Plan and carry out an experiment to determine the concentration of potato cell cytoplasm

Melissa Parvis 4W

Brief

Definition of Osmosis:

“Diffusion of fluid through a semi permeable membrane from a solution with a low solute concentration to a solution with a higher solute concentration until there is an equal concentration of fluid on both sides of the membrane.”

Diagram:

Plant cells have a strong cell wall surrounding them, when they take up water by osmosis they begin to swell up but the cell wall prevents them from bursting. Plant cells become what is called “Turgid” when they are put in dilute solutions, the characteristics of a turgid plant cell are it being swollen and hard, the pressure inside the cell then increases and eventually the pressure because too much and no more water can be absorbed by the cell. This is what makes the stems and leaves of the plant stand up and face the sunlight.

When plant cells are placed in concentrated sugar solutions they lose water because of osmosis, basically the exact opposite happens when the plant cell is turgid and the cell becomes “Flaccid” and the cells shrinks and loses shape.

When a plant cell becomes plasmolsysed this is usually because it has not got enough water. The cell contents peel away from the cell wall and under a microscope (seen below) the cell walls become visible.

<Plasmolysed Cells

< Flaccid Cells

<Turgid Cells

Definitions

Hypertonic: Having the higher osmotic pressure of two solutions.

Hypotonic: Having the lower osmotic pressure of two fluids.

Isotonic point: Situation where a cell can neither swell nor shrink, or, the equal tension point

Variables

A Dependant Variable- The variable which is measured, in this case the mass of the initial and final mass of the potato chunks.

An Independent Variable- This is the variable that is changed. In this case this is the concentration of the sucrose solution.

Controlled Variables

These are variables that will constantly stay the same. These are

The volume of the Solution

The length of the potato chip

Temperature – This is kept the same because all the specimens will be kept in the same environment.

Same species of potato – This is Maris Piper

All the specimens will be kept in their solutions for the same amount of time, this will be two days.

Fair testing

Temperature- All the specimens will be kept at the same temperature. If we left some in a hot environment and some in a cold environment this would not be a fair test. This is because the warmer it is the faster the particles move and this will increase the rate of osmosis. So if I left some in a cold environment and some in a warm environment we would find the specimens left in the warm environment would have a faster osmosis rate because there would be more energy resulting in faster moving water molecules.

Potato species- This is a very simple variable to fix because all we have to do is keep to the same type of potato and in this case we are using “Maris Piper” potatoes. If we were to use all different types of potatoes then we would not get a fair result. This is because different species of potato have different water contents which would mean in the potatoes with the higher water content; (if we were to use different species) they would have a slower rate of osmosis because there would be a smaller difference in the concentrations.

Time left for- The longer the time we leave the potatoes the longer the time we have for osmosis to occur. If we were to leave some specimens for a much longer time than the others, which would mean the potatoes left in their solution longest would have a greater rate of osmosis. To prevent this from happening we are going to take the potatoes out at the same time two days after putting them in their solutions.

Volume of solution- This is fixed by using a measuring cylinder to measure the sucrose solution exactly. If we were to put random amounts of the sucrose solution in with each specimen then we would get completely unreliable results because the specimens with a lot of sucrose solution would have a higher rate of osmosis because there is a larger difference in concentrations and so if there only a little sucrose solution in one specimen then there will be a low rate of osmosis.

Preliminary work

Safety

Safety glasses are not a vital part of safety, since there aren’t any harmful chemicals I will be using in this particular experiment. Although I am using scalpels and glass test tubes so I need to be careful and concentrate when I am using these in case I cut myself. I am also using electricity for the balances so I will need to be careful when I switch them on and off and use the plug socket.

Apparatus list

One potato

One set of cork borers

Three test tubes

A cutting board

A scalpel

Distilled water

Sucrose solution

A balance to measure mass

Method

1. We took three specimens in tubes

2. In the first tube we added 10ml of distilled water (0M)

3. In the second tube we added 10ml of sucrose solution (0.5M)

4. In the third tube we added 10ml of sucrose solution (1M)

5. We took a cork borer and pushed it through a potato three times to make three potato chunks.

6. We cut the ends off ...

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Apparatus list

One potato

One set of cork borers

Three test tubes

A cutting board

A scalpel

Distilled water

Sucrose solution

A balance to measure mass

Method

1. We took three specimens in tubes

2. In the first tube we added 10ml of distilled water (0M)

3. In the second tube we added 10ml of sucrose solution (0.5M)

4. In the third tube we added 10ml of sucrose solution (1M)

5. We took a cork borer and pushed it through a potato three times to make three potato chunks.

6. We cut the ends off and made them all the same size using a ruler and measuring 2cm each time.

7. We then weighed each individual potato chunk and recorded the results

8. We then put 1 potato chunk in each specimen tube

9. We then left them for 2 days.

10. We then took each potato chunk out and dabbed then on paper to remove excess liquid.

11. We then weighed each potato chunk and worked out the percentage difference in mass.

Results

Conclusion

From my results I conclude as the concentration of sucrose solution increases the mass decreases.

