What is the isotonic point of a plant cell?

The aim of this investigation is to determine the isotonic point of a plant cell. First thing to do therefore is to establish what an isotonic point is. This is the point at which there is a balance between the concentration of water on either side of a semi-permeable membrane (see diagrams on reverse for more details). This goal could be approached in any of a number of ways, but I have chosen to perform it as explained below. I would just like to point out first that I will do two experiments on two different potatoes so that I can establish whether the isotonic point varied from one plant’s cells to another’s. I would ideally like to repeated each of these again so that I have a more accurate result, but due to limited time, this will not be an option. Instead, I shall just have to use some common sense, if the results are really close together than I shall assume that the difference is the result of an inaccuracy of the equipment or human error. If the results are fairly far apart however (as I expect them to be (see predictions)) then I shall assume that the isotonic point varies from one potato type to another.

Equipment.

A full list of the equipment I’ll be using is given below. N.B. This experiment, as with many, would be best performed with equipment that would give as accurate a measurement as possible. Unfortunately, also like other experiments, there is a limit on the accuracy of the equipment due to several reasons ranging from cost to current technological levels; the equipment listed below is therefore as accurate as I could get in our labs with the time and resources allotted to me.

Experiment 1

Potato (white)
6 cups
Sucrose solutions in concentrations of

0.2 molars
0.4 molars
0.6 molars
0.8 molars
1.0 molars

Distilled water
Knife
Cutting mat
Paper towels (approximately 4)
Pen
Ruler
Chipping machine
2 Scales (correct to 0.01 grams)
Clock (I originally planned to use a stop clock, but on the day the other class was using them all so we had to make do with a clock correct to the nearest second).
Paper

Experiment 2

Potato (red)
6 cups
Sucrose solutions in concentrations of

0.2 molars
0.4 molars
0.6 molars
0.8 molars
1.0 molars

Distilled water
Knife
Cutting mat
Paper towels (approximately 4)
Pen
Ruler
Chipping machine
2 Scales (correct to 0.01 grams)
Clock (I originally planned to use a stop clock, but on the day the other class was using them all so we had to make do with a clock correct to the nearest second).
Paper

With this equipment gathered, I shall start to perform my experiment. The actual procedure takes about 1 hour and 20 minutes to complete in total, so the bullet point method can be seen below. Please note that the method below is for experiment one but experiment 2 will be done exactly the same way except that I will use the red potato instead of the white one.

Step 1. I will peel the potato and used the chipping machine to cut it into chips.

Step2. I’ll then marked a line straight across the cutting mat with the pen that was parallel to the edge of the mat and 2cm away.

Step 3. I’ll then look for seven chips of approximately the same dimensions and cut them each into three chips of equal length, 2 cm long, using the line on the mat (i.e. I plan to place the chips on the edge of the mat and cut across where the line is.

Step 4. I will then sort these chips into groups of three and weigh the combined weight of each group correct to the nearest 0.01g on each of the scales, record these individual results and then work out the average and record that as well (making sure that the scales are at zero when I place the chips on, and waiting until the reading has stopped fluctuating before I record the results), making sure to keep the piles separate and clearly labelled so as to remove confusion and assure the validity of my results.

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Step 5. I’ll next place each group into a corresponding cup of sucrose solution with the only variation being concentration (see diagram) and record whether they sink or float. The reason for me choosing sucrose solutions ranging from 0 to 1 molar is that the isotonic point must lie somewhere between these because the cells need to have some sugar to survive (so must have more than 0 molar) and can’t have more than 1 molar because this is the maximum concentration of sugar that can be in the solution.. The rest of the concentrations were decided by dividing 1 molar by 5 (because I only had the resources and time to do six cups and we already had one used by the 0 molar solution) and then working my way up from 0 to 1 molar in this interval. N.B. I used three chips in a group so because there wasn’t time to repeat the experiments so, rather than doing three experiments on one chip, adding the values and then dividing by three to get the average weight gain, I simply put the three chips into one experiment, thus getting the average straight away in the time allotted to me.

Step 6. I’ll then wait for half an hour according to the clock (which works out being 30 minutes give-or-take a minute) and then remove the chips, group by group and place them onto the paper towels.

Step 7. Then I’ll lightly pat each group dry with the paper towels and …

Step8. … Finally I’ll weigh each group (again with both scales and again making sure that the scales are at zero when I place the chips on, and waiting until the reading had stopped fluctuating before I record the results) and record the results in the table on the next page.

The reason that this experiment will work is that osmosis will only change the weight of one of the chips when the concentration of sugar in the water is different one side of the cell wall to the other (e.g. when there is a higher sugar to water ratio inside the potato than in the sucrose solution). The potato chips that experience the least change are therefore those in a solution nearest in concentration to that of themselves, and thus are those nearest to their isotonic point. I can then plot a graph using all of the results gathered and the point where the potatoes gained and lost no weight is their isotonic point. Whenever performing an experiment certain safety precautions must be taken to prevent injury to both yourself and others. To this end, I must make sure that:

I only cut away from myself when peeling the potato (to avoid cutting myself by mistake) and made sure to always be aware where the knife was, where it was going and whether anyone or anything was in the way).
I always cut on a cutting mat (to avoid damage to the desk or other surfaces).
I will also try not to spill any solution on the floor and clean up any that I do spill straight away (to stop people from slipping)
I’ll keep the cups in the middle of the table (so that they can’t be accidentally knocked onto the floor).

In order to make sure that the experiment will yield accurate information, I have to make the tests as fair as possible. This includes narrowing the variables so that only the concentration of the sucrose solution varies (i.e. having the same temperature (because higher temperatures mean more energetic particles which mean faster moving molecules and thus a greater rate of osmosis), volume of solution, surface area on chips (so that the area available for particles to transfer through the semi-permeable membrane is the same), etc.), keeping the experiments under watch to avoid tampering (or accidents) and using the same equipment all of the time to keep the results as close to the same as possible. On top of this, I used two sets of scales and then calculated the average reading so as to minimise effects of faulty scales. As it turned out, the scales agreed with each other almost all of the time, but it was worth doing this because it gave a more accurate view of the ones that were measured differently and thus made my work more reliable.

Now is as good a time as any to mention what I think will happen when I perform these experiments. I know from a carrot experiment that I did in class a few weeks ago that a plant cell in sugarless water gains weight, one in slightly sugared water gains less weight in the same amount of time and one in heavily sugared water (near enough at the saturation point) loses weight. This is because weight change through osmosis (the movement of water molecules from a region of high concentration to a lower one via a semi-permeable membrane) only ever occurs if there is a difference between the concentration of sugar outside the cell and inside it; even then only going one way (from a higher water concentration to a lower one). This is because the particles in the fluids (both inside the cell and out) are moving about. The water molecules, unlike the sugar ones, are small enough to pass through the cell membrane and thus gain access to the other side. The side with less glucose molecules would therefore have a larger amount of water molecules moving out because there are fewer holes in the membrane being blocked by them. After this process has gone on for a while however, it gets to a point where the larger amount of water molecules on one side compensates for the fact that there are less holes to get through on that side (due to the glucose molecules) and so the rates of transfer balance. We call this a solution in dynamic equilibrium (see diagram).

Based on this, I think that a graph representing the changes in weight from my experiment should look something like the one below. Because the potato’s internal sugar concentration is never less than the distilled water’s, the chips in there will gain weight whilst the opposite should hold true for those in the 1 molar solution (because they will have a lower sugar concentration than the solution that they are in). Also, the further the solution is from the potato’s isotonic point, the faster the osmosis will occur, regardless of the direction of the waters travel (i.e. from potato to water or from water to potato) because there will be a greater difference between the concentration of water molecules on either side of the semi-permeable membrane (osmotic potential) and thus there will be more water molecules going from side A to B in a second than there would be if A were of a lower value (where A and B are the two sides of the of the semi-permeable membrane (A being the solution).

Conclusion.

The results from my tables show that the chips are clearly changing in mass after the half an hour, but more importantly that they are changing differently and by different amounts depending upon which solution they were placed in.

White potato.

There are two things that can very clearly be seen from the graph regarding the white potato; firstly, the white potato chips in solutions above 0.Nmolars were hypertonic, that is to say that they lost weight, whilst those in solutions below this point, the isotonic point (the point at which the fluid is in a state of dynamic equilibrium), were hydrotonic (that means that they gained weight). The second thing that can be learned from looking at the tables is that the further from the isotonic point a solution was, the faster osmosis occurred and thus the greater the change in weight over a given time.

Red potato.

There are two things that can very clearly be seen from the graph regarding the red potato; firstly, the red potato chips in solutions above 0.Nmolars were hypertonic, that is to say that they lost weight, whilst those in solutions below this point, the isotonic point (the point at which the fluid is in a state of dynamic equilibrium), were hydrotonic (that means that they gained weight). The second thing that can be learned from looking at the tables is that the further from the isotonic point a solution was, the faster osmosis occurred and thus the greater the change in weight over a given time.

