01B_Agar+Experiment

__** Homeostasis: Movement in and out of the cells (Agar Experiment) **__

__**Aims**__ To explore the relationship between surface area to volume ratio and rate of exchanging materials.

__**Hypothesis**__ The smaller surface area to volume ratio will have a higher rate of exchanging materials because diffusion is able to take place more effectively in a smaller area as compared to a larger one.

__**Methodology** Apparatus__
 * 1) Razor Blade
 * 2) White Tile
 * 3) Ruler
 * 4) Glass Rod
 * 5) Stop Watch
 * 6) 3 Beakers
 * 7) 3 pieces of 2 x 2 x 2cm^3 solid agar
 * 8) Data Logger with conductivity probe

__Procedures__ Step 1 - Cut one of the three pieces of solid agar given into 8 cubes of 1cm in length. Do this using the razor blade and on the white tile to minimize injury.

Step 2 - Cut another piece of solid agar into 64 cubes, with a length of 0.5 cm of each side. Do this using the razor blade and on the white tile again.

Step 3 - Leave the last piece of 2 x 2 x 2cm^3 piece of agar alone.

Step 4 - Place the different sizes of agar into 3 different beakers - 8 cubes of 1cm in one, 64 cubes of 0.5cm in another, and the 1 big cube of agar in the last beaker.

Step 5 - Starting with the largest piece of agar block, fill the beaker with 200ml of water and put the conductivity probe into the beaker.

Step 6 - Run the data logger and wait for 2 minutes.

Step 7 - During the 2 minutes, one must stir the water using the glass rod continuously and gently without hitting the conductivity probe in the process.

Step 8 - After 2 minutes, stop the data logger and record the highest conductivity in the period of the two minutes by inferring from the graph on the data logger.

Step 9 - Repeat Steps 5 - 8 using the other two sizes of agar and record the results in a table.

Step 10 - At the end of the experiment, dispose the agar blocks into the bin and the water in the sink. This is to prevent any agar blocks from being stuck in the sink.

__**Results**__ __Data__
 * **No. of pieces of agar cubes** || **Length**
 * (cm)** || **Surface area (cm^2)** || **Volume**
 * (cm^3)** || **Surface area to volume ratio** || **Rate of conductivity change** ||
 * 1 || 2 || 24 || 8 || 3 : 1 || 4.05 ||
 * 8 || 1 || 6 || 8 || 3 : 4 || 2.61 ||
 * 64 || 0.5 || 1.5 || 8 || 1 : 16 || 7.27 ||

__Graphs__ 1 cube - 2cm x 2cm x 2cm  8 cubes - 1 cm x 1cm x 1cm  64 cubes - 0.5cm x 0.5cm x 0.5cm  Note: Graphs do not belong to the experiment done by our pair because we did not manage to save any graphs from the data logger while conducting the experiment.

__**Results Analysis**__ From the results, we can observe that the rate of conductivity change is affected by the surface area to volume ratio of the cubes as the rate of conductivity change increases as the surface area to volume ratio increase too.

1. We kept hitting the conductivity probe during the experiment and we realized that the graph will be affected every time we do that, hence all the graphs that we had could not be used for the report because the data is irregular. **
 * __Errors Made__

**__Discussion__** We tried to make sure that the glass rod do not hit the conductivity probe during the experiment and that the agar cubes are cut into almost equally sized pieces to ensure a fair experiment.
 * __1. What precautions did you take in this experiment?__ **

__2. What can you infer from the results above?__ As the surface area to volume ratio increases, the rate of conductivity increases too. This is because there are limitations to which diffusion can take place effectively in the cells. A smaller surface area to volume ratio will result in a more effective and complete diffusion in the cell as compared to a cell which has a bigger ratio. Therefore, the conductivity rate for the smaller cubes are higher as compared to the one in the bigger cube.

__3. Compare this experiment with the one that you did in Lower Secondary where colored agar were soaked in acid. Which do you think is more accurate?__ The experiment that we are doing now. This is because we are making use of the data logger and the conductivity probe in order to measure the results for the highest conductivity in the different sized agar during the experiment. Hence, we are able to have results that are more accurate as compared to seeing the rate of diffusion through observing the distance in which the acid was able to be diffused into the data because the results observed may be wrong due to human error. __4. How do you relate this to the shape of simple (E.g earthworm) to complex (E.g Tiger) living organisms?__ A simple organism such as the earthworm will have a smaller cell size (E.g 64cube - 0.5cm in length) as compared to a larger and more complex organism such as the tiger (E.g 1 cube - 2cm in length). Hence, complex living organism will need to have a higher surface area to volume ratio for their cells in order to facilitate the diffusion of nutrients in and out of their cells for cell growth, which is relatively bigger as compared to simple organisms which do not need such a big surface area to volume ratio. __5. Once a cell grows to a certain size it becomes to 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?__ The volume of a cell increases faster than the surface area of a cell as the cell grows. Hence __6. Examine the agar cells below and work out the surfaced area to volume ratio of each cell. Which is the most efficient and which is least?__ The most efficient agar cells will be the 64 cubes one with a length of 0.5cm and a surfaced area to volume ratio of 1:16 because it has the smallest surfaced area to volume ratio. The least efficient one would then be the 1 cubed agar with a length of 2cm and a surfaced area to volume ratio of 3:1 because it has the largest surfaced area to volume ratio.

__**Conclusion**__ The surface area to volume ratio of the cell is in relative to the size of the cell. If the size of the cell increases, the surface area to volume ratio of the cell increases too. This is to ensure that diffusion is complete throughout the whole cell.