INTERNATIONAL TURKISH HOPE SCHOOL ACADEMIC YEAR CHITTAGONG SENIOR SECTION BIOLOGY HANDOUT OSMOSIS, DIFFUSION AND ACTIVE TRANSPORT CLASS 9

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1 INTERNATIONAL TURKISH HOPE SCHOOL ACADEMIC YEAR CHITTAGONG SENIOR SECTION BIOLOGY HANDOUT OSMOSIS, DIFFUSION AND ACTIVE TRANSPORT CLASS 9 Name :... Date:... d) Movement of substances into and out of cells Students will be assessed on their ability to: 2.12 understand definitions of diffusion, osmosis and active transport 2.13 understand that movement of substances into and out of cells can be by diffusion, osmosis and active transport 2.14 understand the importance in plants of turgid cells as a means of support 2.15 understand the factors that affect the rate of movement of substances into and out of cells, to include the effects of surface area to volume ratio, temperature and concentration gradient 2.16 describe experiments to investigate diffusion and osmosis using living and non-living systems. Human Biology Students will be assessed on their ability to: a) Recall simple definitions of diffusion, osmosis and active transport. b) Understand that movement of substances into and out of cells can be by diffusion, osmosis and active transport. c) Understand the factors that affect the rate of movement of substances into and out of cells to include the effects of surface area to volume ratio, temperature and concentration gradient. d) Describe how to carry out simple experiments on diffusion and osmosis using living and non-living systems. Cambridge Specification 2.1 Diffusion 2.2 Osmosis 2.3 Active transport Candidates should be able to: (a) define diffusion as the movement of molecules from a region of their higher concentration to a region of their lower concentration, down a concentration gradient; (b) define osmosis as the passage of water molecules from a region of higher water potential to a region of lower water potential, through a partially permeable membrane; (c) describe the importance of a water potential gradient in the uptake of water by plants and the effects of osmosis on plant and animal tissues; (d) define active transport as the movement of ions into or out of a cell through the cell membrane, from a region of their lower concentration to a region of their higher concentration against a concentration gradient, using energy released during respiration; (e) discuss the importance of active transport as an energy-consuming process by which substances aretransported against a concentration gradient, as in ion uptake by root hairs and glucose uptake by cells in the villi. Teacher in Charge: ARIF ULLAH Page 1

2 The movement of water molecules from a region of higher water potential to a region of lower water potential, across a partially permeable membrane, along the gradient, is known as OSMOSIS. Eg: Absorption of water from soil by Root Hair Cells and Turgidity in Plant Cells. Movement of water into cells. The movement of particles from a region of Higher concentration to a region of Lower concentration, along the concentration gradient, is known as DIFFUSION. Eg: Dissolving of ink; Smell travelling across room. The movement of ions from a region of Lower concentration to a region of Higher concentration, against the concentration gradient, using ENERGY (ATP) [from Mitochondria by Respiration] is known as ACTIVE TRANSPORT Examples of active transport include the uptake of glucose in the intestines in humans and the uptake of mineral ions into root hair cells of plants. Concentration gradient: The difference between the water potential of two solutions which causes substances to move from higher to lower regions. Teacher in Charge: ARIF ULLAH Page 2

3 Comparison between osmosis and diffusion Similarities 1. They are both a passive process [do not require energy]. 2. They occur down a concentration gradient [higher to lower]. Differences Diffusion 1. It is the movement of particles. 2. It does not require a semi permeable membrane. Osmosis It is the movement of water molecules. Requires a semi permeable membrane Water Potential: The tendency of a solution to lose water from a higher water potential to a lower water potential region, i.e. Dilute solution to concentrated solution, or hypertonic to hypotonic. Solution: A mixture of a Solute (Sugar) and a Solvent (Water) Solutions are compared using the following terms. Hypotonic: A solution containing a higher water potential. [More water] Hypertonic: A solution containing a lower water potential. [Less water] Isotonic: Same solute and solvent. Wilting: The condition of a plant in the presence of less water in cells which cause the cells to lose turgidity and ultimately lose their shape. Turgidity: Plant cells have cell walls which prevent the cells from bursting when placed in a hypotonic solution. The water inside the cell exerts pressure on the cell wall and becomes swollen. This is known as turgor pressure. The process is known as turgidity. Partially permeable membrane: The membrane that allows only small molecules to enter and exit a cell. Large molecules are prevented to enter through the tiny pores, e.g. cell membrane. Plasmolysis: The shrinkage of the cytoplasm of a plant cell. Crenation: The shrinkage of the cytoplasm of an animal cell. Teacher in Charge: ARIF ULLAH Page 3

4 EFFECT OF OSMOSIS IN PLANT AND ANIMAL CELL. a) Plant cell in hypotonic solution (dilute) -water moves into the cytoplasm which has a lower water potential, from the outside (solution), which has a higher water potential, through the partially permeable membrane by osmosis. -The cell sap in the vacuole has a lower water potential. Therefore water moves into it from the cytoplasm. The vacuole swells and pushes the cytoplasm and the cell membrane. -The cell wall is strong and prevent the cell from bursting. -The cell wall is known as turgid. b) Plant cell in hypertonic solution: (concentrated solution) -water moves out of the cytoplasm, which has a higher water potential, from the outside solution, through the partially permeable membrane by osmosis. -the cell sap in the vacuole has a higher water potential. Therefore, water moves out of it into the cytoplasm. The vacuole and the cell membrane loses shape and shrinks in size. -the cell wall is strong and prevents the cell from collapsing. -the cell is said to be flaccid and plasmolysed. c) Animal cell in hypertonic solution: (dilute) -water moves into the cell by osmosis from outside. -the cell swells in size and bursts, as there is no cell wall to prevent it from bursting. -the cell is said to be plasmolysed. d) Animal cell in hypertonic solution (less water/concentrated): -water moves out of the cell by osmosis, from inside. -the cell loses shape and forms spikes, as there is no cell wall to keep it firm. -the cell is said to be crenated. e) Plant and Animal cell in isotonic solution: -a solution that is isotonic with respect to both cells, there is no net movement of water in or out of the cell. Both cells remain unaffected. Teacher in Charge: ARIF ULLAH Page 4

