Investigating the Water Potential of Potato Cells
The aim of my investigation is to find the water potential of potato
Ensuring a Fair Test
When carrying out the experiment I will need to control a number of
factors that would affect the rate of osmosis:
- Concentrations of sucrose. These must be accurately made during the
experiment, as a different concentration of sucrose will result in a
different concentration gradient, therefore changing the rate of
- Surface area of potato. This must also be kept constant because if
there is more surface area between the cells and the surrounding
sucrose, then there will be more cells in contact with the solution.
This means a larger volume of water can leave or enter the cell at a
given time, thereby affecting the experiment. This must also therefore
be kept constant.
- Time in solution.This is another factor which must be controlled.
This is because if the potatoes are in solution for longer, more water
movement will occur, thus making the experiment unfair.
- Volume of solution. This would affect the experiment in that a
different volume of solution contains different numbers of molecules.
Therefore there may (for example) be more water in one test tube
opposed to another. This means the water potential of that test tube
would be greater, making the experiment inaccurate.
- Chips from different potatoes. This would affect the experiment in
that potato chips from different potatoes may contain cells of a
different starting water potential, and when I calculate the
percentage change, that potato chip
would give me an anomalous result.
This is what I am finding out the water potential of
This will be used to cut up the potato and take the skin off
This will be used when cutting the potato
6 Boiling Tubes
The potato strips and sucrose solution will be placed in these
Boiling Tube Rack
This will be used to hold the boiling tubes during the experiment
This will be used to measure the length/surface area of the potato
1M Sucrose Solution
Solute to be used which will determine potato's water potential
This will be used to make up the various solute concentrations
Will be used when adding water to the sucrose/water solution
Will be used when adding sucrose to the sucrose/water solution
To be used for measuring the change in potato mass
To time how long the potatoes are in solution for
To hold the sucrose and water in before the solutions are made up
Will be used to cut potato into strips of 1cm height and width
Will be used to dry off excess water that hasn't been taken up by the
Below is a diagram showing how I will set up my apparatus
Whilst carrying out this experiment I will aim to do 2 repeats for
each data reading, which will make me end up with 3 readings in total
for each solute potential. This is to allow for any human error in the
procedure, which can then be identified by looking at an average of
the results. If an extreme value does occur it can be considered as an
anomalous result. This will enable me to make accurate conclusions
about the data I have obtained, and seeing how the sucrose
concentration affects the water potential of potato cells
Sucrose Concentration (M)
Volume of Sucrose used (cm3)
Volume of Water used (cm3)
Surface Area of Potato
Initial Potato Mass
Final Potato Mass
Percentage change in mass (%)
Time in solution = 20 minutes
I have discovered from my preliminary results that as I increase the
sucrose concentration, the water potential of the potato cells
decreases. This can be seen in that the potato loses mass and
therefore has clearly lost water from its cells.
I am pleased at the way my experiment went and despite the anomalous
result for 0.8M sucrose, I think that my procedure is a good one which
I will use in the actual experiment. My range is fairly good also. I
have also discovered that somewhere between 0.4M and 0.6M the change
in mass is zero. For this reason in my actual experiment I will not
have such a large solute concentration range, but will stop at about
0.8M of sucrose.
As well as this I carried out an earlier preliminary experiment, in
which I immersed the potato in sucrose solution and then recorded
changes in its flaccidness. This was done by putting a pin through one
end of the potato and then into a cork and then measuring how high the
other end of the potato was on a scale. I then put this potato piece
into some sucrose solution. I found that a high water potential
solution meant the potato was more firm so drooped less. It was
therefore higher up the scale. When the water potential of the
solution was low then the potato would become droopier and was
therefore lower on the scale. However I found this to be a very
inaccurate experiment so I have decided not to do this in my actual
1. Using a knife I will skin the potato and then using the potato
chipper, I will cut the potato into long strips which are 1cm in
height and 1cm in width. Next I will cut the potato strips into
lengths of 5cm. This will give me a surface area of 22cm2.
2. Next I will make up the sucrose concentrations which the potato
will be immersed in. The concentrations I will make are 0, 0.15,
0.30, 0.45, 0.60 and 0.75M. The compositions to make up these
concentrations are shown in the table below.
Concentration of sucrose solution
Volume of 1M sucrose concentration
Volume of Distilled Water
These concentrations will be made up in 6 test tubes using a 20ml
syringe for adding the water and 5ml syringe for adding the sucrose.
3. Next I will weigh each potato piece I have cut using a balance.
4. I will now place one piece of potato to each test tube for a
period of 20 minutes.
5. Once 20 minutes have passed I will remove the potatoes from the
test tubes and lightly dry them using a piece of tissue paper.
