Photosynthesis Analysis

For photosynthesis to take place a plant must be supplied with carbon dioxide and water. It also requires light, suitable temperature and the presence of chlorophyll.

CO2 + H2O CH2O O2

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Carbon water carbohydrate oxygen

dioxide

A plant that grows quickly will need more stomata to obtain more carbon dioxide in order to increase the rate of photosynthesis.

Plant leaves are rather like solar panels, they are broad and flat to trap light from the sun. Leaves have large numbers of tiny holes called stomata. Carbon dioxide enters the leaf through these pores and diffuses directly into the cells that need it, Oxygen escapes by the same route.

Stomata;

Pores surrounded by two specialised cells, guard cells. Unlike other epidermal cells they contain chloroplasts. Most of the cellulose microfibrils of the cell wall run across Most of the cellulose microfibrils of the cell wall run across the width of the cell, this allows the two cells to bend and so open and close the pore. Successive exchange of gases by the leaf depends on stomata being open. These press and also allow water to escape by diffusion into the air (water in the gaps between the cells). It is then free to diffuse out stomata. In hot environments the evaporation of water in this way helps to cool the plant. However water lost has to be replaced and if water were reduced by closing the stomata this also inevitably reduces gas exchange. Stomata are involved in a continual balancing act, as they open and close, according to the external conditions and the needs of the plant.

(A vertical section through stoma, showing also part of the lower surface of leaf. (b) pattern of cellulose microfibrils in guard cell walls.)

A leaf works as a photosynthetic organ. Carbon dioxide he greater is from the atmosphere diffuses through the stomata. Water, drawn up to the leaves from the soil via the conducting tissues of the roots and stem, passes out of the xylem elements in the veins to the surrounding cells. Water is supplied for photosynthesis. With water come mineral salts( nitrates, sulphates and phosphates) required for the synthesis of proteins. Oxygen and excess water vapour diffuses out of the leaf via the intercellular air spaces and stomata. Sugar and other products of photosynthesis are moved to there parts of the plants in the sieve tubes.

When stomata are open the rate of photosynthesis increased as fast as the maximum rate of respiration. Under these circumstances the plant will use carbon dioxide from its respiration for photosynthesis but the bulk of its carbon dioxide must be brought from the atmosphere. If the stomata are closed photosynthesis can still continue, using the carbon dioxide from respiration. An equilibrium can be reached between photosynthesis and respiration, photosynthesis using carbon dioxide from respiration and respiration using oxygen from photosynthesis. However the rate of photosynthesis under theses circumstances will be much slower than an external source of carbon dioxide is available, The stomata cannot remain closed indefinitely, for open stomata are necessary in order to maintain the transpiration stream, which is the only way leaf can obtain water.

Once I processed all my results in a table, to visually see the difference between the leaves, I plotted the result s in a graph.. Analysing the graph I drew I can clearly see that there is a difference between the no of stomata in Ilex, Heptapleurum and hedera showing the white area containing more stomata. However in my prediction I stated the green areas would contain more stomata due it containing chlorophyll a and b, the photosynthetic pigments and therefore it absorbs more light at different wavelengths.

There is a big difference between the no of stomata in the green and white areas for chlorophyta. We should take into account that chlorophyta is a small house plant therefore it is correct when it shows less stomata because then there will be less carbon dioxide available therefore the plants produce less glucose therefore it grows slowly resulting in a small house plant. All the others are large bush plants however heptapleurum is also a houseplant but is a large plant. This results in more stomata, more carbon dioxide uptake and therefore more production of glucose so the plants will grow more quickly into larger plants.

Flowering plants are divided into two groups monocots and dicots. Chlorophytum belongs to the monocot group and the rest pf the species belong to the dicot group. In general the greater number of stomata per unit area, the rate of stomata transpiration, however their distribution is also important. The green areas of dicotyledonous leaves usually have more stomata than the whiter areas and whereas monocotyledonous leaves hive a similar distribution in both areas, observing the graph concludes this.

The aim of my investigation was to see if there is any significant difference between the numbers of stomata in variegated leaves. I have analysed this simply by processing a graph however I will further analyse this using a more complicated method, a statistical analysis by using the student t-test.

My null hypothesis is that is no difference between the white and green areas of the plant.

My alternative null hypothesis that there is a difference between the no of stomata found in the green and white areas of variegated plants.

I need the means of all the data collected for the 4 different species to begin the test.

I will show 1 example of how the mean is found and the rest of the values I will put into a table. Mean =???n

Heptapleurum:

Green area:

42,38,34,36,32,39,41,38,32,39,47,37,27,42,47,48,46,38,38,31,33,43,43,40,41/24= 38.72

White area:

34,37,48,42,46,43,49,45,52,48,44,52,48,44,38,46,41,49,46,50,35,37,39,49/24= 44.25.

To perform the t-test we also need the standard deviation.

