When water potential outside of the cell is equal to the water potential inside of the cell, it is in a state of equilibrium.
From previous experiments, I know that carrots have more sucrose dissolved in them than celeriac and potatoes.
Results for reducing sugar test
When I carried out the Benedict’s reagent test for reducing sugars, such as glucose, I found that when the carrot was heated in a water bath, the solution of Benedict’s reagent turned from the Copper II Sulphate blue, to a brick red precipitate. Whereas when the potato was heated in the water bath, the colour change was from the Copper II Sulphate blue, to an orange precipitate. I also carried out the test on celeriac. This had the least amount of reducing sugar present due. It had a very subtle colour change; from the Copper II Sulphate blue, to a faint green solution.
The results showed that the carrot contained a vast amount of reducing sugar and the potato contained slightly less than the carrot. It also showed that the celeriac had the least amount of reducing sugar present, so I decided to perform the test for non-reducing sugar on the celeriac.
Result from non-reducing sugar test
When the solution was placed in the water bath, yet again there was a very subtle colour change; from the Copper II Sulphate blue, to a faint green solution. This led me to believe that celeriac had very little reducing sugar and also very little non-reducing sugar present. This shows that celeriac must have a less negative water potential than the carrot and potato. I think this because it has much less solutes dissolved in its cells.
After the reducing sugars test, I placed the test tubes into a colorimeter. This measured the percentage transmission of light through the test tube.
Results from colorimeter
The results show that the celeriac let most light through. This means that there was much less precipitate than the carrot or potato.
I also performed the test for starch on all three specimens.
Results from starch test
The result for a positive starch is a colour change from an orange/brown iodine solution to black. My results show that the potato contained the greatest quantity of starch and the celeriac had less than the potato. It also shows that the carrot has the least amount of starch present. Starch is a relatively insoluble molecule, which has little effect on the water potential of a cell.
When performing a recent experiment, I found the water potential of a carrot to be -1060kPa, and the water potential of a potato to be -680kPa. My previous experiments and general knowledge have led me to believe that the celeriac will have a less negative water potential than both the potato and carrot. I think this because it has less solute dissolved in its cells than a potato or carrot. I estimate the water potential of the celeriac to be less than -680kPa.
Method
Firstly, pour about 100ml of 1.00 mol dm-3 sucrose solution into a beaker and about 100ml of tap water into another beaker. Then, place 5 test tubes into a test tube rack and label them 1 - 5.
Using one 10ml graduated pipette for water and one for 1.00 mol dm-3 sucrose solution, add the following volume of liquid into each test tube.
Use the cork borer, cut 5 cylindrical pieces of celeriac from a single piece of celeriac. Cut each cylindrical celeriac piece to a length of 5cm with the razor. Using the scales, measure the mass of each cylindrical piece of celeriac and record the mass in a table. Place one piece of celeriac into each test tube and start the stopwatch. Place 5 paper towels in the vicinity of the test tubes and label them 1-5. After 15 minutes, empty all test tubes over a sink, while keeping the celeriac in the tube. Place each celeriac piece on to its corresponding paper towel. Use the paper towel to dry off any excess solution from the celeriac pieces. Using the scales again, measure the mass of each celeriac piece and record in a table. Subtract the mass before the experiment from the mass after to find the change in mass. Using this formula – percentage change in mass = change in mass/original mass x 100 - calculate the percentage increase or decrease in mass. Plot these values on the y-axis of a graph against the concentration of sucrose solution on the x-axis. Join the points with straight lines and determine the molarity at which the percentage change in mass is zero – where the line intersects the x-axis. Using a calibration graph – the x-axis = concentration of sucrose and y-axis = water potential - calculate the water potential of celeriac. Repeat the above method at least another two times.
Independent variable – we will control the molarity of the sucrose solution. We will use 5 different solutions of sucrose, 0.00, 0.25, 0.50, 0.75, and 1.00 mol dm-3. We will dilute the 1.00 mol dm-3 sucrose solution using water. Each test tube will contain 10ml of liquid.
