The Effects of Surface Area to Volume Ratio in Agar
AP Biology II 11/22/13
Title: The Effects of Surface Area to Volume ratio in Agar
Introduction: What is an efficient way to maximize mass but minimize diffusion time in cell?
Answer: An efficient way to maximize mass but minimize diffusion time in a cell is to increase its surface area.
If you increase the surface area of a cell relatively to its volume, then the diffusion time will decrease.
Materials: Agar cubes, bromothymol blue- pH indicator, vinegar, ruler, spatula, beaker.
Methods: I used agar cubes as cell models. The agar was cut into blocks corresponding to the sizes indicated to us in Table 1. The cubes have been dyed with bromothymol blue- pH indicator. Then they were placed into vinegar, where they began to turn yellow as the vinegar diffuses into the agar. I timed this diffusion process for 3 different sized cells and compared them. Diffusion is complete when the blue color complete disappears from the center of the cell. For part 2 we became the “Intelligent Designer” for a competitive Cell Diffusion Race. Each student/group received an equal size block of bromothymol blue agar and had the opportunity to design a cell to maximize mass but minimize diffusion time.
Results:
Cell Size (cm)
Surface Area
Volume
SA:V
Time for Complete Diffusion
1 x 1 x 1
6
1
6
10:17 minutes
2 x 2 x 2
24
8
3
46:01 minutes
1 x 1 x 8
48
8
6
10:48 minutes
Conclusions: I accept my hypothesis as being correct. I accept it because the data I collected supports it. Increasing the surface area at a faster rate relative to the volume does prove to increase diffusion time. Having a cell with a ratio of 6:1 like cell (1x1x1) diffuses faster than a cell with a ratio of 3:1 like cell (2x2x2). The diffusion times I recorded advocates the validity of this statement. A real world adaption of this concept would be a frog. The frog resembles the (1x1x) cell because it is smaller than most things but its SA: V ratio is
Title: The Effects of Surface Area to Volume ratio in Agar
Introduction: What is an efficient way to maximize mass but minimize diffusion time in cell?
Answer: An efficient way to maximize mass but minimize diffusion time in a cell is to increase its surface area.
If you increase the surface area of a cell relatively to its volume, then the diffusion time will decrease.
Materials: Agar cubes, bromothymol blue- pH indicator, vinegar, ruler, spatula, beaker.
Methods: I used agar cubes as cell models. The agar was cut into blocks corresponding to the sizes indicated to us in Table 1. The cubes have been dyed with bromothymol blue- pH indicator. Then they were placed into vinegar, where they began to turn yellow as the vinegar diffuses into the agar. I timed this diffusion process for 3 different sized cells and compared them. Diffusion is complete when the blue color complete disappears from the center of the cell. For part 2 we became the “Intelligent Designer” for a competitive Cell Diffusion Race. Each student/group received an equal size block of bromothymol blue agar and had the opportunity to design a cell to maximize mass but minimize diffusion time.
Results:
Cell Size (cm)
Surface Area
Volume
SA:V
Time for Complete Diffusion
1 x 1 x 1
6
1
6
10:17 minutes
2 x 2 x 2
24
8
3
46:01 minutes
1 x 1 x 8
48
8
6
10:48 minutes
Conclusions: I accept my hypothesis as being correct. I accept it because the data I collected supports it. Increasing the surface area at a faster rate relative to the volume does prove to increase diffusion time. Having a cell with a ratio of 6:1 like cell (1x1x1) diffuses faster than a cell with a ratio of 3:1 like cell (2x2x2). The diffusion times I recorded advocates the validity of this statement. A real world adaption of this concept would be a frog. The frog resembles the (1x1x) cell because it is smaller than most things but its SA: V ratio is