DIFFUSION LAB
PURPOSE: Investigating the relationship between diffusion and cell size.
QUESTIONS: - What is the relationship between the surface area and the volume of a cell?
- How does this relationship affect the rate of diffusion of materials that pass
in and out of the cell?
INTRODUCTION:
When cells grow to a certain size, their rate of growth slows down until they stop growing entirely. They have reached their size limit. When one of these cells divides into two smaller cells, the rate of growth increases again. In this investigation you will explore one of the factors which limits cell size and growth rate.
Materials needed for cell activity and growth must in some way gain entrance into the cell, and wastes must be eliminated from the cell. It would seem reasonable that the larger the cell the greater the surface area for the passage of materials into and out of the cell. But, on the other hand, the bigger the cell the greater its volume. This greater volume increases the cell’s needs for materials and these materials must pass through the cell’s surface.
MATERIALS NEEDED:
4 cubes of 3% agar-phenolphthalein with sodium hydroxide
0.1cm, 1 cm, 2 cm, and 3 cm on a side
millimeter ruler
100 ml of vinegar
250 ml beaker
plastic spoon
razor blade
paper towel
PROCEDURE:
-Place the 4 agar cubes in the beaker and pour in the vinegar until the cubes are submerged.
-Record the time
-Turn the blocks frequently for the next 8 minutes.
* While you are waiting for the 8 minutes to pass, do the calculations necessary to
fill in TABLE 1.
Surface area of a cube = length x width x number of surfaces
Volume of a cube = length x width x height
Ratio of surface area to volume =
The ratio of surface area to volume may also be written “surface area: volume”. To do this, divide both terms by the second term.
Example: The surface area of a cube is calculated to be 24 cm2.
The volume is calculated to be 8 cm3.
The ratio of surface area to volume is or or 3 : 1
-After 8 minutes, use the spoon to remove the agar blocks from the vinegar.
-Place them on a paper towel and blot then dry.
-RECORD the following data and calculations in TABLE 2
-Measure the depth of the uncolored area in centimeters. This is the extent of diffusion.
-Measure the distance across the colored area.
-Calculate the surface area of the central colored part of each block in square centimeters.
-Express these measurements in simplest ratios as you did in TABLE 1
DATA AND CALCULATIONS:
TABLE 1 Copy into lab manual. Surface area to volume ratios of agar blocks.
CUBE DIMENSION
/ SURFACE AREA / VOLUME / SIMPLEST RATIO3 cm / side
2 cm / side
1 cm / side
0.1 cm / side
TABLE 2 Copy into lab manual.
UNCOLORED AREA /COLORED AREA
CUBE DIMENSION /DEPTH
/ DISTANCE ACRSS / SURFACE AREA /VOLUME
/ SIMPLEST RATIO3 cm / side
2 cm / side
1 cm / side
0.1 cm / side
DISCUSSION:
- Compare the ratios in TABLE 2 to the ones derived from the total area and volume in TABLE 1.
- List the agar cubes in order of side size, from largest to smallest. List them in order of surface area-to-volume ratio, from largest to smallest.
How do these two lists compare?
- Calculate the surface area-to-volume ratio for a cube 0.01 cm/side.
- Which has the greatest surface area, a cube 3 cm on a side or a microscopic cube the size of an onion skin cell?
Which has the greatest surface area in proportion to its volume, a cube 3 cm on a side or a microscopic cube the size of an onion skin cell?
- What evidence is there that vinegar diffuses into an agar block?
- If the agar blocks were living cells and the vinegar were a vital substance, which block would have the most efficient ration of surface area to volume?
- What happens to the surface area-to-volume ratio of a cell as the cell grows?
- When one cube-shaped cell divides into two equal parts, how does the volume of each small cell compare with the volume of one large cell?
Does the surface area change in the same proportion? Explain
- Propose a hypothesis to answer the following questions:
Why does the growth rate of a cell slow down as it gets larger?
How does division affect the cell’s ability to absorb material for growth?
As the cell______, the ratio of ______to
volume______. Thus the cell has proportionately ______
surface area through which materials may ______into the cell and
wastes may ______out of the cell.
Cell division increases the surface area in proportion to the volume of the cells.
Therefore, the cell’s ability to absorb material needed for growth is ______.
Surface area of a cube = length x width x number of surfaces
Volume of a cube = length x width x height
Ratio of surface area to volume =
The ratio of surface area to volume may also be written “surface area: volume”. To do this, divide both terms by the second term.
Example: The surface area of a cube is calculated to be 24 cm2.
The volume is calculated to be 8 cm3.
The ratio of surface area to volume is or or 3 : 1