Concentration is the variable we will be studying this will be done by using sugar as our concentration variable. We will be using potatoes of known masses and lengths, and placing it in a solution of known volume and known solute concentration. However each time the solute concentration will change.
PREDICTION
In my opinion, after researching, I think when the potato is placed in
distilled water without a concentration, the water potential of solute outside the cell will be less than that inside the cell, so the solution will be hypotonic. As always when the concentration/water potential gradient differs osmosis takes place and water will enter the potato. As the protoplast will expand, so will the turgidity. As more water enters the cell in the potato the weight of the potato will increase, when the cells have expanded as much as the can the potato will have reached full turgor.
As the concentration of the solution increases to 0.1molar, the potato will begin to decrease in weight. This is because, the water potential/concentration gradient will be less so, the tendency for the solution to diffuse into the potato will be a lot less. As the water potential/concentration will increase more, the final weight will begin to decrease increasingly, due to the gradient becoming less.
Finally when the solute has a concentration level of 1 molar (a solution of higher concentration), its protoplast will begin to shrink, and turgidity will decrease and eventually the protoplast will break free from the cell wall. Due to this effect, the mass of the potato will be less than before it was put in the solution. This is because the concentration outside the cell will be more than that in the cell; the solution outside the cell is hypertonic. When this happens osmosis takes place, but water will leave the vacuole rather than enter.
(Pg. 49, Biology a Functional Approach, MBV Roberts). (Mastering biology, OFG Kilgour, Pg. 89).
I predict that if you double the concentration level of the sugar solution outside the cell compared to that inside the cell, the potato will decrease in size and weight. This is because the cell will want to lose water in order to correct the concentration gradient, in and out of the cell. The opposite will happen if the potato is placed in a solution where the concentration outside is hypotonic compered to that inside the cell.
PRELIMINARY WORK
In order to do my final method I had to find out how much of every thing I had to use. In order to know this I had to do some preliminary work so that my final method would be a fair test and that I was using the right measurements and amounts of every thing. The first task I set myself was to find out what concentrations of sucrose solutions were to be used.
APPARATUS
Potato
Scalpel
Ruler
Measuring Cylinder (250cm³)
1 molar sucrose Solution
Test Tubes
Chirograph pen
0.5 molar sucrose solution
- molar sucrose solution
METHOD
- Collect potato
- Cut out 6 pieces from the potato using a scalpel. Measure them to the same dimension using a ruler. Write down the length in a table.
- Use a measuring cylinder to place 50cm³ of 1molar sucrose solution in to a sample tube. Label the tube
- Place 50cm³ of 0.5 molar sucrose solution in to another tube and label it.
- Place 50cm³ of 0.1 molar sucrose solution in to another tube and label it.
- Set up repeats for each of these
- Put one piece of potato in each of the 6 sample tubes.
- Leave for 24 hours.
- After 24 hours take the pieces of potato out of the tubes and measure their lengths using a ruler. Write down the new lengths in your results table.
- Pack away all equipment.
RESULTS
The results we received are shown in the table below.
From my results you can clearly see that the length of the potato decreased in the I molar test. However, the length increased in all other tests.
In conclusion, my preliminary work was very useful as it helped me to know how much of each solution is to be used. It also gave a better idea of the kind of results we were going to receive and to adjust the concentrations of the sugar solutions accordingly i.e. increasing the range of concentrations.
The preliminary work also highlighted a very important factor, which was that the results themselves were not accurate enough, as the measurement of the potato was only as accurate as the instrument used. A better method would be to use an electronic balance.
The preliminary work also pointed out that the idea of repeats was definitely good, however a link must be made in order to compare the two results. E.g. percentage increase/decrease and graphing results.
Finally to calculate the water potential of a potato (the aim) the concentration of the potatoes contents can be found from the graph drawn and this figure can be cross matched onto a graph showing water potential at various sucrose solution concentrations.
METHOD
APPARATUS:
- 10 Test tubes,( to put the solution and potato in)
- 200cm3 of 1 molar sugar solution,(from which the required concentrations of solutions will be made.)
- 200cm3 of distilled water,(from which the required concentrations of solution will be made.)
- Measuring cylinder (250cm³) (used to measure the sucrose and distilled water when making the solutions)
- Scalpel,(used to cut the potato to the required length)
- Borer,(used to cut the potato out, will give an even dimension to all the potato pieces cut out)
- Balance,(used to weight the potato pieces before and after, recordings will be done to 2 decimal places.)
- A Potato,
- Chirograph pencil (used to mark the test tubes with information)
METHOD
- Collect all apparatus.
- Label 5 of the test tubes with the following categories using a chirograph pencil:
- 0.5 molar,
- 0.2 molar,
- 0.1molar,
- distilled water,
- Label the remaining 5 test tubes with the following categories using a chirograph pencil.
