Name______________________________________ Date_______________
Lab 6 ME 5 HW: Why are biological cells only so big?
Have you ever wondered why biological cells happen to be around 0.01 mm in size? Why
can’t they be 0.10 mm or 1.0 mm? Some fundamental math is behind it. We’ll find that the
same reasoning explains some other phenomena like why cats usually survive long falls,
why elephants have extra-thick legs, and why hippos have to spend a lot of time in the
water. (Now is a good time to address this question, because we’ll use some of the
mathematical modeling ideas that came up in our study of acceleration and mass.)
A little biology first: A cell needs oxygen, which it takes in through its surface. Oxygen
diffuses into the cell at a rate proportional to the surface area of the cell. The rate at which a
cell uses oxygen is proportional to its volume.
So let’s reason out what happens as a cell gets bigger and bigger, like these spherical cells
with diameters of D0, 2D0, and 3D0:
3D0
2D0
D0
1. Calculate the surface area of each the cells (in terms of D0) and fill in the table. In the last
column, enter the surface area in terms of the surface area of a sphere of diameter D0. For
example, if it’s 6 times more, enter 6.
Diameter D
D0
Surface area A
Surface area in terms of surface area of sphere of diameter D0
1
2D0
3D0
2. So for a cell of double the size (diameter), what can you say about the cell’s capacity to
take in oxygen? (It increases by a factor of ______ .) And for a cell of triple the size, the cell’s
capacity to take in oxygen increases by a factor of ______ .
Note: As the formula for the surface area of a sphere suggests, and your calculations
support, we can say in general that the surface area of a sphere increases as the square of
the size (diameter).
3. Calculate the volume of each of the cells (in terms of D0) and fill in the table. In the last
column, enter the volume in terms of the volume of the sphere of diameter D0. For example, if
it’s 12 times more, enter 12.
Diameter D
D0
Volume V
Volume in terms of volume of sphere of diameter D0
1
2D0
3D0
4. So for a cell of double the size (diameter), what can you say about the cell’s need for
oxygen? (It increases by a factor of ______ .) And for a cell of triple the size, the cell’s need for
oxygen increases by a factor of ______ .
Note: As you might see from the formula for the volume of a sphere, we can say in general
that the volume of a sphere increases as the cube of the size (diameter).
5. Examine your responses to (2) and (4), and explain in your own words the trouble that
arises as cells get bigger and bigger.
6. It’s useful to form the ratio A/V, the “surface area to volume ratio.” Calculate that ratio for
each of the cells (in terms of D0) and fill in the table. In the last column, enter the surface
area to volume ratio in terms of that ratio for the sphere of diameter D0. For example, if it’s
half as much as for the sphere of diameter D0, enter 1/2.
Diameter D
D0
Surface area to
volume ratio A/V
Surface area to volume ratio, in terms of Surface area to
volume ratio of sphere of diameter D0
1
2D0
3D0
7. In general, what functional form describes the relationship between A/V and D? Sketch a
qualitatively correct graph of A/V vs. D.
8. For a cell that’s 10 times the size (10D0), by what factor does the surface area increase?
By what factor does the volume increase?
By what factor does the surface area to volume ratio change?
So, for a cell that’s 10 times the size, there’s only _______ as much surface area to take in
oxygen associated with each unit of volume that requires it.
9. To be concrete, let D = 10 µm. Calculate the surface area, the volume, and the surface area
to volume ratio. (It’s OK to keep µm as your unit.) Interpret the result of your calculation
for the surface area to volume ratio in your own words. (Recall how we interpreted the
result of a calculation like 20 m/4 s = 5 m/s: “5 m is the distance traveled in each second.”)
Repeat for a cell for which D = 100 µm.