A Level Biology Project
This is an experiment to examine how the Surface Area / Volume Ratio affects the rate of diffusion in substrates and how this relates to the size and shape of living organisms.
This is an A-level biology project. It helped me get an A grade for biology many years ago. The whole project is reproduced here for your reference.
- Background Information
- Extension Work
The surface area to volume ratio in living organisms is very important. Nutrients and oxygen need to diffuse through the cell membrane and into the cells. Most cells are no longer than 1mm in diameter because small cells enable nutrients and oxygen to diffuse into the cell quickly and allow waste to diffuse out of the cell quickly. If the cells were any bigger than this then it would take too long for the nutrients and oxygen to diffuse into the cell so the cell would probably not survive.
Single celled organisms can survive as they have a large enough surface area to allow all the oxygen and nutrients they need to diffuse through. Larger multi celled organisms need specialist organs to respire such as lungs or gills.
- Gelatin blocks containing cresol red dye
- 0.1M Hydrochloric acid
- Stop Watch
- Safety glasses
1. A block of gelatin which has been dyed with cresol red dye should be cut into blocks of the following sizes (mm).
5 x 5 x 5
10 x 10 x 10
15 x 15 x 15
20 x 20 x 20
10 x 10 x 2
10 x 10 x 10 (Triangle)
10 x 15 x 5
20 x 5 x 5
The triangle is of the following dimensions. [not reproduced]
The rest of the blocks are just plain cubes or rectangular blocks.
Cresol red dye is an acid / alkali indicator dye. In the alkali conditions of the gelatin it is red or purple but when it gets exposed to acid it turns a light yellow colour.
Gelatin is used for these tests as it is permeable and so it acts like a cell. It is easy to cut into the required sizes and the hydrochloric acid can diffuse at an even rate through it.
I am not using any blocks bigger than 20 x 20 x 20 as a preliminary test found that it was only practical to use blocks of 20mm³ or less as anything bigger than this would take longer than the amount of time that we have to do the experiment.
2. A small beaker was filled with 100cm³ of 0.1 molar Hydrochloric acid. This is a sufficient volume of acid to ensure that all the block sizes are fully covered in acid when dropped into the beaker.
3. One of the blocks is dropped into this beaker and the time for all the red dye to disappear is noted in a table such as the one below.
|Dimensions (mm)||Surface Area||Volume (mm³)||Surface Area / Volume Ratio||Test 1||Test 2||Test 3||Average Time|
4. This test should be repeated for all the sizes of blocks three times to ensure a fair test. Fresh acid should be used for each block to ensure that this does not affect the experiment’s results.
5. The surface area / volume ratio and an average of the results can then be worked out. A graph of Time against Surface Area to Volume Ratio can then be plotted. From this graph we will be able to see how the surface area affects the time taken for the hydrochloric acid to penetrate to the centre of the cube.
I predict that as the Surface Area / Volume Ratio increases the time taken for the hydrochloric acid to penetrate to the centre of the cube will go down. This is because a small block has a large amount of surface area compared to it’s volume so the hydrochloric acid will have a large surface area to diffuse through. A larger block has a smaller amount of surface area in relation to it’s size so it should take longer for the hydrochloric acid to diffuse into the centre of the cube. The actual rate of the hydrochloric acid diffusing through the gelatin should be the same for all the blocks but when the surface area / volume ratio goes up it will take less time for the hydrochloric acid to reach the centre of the cube.
I carried out the above experiment and these results were obtained.
|Dimensions (mm)||Surface Area||Volume (mm³)||Surface Area / Volume Ratio||Test 1||Test 2||Test 3||Average Time|
|5 x 5 x 5||150||125||1.2:1||7.02||6.57||4.53||6.16|
|10 x 10 x 10||600||1,000||0.6:1||10.3||23.25||15.33||16.28|
|15 x 15 x 15||1,350||3,375||0.4:1||29.55||30.22||23.45||28.01|
|20 x 20 x 20||2,400||8,000||0.3:1||53.4||32.44||58.56||48.3|
|10 x 10 x 2||280||200||1.4:1||0.26||0.37||1.58||1.01|
|10 x 15 x 5||550||750||0.73:1||7.2||10.23||10.47||9.3|
|20 x 5 x 5||450||500||0.9:1||3.18||2.58||4.09||3.29|
|10 x 10 x 10
The Surface area to Volume ratio is calculated by
Surface Area To Volume Ratio = Surface Area / Volume
From these rates I was able to plot a graph of the Surface Area to Volume Ratio against time.
In all the blocks of gelatin the rate of penetration of the hydrochloric acid from each side would have been the same but all the blocks take different amounts of time to clear because they are different sizes. As the blocks get bigger it takes longer for the hydrochloric acid to diffuse through all the block and so clear the dye. It takes longer to reach the centre of the cube even though the rate of diffusion is the same for all the cubes.
As the volume of the blocks goes up the Surface Area / Volume ratio goes down. The larger blocks have a smaller proportion of surface area than the smaller blocks. The smallest block has 1.4mm² of surface area for every 1mm³ of volume. The largest block only has 0.3mm² of surface area for each 1mm³ of volume. This means that the hydrochloric acid is able to diffuse to the centre of the smallest block much faster than the largest block. The acid took 48 minutes to diffuse to the centre of the largest block but only 1 minute in the smallest block. A living cell would not survive if it had to wait 48 minutes for oxygen to diffuse through it so living cells need to be very small.
