Osmosis is a form of passive transport, and a specialised form of diffusion. It is the movement of water from a dilute solution to a more concentrated one, through a semipermeable membrane. Hence it is where water moves from a high to a low concentration. I will design an experiment to test this process on a typical example of osmosis’ effect on plant cells, taking as the sample, potato. It is through osmosis and the fluctuating levels of water in the cell that the concentrations of minerals in the plant are regulated.
Expt. 1 Last year, I conducted two experiments in which dialysis tubing was used to simulate the semipermeable membrane. In the first experiment a knot was tied in one end of a length of soaked dialysis tubing and a pipette was used to half fill it with a strong glucose solution. The air bubbles were expelled and a knot was tied in the other end. At this point the tubing was flaccid (limp). When it was then placed in a test-tube full of distilled water, it became turgid (firm). [See diag.1]
Expt. 2 In the second experiment red-dyed sugar solution was placed in a bell-shaped piece of apparatus, with a capillary tube off it and a cellulose film over the bottom [See diag.2]. This was then placed in a beaker full of distilled water and left for half an hour, by which time the level of water in the beaker had gone down, and red dye had moved up the capillary tube. The water in the beaker had not turned red. The experiment was then repeated with a stronger solution. The volume of water in the beaker decreased further; the dye went further up the capillary tube.
These two experiments clearly demonstrated that, while small water molecules pass through a semipermeable membrane, larger molecules (such as glucose) do not. This is why the water in the beaker did not turn red, while the red dye in the capillary tube did become paler. The second experiment also showed that the more concentrated the solution, the more water diffused in by osmosis.
Expt. 3 More recently, I conducted an experiment investigating these processes in a real plant cell. An onion was cut up, and two small pieces of the epidermis (a single layer of cells) were peeled, cut and placed onto microscope slide. Onto one of them was pipetted a few drops of distilled water; onto the other was pipetted a few drops of 1 molar glucose solution. Cover slips were then placed onto the two pieces of onion epidermis. [see diag.3a]. These were then placed under the microscope, and their cells looked identical [see diag.3b]. Fifteen minutes later, however, a marked change had been observed. While the onion cells bathed in distilled water had become turgid [see diag.3c], those bathed in glucose had been plasmolysed [see diag.3d].
Plasmolysis occurs when water diffuses out of the cell vacuole, causing the cytoplasm to have a decreased volume, in turn causing the cell membrane to pull away from the cell wall. It only occurs when the cell is hypoosmotic relative to the bathing solution. That is to say that plasmolysis only occurs when the cell’s solute potential (?s) is lower and therefore water potential (?w) is higher than that of the bathing solution. The bathing solution is hypertonic: this means that there is a higher concentration of solute, and thus lower concentration of water, in the bathing solution than the cell. Totally dilute water has infinite water potential, and zero solute potential.
In Expt.3, the water potential of the onion cells in the glucose solution was greater than that of the onion cells in the distilled water. So, because the onion cells in the glucose solution were hypoosmotic, they were plasmolysed. Whereas the onion cells in the distilled water were hyperosmotic (had higher solute, and therefore lower water, potential), so became turgid. In my investigation of osmosis in potato cells, I hope to be able to determine the water potential of the cells. I will do this by finding the isotonic (equal solute, and therefore water potential) bathing solution.
Key Variables – I will be testing a total of 5 bathing solutions (0.2, 0.4, 0.6, 0.8 mol glucose solutions & distilled water). I will require 10 potato pieces (two for each solution). As they will be cut from a slice of potato 1 cm tall, the height is already constant. I must also ensure that the breadth of each piece is constant: 1cm. It is vital that the breadth and height are constant so that it is possible to measure any change in length. Of course, in order to measure the change, it will be necessary to ensure that the initial lengths are constant, and to ensure that I measure the same sides at the beginning as the end. To do this, I will take a tiny segment out of one side (I will measure the other).
