The process of osmosis also occurs in animal cells, but in an animal cell there isn’t any cell wall to prevent them from swelling. This means if they were to be placed in pure water, they would take in water by osmosis up until they burst, that is called haemolysis, which occurs only in red blood cells.
PREDICTION (HYPOTHESIS): -
I predict that the mass of the apple and the potato tissue may increase (rise) in water and dilute concentrations of sucrose depending on the concentration of sucrose inside the cell. When the concentration of sucrose outside the cell is the same as the concentration inside the cell, the mass will remain the same. When the concentration outside the cell is higher it will decrease the mass of the apple and potato tissue.
EQUIPTMENT: -
- 1x large potato
- 1x large cooking apple
- 1x sharp knife
- 1x metal borer with plunger
- 6 Petri dishes
- Beaker of distilled water
- 50ml of the following molar of the sugar solution: -
0.00M (distilled water) 1.5M, 1.75M, 2.00M, 2.25M, 2.5M (sucrose).
- 1x Electronic balancer (for weighing apple and potato strips)
- 4-5 pieces of kitchen towel
- 1x Wall clock
- Blank table of data (to enter the data information and results)
METHOD STEP BY STEP: -
- Label six Petri dishes 1-6 with the right molar and solution name.
- Use the metal borer to produce 18 strips of apple and 18 strips of potato.
- Use the graph paper to measure out all the length of the strips and make sure they are 20 mm in length.
- Select three apple strips and three potato strips. Starting with the apple strips, weigh the first apple strip. Call this apple strip 1 and record down the weight in grams. Do the same for the other apple strips and then record the information. Then move on to the potato strips and use the same method to find out the weight for each strips, the way it was done for the apple strips.
- Replace both apple and potatoes strips 1, 2 and 3 and pour 40ml of distilled water over them and leaves it for 1hour 45minutes. This Petri dish is the 0.00m sugar solution (e.g. no sugar, 100% water = 0.00m sugar solution).
- Repeat the step number four again with the second, third, fourth, fifth and sixth Petri dishes, but the only difference will be that the concentration will change. Petri dish two, 0.25M, Petri dish three, 0.50M and Petri dish 4, 0.75M, Petri dish 5,1.00M, Petri dish 6, 1.25M sugar solution. Leave all the experiment to run for 1hour 45minutes.
- After doing all this record all the information on the table and while the experiment is running calculate the average initial weights of the potato and apple strips for each sugar concentration (molarity) and record down the data for the average initial weight.
- After 1hour 45minutes, carefully pour out the liquid solution from all the Petri dishes but being careful that the potatoes and apple strips don’t get mixed up.
- With a paper towel remove all the extra excess liquid from each of the strips before weighing them again. Weigh each strip in the same order as the way done right at the start and then record down the data under the column of final weight.
- Calculate the average final weight for each of the sugar solution 0.00M-1.25M.
RELIABILITY OF THE RESULTS
- The sizes of the all the strips must be large enough to see on difference.
- Replicate experiment three times
- It should be done in room temperature, because the sucrose may evaporate because it may affect the rates of osmosis.
- The experiment needed to be timed in order to make it a fair test.
- Same amount of potato and apple strips.
ANALYSING THE RESULTS
The graph for this experiment results very clear information about the mass of both the apple and potato experiment. If I compare both of the results, the apple strip shows more higher percentage in mass and the sucrose concentration rises quite high in 0.80 sucrose molar and then from there it keeps decreasing until it reaches to 2.00 molar concentration. The potato strip however shows a very different result. From distilled sucrose molar solution it keeps dropping (decreasing) until it reaches to 1.60 molar, which is the point from when it goes up nearly 10%.
Apples are sweeter then potatoes so this means it won’t affect the process of Osmosis. The example for this is photosynthesis. The sugar in the process of photosynthesis is stored as starch (which is insoluble), which therefore affect the process of osmosis. It also attracts animals which therefore disperses the seeds.
The starch is used to give energy for the growth of a new potato, and the new plants grow from potato tubers through the dotted eyes marked on the potato skin. In this case it is important that the process osmosis shouldn’t be used as it will affect osmosis.
Endosmosis is the movement of the water into a cell and exosmosis is the movement of the water out of the cell. When both endosmosis and exosmosis are equal it means that the movement of the water into the cell is equal to the movement of water out of the cell. This point is called the equilibrium point. At this point the concentration of the solution surrounding a cell is the equal to the concentration of the solution in the plant cell vacuole, which is the cell sap.
After plotting all the information on the graph, I have identified that the equilibrium point is reached at 0.48 molar when endosmosis is equal to exosmosis for the experiment with the potato strips. However for the apple strips the equilibrium point is at 1.70 molar. This shows that the apple strip reaches equilibrium with higher concentration than to the potato experiment. So therefore this proves that the apple strip contains more concentrated cell sap than the potato tissue. The apple concentration is 1.22 molar more than the potato concentration. However I cannot say it very accurately as there are some errors in this experiment.
EVALUATION
After completing the whole experiment I have gathered a table of result and a best fit line graph which shows what happened at each stages of the experiment. In my table I have identified some errors in the percentage change in mass. One of the figures from both of the result is quite abnormal. In the apple strip result the 0.80molar sucrose concentration has -3.3 in its percentage change in mass which conveys a very low figure compared to the others and so it doesn’t fit into the data. This is why I have taken out this data because its anomalies from the rest of the data. I have done the same for the potato strip result. There was also an anomalies figure in the 2.00molar sucrose concentration, which had -8.8 in the column of the percentage change in mass. Once again I have taken out this data because its anomalies from the rest of the information and doesn’t fit in with them.
I think that the anomalies data may have occurred, due to the amount in which the strips had weighed after it was taken out of the solutions. I have decided that it was probably because the water moved out of the strip which then became plasmalysed.
The actual difference was because the mass was small which increased the error and also because of the increase in the size of the plant strip pieces. This is why I would say that the experiment wasn’t exactly done in a completely fair way, there were some complications.
This experiment has shown quite a variable result in replicates. To make it more accurate and avoid making the errors I could have made more concentrations for both the potato and the apple experiments, for example: -
- Potato: - 0.00-0.80 molar
- Apple: - 1.20-2.00 molar
MORE LEFT TO ADD ON ABOUT THE EQUILIBRIUM FIND OUT WHAT IT MEANS AND EXPLAIN IN DETAIL ABOUT THE DATA USING MY PLAN SHEET.