Osmosis, What is the effect of sucrose concentration on the rate of osmosis in a potato cell? What is the concentration inside a potato?
Investigation - Biology Coursework
What is the effect of sucrose concentration on osmosis in a potato cell?
What is the concentration inside a potato?
PLANNING 2
AIM 2
WHAT IS OSMOSIS? 2
FAIR TEST 3
PREDICTION 4
PREDICTION GRAPH 4
RISK ASSESSMENT 4
APPARATUS 4
DIAGRAM 5
METHOD 6
BLANK RESULTS TABLE 6
PILOT TEST 7
PILOT TEST RESULTS TABLE 7
PILOT TEST ANALYSIS 7
PILOT TEST GRAPH 8
OBTAINING EVIDENCE 9
RESULTS TABLE 9
GRAPH OF RESULTS 9
ANALYSIS 10
SCIENTIFIC EXPLANATION OF RESULTS 10
CONCLUSION 10
PREDICTION CONFIRMED? 10
EVALUATION 11
QUALITY OF RESULTS 11
ANOMALOUS RESULTS 11
QUALITY OF THE METHOD USED 12
METHOD IMPROVEMENTS 12
IDEAS FOR COLLECTING FURTHER EVIDENCE 12
BIBLIOGRAPHY 16
PLANNING
Aim
What is the effect of sucrose concentration on the rate of osmosis in a potato cell?
What is the concentration inside a potato?
What is Osmosis?
Osmosis is the net movement of water from a high water concentration (high water potential, a dilute solution) to a place of lower water concentration (low water potential, a concentrated solution) through a partially permeable membrane.
This is a diagram to show how osmosis works:
When sugar dissolves in water...
This diagram, I sourced from [www.gcsesciencehelp.co.uk/concepts/osmosis]
I have edited it slightly
Hypertonic Solution
Hypotonic Solution
Concentrated Sugar Solution - A low water concentration or low water potential
Dilute Sugar Solution - A high water concentration or high water potential
A low concentration of free water molecules
A high concentration of free water molecules
Sugar molecules pass through the pores more slowly
Water molecules are free to move through the pores in the membrane
Selectively permeable membrane
Sugar molecules attract some of the water molecules and stop them from moving freely. This, in effect, reduces the concentration of water molecules. In the diagram above, the molecules on the left have "captured" about half of the water molecules. There are more free water molecules on the right of the membrane than on the left, so water will diffuse more rapidly from right to left across the membrane than from left to right. This creates a net movement of water from a high water potential to a low water potential.
The partially permeable membrane does not act as a sieve in this case. The sugar molecules can diffuse from left to right but, because they are bigger and surrounded by a cloud of water molecules, they diffuse more slowly than the water
This information was sourced from "A Level Biology" Jones and Jones
Fair Test
Factors affecting osmosis and how I will control them in my investigation:
) Surface Area: The surface area of the potato piece affects the amount of sucrose surrounding the potato. The larger this surface area is, the faster the rate of osmosis occurs. I will control this variable by making all of the potato pieces have the same surface area. I will do this by using a corer to cut the potatoes into tubes instead of cutting them into cubes. This way, all the pieces will have the same cross sectional area. Then I will cut each piece to 3 cm long using a scalpel. The potato pieces will now have an equal surface area. In addition, any traces of skin will have to be removed. The potato skin is not a partially permeable membrane, and therefore water cannot travel through it by osmosis.
2) Mass of Potato Piece: As the potato piece gets larger, more water can go into the potato out of it. Using the cutting method approved above, the mass of the pieces should be approximately equal. However, I feel that another measure should be taken. The potatoes will be weighed before and after the experiment to gain a mass change during incubation. The mass change will vary slightly from potato to potato. Therefore, I will calculate the percentage mass change (Mass Change (g) / Mass Before (g)). This will eliminate any minor errors
3) Time of Incubation: As the time of incubation grows, the rate of osmosis speeds up. For this experiment, all the pieces of potato will be subjected to 30 minutes of submersion. I feel that this is a good time because...
