Question

A baby hamster kidney (BHK) cell has a surface area of 3,400 micrometres². Assume that the endocyte clathrin-coated pits account for 2% of this area and that the surface area of a single coated pit is the same of that of a 100nm diameter clathrin coated vesicle that will be formed from it. If the cell surface has 100,000 transferrin receptors, around 70% are localised within coated pits, work out the following:

Answer 1

If a transferrin receptor has a cross-sectional area of 50 Angstroms X 100 angstroms (5x10nm), how many receptors could be accomodated in a 100nm clathrin-coated vesicle assuming that only 50% of the surface area can be occupied by proteins?

Answer 2

@BrandonGarza777

Answer 3

Was this the only anwser you needed help on because I never had a question like this in bio. I would hate to give you the wrong anwser im sorry.):

Answer 4

yeah, and its okay, i'll just wait till someone else sees this and hopefully knows, and thanks anyways(:

Answer 5

transferrin cross section area = 5 nm x 10 nm = 50 nm

area of a vesicle (assuming spherical) = 4pi(50nm^2) = 31415.926 nm

only 50% of it can be occupied by tranferrins (i think this is because it's budding off the cell and thus only 50% is exposed)... 15707.96 nm

total area available divided by area occupied by single receptor=

15707.96 nm/ 50 nm = 314 receptors per vesicle


The informations given in the first part is useless as they didn't give a number of pits or receptors.

Answer 6

thanks...