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# 2007 Biology C.W AS

Extracts from this document...

Introduction

Plan Aim: To investigate the differences in water potential (?) between swede and potatoes and to find evidence to explain the differences. Water potential is the capacity of a system to lose water, so in pure water the water potential would be zero. The water potential is lowered by the presence of solute molecules in the vacuole of a plant cell, and so the greater the concentration of solutes, the lower the water potential. The water potential can be calculated by adding the solute potential to the pressure potential, as more pressure would cause an increase in the number of solutes diffusing across a semi-permeable membrane into the solution; in other words, and high pressure potential would cause an increase in the rate of osmosis. There are less solutes in a weak solution, which would mean that the swede chips would have a higher water potential in a weaker solution and a low water potential as the concentration of the solution increases. Prediction: From this, I can predict that as the concentration of the solution increases, the water potential will decrease, so the chips of swede will lose mass in a stronger concentration of solution and gain mass when they have a higher water potential. A plant cell surrounded by pure water would have a lower water potential than the water around it because the vacuole contains solutes, giving it a higher solute potential. So, if solutes are present in the vacuoles of the swede cells and these are of a higher amount than that of solutes in the surrounding solution, the water potential will be more negative. So, water enters by osmosis which exerts a pressure on the cell wall, causing it to become turgid. This would mean that the cell in question would have a high pressure potential. I will use my preliminary work to help plan my experiment because by looking at the different rates of osmosis in potatoes and the water potential, I can make a prediction of how the swede results will turn out according to the scientific knowledge acquired. ...read more.

Middle

Also, the fact that the chips were in the solutions for approximately 48 hours could have led to parts disintegrating and small areas of swede breaking off. The conclusion that I have drawn from analysing my results is that the water potential of swede becomes more negative as the concentration of solution that it is suspended in increases. This can be explained by looking at the definition of water potential, as the definition states that water potential is the capacity of a system to lose water, so in pure water the water potential would be zero. However, at a concentration of 1M, the swede had a low water potential as there are were solutes in the strong solution, and the water potential is calculated by adding solute potential to pressure potential. From this, I have proven that my prediction was correct, as I predicted that the chips would lose mass and have a more negative water potential in a stronger solution. The reason that the chips lose mass in a higher concentration of solution can be explained by the definition of osmosis. Osmosis is the diffusion of molecules from a high water potential to a low water potential across a semi-permeable membrane, and because the water potential was more negative in a solution which had a larger amount of molecules in, the rate of osmosis increased, as the water molecules inside the chip diffused out across the semi-permeable membrane into the solution of sucrose which had a higher number of molecules. The chips in the stronger solution would also have a lower solute potential than the chips in the weaker solutions, as a chip surrounded by pure water would have more solutes in it's vacuole than the water outside it would contain, so the osmosis would take place from the higher concentration (the water) to the lower concentration (the chip), causing the chip to become more turgid. ...read more.

Conclusion

Weigh the pin beforehand and weigh the chip when it's still on the pin, subtracting the weight of the pin afterwards. Would improve accuracy of the results as no chip cells would be damaged by removing the pin. 5. The weights of the chips in the same concentration of solution varying. The amount of solution in the tube not being exactly the same for every tube. Use a more precise measuring device and make sure the tubes are completely dry beforehand. Would improve accuracy as the amount of water entering and leaving the chip would be more or less the same. The main sources of error which affect the accuracy of my experiment are the way that the chips were suspended in the solution by the pin and the method that the surface water of the chips was dried off. The effect of these sources or error cause my confidence intervals to fluctuate and make my entire experiment less reliable, as in the end it all alters how my results link back to my original prediction. The amount of solution in each tube varies slightly with every chip as I was only able to measure to the nearest millilitre, but this would not make a great difference to my results as the fluctuations would be so small it would hardly notice. However, even a small difference will affect my confidence intervals. The same applies to only being able to measure length to the nearest millimetre. Reliability and Accuracy I do not believe that my experiment was particularly reliable, considering the variation in results that I discovered. If I repeated my experiment, I would most likely find that my results were quite a lot different to before, regardless of whether the confidence intervals were improved of not. As there are so many extraneous variables involved in the experiment which were very difficult to control, the experiment would have needed to be repeated many more times to able to obtain accurate confidence intervals. ?? ?? ?? ?? Rachel Nash 12.6 ...read more.

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