Cool Diffusion and Osmosis Lab

After
The overall point of this lab was to study the processes of diffusion osmosis in a model membrane system and investigate the effect of solute concentration on water potential as it relates to plant tissues.

Part 1A
In this part of the experiment we used dialysis tubing to represent a cell membrane and we filled it with 15ml of 15% glucose/1% starch. We then put the dialysis tubing into a 250ml beaker that was filled two thirds of the way with iodine. The purpose was to study the effects of diffusion of the cell membrane.

This is the dialysis tubing that we used to represent the cell membrane 

This is the 250 ml cup that we used to represent the outside of the cell 

We tested the 250 ml and the dialysis tubing for presence of glucose. We found that the dialysis tubing contained glucose and the iodine solution did not.


Then we put the dialysis tubing in the iodine solution for thiry minutes to allow diffusion to take place.

After thirty minutes we observed the changes in color of both the dialysis tubing and the iodine solution.

We then tested each part separately to see if it contained glucose. We found that both of the parts contained glucose.

Analysis
During the process of diffusion Iodine had to be entering the dialysis tube and glucose had to be leaving. This is proven because of the fact that the initial color of the dialysis bad was clear and the beaker was orange and at the end of the experiment the dialysis tubing was black and the iodine was a lighter orange. To quantify this experiment, we could have measured a more accurate way to measure the color change and put a value on the concentration of the iodine to water solution. The reason that we didn't account for the starch movement is because starch molecules are far to large to diffuse across a cell membrane.


Part 1B

Within the dialysis bag portion of the lab, we see that there is a direct relationship between mass percent change and the molarity of solution within the dialysis bag. To justify this, we place dialysis bags of different molarities of sucrose (and one of distilled water) into cups of water. The simple explanation comes to the system trying to get close to the net equilibrium state as possible. The bags illustrates a hypotonic solution where the amount (molarity) of water outside of the system is higher than within the solution itself. This causes water to reach equilibrium by passing through the membrane of the bag and balance out the molarity of water in and out of the system. The increase of water being introduced into the dialysis bag causes an increase of mass. The higher the molarity of sucrose in the dialysis bag, for example, the higher the amount of water needed to balance out the system. The purpose of calculating the mass percent change instead of the basic change in mass is to show the ratio of change. Two solutions with different molarities can have the same amount of mass change-as seen in 0.2M and 1.0M-but the ratios significantly differ. This whole concept is osmosis, the movement of water through a selectively permeable membrane. Now let's say we place the dialysis bags into cups filled with 0.4M of sucrose solution. Then what would happen is solutions lower than 0.4M would try it's best to dilute the 0.4M by letting the remaining water inside pass out of the dialysis bag. The mass would decrease and so will the molarity of the outside solution until it reaches equilibrium. This would make it a hypertonic solution. Solutions with molarities above 0.4M would have the  outside water pass into the membrane to dilute the solutions until it reaches 0.4M. This stays as a hypotonic solution. The concept of osmosis still applies.


Part 1C

       In this part of this experiment we were attempting to understand the concept of water potential and it's two components, pressure potential and solute potential. More specifically, we wanted to determine the water potential of potato cells. Before we head into a discussion of what exactly we did, let's establish some background information.

       Water potential is described in terms of pressure potential and solute potential by its equation Ψ=Ψ(p)+Ψ(s).  As mentioned before, water potential has two components, pressure potential and solute potential. Let's look at the former in greater detail.

       Water under standard atmospheric pressure is said to have a pressure potential of zero. As you increase the pressure of a system, you increase its pressure potential and by extension, its water potential. This is because under more pressure, water has the potential to move more and do more work. For example, let's say I have a syringe filled with pure water and a bucket of pure water. In this scenario, we have two reservoirs of water: one in the syringe and one in the bucket. Now, I stick the needle of the syringe in the bucket of water and push the plunger. In which direction does water travel? Out of the syringe and into the bucket. The reason for this is that by pushing the plunger, I'm increasing the pressure of the water inside the syringe. By increasing the pressure, I'm increasing the pressure potential and also the water potential to a number greater than 0. Meanwhile, the bucket of water exposed to atmospheric pressure still has a pressure potential of 0 and a water potential of 0. Water travels from areas of high water potential to areas of low water potential, so water will travel out of the syringe and into the bucket. Now what if I pulled the plunger back instead of pushing it forward? We'd have the same situation in reverse. By pulling the plunger, I'm decreasing the pressure in the syringe. Doing so decreases the pressure potential and the water potential to a number less than 0. Again, water will flow from an area of high water potential to an area of low water potential. Since the water in the bucket still has a water potential of 0, water will be sucked into the syringe. This example is performed under the assumption that both the water in the gun and the water in the bucket were pure, that is, there were no dissolved solutes and the solute potential was 0. When we start dealing with solutes in water, we have to account for them in addition to the pressure of the water.

