Materials Included In Kit

Additional Materials Required

Prelab Preparation

Part I

Prepare the 15% sucrose solution by weighing out 90 g of sucrose and placing it into a graduated beaker or Erlenmeyer flask (600-mL or larger). Fill to the 600-mL mark with DI water and stir to dissolve sucrose. Prepare the 30 % sucrose solution in a similar fashion with 150 g of sucrose and DI water up to the 500-mL mark.

Cut three 6–7" lengths of dialysis tubing and soak in DI water for about 10 minutes.

Part II

The material quantities listed above presume a class of 30 students working in pairs, with each pair of students working with two sucrose solutions. Table 1 chart details the proportions for making each of the sucrose solutions. The procedure for each is: add approximately 500 mL of distilled or deionized water to a 1000-mL volumetric flask, weigh out required amount of sucrose and add it to the flask. Stir or shake to dissolve sucrose. Fill to 1000-mL mark with water and invert several times to mix.

Safety Precautions

Although the materials in this activity are considered nonhazardous, please use all normal laboratory safety precautions. Use caution when cutting. Remind students to wash their hands thoroughly with soap and water before leaving the laboratory. Please review current Safety Data Sheets for additional safety, handling and disposal information.

Disposal

Please consult your current Flinn Scientific Catalog/Reference Manual for general guidelines and specific procedures, and review all federal, state and local regulations that may apply, before proceeding. The sucrose solutions may be disposed of by disposing of down the drain with plenty of excess water according to Flinn Suggested Disposal Method #26b. The leftover potatoes and the cores may be disposed of in the regular trash.

Lab Hints

• Enough materials are provided in this kit for 30 students working in pairs or for 15 groups of students plus one demonstration. Both parts of this laboratory activity can reasonably be completed in one 50-minute class period. The Membrane Diffusion Worksheet may be completed the day after the lab.
• In Part I. Initiate the demonstration at the beginning of class and allow it to continue while you discuss the concepts and principles of osmosis and diffusion.
• In Part I. A drop of food dye placed in the 15% sucrose solution used to fill the dialysis bags will enhance visibility of the liquid levels in the pipets.
• In Part I. Concurrent with this demonstration, fill two beakers with hot and cold water respectively. Allow the water to become still and gently add a drop of food dye to each beaker. Observe as the dye diffuses throughout the water in each beaker. Placing the beakers in front of a white background will make it easier for students to observe the results.
• In Part II. To save class time, use a cork borer to pre-cut cores from the potatoes. Cores should be longer than 4 cm so students can trim them to length. Seal the cores in an airtight bag or small container until class.
• In Part II. To see if sweet potatoes live up to their name, assign three teams of students (covering all six sucrose concentrations) cores from a sweet potato.

Sample Data

Baking Potato

Red Potato

Sweet Potato

Answers to Questions

1. Calculate the Percent Change in Mass for the two samples. The Percent Change in Mass is calculated using Equation 1.
   

   See Data Tables.

2. Obtain data from other members of the class and average all the values for each sucrose solution. Record these values in the data table.
   

   See Data Tables.

3. Plot a graph of Percent Change in Mass vs. Sucrose Solution Concentration. The graph should have positive and negative values on the y-axis with the zero line (no change in mass) serving as the x-axis.
   

   See Discussion for Part II.

4. Review the graph created in Question 3. How did soaking the potato cores in a sucrose solution affect the mass of the potato cores? Is there a correlation between the molarity of the sucrose solution and the change in mass? If so, how was the mass affected. Using the information in the Background section, construct a hypothesis as to what is occurring within the cells of the potato.
   

   See Discussion for Part II.

Discussion

Part I

The demonstration presents the three possible scenarios that an organism immersed in a liquid medium might face. In one case, the surrounding medium has a higher concentration of water; in the second, the surrounding medium has an equal concentration of water; in the third, the surrounding medium has a lower concentration of water.

The results of the demonstration show that the net movement of water is to the location where it is less concentrated. In the beaker of DI water, the net movement is into the dialysis bag, causing the liquid level in the pipet to rise. In the beaker of 30% sucrose, the net movement is out of the dialysis bag, causing the liquid level in the pipet to drop. In the beaker of 15% sucrose, there is little or no net movement evident as water is equally concentrated on both sides of the membrane. The case of the Paramecium is analogous to the beaker of distilled water in that the concentration of water within the Paramecium cell membrane is less than that outside the membrane. Saltwater fish present the opposite case as water is more concentrated in their body fluids than in the surrounding water. Saltwater fish must take in water while they simultaneously rid themselves of excess salts.

