Sneha Patel Page 5/6/2007
To determine the water potential of a potato tuber cell using varying salt solution.
Osmosis is the diffusion of water molecules from a region where it has higher water potential to a region where it has lower water potential through a partially permeable membrane. As osmosis is a type of diffusion the same things that affect diffusion have an effect on osmosis some of theses things are:
- The concentration gradient – the more the difference in molecules on one side of the membrane compared to the other, the greater the number of molecules passing through the membrane and therefore the faster the rate of diffusion.
- The surface area – the larger the area the quicker the rate of diffusion
- The size of the diffusing particles – the smaller the particle the quicker the rate and polar molecules diffuse faster than non-polar ones.
- The temperature – the higher the temperature the more kinetic energy the particles have and so the faster they move.
From the diagram we can see the process of osmosis in a simple expression. On the right side there is pure water, which has the maximum water potential of 0. Water potential is the pressure created by water. As you can see from the diagram the pure water is pushing its way through the semi permeable membrane at a high pressure. This is its water potential. Water potential is measured in kilopascals (kPa) and is represented by the letter Ψ.
I predict that the lower the concentration of salt solution the higher the water potential will be and therefore the higher the % change in mass will be, I predict this because osmosis is the diffusion of water molecules from a region where it has higher water potential to a region where it has lower water potential through a partially permeable membrane. Therefore the distilled water will have the highest water potential and the 1.0 mole of salt solution will have the lowest water potential. This means that the potato cell that is left in the distilled water should gain the most weight according to the definition of osmosis.
There are some things that need to be considered, water potential is affected by solute potential and pressure potential.
When solute molecules are dissolved into water, the concentration of water molecules is reduced, and hence so is the water potential. The solute potential is devised by the amount of solute in the experiment in this case it would be the salt. Solute potential is always in negative numbers as the only positive solute potential is that of pure water and that does not include any solutes. Its symbol is Ψs.
Pressure potential has the symbol Ψp. When water enters plant cells by osmosis, pressure can build up inside the cell, this increases the pressure potential. Pressure potentials are caused by resistance of the tissues to water flow.
To calculate water potential this formula is used:
Water potential Ψ = solute potential Ψs + pressure potential Ψp
The differences in pressure potential of potato cells placed in different concentrations of salt solution, will cause the cell to react in different ways. The diagram below depicts what happens to the cell:
Figure 2 Plasmolysis and water potential in a plant cell
A As you can see from the first picture, the plant cell has been placed in a high concentration of water for example pure water. The water is entering the cell causing a huge amount of pressure thus causing the cell to swell and become turgid. When the cell becomes turgid the protoplast (the outer cell surface membrane, cytoplasm and the tonoplast) starts to swell and is pushed against the cellulose cell wall because of the pressure of water entering the cell. Eventually the pressure increases to a point where the cell resists any more water entering the cell. The protoplast is staying pressed against the cell wall and this point the cell has become turgid. Plant cells always have a strong cell wall surrounding them. When the take up water by osmosis they start to swell, but the cell wall prevents them from bursting. Plant cells become "turgid" when they are put in dilute solutions. Turgid means swollen and hard. The pressure inside the cell rises, eventually the internal pressure of the cell is so high that no more water can enter the cell. This liquid or hydrostatic pressure works against osmosis.
B The middle picture depicts what should occur when the water potential inside and outside of the cell is at equilibrium. At this point the protoplast is no longer being pressed against the cell wall and is beginning to pull away. At this stage the pressure potential equals 0. The plant cells are placed in a solution, which has exactly the same osmotic strength as the cells they are in a state between turgidity and flaccidity, this stage is known as incipient plasmolysis.
C Finally the final picture illustrates what happens to the cell when the water potential outside the cell is lower then inside the cell. Water leaves the cell and as it leaves the protoplast completely pulls away from the cell wall. This causes the cell to shrink and we say it has become flaccid or plasmolysed.
I predict from my preliminary results that my graph showing the relationship between concentration of salt solution and % change in mass will look something like this:
There are 3 basic stages to this graph and these are co ordinate to the stages shown in the Figure 2.
A (between 0.00 and 0.30 mole) = where the cell has become turgid because it has been placed in a high water potential solution. Water is entering the cell because the water potential outside the cell is higher then inside the cell, and as we know water moves from a region of high potential to a region of low water potential. Here the cell will gain mass according to the definition of osmosis. I predict that for the potato cells placed in between the concentrations of 0.00 and 0.40, the cells will gain mass and have a high water potential.
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B (between 0.30 and 0.50 mole) = the amount of water potential outside the cell is equal to that of inside the cell therefore the cell will not lose or gain wait, this point is called incipient plasmolysis. When plant cells are placed in a solution, which has exactly the same osmotic strength as the cells they are in a state between turgidity and flaccidity. We call this incipient plasmolysis. (the pressure potential is equal to zero so the Ψ = Ψs therefore there is no movement of water)
C (between 0.50 and 1.00 mole) = the cell has become flaccid or plasmolysed and has lost water. Therefore the cell has lost weight. I predict that between concentrations of 0.50 and 1.00 the potato cell will lose weight and become flaccid. Flaccid cells are when the cell has lost so much water that the inner membrane (the protoplast) pulls away.
