Name: ______________________________ AP Biology

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Diffusion and Osmosis

Objectives: Describe the physical mechanisms of diffusion and osmosis. Describe how molar concentration affects the process of diffusion. Predict cell outcomes when changing the concentration of

solute in a solution in which the cell is suspended.

Introduction: Many aspects of the life of a cell depend on the fact that atoms and molecules are constantly in motion. This energy results in molecules bumping into and rebounding off each other and moving in new directions. One result of this molecular motion is the process of diffusion.

Cells must move materials through membranes and throughout cytoplasm in order to maintain homeostasis. The movement is regulated because cellular membranes, including the plasmaand organelle membranes, are selectively permeable. Membranes are phospholipid bilayerscontaining embedded proteins.

The cellular environment is aqueous, meaning that the solvent is water, in which the solutes,such as salts and organic molecules, are dissolved. Water may pass freely through the membrane by osmosis or through specialized protein channels called aquaporins. Most ionsmove through protein channels, while larger molecules, such as carbohydrates, are carried by transport proteins.

The simplest form of movement is diffusion, in which solutes move from an area of high concentration to an area of low concentration. Diffusion does not require energy input. Themovement of a solute from an area of low concentration to an area of high concentration requires energy input in the form of ATP and protein carriers.

Water moves through membranes by diffusion; this process is called osmosis. Like solutes,water moves down its concentration gradient. Water moves from areas of high potential (high water concentration) and low solute concentration to areas of low potential (low water concentration) and high solute concentration. In walled cells, osmosis is affected not only by the solute concentration but also by the resistance to water movement in the cell by the cell wall. This resistance is called turgor pressure (the physical pressure exerted on the cell).

The terms hypertonic, hypotonic, and isotonic are used to describe solutions separated by

selectively permeable membranes.

A hypertonic solution has a higher solute concentration and a lower water potential as compared to the other solution; therefore, water will move into the hypertonic solution through the membrane.

A hypotonic solution has a lower solute concentration and a higher water potential than the solution on the other side of the membrane; water will move down its concentration gradient into the other solution.

Isotonic solutions have equal water potential.

Name: ______________________________ AP Biology – Lab 04

Understanding Water Potential (W) In non-walled cells, such as animal cells, the movement of water into and out of a cell is affected by the solute concentration on either side of the plasma membrane. As water moves out of the cell, the cell shrinks; if water moves into the cell, it swells and may eventually burst or lyse. In walled cells, including fungal and plant cells, the presence of a cell wall prevents the cells from bursting as water enters; however, pressure builds up inside the cell and affects the rate of osmosis.

Water potential predicts which way water diffuses through plant tissues and is abbreviated

by the Greek letter psi (W). Water potential is calculated from two major components: (1) the

solute potential (S) – is dependent on solute concentration, and (2) the pressure

potential (P), which results from the exertion of pressure — either positive or negative

— on a solution.

Water Potential = Pressure Potential + Solute Potential

s+ p = w

Water moves from an area of higher water potential to an area of lower water potential. Water potential measures the tendency of water to diffuse from one compartment to another compartment.

The water potential of pure water in an open beaker is zero (w = 0) because both the solute and pressure potentials are zero (s = 0; P = 0). An increase in positive pressure raises the pressure potential and the water potential. The addition of solute to the water lowers the solute potential and therefore decreases the water potential. This means that a solution at atmospheric pressure has a negative water potential because of the solute.

The solute potential (s) = – iCRT, where i = the ionization constant, C = the molar concentration, R = the pressure constant (R = 0.0831 liter * bars/mole * K), and T = the temperature in K (273 + °C).

A 0.15 M solution of sucrose at atmospheric pressure (P = 0) and 25°C has an osmotic potential of -3.7 bars and a water potential of -3.7 bars. A bar is a metric measure ofpressure. A 0.15 M NaCl solution contains 2 ions, Na+ and Cl- (where sucrose stays as one particle); therefore i = 2, and the water potential = -7.4 bars.

