Osmosis is the process by which different water concentrations balance out between cells. The water travels through semi permeable membranes until the concentration is equal on both sides of the membrane.

Biology Coursework

Osmosis

The Oxford Dictionary defines Osmosis as “the passage of a solvent through a semi-permeable partition into another solution.” In Biology, Osmosis is the process by which different water concentrations balance out between cells. The water travels through semi permeable membranes until the concentration is equal on both sides of the membrane. The large particles in the highly concentrated liquid do not pass through the semi permeable membrane, but the water molecules do. Plant cells are surrounded by a cell wall made from cellulose fibres. This wall is fully permeable to allow dissolved substances to pass through it easily. There is a thin layer under the cell wall called the cell membrane, and this is only semi-permeable so only certain substances can leave and enter the cell.

Any substance dissolved in water is called a solute; a solvent is a liquid that is able to dissolve another substance, called a solute, to form a solution. The water content of plant cells varies depending on environmental conditions. The water plays a vital role in the support of tissues and the transport of materials around the plant. If a plant does not have water, it may wilt, and will eventually die. Water is mainly absorbed through the roots, which are covered in specially adapted root hair cells, to create a larger surface area for absorption. It is pulled up the plant through the xylem by a pull resulting from the evaporation of water through the stomata on the leaves. This evaporation is called transpiration. The evaporation of water from the leaves also removes heat energy from the plant and helps to prevent overheating. Transpiration pulls water up the plant stem, but osmosis is the process whereby water is drawn into or out of cells and tissues. Osmosis is the flow of water by diffusion through a semi permeable membrane from areas of high water concentration to regions of low water concentration. The diagram below illustrates this:

Plant cell membranes are semi permeable, which means that they will allow substances with smaller molecules to go through them, but not others. Water can freely penetrate all membranes. The cell wall will allow anything dissolved in water to pass through it.

If water enters the cell by osmosis or not will depend on the balance between the solute concentrations on the inside and the outside of the cell, and the state of the cell. If the concentration is the same on both sides of the membrane, then there will be no movement of water particles. This is called an equilibrium state and it is said that the liquids on either side are isotonic. A liquid that has more particles in it than another is referred to as being hypertonic. The less concentrated solution is hypertonic. The concentration is measured by ...

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A molar solution contains a fixed number of solute particles in a litre of water. To find this number, you have to find out what a mole of those particular particles is. Finding the relative atomic mass in grams, then dissolving it in a litre of water will create a molar solution. A 0.1 molar solution effectively contains a tenth of the particles in a molar solution, so 0.1 is 10% molar, 0.5 is half molar, and 0.75 is three quarters molar.

A cell can only take in water until it pushes against the cell wall however, as this is rigid and will not expand when pushed against. A cell that is full of water is called turgid and cannot expand further as the inward force of the stretched wall balances the outward pressure on the cell wall. This wall pressure is called turgor pressure and the internal outward force on the wall is called osmotic pressure.

As it would be impossible to measure with any degree of accuracy the expansion or contraction of cells on an individual basis we decided to look at the mass of the potato chip. A chip placed in an isotonic solution should show no change whereas one placed in a hypertonic solution will lose mass.

Key factors

There are many key factors that will affect this experiment, and we are only going to be changing one of the, so the others need to remain constant.

Temperature: All of the solutions need to remain the same temperature. Due to kinetic theory reactions speed up in warmer environments, and this could affect our experiment, so if all of the solutions are kept at the same temperature then there should be no problem with temperature affecting the speed of the osmosis in each individual container. As the water we use will be from the same tap at roughly the same time, and all the beakers will be in the same place I think there will be no problem with temperature.

Amount of sucrose used: This will have a very wide affect on the experiment. A strip of potato in sucrose solution will have a very different rate of osmosis than that of one in a dilute water solution. The amount of sucrose used will determine the rate of osmosis. As this is what we are going to be altering we do not need a method to keep this constant.

