Diffusion and Osmosis

Overview

In this laboratory you will investigate the processes of diffusion and osmosis in a model membrane system. You will also investigate the effect of solute concentration on water potential as it relates to living plant tissues.

Introduction

Many aspects of the life of a cell depend on the fact that atoms and molecules have kinetic energy and are constantly in motion. This kinetic energy causes molecules to bump into each other and move in new directions. One result of this molecular motion is the process of diffusion.

Diffusion is the random movement of molecules from an area of higher concentration of those molecules to an area of lower concentration. For example, if one were to open a bottle of hydrogen sulfide (H₂S has the odor of rotten eggs) in one corner of a room, it would not be long before someone in the opposite corner would perceive the smell of rotten eggs. The bottle contains a higher concentration of H₂S molecules than the room does and, therefore, the H₂S gas diffuses from the area of higher concentration to the area of lower concentration. Eventually a dynamic equilibrium will be reached; the concentration of H₂S will be approximately equal throughout the room and no net movement of H₂S will occur from one area to the other.

Osmosis is a special case of diffusion. Osmosis is the diffusion of water through a selectively permeable membrane (a membrane that allows for diffusion of certain solutes and water) from a region of higher water potential to a region of lower water potential. Water potential is the measure of free energy of water in a solution.

Diffusion and osmosis do not entirely explain the movement of ions or molecules into and out of cells. One property of a living system is active transport. This process 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.

EXERCISE I A: Diffusion

In this experiment you will measure diffusion of small molecules through dialysis tubing, an example of a selectively permeable membrane. Small solute molecules 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 movement of a solute through a selectively permeable membrane is called dialysis. The size of the minute pores in the dialysis tubing determines which substances can pass through the membrane.

A solution of glucose and starch will be placed inside a bag of dialysis tubing. Distilled water will be placed in a beaker, outside the dialysis bag. After 30 minutes have passed, the solution inside the dialysis tubing and the solution in the beaker will be tested for glucose and starch. The presence of glucose will be tested with Benedict's solution or Testape. The presence of starch will be tested with Lugol's solution, (Iodine Potassium-Iodide or IKI).

Procedure

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

2. Place 15 mL of the 15% glucose/1% starch solution in the bag. Tie off the other end of the bag, leaving sufficient space for the expansion of the contents in the bag. Record the color of the solution in Table 1.1.

3. Test the 15% glucose/1% starch solution for the presence of glucose using the Benedict's test. Record the results in Table 1.1.

4. Fill a 250 mL beaker or cup two-thirds full with distilled water. Add approximately 4 mL of Lugol's solution to the distilled water and record the color of the solution in Table 1.1. Test this solution for glucose and record the results in Table 1.1.

5. Immerse the bag in the beaker of solution.

6. Allow your set-up to stand for approximately 30 minutes or until you see a distinct color change in the bag or in the beaker. Record the final color of the solution in the bag, and of the solution in the beaker, in Table 1.1.

7. Test the liquid in the beaker and in the bag for the presence of glucose. Record the results in Table 1.1.

Table 1.1

EXERCISE 1 B: Osmosis

In this experiment you will use dialysis tubing to investigate the relationship between solute concentration and the movement of water through a selectively permeable membrane by the process of osmosis.

When two solutions have the same concentration of solutes, they are said to be isotonic to each other (iso means same, -ton means condition, -ic means pertaining to). If the two solutions are separated by a selectively permeable membrane, water will move between the two solutions, but there will be no net change in the amount of water in either solution.

If two solutions differ in the concentration of solutes that each has, the one with more solute is hypertonic to the one with less solute (hyper means over, more than). The solution that has less solute is hypotonic to the one with more solute (hypo means under or less than). These words can only be used to compare solutions.

Now consider two solutions separated by a selectively permeable membrane. The solution that is hypertonic to the other must have more solute and therefore less water. At standard atmospheric pressure, the water potential of the hypertonic solution is less than the water potential of the hypotonic solution, so the net movement of water will be from the hypotonic solution into the hypertonic solution. Label the sketch below to indicate which solution is hypertonic, which is hypotonic, and use arrows to show the initial net movement of water.

