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# Experiment to find the effect of sucrose solution concentration on potato and apple tissue.

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Introduction

Experiment to find the effect of sucrose solution concentration on potato and apple tissue. Introduction: I am going to investigate the effect of sucrose solution on the mass of apple and potato tissue and use the information to find the mean water potential of potato and apple cells. Prediction: I predict that the potato cells will have higher water potential than apples because they contain insoluble starch with will not lower the water potential. Apples contain soluble sugar, which will lower the water potential. Background: Osmosis is the movement of water through a partially permeable membrane, from a region of high water potential to a region of lower water potential. A cell is surrounded by a plasma membrane that is partially permeable, so water can enter and leave by osmosis. When a cell is put in a solution with a higher X than that of the cells, there is a net intake of water in to the cells by osmosis. As water enters the cells their X rises until eventually the X of the cells = X of the solution and there is no further net intake of water. This makes the cells turgid. When a cell is placed in solutions with a lower X than that of the cells, there is a net loss of water from the cells by osmosis. ...read more.

Middle

1.74 1.88 +0.14 +0.14 +8.09 0.4 1 1.74 1.74 0 2 1.77 1.75 -0.02 -0.01 -0.57 0.6 1 1.77 1.62 -0.15 2 1.74 1.58 -0.16 -0.16 -8.83 0.8 1 1.75 1.42 -0.33 2 1.74 1.39 -0.35 -0.34 -19.48 1 1 1.75 1.27 -0.48 2 1.76 1.27 -0.49 -0.48 -27.64 Table to show mean % changes in mass of apple pieces in different concentrations of sucrose solution. Concentration (M) Replicate Mass at 0hrs (g) Mass after 3hrs (g) Change in mass after 3hrs (g) Mean Change (g) Mean % Change 0 1 1.12 1.08 -0.04 2 1.21 1.14 +0.07 -0.06 -3.00 0.2 1 1.27 1.31 +0.04 2 1.18 1.25 +0.07 +0.06 +4.00 0.4 1 1.22 1.26 +0.04 2 1.27 1.27 0 +0.02 +1.00 0.6 1 1.17 1.15 -0.02 2 1.16 1.14 -0.02 -0.02 -1.70 0.8 1 1.11 1.05 -0.06 2 1.11 1.04 -0.07 -0.07 -5.80 1 1 1.28 1.08 -0.20 2 1.17 1.01 -0.16 -0.18 -14.70 Mean Values Potato; 18.02 + 8.09 + (-0.57) + (-8.83) + (-19.48) + (-27.64) 6 = -5.03 Apple; 4.00 + 1.00 + (-1.70) + (-5.8) 4 = -0.625 Conclusion: Graph 1 shows us that between 0 and 0.2M solutions the potato pieces gained mass. This is because water is entering the cells by osmosis. ...read more.

Conclusion

This would also give untrue final weights. Also in the boiling tubes, in the higher concentration the potato pieces will float, meaning that there is less surface area in contact with solution. This will affect the time taken to reach equilibrium, as there is less surface area for water to go in/out of the cells. Averaging the results and turning them into percentages also reduces accuracy as if one of the results is wrong it gets combined with the right one to give an overall inaccurate answer. There are no obviously anomalous results for the potato. The apple, however, had two anomalous results (labelled on the graph) at 0M and at 1M. The 1M result could be explained by; because it had the biggest change in mass it did not have enough time to complete osmosis thus giving an inaccurate result. If repeating the experiment, the potato pieces should be left for longer in order for the 1M solution to complete osmosis. The 0M could be explained by the same reasons, but I still would have expected a positive result. The reason for this could also be that some of the outer cells were damaged when they were being cut, making them weaker and more likely to burst, even with their cell wall. If the outer cells burst it would mean they couldn't take up as much water as they should thus reducing the final mass. ...read more.

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## Here's what a teacher thought of this essay

4 star(s)

Overall, this report describing an experiment to find out the water potential of potato and apple tissue is a very competent piece of scientific writing. It describes clearly the strategy used to calculate the water potentials, and presents some valid results which are processed correctly to yield values for both tissues. The writer uses appropriate terminology throughout and structures the whole report so that it is easy to follow.

To gain the highest grades at A' Level, a few improvements could be made:

[1] A clearer explanation of the methodology which links % change in mass of the plant tissue with sucrose sol. water potential values (linking graph 1 with graph 2);

[2] A more detailed discussion of control variables to reassure the reader that the data collected was valid, i.e. all that was changed was the conc. of sucrose. (IV)

Nonetheless, a good effort.

4 stars

Marked by teacher Ross Robertson 10/05/2013

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