Aim To determine the water potential of a potato tuber cell

Aim To determine the water potential of a potato tuber cell Fig 1- picture of a plant cell

Introduction

Osmosis is the movement of water through a semi permeable membrane, separating solutions of different concentrations. The water passes from a region of high concentration to a region of low concentration, until the two concentrations are equal in concentrations of water.

Many cell membranes behave as semi permeable membranes, and osmosis is a vital mechanism in the transport of fluids in living organisms, for example, in the transport of water from the soil to the roots in plants.

If a cell is in contact with a solution of lower water concentration than its own contents, then water leaves the cell by osmosis, through the cell membrane. Water is lost first from the cytoplasm, then the vacuole through the tonoplast (vacuolar membrane). (Fig 1) The living contents of the cell contracts and eventually pulls away from the cell wall and shrinks, this is known as Plasmolysis.

If you put a plant cell in water, water enters by Osmosis, and then swells up. However, the cell will not burst. This is due to the fact that the cell walls are made from cellulose, which is extremely strong. Eventually, the cell stops swelling, and when this point is reached, we say the cell is turgid. This is important, because it makes plant stems strong and upright.

The formula for water potential is

Water potential = osmotic potential + pressure potential

Preliminary work

For my preliminary work, I cut 6 potato cylinders by using the cork borer. They are all the same length and mass. Then I placed one potato cylinder into each test tube with different concentration. After 24 hours, I re-weight it and record the result.

Here is a table of my preliminary result table.

Improvement

There are a number of factor, which affect the accuracy of my result such as the solution. The reason behind this is because someone else makes the solution, so we cannot prove that it is accurate.

For the improvement of this, I’m going to make the solution by myself. By doing this, I can make sure that my solution is accurate.

Also different type of potato is being used which make the result not reliable. For example, in this experiment, I using 2 different types of potato, one is an old potato and the other one is a new potato. The result will not be reliable as old potato’s cell wall is dried out, so it’s going to gain more water and the new potato will gain less. This will affect the accuracy of the result. For the improvement of this, I’m going to use one potato to produce 6 potato cylinders. Therefore each potato cylinder will have the same strength of cell wall.

The method could be improved by using a more accurate and precise equipment. E.g. use a 10ml of syringe to measure the amount of solution instead of 20ml syringe.

Prediction

Fig 2-cell membrane Fig 3- different type of plant cell

I predict that as concentration of sucrose increases, the weight of the potato chip will decrease. My reasoning behind this, is that the higher the concentration of sucrose in a solution, the lower the water potential. When the potato chip is put into the solution, it will, by osmosis lose some of its water, causing the potato chip reduce in weight and length. This is because osmosis moves from higher water potential to lower water potential. (Fig 2)

However, if a potato chip is placed into a solution of 0 molar concentrations such as distilled water, it should gain weight and become more turgid. (Fig 3) This is because the solution has more water potential (its molecules’ ability to move) than the potato chip, and so water moves from a region of high water potential (the solution) to a region of low water potential (the potato chip.) across a semi permeable membrane. Although the tuber will be heavier, it won’t get significantly heavier than that of the other tubers. This is due to plasmolysis occurring in the cells of the potato, allowing no more water to pass into it, protecting the cells from bursting.

The potato within the 1 molar concentration will decrease in mass because of osmosis as water will move from a high water potential (which is the external solution) to lower water potential (which is the potato cell). Therefore the potato cell will have lower water content that will decrease in mass and size. As there is less water in the potato, the cell will become plasmolysed. (Fig 3)

Here is a predicted graph.

I predict my graphs will look like this because, the potato chip used for 0 molar concentration will gain water, as the water moves from a region of high concentration (the solution) to a region of low concentration (the potato chip.) This result will be an example of turgid cell. However, in the middle of the scale, say 0.2 molar concentrations, the graph will go right down, due to the fact that the solution and potato chip are about equal in concentration. The reason for the minus results for 1 molar concentration is that the solution has a lower concentration of water to that of the potato, and so, by osmosis water moves from the potato to the solution, causing the weight to decrease.

