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Transcript
Agar Lab The Relationship Between Cell Size and Diffusion 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 larger cells divides into two smaller cells, the rate of growth again increases. In this investigation, you will explore one of the factors that limits cell size and growth rate. Materials needed for cell activity and growth must in some way gain entrance into the cell, and at the same time wastes must be eliminated from the cell. It would seem reasonable then, 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 is its volume. The greater volume increases the cell’s needs for materials that must pass through the surface of the cell. 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? MATERIALS: 3 cubes of 3% agar-phenolphthalein (1 cm3, 2 cm3, and 3 cm3) millimeter ruler 75 ml of .1M NaOH 250 ml beaker plastic spoon plastic knife PROCEDURE: 1. Using a plastic knife, trim the agar blocks to make three cubes- 3cm, 2cm, and 1 cm on a side. 2. Place the three agar cubes in the beaker and pour in the .1M NaOH solution until the blocks are submerged. Record the time. Very gently turn the blocks frequently for the next 10 minutes. After 10 minutes, remove the agar blocks to a paper towel with the spoon and blot them dry. 3. Record your hypothesis as to which cube will show a greater amount of diffusion. 4. While you are waiting for the ten minutes to pass, do the calculations necessary to fill in the table on the back of this lab sheet. Surface area (S.A.) of cube = length x width x number of surfaces Volume of cube = length x width x height Ratio of S.A. to volume = surface area/volume The ratio may also be written “surface area: volume”. The ratio must be expressed in its simplest form. To do this, divide both terms by the second term. Ex. The surface area of a cube is calculated to be 24 cm 2. The volume is calculated to be 8 cm3. The ratio of surface are to volume would be: 24/8 or 3/1 or 3:1 5. After 10 minutes, remove the agar blocks from the sodium hydroxide and blot them dry. Then use the razor blade to slice each block in half. Dry the razor blade/plastic knife thoroughly after slicing each so that the NaOH will not be smeared on the surface of the next block. DO SO VERY CAREFULLY! Notice the depth of the colored area. This is the actual extent of diffusion. Measure the distance across the uncolored area with a ruler (do not round off these measurements to the nearest centimeter) Determine the volume of the central uncolored part of each block in cubic centimeters. Calculate the % diffusion by using the formula: Final volume/initial x 100 DATA: CUBE Surface Area of Cube Volume of Cube S.A.: Volume Ratio Final Volume Undiffused Final Volume Diffused 1 cm3 2 cm3 3 cm3 Hypothesis: ANALYSIS QUESTIONS: 1. What evidence is there that NaOH diffuses into an agar block? Is there any evidence that something was diffusing out of the agar? Explain. 2. If the agar blocks were living cells and the NaOH were a “vital substance” that the cells needed, which block would have the most efficient ratio of surface area to volume? How do you know? 3. How do you think the surface area to volume ratio effects how fast diffusion can take place? 4. 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? 5. How does your body use surface area to volume ratio to help exchange gases? 6. How do single celled organisms maintain efficient gas diffusion? 7. Why can’t certain cells like bacteria grow to the size of a small fish? 8. Are multicellular organisms unaffected by the surface area-volume ratio constraint? % Diffusion