Dependant variable – we will measure the mass of the celeriac before the experiment and then again when the experiment is over. I will then subtract the mass before the experiment from the mass after to find the change in mass. Using this formula – percentage change in mass = change in mass/original mass x 100 - I will calculate the percentage increase or decrease in mass. I will then plot these values on a graph against the concentration of sucrose solution.
Constant factors – the temperature will be a constant factor throughout the experiment. It will be kept at room temperature at all times. This is because the temperature affects the amount of kinetic energy. For example if the temperature in test tube A is higher than the temperature in test tube B and assuming every other factor was constant, the water molecules in test tube A will have more kinetic energy and therefore osmosis will occur quicker. This will affect the mass of the celeriac in test tubes A and B because another factor that will be kept constant will be the length of time the cells are in the different solutions. Therefore, test tube A will have a larger percentage increase/decrease in mass than the test tube B.
As I have said above, the length of time will be a second constant factor throughout the experiment. We will carry out the experiment for 15 minutes. For example, if test tube C had a time limit of 10 minutes and test tube D had a time limit of 15 minutes and assuming everything else was constant, then at the end of the experiment, test tube D would have the greater percentage increase/decrease in mass than test tube C. This would be because test tube D was allowed a longer time limit than test tube C, and as a result there was a greater time for osmosis to occur.
The volume of solution in the test tube will also be a constant factor. The volume will be kept at a constant of 10ml. If the volume was to differ in each test tube, more/less osmosis may occur. For example, if test tube E has a larger volume of solution than test tube F and assuming everything else was constant, then at the end of the experiment, test tube E would have the greater percentage increase/decrease in mass than test tube F. This would be because test tube E had more solution molecules in it than test tube F, thus there was a greater number of molecules to diffuse in/out of the celeriac.
Another factor that will be kept constant will be the size and shape of the celeriac we will use. We will ensure constant surface area to volume ratio by using a cork borer, to determine the shape of the celeriac. Then we will cut each celeriac to 5cm in length. This will guarantee surface area to volume ratio is consistent. If the surface area was to differ between test tubes, then this factor would affect the percentage increase/decrease.
For example, if test tube G contained a celeriac of length 5cm and test tube H had one of 10cm, while assuming everything else was constant, then at the end of the experiment, test tube H would have the greater percentage increase/decrease in mass than test tube G. This would be because test tube H had more surface area exposed to the solution molecules than test tube G, thus there were many more channels for the solution molecules to diffuse in/out of the celeriac.
The final factor that will be kept constant will be the celeriac that is used. Cells from different celeriac plants may have a different water potential to each other. So, I will only use one celeriac plant throughout the experiment. If I were to use a range of celeriac plants during the experiment, I cannot guarantee valid results. For example, if test tube J contained a celeriac that was different to that in test tube K, while assuming everything else was constant, the water potential may be different for each celeriac. This would affect the mass of the both celeriac at the end of the experiment. Therefore there would be a greater percentage increase/decrease in mass in one test tube than another. This would be because of the difference in water potential.
Risk assessment
When handling the razor, use with caution and do not leave it lying around. When handling glassware such as test tubes and beakers, handle with care to avoid breaking. When placing test tubes into a test tube rack, ensure they are in a steady position so it is unlikely they will fall out of the rack. Ensure there is a safe working space around you.
Bibliography
1 http://www.visionengineer.com/env/normal_osmosis.gif - this picture helped in my explanation of how osmosis worked.
2 Cambridge Advanced Sciences Biology 1 (Page 56) – this source was useful because it provided a condensed definition of osmosis that was easy to understand.
3 http://web.mala.bc.ca/mcmillan/biology122terms3.htm – this source was useful because it provided a simple definition for water potential.
4 http://dl.clackamas.cc.or.us/ch105-03/osmosis.htm - this source helped to expand my knowledge of water potential and osmosis and introduce new terminology that I now understand and can utilise fluently.
5 Cambridge Advanced Sciences Biology 1 (Page 58) – this source was useful because it showed a diagram (Fig 4.8) of how Plasmolysis occurs and also provided detailed description of the process.
