Diffusion and Osmosis

Concepts Instructor Prep Student Protocol

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Diffusion and osmosis are important concepts that explain how water and other materials that cells need are transported across cell membranes.

Let's talk about diffusion first. It is defined as the net movement of particles from an area of high concentration to an area of lower concentration. The graduated change in concentration between the two areas is referred to as the concentration gradient. Although diffusion is net directional, the particles are constantly moving in both directions due to random motion, so even at equilibrium when the particle density is the same throughout the concentration gradient, particles continue to move in both directions at a constant exchange rate.

Similarly, water moves across cell membranes by diffusion in a process called osmosis, but not everything can freely pass through cell membranes, which is why they are referred to as semi-permeable. This is important, because it means that cells can regulate and maintain different concentrations of solutes inside versus outside their membranes. Depending on the relative solute concentrations of solutions separated by semi-permeable membranes we refer to them as hypotonic, isotonic or hypertonic. Hypotonic is when the solute concentration is greater inside the cell in comparison to outside. Isotonic is when the inside solute concentration equals the outside concentration. Hypertonic is when the outside solute concentration exceeds the inside solute concentration. This can affect the movement of water into and out of the cell as water moves to the area of greater solute concentration. In turn, this can affect the shapes of the cells causing cell bloating in hypotonic solutions, no shape changes in isotonic solutions and the cell shrivels in the presence of hypertonic solutions.

The capacity for water to move into cells is different between plant and animal cells because of the presence of the additional plant cell wall. Cell walls are rigid and only permeable to small molecules. When water moves into plant cells the membrane gets pushed up against the cell wall creating hydrostatic or turgor pressure. This pressure limits the amount and rate at which water can enter the cell.

Diffusion is also a major limiting factor to cell size and helps explain why unicellular organisms are generally very small. Multi-cellular organisms are made up of many small cells giving a greater total surface area to volume ratio and increasing diffusion rates. Many aspects of our physiology, like breathing and digestion rely on diffusion. For example, human lungs have many small alveoli which are like small pockets. This extra surface area makes the lung more efficient at diffusion of gases in and out of the bloodstream.

In this lab, you will use two cell models, agar cube and dialysis tubing to test the principles of diffusion and osmosis.

