0.50m: 1st attempt 1.69 1.62 -4.142011834
2nd attempt 1.78 1.67 -6.179775281
0.75: 1st attempt 1.76 1.6 -9.090909091
2nd attempt 1.71 1.56 -8.771929825
1.0m: 1st attempt 1.74 1.59 -8.620689655
2nd attempt 1.4 1.25 -10.71428571
Percentage change in accordance to the varying solutions
Concentration Average % change
0.00m 4.94128412
0.25m -2.11022273
0.5m -5.16089356
0.75m -8.93141946
1.0m -9.66748768
These preliminary results give me an overall impression on the change in mass gain or loss when placed in varying concentrations of sucrose solution.
Planned method:
A range of sucrose sugar solutions will be prepared with concentrations 0 molar, 0.25 molar, 0.5 molar, 0.75 molar and 1 molar. This will be done by adding varying amounts of distilled water to varying amounts of sucrose solution. Sections of potato will be cut using a scalpel and will be measured using a ruler. This part of the preparation must be done very accurately as a change in the surface area may allow more or less osmosis to occur. The mass of each chip will be measured as well so that more results can be obtained. Three chips will be placed in each test tube each time so that I can take an average for each tube. I will use 10 ml of each concentration of sugar solution and once in the test tubes they each will be labelled. The potato pieces will then be placed in the different test tubes and then left for 30 minutes. Then the potato pieces will be removed, the surface solution removed using paper towels and then they will be re-weighed. If I then have time afterwards I will repeat this experiment again as to obtain a second set of results. This will hopefully produce more accurate results from which I will be able to draw a more accurate conclusion.
Skill Area O: Obtaining evidence
Method:
1. I took two average sized ground potatoes and checked that they were both healthy and hard.
2. Using a standard kitchen knife I peeled the potatoes and cut each one into an even block approximately 7cm by 5cm by 4cm on a white tile.
3. Using a scalpel and ruler I cut the potato into ¡§chips¡¨ which were 5 cm long. I had to be very careful whilst cutting the potato as the scalpel is exceptionally sharp. I then had 15 chips.
4. Taking a test tube rack I placed 5 test tubes and then labelled them 0 molar, 0.25 molar, 0.5 molar, 0.75 molar and 1 molar.
5. Using a measuring cylinder I measured out different amounts of sucrose solution and distilled water which I then poured into the test tubes in a percentage ratio giving me the various molar concentrations.
6. I then weighed every potato chip on an electronic balance and recorded the weights.
7. I swiftly put 3 potato chips into each beaker and then started my stopwatch. 3 chips were used to create an average which gave me a better set of results and more accurate graphs.
8. Whilst waiting I set out some paper towels with which I was going to dry the paper and I drew up a basic table for my results.
9. After 30 minutes I drained out the solutions in the sink and placed all the chips on the paper towel in the order I had put them in the test tubes as to not confuse myself as to which chip came from which solution.
10. I dried each chip with the paper towel and then placed each one on the scales so that I could weigh them.
11. Each potato was measured accurately on the electronic scales and then the weights were recorded.
12. As I had time after doing the first set of results I redid the experiment under exactly the same conditions. This gave me secondary set of results which gave me a more accurate view on the changes.
Precautions:
As was stated in my planning section of the coursework, I had to keep all of the different non-variables the same, to make sure that none of them affected the results of the experiment in any way.
Whilst cutting the potato, extreme care and precision had to be taken with the scalpel as it is very sharp and could easily cause a serious wound.
The measurements for the solutions had to be perfect as to not change the out come of the experiment.