-
(2nd)0.5 molar,
-
(2nd)0.2 molar,
-
(2nd)0.1molar,
-
(2nd)distilled water,
- Use the following table and a measuring cylinder to make the sucrose solutions and pour them into the appropriate test tubes.
- Push the borer into the potato to produce a cylinder. Repeat this until you have 10 cylinders of potato.
- Using a Scalpel and a ruler, measure 2cm of each cylinder and cut it. You should be left with 10 identical potato cylinders.
- Using a balance weigh a cylinder of potato and record it’s weight in a table like the one shown below, place the potato cylinder weighed into the appropriate test tube.
- Repeat step 5 and 6 for all the potato cylinders.
- Leave the test tubes together for 1 day in a room.
- After 1 day take the beakers with the solution and potatoes.
- Remove a potato cylinder one at a time and using a balance weigh a cylinder at a time. Write the final weight in the final weight column in the table.
- Repeat step 10 for the rest of the cylinders
DIAGRAM
SAFETY
Due to this experiment having sharp instruments to handle, safety will have a very important role to play. Whilst the scalpels are not in use be sure to keep them from falling on the floor and hurting someone. Whilst holding a scalpel be sure to not have it facing others, always keep it facing down. Whilst cutting with a scalpel be sure to keep fingers away from the blade and watch where you cut.
Sugar solution can be very oily and slippery which can lead to accidents. Be sure to clean up any spilt solutions so as to avoid other accidents e.g. slipping. Goggles are a must and should be worn throughout the experiment.
FAIR TEST
In order to make this test fair, there are many things that will need to be constantly checked. Some of the key variables that can be changed are:
- Temperature
- Surface area of potato
- Volume of Solution
- Weight of potato
- Concentration.
Firstly, the only thing that should be changed throughout this experiment should be the concentration of the solution and nothing else.
The potato cylinders should be approximately the same in length, which should make them roughly the same in weight.
The temperature is being kept the same in our test for all our beakers by placing them all in the same place. So if there is an increase or decrease in temperature it will be the same for all the beakers and so affect them all.
The skin will be removed of all the potato pieces and the shape will be kept the same by using a borer. This is so to keep the surface area of the cylinders the same.
The volume of the liquid can also affect the outcome so it must also be kept the same. Having repetitions of all the solutions will also make this test more reliable. If all of these features are looked at the experiment conducted will be a fair test.
ON THE NEXT PAGE IS A GRAPH SHOWING MY RESULTS.
CONCLUSION
The results on the last page have formed a line for which I have drawn a line of best fit. This line of best-fit states that there must be a reason for what has occurred in my experiment. It shows the basic trend which is that as the concentration of the solution increases, the percentage change in weight of the potato after osmosis has occurred decreases.
However, there are two results that do not agree with this line of best fit. They are known as anomalous results. They are the results for the solution of concentration of 0.5 molar where the percentage increase worked out to be –23.15%. The second anomalous result was for the solution with a concentration of 1molar, this gave a percentage increase of –45.0% again this did not work in conjunction with the line of best fit.
Besides this the other results seemed to be just right according to the line of best fit. The first result we received was for the solution with a concentration of 0 molar at 26.15%. The second was at 18.6% for the solution with a concentration of 0.1molar. The last result that matched with the line of best fit was for the 0.2 molar test with a percentage of 5.28%.
After researching I know can explain my results. At the beginning when the solution outside the potato was 1 molar you can see the percentage is –45.0%. A minus sign in front of the percentage can only mean the change has reversed. In our case it had. Looking back at our result table, I can clearly see the weight for the potato in a solution of 1 molar has decreased in weight.
This occurred due to osmosis taking place, however, rather than water moving into the cell, water was moving out instead, thus decreasing the weight of the potato. The solution surrounding the cell is said to be hypertonic and so therefore shrinks the cell, which is now known to be becoming flaccid. In this state the cell may even become plasmolysed.
Cells owe many of their properties to the fact that the plasma membrane is selectively permeable. If pure water, or a solution whose solute concentration is lower than that of the contents of the cell surrounds a cell, water flows into the cell by osmosis, and the cell swells up. In this case, the osmotic pressure of the external solution is lower than that of the cell, and accordingly the solution is said to be hypotonic to the cell.
On the other hand, if the cell is surrounded by a solution whose solute concentration and hence osmotic pressure, exceeds that of the cell, water flows out of the cell, which therefore shrinks. (This makes osmotic pressure a variable). In this case, the external solution is said to be hypertonic to the cell. If the solute concentration of the cell and its surrounding medium are the same there is no net flow in either direction and the external solution is said to be isotonic with the cell. The osmotic flow of water into a cell is called endomosis, and the flow out of it is called exomosis (mastering biology, OFG Kilgour, Pg. 89).
