Introduction
Animals adapt in several different ways to their environment, with a significant way of adapting through bodily surface area. Organisms are known to adapt by altering their surface area to volume ratio. This particular ratio is extremely important for animals in their environment, as it dictates how they react to the common features present throughout their habitat. The surface area to volume ratio (abbreviated to SA:V) describes how much surface area of an organism there is (body area that is exposed to an organism’s exterior environment) in comparison to its volume (interior capacity). According to the SA:V ratio, the larger the organism becomes, the less surface area will be available to suit its increasing needs due to the organism’s increasing volume.
Animals living in drastically different environments will consequently have different surface area to volume ratios. For example, animals inhabiting arid and hot environments show a tendency to have a larger SA:V ratio since it allows the animals to stay cool and maximize their heat loss. The African bush elephant (Loxodonta africana) residing in southern Africa (2), for instance, has an adaptation of large ears in order to increase the SA:V ratio and maximize heat loss, preventing them from overheating. On the contrary, animals inhabiting colder and wetter environments will most likely have a smaller SA:V ratio, as they usually have a larger volume but smaller surface area. Most of the volume is not exposed to the organism’s exterior environment, leaving the organism with smaller features (such as the small ears on a polar bear) in order to minimize heat loss by decreasing the amount of exposed surface area. By minimizing heat loss, these organisms may survive and prosper at freezing temperatures. As discussed, this SA:V ratio is exceedingly important for all organisms because it conveniently allows them to survive in their particular environments and exchange nutrients, gasses and waste within their environment due to a process called diffusion.
Diffusion is most commonly defined as “The passive movement of molecules or particles along a concentration gradient, or from regions of higher concentration to lower concentration (1)”. As the life-giving substances are continuously exchanged within an organism, smaller diffusion distances are preferable as it ensures that the diffusion process takes place as quickly as possible. The larger the surface area, the faster molecules can move in and out of an organism whereas the thicker the surface, the longer it will take for particles to diffuse from areas of higher concentration to lower concentration.
The reason why cells need to be so microscopically small is because their surface area to volume ratio is the most efficient at their size, allowing them to survive. Their SA:V ratio is large, allowing them to diffuse substances efficiently and regularly, as cells are constantly interacting with their surrounding environment. If cells were too large, the substances would not be able to diffuse efficiently as the volume of an object increases at a faster rate than its surface area. The surface area would not be great enough to take up the volume held by the cell, meaning the cell would have to divide (3).
Research question
How is the surface area to volume ratio associated with the rate of diffusion in different sized cells?
Aim
The aim of the experiment that shall be performed is to analyze how different sized blocks of agar jelly (resembling cells), with different SA:V ratios can affect the movement of molecules or diffusion. Furthermore, it will also be investigated how quickly material can diffuse through each of the model cells, depending on their size.
Hypothesis
As previously stated, organisms that possess a higher surface area to volume ratio will diffuse substances from areas of high concentration to low concentration at a faster rate than those who posses a lower SA:V ratio. Within the experiment, the agar jelly has been created with phenolphthalein indicator that turns colourless when acid is present, and contrarily, stains the jelly with a pink tint. The hydrochloric acid used in the experiment symbolizes life-giving substances an organism needs, such as nutrients. The faster the substances reach the cell, the higher the chances of the cell’s survival. As the hydrochloric acid diffuses into the agar ‘cells’ the indicator loses the pigmentation and turns colorless. Thus, the agar ‘cells’ with a large surface area to volume ratio will turn colorless quicker than those with a smaller surface area to volume ratio.
It is most likely that Cell A (2 x 2 x 2 cm3) will take the longest time to diffuse due to its low SA:V ratio, whereas Cell E (1 x 1 x 0.5 cm3) shall take the shortest time to diffuse and turn colorless due to its high SA:V ratio. In regards to Cells B-D, Cell B (2 x 2 x 1 cm3) will be the penultimate agar ‘cell’ to diffuse and turn colorless, Cell C (1 x 1 x 2 cm3) will diffuse slower than Cell E (due to its greater SA:V ratio) but faster than Cell B, and Cell D (1 x 1 x 1 cm3) will be the second to completely diffuse and turn colorless, as it still has a small SA:V ratio.
Variables:
Controlling the variables:
Apparatus and Materials
Materials:
Agar jelly with phenolphthalein indicator
Hydrochloric acid (HCL)
Rubber latex gloves
Apparatus:
5x 25ml beakers
Stopwatch
Cutting board
Knife
Spatula
Ruler (cm)
Method
A 4cm length of agar jelly with a cross-sectional area of 2cm2 and a height of 2cm3 was cut.
The block was next cut in half to form two 2 x 2 x 2 cm3 cubes. One of these agar ‘cells’ represented Cell A.
The remaining cube was cut in half again to produce a 2 x 2 x 1 cm3 cube titled Cell B.
Next, the other remaining cube was parted in half to produce a 1 x 1 x 2 cm3 cell titled Cell C.
Subsequently, the remaining part was parted in half once more to yield a 1 x 1 x 1 cm3 cube (Cell D)
Thereafter, the final piece was cut in half to create Cell E, with 1 x 1 x 0.5 cm3 dimensions.
Each agar model cell was submerged in 25ml of 1M hydrochloric acid. As soon as the cubes came in contact with the HCL, a stopwatch was used to measure the time taken for each block to turn completely clear and colorless.
Safety: Gloves were worn to prevent contamination with the hydrochloric acid as it is corrosive. Safety glasses were worn to prevent exposure to the HCL, as contact with eyes can cause permanent blindness.
