Stacy Hernandez Period 1 Mrs. Riley AP Biology Osmosis and Diffusion Lab I. Introduction: Diffusion is vital to many life functions of a cell, it allow the transportation of vitally important nutrients and compounds without the expenditure of excess metabolic energy. To explain diffusion, it is as if a bottle of perfume is opened at one end of the room, then in a short amount of time a person at the other end of the room can detect the scent of the perfume; this is the process of diffusion.
Diffusion is a movement from a higher concentration to a lower concentration.
A special kind of diffusion is known as osmosis, osmosis is fundamental in many industrial processes such a water purification, desalination, and waste material treatment. Osmosis is when water is separated by a selectively permeable membrane, with different solute concentrations on either side of the membrane. In osmosis water is moved from regions of low concentration to regions of high solute concentration.
Osmosis occurs in both plant and animal cells, across a cell’s plasma membrane. Osmosis also has a major impact on living cells.
There are usually more or fewer dissolved particles in one of two compared solutions separated by a membrane, such as cell and the media in which it exists, in environment’s with solute concentrations that match their cytoplasm. When the two solutions on either sides of the membrane are equal and no net movement is detected, the solutions are isotonic. This means that the solutions have the same concentration of solutes. If two solutions differ in the concentration of solutes that each has, the one with more solute is hypertonic. The solution that has less solute is hypotonic.
Water potential determines the direction and rate of osmosis. Water potential consists of pressure potential, which is the exertion of pressure on a solution, and osmotic potential, which is the relative concentration of solutes within the two solutions. II. Hypothesis: Diffusion and osmosis occur between different molar solutions until the solutions are isotonic, effecting the pressure of the plant cells. III. Materials and Procedure: Materials: Per group: In this experiment my group used the following materials: * Four glucose indicator strips * One graduated cylinder * 1 plastic cup, 250 ml dialysis tubing, 7ft. Six plastic cups, 250 ml * Glucose starch solution, 15 ml * IKI (potassium iodide) solution, 1 ml * Sucrose solution, 0. 2 M, 175 ml * Sucrose solution, 0. 4 M, 175 ml * Sucrose solution, 0. 6 M, 175 ml * Sucrose solution, 0. 8 M, 175 ml * Distilled water * Paper towels * Potato * Plastic wrap * Microscope slide * Coverslip * Compound microscope * Scalpel * Forceps Shared materials: * Balance * Compound microscope * NaCl solution * Onion epidermis Procedure: Part A, Diffusion: We began part A of the experiment by first pouring 15 ml of the prepared glucose/ starch solution into a graduated cylinder.
Then we obtained a piece of dialysis tubing that has been soaking in water, and we tied a knot in one end of the tubing. Then we opened the tubing by rubbing the untied end between our fingers, and poured 15 ml of the glucose/ starch solution into the tubing. Then recorded the color of the solution in the bag and noted it in Table 1 in the Analysis section. After, we then determined if glucose was present in the tubing by dipping one of the glucose indicator strips into the solution and then recorded the data in Table 1. Afterward we then carefully tied a knot at the open end of the bag, but left enough space in the bag for expansion.
Then we filled a plastic cup approximately 2/3 full with distilled water and added 1ml of potassium iodide to the cup, then recorded the color of the solution in Table 1. After that we determined if glucose was present by dipping another glucose strip into the solution in the beaker and recorded the data in Table 1. Then we immersed the dialysis bag completely in the solution and waited 30 minutes. Then we removed the dialysis bag from the cup and recorded the final color of the solutions in the bag and the cup in Table 1.
Using glucose indicator strips we then determined the glucose content in both the beaker and the dialysis bag. Finally we recorded the presence of absence of glucose in Table 1. Part B, Osmosis: My group obtained six plastic cups and labeled them as follows: water, 02. M, 0. 4 M, 0. 6 M, 0. 8 M, and 1. 0 M. Then we obtained six pieces of dialysis tubing from the beaker of water and tied a knot in one end of the tubing. Afteer, we opened one piece of dialysis and poured 25 ml of distilled water into the tubing and then tied of the other end securely leaving room for expansion.
Then blotted the tube dry and placed it in the cup labeled “water”. Then we repeated the same process witht eh remaining five pieces of dialysis tubing, adding a different sucrose solution to each bag: 0. 2 M, 0. 4 M, 0. 6 M, 0. 8 M, and 1. 0 M. After, we then weighed each bag and recored each bag’s initial mass in Table 2. Then filled the six plastic cups approximately ? full of distilled water and immersed one bag in each of the the properly labeled cups. After waiting 30 minutes, we removed the bags from the cups, dried them and weighed each bag once again recording the final mass of each bag in Table 2.
