Calculations

December 1, 2006 at 1:35 pm #6469

Participant

Can anyone help me on the following q’s..thanx

A baby hamster kidney (BHK) cell has a surface area of 3,400 μm2. Assume that the endocytic 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 100 nm diameter clathrin-coated vesicle that will be formed from it. If the cell surface has 100, 000 transferrin receptors and 70% are localised within coated pits, work out the following:

3) What is the relative concentration of transferrin receptors in coated pits compared to the rest of the cell surface? How does this value define the concept of receptor-mediated endocytosis?

4) If each transferrin receptor binds 2 transferrin molecules and all the receptors re saturated when the extracellular transferrin concentration reaches 10 nM, how many transferrin molecules will enter in the fluid phase (i.e. not bound to receptors) at this concentration?

5) What concentration of transferrin would be required to ensure that as many transferrin molecules enter the fluid phase of a coated vesicle as are bound to the receptors?

6) If the coated pits have a lifetime of 1 minute, how long does it take for a BHK cell to endocytose the entire surface? What does this mean in terms of the residence lifetime of plasma membrane lipids and proteins and how are such factors degraded or recycled?

December 3, 2006 at 11:41 pm #61757

hotlikefudge

Participant

Hey,
I have got to do this question for an assignment. So far I have these answers, not sure if correct though:
3. Concentration in pits: 70,000 receptors (70% of 100,000) / 68 (2% of 3400- the surface area covered in pits)= 1029 receptors.
Concentration in the rest of cell: 30,000/ rest of cell surface
30,000/3332= 9 recptors.
Thereofre relative concentration is 1029/9 =114:1 receptors. (114 receptors in pit to every 1 receptor on cell surface)

4. 10nM= 10*10^-9M
(10 *10^-9) * Avagadro (6.022 *10^15)
=6.022 * 10^15 – 200,000 (molecules of Transferrin needed)
=6.002 *10^15.

Thats all I’ve done so far.
Do you know where this question originates from?

December 4, 2006 at 1:49 am #61763

canalon

Participant

3) Good but you have forgotten to answer this

quote :

How does this value define the concept of receptor-mediated endocytosis?

4) A mistake in your explanation N Avogadro=6,022*10^23, but the calculation is OK. In the limits of the question which really sucks. They give you a concentration 10 mM. So your calculation is correct if you assume tha your kidney cell is in 1L bottle, and that is not really likely. BUT and that is where the question suck, the question do not state what is the olume considered. But I think reading the rest that you should consider making the calculation for 1 vesicule (100nm diameter) and that would help for 5)

Good luck for teh rest

December 4, 2006 at 1:00 pm #61777

hotlikefudge

Participant

Kewl, thanks for pointing that out!
Think I’ve got the answer to number 5 too.

5. There are 100,000 transferrin receptors, 200,000 are bound.
We need equal concentrations of bound and unbound receptors, therefore 400,000 transferrin receptors.
400,000/ Avagadro 6.022*10^23
=1.33 aM.

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