Formula:

? stands for the sum of

? refers to the individual values of the sample

n is the total number of individual value in the sample

It is much convenient to use the calculator to obtain the standard deviation. Using the

Button.

Species

Total no of stomata counts

Mean no of stomata counts

Standard deviation

Heptapleurum

Green

24

38.72

5.45

White

24

44.25

5.33

Ilex

Green

45

27.89

11.32

White

41

30.63

10.75

Hedera

Green

40

44.05

10.15

White

33

50.30

10.45

Chlorophyta

Green

32

17.53

5.04

White

21

4.71

5.99

Now I have my values to begin one of the t-test. I will show 1 test step by step and the rest I will work out using the formula.

I will use the standard deviations to calculate significance of the difference between two means.

Heptapleurum

1: Work out the mean of the green and white area. (shown above)

2: Subtract the smaller mean from the larger mean.

44.25-38.72=5.53

3: work out the standard deviation of one set of data. Multiply this number by itself (i.e square it) and divide it by the number of species in that set of data.

5.45 x 5.45/24=1.2376

4: repeat for the other set of data:

5.33 x 5.33 /24 = 1.1819

5: Add together the figures you calculated in steps 3 and 4.

1.237 + 1.1819= 2.4819

6: Take the square root of the figure calculated in step 5, what is calculated is called the standard error.

2.4189 = 1.555

7: Divide the difference between the two means (step 2) by the figure calculated in step 6. This is your t value.

5.542/1.555 = 3.564

The degrees of freedom is needed to look up the critical value from the tables provided for which we will test our t-value.

You find the degrees of freedom by counting the total no of counts per specie and then look the value up in the table. As we are unable to find the exact critical value due to the figures not found in the table we look at the critical value before and after and then make our conclusion to see if our t value is significantly different to the value given in the table.

= 24 + 24 = 48 45= 2.01 50 = 2.01

It is obviously that 3.564 is much greater than the critical value therefore we are at least 95% confidence that no of stomata from Heptapleurum in the green and white areas differs. Therefore we reject null hypothesis and accept the alternative hypothesis.

Now for the rest of the species I will be using a the formula:

Ilex:

Mean of X- mean of Y

Standard error

27.89-30.63

2.38 =1.1513

Degrees of freedom

critical value = 45+41= 86 80= 1.99

90=1.99

The critical value is smaller than the t value therefore we accept the null hypothesis conclude there is no difference between white and green areas. We are less than 95% confident that the difference between the means is not significant. Ilex is common and widespread in woods and hedges. Stiff leathery leaves have spiny margins. Height up to 10m, a slow growing plant that shows that there will be a lower rate of photosynthesis taking place. Less stomata will therefore be found so a not much difference between the no of stomata. Plants have stomata to take in carbon dioxide, because carbon dioxide is needed for photosynthesis to take place. Where there is fewer stomata we know the plant grows slowly due to the thick waxy cuticle layer.

Hedera

50.30-44.05

2.426 = 2.576

40 + 33= 73 table value: 70 = 1.99

80 = 1.99

It is obvious that 2.576 is greater than the critical value of t, this means we are at least 95% confident that the mean of no of stomata from green and white differs. Therefore we reject null hypothesis and conclude there is a difference between the no of stomata found in the green and white areas of this plant.

Ivy is a vigorous woody evergreen climber, ivy clings by means of short-term roots. The leathery alternate leaves are stalked and deep green. Grows throughout the region, except for the far north. Its habitat is on trees and rocks in woodland and on cliffs, old walls and buildings, hedgerows, often in dense shade. There is more stomata found in the green areas of the plants because more chloroplast if is found there, therefore photosynthesis takes place at a faster rate and there should be more distribution off stomata here This rejects our null hypothesis and concludes that there is a difference between the no of stomata found in the green and white areas of a hedera plant. .

Chlorophyta

Degrees of freedom: 32 + 21 =53

50= 2.01

60 =2.00

17.53-4.71

1.589 = 8.104

8.104 is much greater than the t value this means we are definitely at least 95% confident that the mean no of stomata from green and white areas differs, so therefore we reject the null hypothesis. Using the graph and the table constructed we can clearly see that there is a significant difference between the number of stomata found in the green area to the white area. As we know light energy is trapped by photosynthetic pigments. Different pigments absorb different wavelengths of light. Chlorophylls reflect green light, which is why plants look green. The green areas contain both chlorophylls a and b. Caratenoids are the other photosynthetic pigments which are found in the white area of the plants but do not absorb light as much as the chlorophylls, which is why there is a higher rate of photosynthesis taking place here resulting into a higher region of stomata in the green areas of the chlorophyta. We can conclude there is a difference between the no of stomata in the green and white areas of chlorophyta specie.

My research and conclusion was drawn from two other modules from the concepts biology module, using the transport and photosynthesis topics and the a level biology book written by Sandra Dawns.

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