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Name _____________________________ Period _________ AP Biology Date ______________________ 1 of 14 LAB 7. DIFFUSON AND OSMOSIS INTRODUCTION The life of a cell is dependent on efficiently moving material into and out of the cell across the cell membrane. Raw materials — such as oxygen and sugars — needed for the production of energy and other cellular products must enter the cell. Whereas waste products — such as carbon dioxide and ammonia — which are generated during energy production and cellular digestion must be removed from the cell. Most of these materials move passively — costing the cell no energy — through the process of diffusion . Diffusion is the movement of molecules from an area of higher concentration of those molecules to an area of lower concentration. A good metaphor for this molecular motion is what happens if you were to open a bottle of hydrogen sulfide gas (H 2 S smells like rotten eggs) in one corner of a room. It would not be long before someone in the opposite corner of the room would smell the odor of rotten eggs. The molecules appear to be moving from an area of higher concentration of H 2 S gas (the bottle) to an area of lower concentration of H 2 S gas (the opposite corner of the room). Eventually a dynamic equilibrium is reached; the concentration of H 2 S gas will be approximately equal throughout the room and no net movement of H 2 S will occur from one area to the other. Since all life takes place in water — either external waters or internal waters — we must also address the special case of the movement of water across cellular membranes. The diffusion of water through a selectively permeable membrane is referred to as osmosis . Water moves from a region of higher concentration of water to a region of lower water concentration. This is often also stated as movement from a region of higher water potential to a region of lower water potential. Distilled water (pure water) has the highest concentration of water or the highest water potential. However, diffusion and osmosis do not entirely explain the movement of ions and molecules into and out of the cell. Some molecules are valuable enough to the cell to expend energy transporting them across the cell membrane. This active transport uses energy from ATP to move substances through the cell membrane. Active transport usually moves substances against a concentration gradient, from regions of low concentration of that substance into regions of higher concentration.

Lab 41 Diffusion and Osmosis 2006

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Page 1: Lab 41 Diffusion and Osmosis 2006

Name _____________________________ Period _________

AP Biology Date ______________________

1 of 14

LAB 7. DIFFUSON AND OSMOSIS

INTRODUCTION

The life of a cell is dependent on efficiently moving material into and out of the cell across the cell membrane. Raw materials — such as oxygen and sugars — needed for the production of energy and other cellular products must enter the cell. Whereas waste products — such as carbon dioxide and ammonia — which are generated during energy production and cellular digestion must be removed from the cell. Most of these materials move passively — costing the cell no energy — through the process of diffusion.

Diffusion is the movement of molecules from an area of higher concentration of those molecules to an area of lower concentration. A good metaphor for this molecular motion is what happens if you were to open a bottle of hydrogen sulfide gas (H2S smells like rotten eggs) in one corner of a room. It would not be long before someone in the opposite corner of the room would smell the odor of rotten eggs. The molecules appear to be moving from an area of higher concentration of H2S gas (the bottle) to an area of lower concentration of H2S gas (the opposite corner of the room). Eventually a dynamic equilibrium is reached; the concentration of H2S gas will be approximately equal throughout the room and no net movement of H2S will occur from one area to the other.

Since all life takes place in water — either external waters or internal waters — we must also address the special case of the movement of water across cellular membranes. The diffusion of water through a selectively permeable membrane is referred to as osmosis. Water moves from a region of higher concentration of water to a region of lower water concentration. This is often also stated as movement from a region of higher water potential to a region of lower water potential. Distilled water (pure water) has the highest concentration of water or the highest water potential.

However, diffusion and osmosis do not entirely explain the movement of ions and molecules into and out of the cell. Some molecules are valuable enough to the cell to expend energy transporting them across the cell membrane. This active transport uses energy from ATP to move substances through the cell membrane. Active transport usually moves substances against a concentration gradient, from regions of low concentration of that substance into regions of higher concentration.

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Name _____________________________ AP Biology

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A. DIFFUSION

In this experiment you will measure the diffusion of small molecules through dialysis tubing, an example of a selectively permeable membrane. Small dissolved molecules (solutes) and water molecules can move freely through a selectively permeable membrane, but larger molecules will pass through more slowly, or perhaps not at all. The size of the minute pores in the dialysis tubing determines which substances can pass through the membrane. We will explore the process of diffusion through a semi-permeable membrane in this activity.

PROCEDURE

1. Obtain a 30cm piece of 2.5cm dialysis tubing that has been soaking in water. Tie off one end of the tubing to form a bag. To open the other end, rub the end between your fingers until the edges separate.

2. Obtain a sample of the 15% glucose/1% starch solution. Test the solution for the presence of glucose using Benedict’s solution. Record the results in Table 1 (Predictions) and Table 2 (Experimental Observations).

3. Obtain a second sample of the 15% glucose/1% starch solution. Test the solution for the presence of starch using Lugol’s iodine (IKI). Record the results in Table 1 and Table 2.

