Investigate the water potential of potato tissue and compare this with the water potential of apple tissue.
Josie Richards 12NB
Investigating Osmosis in plant tissue
Aim: To investigate the water potential of potato tissue and compare this with the water potential of apple tissue.
I am going to find out the water potential of both apple tissue and potato tissue by immersing samples of each type into different concentrations of sucrose solutions. From this I will be able to draw a graph and discover at which concentration of sucrose solutions the two samples of tissue will reach equilibrium. (i.e. the water potential inside the cells will equal the water potential of the solution.) From this, I will be able to find out the solute potential of the concentration of sucrose solution in which the cells reached equilibrium. This in turn will enable me to work out the water potential of the solution and therefore, the plant tissue.
Hypothesis:
I predict that both types of plant tissue will gain in mass when put in solutions with a higher water potential than them. The water potential of the solution outside the plant cells will be higher than the water potential inside the plant cells, therefore water will move into the cells by osmosis through the partially permeable plasma membrane. The plant tissue immersed in concentrations of sucrose with a lower water potential than the plant cells will lose mass as water is lost by osmosis. The plant cells will have a higher water potential than the surrounding solution so water will move out of the cells through the partially permeable plasma membrane by osmosis.
I also predict that the apple tissue will have a lower water potential than the potato tissue and so will lose less water than the potato cells when put in a solution with a lower water potential than the plant tissue. Therefore, in solutions with a water potential higher than that of the plant tissue cells, the apple cells will gain more water than the potato cells. This is because apple cells have a lower water potential than potato cells as they have more solutes in their cytosol. I also predict from this fact that the apple cells and the sucrose solution will be in equilibrium in a lower concentration than the potato cells will be in equilibrium in.
Scientific Background
Osmosis is the net movement of water from a region of high water potential to a region of low water potential across a partially permeable membrane. If a cell is placed in a solution where the water potential is higher outside the cell, water will move into the cell through the partially permeable plasma membrane by osmosis. The cell gains in mass and is said to be turgid. If a cell is placed in a solution where the water potential is higher inside the cell, the cell will lose mass as water moves out of the cell through the partially permeable plasma membrane by osmosis.
In plant cells, the water potential of a cell depends on two factors: the solute potential and the pressure potential. The solute potential of a cell is the amount by which the solutes within the cell lower the water potential. In animal cells this is the only thing that effects the water potential of a cell, meaning that the water potential = solute potential. In plant cells, however, the cellulose cell wall must be taken into account. As water enters a plant cell, the vacuole, tonoplast and plasma membrane push against the cellulose cell wall. This means the cellulose cell wall exerts a pressure on the contents of the cell and pushes water back out through the partially permeable plasma membrane. However, as water moves out of the cell (i.e. in solutions with a lower water potential) the cell wall ensure that the cell does not completely shrivel up. We can therefore say that water potential in a plant cell = solute potential + pressure potential.
I predicted that the potato cells would have a higher water potential than the apple cells. Apples are the fruit of a plant and are primarily used for seed dispersal - organisms eat the fruit and excrete the seeds in other places. Fruits, on the whole, have a sweet taste to attract animals or insects to eat them (and therefore disperse more seeds.) To enable the fruit to taste sweet, plants store sugar in the form of sucrose, which is soluble and sweet tasting. As it is soluble, it affects the osmotic balance of the cell by lowering ...
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I predicted that the potato cells would have a higher water potential than the apple cells. Apples are the fruit of a plant and are primarily used for seed dispersal - organisms eat the fruit and excrete the seeds in other places. Fruits, on the whole, have a sweet taste to attract animals or insects to eat them (and therefore disperse more seeds.) To enable the fruit to taste sweet, plants store sugar in the form of sucrose, which is soluble and sweet tasting. As it is soluble, it affects the osmotic balance of the cell by lowering the water potential.
A potato, unlike an apple, is not a fruit - it is a tuba and its primary job is to produce runners so the plant can expand and produce more plants. Potato plants use potatoes as a store of energy for the new potato plant to sprout in the spring. Potatoes, however, store this energy as starch, which is an insoluble polysaccharide. As starch is insoluble, it does not affect the osmotic balance of the cell and so the water potential is higher than that of the apple.
