Executive Summary for Cookie Business Essays

682 Words Nov 5th, 2012 3 Pages
No.06 Determining water potential of potato tuber cells: the weighing method

If a plant cell is in equilibrium with an external solution of such a concentration that there is not net loss or gain of water, the water potential of the external solution will be equal to the water potential of the cell.

Use of this fact can be made in estimating the water potential of a plant tissue. Samples of tissue are immersed in a range of external solutions of different strengths. The solution that induces neither an increase nor a decrease in the volume or mass of the tissue has the same water potential as that of the cells in the tissue.

The cells to be investigated in this experiment are those of the potato tuber. Changes in mass will be
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The latter, being the independent variable, should be on the horizontal axis; the former on the vertical axis. 7. Calculate the water potential of the potato cells as follows. Find the point on your graph corresponding to a percentage mass change of zero. The molarity of sucrose corresponding to this zero mass change can now be read from the horizontal axis. From table 4.1 find the water potential of a sucrose solution of that molarity. That is the water potential of your sample of potato cells. Express your result in a kPa.

For consideration

1. Criticise this method of finding the water potential of plant cells. How might it be improved? 2. What was the reason for dividing each cylinder into six discs, and why was it necessary to standardise the procedure for drying the discs? 3. With what kind of plant tissue might it be possible to use a change in volume rather than mass for estimating the water potential? 4. How does the value of the water potential differ from the osmotic potential of the solution in the vacuole? 5. In constructing your graph did you join up the points with straight lines or just a smooth curve? Justify whichever technique you used.

|Molarity (mol dm) |Osmotic potential kPa |

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