Results from experiment to show mass of celeriac before and after immersion in various concentration of sucrose solution
Using the raw data I recorded from my experiments, I calculated the change in mass. This was found using the following equation:
Change in mass = mass after experiment – mass before experiment
When I found the change in mass, I used this to calculate the percentage change in mass. This was found using the following equation:
Percentage change in mass = change in mass/original mass x 100
The reason for calculating the percentage change in mass is so that I can plot a graph to work out the water potential of the celeriac.
Average results from experiment to show mass of celeriac before and after immersion in various concentration of sucrose solution
The aim of my experiment was to find out the water potential of celeriac. Using my results to plot a graph of average results, I found the concentration at which celeriac cells were at equilibrium. This is when the water potential of the sucrose solution and the water potential of the celeriac cells are equal; therefore water molecules will not diffuse in or out of the cell. It was found to be 0.57 mol dm-3 sucrose solution. I then used a separate graph to convert the concentration of sucrose solution into water potential. The water potential of celeriac was found to be -1700 kPa. From the results, I can see that as the concentration of sucrose solution increases, the percentage change in mass of the celeriac discs decreases. For example, when the molarity of the sucrose solution is 0.2 mol dm-3, the average percentage change in mass is 14.4%. However, when the concentration is increased to 0.5 mol dm-3, the percentage change in mass has decreased to 4.3%.
Through sections A and B of the graph, the gradient of the graph stays almost constant. The average percentage mass decreases consistently at a similar rate until section C. This is where the gradient of the graph suddenly becomes a lot steeper; therefore, the average percentage change in mass is greater than at sections A and B. The celeriac cells start to lose more mass at section C. During section D, the gradient of the graph becomes flatter; therefore the average percentage change in mass is less than at section C. The final section of the graph, E, the gradient of the graph becomes more flat than at section D. The gradient is similar to that of sections A and B. The outcome of the varying gradients, lead to a sigmoid graph (S-shaped graph), which are typical of graphs that compare percentage change in mass and molarity.
When the celeriac discs are placed in a hypertonic solution such as at point E, 1.0 mol dm-3 sucrose solution, osmosis states that the net diffusion of water molecules will be from a less negative solution to a more negative solution. Therefore the movement of water will be from the celeriac cells, through the partially permeable membrane and into the sucrose solution. The celeriac discs will lose mass and become flaccid, and may even become plasmolysed, i.e. when the protoplast has no contact with the cell wall at any point. My average results show that this occurs. When the discs are placed in a 1.0 mol dm-3 sucrose solution, the celeriac cells lose 12.1% of their original mass.
When the celeriac discs are placed in a hypotonic solution such as at point A, 0.0 mol dm-3 sucrose solution, osmosis states that the net diffusion of water molecules will be from a less negative solution to a more negative solution. Therefore the movement of water will be from the sucrose solution, through the partially permeable membrane and into the celeriac discs. The celeriac discs will gain mass and become turgid, and may even burst due to excess water uptake. My average results show that this occurs. When the discs are placed in a 0.0 mol dm-3 sucrose solution, the celeriac cells gain 19.3% of their original mass.
In conclusion, I have completed the aim of my experiment: to investigate the water potential of celeriac. I found this to be -1700 kPa. I also proved my prediction; as the concentration of sucrose solution increases, the mass of the celeriac discs will decrease. My results proved this because when I put my average results onto a graph, the slope was at a negative gradient.
Evaluation
I think that the method was suitable, when considering the apparatus we used. But it was not as accurate as possible. This was partly due to the apparatus and human error. Regardless of this, the method we used did provide the necessary information to prove the prediction and complete the aim of the investigation.
When performing our experiment, I found a number of sources of error. Firstly, when measuring the depth of the celeriac discs to 2mm, the ruler I used had 1mm gradations. This meant that I could have a potential error of 0.5mm. I also found it difficult to cut such a small depth accurately. This meant that there was a chance of the discs being cut too long or too short, which would have affected the results.