4. Place 15mL of the 15% glucose/1% starch solution in the bag. Tie off the other end of the bag, leaving sufficient space for expansion of the contents in the bag. Record the color of the solution in Table 1 and Table 2.

5. Fill a 250mL beaker two-thirds full of water. Add approximately 4mL of Lugol’s solution to the water and record the color of the solution in both Table 1 and Table 2. This also served as a test of the water for the presence of starch. Record the results in both Table 1 and Table 2.

6. Take a sample of the water and Lugol’s solution from the beaker. Test the solution for the presence of glucose. Record the results in Table 1 and Table 2.

7. Immerse the dialysis bag in the beaker of IKI solution.

8. Complete Table 1. Predictions.

9. Allow the set up to stand overnight. Record the final color of the solution in the bag and in the beaker in Table 2.

10. Test the liquid in the beaker and in the bag for the presence of glucose. Record the results for glucose and for starch in Table 2.

11. Compare your predictions to your actual data. Any surprises? If so, discuss with the class.

12. Complete the Summary Questions for this section of the lab.

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TABLE 1. PREDICTIONS Color of Solution Presence of Glucose Presence of Starch

Initial

contents Initial Predicted

Final Initial

Predicted Final

Initial Predicted

Final

Bag 15% glucose

1% starch

Beaker H2O & IKI

TABLE 2. EXPERIMENTAL OBSERVATIONS Color of Solution Presence of Glucose Presence of Starch

Initial

contents Initial Final Initial Final Initial Final

Bag 15% glucose

1% starch

Beaker H2O & IKI

Use the diagram below to illustrate the contents of the bag and the beaker as well as the movement of molecules during the experiment.

INITIAL STATE FINAL STATE

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B. OSMOSIS

Using the principles you learned in the first exercise, we will now investigate the movement of water in and out of an actual cell rather than merely a model. Water moves from areas of higher concentration of water to a region of lower water concentration. Water is at its highest concentration as distilled water. The concentration of water decreases as solutes are dissolved in the water.

If a model cell (like our dialysis tubing) is filled with a salt solution and placed in a beaker of distilled water, water will diffuse into the cell until a dynamic equilibrium is reached. At that point, no net movement of water will occur between the cell and the beaker. If a model cell is filled with a distilled water and placed in a beaker of salt solution, water will diffuse out of the cell until a dynamic equilibrium is reached. Once again, at that point, no net movement of water will occur between the cell and the beaker.

Let us now use these principles to investigate the movement in and out of plant cells using potato.

PROCEDURE

1. Prepare 1 liter of each of the following sucrose solutions:

0.0M 0.2M 0.4M 0.6M 0.8M 1.0M 2. Pour 100mL of each solution into a labeled 250mL beaker.

3. Slice a potato into discs that are approximately 5cm thick. Use a cork borer to cut four potato cores for each sugar solution beaker. Do not include any skin on any of the potato cores.

4. Determine the mass of the four potato cores together. If there is any wait time for a balance, keep your potato cores in a covered beaker until it is your turn. Record the mass of each set of potato cores in Table 3.

5. Put a set of four cores into the appropriate beaker of sucrose solution. Cover each beaker with plastic wrap to prevent evaporation. Let stand overnight.

6. Remove the cores from each beaker, blot them gently on a paper towel, and determine the mass of each set of four cores. Record the final mass in Table 3. Calculate the percent change in mass for each sucrose solution and record in Table 3.

Final Mass – Initial Mass Percent Change in Mass =

Initial Mass X 100

7. Average the class data.

8. Graph both your individual results and the class average for the percentage change in mass.

9. Complete the Summary Questions for this section of the lab.

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TABLE 3: POTATO CORE INDIVIDUAL DATA

Contents in Beaker

Initial Mass Final Mass Mass

Difference Percent Change

in Mass

0.0M water

0.2M sucrose

0.4M sucrose

0.6M sucrose

0.8M sucrose

1.0M sucrose

TABLE 4: POTATO CORE CLASS DATA

Percent Change in Mass of Potato Cores Contents in

Beaker Group 1 Group 2 Group 3 Group 4 Group 5 Total Class

Average

0.0M water

0.2M sucrose

0.4M sucrose

0.6M sucrose

0.8M sucrose

1.0M sucrose

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Title _________________________________________________________________________

_____________________________________________________________________________

10. From the graph, determine the sucrose molar concentration equivalent to the molarity of the potato cores. __________________________

Explain how you determined this answer.

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C. PLANT CELL PLASMOLYSIS

Cells lose or gain water due to the difference in solute concentrations between the cytoplasm (the intracellular fluid) and the solution surrounding the cell (the extracellular fluid). The movement of water in and out of a cell is governed by the laws of diffusion: water flows from a region of higher water concentration to a region of lower concentration.