At two hours all possible osmosis will have stopped for the test tubes will the highest and lowest concentrations because they will have conducted osmosis the fastest due to their high water potential gradient between the potato cells and the solution. In addition, the concentrations below the two extremes will begin to catch up with the boundaries creating an awkward graph.
At 5 minutes, hardly any osmosis will have taken place. Even a small anomaly in one measurement could result in an experimental failure because the values will be so small.
Therefore, I will conduct my experiment over a period of 30 minutes, a time when none of the above problems will occur.
4) Temperature: As temperature increases, the rate of osmosis increases. As temperature increases, the particles have more energy and move faster. They move faster through the partially permeable membrane and therefore the rate of osmosis speeds up. To solve this problem I will conduct my experiment within one hour on the same day. Therefore, temperature will not fluctuate significantly to ruin the results
5) Potato (age, type, size, plant): Different types of potatoes have different concentrations. Baking potatoes are bitter. They have a low sucrose concentration. New potatoes are sweeter. They have a high sucrose concentration. Small young potatoes are sweeter than larger, older ones. To solve this problem, I will just use a single large potato for all of my potato pieces to prevent shifts in the graph line from occurring as I change to a different potato.
6) Concentration of surrounding solution: The concentration of the solution is the variable that I am measuring. It is my independent ...
This is a preview of the whole essay
5) Potato (age, type, size, plant): Different types of potatoes have different concentrations. Baking potatoes are bitter. They have a low sucrose concentration. New potatoes are sweeter. They have a high sucrose concentration. Small young potatoes are sweeter than larger, older ones. To solve this problem, I will just use a single large potato for all of my potato pieces to prevent shifts in the graph line from occurring as I change to a different potato.
6) Concentration of surrounding solution: The concentration of the solution is the variable that I am measuring. It is my independent variable. Using the results, I will plot Percentage Mass Change against time. I will plot the graph and draw a line of best fit. At the isotonic point (where the line crosses the x axis), I will find the concentration inside the potato.
Prediction
I predict that the two extreme concentrations in this experiment - 0.0M and 1.0M - will create the fastest rate of osmosis. The potato cells have a concentration between 1.0M and 0.0M. I will assume that the concentration is 0.5M. The two concentrations mentioned above show the greatest water potential gradient and therefore the fastest rates of osmosis.
From looking at the pilot test results, I can see that the greatest mass loss took place when the potatoes were in a sucrose concentration of 1.0M. This indicates that the concentration within the potato cell is closer to 0.0M than 1.0M. To expand on this idea, I drew a graph of my results and connected the two average points together. The isotonic point, where the line crosses the x-axis, shows the concentration inside the potato cell. No mass change took place because the water potential inside the cell was equal to that outside it; they were isotonic so no osmosis occurred. From looking at the graph, I can conclude that the concentration inside the potato was approximately 0.38 Moles/Litre. The potato that I use for my real experiment may not be the same variety, size, or from the same plant so 0.38 Moles/Litre is only an estimate.
Prediction Graph
Risk Assessment
The knife and the corer will be used to cut the pieces of potato. These apparatus are sharp. I will use these tools with care to avoid cutting myself.
Apparatus
Number Needed
Apparatus
Reason
Large Potato
Where I will obtain my measurements from
240 ml
Distilled Water
Dilutes the sucrose to provide different concentrations
240 ml
M Sucrose Solution
The solution which surrounds the potato pieces
2
50 ml Measuring Cylinders
Measures the amounts of sucrose and water needed as stated in the concentration table
2
Test tubes
Hold the sucrose/water and the potato cylinders during the experiment
Test tube Rack
Holds the test tubes during incubation
Stopwatch
Measures the time of incubation
Scalpel
Cuts the tubes of potato to the preferred length
Corer
Cuts the tubes of potato from the large potato
Ruler
Measure the length of the cylinders
Tile
Stop the potato pieces from having moisture drawn out of them when waiting to be weighed
Electronic Weighing Scales that records at least 2 decimal places
Weighs the potato pieces before and after incubation. Two decimal places requires the high level of accuracy, which I hope to achieve in this experiment
Some
Paper Towels
Dries excess water/sucrose from the potato when taken out of the solutions after 30 minutes
Diagram
Method
) Set up the apparatus as shown in the diagram without placing the sucrose/water or the potato pieces in the test tubes.
