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 area of its plasma membrane is 5 μm^2.

a) 5 molecules
b) 50 molecules
c) 500 molecules
d) 5000 molecules

Answer 1

The permeability of a lipid bilayer for glucose is 10^-8 cm/sec. A transport protein is required for glucose to enter the cell. In the given scenario, the number of glucose molecules that will enter the cell in 10 seconds is 5000.

Step-by-step explanation:

The permeability of a lipid bilayer for glucose is 10^-8 cm/sec. Given that the cell's glucose transporter is knocked out, glucose cannot pass directly through the cell membrane. Therefore, a transport protein is required for the entry of glucose into the cell.

In the given scenario, the glucose concentration outside the cell is 1 mM. To calculate the number of glucose molecules that will enter the cell in 10 seconds, we need to calculate the rate of diffusion. The rate of diffusion is given by the equation:

Rate = Permeability x Concentration x Surface Area / Thickness

In this case, the permeability is 10^-8 cm/sec, the concentration is 1 mM, the surface area is 5 μm^2, and the thickness of the membrane is assumed to be 6-9 nm. We can convert the concentration to molecules/cm^3 and calculate the rate of diffusion. Once we have the rate of diffusion, we can multiply it by the time (10 seconds) to find the number of glucose molecules that will enter the cell.

The correct answer is d) 5000 molecules.