I had to ensure that every time I handled the potatoes my hands were clean and dry. This was to stop any kind of contamination and made sure that I did not pass on any extra water onto the potatMain Tables of Results for all Concentrations
0.00m concentration mass before (g) mass after (g) change in mass (g) % change in mass
1 1.46 1.52 0.06 4.109589041
2 1.64 1.75 0.11 6.707317073
3 1.72 1.85 0.13 7.558139535
4 1.64 1.89 0.25 15.24390244
5 1.63 1.85 0.22 13.49693252
6 1.75 1.95 0.2 11.42857143
average 1.64 1.801666667 0.161666667 9.757408672
0.25m concentration mass before (g) mass after (g) change in mass (g) % change in mass
1 1.56 1.46 -0.1 -6.4102
2 1.66 1.64 -0.02 -1.2048
3 1.6 1.61 0.01 0.625
4 1.51 1.15 -0.36 -23.841
5 1.63 1.68 0.05 3.06748
6 1.34 1.39 0.05 3.73134
average 1.55 1.48833 -0.0616 -4.0053
0.5m concentration mass before (g) mass after (g) change in mass (g) % change in mass
1 1.65 1.45 -0.2 -12.121
2 1.66 1.52 -0.14 -8.4337
3 1.88 1.71 -0.17 -9.0425
4 1.42 1.36 -0.06 -4.2253
5 1.66 1.51 -0.15 -9.0361
6 1.6 1.52 -0.08 -5
average 1.645 1.51166 -0.1333 -7.9764
0.75m concentration mass before (g) mass after (g) change in mass (g) % change in mass
1 1.64 1.36 -0.28 -17.073
2 1.49 1.34 -0.15 -10.067
3 1.87 1.74 -0.13 -6.9518
4 1.39 1.26 -0.13 -9.3525
5 1.8 1.63 -0.17 -9.4444
6 1.38 1.18 -0.2 -14.492
average 1.595 1.41833 -0.1766 -11.230
1.00m concentration mass before (g) mass after (g) change in mass (g) % change in mass
1 1.52 1.19 -0.33 -21.71
2 1.64 1.39 -0.25 -15.243
3 1.66 1.49 -0.17 -10.240
4 1.61 1.46 -0.15 -9.3167
5 1.66 1.49 -0.17 -10.240
6 1.56 1.42 -0.14 -8.9743
average 1.60833 1.4066 -0.2016 -12.621
This graph shown above gives the line of best fit for the percentage change in mass of the potato chips over the course of the thirty minute experiment. The graph is a curve that slopes downwards and does not go through the origin. Because the line is not straight and does not pass through the origin, it means that the percentage gain and loss in mass and concentration are not directly proportional. However, there is a pattern on my graph, and this is, as the concentration of the solution increases, the percentage change in mass decreases. The graph shows that the percentage gain and loss in inversely proportional to the concentration. The gradient does change in my graph. It gets less steep as X axis gets bigger. This is because the potato chip is becoming as flaccid as it possibly can, and so the change in mass of each molar concentration is becoming closer and closer together. From the line of best fit that has been added in, it can be seen that all of my points were very close to creating a perfectly smooth curve. This shows that my results are fairly reliable. My graph fits in with my prediction of the experiment graph.
It shows that the potato cells increase in mass in solutions with a high water concentration and decrease in mass in solutions with a low water concentration. When the concentration reaches above 0.75 M, there appears to be no further water loss, suggesting that the cell is fully plasmolysed. From the graph an estimate to the concentration of the potato cell can be made as 0.13 M, as this is the point where the potato is not increasing or decreasing in mass, this is known as the isotonic point. This is where no osmosis is taking place, both the potato and the solution have an identical molar concentration. The next point, 0.25 M looses approximately 4.0 %. This shows that the water potential of the salt solution in the beaker is weaker than that of the potato chip. The next, 0.50 M, looses approximately 8.0 % in mass. This shows that the salt solution has an even weaker water potential than 0.25 M and that osmosis took place. This is why the potato lost even more mass, and it shows that the water potential in the beaker is less than that of the potato chip. This pattern carries on through the graph, and even more mass is lost, as more water moves out of the potato into the solution. My results also match with my initial predictions.
This graph of the change in mass helps prove the point of complete plasmolysis, whereby the potato cannot expand and take in any more water. As you can see as the molar concentration increases the change in mass decreases. From right to left the first two points on the graph are very spread out indicating that there was a large change in the mass. This decreases throughout the increasing molar concentration until the change is minuscule (about 0.02g).
This graph above shows a clear indication that there was an overall decrease in mass during the experiment. At the point 0.00 M the line for after the experiment is above the line for before the experiment unlike any of the others. This is because the water potential of the sugar solution is higher than that of the potato chip.