2) Using the measuring cylinders and the concentration table below make up the different sucrose molarities and pour them into the test tubes
Sucrose (ml)
Water (ml)
Sucrose concentration created (Moles/Litre)
40
0
0
32
8
0.2
24
6
0.4
6
24
0.6
8
32
0.8
0
40
.0
3) Take the large potato and cut out twelve tubes using the corer.
4) Remove any pieces of potato skin from the tubes
5) Using the scalpel, make each tube into a 3 cm long cylinder
6) Record the mass of each potato cylinder. When not weighing put the potato pieces on the tile to prevent moisture loss
7) Place the potato cylinders into the test tubes with the solutions and start the stopwatch
8) After 30 minutes, take all of the potato pieces out of the test tubes and place them on the tile
9) Roll each piece once on the paper towels to drain surface moisture and then record their masses.
0) Calculate the mass change of each potato and their percentage mass change.
1) Calculate the average percentage mass changes
2) Create a graph of average sucrose concentration against average percentage mass change
3) Find the isotonic point, where the line crosses the x-axis. That point value on the x-axis is the concentration inside the potato
Blank Results Table
Sucrose Concentration (Moles/Litre)
Mass Before (g)
Mass After (g)
Mass Change (g)
% Mass Change (2 decimal points)
0
0
0.2
0.2
0.4
0.4
0.6
0.6
0.8
0.8
Sucrose Concentration (Moles/Litre)
Average % Mass Change (2 decimal points)
0
0.2
0.4
0.6
0.8
Pilot Test
In the pilot test, only two molarities of sucrose were used. The solutions utilized were created from the following table
Sucrose (ml)
Water (ml)
Sucrose Concentration Created (Moles/Litre)
20
0
.0
0
20
0.0
Pilot Test Results Table
Sucrose Concentration (Moles/Litre)
Mass Before (g)
Mass After (g)
Mass Change (g)
% Mass Change (2 decimal places)
0
3.54
3.74
0.2
5.65%
0
3.45
3.63
0.18
5.22%
3.36
3.06
-0.3
-8.93%
3.45
3.11
-0.34
-9.86%
Sucrose Concentration (Moles/Litre)
% Mass Change (2 decimal places)
0
5.44%
-9.40%
Pilot Test Analysis
The pilot test informs my prediction:
The two extreme concentrations in this experiment - 0.0M and 1.0M - do create the fastest rate of osmosis. The two concentrations mentioned above show the greatest water potential gradient and therefore the fastest rates of osmosis.
From looking at the pilot test results, I can see that the greatest mass loss took place when the potatoes were in a sucrose concentration of 1.0M. This indicates that the concentration within the potato cell is closer to 0.0M than 1.0M. To expand on this idea, I drew a graph of my results and connected the two average points together. The isotonic point, where the line crosses the x-axis, shows the concentration inside the potato cell. No mass change took place because the water potential inside the cell was equal to that outside it; they were isotonic so no osmosis occurred. From looking at the graph, I can conclude that the concentration inside the potato was approximately 0.38 Moles/Litre. The potato that I use for my real experiment may not be the same variety, size, or from the same plant so 0.38 Moles/Litre an estimate.
I cannot see any anomalous results in the pilot test graph, however I cannot be sure, due to there being only two points on the graph to justify this. On the other hand, the difference between original points (before averages were calculated) is minimal and signals to a much more reliable set of data.
In the pilot test, just using 0.0M and 1.0 M, although providing a quite accurate set of results, is not reliable. In the real experiment, I will use other sucrose concentrations between 0.0M and 1.0M concentrations. These, I will create as in the table below:
Sucrose (ml)
Water (ml)
Sucrose Concentration Created (Moles/Litre)
40
0
0
32
8
0.2
24
6
0.4
6
24
0.6
8
32
0.8
0
40
.0
In the pilot test, I did not drain the excess sucrose/water from the outside of the potato pieces. This creates a higher mass than the actual, so the actual percentage mass change is lower. This means that the concentration I discovered, 0.38M, is actually slightly lower.
In the pilot test, the pieces of potato were cut into cubes. This is a very crude way of preparing the experiment. The surface area cannot have been accurate and there was a difference in the masses. In the real experiment, I will use the cutting method shown in the fair test section using a corer and a scalpel.
Pilot Test Graph
OBTAINING EVIDENCE
Results Table
Sucrose Concentration (Moles/Litre)
Mass Before (g)
Mass After (g)
Mass Change (g)
% Mass Change (2 decimal points)
0
2.4
2.52
0.12
5.00%
0
2.56
2.67
0.11
4.30%
0.2
2.53
2.59
0.06
2.37%
0.2
2.52
2.54
0.02
0.79%
0.4
2.39
2.35
-0.04
-1.67%
0.4
2.55
2.52
-0.03
-1.18%
0.6
2.53
2.44
-0.09
-3.56%
0.6
2.64
2.53
-0.11
-4.17%
0.8
2.52
2.35
-0.17
-6.75%
0.8
2.43
2.27
-0.16
-6.58%
2.48
2.24
-0.24
-9.68%
2.51
2.27
-0.24
-9.56%
Sucrose Concentration (Moles/Litre)
Average % Mass Change (2 decimal points)
0
4.64%
0.2
.58%
0.4
-1.42%
0.6
-3.87%
0.8
-6.67%
-9.62%
Graph of Results
The graph for this experiment is shown on the next page
ANALYSIS
Scientific Explanation of Results
From looking at the results, I can see that the greatest mass loss took place when the potatoes were in a sucrose concentration of 1.0M. This happened because the 1M sucrose has lower water potential than inside the potato. This concentration creates the greatest negative water potential gradient in this experiment. Therefore, the potato loses the most water, and consequently mass, to the sucrose by osmosis. The greatest mass gain occurred in 0.0M sucrose (or distilled water). This concentration presented the greatest positive water potential gradient in this experiment. Therefore, the potato gains the most water, and mass.
The isotonic point, where the line crosses the x-axis, shows the concentration inside the potato cell. No mass change took place because the water potential inside the cell was equal to that outside it; they were isotonic so no osmosis occurred. From looking at the graph, I can conclude that the concentration inside the potato was approximately 0.32 Moles/Litre.
This is slightly different to the pilot test prediction for the following reasons:
* The potato that I use for my real experiment may not be the same variety, or from the same plant so the original figure was an estimate.
* The potato used in this experiment was much larger than the one in the pilot test. Larger potatoes have higher water potential than smaller potatoes. Therefore, the potato used in this experiment would have higher water potential, and therefore a lower concentration.
* The potato pieces used in the pilot test did not have excess moisture removed with paper towels before weighing. This creates a higher after mass than the actual, so the actual percentage mass change is lower. This means that the concentration I discovered for the pilot test, 0.38M, is actually slightly lower.
Conclusion
The two extreme concentrations in this experiment - 0.0M and 1.0M - do create the fastest rates of osmosis. The two concentrations mentioned above show the greatest water potential gradients and therefore the fastest rates of osmosis. The sucrose concentration that is furthest away from the concentration inside a potato creates the largest mass change and so the fastest rate of osmosis (In this case - 1.0M). Therefore, I can conclude that the rate of osmosis is influenced by changes in sucrose concentration but at the same time depends on the concentration of sucrose inside the potato, or any other cells being tested.
The concentration inside the potato used in the final experiment was 0.32 Moles/Litre
Prediction Confirmed?