When a cell is in a hypertonic solution, it will experience a net loss of water. A hypertonic solution contains a higher concentration of solutes than the cell and therefore a lower concentration of water. Consequently, water will flow out of the cell from the region of higher water concentration to the region of lower concentration.

When a cell is in a hypotonic solution, it will experience a net gain of water. A hypotonic solution contains a lower concentration of solutes than the cell and therefore a higher concentration of water. Consequently, water will flow into the cell from the region of higher water concentration to the region of lower concentration.

When a cell is in an isotonic solution, it will experience neither a net gain or loss of water. A isotonic solution contains an equal concentration of solutes as the cell and therefore an equal concentration of water. Consequently, water will flow equally into and out of the cell.

Plasmolysis is the shrinking of the cytoplasm of a plant cell in response to diffusion of water out of the cell and into a hypertonic solution surrounding the cell as shown in Figure 2. During plasmolysis the cell membrane pulls away from the cell wall. In this lab exercise, you will examine this process by observing the effects of a highly concentrated salt solution on plant cells.

Figure 2.

HYPOTONIC SOLUTION ISOTONIC SOLUTION HYPERTONIC SOLUTION

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PROCEDURE

1. Prepare a wet mount of an Elodea leaf. Locate a good section of leaf under low power and then observe under high power. Sketch a single Elodea cell in the space below and describe the appearance of the plant cell.

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2. Add 2 or 3 drops of 15% NaCl to one edge of the cover slip. Draw the salt solution across the slide by touching a piece of paper towel to the fluid under the opposite edge of the cover slip. Observe the plant cells in the microscope while you draw the salt water across the slide. Sketch a single Elodea cell in the space below and describe what has happened to the plant cell.

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3. Remove the cover slip and flood the Elodea leaf with fresh water. Observe under high power. Describe and explain what has happened.

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4. Complete the Summary Questions for this section of the lab.

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SUMMARY QUESTIONS

A. DIFFUSION

1. Which substance(s) are entering the dialysis bag and which are leaving the bag? What experimental evidence supports your answer?

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2. Explain the results you obtained by discussing concentration differences and membrane pore size.

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3. Although we didn’t measure it, what other molecule can we assume also moved across the membrane? ________________________________

4. In assessing the movement of glucose, starch and iodine across the membrane in this experiment, you generated qualitative data. Quantitative data use numbers to measure observed changes. How could this experiment be modified so that quantitative data could be collected to show that water diffused into the dialysis bag.

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5. Based on your observations, compare the size of each of the following molecules with the membrane pore size: glucose molecules, water molecules, IKI molecules, starch molecules

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6. What results would you expect if the experiment was set up incorrectly: a glucose and IKI solution was placed inside the dialysis bag and a starch and water solution was in the beaker.

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7. In many animals, glucose, rather than starch, is transported by the blood through the body to all cells. In the digestive system, starches are digested by amylase to yield glucose. Based on the findings of this lab, explain why the digestion of starch to glucose is necessary.

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8. Summarize the process of diffusion.

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B. OSMOSIS

9. Zucchini cores placed in sucrose solutions at 27°C resulted in the following percent changes after 24 hours:

Sucrose Molarity % Change in Mass

0.0M 20%

0.2M 10%

0.4M -3%

0.6M -17%

0.8M -25%

1.0M -30%

Graph the results.

Title _________________________________________________________________________

_____________________________________________________________________________

10. From the graph, determine the sucrose molar concentration

equivalent to the molarity of the zucchini cells. ____________________________________

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11. Why did you calculate percent (%) change in mass of the potato cores rather than use the change in mass directly?

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C. PLANT CELL PLASMOLYSIS

12. What is plasmolysis?

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13. In the winter, icy roads are often salted to remove the ice and make them less slippery. Grasses and other herbaceous plants often die near the side of these roads. What causes this to happen?

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14. When a person is given fluid intravenously (an I.V.) in the hospital, the fluid is typically a saline solution isotonic to human body tissues. Explain why this is necessary.

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15. What if the unthinkable happened at the hospital! A patient was given an I.V. bag with distilled water in it rather than saline solution. Describe what would happen to their red blood cells and explain why this would occur.

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16. Many freshwater one-celled organisms, like Paramecium, have contractile vacuoles. These structures collect and pump out excess water that accumulates in the cell. Explain why these organisms needs such a structure.

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17. Explain why contractile vacuoles would be of little value to one-celled organisms living in salt water.

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18. Popcorn sold at movie theatres is very salty, causing people to become thirsty and to buy soft drinks. Explain why salty popcorn causes this thirst.

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19. Draw a molecular diagram of the cell membrane. Illustrate the following processes and the cellular structures that are involved:

a. diffusion of water molecules across the cell membrane

b. diffusion of polar molecules (i.e., sodium ion) across the cell membrane

c. diffusion of non-polar molecules (i.e., lipid hormone) across the cell membrane