Understanding Diffusion and Osmosis in Plant Tissues Lab
School
Eagle River High School**We aren't endorsed by this school
Course
BIO AP
Subject
Biology
Date
Dec 11, 2024
Pages
13
Uploaded by ColonelTeam13730
Osmosis and Diffusion Lab Introduction and BackgroundIn this laboratory you will investigate the processes of diffusion and osmosis in a model membrane system. You will also investigate the effect of solute concentration on water potential as it relates to living plant tissues.Many aspects of the life of a cell depend on the fact that atoms and molecules have kinetic energy and are constantly in motion. This kinetic energy causes molecules to bump into each other and move in new directions. One result of this molecular motion is the process of diffusion.Diffusion is the random movement of molecules from an area of higher concentration to an area of lower concentration. For example, if one were to open a bottle of hydrogen sulfide (H2S has the odor of rotten eggs) in one corner of a room it would not be long before someone in the opposite corner would perceive the smell of rotten eggs. The bottle contains a higher concentration of H2S molecules than the room does and, therefore, the H2S gas diffuses from the area of higher concentration to the area of lower concentration. Eventually a dynamic equilibrium will be reached; the concentration of H2S will be approximately equal throughout the room and no net movement of H2S will occur from one area to the other.Osmosis is a special case of diffusion. Osmosis is the diffusion of water through a selectively permeable membrane (a membrane that allows for diffusion of certain solutes and water) from a region of higher water potential to a region of lower water potential. Water potential is the measure of free energy of water in a solution.Diffusion and osmosis do not entirely explain the movement of ions or molecules into and out of cells. One property of a living system is active transport. This process uses energy from ATP to move substances through the cell membrane. Active transport usually moves substances against a concentration gradient, from regions of lower concentration of that substance into regions of higher concentration.
Part 1: DiffusionIn part 1, your teacher will demonstrate the diffusion of small molecules through dialysis tubing, which is an example of a selectively permeable membrane. Small solute molecules and water molecules can move freely through a selectively permeable membrane, but large molecules will pass through more slowly, or perhaps not at all. The movement of a solute through a selectively permeable membrane is called dialysis. The size of the minute pores in the dialysis tubing determines which substances can pass through the membrane.Data Table 1In BagIn BeakerSolution ColorInitialbrown/yellowclearFinalGlucose Present?Initial yesnoFinalQuestions for Part 1:1. Which substance(s) is(are) entering the bag? What evidence supports your answer?Water from the beaker is entering the tubing; we know this because the color of the inside of the tubing becomes less intense.2. Which substance(s) is(are) leaving the bag? What evidence supports your answer?The iodine in the iodine solution is leaving the bag; we know this because the color of the liquid in the beaker becomes less clear. 3. Quantitative data uses numbers to measure observed changes. How could this experiment be modified so that quantitative data could be collected to show that water diffuses into the dialysis bag?In order to make a quantitative data set we could measure the amount that leaves/enters the bag. 4. Based on your observations, rank the following by relative size, beginning with the smallest: glucose molecules, water molecules, IKI molecules,membrane pores, starch
molecules.Water, membrane pores, IKI, glucose, starchPart 2: Water Beads1. Select 24 water beads (4 each of the same color.) 2. Place four beads in separate wells of your spot plate 3. Using a pipette fill four wells with each the following concentrations of sugar solutions. (1.0 M, 0.8 M, 0.6 M, 0.4M, 0.2 M, 0M.) Make sure each bead is fully submerged in the solution. Label the wells with the molarity of solution placed in each.4. Let them stand in the solution for 1 hour. 5. After 1 hour, record an average mass for all four beads in each solution.6. Did the beads in different solutions differ in their final size?Data Table 2: Water BeadsSolution0 M0.2 M0.4 M0.6 M0.8 M1.0 M
Final Average Mass (g)0.25 g 0.225 g0.175g0.1475g0.135g0.095 gPart 3: OsmosisIn this experiment you will use dialysis tubing to investigate the relationship between solute concentration and the movement of water through a selectively permeable membrane by the process of osmosis.When two solutions have the same concentration of solutes, they are said to be isotonic to each other (iso means same, -ton means condition, -ic means pertaining to.) If the two solutions are separated by a selectively permeable membrane, water will move between the two solutions, but there will be no net change in the amount of water in either solution.If two solutions differ in the concentration of solutes that each has, the one with more solute is hypertonic to the one with less solute (hyper means over or more than.) The solution that has less solute is hypotonic to the one with more solute (hypo means under or less than.) These words can only be used to compare solutions.Now consider two solutions separated by a selectively permeable membrane. The solution that is hypertonic to the other must have more solute and therefore less water. At standard atmospheric pressure, the water potential of the hypertonic solution is less than the water potential of the hypotonic solution, so the net movement of water will be from the hypotonic solution into the hypertonic solution.Osmosis Procedure1.Each group will be assigned a different solution to investigate. What solution has your group been assigned? Assigned solution:
1.0 M2.Obtain two 30-cm strips of pre-soaked dialysis tubing.3.Tie a knot in the very end of each piece of dialysis tubing.4.Pour 25 mL of your assigned solution into each bag.5.Remove most of the air from each bag by drawing the dialysis bag between two fingers. Tie off the other end of the bag. Leave sufficient space for the expansion of the contents in the bag. (The solution should fill only about one-third to one-half of the piece of tubing.)6.Rinse each bag gently with water to remove any sucrose spilled during filling. Carefully blot the outside of each bag and record the initial mass (in grams) of each bag.7.Fill 3 plastic cups or 1 large beaker two-thirds full with distilled water.8.Immerse each bag in one of the cups/beakers of distilled water and number each cup (1, or 2.) Be sure to completely submerge each bag.9.Let the bags and cups stand for 15 minutes. At the end of 15 minutes remove the bags from the water. Carefully blot the outside of each bag.10.Determine the mass of each bag and record the mass in the Data Table for Part II Group Results.11.Calculate the difference in mass for each bag and record your answers in the Data Table for Part II Group Results.12.Calculate the percent change in mass for each bag and record your answers in the Data Table for Part II Group Results. 13.Record all of the data in your glass spreadsheet and graph the class results for Part II.Data Table 3: Dialysis TubingBagInitial MassFinal MassDifference% Change in MassAverage % Change in Mass128.232.03.813.48%13.23%228.532.23.712.98%Data Table 4: Class Results for Dialysis TubingSolution0 M0.2 M0.4 M0.6 M0.8 M1.0 MAverage % Change in -0.21%1.35%6.7%11.15%11.382%13.23%
Mass14.Graph the class results for the dialysis tubing. Identify the independent and dependent variable:Independent variable is the molarity of the solution Osmosis Questions:1.Identify the independent and dependent variable The independent variable is the concentration of the solutions and the dependent variable is the change in mass of the objects. 2.Describe the relationship between the change in mass and the molarity of sucrose within the dialysis bag.The higher molarity of sucrose within the bag will result in a higher rate of osmosis.3.Predict what would happen to the mass of each bag in this experiment if all the bags were placed in a 0.4 M sucrose solution instead of distilled water. Explain your response.Solution Predicted Change Explanation0 MSolution will go into bagTo create an isotonic solution, the hypertonic solution would need to go into the cell to balance out the lesser than solution inside the bag.0.2 MSolution will go into bagTo create an isotonic solution, the hypertonic solution would need to go into the cell to balance out the lesser than solution inside the bag.0.4 MSolution will not exit or enter the bagBecause the solution is already isotonic , there is no need for transport of solution between the cells. 0.6 MBag will lose sucrose solution As there is a hypotonic solution, the solution will need to leave the bag in order to maintain equilibrium inside and outside the cell. 0.8 MSolution will exit the As there is a hypotonic solution, the solution will
bagneed to leave the bag in order to maintain equilibrium inside and outside the cell. 1.0 MSolution will leave the bagAs there is a hypotonic solution, the solution will need to leave the bag in order to maintain equilibrium inside and outside the cell. ??4.Why did we calculate the percent change in mass rather than simply using the change in massBecause the percent change shows how much of the original sample changed5.A dialysis bag is filled with distilled water and then placed in a sucrose solution. The bag’s initial mass is 20 g and its final mass is 18 g. Calculate the percent change in mass. Show your work.20/18=1.111 1.111/20=0.055550.05555*100=5.55%6.A 1.0 M solution in a dialysis bag is _______________ (hyper, hypo, or isotonic) relative to pure water in a beaker: HypotonicPart 4: Water Potential In this part of the exercise you will use potato chunks in different molar concentrations of sucrose in order to determine the water potential of potato cells. First, however, we will explore what is meant by the term “water potential.”Botanists use the term water potential when predicting the movement of water into or out of plant cells. Water potential is abbreviated by the Greek letter psi (ψ) and it has two components; a physical pressure component, pressure potential ψp, and the effects of solutes, solute potential ψs. ψ = ψp+ ψs
