Why Cells are Small:
Surface Area to Volume Ratios
Biology 11: In Class Activity
Objective: Today’s lab activity is about the size of cells. To understand this, we first
have to understand surface area to volume ratios.
Part I – Determining Surface Area and Volume:
Let’s start with surface area. Review the diagrams below (four cubes of varying size).
The surface area of the cube is calculated by finding the area (length x width) of one
side. But we want to know the whole surface. On a cube, there are 6 sides, so we
multiply that area x 6.
Surface area of a cube = length x width x 6
Volume is another simple calculation – we multiply length x width x height.
The one we’re most interested in though is the surface area to volume ratio. To find
this ratio, simply your surface to volume ratio so that the ratio is per one unit of volume.
Example of Calculating the Surface Area to Volume Ratio:
Surface Area
Volume
= 24 cm 2
= 6 cm 3
Ratio is 24: 6
Simplified Ratio is 4: 1
So for every unit of volume there is 4 units of surface area.
Using the diagrams below, calculate the surface area, the volume, and the surface
area to volume ratio. Fill out the table and answer the questions.
NAME: ____________________________________________
PERIOD: ___________
Surface Area to Volume Questions
1. What happened to the surface area as the size increased?
2. What happened to the volume as the size increased?
3. What happened to the surface area to volume ratio as the size increased?
4. Now imagine you are looking at a very large cell. Next to it is a very small cell.
Which one has a larger surface area to volume ratio?
Part II: Experimental Design and Cell Size
Now you are going to apply what you just learned about surface area to volume ratios.
Your objective is to figure out how this concept applies to biology. The cells you have
looked at in biology have all been with the use of a microscope: they have all been
very, very small. The majority of cells are microscopic. The question is: Why?
You have learned about membrane transport processes, such osmosis, simple
diffusion, and facilitated diffusion. This is how cells obtain almost everything they need:
nutrients, ions, water, etc, and how they get rid of wastes. Will diffusion and osmosis
be enough to support just any cell?
Questions
1. Do you think large cells and small cells carry out diffusion and osmosis at the
same rate? Why or why not?
2. If a cell has a high concentration of something, say, waste, that it wants to get rid
of, which do you predict will be able to get rid of the waste sooner – a smaller cell
or a large one? Write a formal hypothesis that relates cellular transport to size of
the cell.
Now, design an experiment to test your hypothesis from question #2.
Suppose you have large and small potatoes to model cells. Each potato was heated to
the same temperature. Just like in cellular processes, heat diffuses out of areas of high
concentration to areas of low concentration, which makes it a great demonstration for
testing questions like this, since cells are very hard to work with.
If you have a large hot potato and a small hot potato, and two thermometers, how can
you test your hypothesis that one potato (whichever you predicted above) will lose heat
faster than the other? What will you measure and how will you measure it? Describe
your experimental design briefly below, using 1‐2 complete sentences.
Questions
3. Which potato cooled off faster?
4. If the potato was a cell, and the heat was waste the cell was trying to get rid of,
which cell do you think would have an advantage?
5. How might surface area to volume ratios help decide the size of cells?
6. Based on what you have learned today, in what way may multicellular organisms
have an advantage over single‐celled ones?
7. Consider a mouse and an elephant. If both were left in the cold overnight, which
would be in more danger of freezing to death? Why?
PART III – Cells in Salt Solution
PURPOSE: To gain an appreciation of the physical concepts involved in the growth of a threedimensional object, like a cell, and to relate this to the small size of cells.
CHANGES IN CELL SIZE DUE TO SALT SOLUTIONS:
% SALT SOL'N
0
1
5
10
15
20
DIMENSIONS OF LIVING CELLS (m)
RATIO (SA:V)
LENGTH
WIDTH
HEIGHT S. AREA VOLUME
300
30
10
298
30
10
275
27
8.5
250
25
5
200
22
4
150
20
3
FROM THE ABOVE TABLE:
1. Determine the surface area of the cells at each concentration.
2. Determine the volume of the cells at each concentration.
3. Calculate the surface area to volume ratio for each concentration. Reduce the values to the lowest
terms (i.e., V=1)
4. Plot the points for salt concentration vs. surface area on a properly labeled graph ad salt
concentration vs. volume on another. Draw the lines of best fit.
5. Determine the slope of the line for each graph (rise over run).
ANALYSIS:
1. Why was it necessary to determine the slope of each graph in order to compare them?
2. (a) What do the slopes tell you about the change on volume as compared to the change on
surface area?
(b) What does the fact that they are negative slopes mean?
3. What do your ratios tell you about the change in volume vs. the change in surface area as a cell
size changes?
4. What do you suppose your findings have to do with why cells are small? (Think of the surface
area as the membrane and the volume as the cytoplasm.)