This investigation aims to determine what effect an increase in the surrounding temperature has on the plasma membrane of a typical plant cell structure.

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                                                                                                                                                                Manpreet Virdee

Aim

This investigation aims to determine what effect an increase in the surrounding temperature has on the plasma membrane of a typical plant cell structure.

Hypothesis

An increase in temperature will damage and denature the plasma membrane, which would cause the substances contained within the cytoplasm to leak out of the membrane.

 The investigation, which was carried out, was to see the “effect of temperature on membrane permeability”. Different temperatures were used ranging from room temperature to 87 C. Three test tubes were used to give a range of results. They were placed in a colorimeter, and a percentage was recorded that showed how much light has passed through. The following will show a table of the results obtained from the investigation. A mean of the results will be calculated.

From the above results it can be seen that as the temperature increases, the permeability of beetroot membrane becomes more permeable. This shows that the plasma membrane must be denatured. In the cells of a beetroot plant, a substance called anthocyanin is contained within the plasma membrane. It is anthocyanin which gives the beetroot its characteristic blue/purple colour. If a cell is damaged in a beetroot plant and the membrane is broken, the anthocyanin ‘bleeds’ from the cells like a dye. In this experiment the temperature denatures the membrane causing it to bleed the dye out of the cell.

Although the above values are a range of values obtained from the investigation, it does not actually show how the figures are close to one another. This is important as figures need to be analysed to draw correct conclusions. To see how close the ranges of numbers are to one another, I will carry out a standard deviation calculation. The following will show the working out the standard deviation of the results from 53 C.

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The following equation will be used:

S =√ ∑

Key

S = standard deviation  ∑ = sum of

X= individual reading (% transmission) X= average

√= square route  n= number of readings

To work out the standard deviation of the results obtained from 53 C, I will need to work out the average (x). To do this I will need to add all 3 values from the % transmission and then divide by 3. The following will show the workings.

I will now need to minus the % transmission from the average. From the result I can square ...

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