Billy muse AP Bio 4th hour Group number 1 9/17/02 Osmosis/Diffusion Lab 1 Lab 1A - Diffusion I. Background - Diffusion is defined as the random mixing of molecules or atoms due to there constant motion and impact. II. Purpose - To investigate diffusion through a semi-permeable cellulose dialysis membrane and the effect of size on this diffusion. III. Hypothesis – Large polymers, such as starch, will not be able to diffuse (effuse) through the finite size holes in the membrane. Smaller molecules may be slowed down, but not prevented from diffusion through. IV. Procedure - Imbibe a dialysis membrane and tie off one end. Add 15-20 mls of a 15% glucose/1% starch solution to the bag and tie of the end leaving some air space for the tube to expand due to osmosis. Soak the dialysis bag in distilled water containg 3-4 mls of iodine/iodide solution. V. Results – The iodine diffuses into the bag and interacts with the starch to create a blue complex. After 30 minutes no glucose is detected outside the bag nor is their any blue starch complex outside the bag. Data tables Outside bag Inside bag Benedicts assay (out) Starch assay I2 Initial Water, iodine Glucose, starch negative negative 30 mins water Glucose, starch, negative Blue complex in iodine bag 24 hours Some glucose Glucose, starch Slightly positive + 48 hours glucose Glucose, starch positive + VI. Observations - Starch stayed in the bag as no blue complex was noticed outside the bag. Glucose was slowed by the bag, but the benedict’s test showed that some of the glucose left the bag over time. VII. Conclusion (What does all this mean?) – The cellulose dialysis membrane mimics the typical cell wall and lipid bilayer. It selectively lets small molecules out but retains large molecules such as starch. Questions from this section 1) Iodine and water enter the bag. The bag swells slightly and the starch complexes with iodine to form a blue precipitate. Glucose eventually leaves the bag as the benedic’s assay after 24 hours shows a positive test for sugar. 2) The pore size of the membrane must be slightly larger than the glucose. 3) Yes, weigh the bag periodically 4) Rank: small to large water, iodine (IKI), glucose, membrane pores, starch. 5) I would expect the outside water to turn blue as the IKI left the bag. Exercise 1 B: Osmosis Background - Osmosis is defined as the movement of water down a concentration gradient across a semi-permiable membrane Purpose - To observe osmosis in a controlled setting Hypothesis - The bags will shrink in hypertonic solution, but there will be no net change in the isotonic distilled water. Procedure - Add solutions of different concentrations sucrose to 6 dialysis membranes and put the bags into distilled water. Results. The initial mass increased in all bags except the one filled with distilled water. Contents in bag Initial mass Final mass difference % change ddH2O 5g 5g 0 0 0.2 M sucrose 5g 5.35 g .35 g 5% 0.4 5 5.6 g .6 g 10% 0.6 6 6.85 g .85 g 15% 0.8 6 8g 2g 25% 1.0 M 6 10 g 4g 50% Include graph of data here (recommend excel) Observations: All the concentrations above 0 M sucrose were in hypotonic medium when in the beaker of distilled water, thus water rushed in to equalize the water concentrations. Conclusions - Osmosis is a powerful force and can be quantitativally determined by measuring the mass of water change over a given time. Questions from lab handout 1 b 1) The higher the molarity sucrose, the greater the mass change. More water would be needed to equalize the concentrations 2) If the bags were placed in 0.4 molar sucrose instead of distilled water,. Then the distilled water and 0.2 molar bags would shrink as water left them ((hypertonic) , the 0.4 molar bag would stay the same, and the 0.6 – 1.0 molar will still swell. 3) Because all the bags start out at slightly different masses, we looked at the % change to normalize the data to the starting weight. 