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AP Biology Lab 1 Osmosis and Diffusion Lab Activity Student Study Guide BACKGROUND

DID YOU KNOW? Osmosis is fundamental in many industrial processes such as water purification, desalination, and waste material treatment.

If a bottle of perfume is opened at one end of the room, in a fairly short time someone at the other end of the room will be able to detect the scent. How does this happen? It depends on the movement of molecules. All molecules in gases and liquids are in a state of constant, random motion. The molecules in the perfume, released from the bottle, travel through the air. When molecules are closely packed together, they collide with other molecules, their kinetic energy is redirected, and the molecules move away from the point of the collision. As a result, molecules tend to distribute themselves evenly throughout the area they occupy. When the molecules are evenly distributed, they reach dynamic equilibrium. Individual molecules are still in motion, but there is no net movement of molecules; as many molecules move in one direction as they do in the opposite one. This movement from a higher concentration to a lower concentration is known as diffusion.

Figure 1

Water Sugar

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© 2002 WARD’S Natural Science Establishment, Inc. All Rights Reserved

Diffusion also occurs in water separated by a selectively permeable membrane, with different solute concentrations on either side of the membrane; this special kind of diffusion is known as osmosis. In osmosis, water moves from regions of low solute concentration to regions of high solute concentration. Because systems always tend toward equilibrium, this movement occurs to even out the concentrations on the two sides of the membrane (Figure 1). Only a solute’s relative concentration, or water potential, affects the rate of osmosis; the size of the solute molecules does not matter. The higher the concentration of solutes, the faster water will flow through the membrane to equalize the concentration. This is due to the higher water potential of the solution.

Isotonic solution: The solute concentration outside of the cell is the same as the solute concentration inside the cell.

Hypotonic solution: The solute concentration outside of the cell is lower than the solute concentration inside the cell.

Hypertonic solution: The solute concentration outside of the cell is higher than the solute concentration inside the cell.

Osmosis occurs in both plant and animal cells, across a cell’s plasma membrane. The membrane allows the free age of some materials while restricting the diffusion of others. Small, neutrally charged molecules such as oxygen, carbon dioxide, water, and glucose can freely through the plasma membrane. Diffusion is vital to many life functions of a cell, it allows the transport of vitally important nutrients and compounds without the expenditure of excess metabolic energy. Diffusion is also responsible for the transport and exchange of oxygen and carbon dioxide in the lungs. During respiratory gas exchange, oxygen is broken down and carbon dioxide produced. As a result, the concentration of oxygen is always lower inside the cell than outside, while the concentration of carbon dioxide is higher inside. The gases travel down their respective concentration gradients through the body’s intercellular fluid. Oxygen is picked up by the oxygen-poor blood as it es through the lungs and diffuses into the body cells as carbon dioxide diffuses out. In plants, diffusion aids in the transport of water and nutrients in the xylem. A plant’s root cells expend energy to maintain a high solute concentration in the xylem, which in turn forms a concentration gradient within the roots for water; a higher solute concentration equals lower water concentration. Water flows down the concentration gradient, causing it to diffuse into the roots. Water pressure builds up within the xylem, forcing it up the stem. Osmosis allows for the absorption and early transport of water into the root system of plants and, with transpirational pull, helps transport water in the xylem. Osmosis also has a major impact on living cells. 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 solution is hypotonic if it has fewer dissolved particles than the other solution and hypertonic if it has more particles than the other solution. If the number of dissolved particles are the same in the two media, it is called isotonic.


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Figure 2 Plant Cells Hypertonic

DID YOU KNOW? Osmosis was originally discovered by a French physicist, Abbé Jean Antione Nollet, in 1748.

Isotonic

Hypotonic

If a cell is hypertonic to its surroundings, water rushes into the cell, causing it to expand. If this occurs in an animal cell, the cell bursts, or cytolizes. In a plant cell, however, the rigid wall of the cell holds the cell together. Eventually the pressure exerted by the wall on the water in the cytoplasm will equal the osmotic potential of the solution, and the net movement of water will become zero. Water will still be moving, but the amount moving into the cells from osmosis is the same as the amount being forced out by the cell wall pressure. This pressure exerted on the cell wall by the water is known as turgor; it is responsible for ing the stems and leaves of plants. If a cell is hypotonic to its surroundings, it loses water, causing it to shrink. This is called plasmolysis. In plants, the plasma membrane will pull away from the cell wall; the loss of pressure on the cell wall is what causes plants to wilt. Animal cells, when plasmolyzed, lose shape and , and shrivel up. Animals in hypertonic media, like saltwater fish, must find ways to retain their water, while animals in hypotonic media, like paramecia in freshwater ponds, must develop mechanisms to pump out the excess water they absorb. Rates of diffusion and osmosis, and the water potential of a solution, can all be determined through lab activities and calculations. The osmotic potential is calculated using known concentrations of solution. A solution of sugar and water is placed in a tube and the tube is covered by a selectively permeable membrane. If this tube is placed in a beaker of pure water, the water will diffuse into the tube and the level of the water will rise. Determining how high the water rises helps ascertain the water potential of the solution. Water potential determines the direction and rate of osmosis. It consists of two components: pressure potential, which is the exertion of pressure on a solution, and osmotic potential, which is the relative concentration of solutes within the two solutions.


