through the xylem in the plant.B. Methodology1. Prepare pechay leaves with petiole. Cut off 1 cm from the base of the petiole.2. Immerse the petioles in a beaker containing 10 ml of 1% eosin dye solution.3. After 10-15 minutes, remove the leaf and split the stalk longitudinally to measure the length covered by the dye. 4. From another leaf, obtain a cross section of the petiole and examine under the microscope. C. Results

D. DiscussionAfter 15 minutes, the stem all the way to the veins of the pechay leaves have red stain indicating that the eosin dye solution where the stem is immersed have been absorbed and transported by the xylem( as seen in x-section, vascular bundles have bright red stain).

VI. Comparison of cuticular and stomatal transpiration by four leaves methodA. IntroductionTranspiration is the evaporation of water through the stomata of the plant. The stomatal frequency and size is directly proportional with rate of transpiration the more stomata and the bigger their size, the faster the rate of transpiration. Also, the stomata in the abaxial surface contain more stomata which helps in conservation of water. Stomata in the adaxial surface are considerably fewer. By applying petroleum jelly on the different surfaces of the four identical leaves, we can test the resulting dryness of the leaves.B. Methodology1. Prepare four identical leaves and label them A, B, C, D.2. Leaf A is the control. On leaf B, apply petroleum jelly on the upper surface. On leaf C, apply petroleum jelly on the lower surface. On leaf D, apply petroleum jelly on both sides.3. Hang the leaves on a thread where both surfaces are exposed to air. Keep the set-up until the next lab period.C. ResultsAfter a few days upon exposing the four identical leaves in the air, specimen A is the driest, followed by B, C, and D is the least dry.D. DiscussionSpecimen A became the driest because it had no petroleum jelly to cover its stomata. Rate of transpiration remained the same. Leaf B is less dry than A because petroleum jelly covered some stomata on the adaxial side but not the abaxial side which contained more stomata. This decreased the rate of transpiration. Leaf C is less dry than B because petroleum jelly is applied in the abaxial side which covered more stoata, thus reducing even more the rate of transpiration. Leaf D is the least dry because the petroleum jelly covered both abaxial and adaxial sides, reducing the transpiration the most.VII. GuttationA. IntroductionGuttattion is the exudation of water drops through the hydathodes. At night, stomata are closed and there is a high soil moisture. Water will enter the plant roots which will create a slight root pressure. This pressure forces some water to exude through the hydathodes in the leaf margin.B. Methodology1. Plant 5 viable grains in a flower pot with holes at the bottom.2. Immerse the lower portion of the tin can in a shallow dish. Add water from time to time.3. When the seedling are 2-5 cm long, cover the flower pot including the dish with a transparent bell jar. Take note of the droplets formed on the leaf surfaces.

C. Results Guttation is observed in the leaf margin.D. Discussion1. Explain how guttation was induced in the rice seedlings. Adding water from time to time creates a high degree of moisture wherein water enters the roots of the rice seedlings. This will create a slight root pressure. Root pressure forces the water to exude in the leaf margin.2. Differentiate guttation from transpiration. In both guttation and transpiration, plant loses water. Guttation is induced by the slight root pressure because of high soil moisture while transpiration is caused by heat in the plants environment. In guttation, water escapes through the hydathodes and tends to accumulate in droplets in the leaf margin while in transpiration, water escapes through the stomata and evaporates.

References:Botany by Mausethhttp://plantcellbiology.masters.grkraj.org/html/Plant_Cellular_Physiology1-Water_And_Water_Dynamics.htmhttp://answers.yahoo.com/question/index?qid=20061129041445AAaMGDThttp://www.tutorvista.com/content/biology/biology-iv/plant-water-relations/imbibition.php#http://en.wikipedia.org/wiki/Guttationhttp://lariceman.wordpress.com/tag/guttation/

Activity 12Introduction:Photosynthesis is a complex process by which carbon dioxide is combined with water to form carbohydrate. It converts light energy into chemical energy. And during the process, liberation of oxygen occurs. Chlorophyll, an essential pigment, allows photosynthesis to take place.

Objectives:At the end of the exercise, the student should be able to:1. To know the importance of chlorophyll, light, and carbon dioxide in photosynthesis2. To observe the release of oxygen in photosynthesis3. To perform separation of chloroplast pigments; to identify these pigments

I. Hypotheses:A. If the leaf is positive for starch test (IKI solution turns blue-black) then it is photosynthetic.

B. If the part of the leaf which was pre-covered with black paper is not positive to starch test then that particular part has not been photosynthetic.

C. If the color intensities differ then the CO2 concentration in each of the test tubes set up also differ.

D. If the lighted match continues to burn when placed in the test tube then oxygen is present.

E. If the lengths travelled by the pigments differ then they have different rates of solubility.II. Methodology:A. Boil variegated leaf in water. Immerse the leaf in a test tube with 95% ethyl alcohol. Place the test tube in hot water until pigments are extracted. Then wash the leaf with water and using IKI solution, test for the presence of starch.

B. Place a potted plant in the dark for 48 hours. Then select a few leaves and wrap portions of the leaves with black carbon paper. Expose the whole plant to light for 5 hours. Detach the covered leaves and test for the presence of starch.

C. Fill test tubes A, B, & C with previously boiled then cooled water. Place Hydrilla sprigs in the tubes A & B then add a pinch of NaHCO3 in B & C. Expose A, B, & C to bright light. Observe if bubble formation occurs. Remove Hydrilla from A & B. Add five drops of phenol red to each test tube. Then compare color intensities.