Fig 4- table of water potential

Using a table (Fig4) that expresses the relationship between molarity and water potential, I know that if there is no change in weight in my potato tuber cell samples then the water potential in both the regions have reached equilibrium. I predict that the water potential will be within the range of minus 1280 kPa and minus 260 kPa for a normal potato tuber cell. I think this because minus 1280 kPa is the suggested water potential, (for a quantity of 0.45 molar sucrose) for a plasmolysed cell. Minus 260 kPa is the suggested water potential, (for a quantity of 0.1 molar sucrose) for a maximum turgid cell. My potato tuber cell samples will be within these two amounts.

This graph show that as the molarity increase the solute potential will increase. This is because higher concentration causes more water to leave the cell due to the solute potential. Solute potential lower the water potential inside the cell.

Equipment

Percentage error – showing us how accurate the experiment was

Percentage error = (error / reading) x 100

Ruler (15cm)– 0.05/15 X 100 =0.33%

Syringe (10ml) – 0.1/10 X 100 = 1%

Syringe (20ml) – 0.1/20 X 100 = 0.5%

Digital weighing scales (2d.p.)-0.005/1.57 X100= 0.32% ...

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Equipment

Percentage error – showing us how accurate the experiment was

Percentage error = (error / reading) x 100

Ruler (15cm)– 0.05/15 X 100 =0.33%

Syringe (10ml) – 0.1/10 X 100 = 1%

Syringe (20ml) – 0.1/20 X 100 = 0.5%

Digital weighing scales (2d.p.)-0.005/1.57 X100= 0.32%

250ml beaker- 25/250 X100 =10%

Total Percentage error = 1 + 0.5 + 0.33+0.32+10 = 12.15%

Overall the percentage error is quiet high, which suggest that my result would not be 100% accurate.

Method

Collect all the equipment. Then add labels to the beakers indicating the amount of water and solution in each one.
Then I get the sucrose solution from the technician.
Pour the distilled water into one 250ml beaker and the sucrose solution into the other 250ml beaker. Then add a mixture of distilled water and sucrose solution

Here is a dilution table.

Get six potato cylinders from the centre of the potato using a cork borer with a 2-centimetre diameter. Ensure that there is no potato skin on the ends of the cylinders by cutting them off. Use your ruler to measure the length of your potato cylinders, cutting them all to the same length.
Then I will Place each potato cylinders with care onto a separate piece of filter paper. Weight each sample in turn, recording the results in a table. When weighing, make sure not to include the mass of the filter paper in my table figures.
Once weighed, put one sample into each of the labelled tubes and leave them, ensuring they are fully submerged in the solution. Therefore leave them for 24 hours.
After the time is up remove the samples from each tube carefully and remove any surplus fluid from them quickly using filter paper. Make sure to use the same 'standard' procedure for all of them.
Re-weigh the samples and record the new masses of each of the samples.
Then calculate the mass loss in percentages.

Empty Result table

Table of the result of the mass of potato and the percentage loss in mass

Fair Test

The experiment must be a fair test, if it is not we will be obtaining the wrong results. To make sure it is a fair test all aspects of the experiment must be the same except for the concentration of the solutions, which will give us a varied set of results.
The measurement of the solutions has to be exact. The length and width of the potato chips must be as accurate as possible as this could affect the rate of osmosis. The temperature must be constant
through out the experiment that is why I did the experiment at room temperature. The same weighing scales will be used throughout the experiment; this is because the measurements may vary slightly between scales.

Lastly and most importantly the chips should be in compete contact with the solution if not, osmosis may not occur to its fullest potential. When taking the chips out of the solution, drying them and measuring the mass should all is done as quickly as possible.

Safety

Ensure that the laboratory code of conduct is being strictly adhered to.
Be careful when handling the scalpel.
Be careful when using the cork borer.
Report any spills breakages or accidents so they may be attended to.
Ensure that all equipment is handled in a safe manner so not to cause any accidents.
Ensure that safety goggles are worn at all times during the experiment.

Analysis

Table of the result of the mass of potato and the percentage loss in mass

I think my result is reliable as there is no anomalous result. The result is matching my prediction as I said that as the concentration of sucrose increase, the mass of the potato would decrease.