When a cell‘s cytoplasm is separated from pure water (e.g. distilled water) by a selectively permeable membrane, water moves from the surrounding area, where the water potential is higher (w = 0), into the cell, where water potential is lower because of solutes in the cytoplasm (w is negative). It is assumed that the solute is not diffusing (Figure 1a). The

movement of water into the cell causes the cell to swell, and the cell membrane pushes against the cell wall to produce an increase in pressure. This pressure, which counteracts the diffusion of water into the cell, is called turgor pressure.

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Name: ______________________________ AP Biology – Lab 04

Over time, enough positive turgor pressure builds up to oppose the more negative solute potential of the cell. Eventually, the water potential of the cell (not just osmotic potential!) equals the water potential of the pure water outside the cell (w of cell = w of pure water =

0). At this point, an equilibrium is reached and net water movement ceases (Figure 1b)

Figures 1a-b: Plant cell in pure water. The water potential was calculated at the beginning of

the experiment (a) and after water movement reached equilibrium and the net water

movement was zero (b).

If solute is added to the water surrounding the plant cell, the water potential of the solution surrounding the cell decreases. If enough solute is added, the water potential outside the cell is then equal to the water potential inside the cell, and there will be no net movement of water. However, the solute concentrations inside and outside the cell are not equal because the water potential inside the cell results from the combination of both the turgor pressure (P) and the solute pressure (s), as shown in Figure 2.

Figure 2: Plant cell in an aqueous solution. The water potential of the cell equals that of

surrounding solution at equilibrium. The cell’s water potential equals the sum of the turgor

pressure potential plus the solute potential. The solute potentials of the solution and of the

cell are not equal.

If more solute is added to the water surrounding the cell, water will leave the cell, moving from an area of higher water potential to an area of lower water potential. The water loss causes the cell to lose turgor pressure. A continued loss of water will cause the cell membrane to shrink away from the cell wall, and the cell will plasmolyze.

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Name:_____________________________________________________

1. Obtain 6 strips of dialysis tubing. Dip each in a beaker of water for 30 seconds. Using your thumb and pointerfinger, pinch the tubing and rub it back and forth to open. Now, use the string provided to tie an knot in oneend of each.

2. Now, label 6 small cups with the solution names in the table below.3. Pour approximately 15­20 ml of each of the following solutions into separate bags.

Distilled Water 0.4 M sucrose 0.8 M sucrose0.2 M sucrose 0.6 M sucrose 1.0 M sucrose

3. Remove most of the air from the bag (but leave a little bit of space) and tie the other end of the baggie tightly.4. Rinse the baggies under tap water.5. Blot the bags dry with a paper towel. Check the bags for leaks.6. Record the mass of each DRY baggies in the data table.7. Fill six small cups with enough distilled water to cover your bags. Place the APPROPRIATE baggie in each cup.8. Let the bag sit for 20­30 minutes.9. After 20-30 minutes, remove the baggies from the water, and carefully blot dry and record the final weight.

10. To calculate: percent change in mass= (final mass-initial mass)/ initial mass. Then multiply answer by 100.

Contents in Bag Initial Mass Final Mass Mass Difference Time in Beaker % Change in MassDistilled Water0.2 M0.4 M0.6 M0.8 M1.0 M

11. Use "Google Sheets" to create a line graph of the results for your individual data that shows the relationship between %change in mass and the molarity of the solution. The independent variable is on the X axis, and the dependent variable is on the Y axis

Investigation: Diffusion, Osmosis and Water Potential

Hypothesis: What do you think will happen when each baggie (each with a different amount of sucrose) is placed in distilled water? (If..., Then..., Because format)

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ANALYSIS

1. Explain the relationship between the change in mass and the molarity of sucrose within the dialysis bag.

2. Predict what would happen to the mass of each bag in this experiment if all the bags were placed in 0.4 Msucrose solution instead of distilled water. Explain your response.

3. Why did you calculate the percent change in mass rather than simply using the change in mass?

4. A dialysis bag is filled with distilled water and then placed in a sucrose solution. The bag's initial mass is 20 g,and its final mass is 18 g. Calculate the percent change of mass, showing your calculations. Explain thisoutcome.