D. Fill a beaker with of a liter of water. Using plastic straw, blow air gently into the beaker. Cut the ends of Hydrilla under water and insert thistle funnel over the sprigs (cut ends should face the tube of the funnel). Place a test tube filled with water upside down over the tube of the funnel partially dipped in the water of the beaker. Observe for bubbles emerging out from the cut ends. Determine the nature of the gas by placing a lighted match in the test tube.

E. Prepare the leaf extract of mature Hibiscus leaves. Prepare the filter paper strip. Prepare the chromatogram. Place the paper strip on the hook and insert into the test tube containing solvent mixture of 95 parts petroleum ether and 5 parts acetone (solvent level must be below the 1cm mark). Replace cork with the strip and observe flow of solvent until the 2cm mark. Remove the paper from the tube and let it dry.

III. Results:A. Upon dropping IKI solution onto the bleached leaf, blue-black color developed. This denotes the presence of starch which is actually an indicator that the leaf is photosynthetic.

B. Covered parts of the leaves were negative for starch test while the rest of the parts were positive.

C. Color intensities:Test tube A (with Hydrilla but no sodium bicarbonate) - remained red, no bubbles formedTest tube B (with Hydrilla and sodium bicarbonate) - became yellow but returned to red, bubbles were formedTest tube C (NaHCO3 only) turned from red to yellow.

D. Bubbles emerged from the cut end of the Hydrilla. Then when the lighted match was placed near the mouth of the test tube (immediately after it was removed from the water) the flame on the match kept on burning.

E. After the experiment, four pigments were visible: yellow orange (carotene), light yellow (xantophyll), blue green (chlorophyll a) and olive green (chlorophyll b). The chromatogram also showed that carotene travelled the farthest, followed by xantophyll, chlorophyll a and finally chlorophyll b.

IV. Discussion:A. Starch is a product synthesized from chlorophyll and is therefore found in green (photosynthetic) parts of the leaf. Boiling the leaf prior to chlorophyll extraction removes the cuticle (which may prevent the entry of IKI solution), damages cell membranes to make starch (in the cytoplasm & chloroplast) accessible to IKI solution, and arrests all chemical reactions. Boiling aleafinalcoholremoves itschlorophyll, so theleafloses its green color. A leaf positive to starch test indicates that the leaf is photosynthetic obviously because sugar is a product of photosynthesis which is stored as starch in almost all parts of the plant.

B. Light is the key to jumpstart photosynthesis. It provides the energy needed to perform light reactions. Sample A (leaf grown under normal conditions) contained more starch than sample B (leaf with covered portions). First is because Sample B was kept in the dark for 48 hours to make it consume its reserved starch. Second reason is that covered portions were not permitted to perform photosynthesis due to the absence of light. Unlike the exposed parts of the leaf, covered ones were really expected to be more or less negative to starch test.

C. NaHCO3 (Sodium bicarbonate) provided a good source of CO2 for the Hydrilla in test tube B maximizing its photosynthetic ability. This lowered the pH of the water and caused phenol red to turn into yellow (since Phenol red is an indicator of p H). Indirectly, it is an indicator of CO2 concentration because CO2 in water forms carbonic acid which lowers thepHof water. However, the Hydrilla used up the CO2 in the solution and gave off oxygen gas (indicated by the bubbles) which caused the solution to become neutral again (red). Set up A also actually exhibited the use of CO2 in photosynthesis, but since there was no source present (NaHCO3) the reaction was limited. In set-up C (Na2HCO3 only), the solution became acidic and it turned yellow because of the presence of sodium bicarbonate, which dissociated to form CO2 and ultimately, carbonic acid without any plant to use the CO2 up and produce oxygen. Boiling dissolved the gases present in the water.

D. Oxygen is a by-product of photosynthesis which ultimately, comes from the water absorbed by the plant. When the test tube was lifted and a lighted match was placed near its mouth, the match continued to burn. This just indicates the presence of oxygen (because combustion needs oxygen). And since the life the flame on the lighted match was sustained, it showed that oxygen is present in the gas (or bubbles) produced by the Hydrilla sprig.

E. Chloroplast pigments are known as chlorophylls (green pigments) and carotenoids (yellow-orange-red pigments) . Different pigments absorb different light wavelengths. Solubility of the pigments varies according to the rate they are carried by the solvent. The colors visible in the chromatogram are dark green (chlorophyll a), light green (chlorophyll b), and yellow (xantophyll). Chlorophyll b is the least soluble and xantophyll is the most soluble. The solvent (petroleum ether with 5 parts acetone) carries the pigments as it moves up the paper. The pigments are carried along at different ratesdue to their differences in solubility, the less soluble pigments will move slower up the paper than the more soluble pigments.Each pigment has an Rf value, the speed at which it moves over the paper compared with the speed of the solvent.Rf = Distance moved by the pigment Distance moved by the solventDistance of entire solvent = 6.7cmRf value of chlorophyll a= (6cm/6.7cm) = 0.89 Rf value of chlorophyll b = (5cm/6.7cm) = 0.75 Rf value of xantophyll = (6.7cm/6.7cm) = 1References:http://www.nuffieldfoundation.org/practical-biology/testing-leaves-starch-techniquehttp://www.brilliantbiologystudent.com/testing_a_leaf_for_starch.htmlhttp://www.markedbyteachers.com/gcse/science/testing-starch-in-a-variegated-leaf-lab-report.htmlhttp://breakthelight.wordpress.com/tag/test-tube/An Introduction to Plant Biology (Fourth Edition) by James Mauseth