Preliminary work
I carried out some preliminary experiments to test my method and also to ascertain the size of plant tissue I will be using. I also wanted to determine the length of time I would need to leave the experiment to ensure osmosis had fully taken place.
The first factor I tested was the size of plant tissue. From previous experiments I had discovered that using cylindrical shaped pieces of plant tissue was best. I did a pilot test to determine which size cork borer I will use in the real experiment.
Pilot experiment 1: the results from this pilot test showed me the size 3 cork borers produced a more dramatic increase or decrease in mass, in both apple and potato tissues:
Potato
Apple
Size 2
Size 3
Size 2
Size 3
0M conc
M conc
0M conc
M conc
0M conc
M conc
0M conc
M conc
Loss 0.2g
Gain 0.3g
Loss 0.3g
Gain 0.5g
Loss 0.02g
N/A
Loss 0.1g
Gain 0.1g
I will therefore use the size 3 cork borers (3mm diameter) to cut my apple and potato tissue.
I also did a pilot experiment to test the amount of time needed for the experiment to take place completely. The results were interesting; they showed me that if the experiment is left for too long, the effects of osmosis will begin to reverse. Here are the results:
Mass at 0 hours (g)
Mass at 2hours (g)
Mass at 3.5 hours (g)
Potato in 0M
.86
2.35
2.26
Potato in 1M
.90
.65
.46
Apple in 0M
.44
.45
.61
Apple in 1M
.43
.57
.37
There are two anomalous results (in bold) suggest that leaving the potato and apple tissue in the sucrose solutions for too long will effect the osmotic balance. In 0M solutions, the potato cells started to gain in mass and then lost mass. This could perhaps be to do with the pressure potential of the potato cells. (In the 0M solution the potato cells gained water as the ? was higher outside the cells.) As water moves into the cell, pressure is exerted on the cell wall which has a great tensile strength. This pushes the water back into the surrounding sucrose solution, therefore decreasing the mass.
The other anomalous result was the apple sample in 1M solution at 3.5 hours. Here, the mass of the apple also firstly increased and then decreased again suggesting that 3.5 hours is too long for the plant matter to be left in the sucrose solutions.
I therefore thought of leaving my experiment for 3 hours. After this I check if my modification was suitable. The results still showed enough of an increase/ decrease between the mass of the plant matter at the start and at the end, and therefore I decided that in my actual experiment I would leave my experiment for 3 hours.
From other preliminary work, I discovered that the concentrations of sucrose solutions would be the following: 0M, 0.2M, 0.4M, 0.6M, 0.8M, and 1M. This range and these intervals of concentrations will give me sufficient points to plot on my graph at the end to construct an accurate line of best fit.
I also discovered that using Petri dishes to immerse the plant material into sucrose solution in is sufficient only if 40cm3 of solution is used to cover the plant samples completely.
Why sucrose solution?
Sucrose is a naturally occurring disaccharide in plants; therefore it is unlikely to affect the chemical properties of the potato's cells. Sucrose is also a safe material; therefore little precautionary measures need to be taken; only to wear a lab coat to protect clothing, and with regards to the sucrose, goggles are not needed. Sucrose is widely available and hence has been widely tested. The water potential of various concentrations have been found, this means that by finding the line of best fit, the point at which the line intersects the x-axis, the plant material will have an equal water potential to that concentration of sucrose.
Apparatus
Reason for use
2 x pipettes
Allow a reasonable degree of accuracy and are easy to use
2 x 250cm3 measuring cylinders
To enable a little more than the theoretical 192cm3 required in total of each substance for each tissue.
2 x Petri dishes with lids
Petri dishes allow the plant samples to be fully immersed without the complicated method of removing the samples when they are to be weighed (when using boiling tubes.) Also, the lids ensure that no solution evaporates from the dishes, which would therefore affect the results.
Cork borer (3mm in diameter)
To ensure all the plant samples are all exactly the same width.
2 x Knife (one for apple, one for potato.)
To cut the apple and potato samples to the correct length.
Ruler
To measure the above
2 x White tile (one for apple, one for potato)
To cut the plant samples on.
Electronic Balance (measuring to 1 decimal place)
To measure the mass of the plant samples before and after the experiment with a suitable degree of accuracy.
