Joshua's Cell Size and Diffusion Argument

Joshua Everett

AP Biology

Mr. Hammer

December 1, 2014

Cell Size and Diffusion Argumentation

Diffusion is the movement of molecules from high concentration to low concentration. This process is extremely important because it helps living things maintain homeostasis in which nutrients are brought into cells while waste flows out of them. Depending on this, cells will continue to shrink and swell in size. The size of these cells are a great factor in how many molecules move in and out of the cell and at what speed. Many scientists today have wondered why cells are so particularly small in organisms. In this lab activity, my lab partners and I had the responsibility for choosing an explanation to why cells are so small. Explanation 1 stated that cells that have a larger surface area to volume ratio are more efficient at diffusing essential nutrients. Explanation 2 stated that the rate of diffusion is related to cell size in which nutrients diffuse at a faster rate through small cells than they do through large cells. After serious analyzation of the results of this activity the explanation that I decided to argue for explanation two.

My lab partners were successful in proving this statement through an activity performed using agar to construct model cells. Agar is a gel-like substance that is easy to cut into variety of shapes. Agar, in this case, is blue and contains phenolphthalein so when it comes into contact with an acid, it changes to a color similar to being clear. Because agar has properties in which chemicals are able to diffuse through it, I am able to see how far an acid diffuses into my model cell.

Continuing on, I cut the agar into four different rectangular prisms that have different dimensions, two being relatively big and two being relatively small. After we measured out the dimensions of the prisms we proceeded by placing the prisms into a plastic container with a weak acidic solution, called vinegar. As soon as we placed the pieces into the vinegar, a stopwatch was started to record the how long it will take the diffusion process to start and how long the process will take for the model cell to become completely clear. At approximately 7.3 seconds, diffusion and a color change was evident. The prisms were left in the vinegar for as long as time permitted and were taken out at approximately 32.9 minutes. The prisms were in the process of diffusing so in the middle of each of the cells was a leftover blue prism in the center. Anticipating that the dimensions of the inner prism would be useful, we recorded the measurements in data tables.

This is a table that depicts the dimensions of the prisms before diffusion began. 

This is a table that depicts the dimensions of the inner blue prisms after diffusion was stopped.  

After the experiment was completed, I calculated the surface area and the volume for each of the four original blue prisms and each of the inner blue prisms after diffusion was stopped. The surface area equation is as follows: 2(WL+HL+WH) and the volume equation is length times width times height. 

This is a table depicting the surface area for the prisms before the diffusion started.

This is a table depicting the surface area for the inner blue prisms after diffusion was stopped.

This is a table depicting the volumes of the original prisms before diffusion. 

This is a table depicting the volumes of the inner prisms after diffusion was stopped. 

From these two calculations for both sets of four prisms, I was able to form surface area to volume ratios. For the original prisms, prism 1 had a ratio of 10 : 3 (3.33), prism 2 had one of 48.3 : 20.825 (2.32), prism three had one of 46.9 : 19.6 (2.39), and prism 4 had a ratio of 18.6 : 5.415 (3.43).

After I calculated these ratios, I proceeded to calculate the diffusion rates for these cubes. In order to develop the equations for diffusion rates, visualization was key. With the help of Mr. Hammer, my group members and I were able to use equations to find the rate of diffusion for each side of each prism.

This is the diagrams that Mr. Hammer drew for my group in order to see a better view of  what each equations was finding. 

Towards the bottom of the picture, there are three equations that were used for all four prisms. In each equation, the dimensions of the original prisms and the inner prisms are utilized. Subtraction of the either the length, width, or height is to find the distance inner prism to the outer prism depending on the dimensions being dealt with. Dividing by two is necessary because by doing that, the true distance on either side of the prism is found. Overall the process to find the distance between each side of each prism is shown below:

How to Calculate Cube Diffusion Distances

side one= big length - small length/ 2

side two= big width - small width/ 2

side three= big height - small height/ 2  

*repeat for all prisms*

This is a data table depicting all the side distances between each original prism and its inner prism.

Once all the distances of each side was found, the averages of the distances of the sides for each prism were calculated. For prism 1 is was 0.5, for prism two it was 0.53, for prism 3 it was 0.53, and for prism 4 it was 0.57. These averages were then used to calculate the average rates of diffusion for each of the cubes. The time that diffusion stopped was at 32 minutes and 9 seconds which is actual 1,929 seconds. I divided the average of each prism by 1,929 to determine the average rate. The average diffusion rates for each of the prisms is shown below:

This is a data table depicting the average diffusion rates for each of the cubes.

After analyzing my results, I saw that there was a direct relationship with the surface area to volume ratio with the diffusion rate within the original prisms. As the surface area to volume ratio increases, the diffusion rate increases as well, for example, prism 2 and 3 have a surface to volume ratio of 2.32 and 2.39 respectively. The diffusion rate for both prisms is 0.000276 cm/sec. In addition, prism 2 had had 15% of its cell not diffused with vinegar and prism 3 have 23% of its cell not diffused by vinegar. Prism 4 has the highest ratio of 3.43 and has the highest diffusion rate of .000294 cm/sec. Also, it only had 4.2% of its cell not diffused by vinegar. This means that about 96% of the cell was diffused with vinegar which is higher than prisms 2 and 3 which says that prism 4 had a higher diffusion rate than prism 2 and 3 because they only had 85% and 77% of its cell diffused respectively. There is an increase in the diffusion rates as the surface area to volume ratios increase. This means that as the diffusion rates increases, the efficiency of the cell increase as well being it is able to diffuse molecules into or out of the cell faster. A cell having a larger surface area to its volume is highly beneficial not just because it will have a higher diffusion, but its the perks of having a high diffusion rate. With a higher diffusion rate, more nutrients are able to enter the cell while wastes are leaving the cell. A larger surface area lead to a higher diffusion rate because there is more cell membrane that is semi-permeable in which it allows molecules in and out of the cell.This property makes a cell highly efficient because it is able to maintain a cell’s homeostasis in which more nutrients are entering the cell more quickly while also removing unwanted waste from a cell in a timely fashion. An increase metabolism for the cell is extremely important for a cell to continue performing its function within living organisms. Overall, explanation 1 is most valid.

Explanation two states that the rate of diffusion is related to cell size in which nutrients diffuse at a faster rate through small cells than they do through large cells. This is not an acceptable statement because it is not fully supported by the data collected. prism 1, in particular, fits the patterns of that there is an increase in the surface area to volume ratios however, its diffusion rate does fit within the pattern. We would expect the diffusion rate of prism 1 to be somewhere between 0.000276 and 0.000294 but instead its rate it 0.000259. Prism one is considered to be one of the two smaller prisms but its diffusion rate makes it seem that the prism is an extremely large cell when it is not. This finding led to me to conclude that explanation two is not true in all extents. On the other hand, there could have been a human error in the process of handling prism one in the experiment. Overall, explanation two is not acceptable or valid.