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 μm2. please explain where your calculations come from

Answer 1

Surface area = 5um2 = 5 X 10^-8 cm2
Volume/sec = Permeability X Surface area
           = 10^-8 cm/s X 5 X 10^-8 cm2
           = 5 X 10^-16 cm3/s

Avagadro number to get molecules
1mM Glucose converts to moles to Avagadro number
           = 5 X 10^-16 X 6 X 10^17
           = 30 X 10^1
           = 300/S

1 sec 300 molecules transport
10S = 3000 molecules.