Pedro Rodriguez

Mrs. Pita

AP Biology Period 6

30 September 2013

AP Biology Investigation 4: Diffusion and Osmosis

Abstract:

This lab addresses the properties of osmosis and diffusion and their function in maintaining homeostasis in the cell. We used two phospholipid bilayer models to simulate the movement of water and nutrients across a cell membrane and observe osmosis in living tissue. In Part 3, we directly observe osmosis in a living specimen. In all parts of this lab, after performing a guided activity, we are then directed to design their own experiments, to further develop our understanding of the topics explored. Our understanding of these exercises will allow us to explain how cell size and shape affect rates of diffusion, as well as pose scientific questions about the selective permeability properties of cell membranes.

Introduction:

The absorption of nutrients, excretion of cellular wastes, and the exchange of respiratory gases are life processes which depend upon efficient transport of substances into, out of, and throughout living cells. Diffusion is one of the most common and efficient means by which substances are passively transported between cells and their aqueous environment. Diffusion is the movement of a substance (liquid or gas) along a concentration gradient from high to low concentration. Diffusion is vital to many life functions of a cell. Diffusion allows the transport of vital nutrients and compounds without the expenditure of energy.

The cell membrane is the selectively permeable barrier whose total surface area is important to regulating the substances that diffuse into or out of the cell. Small, neutrally charged molecules such as oxygen, carbon dioxide, and glucose can pass freely through the membrane, while the diffusion of other materials is restricted. Materials that cannot diffuse across the membrane or need to be transported against a diffusion gradient can be actively transported across the membrane with the expenditure of energy. Osmosis is a special kind of diffusion that occurs as water is separated by a selectively permeable membrane with different solute concentrations on either side of the membrane. During osmosis, water moves from regions of low solute concentration to regions of high solute concentration without the expenditure of energy.

Organisms rarely exist in environments with solute concentrations that match their cytoplasm; there are usually more or fewer dissolved particles in one of two compared solutions separated by a membrane, such as a cell and the media in which it exists. A hypertonic solution is a solution in which the solute concentration is higher

outside of the cell; therefore, water will flow to the external environment, causing the cell to shrink. Hypotonic solutions consist of a low concentration of solutes outside the cell; therefore, water will flow into the cell, causing cellular expansion. In the case of plant cells with cell walls, expansion is restricted, so pressure builds. This pressure is called turgor pressure.

Isotonic solutions, on the other hand, are solutions in which the solute and solvent concentrations are at equilibrium: there is no net flow of materials across the selectively permeable membrane.

Only a solute’s relative concentration, or water potential (y), affects the rate of osmosis. Water potential consists of two components – pressure potential (y p) the exertion of pressure on a solution; and solute (or osmotic) potential (ys), the relative concentration of solutes within the two solutions.

Purpose:In this lab, you will microscopically observe an Elodea densa plant leaf and explore the effects of different solution concentrations on the cells. You will then use the solutions to determine the water potential of plant tissues, such as white or sweet potato tubers.

Hypothesis: If a plant cell is placed in a solution then the concentration of the solution will determine whether the plant cells will either lose water, causing plasmolysis, or gain water, increasing its turgor pressure.

Materials: 1 Thermometer 1 Graduated cylinder 6 Plastic cups 150 mL Red mystery solution 150 mL Orange mystery solution 150 mL Yellow mystery solution 150 mL Green mystery solution 150 mL Blue mystery solution 175 mL Distilled water 1 Microscope slide 5 Paper towels 1 Pair of forceps 1 Compound microscope 1 Scalpel 1 Coverslip 6 Potato tubers 1 Balance 1 Ruler

Procedure:

Part 3A – Procedure: Structured inquiry

1. Using the forceps, remove an Elodea densa leaf from its stem and place it gently on a clean microscope slide.

2. Add two to three drops of distilled water to the slide and cover with a coverslip.

3. Examine the cell at 40X magnification and note the characteristics of the cells. In your lab notebook, draw several cells that show a good representation of the cells you observed. In your drawing, label all visible structures and organelles.

4. Remove the microscope slide. Choose one of the solutions from Part 2 of this lab. Add two to three drops of this solution across the leaf sample.

5. Allow the slide to sit for two to three minutes in the solution and re-examine the sample under the microscope.

-To speed up the reaction to the cells in solution, place a paper towel on the opposite end of the coverslip to wick your solution through the cells.

6. Note the appearance of the cells. In your lab notebook, draw several cells that show a good representation of the cells you observed. In your drawing, label all visible structures and organelles.

Part 3B– Procedure: guided inquiry

1. Make potato cores with a borer or use pre-made potato bores.

2. Weigh each core and measure the length of each core. Record your data in your laboratory notebook.

3. Place one or more potato cores in each of the mystery sucrose solutions.

4. Record your observations.

5. Wait 30 minutes.

6. After 30 minutes, re-weigh the cores, and calculate the changes in their weight. Record your data in your laboratory notebook.

Results/Data Collection/Analysis:

Part 3B:

Solution Initial Mass (g)

Final Mass (g) Change in Mass (g)

% Change in Mass

Mystery Blue 0.6 0.7 0.1 16.7%Mystery Red 0.5 0.6 0.1 20%Mystery Orange 0.5 0.6 0.1 20%Mystery Yellow 0.3 0.1 -0.2 -66.7%Mystery Green 0.3 0.3 0 0%Mystery Clear 0.3 0.3 0 0%Solution Molar Solute Pressure Water

Concentration Potential Potential Potential Mystery Blue 0.2 -4.54 0 Bars -4.54Mystery Red 0.4 -9.07 0 Bars -9.07Mystery Orange 0.6 13.6 0 Bars -13.6Mystery Yellow 1.0 -22.7 0 Bars -22.7Mystery Green 0.8 -18.21 0 Bars -18.21Mystery Clear 0 0 0 Bars 0Potato Core 0 0 0 Bars 0

Assuming the solute concentration of the solution at equilibrium is .3 M, what is the water potential of the potato core? Show Calculations. Water potential= .3 + 0 = .3

Conclusions and Discussion:

On the whole, our results include for the most part a positive change in mass as well as many negative water potentials. The mass of the cells put in the each of the different solutions generally increased due to diffusion through the semipermeable membrane from the areas of higher concentration (outside the bag) to the areas of lower concentration (inside the bag). The only exception was the cells put in the Albumin/Salt solution: a loss of mass occurred due to diffusion from the area of higher concentration (inside the bag) to the area of lower concentration (outside the cell). Errors that could have taken place include placing the wrong bag in the wrong solution, improperly tying the bag, or even adding foreign substances to the solutions. The results do in fact agree with the hypothesis as the concentration of the solutions changed the mass of the cells through diffusion. Suggestions for improvement may include working more efficiently in the lab stations by having each person doing solely one task such as having one person weighing the cubes, one person timing the process, and the last person preparing the next experiment with the Elodea plant cell.

Questions

1. How does the size of the potato core affect the rate of diffusion?2. Does the kind of potato affect rate of diffusion?3. Does the surface area of the cell affect the way it diffuses?4. What type of potato allows the diffusion to happen the quickest?5. Does the potato skin help slow or quicken diffusion? 6. Does the amount of time the potato cores spend outside affect the rate of

diffusion?7. Does food coloring play a role in the diffusion of the cell?8. How does the water potential affect turgor in the cell?9. How would diffusion occur in other types of solutions?10. How would this diffusion occur in different concentrations but in the same

plant cells?

11. Would the rate of diffusion be different in other kinds of plant cells besides Elodea densa?

12. What other types of solutions would cause the Elodea cells to act differently than they did in this experiment?

Part 3C-Open Inquiry:

Introduction:

                 This experiment was performed to demonstrate the process of osmosis and to show visible as well as quantitative evidence proving that osmosis occurred. Through the tasks of determining the percent concentrations in two different solutions, we were studying the process of osmosis. Osmosis is the best way to perform this experiment because as we went through the experiment, the weight of the beaker/dialysis tubing changed and the only logical explanation was that diffusion of water had occurred. Osmosis is the diffusion of water. Depending on which was heavier (the beaker or the dialysis tubing) after the experiment was performed, the direction of water diffusion was apparent. If the beaker was heavier, then that implies that the water diffused from the dialysis tubing to the beaker. In contrast to this process, if the dialysis tubing was heavier after the experiment, then the water would have diffused from the beaker to the dialysis tubing. We saw both these processes of osmosis occur in our experiments because we had two different solutions.

Materials and Methods:

Lab Procedure:

Materials include dialysis bag, water, sucrose, and beakers1. We filled the dialysis bag with approximately 10 grams of unknown solution.

2. We filled the beaker with about 200 grams of water and 2 grams of sucrose.

3. Then we completely submerged the dialysis bag into the beaker with solution and it remained in there for twenty minutes.

4. Next we removed the dialysis bag from the beaker, dried it off, and measured its mass.

5. We used the mass of the dialysis bag that was obtained, and determined the grams of sucrose and water present in the beaker. We did this by calculating the concentration of the unknown solution, and set up a proportion whereby x was equivalent to the grams of sucrose originally in the dialysis bag.

                After determining the mass of starch and water inside the beaker and the mass of unknown solution inside the dialysis tubing both before and after osmosis, we can perform some data analysis on the results: Because we know that only water can diffuse

through the membrane, the gain in mass of solution inside the dialysis tubing is exactly equal to the loss of water from the beaker.

Thus, based on the before and after masses of solutions A (table A) and B (table B), we have that solution A originally had a percent composition by mass of 1.0% and that solution B originally had a percent composition of 0.9%. However, these results may be inaccurate due to many sources of error that will later be discussed.  

DISCUSSION:

                The experiment was designed to enable us to calculate the percent concentration of starch in solutions A and B.  Using a procedure involving osmosis occurring between the unknown substances and a solution of known concentration, we were able to calculate the percent composition of starch in solutions A and B.  As the results demonstrate, we determined that A originally had a percent composition by mass of 1.0% and that solution B originally had a percent composition of 0.9%.

                When solution A was being tested, it was observed that the dialysis tubing filled with solution A weighed less after osmosis than it did before osmosis.  As the dialysis tubing is semi-permeable and allows only water through, we can conclude that water diffused out of the tubing and into the beaker.  This indicates that solution A had a lower concentration of starch than that of the solution in the beaker—however, this was not supported by our calculations.

                When solution B was being tested, the dialysis tubing filled with solution B weighed more after osmosis than it did before osmosis.  It can be concluded, therefore, that water diffused into the tubing and out of the beaker.  This indicates that solution B had a greater concentration of starch than that of the solution in the beaker.  This was supported by the calculations made.

                However, there were possible factors for error in this experiment.  The fact that the procedure was nonstandard may have had some impact; the amount of time that was needed for osmosis to occur fully was unknown.  The tubing may not have been submerged in the beaker for long enough to reach equilibrium. 

Conclusion:

                The word osmosis refers to the diffusion of water across a selectively permeable membrane in order to evenly distribute concentration levels on both sides of the membrane. In this lab, we have used specific materials and methods to present this process. Our hypothesis, in which we predicted that by figuring out the mass differences we would be able to figure out the percent composition of the starch in both the solutions, was correctly proven. We stimulated our process and were able to correctly form a reasonable conclusion and percent composition. Our process used the aspect of osmosis and diffusion and provided us with a great lab.