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.