Another source of error I found was timing all of the celeriac discs correctly. There were 5 celeriac discs per test tube and 6 test tubes all together. Therefore, when I placed the celeriac discs into one test tube I started the stopwatch. This meant that I had to quickly fill the remaining 5 test tubes with the celeriac discs. I found it difficult to put the discs into the solution accurately due to the small diameter of the test tubes. This meant there was a time delay between the stopwatch being stated and all celeriac discs being placed in the appropriate test tubes. Therefore, the test tube that had celeriac discs placed in it first was immersed in the solution for a longer period of time, which would have affected the results.
A final error I found was when blotting the celeriac discs of excess solution. We were told to blot each disc 10 times. But it was hard to keep the force applied to the discs constant. This means that some discs may have been blotted too much or too little which would have affected the results.
The equipment we used was not as accurate as possible. For example, the ruler we used had 1mm gradations. We had to cut discs with a depth of 2mm. The ruler had a potential error of 0.5mm, therefore dramatically affecting results. Using a ruler was not an accurate method because it was hard to cut so accurately to such a small depth. The inaccuracy of the ruler may have lead to some discs having more or less surface area exposed to the solution. Therefore, some results represented a false percentage change in mass, either losing more mass than they should have or not losing enough mass. The solution would be to have a machine that cut pieces equally, therefore eliminating any human error. This improvement would ensure that the surface area would stay constant on all pieces of celeriac; therefore each disc would show a true representation of the change in mass.
There was only one anomaly in our experiment. This was at molarity 0.0 mol dm-3 sucrose solution, by group 3. The result shows a percentage increase in mass of 47.8%. This could be due to a number of reasons. When finding weighing the discs on the scales, the wrong value may have been plotted in the table of results. Or when removing the discs from the 0.0 mol dm-3 sucrose solution, they may not have been blotted enough or at all, therefore giving an inconsistent and anomalous percentage change in mass. This anomaly demonstrates one of the above sources of error i.e. the force applied when blotting the discs.
In the method, I would alter a number of things. Firstly, instead of cutting up the celeriac discs manually and inaccurately, I would use a machine that cut pieces equally, therefore eliminating any human error. This improvement would ensure that the surface area would stay constant on all pieces of celeriac; therefore each disc would show a true representation of the change in mass. This would increase the reliability of the results. Another change would be to eliminate the time delay involved when placing the discs into the test tubes. This would be done by using a funnel in the test tubes into which I would place some paper to rest the discs on. Then I would simultaneously remove all pieces of paper and the celeriac discs would drop into the solution at the same time. This would eliminate the delay that would normally occur between each test tube and therefore, each celeriac disc would have equal amount of time in the solution. This would allow us to see a true representation of the percentage change in mass of the celeriac discs at different concentrations, which in turn will also increase the reliability. Another improvement would be to find a more suitable way of blotting of the discs to absorb excess solution. The range of results for 0.0 mol dm-3 sucrose solution has the greatest range of results. This means that it is the least accurate and reliable sets of results that we have recorded. The range is consistent, around 21.0% for 3 concentration of sucrose. The lowest range is that of concentration 0.6 mol dm-3 sucrose solution. This meant that this set of results was the most reliable and accurate. These large differences in range lead me to believe the main error factor influencing the results would be the blotting of the discs. If this source of error was to be eradicated, I believe the range would be consistent and reduced to at least 20.0%. When considering the large range, my results do not feel very reliable. The range bar for 0.4 mol dm-3 sucrose solution overlaps the range bar for 0.2 mol dm-3 sucrose solution. This is because the highest percentage change of 0.2 mol dm-3 sucrose solution is 24.3% and the highest percentage of 0.4 mol dm-3 sucrose solution is greater than that of 0.2 mol dm-3 sucrose solution, 25.9%. This leads me to believe that when group 3 blotted their celeriac discs for 0.4 mol dm-3 sucrose solution, they did not apply enough pressure and therefore they left some excess solution on their discs. This means that the discs weighed more than there true mass and it therefore affected the results. Regardless of this, I am confident that my other results are adequate. When finally assessing all the evidence, I still believe that conclusions I drew are true, but they may not be based on reliable evidence.