Scientific calculations - investigating transpiration
A simple method for investigating water loss from plant leaves is to measure their change in mass over a period of time.
Various factors that affect water loss from the leaf can be investigated using this method, for instance:
- air movement - direct a fan on the leaves
- temperature
- obstructing the stomataTiny holes in the epidermis (skin) of a leaf. They control gas exchange by opening and closing and are involved in loss of water from leaves. Singular is stoma., eg with petroleum jelly
Method
- Remove a number of leaves from a bush or tree.
- Find the mass of each leaf.
- Suspend each leaf from a piece of wire or string.
- After a set period of time, re-measure the mass.
| Experiment number | 1 | 2 | 3 | 4 |
| Surface coated with petroleum jelly | Neither | Upper | Lower | Both |
| % decrease in mass in Leaf 1 | 43 | 37 | 5 | 2 |
| % decrease in mass in Leaf 2 | 38 | 38 | 3 | 1 |
| % decrease in mass in Leaf 3 | 37 | 35 | 6 | 3 |
| % decrease in mass in Leaf 4 | 42 | 36 | 4 | 2 |
| % decrease in mass in Leaf 5 | 40 | 34 | 3 | 2 |
| Mean | 40 | ? | 4 | 2 |
| Surface coated with petroleum jelly | |
|---|---|
| 1 | Neither |
| 2 | Upper |
| 3 | Lower |
| 4 | Both |
| % decrease in mass in Leaf 1 | |
|---|---|
| 1 | 43 |
| 2 | 37 |
| 3 | 5 |
| 4 | 2 |
| % decrease in mass in Leaf 2 | |
|---|---|
| 1 | 38 |
| 2 | 38 |
| 3 | 3 |
| 4 | 1 |
| % decrease in mass in Leaf 3 | |
|---|---|
| 1 | 37 |
| 2 | 35 |
| 3 | 6 |
| 4 | 3 |
| % decrease in mass in Leaf 4 | |
|---|---|
| 1 | 42 |
| 2 | 36 |
| 3 | 4 |
| 4 | 2 |
| % decrease in mass in Leaf 5 | |
|---|---|
| 1 | 40 |
| 2 | 34 |
| 3 | 3 |
| 4 | 2 |
| Mean | |
|---|---|
| 1 | 40 |
| 2 | ? |
| 3 | 4 |
| 4 | 2 |
Analysis of results
There may be variation in the decrease in mass of different leaves.
It is important to repeat the experiment and calculate a mean for each set of data.
For experiment 1:
mean percentage decrease in mass = \(\frac{loss~in~Leaf~1 + Leaf~2 + Leaf~3 + Leaf~4 + Leaf~5}{5}\)
\( = \frac {43 + 38 + 37 + 42 + 40}{5} = \frac {200}{5} = 40\)
Question
What is the mean percentage loss in mass in experiment 2?
36.
Calculation:
mean percentage decrease in mass = \(\frac{loss~in~Leaf~1 + Leaf~2 + Leaf~3 + Leaf~4 + Leaf~5}{5}\)
\( = \frac {37 + 38 + 35 + 36 + 34}{5} = \frac {180}{5} = 36\)
Water loss through the stomata
Water is lost through open stomata. Scientists sometimes count all the stomata on a leaf surface, but usually they take a sample. This must be a representative sampleA representative sample is one that accurately represents the whole of the group..
To be representative of the whole leaf, the representative sample must:
- include a sufficient number of counts - not just one or two - of stomata over different parts of the slide
- must be random, and not select areas where there are many or few stomata
A number of random counts of stomata should be made with a microscope. Count the number of stomata in the field of view. Then move the slide slightly and count the number of stomata in a different field of view.
Make at least five random counts, then calculate a mean.
In this field of view, there are 12 stomata - nine open and three closed.
Using this method, and a calibrateTo set an instrument or scale against a standard. eyepiece The eyepiece, or ocular, is the lens at the top of a compound microscope. It has a longer focal length than the objective lens and magnifies the image produced by the objective. graticuleA glass or plastic disc fitted into the eyepiece of a microscope. The graticule has a scale ruled on it and is used to estimate the size of a specimen when viewed with a microscope., you could estimate the number of stomata per millimetre cubed.
The images show the fields of view of a plant leaf viewed with a microscope. Count the number of stomata in each sample.
Image caption, Sample 1
Image caption, Sample 2
Image caption, Sample 3
Image caption, Sample 4
Image caption, Sample 5
1 of 5
For these counts, the mean is:
\(mean\; =\; \frac{sample\: 1\: +\: sample\: 2\: +\: sample\: 3\: +\: sample\: 4\: +\: sample\: 5}{numbers\; of\; samples}\)
\(=\; \frac{14\: +\: 12+\: 11+\: 12+\: 11}{5}\; =\; \frac{60}{5}\; =\; 12\; stomata\; in\; the\; field\; of\; view\)