To determine the water potential of potato tuber tissue

Michael Hutton-Ashkenny

Introduction

Osmosis is the net movement of water down a gradient of water potential through a partially permeable membrane until the water potential gradient is constant. When a tissue looses water, it looses mass, and so we can take a measure of how much osmosis has taken place by weighing the tissue before and after it spends time is solution. If there is no mass change then the potato tissue will have the same water potential as the external solution, and so the two water potentials are in equilibrium, and there is a flat gradient of water potential.

The formula for the water potential of a plant tissue is written: as the water potential equals the solute potential added to the pressure potential (from Biology 1 by OCR):

ψ = (+)ψp + (-)ψs

The units for these values are recorded in KPa. Osmosis occurs until there is no net movement of water, because the water potential is equal on both sides of a partially permeable membrane.

Preliminary Work

I conducted a pilot study before I planed my actual experiment to gauge what sort of length potato to use and to see if my values for sucrose molarity were adequate. For my GCSE coursework I conducted an experiment to see the effect of varying the surface area of potato chip in sucrose solution. I used a 1M sucrose solution and the potato chips lost about on average about 0.6g, for chip volumes varying from 2cm3 to 8cm3, left in solution for 45mins. This is quite substantial so I decided to make my dilutions from 1M and diluting it sufficiently. I found from previous experiments that the water potential for a potato tuber would be about 0.3M, so I chose a range from 0.3M up to 0.5M, using distilled water as a control. From various Internet sources (www.essaybank.co.uk), I found that the water potential can differ allot from different potatoes, and so I could not quote any source for this. The pilot study allowed me to see the reliability of the experiment, and to help me formulate a prediction.

Pilot Study Method

Label 6 boiling tubes from 0.3 – 0.5 molar sucrose
Label appropriate beakers and graduating cylinders with either H2O of sucrose.
Add the correct volumes of water and sucrose; using the appropriate graduating cylinder to make sure that there is no contamination; to the correct boiling tubes.
Take cores of potato of 1.7cm diameter, corer number 10 and cut them to 3cm length until there are 6 cores.
Weigh all of the cores and record the weight.
Start the stop-clock and add a core to a solution at five second intervals (don’t forget to record which core went in which solution).
When the time = 30 min take the cores out at 5 second intervals, and place them on a labelled piece of paper towel.
Gently dab excess solution of the potato and record the mass of the core of each solution.

Table to show relationship between Mass Gain of potato chip and external sucrose solution molarity.

I recorded mass values to 2 d.p because that is the value to which the balances can read and I need my results to this accuracy, as my results tend to be changes of under 0.1g.

I realised that I would needed to go to lower values of solute potential for the external solution if I was to get a proper result for the water potential of the potato tuber tissue. This is because, as the graph shows, mass loss began on my first reading of 0.3M, and everything under that was increased mass loss. So, I decided to get rid of 0.4M and 0.45M and replace them with 0.2M and 0.25M, so I can have solute potentials either side of the expected for the potato tissue.

Prediction

My pilot study results give me a base for a prediction, as they give values for the mass loss in various solute potentials. This means, I already can see a trend forming for my experiment. This is, in short, a water potential of potato tuber tissue around 0.3M sucrose and as the solute potential of the external solution increases, as does the mass loss of the potato chips. I predict that for my experiment that as I increase the solute potential of the external solution from 0KPa to -1120KPa the mass gain will decrease as the solute potential decreases. I expect the line to cross the x-axis when the sucrose solution is at around 0.30M to 0.35M (this is between –800KPa and –900KPa, see table below). I get the estimated value for the water potential of the tissue by looking at my pilot study graph and results. I expect my result to be slightly different than what I obtained from my pilot study, because of the limited amount of time I was able to allow the potato chips to spend in the solutions. This would not have given very accurate results for my pilot study.