Finally we calculated the c=percent change in mass for each of the dialysis bags using the formula: % Change = (Final mass- Initial mass)/ Initial mass x 100 and recorded this data in table 2; also gathering the class average results of the experiment. Part C, Water potential: After our instructor assigned us with one or more sucrose solutions of varied concentration and/or distilled water, and with prepared potato cylinders, we began this experiment by weighing four potato cylinders together and recorded the initial mass of them in Table 3 in the Analysis section. Then we poured 100 ml of one olution we were assigned into a plastic cup and took the initial temperature and recorded it in Table 3. Then we inserted four potato cylinders and cover the cup and let it stand overnight. The following day, we measured the final temperature of the liquid in the cups ad recorded the data in Table 3. After removing the four cylinders from each cup, we blotted them dry and weighed them, recording their final mass in Table 3. We then calculated the percent change in mass for the four cylindrs using the formula: % Change = (Final Mass- Initial Mass)/ Initial Mass x 100, and recorded the data in Table 3.
Part D, Water potential calculation: We determined the solute potential of the sucrose solution, the pressure potential, and the water potential. Part E, Plant cell plasmolysis: We prepared a wet mount slide of onion skin and observed it under a light microscope, the sketched out the results. After that we then added a few drops of NaCl solution, the observed it under a light microscope and sketched out the results. IV. Data: Table 1 Diffusion Color| Glucose Content| Time| Dialysis bag| Beaker| Dialysis bag| Beaker| Start| Clear| Yellow-brown| +| -| 30 minutes| Dark blue| Yellow-orange| +| -|
Table 2 Osmosis Investigation Solution| Dialysis Bag Initial Mass (g)| Dialysis Bag Final Mass (g)| Change in Mass (g)| % Change in Mass| Class % Change in Mass| Water| 25. 69 g| 25. 25 g| 0. 44 g| -0. 54 %| 0. 33 %| 0. 2 M| 27. 17 g| 27. 58 g| 0. 41 g| 1. 50 %| 1. 50 %| 0. 4 M| 27. 73 g| 28. 78 g| 0. 05 g| 3. 7 %| 3. 7 %| 0. 6 M| 28. 84| 30. 62| 2. 22 g| 6. 17 %| 6. 17 %| 0. 8 M| 28. 88 g| 30. 96 g| 2. 08 g| 7. 2 %| 7. 2 %| 1. 0 M| 30. 24 g| 32. 25 g| 2. 01 g| 6. 6 %| 6. 6 %| Percent Change in Mass of Sucrose Solution in Dialysis Tubing Table 3 Potato Cell Water Potential
Solution Temp C| Potato Cylinders| Solution| Initial Temp| Final Temp| Initial Mass| Final Mass| Change in Mass| % Change in Mass| Class Average| Water| 26 C| 18 C| 5. 48 g| 6. 29 g| 0. 81 g| 14. 78 %| 14. 54 %| 0. 2 M| 21. 5 C| 18. 5 C| 5. 14 g| 5. 19 g| 0. 05 g| 0. 97 %| 0. 97 %| 0. 4 M| 23 C| 18 C| 6. 72 g| 5. 39 g| 1. 33 g| 19. 79 %| 19. 79%| 0. 6 M| 20 C| 21 C| 8. 34 g| 6. 37 g| 1. 97 g| -17. 28 %| -25 %| 0. 8 M| Room| 21 C| 5. 73 g| 4. 93 g| 0. 8 g| -14 %| -23%| 1. 0 M| room| 22 C| 5. 73 g| 5. 95 g| 0. 41 g| 0%| -30. 3%| Percent Change in Mass of Potato Cores at
Different Molarities of Sucrose V. Questions: 1. Create a Venn Diagram comparing osmosis and diffusion. Osmosis -molecules go through a semipermeable membrane. -just water Osmosis -molecules go through a semipermeable membrane. -just water Diffusion -molecules spread out over a large area. -Everything but water. Diffusion -molecules spread out over a large area. -Everything but water. -Molecules mover around to create equilibrium. -Molecules mover around to create equilibrium. 2. Part A of the experiment was a demonstration of diffusion. Give an example of diffusion occurring in the setup.
Do you think osmosis occurred in this part of the experiment? If you answered yes, explain why you believe this to be. There is a diffusion of glucose out into the medium. No, I don’t think osmosis occurred in this part of the experiment. 3. Did the dialysis tubing serve as a selectively permeable membrane? Explain your answer. Yes, because it restricted certain molecules or particles to diffuse through its microscopic holes. Based on the tiny size of the holes, water and glucose were small enough to fit through them, whereas starch and potassium iodide weren’t able to diffuse through. . In part B, what caused the mass of the dialysis bags to change? Was there more or less water in the dialysis bags at the conclusion of the experiment? Explain. Because there was no net flow of water molecules going into the bags, the mass of the dialysis after 30 minutes increased. 5. Was the distilled water in the beakers hypertonic or hypotonic in relation to the sucrose solutions found in the dialysis bags? It was hypotonic in relation to the sucrose solutions found in the dialysis bags because it had a lower solute concentration. 6.
