Apparatus:
- Metre ruler
- Quadrat
- Light sensor (lockets)
- Compass
Method:
Firstly we measure 1 metre up the tree so that all the results are from 1 metre up, then we find which way north is (with a compass) and measure the amount of light with a light sensor on the north, south, east and west sides. Then we place a quadrat at the height and count all the squares that have Pleurococcus in them one each of the 4 sides. As there are 100 squares in the quadrat it will be out of a hundred and will easily be converted into a percentage (e.g. 54 out of 100 squares have Pleurococcus so it’s 54%). We will do this on 10 trees.
Fair test:
To makes this a fair test we will make sure that we only measure the same tree species, we will tell if the trees are the same species by getting a leaf off one tree and comparing it to others. All the measurements will be taken 1 metre up each tree. We will use the same type of quadrat (100 squares) and we will use the same type of light sensor each time. We will get the same people to do the same tasks ever time to make this a fair test.
Results:
I found that where there is the most light (on the south side) there is the least amount of Pleurococcus and where there is least light (on the north side) there is the most Pleurococcus. For example tree 4 (shaded in grey) has the most Pleurococcus on the north side and least on the south. This is the only tree that has the most Pleurococcus on the north and least on the south. This gives a fraction of 1/10.
Conclusion:
My prediction was correct; most Pleurococcus grows on the north side because there is least light on the north side (as the sun rises in the east, sets in the west, leaving the south to exposure). Pleurococcus grows where the is less light because it is only a single celled organism and it has no roots, therefore the only way it can collect water is absorb when it is running down the tree when it rains, this is show by the rose diagram. The rose diagram shows that the side with the most light on average which is the south side has the least Pleurococcus. As Pleurococcus is a green plant it photosynthesizes (6CO2 + 6H2O → C6H12O6 + 6O2) and for photosynthesis water (H2O) is needed so if the Pleurococcus is in direct sunlight the water in the Pleurococcus will dry out quicker, therefore there is more Pleurococcus where there is less light. My prediction was correct, this is shown in the averages; on the north side there is least light (85.76%) and most Pleurococcus (45.6%) and on the south there is most light (91.04%) and least Pleurococcus (25.5%).
Evaluation:
I think that the experiment worked out pretty well because from the averages I can see a pattern. The averages were accurate enough but the separate results were not, this is because some of the trees that we tested for Pleurococcus we in a shaded area for example on trees 2,3 and 4 there is a lot of Pleurococcus on the east side because these trees were next to a fence that shaded them on the east side. To make this experiment more accurate I could have counted the amount of Pleurococcus in the individual squares in the quadrat, for example:
I would have counted this as 4 out of 6 but if I had more time I would have counted it as 2.5 out of 6.