5 Effect of OSMOSIS on an Animal Cell (RBC) Effect of OSMOSIS on a Plant Cell Teacher in Charge: ARIF ULLAH Page 5

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7 Factors favoring diffusion Distance (the shorter the better), e.g. thin walls of alveoli and capillaries. Concentration gradient (the bigger the better). This can be maintained by removing the substance as it passes across the diffusion surface. (Think about oxygenated blood being carried away from the surface of alveoli). Size of the molecules (the smaller the better). Surface area for diffusion (the larger the better). Temperature (molecules have more kinetic energy at higher temperature). Teacher in Charge: ARIF ULLAH Page 7

8 EXPERIMENTS related to OSMOSIS and DIFFUSION 1. Selectively permeable membrane - only allows small sized molecules (water, glucose & amino acid) but not the large sized molecules (protein & starch) to pass through. e.g. cell membrane of all living cells, the internal wall of the gut & visking (dialysis) tubing. Dialysis - A physical process by which small sized molecules (water, glucose & maltose) are separated from large-sized molecules (starch & protein) by using a selectively permeable membrane. Food tests: Test for Starch-Iodine test Add few drops of Iodine solution to the food sample. If the colour changes to blue/black, starch is present. If the colour remains brown/yellow, starch is absent. Test for Glucose (reducing sugar)-benedict's test. Add equal amount of benedict's solution to the food sample. Crush the food sample and add water in the food sample, if the food sample is solid. Heat the solution in water bath. If red precipitate is seen, glucose is present. If the colour remains blue, glucose is absent. Teacher in Charge: ARIF ULLAH Page 8

9 2. Experiment to show the selective permeability of visking (dialysis) tubing to starch and glucose Procedure 1. Set up the apparatus as shown above. 2. Wash the filled visking tubing under water before immersion in the distilled water (to remove any starch and glucose on the outer surface). 3. Test for the presence of starch and glucose (reducing sugar) in the distilled water after 1 hour. Result Only glucose (simple sugar) is present in the distilled water outside the visking tubing. Interpretation Glucose molecules are small enough to pass through the tiny pores on the selectively permeable membrane of the visking tubing by diffusion. Starch molecules are too large so they cannot pass through the membrane. 4. Experiment to show the action of saliva on starchy food Procedure 1. Fill one visking tubing with starch solution. 2. Fill the other visking tubing with starch and saliva solution. 3. Tie the other ends of both visking tubings with threads and support them with glass rods. 4. Wash the two visking tubings under water and then put them separately into a beaker of distilled water as shown. 5. After 30 minutes, take samples of distilled water from the two beakers to test for starch and reducing sugar. Teacher in Charge: ARIF ULLAH Page 9

10 Result Solution in dialysis tubing Starch Starch + saliva Distilled water outside the dialysis tubing Benedict's Test Iodine Test 1. No reducing sugar and starch are found in the distilled water surrounding the visking tubing containing only starch solution. 2. Only reducing sugar is found in the distilled water surrounding the visking tubing containing both starch and saliva solution. Interpretation Saliva contains a digestive enzyme, amylase, which can digest starch into smaller molecule - maltose. In the absence of saliva, the starch molecules are too large to pass through the visking tubing. In the presence of saliva, the large starch molecules are broken down into smaller maltose molecules which can then pass through the visking tubing Conclusion The visking tubing represents the internal wall of the small intestine and the distilled water represents the blood. In the presence of digestive enzyme (e.g. amylase) can the large food molecules (e.g. starch) be broken down into smaller molecules (e.g. maltose) and passed through the membrane or absorbed. 5. A potato was set up as shown in the figure below (left-hand side). The investigation was left for several hours. The results are shown on the righthand side of the figure. Teacher in Charge: ARIF ULLAH Page 10

11 1. Describe what happened to a. the water in the disk b. the salt solution in the hollow in the potato. [2 mark] 2. Name the process that is responsible for the changes that have occurred. [1 mark] 3. Explain why these changes have occurred. [3 mark] 4. Where does this process occur in a plant? [1 mark] 5. What is the importance to the plant of this process? [1 mark] Answers 1. a. The volume of water in the dish decreased. b. The volume of salt solution in the potato increased. 2. Osmosis 3. (3 points from) - there was a higher concentration of water in the dish than in the potato - so water moved into the potato. - from a high concentration of water to a lower concentration of water - by osmosis. 4. Root hairs, or in the roots. 5. Osmosis enables the plant to absorb water to maintain cell turgidity (or to replace water lost by transpiration). Teacher in Charge: ARIF ULLAH Page 11

12 Surface area to volume ratio Side = 1cm Area = 1x1 Volume = 1x1x1 SA 6x1x1 Vo 1x1x1 Side = 10cm Area = 10x10 Volume = 10x10x10 SA 6x10x Vo 10x10x10 10 Side = 100cm Area = 100x100 Volume = 100x100x100 SA 6x100x100 _6_ 0.06 Vo 100x100x We can see that, as the cubes become bigger, their surface area increases. But their surface area to volume ratio decreased. Therefore, it would take more time for substances to diffuse into the longer cube. For example v An ant has a small area v An Elephant has a large surface area v An ant has a small volume v An Elephant has a large volume v But the surface area to volume ratio of an ant is greater than that of an Elephant. Teacher in Charge: ARIF ULLAH Page 12