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Biology Co/Wo Q: How the Concentration of Sucrose affects Osmosis in plant cells


BIOLOGY COURSEWORK: How the Concentration of Sucrose affects Osmosis in plant cellsAIM: To investigate how changing the concentration of sucrose can affect the osmosis in plant cellsRESEARCH Definition of Osmosis: Osmosis is the movement of water molecules from an area of high water concentration (weak solution) to an area of low water concentration (strong solution) through a partially permeable membrane. Water moves in both ways in order to balance the concentrations evenly. It is known as a net movement of water into there are where there is less water. Water molecules move random with a certain amount of kinetic energy. Distilled water separated by a partially permeable membrane looks like in the diagram below: Water molecules are moving from one side of the membranes to the other but there is NO NET OSMOSIS. If a substance is dissolved in water, the kinetic energy of the water molecules is LOWERED. This is because some water molecules collect on the surface of the other molecules

In osmosis, the potential of the water molecules to move- is known as the OSMOTIC POTENTIAL. Distilled water has the highest potential (zero). When water has another substance dissolved in it, the water molecules have less potential to move. The osmotic potential is NEGATIVE. Water molecules always move from less negative to more negative water potential (meaning from pure water to a more concentrated solution). Now net osmosis can be understood as: LESS NEGATIVE MORE NEGATIVE (look in diagram below)


Year 11 Biology Coursework

On the other hand, the osmotic potential of a cell is known as the water potential. Cell membranes are completely permeable to water therefore, the environment the cell is exposed to can have a dramatic effect on the cell. When there is a high concentration of sucrose and a low concentration of water- this is called a hypotonic solution. This is where water passes into the vacuole by osmosis since the water molecules in the solution is attracted to the other water molecules in the cell, this is known as net endosmosis occurring and the cell becomes turgid. The vacuole pushes against the cell wall, and this in turn tops the cell from bursting. The cell is turgid.

The cell is flaccid and causes plasmolysis. This is in the units of osmotic pressure

For Water potential of cytoplasm= example: - 50 Osmotic potential of solution= - 20

For Water potential of cytoplasm= example: - 50 Osmotic potential of solution= - 80

When a solution is hypertonic, it contains a low concentration of solute (in this case sucrose) and a high amount of water. In this situation, water is caused to diffuse out of the cell- this process is known as exosmosis. This makes a few changes to the shape of the cell, such as it shrivels and wilts. It becomes flaccid, in which the cell would feel soft, limp and weak. If a lot of water leaves the cell, the cytoplasm starts to peel away from the cell wall (but the cell wall still keeps intact). This is where the cell has undergone plasmolysis.


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If the osmotic potential of the solution is the SAME as the water potential of the cell, which in my case, is where the sucrose and distilled water both have the concentration of 0.5M. The solution is called isotonic.

For Example: Water potential of cytoplasm= - 50 Osmotic potential of solution= - 50 (This is the osmotic pressure)

In an isotonic solution, no net osmosis occurs. The cell is not plasmolysed, but its neither full turgid either.

HYPOTHESISFrom my background research, I predict that the more concentrated the solution (the more sucrose it contains), the lower the osmotic potential is. The plant cell (in this case potato cell) would become flaccid, as water has moved out of the cell causing the mass to have decreased. As concentration of sucrose decreases, and the concentration of dilute water increases, the mass of the potato cell will become turgid and strong. It will gain mass as water diffuses into the cell. I predict that graph will look like this because if the concentration of sucrose is low, then the water potential is lower on 0.5 M of sucrose the outside and higher on the inside, and solution and dilute so water will move in the cell, meaning water the cell will gain in mass. This will cause the points on the graph to be positive when it comes to percentage mass change, and after 0.5 M, the concentration of water becomes more less than the sucrose solution meaning the water potential is high on the outside but low in the inside of the cell, therefore water moves out. The cell loses mass, so the percentage mass change becomes negative- so it will be placed in the negative section of the graph-below zero. At concentration 0.5M, the solution will be isotonic; therefore, the % mass change should be at zero as no net osmosis should occur, therefore no water should diffuse in or out of the cell, and so should not affect the mass of the cell at all. I have planned a simple procedure in order to carry out this investigation: Apparatus/Equipments: Sucrose Solution and Water (two beakers to hold them) Weighing machine


