Experiment 2 will also have a mass increase but preliminary experiments show that the mass increase is fairly low so I believe that the point of isotonicity will be close to this point, so my prediction is that the molarity of the potato cells is between 0.2 and 0.4 moles.
Experiment 3 and experiment 4 will show decreases in mass but the results for experiment 5 and experiment 6 will show large mass decreases because the concentration of the sucrose solution will be 0.8 and 1 Mole respectively and preliminary experiments show that this far exceeds the isotonic point for potatoes.
Predicted Graph
Explanation of prediction.
Osmosis is the net movement of water molecules from an area of high water potential to a region of lower water potential across a partially permeable membrane.
In this experiment water is diffusing into or out of the potato chip cells. The sucrose molecules in the sucrose solutions are too large to pass through the partially permeable membrane of the potato cell, it is only the water molecules that can pass through. The basis of osmosis is that water molecules diffuse through the partially permeable membrane from a region of high sucrose concentration to a region of lower sucrose concentration with the aim of making the solutions isotonic. Random movement of particles achieves this; the water molecules have more space to move in the direction of the sucrose molecules because there are less small water molecules around the large sucrose molecules.
The water moves across the partially permeable plasma membrane of the potato from an area where there is less space to move to a region where there is more space to move this is because when the water molecules randomly rebound of each other and the sucrose molecules there is less chance of them colliding in the hypotonic solution than they are when they are in a hypertonic solution. When they collide with each other and rebound in the direction of the hypotonic solution they go further in that direction because there are fewer particles to stop them moving this way so net movement of water particles is from the hypertonic solution to the hypotonic solution until the amount of particles in each solution is generally the same.
The volume of water transferred is decided by the steepness of the concentration gradient which is basically the difference between the concentrations of the substances on each side of the partially permeable membrane. In experiment 1 the steepness of the concentration gradient is very high because the distilled water has a concentration of 0 sucrose molecules whereas the potato contents will be much more concentrated with sugars and salts. This is similar for experiment 2 but the concentration of sucrose molecules will be higher so their will be less net movement towards the potato cells. The concentration gradient is getting less steep as the sucrose solutions concentration increases, this happens until the isotonic point is reached then the concentration gradient will be reversed because there will be a higher volume of water in the potato sap than there is in the sucrose solution. In this instance water will diffuse out of the potato. The amount of diffusion will increase as the sucrose concentration increases, so there will be more mass lost at a faster rate for the higher concentrations of sucrose.
Apparatus
1 knife
1ceramic tile
12 boiling tubes
1 potato tuber/12 potato chips
2 boiling tube racks
2 250ml beakers
1 M sucrose solution
Distilled Water
1 Balance
Blotting Paper
1 Potato Borer
2 10ml syringes
2 25ml Syringes
1 15cm Ruler
Method
- Measure out 2 solutions each for the volumes given below. Using the syringes dispense the solutions into 12 labelled test tubes.
2. Use the knife to skin the potato and the potato borer to make smaller potato chips.
- Use the knife to cut the chips to the size 50mm x 10mm x 10mm on the ceramic tile.
- Using blotting paper blot each side of each chip for 3 seconds.
- Weigh each chip on the balance; take a reading to 2 significant figures.
- Place 1 chip in each boiling tube; record the destination of each chip with its initial mass.
- After 30 minutes remove the chips from their solutions and blot each side of each chip for 3 seconds.
- Weigh each chip again and calculate the average mass change per gram for each concentration.
Diagram of apparatus
Safety
Care must be taken when using the knife as it is very sharp and can be dangerous.
No pieces of potato will be dropped as they can make the floor slippery and hazardous.
The worktop must be cleaned after use as it may become sticky and damage other users work.
Any spillages must be cleaned promptly to prevent injury from loss of balance.
Explanation of method
The experiment will be carried out twice, so the average mass change per gram can be taken from two different chips, this can show up any anomalies and make the results more accurate. A knife is used because it is sharp enough to cut the potato and I am responsible enough to use it safely. Preliminary experiments have shown that the size I have selected is sufficient to show a significant mass change, the potato borer is used as it is easier to make smaller chips after using this. Any excess cell sap can be removed by blotting, because otherwise the results may be inaccurate, the chips must not be squeezed because this will reduce the mass and change the results. The final mass must be measured so the average mass change per gram can be calculated then I will be able to plot these results on a graph of water potential and mass change, and then determine the water potential of the cell contents. This is determined by reading the sucrose concentration when the y-axis value is 0.
Results
Tables to show changes in mass of potato chips in different solutions of sucrose.
Table 1
Table 2
Table 1 shows the initial mass in grams of each chip which went into each solution and the final mass of the chips from each solution after 30 minutes immersed in the sucrose concentrations.
Table 2 shows the mass changes in grams of each chip from their respective solutions and also the mass change per gram which was worked out by dividing the mass change by the initial mass of the chip. The percentage mass change was also worked out by multiplying the mass change per gram by 100 and is shown in the table.
Graph 1 shows the average mass change of the chips in grams against the different sucrose concentrations in Moles.
Graph 2 shows the changes in mass of the chips in grams against the different sucrose concentrations in Moles for the first attempt and the second attempt, I used it to determine whether there were any anomalous results.
Graph 3 is a reference graph of Sucrose Concentration in Moles against Water Potential in Kilo-Pascals from Biology: a functional approach, 2nd edition, by M. B. V. Roberts and T. J. King; which I used to find out the concentration of the potato cell contents.
Conclusion
The results show that my initial prediction was correct and the predicted graph I drew was also correct as it showed an increase in mass and then a decrease after the isotonic point was reached. My prediction that the mass would increase most at 0M was correct. The idea shown in my preliminary results that the molarity of the potato contents would be between 0.2 and 0.4 M was accurate. This is shown by the fact that 0.2 M was the last point at which the mass of the potato chips increased. The results from 0.4 M were mass decreases and this trend continued down to 1.0 M. After the isotonic point has been reached the mass of the chips decreases more. This idea of inverse proportionality between the variable has been shown on the graph. The largest mass gain was when the sucrose concentration of the solution surrounding the potato chips was pure water and had a water potential of 0, this is because the concentration gradient between the two solutions was steep and diffusion occurred to balance out the concentration of the solutions. The largest decrease in mass occurred when the solution surrounding the potato chips had a sucrose solution of 1M, this is because the solution surrounding the potato chips was hypertonic to the contents of the potato chip so net movement of water was out of the chip and into the surrounding solution.
The reason this occurs is from osmosis of water into and out of the cells. When the solution outside the potato cells has a water potential of 0 then this is higher than the potato cell contents and the net movement of water will be out of the potato and into the surrounding solution, so the solution surrounding the potato chips will be hypertonic to the potato cell contents. The opposite of this occurs when the surrounding solution has a sucrose concentration of 1.0 M, this is because there is a lower concentration of sucrose inside the potato in this case the potato cell contents are hypotonic to the surrounding solution and water will enter the potato until equilibrium has been reached. The isotonic point is reached when the surrounding solutions concentration is exactly the same as the potato cell concentration and the net movement of water into and out of the chips is relatively constant. On the graph this point occurs when the value of the y- axis, which is the mass change of the potato chips, is 0. The X-axis value on my graph for this point was 3.667 M. This result is the concentration of the sucrose solution when the mass of the potato does not change and it is also therefore the concentration of the potato chips contents. If I use the reference graph of water potential against sucrose concentration it shows that the water potential of the potato cells is 975Kpa.
Evaluation
I believe that the method I used was acceptable as it provided me with a valid result for the water potential of a potato chip. I do believe that the method had many flaws.
The flaws included the number of repeats and the timing and cutting of the chips. I only carried out two repeats, I feel this was not enough because if one result did not match the other result then I would have to plot both results on the graph to see which result fitted in more closely with a best fit line, and even after this I would only have one result for this particular concentration of sucrose solution which is very unreliable. It would have been more accurate to do 3 or more experiments for more accuracy.
Cutting the chips to the required size proved more difficult than I had first anticipated, I think this could be resolved by either increasing the mass of the chips so it would be easier to cut the chips or simply using a chip borer to obtain chips with a constant shape and mass.
There are possible improvements to the method I believe leaving the chips in the solution for longer would increase the amount of mass change which would give a more defined graph and results. I believe that using a very strong sucrose solution would completely plasmolyse the potato chips. If I tried this and worked out the difference in mass between the turgid potato chip and the plasmolysed chip I could also work out the volume of water the potato chips can hold and the volume of water in the potatos at standard temperature and pressure. To know which results were the ones for the plasmolysed and turgid chips I would have to wait until the graph gave a horizontal line which indicates that there was no change in mass over two consecutive results.
The first result for 0.2 M sucrose solution seems to be anomalous as it does not fit in very well with the other results as shown by the graph. Both of the sets of results show a steep decrease in mass at 0.6M so I decided that this was not an anomaly. Apart from this anomaly I believe that the results are reliable and I believe that my conclusion that 975KPa is the water potential of the potato cells is accurate.
There are many further investigations I could carry out, such as investigating the water potentials of carrots and other root vegetables, measuring the extension and reduction of the potato chips length after immersion in the sucrose solution. There is also the alternative of using sodium chloride instead of sucrose, though there may not be very much difference.