The initial lengths will be 2.5 cm. Obviously, it is hoped that the potato will have an even density, but in case it does not, the individual masses will be measured beforehand as well as afterwards. Through the use of such a precise piece of apparatus, I hope to be more accurate in my final readings. Also, this way the comparison between increase/ decrease in the lengths of potato pieces may be reinforced by a comparison between the increase/ decrease in their masses. So the readings will not only be more accurate, but also more reliable.
The initial volumes of the bathing solutions must be constant, in order to better compare any changes in volumes. The initial volumes will be 20ml. The amount of time during which the potato pieces are left in their bathing solutions must be constant. The time set will be 30 minutes. The only variable left is the strength of the bathing solution. These are: 0.2, 0.4, 0.6, 0.8 mol glucose solutions & distilled water. I am using these solutions because I want to show some potato cells become turgid and be plasmolysed: and I will need this sort of range to do so, because potato has a lower water potential than onion (and will therefore need a stronger solution than that used in Expt.3).
Aim – To discover the water potential of the average potato cell in terms of the relative isotonic bathing solution.
Hypothesis – As demonstrated in the three experiments in the Introduction, osmosis determines that water travels from high to low concentration over a selectively permeable membrane. In the planned experiment, when the bathing solution is hypertonic (lower water potential, higher solute potential) relative to the potato cell, water will flow out of the vacuole into the bathing solution (plasmolysis, Expt.3). This will cause the potato cells to become flaccid, shrinking and losing weight. The potato pieces that lose mass, and decrease in length will be the ones with high water potential, relative to the bathing solution. I think this will happen in the distilled water, because completely distilled water has infinite water potential.
When the bathing solution is hypotonic (higher water potential, lower solute potential) relative to the potato cell, the cell will become turgid because water will flow from the bathing solution into the vacuole. The potato pieces that increase in length and gain mass will be the ones with low water potential, relative to the bathing solution. I predict this for the strong glucose solutions (0.4, 0.6, 0.8 mol), because I have seen from experiments 1&2 that strong glucose solutions have low water potential.
When the bathing solution is isotonic (equal water potential, equal solute potential) relative to the potato cell, the cell will not change. The potato pieces that neither increase nor decrease in length and neither lose nor gain mass will be the ones with the same water potential, relative to the bathing solution. I believe this will occur in the weakest glucose solution (0.2mol) because, although a 0.1mol solution plasmolysed the onions in Expt.3, potatoes have a lower water content than onions (85% as opposed to 89%), meaning that they will also have a lower water potential.
I am therefore predicting that the water potential of the average potato cell is equal to that of a low strength glucose solution. I say this because I know that, in distilled water, the cells become turgid; and in strong glucose solutions, the cells plasmolyse.
Apparatus – slab of potato 1cm thick, scalpel, white tile, solutions (0.3, 0.6, 0.9 mol glucose), distilled water, 5 boiling tubes, clock, ruler (mm), electric balance (to 0.01g), and a measuring cylinder, forceps, boiling tube rack, paper towels.
Planned Method – First I will cut out the 10 potato pieces (10x10x25mm) on the white tile with the scalpel, taking a chunk out of one side [see diag. 4a]. Then I will measure their masses on the electric balance. Then I will measure out the volumes of the bathing solutions (20ml each) in a measuring cylinder (the measuring cylinder will be dried after each measure) [see diag. 4b]. Then I will pour the bathing solutions into the 5 boiling tubes, which I will have labelled, and two potato pieces will be placed into each boiling tube and the timer started [see diag. 4c]. 30 minutes later, I will remove the 10 potato pieces, rinse (briefly) and dry them, and measure their lengths and masses. These will then be recorded. Then the volumes of the bathing solutions will be recorded (again, the measuring cylinder will be dried after each measure).
Safety Precautions – As with all substances in a science laboratory, it is vital that I do not taste the potato, the solutions or the distilled water. I must not touch my face or licking my fingers before washing my hands, but I should attempt to avoid getting the liquids if possible, as it risks contaminating our results. Evidently, care must also be taken with the scalpel in the cutting of the potato, as it is a sharp instrument.