Surface area of the potato: If one strip has a large surface area then more cells can be involved in the reaction quicker than a strip of potato with a smaller surface area. If the surface area is greater then the rate of osmosis will increase. We will use a potato borer to keep the surface area of the potato chip constant.

Size of the container’s opening: The rate of evaporation of the water will be dependant on the surface area of the container. If it has a large surface area then the rate of evaporation will be quicker than that of a container with a small surface area. Because of this we have decided to use 250ml beakers for all of the solutions.

Plan

To generate a wide variety of results, I will put the potato chippings into 5 different concentrations of sugar, 0.00 M 0.25 M 0.50 M 0.75 M, and 1.00 M sucrose solution. Using a potato chipper, chips were produced that were 7x7x50mm. These chips are a good size, as they are rigid enough not to break, whilst being thin enough for osmosis to take place in all the cells by the end of the experiment. If the chips were the same mass, but cuboids, then the cells in the centre may not have had the solution reach them in the twenty minutes we allowed for the experiment. If there are no anomalies, I think my results should be very accurate, as a potato chipper creates chips that are far more even that those that could be made by the human hand. I will use the same potatoes, as this will eliminate as far as possible any variation resulting from different treatment or source of supply. The potato chips will be immersed in each solution for 20 minutes to ensure that all cells, including those in the central core, have had time to react to the solute. I will measure the change in mass, to find out where the isotonic concentration lies.

I am going to place the potato chips into 250ml beakers, and have 100ml of solution in each one. This is enough to fully submerge the potato chips, to make sure that the whole of each one is completely covered at all times. I have decided against using boiling tubes, for the simple reason that the potato will be harder to get out of the tube than a beaker.

We measured out the sucrose solutions and poured them into the beakers. We worked out how much one mole of sucrose is with the formula for sucrose C12H22O11. We then found out the relative atomic mass for sucrose, which is equal to 342ml. Using this information, we then created our concentrations at 0.00 M 0.25 M 0.50 M 0.75 M, and 1.00 M accordingly. We then weighed each of the potato strips and placed them in their solutions. After twenty minutes had passed we removed them, and weighed them again. We converted the figures into percentages to show the mass lost.

Prediction

I predict that I will find the isotonic concentration to be between 0.5 and 0.75 moles. I also predict that the potato chips placed in the saline solution will change in mass as shown in this graph:

Results

The tables below show the results for potato tubes placed in varying molarities of sucrose solution for 20 minutes:

Conclusion

The results show that, in accordance with my prediction, the cylinders will lose weight when solute concentration is low (high water concentration), and gain weight in strong solutions (low water concentration). This is due to osmosis, where water passes from weak solutions to strong solutions across the cell membrane. The graph of average % change crosses the x-axis at about 0.65M. This concentration is where the plant cells will be isotonic, where the concentrations inside and outside the cell are in balance with. Because of the low number of solutions sampled however, it would be wrong to use the graph to find out what the weight percentage change would be for other concentrations of sucrose, yet I have had to do this to approximate the isotonic solution.

Evaluation

Although the results of the sucrose experiments support my prediction, there is one major anomalous result, and a few smaller ones. These could be improved by altering the experiment, for example by keeping the test-tubes in a water bath at a set temperature, or by measuring the sizes of potato cylinders before and after with a more accurate method, e.g. accurate weight measurement. The test might also be more accurate if the potato cylinders were left in the solutions for a longer period of time to allow the solution to reach the core of the potato cylinders. I could have also repeated the experiment many times and found the average results, as an average would have displayed much more accurate results. Another factor is the actual specimen from which our potato cylinders were cut. We used two different potatoes, and they may have been from different plants, thus creating anomalies.
If we were to repeat the experiment, we could extend the experiment to use samples from other plants, to discover whether the same would also happen in other plants, or we could use many more concentrations so we could exact our predictions for other molarities of sucrose solution, thus finding the isotonic solution a lot easier.