David Huang! AP Biology! Oct. 4,2013! AP Biology Osmosis Laboratory Analysis! Introduction:!! There are several different methods for the

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1 David Huang AP Biology Oct. 4,2013 AP Biology Osmosis Laboratory Analysis Introduction: There are several different methods for the transportation of molecules across the phospholipid bilayer. These transportation methods can either be downhill, which requires no input of energy (in the form of ATP and other metabolic energies), or uphill, which requires the input of energy. In this particular lab, we will examine the former type of transport. Osmosis and diffusion, although both are downhill transports, bear some minute differences. Both of these types of transports travel down the electrochemical gradient meaning they do not require energy. However, osmosis occurs when there is a pressure difference, while diffusion occurs when there is a concentration difference of a solvent (usually water). In osmosis, the difference in pressure, or osmotic pressure, is caused by the difference in solute concentration. Both of these types of transport are important to the biological processes of organisms. The alveoli in our lungs is dependent on the diffusion of gasses. When there is a high concentration of oxygen and low concentration of carbon dioxide in the alveoli and a high concentration of carbon dioxide and low concentration of oxygen in the capillaries, diffusion starts to occur. The oxygen from the higher concentration will begin to diffuse into the capillaries while the carbon dioxide is diffused out. Water potential, expressed as, is the potential of water to diffuse from one area to another. can only be either zero or negative, with pure water being zero and any other solvents being negative. The more negative a solvent is the lower the is. Our experiment with the potatoes is an example of this. When the potato cube is submerged in the deionized water it gained mass because the potato itself had a lower water potential than the deionized water, thus in an attempt to reach equilibrium the deionized water will osmose into the potato. Similarly, if the potato is placed into a solution that has a lower water potential than it does, the potato will lose mass. An example of this would be the solutions in the experiment that had 0.6m or higher concentration of sucrose (there is a higher concentration of solutes in the solution and a lower concentration of water thus causing water to osmose out of the potato). Procedures: (Exercise 1) After all the materials needed for the lab are gathered, the grams of sucrose that are to be dissolved into the deionized water must be calculated. The following is an example of the calculations, if the desired molarity of sucrose was 0.2M, the calculation would be 0.02 moles of sucrose * 342.3= 6.8g which would mean 6.8 grams of sucrose must be added into the solvent. To clarify, the 0.2M of sucrose equals to 0.02moles per 1 liter of water and the is the molecular mass. For the 0.4M solution it would be changed from 0.02 moles of sucrose to 0.04, these changes must also applied to the other respective solutions. After the sucrose solutions have been made, 10ml. of each solution (including plain deionized water) are added to the dialysis bags. When the dialysis bags are rinsed and dried, they are weighed and the weight recorded. Then all of the dialysis bags are immersed in separate beakers containing 250ml of distilled water. After the bags have rested for minutes, they are taken out and dried. The bags must then be weighed and recorded. (The percent change in mass can be found through the follow equation: [(final mass - initial mass)/ initial mass]*100.

2 (Exercise 2) The left over solutions from the first exercise are poured into their own respective labeled beakers. A potato is sliced into 6 equal cubes and are weighed as well as recorded. These potato cubes are then put into the beakers containing the sucrose solutions. After one day of submerging the potatoes in the solution, they are taken out, dried, and weighed. The mass difference is calculated as well as the percent change in mass using the given equation: [(final mass - initial mass)/ initial mass]*100. Analysis Questions: 1. Explain the relationship between the change in mass and the molarity of sucrose within the dialysis bag. As the molarity of sucrose increases in the dialysis bags( e.g. from 0.2M to 0.4M), so will the mass difference. Since the concentration of sucrose is higher and the water concentration is lower than the solution it is immersed in, the water will diffuse into the dialysis bag, in an attempt to reach equilibrium. 2. Predict what would happen to the mass of each bag in this experiment if all the bags were placed in 0.4M sucrose solution instead of distilled water. Explain your response. The bag filled with the distilled water will decrease in mass as will the dialysis bag with 0.2M of sucrose. The decrease in mass is due to the concentration of the solutes being higher in the solution outside of the dialysis bag, which will cause water to flow out. The dialysis bag filled with 0.4M sucrose solution will see very little change as the concentration of solutes are the same in and out of the dialysis bag (the solution is isotonic in terms of the dialysis bag). Any of the bags containing a sucrose solution of 0.4M or higher will see a increase in mass due to the solutes inside the dialysis bag is higher than that of the outside (this will cause water to flow into the bag). 3.Why did you calculate the percent change in mass rather than simply using the change in mass? Calculating the percent change is more accurate when we are looking for a cross comparison between multiple sets of data. Finding the mass difference is only relevant for sets of data that started with the same mass. In our experiment, however, all of our dialysis bags had different masses in the beginning. If we found the mass difference it would have little use to us since it s not relative to these sets of data (which had different starting masses). 4. A dialysis bag is filled with distilled water and then placed in a sucrose solution. The bag s initial mass is 20g, and its final mass is 18g. Calculate the percent change of mass, showing your calculations.

3 Tables and Graphs for Exercise 1 (Osmosis Lab): Contents in Bag Initial Final Difference % Change in Class Average Distilled Water 10.56g 10.58g 0.02g M 10.02g 11.03g 1.01g M 10.47g 12.52g 2.05g M 10.82g 12.01g 1.19g M 10.91g 12.37g 1.46g M 10.92g 13.13g 2.21g Osmosis as Observed in Sucrose-filled Dialysis bags 22.5 % Change in M 0.2M 0.4M 0.6M 0.8M 1.0M Sucrose Molarity Individual Results Class Average

4 Tables and Graphs for Exercise 2: Water Potential of Potato Cells Contents in Beaker Initial Final Difference Distilled Water 1.20g 1.29g 0.09g 7.5% % Change in 0.2M 1.08g 1.21g 0.13g 12.04% 0.4M 1.60g 1.68g 0.08g 5% 0.6M 1.23g 1.05g -0.18g % 0.8M 1.51g 1.24g % 1.0M 1.15g 0.8g % 20 Water Potential as Observed in Potato Cells % Increase/Decrease in M 0.2M 0.4M 0.6M 0.8M 1.0M Sucrose Molarity Individual Data

5 Conclusions: Although the most of the steps were followed correctly, there was one mistake that should be addressed, which was leaving the dialysis bags in the solutions over night. However, this did not seem to cause any errors in the experiment as the results corresponded with the theoretical results; that is, the results hypothesized using previously known ideas such the results of cells being immersed in hypotonic solutions, isotonic solutions, and hypertonic solutions. Our experiment, Exercise 1, reflects the theoretical results as the dialysis bags filled with 0.2M, 0.4M, 0.6M, 0.8M, and 1.0M sucrose gained mass while the bag filled with distilled water had relatively little change. The only deviating result, when compared to the class, that we had was the distilled water dialysis bag. However, I suspect this is due to, mainly, human error but it could also be due to the fact that water is constantly flowing in and out of the dialysis bag to keep equilibrium. The results for exercise 2 were mostly accurate except the for the potato submerged in distilled water. This potato only gained a mass of 7.5% which, in theory, should be higher than the 12.04% of the potato submerged in the 0.2M sucrose solution. Distilled water, which has less solutes than the 0.2M sucrose solution, should have made the potato gain much more weight as the difference in solutes is higher. Aside from this unexpected result, the other samples appeared to be quite persistent with the intended results.