Skill Area E: Evaluation
The experiment was very successful in my opinion. I obtained a large quantity of very accurate results from which I was able to create informative graphs. I think I took easily enough results for the amount of concentrations that I was using, and the time that I used for the experiment to last was enough to allow sufficient osmosis to occur. However if I was to repeat the experiment I might well increase the time of the result to allow more osmosis to happen and possibly find out the saturation point of the chips. The range of concentrations was adequate but I would possibly create more concentrations if I repeated the experiment so that I would have more varied results, i.e. 0.10m, 1.15m, 1.20m, and so on. This way would have allowed me to also find out the isotonic point far more accurately as the one that I estimated is very approximate.
The cutting of the potatoes was the most difficult part of the experiment as although I was recording my results by mass, it could well have affected the surface area and so the overall rate of osmosis. If I were to repeat the experiment I would have possibly found a machine to cut the potato as it would ensure that all potatoes would be the same weight and dimensions. As well as the potato I could have found a more accurate way to measure out the solutions and to determine the molar concentrations. Perhaps I could have used a burette. This would ensure that I have an accurate amount of fluid in each test tube. I could also weigh each chip on a more accurate scale, e.g. not to 0.00g but to 0.0000g.
There were not any out of the ordinary results, but some were not as close to the line as others. This may have been caused by human. When the potato chips were removed from the test tubes and dried I may well have dried some potatoes more thoroughly than others and so some would have more excess water, which would add to the mass. If the experiment was repeated I could find another way to dry the potatoes that would ensure that all were dried in the same way for the same time. However with all this said I think that the experiment was truly successful and I was very pleased with the complete comparison of my results with my initial prediction.
Osmosis and Plant Cells
The aim of this experiment is to investigate the movement of water into and out of plant cells by osmosis. The cells chosen for study will be taken from potato tubers as they provide a ready supply of uniform material.
Any substance dissolved in water is called a solute; a solvent is a liquid that is able to dissolve another substance, called a solute, to form a solution.
The water content of plants varies depending on environmental conditions. In Land plants this water plays a vital role in the support of tissues and the transport of materials around the organism. Lack of water leads to wilting and eventually death. Water is mainly absorbed through the roots which are covered in specially adapted root hair cells, with large surface areas and thin cell walls to aid absorption. It is drawn up the plant through xylem vessels by a pull resulting from the evaporation of water through the stomata on the leaves. This evaporation is called transpiration and the xylem flow resulting is called the transpiration stream. Soluble food substances formed during photosynthesis are transported around the plant in the phloem tubes.This movement of water through the plant in the xylem vessels or phloem tubes is similar to the flow of blood in humans as it transports soluble mineral salts, nutrients and auxins, (plant hormones), from place to place. The evaporation of water from the leaves also removes heat energy from the plant and helps to prevent overheating.
Transpiration pulls water up the plant stem but osmosis is the process whereby water is drawn into or out of cells and tissues. Osmosis is the flow of water by diffusion through a differentially permeable membrane from areas of high water concentration to regions of low water concentration. The diagram below illustrates this:
All plant cell membranes are differentially permeable, which means that they will allow some substances to penetrate them but not others. Water can freely penetrate all membranes. The cellulose cell wall does not act as as differentially permeable membrane and will allow most substances that are dissolved in water to freely pass through it.
During photosynthesis,carbon dioxide and water are combined to from glucose sugar. This involves converting light energy into chemical energy stored in the bonds of the sugar molecules. The glucose may be stored in the photosynthesising cell’s vacuole as a sugar or converted into starch. The advantage of using the cell vacuole as a storage organ is that the concentration of solutes in the cytoplasm can be kept within levels that allow the cell to function best. Water entering a cell then has to pass through two membranes to reach the solutes in the cell vacuole.
Whether water enters the cell by osmosis or not will depend on the balance between external and internal solute concentrations and the state of the cell. If the solutions on each side of the differentially permeable membrane are equally concentrated then there will be no net movement of water across the membrane. This is called an equilibrium state and the solutions are referred to as being isotonic. A solution that contains more solute particles than another, and is hence more concentrated, is referred to as being hypertonic. The less concentrated solution is hypotonic. This concentration of solute particles is usually described as a molarity.
A molar solution contains a fixed number of solute particles in a litre of water. The easy way of measuring out this fixed number of particles is to use a mole of particles. One mole contains 6 x 10 23 particles, (600,000,000,000,000,000,000,000). Measuring out the Relative Atomic Mass, (R.A.M.), or Relative Molecular Mass, (R.M.M) of the solute in grams and dissolving this in a litre of water will make a 1 molar solution. A 1 molar saline solution, (sodium chloride dissolved in water), would contain .......
