Determination of the water potential of root/tuber cells by the weighing method.
Determination of the water potential of root/tuber cells by the weighing method
Introduction
When a plant cell is bathed in a solution of the same water potential, its mass and volume remain the same, because water enters ad leaves at the same rate.
If samples of tissue are immersed in a range of solutions of different concentrations (molarities), the cells will gain water by a method known as osmosis and therefore, mass, in solutions of higher water potential and lose water and mass in solutions of lower water potential. The water potential of the tissue is equal to that of the solution in which it neither gains nor loses mass. The purpose of this practical is to estimate the water potential of the potato tuber cells by this method.
In practice, none of the experimental solutions is likely to have exactly the same water potential as the cells but the solution in which there would have been no gain or loss in mass can be estimated from the graph.
Root vegetables and potato tubers are widely available in bulk, but in principle this technique could be applied to any plant tissue.
Apparatus
* Boiling tubes x6
* Boiling tube rack
* Wax pencil
* Cork borer
* Razor blade or sharp knife
* White tile
* Root/tuber
* Filter papers
* Forceps
* Balance to 0.001g
* Distilled water
* 1.0 mol sucrose solution
* 2x 25cm3 measuring cylinders
* Stop clock
Procedure
. Label 6 boiling tubes "DW" (distilled water), 0.2, 0.4, 0.6, 0.8 and 1.0 mol dm3. Place 20ml of distilled water in the first tube and the appropriate amounts of sucrose solution and water in each of the other tubes to make up the correct concentrations.
2. Using a cork borer and a razor blade or sharp knife, prepare 6 root/potato cylinders, each about 10mm diameter and 50mm long. Place each on a separate piece of filter paper on which you have labelled with the figures 0, 0.2, 0.4, 0.6, 0.8 and 1.0 respectively.
3. Take the cylinders and the boiling tube rack to a balance. For each cylinder, record its mass on the filter paper, transfer it to the boiling tube with forceps, make a note of which tube it is in and record the mass of the filter paper on its own. Calculate the initial mass of each cylinder once you have transferred them all to the tubes. Note the time.
4. After at least 30minutes, remove the cylinders from the tubes in turn and in the same order that you inserted them. Remove any surplus fluid quickly and gently with the filter paper. Then reweigh each cylinder and record its mass.
5. Work out the percentage change in mass of each cylinder (change in mass x 100 ÷ original mass) Plot this against the molarity of the sucrose solutions. Your vertical axis will have increased mass at the top, no change in mass in the middle and decrease in mass at the bottom.
6. Calculate the water potential of the potato cells as follows. Find where your line crosses the place on the vertical axis corresponding to no change in mass. Read off the horizontal axis the molarity of sucrose at this point. From the table below, find the water potential of a sucrose solution of that molarity. That is the water potential of your sample of potato cells.
Table: relationship between molarity and solute potential of sucrose solutions
Molarity (mol dm-3)
Solute potential (Kpa)
0.05
-130
0.10
-260
0.15
-410
...
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6. Calculate the water potential of the potato cells as follows. Find where your line crosses the place on the vertical axis corresponding to no change in mass. Read off the horizontal axis the molarity of sucrose at this point. From the table below, find the water potential of a sucrose solution of that molarity. That is the water potential of your sample of potato cells.
Table: relationship between molarity and solute potential of sucrose solutions
Molarity (mol dm-3)
Solute potential (Kpa)
0.05
-130
0.10
-260
0.15
-410
0.20
-540
0.25
-680
0.30
-860
0.35
-970
0.40
-1120
0.45
-1280
0.50
-1450
0.55
-1620
0.60
-1800
0.65
-1980
0.70
-2180
0.75
-2370
0.80
-2580
0.85
-2790
0.90
-3000
0.95
-3250
.00
-3500
Hypothesis A.1a
I believe that the potato cylinders placed in the distilled water will be hypo-osmotic, meaning the water will diffuse by osmosis into the potato from an area of high concentration down the concentration gradient; this will result in a gain of mass for the potato.