I predict this to happen because of osmosis. If the two water potentials of two systems either side of a partially permeable membrane are the same, then there will be equilibrium and there will be no net movement of water molecules. This means that the potato chip will neither gain nor loose mass (thus the y-value is 0 on a graph). However, when the solute potential of the solution on the outside of the potato chip falls (becomes more negative) then the equilibrium of the systems changes:

ψ = ψp + ψs

Thus the water potential of the solution becomes more negative as its ...

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ψ = ψp + ψs

Thus the water potential of the solution becomes more negative as its solute potential becomes more negative, so the gradient of water potential out of the potato tuber becomes steeper. Under controlled time, the potato tuber in the solution with the most negative solute potential will loose the most mass, and the one in the solution with the least negative solute potential will loose the least mass. In this case, the solute potential of 0 actually relates to a mass gain. The measuring of mass in this experiment is an attempt to measure the gradient of water potential. This explains why the mass change and solute potential are in direct proportion; the water potential gradient has to be directly proportional to the difference in water potential. However, if I increased the time the chips spent in solution, and put on some higher molarities, I would expect to see the graph plateaux at each end. This is because as the tissue becomes closer to fully turgid or fully plasmolysed, it is approaching a finite state, where mass loss (or gain) can no longer change.

Table to show relative solute potentials from sucrose molarities Biological Sciences 2 by Cambridge.

Variables

Dependant Variable

My dependant variable is the mass change of the potato chips while they are in solution. I will measure this by weighing the chips before and after a set time period in the solution. This then gives me a relation to how much water is lost to the solution.

Independent Variable

The independent variable is the molarity, or solute potential, of an external sucrose solution. These vary from 0KPa to –1120KPa.

Control Variables

Time: If a two chips are left in the same solution, both loosing water through osmosis, then if one is taken out before the other, then it will be osmosing for a longer period of time, and thus loose more water, and mass. I will control this by timing the time of each potato chip spends in solution, and keeping it the same for each chip. Time is measured to the nearest minute, using a stopwatch.

Volume and Surface Area: If there is a higher surface area then there will be more places for water to leave the tissue, so the rate of osmosis will increase. If there is a larger volume, there will be more water to be lost from the tissue of the plant, and so a higher mass loss is potential. To control these variables I use the same diameter potato corer and cut all chips to the same length at all times. I will cut to the nearest millimetre, using a 15cm ruler.

Temperature: When molecules are heated, they move faster. When a chip in solution becomes warmer, then its molecules move faster, and so the rate at which water molecules are leaving the chip increases. To keep the temperature constant, I will keep the windows near me closed at all times, and keep my experiment in the same place in the lab as much a possible. A thermometer in the lab read 21oC throughout the afternoon, and all the solutions will have had time to equilibrate at room temperature.

Pressure: The pressure exerted on the potato must be kept constant so that the equation ψ = ψp + ψs is not disturbed by fluctuating pressure potentials. The only thing that should affect pressure is the volume of sucrose solution placed on the chips, so I will keep this the same by having a constant final volume for all my dilutions.

Potato Type or age: Different potatoes may have different water potentials in their tuber tissue, and to make sure I am not trying to measure different water potentials I will have to use potatoes with approximately the same age, and same breed.

Method

Apparatus List

Reason For Apparatus Choice

Dilutions Table for different solute potential solutions.

The range of my independent variable is based around the water potential of the potato tuber tissue. It would be difficult to make the solutions accurate to more than 0.05M, and so I chose two readings above and below the expected 0.3M, and used distilled water as a control. I shall repeat my readings two times each, so that I can see if there is a pattern, and to see if there are any anomalous readings. For example, if two readings are the same and another is wildly different. If I encounter a result, which is far away from two others in a solute potential range, then I will not include it when I average my results. To be wildly different I would expect it to be 0.1g to 0.15g away from the other two readings. I am looking for consistency within my readings. If all three readings are wildly different then I will repeat that range again to get a new set of results, and take the closest matching results for an average.

