Homeostasis+4B+(Agar)

Homeostasis Exploration 4B Movement In and Out of Cells (Agar)

Aim: To explore the relationship between surface area to volume ratio and rate of exchanging materials.

Results:
 * No. of pieces of agar cubes || Length/cm || Surface area/cm^2 || Volume/cm^3 || Surface area to volume ratio || Rate of conductivity change/mS ||
 * 1 || 2 || 24 || 8 || 3 : 1 || 1.07 ||
 * 8 || 1 || 48 || 8 || 6 : 1 || 1.21 ||
 * 64 || 0.5 || 96 || 8 || 12 : 1 || 2.59 ||

Discussion questions: Q1) What precautions did I take in this experiment? Ans: I made sure that the dimensions for the agar cubes were correct and and that there were the correct number of them. Also I made sure that about of solution was the same. I also ensured that the time was exactly 120 seconds.

Q2) What can you infer from the results above? Ans: I can infer that the surface area greatly affected the rate of conductivity. Between 8 and 64 cubes of agar of the same total volume, the rate of conductivity jumped from 1.21 mS to 2.59 mS.

Q3) What do the graphs reveal to you about the rate of diffusion and the surface area of the agar? And: The graphs show that the higher the total surface area the agar has, the higher the rate of diffusion is.

Q4) Combine the results to obtain a class data. Is there a difference? Ans:

Q5) Compare this experiment with the one that you did in Lower Secondary where coloured agars were soaked in acid. Which do you think is more accurate? Ans: This would be more accurate as the coloured agars had food colouring, and that would affect conductivity due to the extra molecules there.

Q6) How do you relate this to the shape of simple (e.g. earthworm) to complex (e.g. tiger) living organisms? Ans: When everything is simple, in the experiment's case the uncut block of agar, the conductivity is low as compared to much more complexed animals, in the experiment's case the 64 small cubes of agar.

Q7) Once a cell grows to a certain size it becomes too large for the complete diffusion of needed substances throughout its cytoplasm. As a cell grows, is the surface area of the cell membrane as efficient relative to the volume of the cell? Ans:

Q8) Examine the agar cells below and work out the surfaces area to volume ratio of each cell. Which is the most efficient and which is the least? Ans: The agar that was cut into 64 cubes was the most efficient and the agar cube that was left uncut was the least efficient.

Conclusion: The larger the total surface area to volume ratio is, the faster the rate of diffusion.