A 0.1 molar solution contains ten times fewer solute particles than a 1 molar solution. I will use serial dilution to obtain the molarities for each test.
Even if the solute concentration external to the cell is hypotonic to the vacuole contents the cell will not continue to take in water by osmosis for ever. The cellulose cell wall provides a rigid barrier to uncontrolled expansion. A cell that is full of water is called turgid and cannot expand further as the outward pressure on the cell wall is balanced by the inward force of the stretched wall. This wall pressure is called turgor pressure and the internal outward force on the wall is called osmotic pressure. At the other extreme, a cell placed in a solution that is hypertonic to its contents will lose water by osmosis. The cytoplasm will cease to exert a pressure on the cellulose cell wall and the cell, described as flaccid, will lack support. Water loss can continue to such an extent that the cytoplasm, and attached cell membrane, contracts and detaches from the cell wall. A cell in this condition is said to have undergone plasmolysis. This very rarely, if ever happens in nature.
As osmosis is the diffusion of water molecules and as diffusion is the random movement of particles from areas of high concentration to low concentration it might be expected that any factors that speed up or slow down the movement of these particles will affect the rate of osmosis.
Using knowledge of the process of osmosis and with a good understanding of molarity I should be able to determine the solute concentration of the vacuoles in potato tuber cells. As it would be impossible to measure with any degree of accuracy the expansion or contraction of cells on an individual basis I have decided to look at gain or loss of water in terms of increase or decrease in mass. A cell placed in an isotonic solution should show no change whereas one placed in a hypertonic solution will lose mass.
To generate reliable and precise results I will use solutions whose molarity has been determined to two decimal places and measure the mass of my potato samples in grams to the same degree of accuracy. I will use two centimetre long tubes cut from New Potatoes with a cork borer as this length fits easily into a boiling tube and gives me a mass between 1 and 2 grams. My results are therefore likely to be accurate to within ± 0.005 g which represents a mass 100th the size of my sample. This mean that I will be 99% confident my measurements are precise.
The solutions I have chosen for my initial tests into cell vacuole solute concentration are:
0.00 M 0.25 M 0.50 M 0.75 M 1.00 M saline solution, and
0.00 M 0.25 M 0.50 M 0.75 M 1.00 M glucose solution.
To eliminate, as far as possible, any errors in my procedure I intend to set up six samples at each concentration. This will hopefully allow me to increase the reliability of my data by using an appropriate number and range of samples.
I intend to use material from the same batch of potatoes as this will eliminate as far as possible any variation resulting from different treatment or source of supply. New Potatoes should have been freshly harvested and this will reduce any affects that long storage might have on the tubers. I obviously cannot determine whether the potatoes came from the same plant but can be fairly sure they are from the same variety. The potato tubes will be immersed in each solution for 24 hours to ensure that all cells, including those in the central core, have had time to react to the external solute concentration. I expect to find a solute concentration of between 0.2 M and 0.4 M to be isotonic with my samples based partly on taste but also on the fact that most plants must tolerate a reasonable salt concentration in their surroundings. Both these reasons for my prediction are very subjective and therefore I will run my initial tests to a 1 molar concentration.
Particles in liquids and gases have kinetic energy. They move about, at speed, in all directions. The particles move about randomly. In an area of high concentration, some of the particles collide with each other, lose energy and slow down. Others will escape from the concentrated area to places where there are fewer or none. Very few particles leave an area of low concentration to go to an area where the concentration is higher. This creates a diffusion gradient - the result is that particles diffuse from an area of high concentration to an area of low concentration.
- Some membranes in plant and animal cells allow certain particles to pass through them and not others. They are selectively permeable.
- The diffusion of water through a selectively permeable membrane is called osmosis. The rate at which osmosis takes place is affected by the concentration of water in the two solutions on each side of the membrane.
- If a selectively permeable membrane separates the two solutions, water moves through it in both directions at the same time. However, more water leaves a dilute solution (high water concentration) and passes into a more concentrated solution (low water concentration) than enters it. Although the water appears to move across the membrane in one direction, it is in fact moving in both directions but more one way than the other.
- When the concentration of water is the same on both sides of the membrane, the movement of water will be the same in both directions. At this point, the net exchange of water is zero and the system is in equilibrium.