Biologists have terms for each of the three scenarios detailed above. In the case where the fluids inside and outside the cell are equally (or nearly so) concentrated, the surrounding fluid is said to be isotonic to the cell cytoplasm. If the surrounding fluid has a higher concentration of water (therefore a lower solute concentration) it is hypotonic to the cell cytoplasm. If the surrounding fluid has a lower water concentration (therefore a higher solute concentration), it is hypertonic to the cell cytoplasm. The series of diagrams below represents a graphic summary of what we’ve discussed. Each circle represents a cell and the arrows show the movement of water into and out of the cell. The double arrow represents a high rate of water movement, the single arrow a low rate. In the first diagram the net movement is into the cell, in the second there is no net movement, and in the third the net movement of water is out of the cell. With no corrective mechanism what is the ultimate fate of each cell?

This far we’ve spoken mostly in terms of water concentration. The concentration of water (the solvent) is dependent on the concentration(s) of materials (solutes) dissolved in it. A solution that is 10% salt and 90% water has a higher salt concentration and a lower water concentration than pure (i.e., 100% water). Chemists, by convention, look at, and define, solutions with respect to solute concentration and, as such, never speak in terms of water concentration. As defined by a chemist, osmosis is the movement of water (or other solvent) through a membrane in such a direction as to equilibrate the concentration of solutes on either side of that membrane.

Part II

Part II shows the effects of a range of solutions of different concentrations on a natural membrane system. The range of solutions was selected so that some would be hypotonic to the potato cores and some would be hypertonic to the potato cores (except the sweet potato). Some would gain water (those showing a positive change in mass) and some would lose water (those showing a negative change in mass). If you draw a best-fit line on the graph (from Post-Lab Question 3), you will see that there is a point where the line crosses the x-axis (the zero line, representing no change in mass). This value (in molarity of sucrose solution) represents the isotonic point (i.e., the point at which the concentration of water is equal inside and outside of the potato cells). Presumably a potato placed in this concentration of sucrose will neither gain nor lose water.

It is possible to precisely measure the length of the potato cores before and after soaking and see the same affect. If the potato cells are losing water or gaining water, the length of the cores should change as the cells shrink or expand. By graphing change in length, a similar value for the isotonic point should be found.

It may also have been observed that the texture of the potato cores varied, depending on which solution they were soaked in. Cores in hypotonic solutions (lower sucrose concentration) gained water and remained firm and crisp. Cores in hypertonic solutions (higher sucrose concentration) lost water and became shriveled and limp. Plant cells rely on absorbed water to retain their shape. This provides insight into why plants wilt if they are deprived of water. Plants do not have skeletons like we do and actually rely on water pressure (turgor in plant cells) to remain erect and keep their leaves oriented to the light.

References

Abramoff, P.; Thomson, R. G. Laboratory Outlines in Biology–V; W. H. Freeman: New York, 1991; pp 111–118.

Stern, K. R. Introductory Plant Biology, 7th ed.; Wm. C. Brown: Dubuque, IA, 1997; pp 144–146.

Green, N. P. O.; Stout, G. W.; Taylor, D. J. Biological Science: Vol. 2, 2nd ed.; Soper, R., Ed.; Cambridge University: Cambridge, 1990; pp 931–932.

Introduction

Osmosis is defined as the diffusion of water through a selectively permeable membrane from an area where water is more concentrated to an area where water is less concentrated. The enormous implications of this statement—for the simplest organisms to the most complex—make a working knowledge of the concept of osmosis essential for all students of biology.

Concepts

• Concentration gradient

• Osmosis
• Diffusion
• Permeable membrane

Background

Diffusion can be effectively demonstrated by observing the action of a drop of dye in a glass of water. The water molecules in a glass of water are in constant, random motion. If a drop of blue food dye is added to the glass, the dye begins to slowly diffuse throughout the water. The individual dye molecules disperse, and compelled by collisions with the moving water molecules, eventually become evenly distributed throughout the glass. The water now appears to be a uniform shade of blue. This process can be accelerated by warming the water—which increases the velocity of the water molecules, in turn increasing the rate of collisions and the speed with which the dye molecules diffuse. Cooling the water has the opposite effect.

The movement of individual molecules in the system just described is indeed random. The net movement of the dye molecules is directional, in the sense that they (initially) move from a region where they are highly concentrated (the dye droplet) to where they are less concentrated (the surrounding water). Thus, diffusion (the term typically refers to the net movement of molecules) is said to occur down a concentration gradient—from high to low. When the dye has become evenly distributed, and the dye and the water are at equilibrium, net movement of the dye molecules appears to have stopped. It is important to realize that the random movement of individual dye and water molecules continues.

