Cell A has a surface area of 50\, \mu\text{m}^250μm 2 50, mu, start text, m, end text, squared and a volume of 10\, \mu\text{m}^310μm 3 10, mu, start text, m, end text, cubed. Cell B has a surface area of 14\, \mu\text{m}^214μm 2 14, mu, start text, m, end text, squared and a volume of 7\, \mu\text{m}^37μm 3 7, mu, start text, m, end text, cubed. Using the information above, determine the surface-area-to-volume ratio of each cell. What is the SA:V ratio of the cell that will exchange materials with its environment at the fastest rate of diffusion?

Explanation:

1. Show the data properly:

Cell A:

Surface area:

Volume:

Cell B:

Surface area:

Volume:

2. Find the surface-to-volume ratio of each cell.

Cell A:

SA: V

Cell B:

SA:V

3. Compare

The cell that will exchange materials with its environment at the fastest rate of diffusion is that with the greater SA:V ratio. That is cell A.

The reason is that a greater surface  in relation to the volume the cell contain means that it has a bigger area to permit the exchange of the materials.

actually, no it isn't. when i was in junior high school, it was a blast. considering that it was a pre international baccalaureate school.

explanation: