DETERMINATION OF THE WATER POTENTIAL OF POTATO TUBER CELLS. Method. Five sucrose solutions with varying molarity and one control containing distilled water were prepared and poured into test tubes. The potato discs were dried, weighed and added to the test tubes.
The discs were then weighed again after a period of 24 hours. The percentage change in mass was then calculated. Apparatus. Specimen tubes with stoppers x 6 1 cm 3 diameter cork borer razor blade filter papers balance distilled water sucrose solutions with varying concentrations potato cut into small discs Results. (Fig 1. 0) Sucrose concentration (M) Initial mass (kg) Final mass (kg) Change in mass (%) 0.
10 0. 95 1. 25 31. 570. 20 0. 94 1.
13 20. 210. 30 1. 03 1.
15 11. 650. 40 0. 95 1. 05 10. 520.
50 0. 88 0. 86 -2. 270.
60 0. 93 0. 84 -9. 67 Control 0. 99 1.
40 42. 41 Discussion. Osmosis is the passive diffusion of water molecules across a selectively permeable membrane from a down a concentration gradient. The water potential of a system is the tendency for water to exit the system. In this experiment the aim was to measure the tendency for water to leave the tuber cells.
As the water potential of pure water is zero the concentration of sucrose in solution will have an effect on the water potential, this is called the solute potential. The greater the concentration of sucrose the more negative the water potential, because water moves from a high to low water potential. When the potato is put into water it contains solute molecules which draw water in providing the external solute concentration is lower. The more solute molecules present the lower the water potential such change is referred to as the solute potential. To find the water potential of the cells we need to find out at which concentration of sucrose solutions was a state of equilibrium obtained, i. e.
The Essay on Water Potential of a Potato
... water potential of a potato I will be placing fixed volumes of potato, with the same surface area, into different concentrations of sucrose ... to relationship of water potential, solute potential and pressure potential: = s + ... Solute Potential. As the presence of a solute lowers the water potential, the more concentrated the solute is, the lower the water potential will be. Pressure Potential. ...
the solution in which there was no change in the volume or mass of the tissue; which is equal to that of the cells. When this happens we know that the solute potential and water potential of the potato and the sucrose solution will be the same. We know this because if this weren’t the case then there would be a change in mass. Figure 1. 1 shows the relationship between the molarity of the solution and the percentage change in mass of the tuber sample.
This was used to find the concentration at which there was no net exchange of water. I found this to be 0. 46 mol. dm-3. When I applied this value to figure 1. 2 (which displays water potential against sucrose concentration) I found the water potential for 0.
48 to be -1420 k Pa, which as explained is the same as the water potential for the potato tuber cells. The graph plotted in figure 1. 1 showed the kind of downward curve I expected to see. At 0. 0 M the % change in mass is 42. 41%, what’s happened here is the cells have taken up a lot of water and become almost fully turgid.
This happens because their water potential is more negative than that of distilled water, which is zero k Pa. When a cell becomes fully turgid the pressure exerted on the cell wall prevents anymore water from entering. This technique can be observed in guard cells undertaking stomatal opening. A stoma opens when its guard cells actively take up K+ and water from surrounding cells enters by osmosis. When the vacuoles in the guard cells gain water, the cells become turgid and swell. This causes them to buckle outward increasing the size of the gap (stomatal aperture) between them.
When the stoma is open the plant takes in CO 2 during the day for photosynthesis. However turgidity is not only used in stomatal activity. It is an important part of the plant’s structure, giving it strength to stand on end without which the plant would simply wilt. From 0.
The Essay on Dconcentrations Of Solutions Determine The Mass Of A Potato
Introduction: The way to get the full results of this lab was through the process of osmosis. Osmosis is the movement of water across a membrane into a more concentrated solution to reach an equilibrium. When regarding cells osmosis has three different terms that are used to describe their concentration. The first of these words is isotonic. Cells in an isotonic solution show that the water has no ...
0 M to 0. 3 M the percentage change in mass decreases which is to be expected as the water potential of the solution is becoming more negative, which means that less water will transfer from the cells into solution. At 0. 3 M to 0. 4 M there is a slight anomaly as the change in mass goes down by only 1%, which can easily be seen as the gradient of the line alters dramatically. However between 0.
4 M and 0. 5 M the gradient returns to that observed between 0. 0 M and 0. 3 M, the change in percentage change also returns to normal. This anomaly is most likely due to an error made when weighing the sample either before or after being put into the solution. It could also have happened because the sucrose solution may have been contaminated, or prepared inaccurately, e.
g. if the molarity had been closer to 0. 2 M then the percentage change in mass would have been greater, that would explain the result found. From 0. 5 M to 0. 6 M the gradient of the line alters slightly which could be a sign that plasmolysis is starting to occur.
Plasmolysis occurs when the external solution has a more negative water potential than the internal solution of the cell. As a result water is drawn out of the cell and the plasma membrane loses contact with the cell wall, and hence the pressure potential falls to zero. This point is referred to as incipient plasmolysis, and when it occurs the cell shrinks and becomes flaccid. This can also be observed in stomata during the night. There is no need for CO 2 so the stomata are kept closed to prevent the loss of water. To close the guard cells lose K+, which incurs a loss of water.
The cells become flaccid an sag together, closing the space between them. This use of ions to control water uptake can be reversed; and water can be used to encourage the uptake of mineral ions from the soil into the roots of plants. This experiment was quite limited as the degree of experimental error that could occur gives rise to scepticism over the validity of the results. The cutting up of the potato into pieces of equal surface area and mass was extremely difficult given the equipment available. If error had occurred then this would have an effect on the amount of water moving to and from the cells. However this may have had little influence in this case, as the potato discs were meticulously weighed and cut out.
The Essay on Change In Mass Solution Potato Cell
... Formula = Change in mass (+ or -) x 100 Original mass. Results table Solution strength (m) Original mass (g) Final mass (g) Change in mass (g) Calculate percentage changing in mass. Beaker 1. ... happens when a plat cell is placed in a solution weaker then the solution in a cell? Osmosis takes place. Water diffuses into the cytoplasm ...
Another area of likely error was the drying out of the discs. There was a degree of difficulty in deciding if the discs were dry enough, and the length of drying time each disc received. If the discs had not been dried sufficiently then the percentage change in mass recorded would have been greater than before. Such a change would result in a more negative water potential, making the results less accurate. If given the chance to repeat this experiment I would like to use more solutions ranging from 0. 30 M to 0.
60 M to give a more accurate graph, thus giving a more accurate determination of the water potential. I would also like to determine with the same degree of accuracy the point of incipient plasmolysis.