       In most high school freshman biology courses, we learn that solutes and water naturally diffuse from areas of high concentration to areas of low concentration. This concept is still the same for water potential, we're just putting a mathematical spin on things. Just like we said that water at atmospheric pressure has a pressure potential of 0, water without any solutes in it has a solute potential of 0. As you add solutes to water, you decrease the solute potential and by extension, the water potential. Since you can't remove solutes from water that has no solutes, you'll never see a solute potential greater than 0. This makes sense. If you have a bucket of pure water and a bucket of water and sand, the pure water will move more. It has a greater potential to do work. Now we won't be dealing with things as big as sand. We'll typically be dealing with solutes on a much smaller scale, but the concept still holds. Let's take a look at an example. Let's say I have a classic U-tube. Everyone's favorite. The tube is at atmospheric pressure so that the pressure potential in the entire system is always 0; that way we can just isolate the difference that solute potential makes. In the left arm of the U tube is water with a few solutes in it; the solute potential is just slightly less than zero. In the other is water with a lot of solutes in it; the solute potential is extremely less than zero. It doesn't matter what solutes they are, just that they're there. Separating the two arms is a membrane permeable only to water; that is to say that the solutes can't move, only the water. As our years of biology experience will tell us, the system naturally wants to reach equilibrium. It wants to make the solute concentrations of both sides equal. If it can't move the solutes in order to achieve this, then it will move water. We should already know that, based on osmosis, water moves from areas of high concentrations to low concentrations, so water will move from the left to the right. From a water potential standpoint, we can arrive at the same conclusion. The left side has less solute than the right side. Therefore, its solute potential is greater (less negative). Therefore, the left side's water potential is greater than that of the right side, so again we can say that water will move from the left to the right.

Now that we've set straight some background information, we can move onto our process for determining the water potential of the potato cells. In order to do so, we placed identical slices of potato into solutions of varying concentrations of sucrose and observed the change in mass of each slice due to the movement of the sucrose solutions and water in to or out of the potato cells. Here are our findings in a graph of percent change in mass dependent on the sucrose concentrations of the solutions.

Note the second point, where the potato slice was placed in solution in which the solute (sucrose) concentration was 0.2 M. This potato slice had no change in mass. You know what that means? There was no net movement of water in to or out of the potato cells, so the water potential of the potato slice must have been the same as the water potential of the solution. Let's look at pressure potential. Since this was an open system (after all, none of this experiment was performed in an air tight chamber; we just had plastic cups), both the potato slices and the sucrose solutions were exposed to atmospheric pressure. Therefore, the pressure potentials of both were zero. If the water potential of the potato and the solution were the same and the pressure potentials of both were 0, then it follows that their solute potentials, and by extension, their solute (sucrose) concentrations were equal as well--0.2 M. In the next section, part 1D, we used this knowledge to determine what the potato's water potential was, but for right now, it will have to suffice knowing that the the potato's sucrose concentration was 0.2 M.


Part 1E
Plasmolysis is the shrinking of the cytoplasm of a plant cell in response to diffusion of water out of the cell and into a hypertonic solution surrounding the cell. The cell membrane pulls away from the cell wall and thus leads to the shrinking of the cytoplasm. The onion cells plasmolyze because the cell was placed in a hypertonic solution of NaCl. So as a result, the water wants to leave the cell and thus causes it to shrink. This experiment can be related to real life because in winter grass often dies near roads that have been salted to remove ice. The salt concentration out of the cell is greater out of the cell-creating a hypertonic solution. This causes water to move out of the grass cell causing plasmolysis and the cell shrivel out and die due to lack of water