5. Did you results match your predictions? Propose an explanation for why your results (either overall or anindividual bag) may have differed from what you were expecting.

SAMPLE SET UP

Materials

ForcepsMicroscope Slide w/coverslip 1.0M Sodium chloride solution Distilled WaterOnion

1. Obtain a piece of onion. Using fine point forceps or tweezers peel off a small piece of theonion epidermis, Start at one of the corners of the onion piece and peel gently away fromthe onion. (see below) Place it flat on a microscope slide.

2. Add a drop or two of distilled water to the onion epidermis and gently place a coverslip onthe microscope slide. Try not to trap any air bubbles under the coverslip.

3. Examine the onion cells under 40x magnification. SKETCH the onion cells in the spacebelow.

4. Prepare another section of onion epidermis and place it on a new microscope slide.5. Add a drop or two of 1.0M sodium chloride to the onion epidermis and gently place a

coverslip on the microscope slide.6. Examine the onion cells under 40x magnification. Note if they appear different as

compared to the cells in distilled water. SKETCH the onion cells in the space below.7. CLEAN UP and dispose of all materials according to your instructor.

Analysis

1. When you prepared onion cells on a microscope slide containing the sodium chloride solution, how did they compare to the onion cells that were prepared on a slide containing water? What does this indicate about the direction of water movement? PLEASE explain using observations obtained from your activity and your knowledge of osmosis.

Procedure

Distilled Water Sketch Sodium Chloride Sketch

EXERCISE 3 ­ Determining the Water Potential of Potato Cells

In animal cells, the movement of water into and out of the cell is influenced by the 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 the cell, the cell may swell or even burst. In plant cells, the presence of a cell wall prevents the cells from bursting, but pressure does eventually build up inside the cell and affects the process of osmosis. When the pressure inside the cell becomes large enough, no additional water will accumulate in the cell even. So movement of water through the plant tissue cannot be predicted simply through knowing the relative solute concentrations on either side of the plant cell wall. Instead, the concept of water potential is used to predict the direction in which water will diffuse through living plant tissues.

In a general sense, the water potential is the tendency of water to diffuse from one area to another. Water potential is expressed in in bars, a metric unit of pressure equal to about 1 atmosphere and measured with a barometer.

Consider a potato cell is placed in pure water. Initially the water potential outside the cell is 0 and is higher than the water potential inside the cell. Under these conditions there will be a net movement of water into the cell. The pressure potential inside the cell will increase until the cell reaches a state of equilibrium.

Directions:

1. Pour 100 mL of each solution (the six sucrose solutions listed above in the first activity) into separate beakers/cups. Slice a potato into 2 equal cylinders using the potato borer.2. Blot the cylinders dry and determine the initial mass of both potato cylinders "together" for each solution and record.3. Place the cylinders into the beaker with the appropriate solution and cover with plastic wrap. Leave overnight.4. Remove the cylinders from the beakers, blot dry and record the final mass. Determine the temperature of the room. ________ and the % change in mass. (Final - Initial Mass/Initial Mass = % change)5. Complete the table and create a line graph of your results in "Google Sheets". 

Solutions Initial Mass Final Mass Mass Difference %Change in MassDistilled Water0.2 M0.4 M0.6 M0.8 M1.0 M

6. Determine the molar concentration of the potato cores. This would be the sucrose molarity in which the mass ofthe potato core does not change. To find this, draw the straight line on your graph that best fits your data. Thepoint at which this line crosses the x axis represents the molar concentration of sucrose with a waterpotential that is equal to the potato tissue water potential.

What is the Molar concentration of the cores? ___________

7. Calculate the solute potential ( Ψs ) for each sucrose solution and the potato using the formula below.

Solute Potential Formula:        Ψs = ­iCRTi = ionization constant (sucrose does not ionize in water) C = molar sucrose concentration R = pressure constant (0.0831 liter bar/mole °K )T = temperature °K (273 + °C )

8. Calculate the water potential for each sucrose concentration and the potato.

9. Explain how your calculations would explain the direction of water flow between the potato and the assignedsucrose solution. Did your % change in mass agree or disagree with these expected results? Explain.

SAMPLE SET UP