From the table, I found that the mass of the potato cylinders put in high concentrations of water, increased. For example, when a potato cylinder was put in a concentration of 0M (no sucrose at all), the average mass increased by 25%. However, the mass of the potato cylinders put in low concentrations of water, decreased. For example, when a potato cylinder was put in a concentration of 1.0M, (pure sucrose solution), the average mass decreased by 44.67%. I also found by looking at the line graph drew that the mass of most potato cylinders decreased, even with a concentration of 0.36M. A pattern can easily be seen in the line graph, the pattern is as the concentration of sucrose molecules increases, the percentage change in mass decreases.

By looking at my line graph, I have also found that equilibrium would have been reached at a concentration of 0.36M if I had used that particular concentration. I also found some information by not looking at the line graph, but by looking at the potato cylinders after the experiment. I found that the potato cylinders put in high concentrations of water, became fat and firm, while the potato cylinders put in low concentrations of water became soft and thin.

In my results I have identified several features. Every sample of potato that I used in my experiment followed the water potential gradient by either loosing or gaining weight. For example, as shown in the table, the most concentrated sucrose solution - which in this case was 1 mole, more weight was lost. Evidently this is due to the fact that the concentration of the external solution was higher than that of the internal solution of the potato cells and therefore the water moved from the region of higher water potential to the region of lower water potential through a partially permeable membrane by osmosis. In the beaker containing the least concentrated sucrose solution osmosis also occurred, however the opposite happened. The potato sample gained weight. The internal solution was of a higher concentration than the external solution and so the water moved down the water potential gradient - hence into the potato cell. The cell became fully plasmolysed over the 24 hour period in which the investigation took place.

Plasmolysis occurred in the experiment with the 0.6, 0.8, 1.0 mole concentration of the external solution due to the fact that water was leaving the cell by osmosis because the cell was placed in a solution of a lower water potential. The protoplast of the cell gradually shrunk and began to pull away from the cell wall. This process is called plasmolysis and only occurs in plant cells. When a plant cell is plasmolysed the protoplast shrinks away from the cell wall, the external solution has passes through the cell wall and becomes in direct contact with the shrunken protoplast. As you can see from the graph and the results show a clear negative correlation, a very obvious inversely proportional trend. This is because the concentration of water was higher in the potato cylinders and lower in the solution. The mass of each potato cylinder put in a low concentration of water, decreased because more water molecules were getting out of the potato cylinders, into the solution. The potato cells became flaccid, in low concentrations of water. This is clear because each potato cylinder put in low concentrations of water became soft and thin. They became flaccid because the potato cells released water molecules, through the selectively permeable membrane, into the solution.

Another feature that I identified whilst processing my results was that the potato cell that was placed in the sucrose solution of 0.4 mole concentration only lost 0.05 grams in weight over the 24 hour period in which my investigation took place. This shows that the potato cell reached equilibrium. Equilibrium is reached when the water potential in one region is equal to the water potential in another region. This particular potato cell was in equilibrium with the external solution of 0.4 moles. During the 24 hour period there would have been very little net gain or loss of water and this is shown in my results. There was evidently some movement of water however only very little and therefore the potato cells in this concentration of solution only lost a mere 0.05 grams of weight.

In the potato cylinders, osmosis was affected by the amount of water molecules diffusing through the selectively permeable membrane. The concentration of water inside the potato cylinders and the
concentration of water outside the potato cylinders (in the solution) controlled the amount of water molecules moving through the selectively permeable membrane (potato cylinders). So when potato
cylinders were put in high concentrations of water, (for example, a concentration of 0M), water molecules moved into the potato cylinders. This is because the concentration of water was higher
outside the potato cylinders (in the solution) and lower inside the potato cylinders. The mass of the potato cylinders put in high concentrations of water increased because more water molecules were
moving into the potato cylinders from outside the potato cylinders. The potato cells became turgid, in high concentrations of water. This is clear because each potato cylinder put in high concentrations of
water became firm and fat. They became turgid because the potato cells took in water molecules, through the selectively permeable membrane.

It is clear using the results plotted on the line graph that equilibrium would have been reached at a concentration of 0.36M, if the concentration 0.36M had been used. This is because the line of
best fit, on the line graph goes through 0.36M, when there was no percentage change in mass. Equilibrium would have been reached at this point because the number of water molecules inside the potato cylinder would have become equal to the number of water molecules in the solution, (outside the potato cylinder).