Prediction

I predict when the concentration changes, the mass will also change. When the solute outside the potato is higher than the concentration inside the potato, the overall mass of the potato chunk will decrease. When the solute outside the potato is lower than the concentration inside the potato chunk, the mass of the potato will increase. However, if the concentration of the water is the same inside and out of the potato chunk then the mass will stay the same.

I predict the potato chunk will lose the most water when there is a higher concentration of sucrose solution and I also predict the potato will gain more water when there is a lower concentration of sucrose solution. The diagram below helps to explain this:

When the two solutions are divided by a semi permeable membrane the water moves from the area of high concentration to an area of low concentration until both sides are equal.

I predict in above 0.2M of solution there will be a higher concentration than the water it is in so the mass will decrease because water moves out of the potato. I also predict that in 0.2M of solution there will be a lower concentration in the potato than in the water so the mass will increase. I also predict the isotonic point will be around 0.5M as it shows in my graph below which I have made from my preliminary results.

We can see in my sketch graph which I made below that above the x axis of the graph the cell is turgid and the cell is most turgid at 0M. This is because it acts as a hypotonic solution because since the sucrose solution has a higher concentration, this means it has lower water potential as there are a larger number of solute molecules and so there is less number of freely moving water molecules, as they form clusters around the solute. On the other hand, the water in the potato cell has a higher water potential because it has a weaker concentration, which means there are less solute molecules and thus, more freely moving water molecules. As a result, the water in the potato cell acts as the hypotonic solution.

We can also see in my sketch graph below that below the x axis of the graph the call is Flaccid and the cell is most flaccid and 1M. This is because it acts as a hypertonic solution this is because when the potato chip is immersed in a sucrose solution of higher concentration, in this case 1M of sucrose solution, there is a greater difference in the concentration between the water in the potato cell and its external solution so that is a greater concentration gradient. Since the water in the potato cell is of the same concentration and its external solution it becomes more concentrated, as the external solution has a greater number of solute molecules and so less water potential. In other words, more water molecules are required to move out of the potato cells to the external solution. Since more water leaves the cells, the potato chips decrease more in mass.

Method

Apparatus

One potato

One set of cork borers

Twelve test tubes

A cutting board

A scalpel

Distilled water

Sucrose solution

A balance to measure mass

Modifications: I will be using twelve specimens this time instead of just three, and I will also have three specimens for each molar we test. I will also use a sieve and tweezers to help me pick the potatoes out which will make it easier.

Method

1. We will take twelve specimens in tubes.

2. In the first tube we will add 10ml of sucrose solution. (0.2M)

3. In the second tube we will add 10ml of sucrose solution. (0.2M)

4. In the third tube we will add 10ml of sucrose solution. (0.2M)

5. In the third tube we will add 10ml of sucrose solution. (0.4M)

6. In the third tube we added 10ml of sucrose solution. (0.4M)

7. In the third tube we will add 10ml of sucrose solution. (0.4M)

8. In the third tube we will add 10ml of sucrose solution. (0.6M)

9. In the third tube we will add 10ml of sucrose solution. (0.6M)

10. In the third tube we will add 10ml of sucrose solution. (0.6M)

11. In the third tube we will add 10ml of sucrose solution. (0.8M)

12. In the third tube we will add 10ml of sucrose solution. (0.8M)

13. In the third tube we will add 10ml of sucrose solution. (0.8M)

14. We will take a cork borer and push it through a potato twelve times to make twelve potato chunks.

15. We will cut the ends off all the potato cylinders and made them all the same size using a ruler and measuring 2cm each time.

16. We will then weigh each individual potato chunk and record the mass.

17. We will then put 1 potato chunk in each specimen tube.

18. We will then leave them for 2 days.

19. We will then take each potato chunk out with tweezers and a sieve and dab then on paper to remove excess liquid.

20. We will then weigh each potato chunk again and record the results.

Using a cork borer

Results

Before

After

Conclusion

I conclude that as the concentration of sucrose solution increases the mass of the potato decreases, I also conclude the isotonic point is around 3M.

