Surface Area to Volume Ratio and the Relation to the Rate of Diffusion.

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Surface Area to Volume Ratio and the Relation to the Rate of Diffusion

Aim and Background

This is an experiment to examine how the Surface Area / Volume Ratio affects the rate of diffusion and how this relates to the size and shape of living organisms.

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 organs to respire such as lungs or gills.

Method

The reason I chose to do this particular experiment is because I found it very interesting and also because the aim, method, results- basically the whole experiment would be easily understood by the average person who knew nothing about Surface Area/Volume Ratio. The variable being tested in this experiment is the rate of diffusion in relation to the size of the gelatin cube. Another experiment one could do to determine the surface area to volume ratio is to construct a set of cubes out of construction paper- 1 x 1, 2 x 2, 3 x 3 and 4 x 4 (cm).Then use this formula to determine the surface area- L x W x 6 and compare it with the volumes. The formula to determine volumes of cubes is L x W x H. Although that type of experiment will show no insight into SA/V ratio in relation to the rate of diffusion.

Equipment

1. Agar-phenolphthalein - sodium hydroxide jelly

2. O.1 M hydrochloric acid

3. Ruler (cm and mm)

4. Razor blade

5. Paper towel

6. Beaker

Method

1. A block of gelatin which has been dyed with phenolphthalein 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

30 x 30 x 30

20 x 5 x 5

Phenolphthalein 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 almost colorless.

Gelatin is used for these tests because it is permeable which means 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.

2. A small beaker was filled with about 400ml of 0.1 molar Hydrochloric acid. This is a sufficient amount 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, left for 10 minutes, then removed, dried, and cut in two to measure the depth of penetration. This test should be repeated for all the sizes of blocks three times to ensure an accurate test. Fresh acid should be used for each block to make sure that this does not affect the experiment's results.

4. The Surface Area/Volume Ratio and an average of the results can then be worked out. A graph of Surface Area to Volume Ratio can then be plotted along with percentages left colored and uncolored . From this graph we will be able to see how surface area affects the rate of diffusion of materials into the cubes.

Results

I carried out the above experiment and these results were obtained.

Dimensions (mm) Surface Area Volume (V) (mm) Surface Area / Volume Ratio Test 1 Test 2 Test 3

5 x 5 x 5 150 125 1.2:1 1mm 1mm 1mm

10 x 10 x 10 600 1,000 0.6:1 1mm 1mm 1mm

20 x 20 x 20 2,400 8,000 0.3:1 1mm 1mm 1mm

30 x 30 x 30 5,400 27,000 0.2:1 1mm 1mm 1mm

The Surface Area to Volume Ratio is calculated by

SA = cm

From these results I was able to make a graph of the volume still coloured along with the percentages left coloured and uncoloured.

Dimensions (mm) Volume left coloured 3(mm ) Percentage coloured compared to original volume Percentage penetratedby the acid

5 x 5 x 5 3mm 60% 40%

10 x 10 x 10 8mm 80% 20%

20 x 20 x 20 18mm 90% 10%

30 x 30 x 30 28mm 93.3% 6.7%

Length of side not penetrated = (s - 2x)

3

Volume left coloured (Vc) = (s - 2x)

Percentage still coloured (C%) = Vc x 100

V 1

Percentage of cube penetrated = 100 - C%

Interpretation

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 cubes have different percentages still coloured because they are different sizes. As the blocks get bigger the hydrochloric acid to diffuses smaller percentages of the cubes. It would take longer to totally diffuse the largest 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 surface area than the smaller blocks. The smallest block has 1.2mm squared of surface area for every 1mm cubed of volume. The largest block only has 0.2mm squared of surface area for each 1mm cubed of volume. This means that the hydrochloric acid is able to diffuse the smallest block much faster than the largest block.

When the Surface Area/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.

The single error or limitation I encountered was the impossiblity to precisely measure the size of gelatin block. I measured the sizes to the nearest mm so the sizes of block that I used should be correct to the nearest mm.

Discussion

It is important that cells have a large surface area to volume ratio so that they can get enough nutrients into the cell.

Single celled organisms 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.

Here is a diagram of a standard leaf:

Their are openings within a leaf called stomata. These allow for the gases to flow in and out of the leaf. Leaves of plants have a large surface area, and the irregular-shaped, spongy cells increase the area even more meaning a larger amount of gas exchange. An example of surface area to volume ratio in a real world context would be something such as the example that was just explained.

Therefore, by increasing the surface area the rate of diffusion will go up.

Appendices

(2002) Biology: The Surface Area to Volume Ratio of a Cell [Web document] http://www.geocities.com/CapeCanaveral/Hall/1410/lab-B-24.html

This piece of information was a good start for the investigation of Surface Area to Volume Ratio investigation. Even though it has no mention about rate of diffusion in relation to SA/V ratios, its relevance to my investigation was crucial.

(2002) Encyclopedia Britannica: Biology- Surface Area to Volume Ratio

[CD-ROM]

I found this source of information to be very reliable. The Encyclopedia Britannica is a popular and credible way to gain information. It covers the whole range of factors relating to SA/V ratios as well as the rate of diffusion. It was very appropriate for my investigation.

(2000) Sizes of Organism's: Surface area to Volume ratio [Web document] http://www.tiem.utk.edu/~mbeals/area_volume.html

This document had an in depth discussion about the relation between Surface Area and Volume Ratio's. It used plenty of examples to get the point across more clearly. It also touched on Surface Area to Volume Ratio's of sphere's.