Brae burn apple
red potato
Paper towels (damp)
These are to place the plant samples on before they are put into the sucrose solutions to ensure water is not lost by evaporation.
Method:
. Make up the sucrose solutions into the Petri dishes using the dilution table below using the pipettes. Label the Petri dishes with small stickers using the type of plant material, and the concentration of sucrose solution.
Concentration sucrose solution (M)
Volume distilled water (cm3)
Volume sucrose (cm3)
0
40
0
0.2
32
8
0.4
24
6
0.6
6
24
0.8
8
32
0
40
2. Cut 12 samples of apple and 12 samples of potato using the cork borer. Ensure they are all 3cm long. Place between two damp paper towels to ensure water is lost from the cells by evaporation.
3. Weigh and record the mass of each plant sample. Make sure the samples do not get mixed up by numbering the samples on the damp paper towels and recording their masses in a table.
4. Pair up samples of apple with similar masses, and do the same for the potato samples. Again, record which number sample is going in which concentration of solution.
5. Place the 24 sets of pairs in the Petri dishes (i.e. two to a dish.)
6. Write which replicate is which next to the Petri dishes.
7. Leave for 3 hours.
8. After 3 hours, one by one re-weigh the samples and record. Before placing on the balance the samples must be blotted to remove any water on the outside of them. Then weigh, blot again, and then re-weigh to ensure all the excess water is removed.
Safety:
Although this is not a particularly dangerous experiment, precautions must still be taken. Potato juice is an irritant and so eye protection must be worn throughout the experiment. Also, sucrose solution is very sticky and so lab coats should be worn to protect clothing.
Conclusion
The graph showing percentage change in mass in the two different plant materials has a very definite trend; for both plots, the points are above the X-axis in concentrations higher than the point of equilibrium (where the line crosses the x-axis.) This shows that in these concentrations (lower than 0.26M for apple and lower than 0.36M for potato) both types of plant cell gained water. This was because the cells gained mass due to water entering the cells by osmosis. In solutions with a concentration higher than 0.26M (for apple) or 0.36M (for potato), the percentage change in mass was increasingly negative. This was because the cells lost mass as water moved out of the cells by osmosis.
Osmosis is the movement of water from an area of high water potential to low water potential through a partially permeable membrane. The amount that the solute molecules lower the water potential is called to solute potential. In plant cells, there is an extra factor which affects the water potential of the cell - pressure. If a plant cell is in a solution with a higher water potential, water will move into the cell by osmosis. As the cell takes up more and more water, the pressure on the cellulose cell wall increases and pushes water back the other way until the two solutions are in equilibrium. (The cell wall has a great tensile strength, therefore does not just burst.) We can therefore say that, in plant cells, water potential = pressure potential + solute potential.
More water is needed to reach equilibrium for bigger differences of water potential between the sucrose solution and the plant cells. In the more dilute solutions (i.e. from 0M to 0.36M) more water was gained by the potato than the apple cells, as there was a bigger difference between the water potential of the cells and the solution, than there was between the apple cells and the sucrose solution. The potato cells also lost more water than the apple cells by osmosis. Again this is due to the greater difference of water potential between the sucrose solution and the potato cells.
From my graphs I can conclude that, as I predicted, the water potential of the apple (-700kPa) was much higher than that of the potato (-1000kPa.) This is because apples have much more soluble solutes in their parenchyma i.e. sucrose. This means there are less water molecules relative to the solute molecules making the water potential low. The potato had a higher water potential as its stores starch rather than sucrose. Starch is insoluble and therefore is not a solute in the cytosol of the potato parenchyma cells. This means that the water potential is not brought down and so it is higher than the water potential of the apple cells.
I can also see that, due to the above reason, the apple cells were in equilibrium with the sucrose solution at 0.26 M. This is lower than the potato, which was in equilibrium with the sucrose solution at 0.36 M. In a 0.26M solution, there is less water relative to a 0.36M solution (i.e. the sucrose is more concentrated.) This means that a 0.26M solution has a lower water potential than a 0.36M. As the plant cells' water potential is equal to the water potential of the sucrose solutions in equilibrium, we can conclude that the water potential of the potato was higher than that of the apple by looking at these above figures.