This will be done by gently wiping the potatoes on the tissue
paper. This is to remove water not taken in by the cells of the
6. I will then weigh each potato piece again and record this in a
table. I will then be able to calculate the percentage change in
7. I will then repeat the experiment twice to account for any
experiment inaccuracy and to identify any anomalous results.
As I stated earlier, and have seen from my preliminary work, I predict
that the mass of the potato will decrease most when the sucrose
concentration is highest- i.e. when the water potential is smallest. I
also predict that when the concentration of the sucrose solution is
0%-i.e. the water potential is greatest; the potato will increase in
Osmosis is the passive net movement of water from a high water
potential to a low water potential across a semi-permeable membrane.
This process can be sped up by several factors, including temperature,
pressure and the water potential gradient.
The potato strip originally contains a certain amount of water in its
cells. Potatoes are storage cells which are packed with cellulose, so
for this reason I predict the potato will have a low water potential.
When it is placed in a solution of a different water potential, water
moves either into or out of its cells to create an osmotic balance. As
a result the potato can become turgid- meaning its cell membrane is
pushing out against the cell wall due to much water having entered its
cells, or it may become plasmolysed- the cell membrane shrinks away
from the cell wall due to the volume of the cell decreasing (i.e.
water moving out of the cell).
When the potato is placed in distilled water, there is a water
potential difference between the cell and the surrounding solution. As
a result water moves down the water potential gradient and enters the
cell in order to create an osmotic balance. This is shown in the
Distilled Water (in test tube) Potato Cells
[IMAGE] Net movement of water
[IMAGE] Water molecule
Partially Permeable Membrane
As you can see in the diagram above, there are more water molecules in
the test tube than in the potato cells. As stated earlier osmosis is
the net movement of water from a high water potential to a low water
potential. It is therefore clear that water molecules will move from
the left side of the diagram to the right- i.e. from the test tube to
the potato until there is an osmotic balance. This is why I predict
the potato will increase in mass when the potato is immersed in pure
water. I predict this will continue to happen until the potato is
immersed in a sucrose concentration, which has a lower water potential
than itself. This is when water will start to move out of the cells of
the potato and the potato will therefore lose mass.
Eventually the cells in the potato will become turgid, and at this
point very little water will move into the potato. The diagram below
shows the structure the cell will have at this point.
 Diagram showing the appearance of a plant cell when it is turgid
[IMAGE]The diagram shows that all the cell membrane is in contact with
the cell wall, which is what a turgid cell looks like. It is therefore
clear how the cell becomes turgid, in that the cell membrane pushes
out against the cell wall to allow more water into the cell. However
due to the cell wall, the cell would eventually stop taking in water,
as the membrane can move no further out. The effect is like blowing up
a balloon in a box. When the balloon takes in so much air that it is
all touching the sides of the box, the box exerts pressure on the
balloon, and so it becomes increasingly difficult to blow the balloon
up any further. This same thing applies to water entering the cell,
and is the reason why plant cells to not burst when they become
turgid, but animal cells do (as they lack a cell wall).
For these reasons I think the cell will become more turgid when the
water potential of the sucrose solution is greater than the water
potential in the potato. I predict the mass of the potato will
increase most when the water potential gradient is steepest. The less
negative the water potential is, which the potato is immersed in, the
faster the potato will uptake water, and therefore increase in mass.
I must now consider what will happen when the water potential of the
potato is greater than the water potential of the sucrose solution.
When the potato is immersed in a solution of lower water potential
than its own cells, water moves out of the cells by osmosis. The
solute potential will be higher in the sucrose solution than the cells
of the potato, so therefore the water potential of the solution is
lower. This means the water potential is more negative in the sucrose
solution than that of the potato so water will move out of the
potatoes and into the solution.
Potato Cells Sucrose Solution
The diagram above illustrates the point I have just made in that there
is higher water potential in the potato than there is in the sucrose
solution. This means the water potential is less negative in the
potato than it is in the sucrose solution, so water must move out of
the potato to form an osmotic balance. The solute (sucrose) in the
sucrose solution lowers the water potential, making it more negative,
which causes the water potential gradient between the potato and the
solution to be steeper.
This diagram shows the appearance of a plasmolysed cell. This time
none of the cell membrane is in contact with the cell wall. This means
that there is less fluid in the cell than the cell would like, as
water has moved out of it by osmosis. However, just as a turgid cell
can only take in so much water, the cell can only lose so much water.
When the cell has lost all the water inside it, it can not give out
more water. The graph, which is shown later, shows this effect of
extreme water potential against percentage change in mass
For this reason I believe that increasing the solute potential will
cause the mass of the potato to decrease. The higher the solute
potential the steeper the water potential gradient, and therefore the
faster water will move out of the potato. For this reason I believe
that the mass of the potato will decrease most when the sucrose
concentration is highest.