However in the 1.0M solutions I believe that the opposite will occur as there will be a higher concentration of water within the potato than the solution, containing lots of solute. This will effectively make the potato hypo-osmotic and will lose its water content as the water diffuses out.
The cells will not reach equilibrium (become iso-osmotic), because plants have a cellulose cell wall which cannot expand beyond a certain point as it is turgid, thus the contents will never be able to be equal to the outside or it will not be able to give out all its water as it has a rigid structure. It is the pressure created by the cell wall, which stops the cell reaching equilibrium.
Analysis
Table of results: A.3a
Molarity (M) of sucrose
Weight before (g)
Weight After (g)
Difference in weight (g)
% change in mass of potatoes
0.0
0.75
0.92
+0.17
23
0.2
0.85
.00
+0.15
8
0.4
0.97
.12
+0.15
5
0.6
0.87
0.99
+0.12
4
0.8
0.93
0.98
+0.05
5.4
.0
0.94
0.93
-0.01
-1.1
A graph of results is shown on the following page...
The results clearly show that the reaction known as osmosis took place, between the sucrose concentration and the potato cylinder. It is evident that equilibrium was achieved when the 0.40 Mole sucrose concentrations were utilised, as there was no change in the size of the potato cylinder.
One can clearly see that the lowest concentration of sucrose (0 Mole) possessed the largest increase in size of potato cylinder. This was due to the movement of water by osmosis and caused a percentage change in the mass; this shows that the potato cells were not fully turgid at the beginning of the experiment.
Where the sucrose solution was stronger than the cell sap of the potato cells, the cells lost water by osmosis. This is clearly shown by the results from the 1.0 Mole concentration; the potato cells are said to be 'plasmolysed'. A.1b
The graph gives an adequate set of data, showing that in the distilled water the potato gains in mass by approximately 23% then it gains 18% in the 0.20M solutions. In the 0.60 M solution there is a change, the gradient of the graph decreases steeply and the potato loses mass to 14% then in the 0.80M solutions the gradient decreases as it loses yet more mass to 5.4%.
The first observation accounted was after the potato cylinders were placed in their solutions, one could see a difference; the ones in the 0.0M and 0.20M solutions were floating and the potatoes in the 0.40M, 0.60M, 0.80 and 1.0M solutions were at the bottom of the boiling tube, this lead to the compilation of the next conclusion.
The graph shows that the potato in the 0.0M solutions and in the 0.20M is hypo-osmotic, as stated in the initial hypothesis; this suggests that there is higher water potential in the distilled water and 0.20M solutions than in the potato, which has a high concentration of solutes. Thus, this is why the water diffuses by osmosis down the concentration gradient from high concentration to low concentration, resulting in the potato gaining mass. A.3b
Quite the opposite occurs in the solutions where the molarity is higher 0.60M, 0.80M and 1.0M; the potato in these is hypo-osmotic due to the fact that there is a higher concentration of water inside the potato than in the solutions. Therefore, the water diffused out of the potatoes by osmosis, down the concentration gradient and into the solutions outside, resulted in the loss of mass. A.5b
The graph also indicates that the potato was not able to take in any more water or lose any more to a certain extent. As stated in the hypothesis, the cell wall causes the pressure that prevents this from happening, thus the cell will never become iso-osmotic.