When the surface area to volume ratio goes down it takes longer for the hydrochloric acid to diffuse into the cube but if the ratio goes up then the hydrochloric acid diffuses more quickly into the block of gelatin. Some shapes have a larger surface area to volume ratio so the shape of the object can have an effect on the rate of diffusion.
It is important that cells have a large surface area to volume ratio so that they can get enough nutrients into the cell. They can increase their surface area by flattening and becoming longer or by having a rough surface with lots of folds of cell membrane known as villi. [picture not reproduced]
The villi vastly increase the surface area of the cell whereas the cell which is round only has a small surface area in relation to it’s volume. Both cells above have an volume of 1cm³. The cell on the left has a surface area of 3cm² but the cell on the right with villi has a surface area of 10cm². The cell membrane is made up of a lipid bi-layer with many proteins integrated into it. [picture not reproduced]
Oxygen can diffuse easily through the membrane and Carbon Dioxide and other waste products can easily dissolve out. The concentration of oxygen in the cell is always lower than outside the cell which causes the oxygen to diffuse in. Gases will always dissolve from an area of high to low pressure. The concentration of carbon dioxide outside the cell is lower than the concentration in the cell so the carbon dioxide will always dissolve out of the cell.
Single celled organisms such as amoebas have a large surface area to volume ratio because they are so small. They are able to get all the oxygen and nutrients they need by diffusion through the cell membrane.
Larger organisms such as mammals have a relatively small surface area compared to their volume so they need special systems such as the lungs in order to get enough oxygen. Surface area to volume ratio is very important in lungs where a large amount of oxygen has to get into the lungs. The lungs have a very large surface area because they contain millions of sacs called alveoli which allow oxygen to diffuse into the bloodstream. By having millions of these alveoli the lungs are able to cram a very large surface area into a small space. This surface area is sufficient for all the oxygen we need to diffuse through it and to let the carbon dioxide out.
By increasing the surface area the rate of diffusion will go up.
a) All the gelatin used should be taken from the same block to ensure that all the blocks are made up of the same materials.
b) All the tests should be done at room temperature to ensure that the blocks of gelatin do not melt.
c) The same volume of acid should be used for all the tests to ensure that the rate of diffusion can not be affected by the pressure of a larger volume of acid.
d) Safety glasses should be worn to protect your eyes from the hydrochloric acid.
To help make this experiment more accurate, I repeated it three times for each block size and then used the average of all the results to plot a graph with a line of best fit. I tried to keep all the variables except for the size of the gelatin blocks the same for all the experiments. However, in reality it is impossible to keep all the variables precisely the same. For example:
a) It is also impossible to precisely measure the size of gelatin block each time. I measured the sizes to the nearest mm so the sizes of block that I used should be correct to the nearest mm.
b) When the gelatin blocks are dropped into the beakers the base of the block comes into contact with the bottom of the beaker which reduces the surface area of the block that comes into contact with the hydrochloric acid.
c) The results will be slightly inaccurate as the moment when the gelatin block has lost all it’s dye is a matter of opinion and not something that can be measured precisely.
d) Due to the fairly slow speed of our reactions it is only possible to measure the time of the reaction to the nearest 0.1 second even though the stopwatch shows the measurements to the nearest 0.01 second.
The graph produced shows a smooth curve with a decreasing gradient as the surface area to volume ratio goes up. The only anomaly is the result for the 5 x 5 x 5 block. The result here is higher than the curve of best fit for the graph. The results for the 5 x 5 x 5 block ranged from 4.53 to 7.02 seconds with an average of 6.15 seconds. The line of best fit for the graph suggests that the average should be around 3 seconds. The anomalous result was probably due to experimental error as a result of this being the first block size that I used in the experiment. The most likely explanation is that I was unsure of how to judge when all the dye had disappeared and as a result delayed pressing the stop button of the stop watch. As the experiment progressed with the other block sizes I probably got better at making this judgement.
This experiment could be improved in a number of ways.
1) It could be repeated more times to help get rid of any anomalies. A better overall result would be obtained by repeating the experiment more times because any errors in one experiment should be compensated for by the other experiments.
2) Using more shapes and sizes of gelatin block would have produced a better looking graph.
3) Variables that might affect the rate of diffusion could be investigated. The rate of diffusion may also be affected by temperature, strength of acid and volume of acid
4) The block could be suspended in the hydrochloric acid so than none of it’s surfaces are in contact with the wall of the beaker. A small cradle could be used to suspend the blocks in the acid which would mean that all six sides of the cube should be in contact with the acid. This would ensure that diffusion could occur evenly through all the sides of the cube.
This is a real A-level school project and as such is intended for educational or research purposes only. Extracts of this project must not be included in any projects that you submit for marking. Doing this could lead to being disqualified from all the subjects that you are taking. You have been warned. If you want more help with doing your biology practicals then have a look at 'Advanced Level Practical Work for Biology' by Sally Morgan. If you want more detailed biology information then I'd recommend the book 'Advanced Biology' by M. Kent.