My prediction was confirmed.
* This conclusion confirms the prediction from looking at the pilot test results, indicating that the concentration within the potato cell is closer to 0.0M than 1.0M.
* Taking into account the points mentioned at the end of the previous section, 0.38 was a good estimate.
* The two extreme concentrations in this experiment - 0.0M and 1.0M - did create the fastest rate of osmosis. They did this because they both show the greatest water potential gradient and therefore, the fastest rates of osmosis.
EVALUATION
Quality of Results
My results are all very near or on the line. None of my results were extremely anomalous so all were included in the graph and in my calculations.
The levels of inaccuracies within my measurements are very low:
* Sucrose concentration - to the nearest 0.5 ml
(Made up of 1M sucrose and water measurements)
* Time - to the nearest 10 seconds
(It took ten seconds to remove the potato pieces from their solutions)
* Size - Cross sectional area was perfect due to the use of the corer
* Length - It was difficult to know where to place the scalpel on the potato cylinder even with the aid of a ruler. Errors could be to the nearest mm
* Mass - to the nearest 0.01g
* % Mass Change - to 2 decimal places
Anomalous Results
The two percentage mass changes from 0.2M sucrose solutions are very different to each other - 2.37% and 0.79% - a gap of 1.58%! However, when taking the averages to plot the real graph, that point lies perfectly on the line of best fit
For the highest point in the circle on this graph, the cylinder was cut in half accidentally. This resulted in that particular tube having a larger surface area than the others, and therefore a faster rate of osmosis. There was not enough space in the potato to cut out another tube, so rather than breach another fair test rule (Potato age, size and type) I decided to go with what I had. Thankfully, due to an anomaly for which the reason is unknown, the average was balanced and the point now lies perfectly on the line of best fit.
The graph below shows both percentage mass changes plotted against sucrose concentration.
Quality of the Method Used
The controlled variables of this experiment were controlled with different proportions:
) Surface Area: This is the greatest cause of harm to the results. I removed the potato skin successfully and cut the tubes to the correct length. However, I had some problems with the corer. After eleven tubes, there was hardly enough space for one more. I pushed down and slipped, slicing the tube in two increasing the surface area of this particular tube. This is explored more in the anomalous results section.
2) Mass of Potato Piece: I can be quite sure that this factor did not damage my results. The mass change would have varied slightly from potato to potato due to mass differences. However, I calculated the percentage mass change, eliminating any minor mass differences.
3) Time of Incubation: After half an hour I removed the potatoes from their solutions. There was only ten seconds between emptying the first test tube and the last test tube. I do not think that this gap is significant enough to invalidate my results.
4) Temperature: This experiment was conducted on the same day and in the same classroom so there was not a significant change in the temperature. In addition, if there were a sudden change in temperature, all of the potato pieces would have been subjected to it, keeping the test fair.
5) Potato (age, type, size, plant): I used the same potato for all of my potato cylinders. This point did not affect my results at all
My own method was made inaccurate by many things these are listed below and solved in the next paragraph:
) The pieces of potato may not have had the same surface area due to overlapping of holes
2) When measuring out the water and the sucrose, one could not be sure when to stop pouring due to the dip at the top of measuring cylinder
3) The paper towel drying method is not reliable or fair as variable pressure could be applied to the potato cylinders
4) It was difficult to cut the potato tube at the correct length. Even with the aid of a ruler, the lengths of the cylinders could be out by 1 mm
Method Improvements
My method could have been improved in this experiment. These points counteract the problems found in the previous paragraph. The ways I have thought of are listed below:
) Use a micrometer to measure the surface area of the potato pieces
2) Measure out the sucrose and the water by weight instead of volume
3) Place the potato pieces standing on a tile in the room for ten minutes to air-dry the pieces.
