Potential Potential Potential
In the case that the external solution has more negative water potential than the internal solution water is drawn out of the cell by osmosis and the protoplast shrinks. As it does so pressure potential decreases. A point is reached where the protoplast no longer presses on the cell wall, and hence the pressure potential falls to 0. This point is called Incipient Plasmolysis and the cell is said to be flaccid. This would seem to be a good way of measuring the water potential of a cell, but it is impossible to tell when the cell membrane has just lost contact with the cell wall so in practice we say that incipient plasmolysis is reached when 50% of the cells are plasmolysed. This means that the water potential of both the external and internal solutions will be equal.
Looking back to the equation, at incipient plasmolysis the pressure potential will be 0 so:
ϕcell = ϕs
Water Solute
Potential Potential
So by using the method of incipient plasmolysis we are also working out the solute potential of the solution, as it is equal to ϕcell.
Water potential of potato tuber cells – the weighing method
Method
- Gather all the equipment required:
- 6 Specimen Tubes
- Cork Borer
- Potato Peeler
- Ruler
- Razor Blade
- Labels
- Pen and Paper
- Filter Paper
- Digital Weight Measurer
- Test Tube Wrack
- Label the six specimen tubes: distilled water, 0.1M, 0.2M, 0.4M, 0.6M and 0.8M. Pour 30mm of distilled water into the first tube and an equal volume of each of a series of sucrose solutions of different concentrations in the remainder, each tube should be firmly stoppered.
- Using a cork borer and a razor blade prepare six solid potato cylinders. Using the same cork bore every time to ensure they are all the same width and this ensures a fair test.
- Then measure each cylinder and check that they are all 12mm in length. Yet again using the same ruler just to ensure a fair test.
- Then to obtain a larger surface area for the osmosis to take place on cut each of the 6 cylinders into 6 discs each one measuring 2mm length ways.
- Get a circular piece of filter paper spitting it into six equal pieces then place each set of discs from each cylinder with the designated piece of filter paper.
- Weigh each group of discs on the digital weight measurer with the filter paper, then weigh the filter paper alone and subtract its weight from the weight of it and the discs to obtain the weight of just the discs.
- Record the mass of each of these groups into your table.
- Then place one set of discs into the distilled, one set into the 0.1M and so on.
- After 24 hours remove the discs from each test tube. Remove any surplus fluid from them quickly and gently with some filter paper, using a standardised procedure for all of them.
- Now re-weigh each set of discs recording the results in the same table with the appropriate initial weights.
- Then work out the percentage change in mass for each of the different moralities. (Percentage change in mass = change in mass * 100 / original mass)
- Graph the results by plotting the percentage change in mass against the molarity of the sucrose solution. A line of best ft should be drawn and ten where the line crosses the x-axis this is what the water potential of the potato tubers is.
Water potential of potato tuber cells – the weighing method
A table showing m results for different concentrations of sucrose solution on osmosis in potato tubers
Water potential of potato tuber cells – the weighing method
Interpretation
I recorded m data in an easy to read table along the top row I have given the name/heading for each column. Making it ea for a reader to quickly under stand what each column is about. I have then, down the left hand side, given the different concentrations and then in the middle of the table there are all the values, which correspond to the heading and the concentration. This makes the table very easy to read of and use for constructing the graph. I have also giving m table an appropriate heading do that straight away you know what the data which has been tabulated is about.
I then recorded a graph to show the percentage change against molarity of the potato tuber cells, as well as the actual concentration of the tubers. Along the x-axis I have shown the molarity and along the –axis I have shown the percentage change in mass. Where the line cuts the x-axis is where we assume is the water potential of the potato tubers. By using the graph I can look at any trends or extremities in m results. Using this graph I can also use m line of best fit to tell what the percentage of plasmolysed cells would be at any concentration.
By looking at my results I was able to see that when the molarity of the sucrose solution increased the mass of the potato tubers decreased. The graph showed this as having a negative strong correlation. Most of m plotted points were lying close to the line of best fit except for one meaning m results where quite accurate and I could be happy with them.
I will no look at m results for each concentration and describe what is happening and why:
Distilled water had water potential of 0 and contained no sucrose molecules …..meaning the solution was very dilute. As the solution was very dilute it had …..more water molecules than that of the potato tuber and therefore b the …..laws of osmosis the water moved from an area of high concentration …..(Surrounding solution) to an area of low concentration (Potato …..tuber)causing the potato tubers mass to increase by an average of 11%.
- Molar solution had a water potential of –260kPa which meant that this was not a significant enough concentration to be able to force water to move from the potato tuber to the solution as the tuber still had a higher concentration of sucrose molecules in comparison to the 0.1M solution. So the tuber’s mass increased by an average of 5%.
- Molar solution had a water potential of –540kPa but still the potato tuber took in water via osmosis as the concentration of the surrounding solution still must not have had a high enough concentration of sucrose in comparison to the potato tuber meaning the tuber’s average mass increased b 3%.
- Molar solution had water potential of –1120kPa and thus the potato tubers had an average mass decrease of 18%. This must mean that the 0.4M of sucrose is greater than the concentration of sucrose in the potato cells. So the concentration of the potato cells must lie between 0.4M and 0.2M.
0.6 Molar solution carries on the pattern, the potato tuber had an average …..mass change of –26%. As the concentration of sucrose outside the tuber …..is much greater than that inside the tuber the water molecules move out of …..the tubers causing a loss in mass.
0.8 Molar et again has an increase in the concentration of sucrose causing …..the tuber to lose even more water molecules by osmosis, causing an …..average mass change of –38%.
Osmosis Details
Osmosis is the movement of water molecules across a partially permeable membrane from a region of high water concentration to a region of low water concentration. The membrane in this case is that of the potato tubers. The potato tubers have a plasma membrane composed of lipids and proteins. Each eukaryotic cell has as its boundary to the outside a cell membrane (7.5 to 10 nm in thickness) that envelopes the containing specialized membrane-bound components called organelles. The cell or plasma membrane is a lipid bilayer containing proteins, cholesterol, and oligosaccharides that functions as a selective barrier for entry and exit of substances. The plasma membrane, by limiting the transport of some things and facilitating the movement of others helps to maintain the internal environment of the cell, which is different from the extracellular fluid. Under an electron microscope membranes appear to have a trilaminar structure. This is because the lipid bilayers are arranged such that hydrophilic phospholipid groups are oriented toward the outside of the membrane while the more hydrophobic lipid fatty acid chains form the middle of the trilaminar structure. The molecular make up of each half of the membrane is different in that different lipids and proteins are more abundant in one side over the other.
Proteins are a very important part of the cell membrane. Basically they can be classified into two groups based on physical distribution. Integral proteins are embedded within the cell membrane and may in fact pass multiple times through the membrane. Peripheral proteins are loosely associated with membrane surfaces. Carbohydrate portions of glycoproteins and glycolipids are found on the external surface of the cell membrane where they are important parts of receptor molecules. are necessary to cellular signalling, adhesion, and recognition. Most proteins are fixed in place within the cell membrane by interactions with the . However, some integral proteins can move about and sometimes will accumulate on one region of the membrane in a process called capping.
The cell membrane is not static. It is remodeled by the addition of new membrane vesicles from the while removal takes place in the form of endocytotic, phagocytotic and pinocytotic vesicles being formed and then fused with for processing. Membrane receptors and membrane are often conserved and recycled to the plasma membrane. This membrane trafficking is important in the cell economy.
Here is a diagram of the cell membrane:
Osmosis will also occure in plant cells.