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DID YOU KNOW? It is because of diffusion that just one quart of oil from an oil spill can contaminate up to 250 thousand gallons of drinking water or cause an oil slick nearly 2 acres in size.

The water initially enters the tube because there is a negative osmotic potential in the sugar–water solution. However, the force of gravity begins to exert pressure on the rising column. When the force of gravity (pressure potential) equals the osmotic potential, the column stops rising. The water potential is at zero and an equilibrium has been established. The pressure potential can be determined from the height of the column. With the water potential and the pressure known, osmotic potential can be easily determined. Water Potential (ψ) = Pressure Potential (ψp) + Osmotic Potential (ψπ) The water potential of a solution is determined by calculating the osmotic potential when the pressure potential is zero. ψ = ψp + ψπ; if ψp = 0, then ψ = ψπ


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OBJECTIVES • • • • •

Demonstrate diffusion across a semi-permeable membrane Measure the effects of various concentrations of solute on the process of osmosis Differentiate between hypotonic, isotonic, and hypertonic environments Measure and calculate water potential Examine the effects of osmosis on plant cells

MATERIALS MATERIALS NEEDED PER GROUP 4 1 1 6

DID YOU KNOW? Frogs never drink water. They absorb water from their environment through the process of osmosis.

Glucose indicator strips Graduated cylinder Plastic cup, 250 ml dialysis tubing, 7 ft. Plastic cups, 250 ml Glucose starch solution, 15 ml IKI (potassium iodide) solution, 1 ml Sucrose solution, 0.2 M, 175 ml Sucrose solution, 0.4 M, 175 ml Sucrose solution, 0.6 M, 175 ml Sucrose solution, 0.8 M, 175 ml Sucrose solution, 1.0 M, 175 ml Distilled water, 175 ml Paper towels Potato Plastic wrap Microscope slide Coverslip Compound microscope Scalpel Forceps

SHARED MATERIALS Balance Compound microscope Cork borer with plunger NaCl solution, 15% Onion epidermis


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PROCEDURE As a general laboratory practice, it is recommended that students wear proper protective equipment such as gloves, safety goggles, and a lab apron to avoid staining any clothing or skin.

A. Diffusion 1. Pour 15 ml of the prepared glucose/starch solution into a graduated cylinder.

NOTE

Solute: The constituent of a solution that is dissolved in the solvent.

The pores in the dialysis tubing are extremely small, and can be easily clogged by any oil or dirt on your fingers and hands. Wash your hands before handling the dialysis tubing, and keep physical with the tubing to a minimum.

2. Obtain a piece of dialysis tubing that has been soaking in water. Tie a tight knot in one end of the tubing, or use a piece of string to tie off the end. Solvent: The constituent of a solution that dissolves the solute.

NOTE

3.

Concentration: The ratio of the mass or volume of a solute to the mass or volume of the solvent.

If you choose to tie off the end of the dialysis tubing with string, tie two knots, about 1/4” apart, to prevent leaking.

Open the tubing by rubbing the untied end between your fingers. Pour 15 ml of the glucose/starch solution into the tubing.

4. Note the color of the solution in the bag. Record the color in Table 1 in the Analysis section. 5. Determine if glucose is present in the tubing by dipping a glucose indicator strip into the solution. Record the data in Table 1. 6. Carefully tie a knot in the open end to form a bag. Be sure to leave enough space in the bag for expansion. 7. Fill a 250 ml beaker or a provided plastic cup approximately 2/3 full with distilled water. Add 1 ml of potassium iodide (IKI) to the beaker.

!

CAUTION

The IKI solution is an irritant; it affects skin and eyes, and can stain clothing. Handle the solution with caution. Wash off spills and splashes with water.

8. Note the color of the solution. Record the color in Table 1.


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9. Determine if glucose is present in the beaker by dipping a glucose indicator strip into the solution in the beaker. Record the data in Table 1. 10. Completely immerse the dialysis bag in the solution in the beaker. 11. Wait 30 minutes.

NOTE

Though a color change may become apparent after a few minutes, be sure to allow the tubing to remain in the beaker for 30 minutes to allow the glucose to diffuse out into the medium.

