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Difference in number of Stomata in different leaves

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Introduction

Biology Coursework An investigation will be carried out to see how the numbers of stomata differ in different plant species. Stomata are tiny openings found on the underside of the plant leaf. [1] The pore is formed by a pair of cells called 'Guard Cells'. Guard cells control the opening of the stomata. These pores on the underside of the plant leaf allow carbon dioxide to enter. Carbon dioxide is used in photosynthesis. At the same time when the stomata are open Oxygen and water vapour escape from inside of the plant. Water vapour leaving the plant is called Transpiration. This occurs mostly in leaves of the plant but can be from the flower and the stem. Hypothesis The leaves of different plants and the sections of leaves will have different number of stomata. The amount of stomata on the plant leaf is dependant on the environment. The plants that are in temperate environment will have many more stomata than the plants that are in Tropical environments. Null Hypothesis There is no difference in the number of stomata in any plant leaf or any section of the plant. Any difference is purely due to chance I will investigate how to find out the different number of stomata in a plant leaf. Experiment Apparatus Nail Varnish Microscope Microscopic slide Lab coat Tape Plant leaf Step by Step 1- Get apparatus 2- Take a leaf from a plant and mark 3 sections, top, middle and bottom (to later investigate under the microscope) ...read more.

Middle

= 1.45 To work out the standard deviation I need to use the equation: ? (x - x)2 = 0.3 n-1= 3-2 = 2 Square root of 0.3/2 = V0.15 = 0.39 Plant used was Snowball Area of plant leaf (n) Number of Stomata (x - x) (x - x)2 Count 1 Count 2 Count 3 Average (x) Top 3 4 4 3.67 0.44 0.19 Middle 5 4 7 5.33 -1.22 1.49 Bottom 3 4 3 3.33 0.78 0.61 Mean (x) = 4.11 To work out the standard deviation I need to use the equation: ? (x - x)2 = 2.29 n-1= 3-2 = 2 Square root of 2.29/2 = V1.145 = 1.07 Plant used was Holly Area of plant leaf (n) Number of Stomata (x - x) (x - x)2 Count 1 Count 2 Count 3 Average (x) Top 19 22 18 19.67 1.55 2.40 Middle 23 23 18 21.33 -0.11 0.01 Bottom 28 19 21 22.67 -1.45 2.10 Mean (x) = 21.22 To work out the standard deviation I need to use the equation: ? (x - x)2 = 4.51 n-1= 3-2 = 2 Square root of 4.51/2 = V2.255 = 1.50 Plant used was Privet Area of plant leaf (n) Number of Stomata (x - x) (x - x)2 Count 1 Count 2 Count 3 Average (x) ...read more.

Conclusion

(O-E)2/E = 77.575 The degree of freedom here is (n-1) 2-1 = 1 so the number of classes is 2 The mean chi-squared value is 77.575 which is a very high number. This clearly rejects the null hypothesis. So the number of stomata in the plants from different environments differ is not due to pure chance but the plants have adapted to the environment over the years. Evaluation Even though the results were reliable and rejected the null hypothesis when grouped by the environment of the plant there were some limitations to how accurate the results were. Human error can cause limitations as some plant leaves had a high amount of stomata present and can be counted wrong. Accuracy of the microscope can be a limitation, some places only half of the stomata were shown or less which couldn't be counted as it wouldn't show a valid result. Improvements This experiment can be improved if more results are taken and more different types of plant leaves are used. Instead of 3 counts more counts from the same type of plant leaf can make the results more reliable. Measure the area of plant leaf used and use the same area for all the plant leaves. This will increase the reliability of the results as you will be counting in a fixed area. An electron microscope can be used to count more accurately the number of stomata located in a certain area of the leaf. Reference [1] = http://en.wikipedia.org/wiki/Stomata Visited on 14/04/08 at 6:47pm ...read more.

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Here's what a star student thought of this essay

4 star(s)

Response to the question

Overall a good example piece of coursework. Introduction is adequate, but could explain transpiration and the process what occurs a bit better. Sets up the scientific background for the experiment well. The hypothesis is good apart from the fact that ...

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Response to the question

Overall a good example piece of coursework. Introduction is adequate, but could explain transpiration and the process what occurs a bit better. Sets up the scientific background for the experiment well. The hypothesis is good apart from the fact that the candidate could explore why plants in tropical and temperate environments may differ in the number of stomata. Conclusion and the rejection of the null hypothesis are good. Response to the question in general is to a high level, although the science behind the experiment could be mores solid. Meaning and communication of the experiment is clear.

Level of analysis

The level of analysis is good as two different types of experiments are used and two different methods of analysis. However, the reason for using the chi squared test or standard deviation is not stated, so I am not sure why these tests were used. The amount of plants used from each climate is a good sample, but if the test was repeated there should have been a much wider range used. Improvements suggested were good, and reflected on the experiment well.

Quality of writing

Incorrect use of capitals in some places. Spelling, grammar and punctuation adequate.


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Reviewed by skatealexia 06/03/2012

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