2B_Agar+Experiment

**Exploration 4A** **Beetroot Experiment**
 * Homeostasis- Movement in and out of cells.**

__**Aim**__ To explore the factors which affect the movement of materials in and out of the cells.

__**Results**__ __Qualitative Results for solutions__ Test tube A: Faint Pink Test tube B: Light Pink Test tube C: Dark Pink Test tube D: Faint Pink, slightly darker than A Test tube E: Faint Pink, about the same as A

__Quantitative Results__

**Test Tube** || **Percentage of light that passed through solution** || A || 97.59% || B || 56.80% || C || 18.45% || D || 88.78% || E || 94.21% ||

__**Analysis**__ Test tube A allowed the most amount of light to pass through and test tube C allowed the least.

Test tube C allowed more light to pass through than test tube B.

Test tube D allowed more light to pass through than test tube A.

Test tube E allowed more light to pass through than test tube A. __**Discussion //(according to the questions)//**__
 * 1) The control in this experiment was the volume of the beetroot.
 * 2) By comparing the the different setups that had the same control, independent and dependent variables. For example, you can compare Tube A, B, C and D because the tubes have the same independent variable, which is the different content of the water and also they have the the same control variables.
 * 3) When the beetroot are cut, the cells are cut too and the pigment flows freely from the cells. We need to wash the beetroot thoroughly to get rid of the free flowing pigment first so that we can ensure the colouring of the water was from the cell through osmosis and not from the free flowing pigments.
 * 4) I ensured that all the beetroots were from the same stalk. I also measured them very carefully.
 * 5) Alcohol indeed has an effect on the the beetroot cell membranes. The concentration of the alcohol was a factor too. The more the alcohol content of the solution, the darker the solution will be at the end of the experiment. This could be because of the water potential of the solution. When the solution has a higher alcohol concentration, the water potential of the solution is lowered and thus, lower than that of the cell. Thus, we can say that the solution is hypertonic compared to the cell and thus, more water will diffuse out of the cell through the cell’s partially permeable membrane in a process called osmosis. The pigment will diffuse out of the cell together with the water in the cell. Therefore, the alcohol solution is significantly darker in colour compared to water and the solution with more alcohol content is darker than the one with less alcohol content.
 * 6) D allowed less light to pass through than A as D contained hot water. An increase in temperature of the beetroot’s surroundings, which in this case is the hot water solution, will damage and denature the plasma membrane and cause the cytoplasm and other substances contained within the membrane to leak out. Thus, the pigments in the cells flowed out more easily than those at A. E allowed less light to pass through that A as E has a larger surface area to volume ratio that A thus, more of the plasma membrane of the cells were exposed to the water. More water in the cell diffused out of the cell through the cell’s partially permeable membrane in a process called diffusion as the water was a hypertonic solution.
 * 7) The presence and percentage of alcohol content in the solution. The temperature of the solution. The surface area to volume ratio of the beetroot in the solution.

**Homeostasis- Movement in and out of cells.** **Exploration 4B** **Agar Experiment**

__**Aim**__ To explore the relationships 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 ||  0.69 || 8 ||  1 ||  48 ||  8 ||  6:1 ||  1.64 || 64 ||  0.5 ||  96 ||  8 ||  12:1 ||  4.67 ||

__**Analysis**__ The 64 cubes set up had the greatest conductivity change, followed by the 8 cubes set up then lastly the 1 cube set up.

The surface area to volume ratio of the setups were such that the 64 cubes set up had the greatest ratio followed by the 8 cube set up and lastly the 1 cube set up.

__**Discussion**__
 * 1) Some precautions I did was measure and cute the agar cubes carefully to ensure standardization and correct measuring of the surface area. Another precaution we took was to have the same person stir all 3 experiments so the frequency of the stirring could better be standardized.
 * 2) From the results above, I can infer that the larger the surface area to volume ratio of the cubes, the higher the rate of conductivity change.
 * 3) The larger the surface area, the faster the rate of diffusion.
 * 4) Class data not available during the time in which I did this report. (I sleep early.)
 * 5) I think this experiment is more accurate as more sophisticated equipment were used so I can accurately find the amount of agar that diffused in comparison to the other agar.
 * 6) The shape of living organisms are always structured such that the surface area to volume ratio is maximized. This can be seen in organisms such as the earthworm where it is thin and long.
 * 7) As a cell grows, the surface area if the cell membrane is less efficient relative to the volume of the cell. This is because as the cell grows, the volume of the cell will inadvertently grow more than the surface area of the cell will expand, thus the surface area to volume ratio of the cell will be lesser. Thus, the cell membrane will be less efficient compared to the volume of the cell.