Suppose the dialysis bags were placed in beakers containing 0. 6 M sucrose solution as opposed to distilled water. How do you think your results would change? Sketch a graph below to show how the mass of each of the bags would be affected. My results would greatly change depending on the sucrose concentration in the dialysis bag, the direction in which diffusion occurs will depict the mass change. If the dialysis bag has the same concentration of sucrose as the solution, there would be no mass change because it is at equilibrium. Diffusion of Sucrose Molecules . Study the graph you have plotted for Part C of the experiment. What is important about the point where the best fit line crosses the x-axis? What is the concentration of sucrose in your potato? It is at equilibrium. At that point, the initial mass and the final mass are the same and there is no net flow of water molecules or sucrose molecules. 8. Fill in the blanks: If a cell is placed in a hypertonic solution, it has a less solute in solution than the surrounding fluid, and will therefore experience a net loss of water to its surrounding.
This cell has a low water potential since there is a great deal of osmotic pressure causing water to leave the cell. Conversely, a cell sitting in a hypotonic solution has a high water potential, and since it will experience a net gain of water, there will be little osmotic pressure causing pressure to enter the cell. 9. Using the data from Part C of this activity and the formula for water potential from part D, calculate the osmotic potential of the sucrose solution in bars. ?= -iCRT ? = -1(0. 26 M)(0. 0831 liter bars/ mole Kelvin)(295 K) ? = -6. 37 bars 10. The water potential (?? of a solution is equal to the osmotic potential (? ) plus the pressure potential (? p). Since there is no differential pressure acting on the solution, the pressure potential is equal to zero, making the water potential equal to the osmotic potential. If the equilibrium point between the solutions and the potato cylinders indicates the point where the two water potentials are equal, what is the water potential of the potato cells? ? = ?? + ? F ? = -6. 37 + 0 ? = -6. 37 MPa 11. Would the water potential of the potato cells change if the cylinders were allowed to dry out? In what way?
They would not change if the cylinders were allowed to dry out because they are not in any kind of solution where there’s a difference in concentration. 12. What are the effects on cells when they are placed in a hypotonic solution, a hypertonic solution, and an isotonic solution? When cells are placed in a hypotonic solution where solute concentration is low, they will experience swelling. 13. Why can’t humans drink salt water for hydration? Humans can’t drink salt water for hydration because the water from our cells will diffuse outside where there is a high concentration of solute and low water potential.
Consequently, we would become more dehydrated as the cells lose more water. VI. Error Analysis Lab 1a: Perhaps if the tied knot of the dialysis tube was too tight. Or maybe there was a leak or hole in the dialysis tube, then the data would be inaccurate. Lab1b: The data would be inaccurate if the first step was done wrong, if the person holding the dialysis tube had lotion on their hands, if there was something that blocked the pores of the tube. Lab1c: The data would be inaccurate if the mass of the potatoes was recorded incorrectly.
Lab1d: The data would be inaccurate if the calculations weren’t calculated correctly. Lab1e: The data would be inaccurate if there was too much NaCl added. VII. Conclusion: In this lab, we observed both diffusion and osmosis. I learned that starch molecules are far too big to diffuse out of the dialysis bag and into the water surrounding it because I learned that they were far too big to exit through the semi-permeable dialysis tubing. I learned that water movement towards the potatoes was high. Also, once the onion sample had NaCl added to I, it showed how the cells became hypertonic and shriveled up.
Water potential and concentration gradients are the two phenomenon’s that affected the results of the experiments. There are many important facts pertaining to water potential. Water potential is used by botanists to determine the movement in and out of a cell. It is affected by two components, pressure and solute potential. Water moves from areas of higher water potential (higher free energy and more water molecules) to areas of lower water potential (lower free energy and less water molecules). Water diffuses down a water potential gradient.
Pure water has an atmospheric pressure of zero which is important when using the formula ? s = -iCRT. Water potential is inversely proportional to solute potential. These facts led to or affected the results gained in each section of the lab. In plant and animal cells, loss or gain of water can have different effects. In an animal cell, it is ideal to have an isotonic solution. If the solution is hypertonic, the cell will shrivel from lack of water intake. Inversely, if the solution is hypotonic the cell could take in too much water and the cell will lyse and break open.
For a plant cell, the ideal solution is a hypotonic solution because the cell takes in water increasing turgor pressure. Turgor pressure is important for plant support and maintaining shape. If the solution is hypertonic, the cell will plasmolysis and died from lack of water. In an isotonic solution, the plant cell does not have enough turgor pressure to prevent wilting and possible death. The information gained through this lab is important in understanding the effects of different solutions on organisms in our environments, including ourselves.
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