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.

Diffusion is a “downhill”, (down a concentration gradient) , increasing entropy (randomness), lowering the overall free energy in the system.

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 moves substances against a concentration gradient, from regions of low concentration of that substance into regions of higher concentration. Other methods of molecular transport are facilitated transport,  endocytosis (phagocyctosis and pinocyctosis), and coupled transport.

EXERCISE I A: Diffusion

In this experiment you will measure diffusion of small molecules through semipermeale plastic memebrane. We will use a sandwich bag instead of more expensive dialysis tubing. Small solute molecules and water molecules can move freely through a selectively permeable membrane, while larger molecules will pass through more slowly, or not move at all. The movement of a solute through a selectively permeable membrane is called dialysis. The size of the minute pores in the plastic bag determines which substances can pass through the membrane.

A solution of glucose and starch will be placed inside a bag of dialysis tubing. Tap water with I₂KI added will be placed in a beaker, outside the bag. After a period of time, the solution inside the dialysis tubing and the solution in the beaker will be observed and/or tested for glucose and starch. The presence of glucose will be tested with Benedict's solution.

Procedure

1. Obtain a  store brand sandwish bag. Tie off one end of the tubing to form a bag. 2. Place 20-30 mL of the 20% glucose in the bag.   Add 20-30 ML 1% starch solution in the bag. Tie off the bag tightly closed with soft string., leaving a small air pocket for expansion of the contents in the bag. Record the color of the solution in Table 1.1. 

2. Half fill a 250 mL beaker or cup two-thirds full with tap water. Add enough of  I₂KI solution to the tap water to turn the water the color of tea  Complete  Table 1.1.

3. Immerse the bag in the beaker of solution.

4. REVIEW - Describe the test procedure and results to test for glucose in solution. ______________________________________________________________________________________________________________________________________________________________________________

REVIEW - Describe the test and results to test for starch in solution. ____________________________________________________________________________________________________________________________________________________________________

5. Allow your set-up to stand for approximately 30 minutes or when your teacher tells you.   Record the final color of the solution in the bag, and of the solution in the beaker, in Table 1.1.

Table 1.1  Describes initial experiment set up and results after diffusion

Analysis

1.  What can be concluded about the relative molecule sizes of glucose and starch?

2.  What previous information do you have to support this conclsuison?

3.  How might this information affect organisms that rely on dietary starch as a energy source.

EXERCISE 1 B/C: Osmosis and Water Potential  

OSMOSIS

In this experiment you will use potato cores  to investigate the relationship between solute concentration and the movement of water through a selectively permeable membrane by the process of osmosis.  You will also determine the water potential of  the potato cores.

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 18 potato cores that are 3 cm long.

2. Label 6 cups with the following labels.

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

3.  Add 100 ml of each solution to the proper cup.

4.  Accurately mass 3 cores at a time and add 3 cores to each cup.  Record the starting mass of each group of cores in the corresponding cup on Table II. Let cores sit over night.

24 hours later

5. Remove the cores from the beakers, blot them gently on a paper towel and. repeat mass determinations.   Calculate percentage change. Do this for both your individual results and class average.

WATER POTENTIAL

Read Campbell pp 750-753  to Aquaporins for good explanation of water potential

In this part of the exercise you will also determine the water potential of potato cells. What is  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 has two components, a physical pressure component, ( pressure potential psi_{p =} y _p)  and the effects of solutes, solute potential psi_{s =} y _{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 solute concentration which is determined by how much solute is in solution. The addition of solute lowers the water potential (-). The other factor is pressure potential (physical pressure). An increase in pressure raises the water potential ( + ).  A system open to the atmosphere has a y _p of zero.

By convention, the water potential (y ) of pure water at atmospheric pressure is defined as zero (y = 0). Example -  A  0.1 M sucrose solution at atmospheric pressure (psi_p = 0) has a water potential of -2.3 bars due to the solute (psi = -2.3, a precalculated value). [note: a bar is a unit of pressure, measured with a barometer, that is about the same as 1 atmosphere

Movement of H₂0 into and out of a cell is influenced by water potential 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. This results in an increase in pressure on a cell wall. It is important to realize that overall 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 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. Water will move across a membrane in the direction of the lower water potential. An increase in pressure potential increases water potential  while a decrease in pressure potential (tension or pulling) decreases water potential.  In contrast, solute potential is always negative.  Since pure water has a solute potential of zero, adding solutes will decrease water potential. Therefore, increasing solutes decreases water potential and increasing pressure potential raises water potential.

To illustrate the concepts discussed above, look at a sample system using the figures below. When, i a potato cell, is separated from pure water by a selectively permeable cell membrane, water will move (by osmosis) from the surrounding water ( greater y , into the cell where water potential, y,  is lower (more negative) due to the solute potential (psi_s = -3). In Figure 1.1 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 (y of cell = y of pure water = 0), ( 0=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 outside 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 after adding solute
                                     psi = psi_p + psi_s
                                      -12 = 0 + (-12)

                                    Plasmoylzed potato  cell after being placed in solution
                                    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).

Table II. Effect of Varying Sucrose Molarities on Mass of Potato Cores Over 24 hrs.

Analysis

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

2. 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.

3. Calculate the solute potential (y _p) of the potato cells.  See instructions below..  Show your problem set up. Define each  each term  in the equation to solve for solvent potential.

4.  Make a diagram of the sample application at the end of the lab and answer the question..  Explain your answer

5.  Write the equation for water potential.  Define all terms and tell how solvent and pressure potential affect water potential.

6.  In cold climates towns often salt their roads during the winter.  In spring, roadside grass is often brown and does not begin spring growth.  Explain the relationship between these two events using water potential terminology..

7. Household owners often spread lawn fertilizer in the spring.  If granulated fertilizer spills on the lawn, a brown spot will develop. Explain this observation in relation to this lab.  Use water potential terminology.

8. Salting foods is a way of preserving foods.  Explain how this works as a method of preservation ( do some outside research ).

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

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.

Sample Application.

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?

THIS WILL NOT BE A FORMAL LAB WRITE UP.  SUBMIT YOUR PRELAB, GRAPHS AND ALL ANALYSIS QUESTIONS ANSWERED.  KNOW HOW TO CALCULATE WATER POTENTIAL AND BE ABLE TO EXPLAIN HOW THE POTATO CORE

GRAPH COULD BE USED TO DETERMINE WATER POTENTIAL IN THE POTATO CORES.