On my prediction, I predict that as concentration of sucrose increases, the weight of the potato chip will decrease. The evidence/results I obtained while doing my experiment matches the prediction I made quite well. This is because it is clear by looking at my table of average results and my line graph that the average change in mass, in the potato cylinders, decreased in lower concentrations of water and increased in higher concentrations of water.

From my results I am able to conclude that for the particular type of potato tuber cell I used, the water potential is (0.36 moles) minus 997.71 kPa. I am able to determine this because of several reasons all identified in my results.

Firstly, the potato tuber cell in the sucrose solution of 1 mole concentration lost weight. This means that water moved into the potato cell following the water potential gradient. This shows that the water potential of the potato cell was less than 1 mole.

Secondly, the potato tuber cell in the sucrose solution of 0 mole concentration gained weight. This means that the water moved out of the potato cell and this shows that the water potential of the potato tuber cell is higher than 0 mole (approx. minus 130 kPa) yet lower than 1 mole (minus 3500 kPa).

Lastly, from my results I was able to identify that the potato cell in the sucrose solution of a concentration of 0.36 moles had reached equilibrium with the external solution thus meaning that the water potential of the potato tuber cells used in my investigation is minus 997.71kPa. I find this out by using the solute table on my prediction.

Water potential = solute potential + pressure potential

Ψ = ψs + ψp

As there is no pressure being applied we can therefore presume that the pressure potential is 0 so we can remove the pressure potential from the above equation to give us:

water potential= solute potential

However, in the solute table there is only has the molarity of 0.35. So I use this formula to find out:

-970/0.35=Ψ/0.36

Evaluation

Minor error- is an error that could cause anomalous result.

Major error is an error that can affect all the other result.

In general although there may have been many potential and definite errors within this investigation, my prediction was proved to be right and graph showed my results to be accurate and reliable highlighting that my method was simple to follow showing that this investigate was quite a success. E.g. I had predicted that as the sucrose concentration increases, the mass of the potato would decrease. This was shown to be correct in my results.

In this investigation in which I obtained these results, had several limitations. The most significant limitation is the fact that the potato was squeezed during cutting, causing water to be lost. The cork borer required quite a substantial force to cut through the potato. If a method was used, which did not involve the potato being handled as much during the preparation process a minimal amount of water would be lost and so the results obtained would be nearer their actual values and perhaps more accurate. This could be achieved by using a machine to prepare the potato samples, however a machine as such was not available when carrying out my investigation.

Another less significant limitation during the preparation of the samples was the fact that the sample could have unintentionally been placed on a drop of water or a wet paper towel, while weighing it. This extra water may have caused the sample to have less solute in it than the solution, making it less dense. This extra water also had an effect on the results due to the fact that more water would have been lost by osmosis.

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.

My results are accurate and reliable. There is no anomalous result as you can see on the graph that all the points are laid on the line of best fit. This suggests that my result is accurate. Nevertheless there were several aspects of my investigation that I would definitely change if I ever were to re-do this investigation in the future. My accuracy of observations and noting down any other significant information down efficiently during the experiments could be improved. I did not record the actual type of potato I used in my investigation, this was important as there are hundreds of types of white potatoes.

Due to the fact that there were several sources of error in my investigation there are uncertainties in my results and therefore also uncertainties in the validity of my conclusions. However despite the possible improvements my results justify my prediction, which was based on scientific knowledge. Before carrying out the investigation I predicted that the water potential of the potato tuber cells would be within the range of minus 1280 kPa and minus 260 kPa. My prediction was correct and my results justify this despite the accuracy factors that could be improved. This shows that the uncertainties in the evidence I collected are not a significant problem; in fact they did not affect the validity of my results at all.

Reliability: according to my results I was able to get clear results to make statements, which proved my predictions correct. To show how reliable my experiment was, I have drawn a table to display the concentration of sucrose solution and the mass of the potato after 24hours. This table will help me realise how clear this experiment is in reliability.

This table show us that there is a result that is not reliable and all the other results are really reliable which tell us this experiment was done well as the two sets of result is quiet reliable.

Reference

Fig1 - -

Fig 2, 3-

Fig 4-- - page 36