Jazmean's Cell Size and Diffusion Argumentation

Cell Size and Diffusion Argumentation

Diffusion is the random motion of atoms or molecules from an area of high concentration to an area of low concentration. Diffusion is a mode of passive transport for cells because no energy is required for it to happen. Cells use diffusion in order to move atoms or molecules in and out of the cell. Through this process, a cell can expel waste or gain vital nutrients and at the same time, grow or shrink in size. Despite this, cells are never bigger than a grain a salt (which is 0.5 mm), but can be as small as 6.5 nano meters (which is the size of hemoglobin). So this begs the question, "Why are cells so small?". This is what my group members and I investigated in class. We were presented with two explanations as to why cells are small.  Explanation 1 stated that cells that have a larger surface area to volume ratio are more efficient at diffusion essential nutrients. Explanation 2 stated that the rate of diffusion is related to cell size and that nutrients diffuse at a faster rate through small cells than they do through large cells. Due to the data that was collected, there is no clear explanation that is the most valid or acceptable. 

Our group had the follow materials available for us to use: Agar (a green/blue gelatinous substance that is obtained from algae), weak acid (vinegar), beakers, a stopwatch, rulers, and plastic knives. The agar was in a container. We took the agar from the container and cut the agar into cube-like shapes (in reality they are rectangular prisms).  

Displayed are the measurements of the cubes: 

Cubes (Before Diffusion) (Outer Cube)	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

We then placed the cubes into a container filled with vinegar at the same time. When the cubes hit the vinegar, the stopwatch started to keep time. At 7.3 seconds, the color started to change. Due to time restrictions, we were not able to have our cubes be fully diffused. At 32 minutes and 9 seconds, we stopped the experiment and pulled the cubes out.

Because of the time restrictions and our cubes not being fully diffused, we had to take measurements of the cube not diffused yet. 

Not Diffused Cube Measurements (inner cube)	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

We then had to calculate the diffusion rates. We had to calculate the distance of the length, width, and height out each outer and inner cube by subtraction the big length from the small length and dividing that number by
two. 

Explanation given to my group by Mr. Hammer

How to Calculate Cube Diffusion Rates (Repeat for all Cubes)

side one= big length - small length/ 2

side two= big width - small width/ 2

side three= big height - small height/ 2  

Displayed below is the results.

Cubes	Length	Width	Height
1	0.5	0.6	0.4
2	0.5	0.5	0.6
3	0.75	0.5	0.35
4	0.5	0.7	0.5

We then averaged the length, width, and height in order to make the process easier. And to find the rate, we took the averages and divided it by the time the cubes stayed in the vinegar. The time was changed from minutes into seconds. So instead of it being 32.9 minutes, the time was changed to 1,929 seconds. 

Averaged Diffusion Rates

	Rate of Diffusion
Cube 1	.000259 cm/sec.
Cube 2	.000276 cm/sec.
Cube 3	.000276 cm/sec.
Cube 4	.000294 cm/sec.

In addition to calculating the rates of diffusion of each cube, we also calculated the surface area and volume of each cube. If you compile all the data into one chart, this is what it looks like:

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

As you can see by our final data table, we have conflicting data. For Explanation 1 to be the most valid, the cubes with a larger surface area to volume ratio should have a faster rate of diffusion. This would be true if Cube 1 was not apart of the data table. Cube 1 has a surface area to volume ratio of 10 : 3 and has a rate of diffusion of .000259 cm/sec. However, Cube 2 and 3 have a ratio of 48.3 to 20.825 and 46.9 to 19.6 (respectively) and a rate of diffusion of .000276 cm/sec. Cube 2 and 3's ratios are smaller than Cube 1's but their rate of diffusion is faster than that of Cube 1's. This means that Cube 2 and 3 would be more efficient in diffusion nutrients to the center of the cell than Cube 1. Despite this, Cube 2 and 3 have a smaller surface area to volume ratio and a slower rate of diffusion compared to Cube 4. Cube 2, 3, and 4 support Explanation 1. 

For Explanation 2 to be the most valid, the smaller cubes would have a faster diffusion rate compared to the larger cubes. Cube 4 is the one of the small cubes and it does have a faster diffusion rate compared to the larger cubes. However, Cube 1 is also one of the small cubes but it does not have a faster diffusion rate compared to the larger cubes. 

If you were to ignore Cube 1, Explanation 2 would be the most valid explanation. Cube 3 has a larger surface area to volume ratio, but has the same rate of diffusion has Cube 2. This would disprove Explanation 1. However, because the difference between the ratio is not drastic, one could infer that the small difference would not effect the rate of diffusion drastically either. Because of the conflicting data, I would not feel right as a scientist to call for one side. Further trials need to be conducted in order to solidify the data in order to pinpoint which Explanation is the most acceptable. 

Further Thoughts

These explanations can go hand in hand if you think about it. If you increase the surface area to volume ratio of a cube for example, the size of the cube gets decreases. This is the same for a cell. If you increase the SA to V ratio of a cell, its size decreases. A large SA to V ratio limits how big a cell can get. If a large SA to V ratio makes diffusion more efficient for a cell, it might make sense as to why people think smaller cells diffuse faster than larger cells; it's because their surface area to volume ratio is large.