Question

data table 1: diffusion in agar cubes
cube size\ttotal volume (cm³)\tunpenetrated volume (cm³)\tpenetrated volume (cm³)\tpercent of cell covered by diffusion (%)\tsurface area (cm²)\tsurface area to volume ratio
0.5 cm\t0.125 cm³\t0 cm³\t0.125 cm³\t100%\t1.5 cm²\t12:1
1 cm\t1 cm³\t0.512 cm³\t0.488 cm³\t48.8%\t6 cm²\t6:1
2 cm\t8 cm³\t4.096 cm³\t3.904 cm³\t48.8%\t24 cm²\t3:1
percent of cell covered by diffusion = \frac{penetrated volume}{total volume} x 100
data analysis and discussion:

1. in this investigation, the agar cubes represent living cells. what does the sodium hydroxide (naoh) solution represent in relation to the cells?
2. which cube size demonstrated the most effective diffusion? use your data from the three different sized cubes to explain which \cell\ would survive and which would not. be sure to reference specific numbers from your results.
3. what is the relationship between the size of the cell, the distance the naoh solution traveled, and the rate of diffusion? which cell size would be the most efficient in transporting nutrients throughout the cell via diffusion?
4. why do organisms grow by increasing the number of cells rather than simply increasing the size of their existing cells? consider what happens to the efficiency of diffusion as a cell increases in size.
5. using the data from your experiment, create a graph that shows the relationship between the length of a side of the agar cubes and their surface area - to - volume ratio. once your graph is complete, observe the trend and explain how the surface area - to - volume ratio changes as the size of the cube increases. how does this trend relate to the efficiency of diffusion in cells? you need to use graph paper for this graph (no exceptions). i have some graph paper for you if you do not have your own. what will go on the x - axis? what will go on the y - axis?

Explanation:

1. The NaOH solution represents substances like nutrients or oxygen that need to diffuse into cells.
2. The 0.5 - cm cube demonstrated the most effective diffusion as it had 100% penetration. A cell with high diffusion efficiency like the 0.5 - cm cube is more likely to survive as it can get necessary substances throughout. Larger cubes with lower penetration percentages may not survive well as they can't get enough substances to all parts.
3. As the cell size increases, the distance the NaOH solution has to travel increases, and the rate of diffusion becomes less efficient. Smaller cell sizes are more efficient in transporting nutrients via diffusion due to a higher surface - area to volume ratio.
4. As a cell increases in size, the surface - area to volume ratio decreases, reducing the efficiency of diffusion. By increasing the number of cells, organisms can maintain a high surface - area to volume ratio for efficient exchange of materials.
5. The length of a side of the agar cubes goes on the x - axis and the surface area - to - volume ratio goes on the y - axis. As the side length increases, the surface area - to - volume ratio decreases. This trend shows that smaller cells (represented by smaller cubes) have a higher surface area - to - volume ratio, which is beneficial for efficient diffusion.

Answer:

1. Substances like nutrients or oxygen that diffuse into cells.
2. The 0.5 - cm cube; the 0.5 - cm cube had 100% penetration and is more likely to survive, while larger cubes with lower penetration may not.
3. As cell size increases, diffusion distance increases and rate becomes less efficient; smaller cell sizes are more efficient.
4. Because as cell size increases, diffusion efficiency decreases due to a lower surface - area to volume ratio; increasing cell number maintains a high ratio.
5. X - axis: length of a side of the agar cubes; Y - axis: surface area - to - volume ratio.