For the reasons specified I predict my graph will look like the
following or will portray a similar trend- i.e. as sucrose
concentration increases, percentage change in mass of the potato
decreases. As well as this my preliminary results which I have
explained above show this.
Percentage Change in Mass
[IMAGE] Solute Concentration
I think the graph will look like this for several reasons. When the
solute potential is 0 or only slightly negative, the potato will have
a lower water potential than the solution so water will move into the
potato. The cells will then become turgid. At this point the cells
will no longer be able to take in water, which is why the line is
curve and not a straight line. When the solute potential becomes
greater than the cells of the potato, water will move out of the
potato and into the solution, causing it to become plasmolysed.
However once all the water has left the cell no more water will be
able to move out, and therefore the mass of the potato will not
change. This is why the graph levels off. Where the line intersects
the x axis, this is when there is no change in mass. The potato and
solution have equal water potential so there is no net movement of
water- i.e. insipient plasmolysis. At this point the cells will appear
like the one below
To ensure my experiment is safe I will make sure that I don't drop any
potato pieces on the floor which could be a potential hazard for
someone to slip on. I will also take extra care when using the knife
to skin and cut the potato. I will also wear safety spectacles to
protect my eyes.
Add water to make up the sucrose solution
I have chosen this as it is quite accurate. It is also easier to add
the water, as well as it being more accurate, to add water from a
burette rather than a pipette
Add sucrose to make up the sucrose solution
I have chosen this as opposed to a larger syringe as it is more
accurate as it has a smaller percentage error.
To weigh the Potatoes at the start and end of the experiment
I have chosen a balance as opposed to measuring the flaccidness of the
potato, as it is more accurate. I saw from my preliminary results that
measuring droopiness was very inaccurate
The table above shows why I have chosen to use some apparatus as
oppose to others. The apparatus I am generally using will be more
accurate than using another sort, which will therefore make the
experiment more accurate.
Concentration of Sucrose Solution
Volume of 1M Sucrose Used (cm3)
Volume of Water Used (cm3)
Surface Area of Potato
Potato time in solution (minutes)
Mass of Potato
Percentage Change in mass of potato
Table showing the percentage change in mass of potato strips when they
are placed in solutions of different water potentials
Indicates an anomalous result
From my results I can see that increasing the solute potential, and
therefore decreasing the water potential, causes the percentage change
in mass of the potato to decrease. This can basically be explained in
that immersing a potato strip in a solution with a lower water
potential than itself causes water to move out of the potato in order
to create an osmotic balance. For this reason the potato decreases in
These results agree with my prediction in that the potato decreases in
mass once the water potential of the solution it is in becomes less
than its own water potential. In my prediction I claimed that when
this occurred water must move out of the cell to create an osmotic
From my graph it is clear that when the sucrose potential of the
solution is less than the sucrose potential of the potato, and
therefore the potato has a lower potential, the potato chip increases
in mass- i.e. when the sucrose concentration is less than 0.3M. Above
this concentration, however, the potato has a lower sucrose
concentration than the solution and water must therefore move out of
The processed evidence also supports the fact that the percentage
change in mass of the potato is not a linear relationship with the
sucrose concentration. This can be seen in that my graph curves off at
both ends. This proves the fact that only a certain volume of water
can be taken into the cells before the cell wall exerts such a
pressure that no more can be taken in. When the sucrose potential was
high the water only lost a certain amount of water before it could not
lose anymore. Both these conclusions agree with what my results
From the graph I can now also determine what the water potential of
the potato was. Using a sheet handed out by the teacher showing a
graph relating sucrose concentration and water potential at those
various concentrations I can find out what the water potential of my
potato was. Where the line intersects the x axis indicates when there
is no net movement of water between the potato and the solution it is
in. Here the cells are in the incipient plasmolysis change. Due to the
fact that there is no net movement of water at this point, it can be
said that the two have equal water potential. Therefore I can say that
the sucrose concentration of my potato pieces was 0.2625M. Using the
graph handed out by the teacher I can now read off from the graph what
the water potential is for this solute concentration.
After reading off the graph I have found the water potential of my
potato pieces to be -714kPa. Using my graph for the results I have
obtained I can calculate the rate of reactions (osmosis) at various
From the processed evidence I can see that percentage change in mass
does relate to the concentration of the solution the potatoes are
immersed in. The graph clearly shows this in that the general trend is
percentage change in mass decreases as the sucrose concentration
The graph shows that my prediction is true, in that osmosis has
limitations and that placing a cell in a very low water potential
solution will force water out of the cell and vice versa. Likewise the
cell can only lose or gain a certain amount of water which is also
shown in the graph when the best fit line curves off. The graph shows
that when the solution becomes a lower water potential than the cell,
only a small amount of water leaves the cell. This can be seen in that
when the percentage change in mass of the potato is negative the line
quickly curves off. However when the potato is placed in a solution
which has a higher water potential than itself it readily gains water.