With the information from the diagram my preliminary graph and scientific knowledge, I predict that the potato that goes into the 0.00 concentration, with the highest water potential will gain mass as more water will be entering the cell and I believe that the cell will then swell and become turgid because the water potential outside the cell is more then inside the cell therefore water will enter the cell as the theory of osmosis states water molecules from a region where it has higher water potential to a region where it has lower water potential and so I expect the potato cell that goes into the 1M solution to lose water and become flaccid as the water potential of the cell is higher then outside the cell so I predict that water will leave thus leaving the potato cell flaccid. I predict that between 0.00 and 0.35 the potato cell will gain mass and between 0.50 and 1.00 it will lose mass. I predict that between 0.35 and 0.50 concentrations incipient plasmolysis will occur giving a water potential of between –970Ψ and –1450Ψ.
From the preliminary these are the changes that have been made to insure an even more reliable and accurate experiment:
- In the preliminary the potato pieces were not cut in accurately sized pieces. For this experiment I will cut them accurately using a knife and ruler. I will also make sure they are all the same as the surface area is a variable that may distort the results.
- Another thing that will be changed is the sucrose solution. The solution is a good environment for bacteria to grow and therefore made the preliminary experiment inaccurate. The sucrose solution was also quite thick so when I weighed the potato pieces some sucrose solution that was still on the potato piece could have misrepresented the final mass of the potato. Therefore in this experiment I will use a salt solution that is of 2M.
- In the preliminary the experiment was not repeated so if something had gone wrong there was nothing for it to be compared to. This made it difficult to point of anomalous points. Therefore in this experiment I will repeat it so that it is a reliable experiment.
- Another point is using a syringe to measure out the salt solution instead of a measuring cylinder, which has a higher percentage error than a syringe.
To insure a safe experiment the following care will take place even though no harmless chemicals are being used:
- Normal laboratory procedure should be in order.
- First of all I will wear goggles, as I will be using a knife to cut the potato. Obviously extra care needs to take place while using the knife, as it is very sharp.
- Secondly a clean tidy environment is needed so that nothing is knocked over or spilt.
- Finally long hair will need to be tied back so that it does not interfere with the experiment especially when cutting the potato pieces.
This is where I will record my results.
A table that shows the change in mass between potato cells placed at different salt solutions, in two runs and then finally showing the average change and average % change in mass between the potato cells.
- First of all of the safety measures should take place. Clean tidy environment, hair tied back and goggles. Also all stools should be tucked in and any clothes and bags kept out of the way.
- Make sure a results table like the one above is ready before the experiment begins, also dilution table like this one should be prepared:
- Make sure the 12 test tubes are labelled with the concentration amount going from DW to 1M.
- Now set up the apparatus in the way shown in the diagram below:
- The dilutions are the first thing that you should do as if you cut the potatoes first they will be left outside and could get contaminated. Therefore you should cut the potatoes jus when it is necessary.
- Make sure all the boiling tubes are in the boiling tube racks.
- Take the bottle of distilled water and empty some out into one of the beakers. Then take the salt solution beaker and make sure you don’t forget which one is which. Now using the 20ml syringe take out 10ml of water and put it into the boiling labelled distilled water.
- For the rest of the boiling tubes, using the dilution table take out the required amount of salt solution and the required amount of distilled water until of the boiling tubes have been filled.
- Now get the potato and cork borer and cut the potato pieces.
- Using the ruler, knife and forceps cut the potato pieces in equal 2 cm long pieces, putting each one on a sheet of filter paper when finished.
- After that begin to weigh each potato piece, weighing the filter paper first and then the potato piece and the filter paper together. Then use this equation to calculate the mass of the potato piece on its own:
Mass of potato piece — Mass of filter paper = Mass of potato piece
and filter paper.
Record these results on to the results table.
- Now place one potato piece into each of the boiling tubes closing it will the stopper. Note: remember which one you started with, as you will have to remove them in the same order.
- After this leave the potatoes in the boiling tubes for 45minutes.
- When the potatoes have been left for 45 minutes take them out in the order you put them in. to do this you will have to remove the excess solution into a beaker.
- Put each potato slice onto the filter paper it was on before and then weigh them like before.
- Record these results also on the results table.
- The experiment is now complete. Tidy the area and make sure everything is put away properly
From my graph and results I conclude that my prediction was relatively correct. There is a simple trend that can be seen from my graph, which is: the lower the concentration the salt solution the higher the water potential and the higher the amount of mass gained. The potatoes placed in a high water potential have gained mass and the potatoes placed in a low water potential have lost mass.
