Question

The permeability of a lipid bilayer for glucose is 10^-8 cm/sec. Consider a cell whose glucose transporter is knocked out. If it is placed in a solution that contains 1 mM glucose, how many glucose molecules will enter the cell in 10 seconds? Assume that the cell contains no glucose and the surface areaof its plasma membrane is 5 micrometers.

Answer 1

Approximately 5 x 10^-8 moles of glucose molecules will enter the cell in 10 seconds.

Step-by-step explanation:

To calculate the number of glucose molecules that will enter the cell in 10 seconds, we need to consider the permeability of the lipid bilayer and the concentration of glucose in the solution. The permeability is given as 10^-8 cm/sec.

We can use Fick's first law of diffusion to calculate the rate of diffusion: rate of diffusion = permeability x surface area x concentration gradient. The surface area of the plasma membrane is given as 5 micrometers and the concentration of glucose is given as 1 mM. To convert the concentration to moles, we divide by Avogadro's number.

The concentration gradient is then the initial concentration of glucose in the solution.

Plugging in the values and solving the equation, we find that the rate of diffusion is 5 x 10^-9 moles/sec.

Multiplying this by 10 seconds, we find that approximately 5 x 10^-8 moles of glucose molecules will enter the cell in 10 seconds.