Chigozie's Cell Size and Diffusion Argumentation

Cell Size and Diffusion Argumentation

       Almost all living cells depend on the process of diffusion to obtain essential nutrients that are needed for survival. As these nutrients are taken in by the cells, they are broken down. The resulting energy and molecular building blocks from the nutrients are used to make more cellular components. Because of this, a cell would grow by increasing in size. However, cells never get too big; they are always small. So, why are cells so small? In this lab, my lab group and I worked towards answering this one question. We were given two potential answers to this question. Explanation 1 stated, “Cells that have a larger surface area to volume ratio are more efficient at diffusing essential nutrients”. Explanation 2 stated, “The rate of diffusion is related to cell size. Nutrients diffuse at a faster rate through small cells than they do through large cells. After collecting and analyzing data with my lab group, I concluded that although our data seemed to support both explanations, Explanation 2 was more valid.

My lab group and I tested the validity of these different explanations by constructing a model cell using agar. Agar is a gel-like substance that we could easily manipulate and cut into a variety of shapes. In this lab, phenolphthalein (a chemical indicator that changes color when it comes in contact with an acid) was added to agar so that we could see how far the acid (in this case, vinegar) diffused into the model cell. The agar was blue and would turn a yellowish-clear color when the phenolphthalein in it came in contact with the vinegar. My group cut four different rectangular prisms from the agar. Two were relatively small, and two were relatively big. We measured the dimensions of the cubes and then placed them in a plastic container filled with vinegar.

Rectangular Prism Dimensions Before Diffusion

Rectangular Prism	Length	Width	Height
1 (Small)	2 cm.	2 cm.	1.5 cm.
2 (Big)	3.5 cm.	3.5 cm.	1.7 cm.
3 (Big)	3.5 cm.	3.5 cm.	1.6 cm.
4 (Small)	1.9 cm.	1.9 cm.	1.5 cm.

Rectangular Prisms of Agar before diffusion

 

We started timing the diffusion process with a stopwatch as soon as the cubes were placed in the vinegar. The cells began to change color (from blue to yellowish-clear) at approximately 7.3 seconds. We left the rectangular prisms in the vinegar for as long as time permitted us, which was 32 minutes and 9 seconds. At this time, we pulled out the rectangular prisms. Because of this time restriction, the vinegar was not able to diffuse all the way through the rectangular prism and we had to measure the dimensions of the part of the rectangular prism that the vinegar had not touched.

Dimensions of the Rectangular Prism Not Touched by Vinegar

Rectangular Prism	Length	Width	Height
1 (Small)	1 cm.	0.8 cm.	0.7 cm.
2 (Big)	2.5 cm.	2.5 cm.	0.5 cm.
3 (Big)	2 cm.	2.5 cm.	0.9 cm.
4 (Small)	0.9 cm.	0.5 cm.	0.5 cm.

Rectangular Prism after diffusion

 

            Afterwards, my lab group and I calculated the diffusion rates. Instead of just averaging the dimensions (length, width, height) and putting that average distance over time to find the rate, we had to find the dimensions of the space that had been touched by vinegar since the vinegar did not diffuse all the way through the rectangular prism. Mr. Hammer helped us greatly with the visualization and explanation displayed below. In order to find the dimensions of this space, we had to take the dimensions of the rectangular prism not touched by the vinegar, subtract it from its respective dimension of the original rectangular prism, and divide by two. For example, if the original length was 14 and the length of the rectangular prism not touched by vinegar was 5, we would do 14-5, which is 9, and divide by 2 and get 4.5 as one of the new dimensions that we will use to find the rate of diffusion.

How to Calculate the Sides Needed for the Calculation of the Rates of Diffusion

Side one= (original length minus non-diffused cube length) ∕2

Side two= (original width minus non-diffused rectangular prism width) ∕2

Side three= (original height minus non-diffused rectangular prism height) ∕2

Mr. Hammer's very useful diagram

 

Results of Calculations of Sides Used for Rates of Diffusion Calculations

Rectangular Prism	Side One	Side Two	Side Three
1 (Small)	0.5 cm.	0.6 cm.	0.4 cm.
2 (Big)	0.5 cm.	0.5 cm.	0.6 cm.
3 (Big)	0.75 cm.	0.5 cm.	0.35 cm.
4 (Small)	0.5 cm.	0.7 cm.	0.5 cm.

           

To find the rate of diffusion, we had to divide total distance travelled by the vinegar into the rectangular prism by the total amount of time the rectangular prisms spent in the vinegar. First, we had to average out the dimensions of in the table above for each rectangular prism (sum of length, width, and height divided by three). We then converted the total time, 32 minutes and 9 seconds, into seconds and got 1929 seconds in total. We divided each average dimension by 1929 seconds, the total time, to get the rate of diffusion for each rectangular prism.

Rates of Diffusion for Each Rectangular Prism

Rectangular Prism	Average Dimension	Rate of Diffusion
1 (Small)	.5 cm.	.000259 cm/sec.
2 (Big)	.53 cm.	.000276 cm/sec.
3 (Big)	.53 cm.	.000276 cm/sec.
4 (Small)	.57 cm.	.000294 cm/sec.

           

My lab group and I also calculated the surface area, volume, and ratio of surface area to volume.

All Calculations

Rectangular Prism	Surface Area (SA)	Volume (V)	Ratio of SA to V	Decimal Representation of SA to V Ratio	Rate of Diffusion
1 (Small)	20 cm.2	6 cm.3	10 : 3	3.33	.000259cm/sec.
2 (Big)	48.3 cm.2	20.825 cm.3	48.3 : 20.825	2.32	.000276cm/sec.
3 (Big)	46.9 cm.2	19.6 cm.3	46.9 : 19.6	2.39	.000276cm/sec.
4 (Small)	18.6 cm.2	5.415 cm.3	18.6 : 5.415	3.43	.000294cm/sec.

 

After analyzing all of this data, I concluded that Explanation 2 was more valid. Again, Explanation 2 states, “The rate of diffusion is related to cell size. Nutrients diffuse at a faster rate through small cells than they do through large cells”. At first, it looked like the data that we collected supported both explanations. Explanation 1 states, “Cells that have a larger surface area to volume ratio are more efficient at diffusing essential nutrients”. For Explanation 1, the rectangular prism with the largest surface area to volume ratio also had the highest rate of diffusion. However, rectangular prism 1 refutes this statement because it has the second largest surface area to volume ratio but the lowest rate of diffusion. For Explanation 2, the smallest cell (rectangular prism 4) had the fastest rate of diffusion. However, the other small cell (rectangular prism 1) also had the slowest rate of diffusion, so that rectangular prism also disproved Explanation 2. Because of this, I decided that rectangular prism 1 should not be totally considered for this conclusion. It is the only rectangular prism that refutes both explanations. After taking all of this into consideration, I noticed that the surface area to volume ratios for rectangular prisms 2 and 3 are different, but their rates of diffusion are not. One would expect them to have different rates of diffusion since their surface area to volume ratio is different. However, they have the same rate of diffusion. Because of this key piece of evidence, Explanation 2 is more valid. Rectangular prism 2 and 3 are both the two pieces of agar that we made relatively big, so they support Explanation 2 in that the bigger cells have the same rate of diffusion and also a lower rate of diffusion than that of the smallest cell.

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.