Question

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?

Answer 1

Answer:

• Cell A:

       $SA:V=5\mu m^2/\mu m^3$

• Cell B:

     $SA:V=2\mu m^2/\mu m^3$

• The SA:V ratio of the cell that will exchange materials with its environment at the fastest rate is 5μm²/μm³

Explanation:

1. Show the data properly:

Cell A:

Surface area:

      $SA=50\mu m^2$

Volume:

       $V=10\mu m^3$

Cell B:

Surface area:

     $SA=14\mu m^2$

Volume:

      $V=7\mu m^3$

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

Cell A:

SA: V

        $50\mu m^2/10\mu m^3=5\mu m^2/\mu m^3$

Cell B:

SA:V

        $14\mu m^2/7\mu m^3=2\mu m^2/\mu m^3$

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.