Here is a step-by-step method for my experiment:

Label boiling tubes from 0 – 0.4 molar sucrose solution.
Label appropriate beakers and graduating cylinders with either H2O of sucrose.
Add the correct volumes of water and sucrose; using the appropriate graduating cylinder to make sure that there is no contamination; to the correct boiling tubes. When reading a volume off the graduating cylinder, make sure the cylinder is at eye level, and read from the bottom of the meniscus.
Take cores of potato of 1.5cm diameter. Cut off the skin on the ends and cut them to 3cm length until there are 18 chips. When cutting the chips to length, make sure to cut looking from above, so to avoid parallax errors. Try to cut as straight as possible, to avoid further error.
Weigh all of the chips and record the weights.
Start the stop-clock and add a core to a solution at five-second intervals (don’t forget to record which core went in which solution).
Label a piece of paper town and arrange it on a cutting board, so when the chips come out of solution they are ordered (see diagram below)
When the time = 90 min take the chips out at 5 second intervals, and place them on the labelled piece of paper towel.
Gently dab excess solution of the potato, on the side and ends, then weigh and record the mass of the chips of each solution. Try to be consistent with time and pressure while drying chips; this means the water removed from the chips should be kept consistent and so minimising error.
Work out the mass change in all the chips, and then the percentage mass change of all the chips.

Safety

Glassware should be handled with care at all times, and be kept away from edges. This is so glass is not broken, and does not then cut anybody, and becoming a danger. Care should be taken to always cut down into the cutting board and keep all body parts away from the cutting implements. This is to avoid needless damage to oneself, as the blade and corer are sharp.

Results

Table of results of mass changes in my potato chips for different external solute potentials

The % mass changes are worked out like this:

100*(Mass Difference / Start Mass)

= % Mass Change (Gain)

Conclusion And Interpretation

From my graph, it can be concluded that the water potential of the potato tissue is about 0.31M sucrose or about –850KPa.

At 0KPa there is a high mass gain, average of 8.86%. From this there is a negative correlation as the water potential of the outside solution increases. At

-540KPa there is a change of 3.29% and then at –680KPa 1.81%. At –820KPa the mass change is very near zero, and the best fit line goes through the axis at –850KPa. After this the mass continues to decrease as the solute potential is increased. At –970KPa there is a change of –0.91% and at –1120KPa a change of –3.09%. These values all correspond well on a graph to a strong negative correlation.

At 0KPa, there is distilled water outside to tissue, and so the net movement of water is into the potato tissue quite quickly. The solute potential of the tuber cells will come closer to 0 in this solution, as the contents become more dilute and the pressure potential will increase, as the cells become more turgid. At –540KPa, essentially the same process is taking place, except the gradient of water potential will not be as steep, and so at the end the solute potential will not be as low as at 0KPa and the pressure potential will not be as high. For –680KPa and –820KPa this trend continues, with increasing solute potentials at the end of the chip’s time in the solution and decreasing pressure potentials as the cells are not as turgid as at lower molarities of external solutions. At –850KPa we have the water potential of the external solution and the potato tuber tissue being the same. This means that there is no net movement of water, and so there is no change in the solute potential of the cells, and the cells will neither gain nor loose water, so they will remain as turgid as they already are, so keeping pressure potential constant. At higher external water potentials like –970KPa there will be a net movement of water out of the tuber cells. The pressure potential will decrease as there will be plasmolysis as the cells loose water, and there will be an increase in the solute potential because the concentration of the cells will increase as water leaves. The extent of this increases at –1120KPa as there will be a steeper water potential gradient, there will be more plasmolysed cells, and there will be higher concentrations of solute inside the cells. This means there will be a much lower pressure potential and a much higher solute potential; almost the opposite to 0KPa, or distilled water.

Evaluation

I do not consider there to be any anomalous results in my experiment at all. The largest deviation within a range is only 0.06g, which is not very large at all, definitely not large enough for any of my data to be considered anomalous. Because of that, I included all readings in my averages.