Procedure

1. Obtain six 30 cm stops of presoaked dialysis tubing.

2. Tie a knot in one end of each piece of dialysis tubing to form six bags. Pour approximately 25 mL of each of the following solutions into separate bags:

a) Distilled water
b) 0.2 M sucrose
c) 0.4 M sucrose
d) 0.6 M sucrose
e) 0.8 M sucrose
f) 1.0 M sucrose

Remove most of the air from each bag by drawing the dialysis bag between two fingers. Tie off the other end of the bag. Leave sufficient space for the expansion of the contents in the bag. (The solution should fill only about one-third to one-half of the piece of tubing.)

3. Rinse each bag gently with distilled water to remove any sucrose spilled during filling.

4. Carefully blot the outside of each bag and record in Table 1.2 the initial mass of each bag, expressed in grams.

5. Fill six 250 mL beakers or cups two-thirds full with distilled water.

6. Immerse each bag in one of the beakers of distilled H₂0 and label the beaker to indicate the molarity of the solution in the dialysis bag. Be sure to completely submerge each bag.

7. Let them stand for 30 minutes.

8. At the end of 30 minutes remove the bags from the water. Carefully blot and determine the mass of each bag.

9. Record your group's data in Table 1.2. Obtain data from the other lab groups in your class to construct a for the class data.

Table 1.2: Dialysis Bag Results. Individual Data

*To calculate: Percent Change in Mass = [Final Mass - Initial Mass x 100]/ Initial Mass

10. Graph both your individual data the class average.

EXERCISE 1 C: Water Potential

In this part of the exercise you will use potato cores placed in different molar concentrations of sucrose in order to determine the water potential of potato cells. First, however, we will explore what is meant by the term "water potential."

Botanists use the term water potential when predicting the movement of water into or out of plant cells. Water potential is abbreviated by the Greek letter psi and it has two components, a physical pressure component, pressure potential psi_p and the effects of solutes, solute potential psi_s. Water will always move from an area of higher water potential (higher free energy; more water molecules) to an area of lower water potential (lower free energy; fewer water molecules). Water potential, then, measures the tendency of water to leave one place in favor of another place.

Water potential is affected by two physical factors. One factor is the addition of solute which lowers the water potential. The other factor is pressure potential (physical pressure). An increase in pressure raises the water potential. By convention, the water potential of pure water at atmospheric pressure is defined as being zero (psi = 0). For instance, it can be calculated that a 0.1 M solution of sucrose at atmospheric pressure (psi_p = 0) has a water potential of -2.3 bars due to the solute (psi = -2.3). [note: a bar is a unit of pressure, measured with a barometer, that is about the same as 1 atmosphere. Another measure of pressure is the megapascal (Mpa). 1 Mpa = 10 bars.]

Movement of H₂0 into and out of a cell is influenced by the solute potential (relative concentration of solute) on either side of the cell membrane. If water moves out of the cell, the cell will shrink. If water moves into an animal cell, it will swell and may even burst. In plant cells, the presence of a cell wall prevents cells from bursting as water enters the cells, but pressure eventually builds up inside the cell and affects the net movement of water. As water enters a dialysis bag or a cell with a cell wall, pressure will develop inside the bag or cell as water pushes against the bag or cell wall. The pressure would cause, for example, the water to rise in an osmometer tube or increase the pressure on a cell wall. It is important to realize that water potential and solute concentration are inversely related. The addition of solutes lowers the water potential of the system. In summary, solute potential is the effect that solutes have on a solution's overall water potential.

Movement of H₂0 into and out of a cell is also influenced by the pressure potential (physical pressure) on either side of the cell membrane. Water movement is directly proportional to the pressure on a system. For example, pressing on the plunger of a water- filled syringe causes the water to exit via any opening. In plant cells this physical pressure can be exerted by the cell pressing against the partially elastic cell wall. Pressure potential is usually positive in living cells; in dead xylem elements it is often negative.

It is important for you to be clear about the numerical relationships between water potential and its components, pressure potential and solute potential. The water potential value can be positive, zero, or negative. Remember that water will move across a membrane in the direction of the lower water potential. An increase in pressure potential results in a more positive value, and a decrease in pressure potential (tension or pulling) results in a more negative value. In contrast to pressure potential, solute potential is always negative; since pure water has a water potential of zero, any solutes will make the solution have a lower (more negative) water potential. Generally, an increase in solute potential makes the water potential value more negative and an increase in pressure potential makes the water potential more positive.