All cells are enclosed within membranes that regulate the movement of material into and out of the cell. Some molecules are small enough to pass freely in and out of the cell. A Paramecium, like nearly all animal cells, requires oxygen to survive. The Paramecium has evolved no special apparatus (such as gills or lungs) and has no oxygen-binding compound (such as hemoglobin) to assist it in taking up oxygen. It relies solely on passive diffusion to replenish the oxygen it consumes. The oxygen diffuses from the surrounding water into the Paramecium, through its cell membrane. Permeability is something of a double-edged sword, however. In the case of the Paramecium, water is present outside the cell at a much higher concentration than inside the cell. Due to this concentration gradient, there will be a net movement of water molecules into the cell (water molecules are moving in both directions, but the movement into the cell is at a much higher rate). With no corrective mechanism, the Paramecium would quickly fill up with water and burst like a balloon. Fortunately, Paramecium and similar organisms have evolved a system of contractile vacuoles, which serve to collect and actively expel the incoming water.

The cell membrane of the Paramecium is said to be selectively permeable. A selectively permeable membrane is one that allows the passage of some molecules, while excluding others. Molecules can be excluded on the basis of size (they are simply too large to pass through the pores of the membrane), electrical charge or polarity. In Part I of this exercise we will use dialysis tubing—an artificial membrane made of cellulose—as a model for an actual cell membrane. Dialysis tubing is selectively permeable in that water and other small molecules can pass freely through its pores while sucrose and other large molecules cannot.

Experiment Overview

The following demonstration and experiment capitalize on and highlight the selective permeability of natural and artificial membranes to convey the fundamental principles of diffusion and osmosis. The laboratory activity demonstrates the osmosis of water across a live permeable membrane.

Materials

Safety Precautions

Although the chemicals used in this experiment are considered nonhazardous, please follow all laboratory safety guidelines. Scalpels and razor blades are very sharp. Use caution while cutting, cut away from yourself and others. Wash hands thoroughly with soap and water before leaving the laboratory.

Procedure

Part I. Demonstration

1. Arrange the support stands and clamps to support the pipets as shown in Figure 1.
2. Label and fill one beaker with DI water, a second with 15% sucrose solution, and a third with 30% sucrose solution. Fill each approximately one inch from the top.
3. Knot one end of a length of dialysis tubing (or overlap one end and tie with string). Open the other end by rubbing firmly between thumb and forefinger. Use a pipet to fill the tubing with 15% sucrose solution.
4. Wrap the rubber band tightly around the serological pipet and slide the rubber band within two inches of the tapered end of the serological pipet.
5. Insert the tapered end of the pipet into the open end of the tubing and secure the dialysis bag to the pipet with the rubber band. This step may take some practice and it is important that the rubber band be tight in order to hold the tubing and prevent leaks. Be careful to avoid trapping air bubbles in the tubing.
6. Repeat steps 3–5 with additional tubing, 15% sucrose solution, and the remaining two serological pipets. As each is completed, the pipet can be secured in a clamp with the tubing suspended in air. Blot the outside of the dialysis bags dry with paper toweling and check for leakage.
7. When ready to initiate the demonstration, adjust the level of liquid in each pipet by adding 15% sucrose solution up to the 2.0-mL mark (roughly the midpoint).
8. Lower a bag/pipet into each of the three beakers from Step 2. Note changes in the liquid level in each pipet over a 20- to 40-minute period.
9. (Optional) Use a syringe to add a small drop of food dye to the top of the sucrose solution within the serological pipet. This drop of dye will help the students visualize the level of the sucrose solution within the pipet.

Part II. Experiment

1. Obtain the two solutions assigned, placing 200 mL of each in separate 250-mL beakers or cups. Label the beakers or cups with the solution concentrations and your group’s initials.
2. Use the cork borer or French fry cutter to cut four potato cores from your assigned potato.
3. Trim each of the four cores to 4 cm in length. Note: Make sure to remove any potato skin from either end. Be as precise as possible.
4. Divide the potato cores into pairs and, using the electronic balance, mass each pair of cores to the nearest 0.01-g. Record the mass of each pair in the data table on the Membrane Diffusion Worksheet.
5. Place the first pair of cores in one of the two beakers. Record the concentration of sucrose solution in the beaker. Repeat for the second pair of cores. Note any observations.
6. Cover the beakers with plastic wrap or Parafilm.
7. Allow the cores to soak overnight.
8. The next day remove the cores from the first beaker with the forceps. Blot them dry by gently rolling them (no squeezing!) on a paper towel.
9. Mass them on the balance and record the final mass.

Student Worksheet PDF

10116_Student1.pdf