My graph shows above my x axis the potato is turgid, this is because there is a high concentration of water where as the potato has a low concentration of water, osmosis is the movement of water from a high concentration to a low concentration so the potato sucks up a lot of the water. This relies on the fact of osmosis; if a cell is placed in a less concentrated solution, then in order for equalization to occur, that is for the concentration to be equal on both sides, then water must move into the cell so that its water becomes diluted and so that the concentration gradient is eliminated. In this case, since the potato cells inside the chip obviously have cell sap, a mixture of salts and sugars with water, it forms a weak solution. Even though it is weak, it is still more concentrated than the distilled water. Thus, in order for equalization to occur, some of the distilled water must move across the cell membrane of the potato cell and enter it. In this case, the cells will become bloated or turgid; they become firm and full. Due to this, the potato chip will increase in mass as the potato cells have become firm and turgid; they occupy more space and so the chip increases in mass.

My graph shows on the x axis there was the same amount of water in the potato and the solution it was in. in other words it reached its isotonic point. This means osmosis did not occur. My graph also shows below the x axis the potato became flaccid, this is because there is a higher concentration of water in the potato than in the tube. So the water moved through the potatoes semi permeable membrane to the container.

I predicted all of the above in my prediction. In my prediction I predicted as the concentration changes, the mass will also change. I also predicted when there is a higher concentration outside the potato then the concentration inside the potato will increase. So, this means if there is a lower concentration outside the potato then the concentration inside the potato will decrease. My sketch graph in my preliminary had a different result to my official results, but that is probably because I had no repeats. Although saying that, the principal was the same, the mass increased when the concentration of sucrose solution was 0 M. As the concentration o sucrose solution increased, the mass of the potato gradually decreased.

Evaluation

From my experiment I found out the concentration of a potato cell cytoplasm. I used a good range of sucrose solutions (0M, 0.2M, 0.4M, 0.6M, 0.8M) and each with three repeats, this made It a fair test which gave accurate results and my overall results and method were very good. Overall my results were of a good quality because I have no anomalies. Although I have good results, I did make some errors.

Errors

Length of potato chip – My ruler that I used may not have been accurate or I may have not accurately measured it because it was very difficult to get the measurement to the exact millimetre because the chip was a funny shape and it was awkward.

Volume of Solution – This may have not been exactly the right measurement because it was difficult to measure out the solution and I tried to do it quickly towards the end because I was running out of time. This may have provoked an error. I also used different volume cylinders which may have given me faulty results.

Time in Solution – Not all potato chunks has the same amount of time in its solution, this is because as we did the first measurement we put the first potato chip in with it and then went on to the next so this means one potato chip may have had up to 15 more minutes in the solution than the last potato.

Top pan balance – When I first weighed the potatoes I used a different top pan balance than when I took the potatoes out and re-weighed them. This left room for error because each top pan balance is slightly different so it might have given me 0.01g difference in results.

Improvements

Length of potato – I could use a special potato cutter, or alternatively I could use callipers to help me get an accurate result.

Volume of solution – I could use the same cylinder volume or I could get down to the level of the cylinder and make sure its at the same meniscus.

Time of solution – I could get a machine to make sure each potato is submerged in its solution for the same amount of time. Or I could get twelve different people to put each potato in its solution as the same specific time.

Top pan balance – I could make absolute sure that I use the same top pan balance throughout.

Analysing Evidence

I could say that I do not have enough evidence to support my conclusion fully although I do not have any anomalies. This is because when I look at my results, I could do with a wider range of results. Looking at my graph, it looks reasonably smooth; there are a few points where the line goes off course but not so much as to call them anomalies. To give myself sufficient evidence to support my conclusion I could have a larger range of solutions, for instance (0M, 0.2M, 0.4M, 0.6M, 0.8M, 1M, 1.2M, 1.4M, 1.6M, 1.8M, 2.0M.) and have at least 5 repeats for each measurement. So, because I have no anomalies I have suitable results and data at the moment but I cannot be 100% sure because I need more repeats and a larger range of solutions.

Further Work

If I wish to do the experiment again, to improve the method and results and to give more evidence to support my conclusion and my theory for osmosis, I could try the same experiment with a range of different types of potatoes to if I get similar results, types of potatoes such as:

-Kind Edwards

-Russet Potato

-New Potato

-White round potato

-Yukon gold potato

I could try using different fruits and vegetables, to see if the osmotic theory is similar with them, I could use Fruit and vegetables such as:

-Carrots

-Cucumber

-Bananas

-Apples

I could also try different types of solutions to see if the osmotic theory only works with the sucrose solution and to see if osmosis works better with these solutions.

-Salt

-Orange Juice

-Lemonade

-Coca Cola

When doing this, I would incorporate more repeats, at least five for each reading. I will use a larger range of solutions, being 0M, 0.2M, 0.4M, 0.6M, 0.8M, 1M, 1.2M, 1.4M, 1.6M, 1.8M, 2.0M.

