Question

key idea: the surface area to volume ratio decreases as cell volume increases.

1. use your calculations from the activity diffusion and cell size (q1) to create a graph of the surface area against the volume of each cube, on the grid on the right. draw a line connecting the points and label axes and units.
2. which increases the fastest with increasing size: the volume or the surface area?
3. explain what happens to the ratio of surface area to volume with increasing size.
4. the diffusion of molecules into cells of varying sizes can be modelled using agar cubes infused with phenolphthalein indicator and soaked in sodium hydroxide (naoh). phenolphthalein turns pink in the presence of a base (naoh). as the naoh diffuses into the agar, the phenolphthalein changes to pink and indicates how far the naoh has diffused into the agar. agar blocks are cut into cubes of varying size, so the effect of cell size on diffusion can be studied.

(a) use the information below to fill in the table on the right:

	cube 1	cube 2	cube 3
2. volume not pink (cm³)
3. diffused volume (cm³) (subtract value 2 from value 1)
4. percentage diffusion

(b) diffusion of substances into and out of a cell occurs across the plasma membrane. for a cuboid cell, explain how increasing cell size affects the ability of diffusion to provide the materials required by the cell:

Explanation:

Step1: Calculate volumes of cubes

For Cube 1 with side - length $a_1 = 1$ cm, volume $V_1=a_1^3=1^3 = 1$ $cm^3$. For Cube 2 with side - length $a_2 = 2$ cm, volume $V_2=a_2^3=2^3 = 8$ $cm^3$. For Cube 3 with side - length $a_3 = 4$ cm, volume $V_3=a_3^3=4^3 = 64$ $cm^3$.

Step2: Assume diffusion depth

Assume a constant diffusion depth (say 0.5 cm). For Cube 1, since the side - length is 1 cm, the volume not pink is 0 $cm^3$. For Cube 2, the non - pink part is a smaller cube with side - length $2 - 2\times0.5=1$ cm, so volume not pink $V_{np2}=1^3 = 1$ $cm^3$. For Cube 3, the non - pink part has side - length $4-2\times0.5 = 3$ cm, so volume not pink $V_{np3}=3^3=27$ $cm^3$.

Step3: Calculate diffused volume

For Cube 1, diffused volume $V_{d1}=V_1 - 0=1$ $cm^3$. For Cube 2, $V_{d2}=V_2 - V_{np2}=8 - 1=7$ $cm^3$. For Cube 3, $V_{d3}=V_3 - V_{np3}=64 - 27 = 37$ $cm^3$.

Step4: Calculate percentage diffusion

For Cube 1, percentage diffusion $P_1=\frac{V_{d1}}{V_1}\times100\%=\frac{1}{1}\times100\% = 100\%$. For Cube 2, $P_2=\frac{V_{d2}}{V_2}\times100\%=\frac{7}{8}\times100\% = 87.5\%$. For Cube 3, $P_3=\frac{V_{d3}}{V_3}\times100\%=\frac{37}{64}\times100\%\approx57.8\%$.

Step5: Answer questions 2 and 3

The volume of a cube is $V = s^3$ and surface area is $SA = 6s^2$. As $s$ increases, the volume increases faster. The surface - area - to - volume ratio is $\frac{SA}{V}=\frac{6s^2}{s^3}=\frac{6}{s}$. As $s$ (size) increases, the surface - area - to - volume ratio decreases.

Step6: Answer question 4(b)

As cell size increases, the surface - area - to - volume ratio decreases. Diffusion occurs across the plasma membrane (surface area). With a lower surface - area - to - volume ratio, the rate of diffusion may not be sufficient to supply all the materials required by the cell's larger volume.

Answer:

	Cube 1	Cube 2	Cube 3
2. Volume not pink ($cm^3$)	0	1	27
3. Diffused volume ($cm^3$)	1	7	37
4. Percentage diffusion	100%	87.5%	57.8%

1. The volume increases the fastest with increasing size.
2. The ratio of surface area to volume decreases with increasing size.

4(b) As cell size increases, the surface - area - to - volume ratio decreases, and diffusion may not be sufficient to supply all the materials required by the cell.