There were many errors of my method, and in it also. The main error I encountered was the change in mass due to excess liquid on the chips and then the problem of blotting excess liquid off the chips. Other errors could have arisen, such as errors to do with cutting the potato. It was very difficult to cut down exactly straight, and it was difficult to cut the chips to exactly the correct length, as they were cylindrical and the ruler could be pushed up to no definite edges. I could have used a small guillotine contraption with a ruler by it to cut the chips straight, but one was not readily at hand.

The main error was, in my mind, the blotting of the potato chip on paper towel. It was difficult to have consistent blotting on my potato chips, and so different chips would have lost different masses of water, thus it is difficult to be too accurate about the effect of the independent variable on the mass change. Improving this would be difficult. There would need to be a common consistency throughout the blotting. For example: gently touching each end of the potato chip onto a dry piece of paper towel, and then rolling the chip as gently as possible along the paper. If the pressure exerted could be kept constant, by just pressing as lightly as possible each time, then there would be less error produced from this process. Also, I should time how long the roll takes and keep it the same for each chip, and keep rolling at constant speed. For example, make one rolling along paper towel last 3 seconds for each chip, and place the chip on its ends for one second each.

The apparatus used in each part of the experiment was all fairly accurate. The top pan balance needed to be to 2 d.p, as is clearly shown in my results where the mass differences came as low as 0.02g. I could have used a more accurate balance, but the top pan, I believe, was just the right accuracy for this experiment. There is a completely different way to assess how much osmosis had taken place. This would be to look at a section of the tissue under the microscope and to count how many cells, in a field of view, were turgid, incipient plasmolysed and also fully plasmolysed. This would take out any errors from weighing and over blotting the potato chips. When cutting my chips, as I have already discussed, it was difficult to obtain a straight edge. The blade we used was quite small, and I found it hard to keep it exactly at right angles with the chip. To create straighter edges, I could have used a non-serrated knife. Another idea would be two wooden supports with a blade inside them, to keep the blade straight. This would increase the accuracy of my cutting. Measuring the chips length was only accurate to the nearest millimetre, and it was difficult to measure the chip anyway. The ruler had to be pushed under the chip to measure it properly, and so the ruler’s markings were slightly obscured. I think this was one of my more problematic errors, and a very difficult one to correct. The stop-clock used in my experiment was more than accurate. It measured to the nearest second, and for my experiment, where I left chips in the solution for 90mins, I really only needed to measure to the nearest minute. The final piece of equipment where errors may have arisen was measuring volumes of sucrose and water for my dilutions. The graduating cylinders only measured to the nearest 1cm3, which is not very accurate, but should not have affected my results too much, as I was using large volumes. However, I could have used pipettes, it would have been much more accurate, to the nearest 0.05cm3, this would have made my measurements much more accurate.

I did two repeats for each value of solute potential for my results, and I believe this to be sufficient, as my results were all consistent between readings within a molarity range. The change in the % mass change within a range is a maximum difference of 1.3% (see results table). The maximum difference in the -970KPa range is only 0.1%, which is very small, and so seems reliable. Vertically, the results change at a seemingly constant rate, which can be seen from my graph, comparing results with my best fit line, which is what I predicted, and this supports my prediction well.

All of the errors in my experiment would have affected my results, however, I believe that the blotting error was the main error overall. It needed to be controlled, because it could significantly affect the mass of my chips if there was excess liquid, or if too much had been taken out while blotting. I tried to have consistent blotting, and so any error due to too much or too little blotting is likely to be minimal given that the error is quite likely to be consistent. The improvements I can put to my method, especially to the blotting error would benefit the experiment. With blotting at exactly the same extent the excess water will not contribute to the mass change and so the results will more truly reflect my independent variable. There uncertainties certainly affect my results, but the question is, how valid can a conclusion drawn from them be? The errors will obviously affect my results, and make them harder to draw conclusions from. However, the results agree with the preliminary work I conducted and are consistent with each other, and I think I can be quite firm in my conclusions.