This also proves my prediction in that potatoes store large quantities
of sucrose. It all ready only contains a small amount of water so can
only lose a small amount of water. It can however gain much water. All
these conclusions are consistent with my processed evidence.
Net movement of water
Partially Permeable Membrane
The diagram above can be used to explain my processed evidence. The
graph shows a high percentage change in mass when the sucrose
concentration is low. This is illustrated in the diagram above.
Naturally water must move out of the solution to form the osmotic
balance which explains the increase in mass of the potato.
When the concentration of the sucrose solution increases, the
percentage change in mass of the potato is lessened, which can clearly
be identified on the graph. The science behind this is that the rate
of osmosis is dependant on the water potential gradient. As the
sucrose concentration in the solution increases the potato and
solutions' water potential gradient becomes shallower, causing water
to move across the cell membrane at a slower rate.
The graph shows that at 0.263M sucrose concentration the line
intersects the x axis. This can be explained in that here there is no
net movement of water. As a result the potato does not change in mass.
Here incipient plasmolysis is taking place.
The graph then shows that increasing the concentration further causes
the potato to decrease in mass. This is simply due to the fact that
the potato now has a higher water potential than the solution so water
must move out of the potato. Eventually sucrose concentration has no
effect on osmosis, which can be seen with the graph levelling off.
This occurs because there is no more water in the cells for the potato
The graph relating water potential and sucrose concentration also
shows that increasing the solute potential decreases the water
Water Potential = Solute Potential + Pressure Potential
The equation above shows that increasing the solute potential or
pressure potential will cause the water potential to increase. This
can be seen in the handout graph showing sucrose concentration against
I can therefore deduce that a very low solute potential, or one which
is less than the potato, causes the percentage change in mass to
increase. The cells take in water by osmosis in order to reach an
osmotic balance (incipient plasmolysis) and this causes the mass to
increase. However at some point the uptake of water in a plant cell
will decrease and eventually stop due to the pressure potential factor
exerted on the cell wall by the protoplast. This occurs in order to
prevent the cell from bursting, which is not so in an animal cell.
I can also deduce that a high solute potential, or one which is
greater than the potato, causes the percentage change in mass of the
potato do decrease. This, again, occurs due to water leaving the cell
to create an osmotic balance. The protoplast ends up shrinking away
from the cell wall and therefore pressure potential at this point is
Finally I can deduce that at one specific solute potential there is no
percentage change in mass of the potato. At this point there is an
equilibrium reached between the cells and the solution around them. At
this point the pressure potential has just reached zero. For my
experiment this occurred when the sucrose concentration was
approximately 0.26M, which when read from the graph gave my potato a
water potential of -714kPa.
Overall I think my experiment went very well. I managed to find out
what the water potential was of my potato piece, which means my
procedure was suitable. I obtained only two anomalous results which
also indicates my experiment was carried out accurately.
During the experiment there were a few things which made the
experiment inaccurate. Drying off the excess water from the potatoes
was probably the biggest problem I faced as the potato pieces would be
dried by different amounts. This therefore means that the experiment
is not truly a fair test.
The equipment I used during my experiment was quite accurate and
overall only gave me a small percentage error. I used appropriate
sized syringes for extracting volumes of solution as to reduce the
error margin I would obtain. The balance also read to 2 decimal places
so offered a low percentage error. I believe therefore that the main
cause of my anomalous results was due to drying off the excess
solution from the potato.
The biggest limitation to my procedure was drying the excess water
from the potato. I believe this because in doing so I dried up
different volumes of water from each potato piece. This would
therefore have a large effect when I would calculate my percentage
change in mass.
The second biggest limitation to my procedure would probably be making
up the concentrations. Whilst doing this it is very easy to draw up
slightly too much solution which would therefore cause my results to
be inaccurate and lead to anomalous results.
To improve this experiment I would have used larger pieces of
Percentage error = (error / reading) x 100
The above equation shows how percentage error is calculated. I
therefore believe that increasing the size of potato would decrease
the percentage error. For example if we consider weighing the potatoes
then it is clear that dividing by a larger number gives us a smaller
error. This is shown below in an example
Mass of potato in this example = 5g
Therefore percentage error = (0.005 / 5) x 100
This is still a very small percentage error but can easily be reduced
by using a heavier potato
Mass of potato in this example = 20g
Therefore percentage error = (0.005 / 20) x 100
Therefore simply doing this will greatly reduce the inaccuracy in this
However despite all this I think that my results are very reliable as
they are how I predicted them to be and I am able to explain them
using scientific knowledge. I only obtained two anomalous results
also. The equipment I used was also fairly accurate and would be
difficult to improve it.
 Diagram taken from Biology 1 Advanced Sciences