Print Procedure Steps

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Diffusion in Agar
- NOTE: In this exercise, you will be given agar containing an indicator chemical called phenolphthalein. When phenolphthalein is exposed to the normal alkaline conditions in the agar, it will look pink. But when it is exposed to neutral or acidic conditions, it changes from pink to clear. You will make different size and shaped agar cubes as a model for cells to study the impact of cell size and shape on diffusion rate.
- To make the first set of cells, measure out and cut a small cube of agar where each side measures one centimeter.
- Next, measure and cut out a medium cell cube with sides of 2 cm and a large cell of 3 cm on each side.
- Knowing the length of the sides of your cube cells, calculate their surface area using this equation, where lower case a represents the length of the sides: Surface Area = 6a^2
- Record these values in the appropriate column in Table 1. Click Here to download Table 1
- Then, use the same length value and the equation below to calculate the volume of each cube and add these to the table: Volume = a^3 HYPOTHESES: The experimental hypothesis might be that the acid will diffuse completely to the center of the small cell faster than the medium and large cells. The null hypothesis could be that the acid will diffuse to the center of the small and two larger cubes at around the same time.
- Add 100 mL of 0.1 M HCl to each of the three 400 mL beakers to make the diffusion baths.
- Working in a team, have one experimenter ready with the timer and the second and third experimenters ready to drop each cube into one of the beakers.
- When the first experimenter says go, simultaneously drop all three cubes into their respective beakers and start the timer.
- Observe carefully until one of the cubes becomes completely clear or 10 min have passed.
- Stop the timer, remove the agar cubes from the beakers and place the cubes into a Petri dish.
- Make a note of which of the three cells became clear or had the smallest remaining pink area. Then, also note which cell had the most remaining pink agar.
- Next, in Table 1, calculate the surface area to volume ratio for each cell. Surface Area: Volume Ratio = (surface area)/volume
- As the cell size increases, note whether the surface area to volume ratio increases or decreases. Also consider whether this correlates with your observation of the depth of diffusion into the agar cells. If cells rely on diffusion to deliver essential nutrients and molecules to the whole cell, discuss with your group if it would be better to have a smaller or larger surface area to volume.
- Now, with the remaining agar, cut three rectangular shaped blocks of different sizes and record their length, width, and height. This will test what happens when the shapes of cells are different.
- Calculate the surface area of your rectangular cells using the formula below, where length is l, width is w, and height is h. Surface Area = 2lw + 2lh + 2wh
- Then, calculate the volume of your rectangles using this formula: Volume = l * w * h
- Repeat the experiment by dropping the new shapes into the hydrochloric acid solution for 10 min or until one cube becomes completely clear.
- Remove the cell shapes from the solutions and observe the depth that the hydrochloric acid diffused into each of these cells, and which shapes have the smallest and largest remaining pink areas not reached by the solute.
- Using the surface area and volume data you recorded for your rectangular shapes, calculate the surface area to volume ratio of these cells. Surface Area: Volume Ratio = (surface area)/volume
- Consider whether these values correlate to which cells had the most and least complete diffusion. Additionally, discuss with the group whether these rectangular cells displayed a similar or different pattern of diffusion to that observed with the cube shaped cells, and what this might mean.
Movement of Molecules Across a Semi-Permeable Membrane
- Before beginning the experiment, add 250 mL of distilled water to each of four 1 L beakers.
- Then, label the beakers from 1-4, and add 0.5 mL of iodine to the first beaker. HYPOTHESES: In this experiment, the experimental hypothesis is that some of the solutes will be able to pass through the dialysis tubing membrane and others will not. The null hypothesis is that there will be no difference in the ability to diffuse through the dialysis tubing membrane between the solutes.
- To prepare the dialysis tubing, remove the pieces one at a time from the distilled water bath and tie a tight knot at one end of each tube. These tubes, when filled, will act as model cells with the dialysis tubing acting like the semipermeable membrane.
- Add 10 mL of starch solution to the first tube and tie off the open end, making sure to leave space in case the tubing expands during the experiment.
- Then add 10 mL of the NaCl and dextrose solutions to the second and third pieces of tubing, respectively, and tie off both tubes, again, leaving space in case of expansion.
- After adding 10 mL of distilled water and tying off the fourth tube, weigh each of your model cells.
- Record the initial weight values in grams and the colors of the starting solution in each tube in the appropriate columns of Table 2. Click Here to download Table 2
- After quickly rinsing the outside with tap water, place each piece of tubing in its corresponding beaker for 1 h at room temperature. NOTE: For example, the starch solution tube should be placed into the beaker containing the iodine.
- At the end of the diffusion period, weigh the tubes again.
- Then, observe the tubes carefully, noting any color changes.
- Record all of these data in Table 2.
- Next, to perform a Benedict's Reagent test for simple sugars, make a water bath by adding 250 mL of water to a 600 mL beaker and placing it onto a hot plate.
- Set the plate to high, to boil the water.
- Label two new glass test tubes as H₂O and dextrose, respectively.
- Use a graduated cylinder to transfer 1 mL of solution from the water and dextrose beakers into the corresponding test tubes.
- Then, add 2 mL of Benedict's Reagent to each tube.
- Once the water is boiling, place each test tube into the water bath for 3-5 minutes.
- After this time, note the color of the solution in each tube.
- Then use this key to assess whether the test is positive or negative and record these data in the appropriate column in Table 2. Click Here to download Figure 1
Results
- First, look at the mass of your four dialysis tube cells at the beginning versus the end of the experiment. Calculate the change in mass for each of the four cells and plot it onto a bar chart.
- Note which cells demonstrated the most change, and whether any of the cells appeared visibly different in size.
- For the experiment with the starch and iodine indicator, note whether there was a color change in the fluid in the artificial cell. Also consider whether there was a color change in the water in the beaker, and what both of these observations say about the properties of the dialysis tubing membrane.
- Finally, in the Benedict's Reagent test for dextrose, note whether this simple sugar was able to pass through the semipermeable membrane of the “cell” into the water in the beaker. Discuss with the class which of the molecules you think could and could not pass through the semipermeable membrane.