12. Remove the dialysis bag from the beaker. Record the final color of the solutions in the bag and the beaker in Table 1. 13. Determine the glucose content in the beaker and in the dialysis bag using glucose indicator strips. To test the solution in the bag, make a small cut in the bag with a pair of scissors, and insert the indicator strip through this hole. Record the presence or absence of glucose in Table 1. DID YOU KNOW? Several sea-going birds, such as penguins and sea gulls, can drink salt water. They have special glands in their heads that remove and excrete salt, making it useful for consumption.

NOTE

Be sure to wash your hands thoroughly before leaving the laboratory.

B. Osmosis 1. Obtain six plastic cups and label them as follows: water, 0.2 M, 0.4 M, 0.6 M, 0.8 M, and 1.0 M. NOTE

The pores in the dialysis tubing are extremely small, and can be easily clogged by any oil or dirt on your fingers and hands. Wash your hands before handling the dialysis tubing, and keep physical with the tubing to a minimum.

2. Obtain six pieces of dialysis tubing from the beaker of water. Tie a tight knot in one end of the tubing, or use a piece of string to tie off the end. 3. Open one piece of tubing by rubbing the untied end between your fingers. Pour 25 ml of distilled water into the tubing and carefully tie a knot in the open end to form a bag. Be sure to leave enough space in the bag for expansion. Blot the sealed tubing dry and place it in the cup labeled “water”. NOTE


If at any time the dialysis tubing becomes too dry during the procedure, simply rewet the tubing by placing in water for several seconds and continue setting up your experiment. 7


4. Repeat this procedure with the remaining five pieces of dialysis tubing, adding a different sucrose solution to each bag: 0.2 M, 0.4 M, 0.6 M, 0.8 M, and 1.0 M. Be sure to blot each piece of tubing dry and place it in the proper cup before proceeding with the next concentration. NOTE

5.

Be sure to tie off each of the bags so that they are all approximately the same length.

Carefully blot the bags dry and weigh each one. Record each bag’s initial mass in Table 2 in the Analysis section.

6. Fill the six plastic cups approximately 3/4 full of distilled water. Immerse one bag in each of the cups. Be sure each bag is in the properly labeled cup. 7.

DID YOU KNOW? Osmosis is a reversible thermodynamic process. In other words, the direction of water flow through a membrane can be reversed at any time by proper control of the external pressure on the solution.

Wait 30 minutes. Remove the bags from the cups. Blot them dry, and weigh each one again.

NOTE

To ensure accurate readings, the bags must be as dry as possible.

8. Record the final mass of each bag in Table 2. 9. Calculate the percent change in mass for each of the dialysis bags using the following formula: % Change = (Final Mass – Initial Mass)/Initial Mass x 100 Record this data in Table 2. 10. Graph both your individual and class average results on the graph paper in the Analysis section.

C. Water Potential NOTE

1.

Your instructor will assign you one or more sucrose solutions of varied concentrations and/or distilled water with which to perform the experiment. Your instructor may have prepared the potato cylinders in advance. If this is the case, begin with Step 2.

With the cork borer, cut four cylinders from a potato for each solution you will be using. Cut each cylinder to a length of 3 cm for greater accuracy; remove any skin from the cylinders. Place the cylinders in a covered cup or beaker.


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2. Weigh four cylinders together. Record the initial mass of the cylinders in Table 3 in the Analysis section. 3.

Pour 100 ml of one solution you have been assigned into a beaker or plastic cup. Take the initial temperatures and record in Table 3. Insert four potato cylinders and cover the beaker or cup with plastic wrap.

4. Repeat the procedure for the rest of the solutions you will be using, if any. Let the beaker(s) stand overnight. 5. On the following day, measure the final temperature of the liquid in the beaker(s). Record the temperature(s) in Table 3. 6. Remove the four cylinders from each beaker. Blot them carefully with a paper towel and weigh them. Record the final mass of the four cylinders in Table 3. DID YOU KNOW? In 1885, Jacobus Henricus van’t Hoff developed a formula to prove that thermodynamic laws are not only valid for gases, but also for dilute solutions. His pressure laws are considered the most comprehensive and important in the realm of natural science.

7. Calculate the percent change in mass for the four potato cylinders using the following formula: % Change = (Final Mass – Initial Mass)/Initial Mass x 100 Record this data in Table 3. 8. Graph both your individual and class average results. Place the percent change in mass on the Y-axis and sucrose molarity on the X-axis. Using the graph, determine the molar concentration of the potato cylinders. This is equivalent to the sucrose molarity in which the potato core mass is constant.