On the graph of results, the point where the graph line crosses the place on the axis where the potato neither loses mass nor gains mass has been indicated. This happens at 0.48M; therefore the content of the potato cells in molar strength is 0.48M. A.7a/A.7b
Calculating the percentage change in mass of each cylinder A.5a
(Difference in mass ÷ original mass) x 100
NOTE: mass of cylinders after - mass of cylinders before = difference in mass
0.0M sucrose:
(0.17g ÷ 0.75g) x100 = 23%
0.2M sucrose:
(0.15g ÷ 0.85g) x100 = 18%
0.4M sucrose:
(0.15g ÷ 0.97g) x100 = 15%
0.6M sucrose:
(0.12g ÷ 0.87g) x100 = 14%
0.8M sucrose:
(0.05g ÷ 0.93g) x100 = 5.4%
.0M sucrose:
(-0.01g ÷ 0.94g) x100 = -1.1%
Evaluation
The experimental procedures were very acceptable in the sense that accurate results were obtained, reflecting the theoretical ideas of the initial hypothesis. The procedures were simple to follow, enforcing safety of conductivity and in some cases, allowing you to achieve accurate data. E.1a
Although results obtained were substantial enough to draw a conclusion, there were some that were anomalous. For example: the percentage change in the mass of potato for 0.6M concentration of sucrose was anomalous; the point on the graph that indicates this does not fit to the line of best fit. E.1b
To keep the experiment fair, all variables were controlled apart from one, the one that was tested; the concentration of sucrose. However and unfortunately in practice, it is impossible with the basic apparatus we had to keep all the measurements precisely the same. It is also impossible to precisely measure out the amounts of sucrose and the distilled water each time. As the scale on the measuring cylinder shows the measurement to the nearest cm3, the solutions that we used should be correct to the nearest mm3. Also, when the excess water was dried off on the potato cylinders after the experiment and before it was weighed, filter paper was used. This might have either taken some water out of the potato or it might have left some excess water on the potato. The potato itself was not entirely from the same potato and was not exactly the same size, this could have affected the amount of water gained or lost in the experiment. Another suggestion, which might explain the anomalous, was due to the limitations of the cork borer. When the potato became stuck it was necessary to force it free, one of the consequences of which was that the potato cylinder was compressed and other variables may have affected my results, such as the pH and temperature. Having more than one potato cylinder in a pot may have reduced the accuracy of the results obtained, as this may have reduced the surface area required for the reaction to take place. E.3a/E.3b
A method of improving the results would have been to leave the experiment running for a longer period of time, this would have enabled one to discover the saturation point (the point at which the potato can no longer take in any more water) and dehydration point (the point at which the potato cannot lose any more water) and thus, get a more accurate result. E.5a/E.5b
The evidence obtained, i.e. the results, was fairly reliable in the sense that a conclusion can be drawn, as stated in the analysis. It is evident that equilibrium was achieved when the 0.4 Mole sucrose concentrations, as there was no change in the size of the potato cylinder. E.5b
To improve the investigation, in terms of increasing the reliability of the evidence and minimising the sources of error, you could try more accurate ways of conducting the experiment so excess water is not lost and the reading is therefore more accurate. Ways of improving the investigation are as follows; the use of a burette, which is far more accurate than a measuring cylinder and hence it would give more precise readings and therefore the experiment would be more accurate. One final thing that you could do is to conduct the experiment all over again and to make an average of the two averages. This will insure great accuracy, but results obtained from the initial investigation have sufficient evidence to support a firm conclusion regardless of the fact that an anomalous result was obtained. You could also investigate on the other variables, temperature and pH. To enable this experiment to be completed as accurately as possible, one can repeated it three times and then use an average of all the results to best plot a graph with a line of best fit. Also, to extend the experiment to a greater degree, you could place the potato cylinders under a microscope, enabling you to observe the cells in greater detail and draw some more observational results.
Given the opportunity to conduct the experiment again, you could test each concentration three times, then record the average reading and plot it against the graph; this helps to enhance both reliability and accuracy of the data collected. E.7a/E.7b
Bibliography
. Michael Roberts, "Nelson Science Biology", Nelson, 1995
2. Mary Jones, Richard Fosbery, Dennis Taylor, "Biology 1", Cambridge University Press, 2000
3. McCahill, T.A: Biology Basic Facts, Harper-Collins, Glasgow, 1996
4. website: http://purchon.com/biology/osmosis.htm
AS Biology coursework
By Steven Duong