1. I will take 220, 2cm specimens of carrot and put them in 55 individual test tubes without solution for the moment.

2. I will take 220, 2cm specimens of Banana and put them in 55 individual test tubes without solution for the moment.

3. I will take 220, 2cm specimens of cucumber and put them in 55 individual test tubes without solution for the moment.

4. I will take 220, 2cm specimens of apple and put them in 55 individual test tubes without solution for the moment.

5. I will divide the 220 different specimens of carrot into 4, so I will get 55 in each group. I will then put 5 specimens in for each reading (readings are: 0M, 0.2M, 0.4M, 0.6M, 0.8M, 1M, 1.2M, 1.4M, 1.6M, 1.8M, 2.0M.)

6. I will measure out 10ml 5 times of coca cola solution for each reading.

7. Using the second lot of 55 specimens, I will then put 5 specimens in for the next lot (readings are: 0M, 0.2M, 0.4M, 0.6M, 0.8M, 1M, 1.2M, 1.4M, 1.6M, 1.8M, 2.0M.)

8. 6. I will measure out 10ml 5 times of lemonade solution for each reading.

9. Using the third lot of 55 specimens, I will then put 5 specimens in for the next lot (readings are: 0M, 0.2M, 0.4M, 0.6M, 0.8M, 1M, 1.2M, 1.4M, 1.6M, 1.8M, 2.0M.)

10. 6. I will measure out 10ml 5 times of orange juice solution for each reading.

5. I will divide the 220 different specimens of Banana into 4, so I will get 55 in each group. I will then put 5 specimens in for each reading (readings are: 0M, 0.2M, 0.4M, 0.6M, 0.8M, 1M, 1.2M, 1.4M, 1.6M, 1.8M, 2.0M.)

6. I will measure out 10ml 5 times of coca cola solution for each reading.

7. Using the second lot of 55 specimens, I will then put 5 specimens in for the next lot (readings are: 0M, 0.2M, 0.4M, 0.6M, 0.8M, 1M, 1.2M, 1.4M, 1.6M, 1.8M, 2.0M.)

8. 6. I will measure out 10ml 5 times of lemonade solution for each reading.

9. Using the third lot of 55 specimens, I will then put 5 specimens in for the next lot (readings are: 0M, 0.2M, 0.4M, 0.6M, 0.8M, 1M, 1.2M, 1.4M, 1.6M, 1.8M, 2.0M.)

10. 6. I will measure out 10ml 5 times of orange juice solution for each reading.

11. I will divide the 220 different specimens of cucumber into 4, so I will get 55 in each group. I will then put 5 specimens in for each reading (readings are: 0M, 0.2M, 0.4M, 0.6M, 0.8M, 1M, 1.2M, 1.4M, 1.6M, 1.8M, 2.0M.)

12. I will measure out 10ml 5 times of coca cola solution for each reading.

13. Using the second lot of 55 specimens, I will then put 5 specimens in for the next lot (readings are: 0M, 0.2M, 0.4M, 0.6M, 0.8M, 1M, 1.2M, 1.4M, 1.6M, 1.8M, 2.0M.)

14. 6. I will measure out 10ml 5 times of lemonade solution for each reading.

15. Using the third lot of 55 specimens, I will then put 5 specimens in for the next lot (readings are: 0M, 0.2M, 0.4M, 0.6M, 0.8M, 1M, 1.2M, 1.4M, 1.6M, 1.8M, 2.0M.)

16. 6. I will measure out 10ml 5 times of orange juice solution for each reading.

17. I will divide the 220 different specimens of apple into 4, so I will get 55 in each group. I will then put 5 specimens in for each reading (readings are: 0M, 0.2M, 0.4M, 0.6M, 0.8M, 1M, 1.2M, 1.4M, 1.6M, 1.8M, 2.0M.)

18. I will measure out 10ml 5 times of coca cola solution for each reading.

19. Using the second lot of 55 specimens, I will then put 5 specimens in for the next lot (readings are: 0M, 0.2M, 0.4M, 0.6M, 0.8M, 1M, 1.2M, 1.4M, 1.6M, 1.8M, 2.0M.)

20. 6. I will measure out 10ml 5 times of lemonade solution for each reading.

21. Using the third lot of 55 specimens, I will then put 5 specimens in for the next lot (readings are: 0M, 0.2M, 0.4M, 0.6M, 0.8M, 1M, 1.2M, 1.4M, 1.6M, 1.8M, 2.0M.)

22. I will measure out 10ml 5 times of orange juice solution for each reading.

23. Before putting the chips into the solution I will weigh them all.

24. After all the following, I will get 880 people to put the right chip in its solution at the same time.

25. I will leave the chips for two days

26. I will then get the 880 people to take the chips out at the same time and then re-weigh them on the same top pan balance and record the results.

Bibliography