4) 20-18/20 x 100 = 10% Exercise 1c Water potential Background - The driving force behind osmosis is water potential. Water tends to go from high concentration to low concentration. Purpose - to determine the water potential in a potato cell. In other words to determine the iso-osmolarity within the cell to outside conditions. Hypothesis - The water potential in the cell will be greater than a 0.2 M sucrose solution, but less than a 1 M sucrose solution. In hypertonic medium the potato cells will shrink as water leaves them. In hypotonic medium the cells will swell. In isotonic media there will be no net flow and this will be the concentration to calculate the water potential from. Procedure - Each lab group makes up a slightly different concentration of sucrose water. Each group gets a chunk of potato containing many thousands of cells to soak in their concentration of sucrose water. Results - Class data looked pretty rocky - (see Muse’s handout). This is probably due to Class only soaking for ½ hour or less. Use and plot Muse’s data from the handout. Observations – The point at which the plot of solution concentration vs mass change crosses the 0 line (y intercept) is 0.37 M Conclusion - The concentration of water in the potato cell is equal to the concentration of water in 0.37 M sucrose solution. This is the iso-osmolar point. Exercise 1D Background – The formula to calculate solute potential is s = - iCRT Water potential and solute potential will be equal. Purpose – To use the data from 1C to calculate water potential from a number of different cells. Hypothesis – I believe that different types of cells will have similar water potentials. Results - for potato the iso-osmolar concentration was solute to 0.37 molar s = - iCRT ( - 1) (0.37 M) (0.0831 liter bar/mole K) (296 K) s = - 9.1 bars Question 1 from 1d: Would water potential increase or decrease if potato was dehydrated? It would decrease because the concentration of solute rises as water decreases 2. Hypotonic to its environment, it will swell. 3. p = 0 4. beaker with 0.4% sucrose has lower water potential 5. Water will diffuse out of the bag Zucchini is 0.37 iso-osmolar. Molar concentration of solutes = 0.37 M Question 8 - same as calculation of potato shown above s = - iCRT ( - 1) (0.37 M) (0.0831 liter bar/mole K) (296 K) s = - 9.1 bars Question 9 Adding more solute to the Zucchini cells will decrease the water potential of the Zucchini plugs. Question 10 a. ddH2O has a higher concentration of water molecules 10 b. ddH2O has a higher water potential 10 c. The RBC would swell and burst if put in distilled water. Water would rush into it. Exercise 1E Onion Cell Plasmolysis Background – Onion cells can be easily seen under 20 -100 fold magnification. The cellulose cell wall will remain constant sized but the internal membrane can shrink. Purpose – To visibly watch osmosis in a live cell Hypothesis - Onion cells will shrink in 15% salt solution (hypertonic) Procedure – Take a thin slice of onion and put a cover slip over it. Focus a cell at 100 x magnification and watch membrane. Put a droip of 15% salt on the edge of slide and draw it in with capillary action with a paper towel. Results: 1. Cell wall and internal membrane very close to each other 2. The internal membrane has peeled away from cell wall and cell has shrunk. 3. The membrane returns to close to wall when fresh water is added. The cell has swelled. Over time, the cell and wall bursts. Analysis of results questions: 1. What is plasmolysis? It is the pulling away from the wall of the plasma membrane due to the osmotic shrinking of the cell. 2. The onion cell when added to ddH2O was forced to swell because the water potential inside was less than that outside. Water followed an osmotic gradient and burst the cell. 3. Grass that is in a salty environment cannot overcome the tendency of the water to leave the roots by going from lower water potential to higher water potential. The salt effectively makes the soil hypertonic compared to the inside of the grass roots. The cells shrink and the plants die. What have I learned from Lab exercise 1a – 1e. Water tends to move from areas of high concentration (potential) to low concentration. Plasma membranes are semi-permeable. They let water cross easily, but not large solutes such as starch polymers. Small molecules such as glucose or salt ions can cross, but not as fast as water. Some plant cells have internal solute concentrations equal to 0.37 molar. Therefore, their water concentrations are less than that of ddH2O and will swell if put into fresh water.