Results
A table demonstrating the surface area, volume and surface area to volume ratio (SA:V) for each agar ‘cell’.
A table displaying the time taken for each agar ‘cell’ to lose the pink pigmentation and turn colorless..
Figure 1: A photograph demonstrating each agar ‘cell’ in consecutive order (A-E) during the experiment. Discussion
Based on the data acquired from the experiment, it can be evaluated that Cell E took the shortest time to fully diffuse and turn colorless at an exceptional time of 1.34 minutes.
The agar ‘cell’ that took the longest time to diffuse was Cell A, with a time of 27.38 minutes.
It can be state that Cell A took the shortest to diffuse due to its exceedingly high surface area to volume ratio of 10, meaning that the shorter distance allowed the agar block to lose its pink pigmentation as the HCL reached it before the others. The Cell’s high SA:V ratio meant that it could diffuse the substance more rapidly and efficiently than the other agar jelly cubes. Cell A lost its pink tint over an extended period of time due to its low SA:V ratio of 3, meaning the particles took longer to diffuse from the areas of high concentration to low concentration.
Through the data it may also be discerned that Cell B, though only having a 4cm2 difference of surface area between Cell A, took a significantly less time to diffuse; 12.07 minutes in comparison the Cell A’s 27.38 minutes. This result was somewhat unexpected as the cube was still to some extent smaller than Cube A, however the overall pattern seems to make complete sense according to the predicted outcomes of the experiment, and its relation to diffusion and surface area to volume ratios.
According to the results from the investigation, the method appears to be extremely valid for the experiment. The experiment was conducted exactly as it was expected to, with the results being a strong reflection on the knowledge previously discussed in the paper. The results were expected according to the SA:V ratio relation to the diffusion of substances. Nevertheless, there was one anomalous outcome of the experiment, relating to time taken for Cell B and Cell C to diffuse.
Cell B had a surface area of 20cm2 and a volume of 4cm3 leading to a surface area to volume ratio of 5. Likewise, Cell C also had a SA:V ratio of 5. Based on these ratios, it would be expected that both cubes would result in the same outcomes relating the time taken for each to diffuse. However, this was not the case, according to the results obtained.
Cell C took 8.42 minutes to completely diffuse whereas Cell B took 12.07 minutes, showing a clear difference between the two times as Cell B took an extra 3.65 minutes to diffuse. This result was unexpected as both agar ‘cells’ share the same SA:V ratio of 5. This error may have been most likely caused by imprecise measuring of the agar jelly cubes.
An improvement that would benefit the investigation and allow results to be more accurate would be to ensure each cube was more accurately measured using the ruler, in order to eliminate irregular results, such as the peculiar difference between Cell B and Cell C’s diffusion time periods in relation to their identical surface area to volume ratio. Perhaps, the diffusion time for Cell B and Cell C would equalize after being more precise. Moreover, another improvement to the method would be to ensure all timings were conducted with precision. There would need to be a lab assistant monitoring the agar cubes at all time to ensure results are recorded as soon as the particles completely go translucent. By being precise with the timing, the reliability of the data from the experiment increases as results become more accurate than those that are not properly timed. However, the group took great care with timing throughout the experiment, but a bit more precision could improve the overall results.
A proposed extension to the experiment is to investigate the surface area to volume relationship within a wide range of shapes, such as triangular-based prisms and spheres. Additionally, more cube sizes could be incorporated to provide us with a bigger picture on the relationship between the SA:V ratio and diffusion, allowing us to explore the results even more.
The reason why single-celled organisms can never become very big can be obtained using the results from the experiment, and some background knowledge. As cells increase in size, their volume increases faster their surface area, leading to a smaller surface area to volume ratio for the cell. This means there is less surface area available to suit the cell’s increasing needs. Single celled organisms do not possess transport mechanisms that humans have, such as the circulatory system. The cells depend on the constant exchange of these substances within their environment and hence may not receive enough of the life-giving substances and therefore die. These conclusions are evident within the results of the experiment, as it can be seen how the SA:V ratio decreases from Cell E to Cell A. The diffusion distance is increased and is not efficient enough to provide the cell’s with their increasing needs due to their increasing volume. Thus, there is a clear limit as to how large single cells organisms can be, as the volume (if it keeps increasing at a constant rate) will be even larger than the surface area (4).
As cells become too large, the surface area to volume ratio increases at a fast rate. Therefore, cells needs to either divide into smaller-sized cells or worst case, die (5). The decreased SA:V ratio results in an increased diffusion distance becoming too lengthy for the cell, causing larger cells to die. The surface area is not large enough to exchange the life-giving substances in and out of the cells. We can see an example of a small SA:V ratio in Cell A; 3. The model ‘cell’ in this example demonstrates how the smaller ratio decreases the rate of diffusion.
Conclusion
Based on the data obtained from the experiment, it can be concluded that the surface area to volume ratio is correlated to the rate of diffusion, as organisms with a higher surface area to volume ratio diffuse substances much faster than organisms with a low surface area to volume ratio, due to the shorter diffusion distance. It is evident that my hypothesis is supported by the reliable data provided in the experiment because I predicted which block would diffuse first and why, with detail.
Bibliography (will be edited with correct referencing)
http://www.biology-online.org/dictionary/Diffusion
11th march 2017
https://en.wikipedia.org/wiki/African_bush_elephant
11th march 2017
http://morningsidemicro.wikidot.com/why-cells-are-so-small
11th march 2017
http://science.blurtit.com/297837/why-is-single-celled-organism-have-a-limit-to-the-size-they-can-become
https://www.quora.com/What-happens-when-a-cell-becomes-larger