Question

Imagine a spherical cell with a radius of 10 pm. What is the cells surface area in pm2? Its volume, in pm3? What is the ratio of surface area to volume for this cell? Now do the same calculations for a second cell, this one with a radius of 20 pm. Compare the surface-to-volume ratios of the two cells. How is this comparison significant to the functioning of cells? (Note. For a sphere of radius r, surface area = 4rcr2 and volume = rrr3. Remember that the value of it is 3.14.)

Answer 1

Answer:

comparing the two cells, we can see that the surface area to volume ratio gets smaller as the cell gets bigger, making it harder for optimum amount of material that is needed to sustain the cell to cross the cell membrane. If a cell should grow beyond a certain limit, enough material will not be able to cross the membrane to nourish the cellular volume.

Step-by-step explanation:

For the cell with radius r = 10 pm

The surface area is from the area of a sphere

Area A = $4\pi r^{2}$

A= 4 x 3.14 x $10^{2}$ = 1256 pm^2

volume is calculated from the volume of a sphere

Volume V = $\frac{4}{3}\pi r^3$

V = $\frac{4}{3}$ x 3.14 x $10^{3}$ = 4186.67 pm^3

Ratio of surface area to volume = 1256/4186.67 = 0.29

For the second cell with radius r = 20 pm

The surface area of the cell is

A =  $4\pi r^{2}$ = 4 x 3.14 x $20^2$ = 5024 pm^2

Volume =  $\frac{4}{3}\pi r^3$ =  $\frac{4}{3}$ x 3.14 x $20^{3}$ = 33493.3 pm^3

Ratio of surface area to volume = 5024/33493.3 = 0.15

comparing the two cells, we can see that the surface area to volume ratio gets smaller as the cell gets bigger, making it harder for optimum amount of material that is needed to sustain the cell to cross the cell membrane. If a cell should grow beyond a certain limit, enough material will not be able to cross the membrane to nourish the cellular volume.