All this can be demonstrated with human red blood cells. A solution of 0.9 % sodium chloride is isotonic with human cells, and if red blood cells are placed in such a solution they will neither shrink nor swell, such as in plasma. However, if they are placed in a stronger (i.e. hypertonic) salt solution (say 1.2%) they shrink unceasingly, which makes the cell membrane crinkle as shown in diagram 2. This is called laking or crenation. If the red blood cells are placed in a weaker (i.e. hypotonic) solution they swell and may even burst, this phenomenon is known as haemolysis which literally means blood- splitting.
Diagram 2
(Letts Study Guide, Biology, Chapter 7, Pg. 72, Julian Ford- Robinson)
It follows that if a cell is to maintain its normal size and shape it must either exist permanently in an isotonic environment or, failing that, it must have special mechanisms enabling it to survive in a hypertonic or hypotonic medium. These special mechanisms are all part of the business of osmoregulation, which is immensely important in the lives of organisms.
The unicellular protist Amoeba has both marine and fresh water species. The marine species are isotonic with seawater with the result that there is no loss or gain of water by the cell. The fresh water species on the other hand, are markedly hypertonic to there surrounding medium. The result is that water continually passes through into the cell across its selectively permeable membrane. Undoubtedly the organism would swell up and burst, just like our red blood cell in distilled water, were it not for the contractile vacuole, a spherical sac which collects water as fast as it enters, and periodically discharges it to the outside. By this means the cell is prevented from swelling, and its solute concentration, and hence its osmotic pressure, are kept constant.
At standard temperature and pressure, pure water is given a water potential of zero. Adding solute molecules to the water has the effect of lowering the water potential, i.e. making it negative. This is because the solute molecules attract the water molecules, reducing the number of water molecules, which can diffuse freely. If you go on adding solute, the water potential keeps getting lower and lower, i.e. more and more negative. The component of the water potential, which is due to the presence of solute molecules, is called the osmotic potential or solute potential. (So solute molecules can be a variable according to which direction osmosis occurs in)
Plant cells (including potatoes) generally have a solute concentration that is markedly higher than that of their surroundings. The solutes are located mainly in the sap vacuole, which is surrounded by two selectively permeable membranes, the plasma membrane and the tonoplast. When the cell is surrounded by a weak solution, water is drawn in to the vacuole by osmosis. Diagram 3. The result is that the cell swells. However, it does not burst because of the cellulose wall, which, once fully stretched, resists any further expansion of the cell.
Diagram 3
Diagram 3- showing a summary of the events that pursue if a partially turgid plant cell is placed in (A) a solution of weak solute concentration and (B) a solution of strong solute concentration. The protoplast acts as a selectively permeable membrane, the cellulose wall as fully permeable. Because the cell wall is fully permeable to water and solute, the fluid that fills the space between the cell wall and the protoplast during plasmolysis is isotonic with the external solution. (This shows that concentration is a variable)
As water flows into the vacuole by osmosis, an internal hydrostatic pressure develops which presses the protoplast against the cell wall. This is called turgor pressure and it is opposed by the back pressure exerted by the cell wall against the cells contents. This back pressure tends to force water out of the cell and is called the pressure potential. Obviously, it opposes the continued uptake of water into the cell by osmosis. Turgor pressure reaches its maximum when the cell wall can be stretched no more. At this point full turgor is said to be achieved, or, to put it another way, the cell becomes fully turgid. Turgidity plays a very important part in supporting plants and maintaining their shape and form. The tem of herbaceous plants are held erect by being filled with fully turgid cells packed tightly together. Turgor is also responsible for holding leaves in a flat horizontal position.
However, when a plant cell is placed in a solution, which has a higher solute concentration than the cell sap, something completely different happens. Firstly, the volume of each cell decreases as the water passes out of its sap vacuole. Within a matter of minutes the protoplast shrinks to such an extent that it pulls away from the wall, leaving a gap between the wall and the plasma membrane. This shrinkage of the protoplast from the cell wall is called plasmolysis. Plasmolysis sometimes happens to plants exposed to extremely salty water, but otherwise it rarely occurs in nature. (Pg. 49, Biology a Functional Approach, MBV Roberts).
Water can be removed from plant cells by osmosis. It can also be removed by evaporation. If the cells in the stem and leaves of a plant lose more water by evaporation than they can absorb, turgor is reduced and the plant visibly droops. This is called wilting. It can often be observed on hot dry days. The plants usually recover at night when evaporation is reduced by closure of the stomata, but if the water supply to the roots is inadequate then permanent wilting occurs and the plant dies.
Wilting is not to be confused with plasmolysis. Both involve the cells losing water, but for different reasons. In plasmolysis only the protoplast shrinks, leaving the cell wall behind. In wilting the whole cell, including the wall, shrinks and no gap develops between the plasma membrane and the cell wall. (Mastering biology, OFG Kilgour, Pg. 89).