Water potential = Pressure potential +Solute potentialWater will always move from an area of higher water potential (higher free energy; more water molecules) to an area of lower water potential (lower free energy; fewer water molecules.) Water potential, then, measures the tendency of water to leave one place in favor of another place. You can picture the water diffusing “down” a water potential gradient.Water potential is affected by two physical factors. One factor is the addition of solute which lowers water potential (fewer water molecules.) The other factor is pressure potential (physical pressure.) An increase in pressure raises the water potential. By convention, the water potential of pure water at atmospheric pressure is defined as being zero (ψ = 0.) For instance, it can be calculated that a 0.1 M solution of sucrose at atmospheric pressure (ψp= 0) has a water potential of –2.3 bars due to the solute (ψ = -2.3). (Note: A bar is a metric measure of pressure, measured with a barometer and is about the same as 1 atmosphere. Another measure of pressure is the megapascal (MPa.) 1 MPa = 10 bars.)Movement of water into and out of a cell is influenced by the solute potential (relative concentration of solute) on either side of the cell membrane. If water moves out of the cell, the cell will shrink. If water moves into an animal cell, it will swell and may even burst. In plant cells, the presence of a cell wall prevents cells from bursting as water enters the cells, but pressure eventually builds up inside the cell and affects the net movement of water. As water enters a dialysis bag or a cell with a cell wall, pressure will develop inside the bag or cell as water pushes against the bag or cell wall. The pressure would cause, for example, the water to rise in an osmometer tube or increase the pressure on a cell wall. It is important to realized that water potential and solute concentration are inversely related. The addition of solutes lowers the water potential of the system. In summary, solute potential is the effect that solutes have on a solution’s overall water potential.Movement of water into and out of a cell is also influenced by the pressure potential (physical pressure) on either side of the cell membrane. Water movement is directly proportional to the pressure on a system. For example, pressing on the plunger of a water-filled syringe causes the water to exit via any opening. In plant cells this physical pressure can be exerted by the cell pressing against the partially elastic cell wall. Pressure potential is usually positive in living cells: in dead xylem elements it is often negative.It is important for you to be clear about the numerical relationships between water potential and its components, pressure potential and solute potential. The water potential value can be positive, zero, or negative. Remember that water will move across a membrane in the direction of lower water potential.An increase in pressure potential results in a more positive value, and a decrease in pressure potential (tension or pulling) results in a more negative value. In contrast to pressure potential, solute potential is always negative; since pure water has a water potential of zero, any solutes will make the solution have a lower (more negative) water potential. Generally, an increase in solute potential makes the water potential value more negative and an increase in pressure potential makes the water potential
more positive.To illustrate the concepts discussed above, we will look at a sample system using the figures below. When a solution, such as that inside a potato cell, is separated from pure water by a selectively permeable cell membrane, water will move (by osmosis) from the surrounding water where water potential is higher, into the cell where water potential is lower (more negative) due to the solute potential (ψs). In picture athe pure water potential is 0 (ψ=0) and the solute potential is –3 (ψs= -3.) We will assume, for purposes of explanation, that the solute is not diffusing out of the cell.By the end of the observation, the movement of water into the cell causes the cell to swell and the cell contents push against the cell wall to produce an increase in pressure potential (turgor) (ψ=3.) Eventually, enough turgor pressure builds up to balance the negative solute potential of the cell. When the water potential of the cell equals the water potential of the pure water outside the cell (ψ of cell = ψ of pure water = 0), a dynamic equilibrium is reached and there will be no NET movement of water (picture b.) If you were to add solute to the water outside the potato cells, the water potential of the solution surrounding the cells would decrease. It is possible to add just enough solute to the water so that the water potential outside the cell is the same as the water potential inside the cell. In this case, there will be no net movement of water. This does not mean, however, that the solute concentrations inside and outside the cell are equal, because water potential inside the cell results from the combination of both pressure potential and solute potential. If enough solute is added to the water outside the cells, water will leave the cells, moving from an area of higher water potential to an area of lower water potential. The loss of water from the cells will cause the cells to lose turgor. A continued loss of water will eventually cause the cell membrane to shrink away from the cell wall (plasmolysis.)Procedure for Water Potential1.Each group will be assigned a different solution to investigate. What solution has your group been assigned? Assigned solution:
1.0 molar solution2.Obtain 2 glass containers with lids and fill them ⅔full of your assigned solution. Obtain two types of potato chunks. Keep the potato chunks in a covered beaker or cup until you are ready to mass the chunks. 3.Cut two of each type of potato into 1cm by 1cm cubes. Determine the mass of two potato chunks together. Place these two chunks in a glass container with sugar solution. Place the other two in a second glass container. Write the mass of the potato chunks on the lid of the container. Also record the mass of the chunks in the Data Table for Part III Group Results. Repeat for each of the other cups. Place a lid on each container and let them stand overnight. 4.Use a spatula to remove the potato chunks from the first cup. Tap the probe or forceps on the side of the cup several times to remove the excess solution from the potato chunks. 5.Determine the mass of the potato chunks from the first cup. Record the mass on the cup and in the Data table for Part III Group Results. Repeat for the potato chunks in the other two cups. 6.Calculate the change in mass for the potato chunks in each cup. Record your results in the Data Table for Part III Group Results. 7.Calculate the percent change in mass for the potato chunks in each cup. Record your results in the Data Table for Part III Group Results.8.Calculate the average percent change in mass and record your answer in the data table. 9.Record the average percent change in mass for your solution and potato chunks in the Data Table for Part III Class Results.10.Put all of your data in the class spreadsheet and graph the class results for percent change in mass.Data Table 3: Group ResultsTuber TypeInitial MassFinal MassDifference% Change in MassSweet potato(regular shape1.6 g1.3.3-18.75%Sweet potato (irregular)1.1.8.3-27.27%
Wight potato (no skin)1.2.9.3-25%White potato (skin)1.4.7.7-50%Data Table 3: Class ResultsSolution0 M0.2 M0.4 M0.6 M0.8 M1.0 MAverage % Change in Mass-30.2611.Determine the molar concentration of the potato chunks. This would be the sucrose molarity in which the mass of the potato chunks does not change. To find this, locate the point at which this line crosses the x-axis represents the molar concentration of sucrose with a water potential that is equal to the potato tissue water potential. At this concentration there is no net gain or loss of water from the tissue. Indicate this concentration of sucrose in the space provided below. The molar concentration of the potato chunks would be around a .3-.4 molar solutionWater Potential Questions1.If this looks scary, watch this video first. The solute potential of this sucrose solution can be calculated using the following a scary looking equation that actually is pretty simple to use.Knowing the solute potential of the solution (ψs) and knowing that the pressure potential of the solution is zero (ψp = 0) allows you to calculate the water potential of the solution. ψs = -iCRT i = Ionization constant (for sucrose this is 1because sucrose does not ionize in water for NaCl it would be 2because it ionizes into two elements in water)C = Molar concentration (determined from the graph of our class data) R= Pressure constant (R = 0.0831 liter bars/mole K) T= Temperature in K (273 + degrees C of the solution)The units of measure will cancel as in the following example: ψs = (-1)( 1.0 mol/L)( 0.0831 liter)( 295K)
ψs = -24.51 bars 2.The water potential of the solution at equilibrium will be equal to the water potential of the potato cells. What is the water potential of the potato cells from Part III?The water potential of the potato cells in part 3 would be -24.51 bars.3.Water potential values are useful because they allow us to predict the direction of the flow of water. Recall from the discussion that water flows from an area of higher water potential to an area of lower water potential. For the sake of discussion, suppose that a student calculates that the water potential of a solution inside a bag is –6.25 bar (ψs = -6.25, ψp = 0) and the water potential of a solution surrounding the bag is –3.25 bar (ψs = -3.25, ψp = 0.) In which direction will the water flow? Explain your answer. Because the water potential of the inside of the bag is less than the water potential of the surrounding solution, water will move into the bag. 4.If a potato is allowed to dehydrate by sitting in the open air, would the water potential of the potato cells decrease or increase? Why? The water potential of the potato would decrease because the concentration of “solute” inside the cell will increase as water leaves the cell. 5.If a plant cell has a lower water potential than its surrounding environment and if pressure is equal to zero, is the cell hypertonic (in terms of solute concentration) or hypotonic to its environment? Will the cell gain or lose water? Explain your answer. The cell would be hypertonic to its environment because it will have a low water potential and thus a higher solute concentration to its environment. 6.Zucchini cores placed in sucrose solutions at 27 degrees C resulted in the following percent changes after 24 hours, graph the data in the table below. What is the molar concentration of solutes within the zucchini cells? Click on the drawing below to add your data.
7.Calculate the water potential for the zucchini data. First, calculate the solute potential (ψs) of the sucrose solution in which the mass of the zucchini cores does not change. Second, calculate the water potential (ψ) of the solutes within zucchini cores. Solute potential when zucchini did NOT change- -9.97 bars which is the same from the zucchini since it dont change