D. Water Potential Calculation Osmotic potential can be calculated using the following formula: ψπ = –iCRT i = ionization constant (1 for sucrose) C = osmotic molar concentration (determined in part C) R = pressure constant (R = .0831 liter bars/mole °Kelvin) T = temperature (°Kelvin)

E. Plant Cell Plasmolysis 1. Cut an onion in half lengthwise, from top to bottom, and remove the center. Note the layers of the scale leaves in the onion. The epidermal layer, only one cell thick, between each scale is the layer that will be used in the experiment.


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2. Make several shallow cuts approximately 1 cm apart on one of the scale leaves. Make more cuts perpendicular to the first set, resulting in 1 x 1 cm squares. 3. Using forceps, carefully remove the epidermal layer from one of the squares. If the layer tears while being removed, take another layer from one of the other squares.

DID YOU KNOW? In 1901, Jacobus Henricus van’t Hoff received the first Nobel Prize in Chemistry for his work in chemical dynamics and osmotic electrical conductivity.

4. Place the layer in two to three drops of distilled water on a microscope slide. Place a coverslip on top. 5. Examine the cell layer under 100X magnification and note the characteristics of the cells. Draw several representative cells in the proper space in the Analysis section. 6. Remove the slide from the microscope and add two to three drops of 15% NaCl solution to one side of the coverslip. 7. Carefully touch the edge of a paper towel to the other side of the coverslip in order to draw the NaCl solution across the sample. 8. Allow the slide to sit for two to three minutes in the salt solution and re-examine the sample under the microscope. 9. Note the appearance of the cells. Draw several representative cells in the proper space in the Analysis section.


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WARD’S AP Biology Lab 1 Osmosis and Diffusion Lab Activity

Name: Group: Date:

ANALYSIS Table 1 Diffusion Color Time

Dialysis Bag

Glucose Content Beaker

Dialysis Bag

Beaker

Start 30 minutes

Table 2 Osmosis Investigation Solution

Dialysis Bag Initial Mass (g)

Dialysis Bag Final Mass (g)

Change in Mass (g)

% Change in Mass

Water 0.2 M 0.4 M 0.6 M 0.8 M 1.0 M


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Table 3 Potato Cell Water Potential Solution Temperature °C Solution

Initial Temp

Potato Cylinders

Final Temp Initial Mass (g) Final Mass (g)

Change in Mass (g)

% Change in Mass

Water 0.2 M 0.4 M 0.6 M 0.8 M 1.0M

Plant Cell Plasmolysis

Cells in Distilled Water


Cells in 15% NaCl

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WARD’S AP Biology Lab 1 Osmosis and Diffusion Lab Activity

Name: Group: Date:

ASSESSMENT

1.

Create a Venn diagram comparing osmosis and diffusion.

2.

Part A of the experiment was a demonstration of diffusion. Give an example of diffusion occurring in the setup. Do you think osmosis occurred in this part of the experiment? If you answered yes, explain why you believe this to be.

3.

Did the dialysis tubing serve as a selectively permeable membrane? Explain your answer.


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4.

In part B, what caused the mass of the dialysis bags to change? Was there more or less water in the dialysis bags at the conclusion of the experiment? Explain.

5.

Was the distilled water in the beakers hypertonic or hypotonic in relation to the sucrose solutions found in the dialysis bags?

6.

Suppose the dialysis bags were placed in beakers containing a 0.6 M sucrose solution as opposed to distilled water. How do you think your results would change? Sketch a graph below to show how the mass of each of the bags would be affected.

7.

Study the graph you have plotted for part C of the experiment. What is important about the point where the best fit line crosses the X-axis? What is the concentration of sucrose in your potato?


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8.

Fill in the blanks in the statement below using the following list: net gain, low, hypertonic, exit, osmotic, enter, less, hypotonic, high, more, net loss, isotonic (not all words are used) If a cell is placed in a ____________ solution, it has ____________ solute in solution than the surrounding fluid, and will therefore experience a ____________of water to its surroundings. This cell has ____________ water potential since there is a great deal of ____________ pressure causing water to leave the cell. Conversely, a cell sitting in a ____________ solution has a ____________ water potential, and since it will experience a ____________ of water, there will be little osmotic pressure causing water to ____________ the cell.

9.

Using the data from part C of this activity and the formula for water potential from part D, calculate the osmotic potential of the sucrose solution in bars.

10. The water potential (ψ) of a solution is equal to the osmotic potential (ψπ) plus the pressure potential (ψp). Since there is no differential pressure acting on the solution, the pressure potential is equal to zero, making the water potential equal to the osmotic potential. If the equilibrium point between the solutions and the potato cylinders indicates the point where the two water potentials are equal, what is the water potential of the potato cells?

11. Would the water potential of the potato cells change if the cylinders were allowed to dry out? In what way?


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12. What are the effects on cells when they are placed in a hypotonic solution, a hypertonic solution, and an isotonic solution?

13. Why can’t humans drink salt water for hydration?


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