Year 11 Biology Coursework

cubed) Borer (to cut the potato cell) Stop Watch and Test tube rack 10 test tubes for each of the ten concentrations Method

2 Measuring Cylinders (small ones- with the range from 0-1cm

1. Put lab coat on and collect all equipments. 2. Firstly, we will experiment with the concentration of 0.1 of sucrose solution. Therefore pour, 0.1cm cubed of using a measuring cylinder of the sucrose, and pour in 0.9cm cubed of water into a beaker. 3. Cut out using the borer, 10 (same size) blocks of potato pieces (and make sure you know which one is for which test tube- for this it would be a good idea to put it into chronological order) 4. Weigh all twenty potato blocks and record their mass 5. Place them all into their supposed test tube with the right concentration into at the same time, and allow half an hour for osmosis to carry out. 6. After half an hour, dry the potato blocks/cells 7. Weigh them and record the results The evidence I will collect is the measurement of the mass loss/gained. To do this I will record the weight of the plant cell at first and then after adding it into the concentration, I will leave it there for a certain amount if time and then record the mass of the plant cell afterwards. I will then subtract the initial mass from that which will give me the difference. This will either be the mass gained or loss (resulting a value either negative or positive). I will also find out the % mass change, in which I will divide mass change by initial mass and then multiply by a hundred. Variables (Dependant): Percentage change in mass Controls (Independent Variables): Shape of sample potato Concentration of surface- range Container- pressure of fluid Temperature Time Method of drying Volume of solution used

I will record my results in a table, and then represent them on a graph for visual evidence; this will allow me to distinguish any patterns of such. The range of concentration I am going to test is from pure water with the concentration of zero molars, to one molar- complete sucrose solution, testing each one: 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9 and finally 1 molar. I will use the measuring cylinders to dilute the sucrose solution according to the amount needed to make that particular concentration. (E.g. I will dilute 0.6M with distil water to make a sucrose solution of 0.4).


Year 11 Biology Coursework

PRELIMINARY WORKTo get an idea of what the experiment is like at first hand I did some preliminary work. This way I can find what apparatus we would be using, be more efficient in the real experiment and finally to help us make a decision such as how long to leave the cell potatoes in the solution and the size of potato cell. We can be more prepared in the real experiment we would be more prepared, more in control, and to see if we wanted to change anything, such as perhaps use a different range of concentration and maybe better equipment that make the results more accurate. Even before the preliminary, I prepared a table to record the results and of what I will be measuring:Concentration (M) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Repeat 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 Mass Before (g) Mass After (g) Mass Change (g) % Change in Mass Average %

During the preliminary, my partner collected the ten test tubes (as we were testing ten different concentrations), along with a test tube rack. One of us filled each test tube with the correct amount of distilled water, while the other one did the amount of sucrose. Once that was sorted, grabbing one potato, we first made the mistake of cutting the potato in half and slicing it into cubes. We did not realise later on that we were supposed to use a special tool (NAME) that would cut out cylinder tube-like pieces of potato. So using another potato, we skinned the skin (since the skin is like a barrier and so getting rid of it would allow the process of osmosis to occur more efficiently) and placing them in chronologic order on a tile, we weighed each one.


Year 11 Biology Coursework

Recording the measurement, we placed them in the test tubes which were also placed in the same chronological order (so we dont get mixed up) as fast as we could as we wanted all the potato blocks/cells to have the same amount of time to be in the solution. Once we did that, we left it for half an hour.

Concentratio n 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Weight Before 1.30 1.20 1.25 1.28 1.22 1.24 1.11 1.18 1.20 1.28

Weight After1.29 1.18 1.17 1.18 1.10 1.09 0.98 1.03 0.95 1.08

Mass Change-0.01 -0.02 -0.08 -0.1 -0.12 -0.15 -0.13 -0.15 -0.25 -0.2

% Change-0.77 -1.67 -6.40 -7.