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.

Rameia's Cell Diffusion Argumentation

Cell Diffusion Argumentation

Rameia Ramsey

   In this experiment, my partners and I were challenged with how the size of a cell effects the rate of diffusion within it.  We were given two possible explanations and asked to choose one and use our experiment to confirm why the statement we chose was the correct explanation and also use our experiment to refute the other explanation.  Therefore, cells are small because cells that have a larger surface area to volume ratio are more efficient at diffusing essential nutrients.

    In order to prove this explanation, we cut out rectangular prisms of agar to represent our cells.  We cut out two small rectangular prisms and two large rectangular prisms of agar.  The small rectangular prisms had the dimensions of 2cm for the length, 1.5 cm for the width, and 1.5 cm for the height.  The large rectangular prisms had the dimensions of 4cm for the length, 1.5cm for the width, and 2.5cm for the height.  The volume of the small rectangular prisms was 4.5cm³ and the surface area was 16.5cm^2.  The volume of the larger rectangular prisms was 15cm³ and the surface area was 39.5cm^2.  After measuring and calculating all the dimensions of the prisms we then submerged them into a container of weak acid, in this case our weak acid was vinegar.  In the making of the agar to construct the cells a chemical called Phenolphthalein was added.  This chemicals allows us to see the diffusion of the vinegar into the prism as time went on.  The prisms soaked in the vinegar for a total of 27 minutes.  After taking the cells out of the container we measured how much of the blue colored section of the agar remained.  In the small cells what was left was 1cm in length, 0.5cm in height, and 0.5cm in width.  In the larger prisms the blue that remained was a length of 3cm, width of 1cm, and a height of 0.5cm.  The new volume of the small prism would now be 0.25cm³ and the surface area would be 2.5cm^2.  The volume of the large prisms would now be 1.5cm³ and the surface area is now 10cm^2.  By using these volumes of  the cells we see that in the small cell about 5.6% of the cell did not diffuse while about 94.4% of the cell did diffuse.  In the larger cell we see that 10% of the cell did not diffuse and 90% did diffuse.  As shown in the numbers the larger cell diffused less than the smaller cells which supports our explanation.  The smaller cells were able to diffuse more because they have a larger surface area to volume ration which was a ratio of 3.67 while the larger cells had a smaller surface area to volume ratio of 2.63, in turn making the diffusion in smaller cells faster.

    Our evidence refutes explanation two which states that the rate of cell diffusion is related to cell size.  Nutrients diffuse as a faster rate through small cells than they do through large cells.  In order to calculate the rate of diffusion you divide distance of diffusion by the time it took to diffuse.  The large cells and small cells had the same distance of diffusion which was 0.5cm, and when divided by the total time of submergence which is 27 minutes the rate of diffusion for all the cells was 0.019.  If all of the cells have the same rate of diffusion then the size of a cell can not be attributed to how it diffuses materials.  In conclusion,  cells are their small size because having a larger surface area to volume ratio is more beneficial to diffusing nutrients.  

 The cells made of agar placed into the container with weak acid(vinegar)

 Side view of the cells in the container with the weak acid

The cells after being submerged for 27 minutes.  The yellow parts of the cells show where diffusion has occurred. The blue part shows where diffusion has not reached

Rana's Cell Size and Diffusion Argumentation

Rana Srouji

AP Biology

Mr. Hammer

December 1, 2014

Cell Size and Diffusion Argument

Diffusion is the movement of molecules from a high concentration to a low concentration.  Almost all living organisms have cells that are dependent on diffusion to obtain the essential nutrients needed to survive.  Once the cells take in these nutrients, they break them down to create more components for the cell.  The cell size will then increase, but cells always remain small.  An explanation for this is, cells that have a larger surface area to volume ratio are more efficient at diffusing essential nutrients.

My lab partners and I proved this explanation by creating and testing models of cells using agar.  Originally, the agar was blue in color.  We created four model cells of agar, two of which represented small cells while the other two represented larger cells.  The cells were cut into rectangular prisms for easy calculations.  The small cells were cut with length of 2 cm, width of 1.5 cm, and height of 1.5 cm.  Their volumes were 4.5 cm cubed, and their surface areas were 16.5 cm squared.  The larger cells were cut with length of 4 cm, width of 1.5 cm, and height of 2.5 cm.  The volume of the large cells were 15 cm cubed and their surface area were 39.5 cm squared.  The surface area to volume ratio for the small cells is 3.67 while the ratio for the large cells is 2.63.  Afterwards, all of the cells were then placed into a plastic container filled with vinegar.   All of the cells were placed in the vinegar for 27 minutes.  Within seconds the acid began to diffuse into the agar.  Once the cells were taken out of the vinegar, we took the measurements of the blue colored agar remaining, which is the amount that had not been diffused.  The remaining blue agar in the small cells had a length of 1 cm, width of 0.5 cm, a height of 0.5 cm, and a volume of 0.25 cm cubed.  The amount remaining in the large cells had a length of 3 cm, width of 1 cm, height of 0.5 cm, and volume of 1.5 cm cubed.  In addition, using the volumes, we found that of the small cells only 5.6% of each cell was not diffused within the 27 minutes, and 94.4% of each cell was diffused.  Of the larger cells, 10% of each cell was not diffused, while 90% was diffused.  

Additionally, we used the agar in our experiment because it is a gel-like substance that chemicals can diffuse through, and that can be cut into different shapes.  We created the model cells in two different sizes to show the difference in the amount diffused within the same amount of time, and the difference in the surface area to volume ratios.  It was to show that the diffusion was directly related to the ratio.  The larger cells have a smaller ratio, while the smaller cells have a larger ratio.  Also, we created two cells per size for both the small and the large cells in case any error were to occur.  Phenolphthalein was previously added to the agar before we used it.  The phenolphthalein is a chemical indicator that changes color when coming into contact with acid, so that the amount of acid diffused through the agar would be easily seen.  Once the cells were placed in vinegar, their color would change from blue to yellow because vinegar is an acid which will diffuse through the cell and change the color once it comes in contact with the phenolphthalein.  In conclusion, cells are small because they have a large surface area to volume ratio, which allows them to diffuse most efficiently.  The larger cells had a less amount diffused, only 90% of each large cell diffused, and a smaller surface area to volume ratio, (a ratio of 2.63).  Unlike the small cells which had a larger surface area to volume ratio, of 3.67, and diffused 4.4% more than the large cells, within the same amount of time.  Thus, the cells that have a larger surface area to volume ratio are more efficient at diffusing essential nutrients.  

However, the reasoning for why cells are so small cannot be because the rate of diffusion is related to cell size.  Nutrients do not diffuse at a faster rate through small cells than they do through large cells.  This is because, as my group tested the agar cells, we collected data that showed the distance of diffusion through the cells.  We gathered this data in order to find the rate of diffusion, since the rate of diffusion is the smallest distance of diffusion per cell divided by the amount of time diffused.  Both the small and large cells had the same distance of diffusion, which was 0.5 cm.  The rate of diffusion then is 0.5 divided by the 27 minutes, which then equals 0.019.  Thus, the rate of diffusion cannot be related to the cell size since both the small and large cells have the same rate of diffusion.  

This is an image of the "cells"/agar as soon as they were placed into the container of vinegar.  The acid began to diffuse into the cells as seen by the light yellow coloring on the outer edges of the agar.  

This is an image of the cells/agar after the 27 minutes, once they have been removed from the acid.  The blue inside is the amount that has not been diffused, the yellow coloration is the acid that diffused though the agar and has come into contact with the phenolphthalein.