The results of the two potatoes show that there is a large difference between the isotonic points of the red and white potato (large when considered in the scale at which the results are being considered). This would indicate that there isn’t a universal isotonic point for plant cells but rather that different types of plants have different isotonic points. This is in keeping with my prediction that I made right at the start in the introduction; after all, it stands to reason that if some plants make more sugar than others (e.g. sugar cane) then they would have different isotonic points because of their different sugar to water ratios. The results also agree with my prediction that the chips nearer to the isotonic point lost/gained less weight than those at a distance in the time allotted, thus indicating a slower rate of osmosis (due to the lower amount of water molecules waiting to pass through from one side to the other compared to those on the other side). I also predicted what the graph would look like with a large degree of accuracy, even down to the rough gradient and sideways S shape.

Evaluation.

In my opinion, the experiment went very well, almost exactly as planned in fact. One of the only problems that I encountered whilst performing my experiment was that the other class had taken all of the stop clocks so we had to use a clock instead. This had no alarm and we couldn’t measure the seconds accurately enough because it was mounted on the wall and we couldn’t see it properly from where we were, and were unable to move because of our experiment. This meant that the results could actually be one minute off either way. There were, however, a few problems that were inherent in my plan, that either escaped my attention or were ones that I couldn’t solve with my limited resources. These problems included:

My inability to properly measure the weights of the chips, due to the lack of sufficiently accurate scales. As long as there were readings that jumped up and down, the scales were inadequate for the task at hand.
Some of the chips staying in the solution longer than others because I didn’t have the ability to remove 18 chips from 6 different cups at the same time.
The fact that the pressure I applied to the chips as I removed them from the solutions varied, meaning that some may have had more water squeezed out of them than others. In retrospect, this could have been solved with the use of a spoon to remove all of the pieces, thus reducing the force to that of gravity pulling them against the spoon.
Some of the chips would have had longer to dry or would have been dried with more pressure by the paper towels.
I didn’t have enough time or potatoes to select chips that all had equal mass and density, as well as surface area. The unequal masses of the chips became evident when two chips in a solution floated whilst the other one sank. Had I had more time, I would have selected far more similar chips.
There was insufficient time for me to repeat the experiment. Yet again, this was a problem with time. I would Ideally have liked to have measured each of the three chips individually for posterities sake and then performed each one, and each cup, one at a time. This was not to be however, and my results may be slightly off because of it.
The small variations in chip size that I was due to my errors with cutting more than anything else.
Some chips floated and others sank, meaning that whilst some were completely covered, others were only covered on a few sides (affecting the surface area and thus the rate of osmosis).
I couldn’t control the temperature of the room over the time taken to perform the two experiments, meaning that one may have had a higher temperature than the other, and thus a higher rate of osmosis.

It is important to remember though that even though this seems like a lot of problems and uncontrolled variables, even the most advanced technologies in the world can’t control all of the variables, let alone in the time allotted and with the resources allocated. It is therefore better to compare these results with those of other people who had to perform the same experiments under the same restrictions. Compared with these people, my method is good and I seem to have removed more variables than some of those I spoke to (e.g. using two scales to minimise chances of inaccuracy). On top of this, it would appear that my results were very accurate, especially when compared to some of the other people. The other thing that stands out in my experiment is the accuracy of the predictions that I made; I predicted not only that the 1 molar solution chips would lose weight, but also the general S shape of the graph, something that quiet a few people didn’t predict. As a result, there were no readings that I would consider to be anomalous in my work.

Two points that Ii would like to make here are:

Bearing in mind that the question calls for finding the isotonic point of a plant cell, it is impossible to answer because no two types of plant cells necessarily have the same isotonic point. In an ideal world, we would answer the spirit of this question by performing it with every plant type on earth, but obviously this is impractical and
Should I have more time and resources in which to discover the isotonic point of any plant cell, I would first perform the experiment as above for each plant cell and then would narrow the range of sucrose solutions accordingly. (so in the case of the white potato, I would repeat the experiment with sucrose solutions ranging from 0.N1 molars and 0.N9 molars).this would allow me to more accurately pinpoint the isotonic point of the cell/cells in question and, if time were still available, I could narrow it still further by using a still thinner range of sucrose solutions (e.g. 0.201molars - 0.209molars).

Bibliography.