This trend can be seen throughout the graph:
At the point 0.00M the % change in mass is 14.71g, here the potato piece has gained mass. This is because the theory of osmosis states that water molecules will move from a region where there is higher water potential to a region where there is lower water potential. This means that, as I had predicted the water potential within the cell was less than outside the cell therefore water entered the cell. The water has entered the cell and caused the protoplast to press against the cell wall this is how the cell has swollen and turgid, this effectively increased the mass of the cell. At 0.00M the protoplast has been pushed against the cell wall:
And until 0.34M the graph shows a positive % change in mass. This means that until 0.34 the potato cells continued to gain water because the water potential inside the cell was lower then outside the cell an as the theory of osmosis states water travels from a region of high water potential to a region of low water potential.
At point 0.34 the pressure potential was zero. When the pressure potential equals zero the solute potential is equal to the water potential.
Ψ = Ψs + Ψp
Therefore if Ψp = 0 then Ψ = Ψs
At the point 0.34 incipient plasmolysis occurred. Incipient plasmolysis is when there is no net movement of water into or out of the cell, the protoplast is no longer pressed against the cell wall and so the Ψp equals zero using this number and the formula I calculated my water potential using my reference graph.
My reference graph shows the solute potential against the concentration. Using this graph I was able to read of the actual value of the solute potential and therefore the water potential. Using the reference graph I read of at 0.34M where my incipient plasmolysis had take place, I then concluded that the water potential of the cell was –950. I had originally predicted that it would be between –970 and –1450. I think my prediction is relatively close to the result.
At the 1.00M point the % change in mass is –24.24. Here the cell has become flaccid as I predicted. The inside cell has a higher water potential then the outside so the water has left the cell causing it to shrink. The graph shows that when the water potential inside the cell is higher than the outside i.e. after 0.34M, the cell begins to loose water instead of gaining it because the water molecules travel from a region of high water potential to a region of low water potential. Water has left the cell and as it leaves the protoplast has completely pulled away from the cell wall. This caused the cell to shrink and we say it has become plasmolysed.
Constant error table: (Errors which affect all the points)
Single Error table: (Errors that affect one or more single points. could be a cause for anomalous points.)
I think my experiment was quite accurate although it seems as though most of equipment was quite accurate the % error does show some signs that it wasn’t.
The % error is calculate using this formula:
Divisions of equipment 100 = %error
Total capacity of equipment
The % error for the apparatus I used was:
Syringe 5ml with 1ml divisions:
1/5 x 100 = 20%
Syringe 20ml with 2ml divisions:
2/20 x 100 = 10%
Weighing scale with 0.001g divisions:
0.01/1.7 = 1.7%
To calculate the overall % error for the whole experiment I have to add up all the % errors/ All together the % error for this experiment was = 31.7% which is a considerably large number. This means the experiment was not as accurate as I had predicted it would be. Using better equipment with a lower % error could have made the experiment more accurate.
Overall the experiment was accurate as the point of incipient plasmolysis
I had one anomalous point which was off the line of best fit this shows that one or more of the single errors must have distorted the result. The anomalous point that has been circled on my original graph shows this error bar graph however shows that although I have only circled one anomalous point originally with the error bar graph I can see that only 3 out of the 6 points were within a 5% margin of error.
Although I have only circled one anomalous point originally with the error bar graph I can see that only 3 out of the 6 points were within a 5% margin of error. The error bars are worked out by standard error. Standard error is the measure of error that could have occurred. The smaller the standard error the smaller the error bars. The smaller the error bars the more reliable the experiment. As you can see most of the error bars are quite small making my experiment accurate. The error bars show how accurate the experiment was. These error bars show the accuracy within 5%. So if the point was affected by some errors where it could have been. Therefore the smaller the error bars the smaller the error.
Suitability of Method
My method was quite suitable for the experiment it did allow me to gain a suitable and sufficient result. There were some limitations such as time and the range of equipment available such as buffer solution, but overall the method fulfilled its purpose to help me find out the water potential of the potato pieces.
I think my experiment was reliable to an extent as I did conclude a result that satisfied my prediction. I think my experiment was reliable because most of my equipment was consistent as I didn’t have any problems with any of the apparatus, also the fact that the experiment was repeated and similar results were achieved both times showed it was a relatively steady experiment. From the results I can see that only one of the point is an anomalous point this could have been caused from the fact that the experiment didn’t have a very good range of concentrations. I think I should have used a wider range of concentrations to see what would happen after 1Mole as the solution that was given was of 2Mole. Also as my constant error table shows, my I believe my most significant error was that of the 2Mole salt solution that was provided having said this my results are similar to others so the experiment must have been reliable.
To gain an even further understanding into water potential I could do experiments that change water potential like the affect of temperature on osmosis and the effect of different types of potatoes. I think this would give more accurate and reliable evidence to the conclusion of my experiment.
Essential AS Biology- Glenn and Susan Toole chp.4.3 pg 60
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Essential AS Biology- Glenn and Susan Toole chp.4.2 pg 59
Essential AS Biology- Glenn and Susan Toole chp.4.3 pg 60
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Essential AS Biology- Glenn and Susan Toole chp.4.4 pg 63
Here's what a teacher thought of this essay
***** An excellent account with a very high level of detail and correct use of A level biological terminology throughout.