The shape of a cell affects its surface area to volume ratio. This can be significant in the cell's ability to absorb nutrients or survive toxins. You are given that the volume of a sphere and cylinder are (4/3)pr3 and pr2h, respectively, where r is the radius and h is the height. The surface area for a sphere and a cylinder are 4pr2 and (2prh + 2pr2), respectively.
a. Complete the following table (from Biology 350), which examines cellular geometry.
Organism
|
Shape
|
Diam
mm |
Height
mm |
Volume
mm 3 |
Surface
mm 2 |
S.A.:Vol.
Ratio |
Mycoplasm
|
Sphere
|
0.3
|
--
|
|
|
|
Coccus
|
Sphere
|
1.5
|
--
|
|
|
|
E.
coli
|
Cylinder
|
0.75
|
4.0
|
|
|
|
Yeast
|
Cylinder
|
5.0
|
8.0
|
|
|
|
Diatom
|
Cylinder
|
20
|
60
|
|
|
|
b. Suppose that the Coccus bacteria and E. coli satisfy the discrete Malthusian growth equation
where P0 is the initial population and the doubling time for the population is 25 min. Find the value of k (to at least 4 significant figures) and write the general solution, then determine how long it takes for the total surface area of each of these growing populations to reach 1 m2. (Recall that 1 mm = 10-6 m.)
c. Assume the same population dynamics as given above. Determine how long it takes for each of the populations to grow to where their volumes occupy 1 cm3. (Recall that 1 cm = 10-2 m.)
d. Michael Crichton in the
Andromeda Strain
(1969) states that