4) Use a computer to measure the length of, and cut the potato cylinders
Ideas for Collecting Further Evidence
Here is a detailed plan for how to broaden and extend the experimental results, or evidence, to support my conclusion by using equipment that is more sophisticated and investigating the independent variable more fully:
Aim
The aim of this experiment is to find out if the trend found in the previous experiment continues for sucrose concentrations over 1 Moles/Litre.
In this experiment, the sucrose concentration will be varied. It will be my independent variable. The factor I will measure is percentage mass change. I will calculate the mass before, the mass after and work out the mass change. Then I will calculate the percentage mass change and plot the average points on a graph against sucrose concentration.
Prediction
I predict that the trend will continue to be valid for and extended range. However, I think that the graph will level off eventually due to the concentration of sucrose being so high that it takes all of the water out of the potato before the end of the incubation time. The sucrose with the concentration furthest away from the concentration inside the potato will have the highest water potential gradient, and therefore, the fastest rate of osmosis.
Prediction Graph
Risk Assessment
The knife and the corer will be used to cut the pieces of potato. These apparatus are sharp. I will use these tools with care to avoid cutting myself.
Do not put water or sucrose on the computer or electrocution could result
Apparatus
Number Needed
Apparatus
Reason
Large Potato
Where I will obtain my measurements from
240 ml
Distilled Water
Dilutes the sucrose to provide different concentrations
240 ml
2M Sucrose Solution
The solution which surrounds the potato pieces
Micrometer
To measure the surface area of the potato pieces
2
Test tubes
Hold the sucrose/water and the potato cylinders during the experiment
Test tube Rack
Holds the test tubes during incubation
Stopwatch
Measures the time of incubation
Computer
To measure the length of, and cut the potato cylinders
Corer
Cuts the tubes of potato from the large potato
Tile
Stop the potato pieces from having moisture drawn out of them unfairly (from the surface of the table) when waiting to be weighed and when air-drying
Weighing Scales
Weigh the potato pieces before and after incubation. Weighs the water and the sucrose
Beaker
Holds the water/sucrose when being weighed
Diagram
Method
) Set up the apparatus as shown in the diagram without placing the sucrose/water or the potato pieces in the test tubes.
2) Using the weighing scales and the concentration table below make up the different sucrose molarities and pour them into the test tubes
Sucrose (ml)
Water (ml)
Sucrose concentration created (Moles/Litre)
20
20
24
6
.2
28
2
.4
32
8
.6
36
4
.8
40
0
2.0
3) Take the large potato and cut out twelve tubes using the corer.
4) Remove any pieces of potato skin from the tubes
5) Using the computer, make each tube into a 3 cm long cylinder
6) Use the micrometer to make sure that every potato has the same surface area.
7) Record the mass of each potato cylinder. When not weighing put the potato pieces on the tile to prevent moisture loss
8) Place the potato cylinders into the test tubes with the solutions and start the stopwatch
9) After 30 minutes, take all of the potato pieces out of the test tubes and place them on the tile
0) Leave them for the surface moisture to evaporate for 10 minutes
1) Calculate the mass change of each potato and their percentage mass change.
2) Calculate the average percentage mass changes
3) Create a graph of average sucrose concentration against average percentage mass change
Blank Results Table
Sucrose Concentration (Moles/Litre)
Mass Before (g)
Mass After (g)
Mass Change (g)
% Mass Change (2 decimal points)
0
0
0.2
0.2
0.4
0.4
0.6
0.6
0.8
0.8
Sucrose Concentration (Moles/Litre)
Average % Mass Change (2 decimal points)
0
0.2
0.4
0.6
0.8
After drawing a scatter graph for these results I will analyse them and see whether or not this analysis confirms my prediction.
I feel that I have investigated the independent variable fully and in the future, I would like to conduct a similar experiment for one of the controlled variables (type of cells, etc.) mentioned at the beginning of this coursework.
Bibliography
Here is a bibliography stating the sources of information in this coursework:
* Microsoft Encarta
* CGP - GCSE Biology - Complete Revision and Practice
* Various Internet Websites
* A Level Biology by Jones and Jones
* GCSE Biology by Mackeen