To illustrate the concepts discussed above, we will look at a sample system using the figures below. When a solution, such as that inside a potato cell, is separated from pure water by a selectively permeable cell membrane, water will move (by osmosis) from the surrounding water where water potential is higher, into the cell where water potential is lower (more negative) due to the solute potential (psi_s). In Figure 1.2a the pure water potential (psi = 0) and the solute potential (psi_s = -3) . We will assume, for purposes of explanation, that the solute is not diffusing out of the cell. By the end of the observation, the movement of water into the cell causes the cell to swell and the cell contents push against the cell wall to produce an increase in pressure potential (turgor) (psi_p = 3). Eventually, enough turgor pressure builds up to balance the negative solute potential of the cell. When the water potential of the cell equals the water potential of the pure water outside the cell (psi of cell = psi of pure water = 0), a dynamic equilibrium is reached and there will be no NET water movement

If you were to add solute to the water outside the potato cells, the water potential of the solution surrounding the cells would decrease. It is possible to add just enough solute to the water so that the water potential outside the cell is the same as the water potential inside the cell. In this case, there will be no net movement of water. This does not mean, however, that the solute concentrations inside and outside the cell are equal, because water potential inside the cell results from the combination of both pressure potential and solute potential.

Solute in water
psi = psi_p + psi_s
-12 = 0 + (-12)

Potato cell
psi = psi_p + psi_s
12 = 3 + (-15)

If enough solute is added to the water outside the cells, water will leave the cells, moving from an area of higher water potential to an area of lower water potential. The loss of water from the cells will cause the cells to lose turgor. A continued loss of water will eventually cause the cell membrane to shrink away from the cell wall (plasmolysis).

Procedure:

1. Pour 100 mL of the each solution into a labeled 250 mL beaker.

2. Use a cork borer (approximately 5 mm inner diameter) to cut potato cylinders. Slice the cylinders with a scalpel to prepare potato disks of 0.5 cm thickness. Do not include any skin on the disks. You need twelve potato disks for each beaker.

3. Keep your potato cylinders in a covered beaker until it is your turn to use the balance.

4. Determine the mass of the twelve potato disks together and record the mass in Table 1.4. Put the twelve disks into one of the beakers.

5. Cover the beaker with plastic wrap to prevent evaporation.

6. Let it stand overnight.

7. Remove the cores from the beakers, blot them gently on a paper towel, and determine their total mass.

8. Record the final mass in Table 1.4 and calculate percentage change as in Exercise IB. Do this for both your individual results and class average.

9. Graph both your individual data and the class average for the percentage change in mass in Table 1.4.

Table 1.4

Contents of Beaker	MassInitial	MassFinal	MassChange	Percent Change in Mass	Class Average Percent Change in Mass
a) distilled water
b) 0.2 M
c) 0.4 M
d) 0.6 M
e) 0.8
f) 1.0 M

Calculate the molar concentration of the potato core. This would be the sucrose molarity in which the mass of the potato core does not change. The x-intercept of your graph represents the molar concentration of sucrose with a water potential that is equal to the potato tissue water potential. At this concentration there is no net gain or loss of water from the tissue.

EXERCISE ID: Calculation of Water Potential from Experimental Data

1. The solute potential of this sucrose solution can he calculated using the following formula

psi_s= -iCRT

where	i = ionization constant (for sucrose this is 1.0 because sucrose does not ionize in water)
	C = Molar concentration (determined above)
	R = Pressure constant (R = 0.0831 liter bars/mole oK)
	T = Temperature oK (273 + oC of solution)

2. Knowing the solute potential of the solution (_s) and knowing that the pressure potential of the solution is zero (psi_p = 0) allows you to calculate the water potential of the solution. The water potential will be equal to the solute potential of the solution.

psi = 0 + psi_s or psi = psi_s

The water potential of the solution at equilibrium will be equal to the water potential of the potato cells. Calculate the water potential of the potato cells?

3. Water potential values are useful because they allow us to predict the direction of the flow of water. Recall from the discussion that water flows from an area of higher water potential to an area of lower water potential. For the sake of discussion, suppose that a student calculates that the water potential of a solution inside a bag is -6.25 bar (psi_s = -6.25, psi_p = 0) and the water potential of a solution surrounding the bag is -3.25 bar (psi_s = -3.25, psi_p = 0). In which direction will the water flow? Water will flow into the bag. This occurs because there are more solute molecules inside the bag (therefore a value further away from zero) than outside in the solution.