A similar result was received at 0.5 molar where again the average % increase was a negative figure.(-23.15%). The same principle of water having a high potential inside compared to that outside so water movement goes out of the cells.
The opposite to plasmolysis occurs in the next result that we received which was at a solution concentration of 0.2 molar where the result we received was 5.28% average increase. Here osmosis is taking place, however, it is taking place differently to before. When the potato is dissolved in a solution of 0.2 molar concentration, the liquid surrounding the potato cell is of a less concentration than the solution inside the vacuole. Due to there now being hypotonic solution surrounding the cell the cell begins to swell up because of the water entering the potato.
The same thing occurs again in our next result, at a concentration of 0.1 molar the percentage increase that occurred was 18.6%. Again this means the potato increased in size due to water entering the potato cell in order to correct the concentration gradient.
As the concentration of the solution outside the cell weakens again, the potato again increases in size up to 18.6% in order to correct the concentration of solution inside the cell to match the concentration outside the cell.
This procedure takes place again in our last experiment where in distilled water the percentage increase after two days is 26.15%. The most amount of water is taken into the cell in this experiment because there is no solute mixed into the solution. Due to this the cell has to change the solution inside the vacuole in order to match the concentrations. This can only be done by diluting (or adding more water) to the solution inside the cell vacuole. Due to it having to take in water the cell becomes turgid.
(Pg. 49, Biology a Functional Approach, MBV Roberts). (Mastering biology, OFG Kilgour, Pg. 89). (Letts Study Guide, GCSE Biology, Chapter 7, Pg. 72, Julian Ford- Robinson)
In order for no net movement to take place between the selectively permeable membrane of the potato and the solution outside it the concentration on both sides of the membrane must be equal. In order for this to happen, we must know what the concentration of solution inside the vacuole is. By knowing this we can change the concentration of solution outside the cell and have no net movement.
The concentration of this particular potato can be found from our results graph. The point at which the line of best fit crosses the horizontal axis is the concentration of solution inside the potato. By doing another experiment using the concentration found (0.3 molar) from the graph we can investigate the process of no net movement.
From my conclusion, I see that my prediction that I made before the experiment turned out to be correct. The fact that if the solution outside the potato is double the concentration of the solution inside, the potato will increase in weight turned out to be absolutely correct as you can see from my results.
If I now take the no net water movement value and plot it against the graph which shows the water potential at various sucrose solution concentrations (found in the appendix), I find that the water potential of a potato is –850kPa. My investigation is complete.
EVALUATION
The anomalous results that I have already explained in the conclusion could have been caused for a number of reasons. It could have been caused by the inaccurate measurement of solution when placed in the beaker with the potato. Inaccurate concentrations could also have affected anomalous results. During the practical, we might have read the balance wrong this could also have given the wrong result.
We know our results can be very trust worthy or even very reliable. This is shown by the fact that every two results are similar in my table of results when looking at the percentage change in weight. This shows that all our results were reliable apart from the results received from the experiments conducted with sugar solution of concentration 1 molar. In this the results we received for the % change in weight were -50% and -40% as you can see there is a great difference between these results so they cannot possibly be the same. The fact that we did not repeat our experiments more than just twice makes our results less reliable.
In our experiment, controlling the variables was very limited. For example temperature was not perfectly controlled, using a water bath would have been more accurate.
Controlling the weight of the potato before was all very well however, afterwards measuring it’s increase or decrease in weight was limited as not drying the potato cylinders properly could affect the result. If not dryed properly water would remain on the potato and so the reading would be greater than it’s actual mass. All of this could have concluded to giving us anomalous results.
Besides this, I think the method was performed accurately and on the whole we did not make any errors in our experiment however, there is still room for improvement. Firstly, I could have measured the individual cells in the potato using a microscope. This could have included the length and the area or volume of the cells this would have made the experiment very accurately as to how much water enters the cells. Not only this but it would also have made it very accurate in finding the difference before leaving the potato in the solution and after.
To make an investigation of this, we could have done some further work, this could have included investigating what effect this had on different cells, such as animal cells or even to see what effect this would have on the cells of a tree’s bark. We could have even used different vegetables and even roots of different flowers to see what happened.
Another way to investigate further would be to use a different range of sugar solutions such as
- 0.1molar
- 0.2molar
- 0.3molar
- 0.4molar
- 0.5molar
- 0.6molar
- 0.7molar
- 0.8molar
- 0.9molar
- 1 molar
A wider range of sugar solutions will achieve a wider range of results and in turn will induce a more reliable conclusion.
BIBLIOGRAPHY
- Biology a Functional Approach, MBV Roberts
- Mastering biology, OFG Kilgour,
- Letts Study Guide, A Level Biology, Julian Ford- Robinson
- World Book Multimedia Encyclopaedia CD-ROM