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qew qew
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12 years ago
My background in Lipid, Protein, & Bilayers in Red Cells is not well versed.  However, I'm trying to calculate the following:

1) I have a (human) red cell membrane approx. 8nm thick
2) The cell contains 5.2 x 10^-13 grams of lipid (10^-13 = ten to the minus thirteen)
3) Also, the cell contains 6.0 x 10^-13 grams of protein.
4) The lipid has primarily phospholipid (MW 800) & cholestrol (MW386) in a 1:1 molar ratio.
5) Each phospholipid occupies a surface area of  0.55nm^2 per molecules
6) While, the cholesterol occupies a surface area of 0.38nm^2 per molecules
7) The red cell has a surface area of 145um^2.

Questions:

1) If we assume an average protein has a molecular weight of 50,000, how many molecules of protein are in a single red cell membrane?

2) What is the ratio of lipid molecules to protein molecules in the red cell membrane??

3) What proportion of the total surface area of the membrane is occupied by a lipid bilayer???
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12 years ago
Most of the published area figures (as I assume those above are) just deal with the area occupied by, say, the head group of a phospholipid, looking from above.  This is no good, as it misleads one into thinking X molecules with a surface area of Y will occupy (X * Y) of membrane.

This is untrue - You need to establish the area occupied by 1 molecule IN A BILAYER, i.e. the total area will be half (X * Y) as the monolayer area that would be covered is effectively folded in half to form a bilayer.

This may be blindingly obvious, but has caught me out a few times (it also seems to have done so in a few literature cases aswell!).   By the way, I use a figure of 39Angstroms^2 for the s.a. occupied by a PL (R.R.C. New, Liposomes - A Practical Approach, IRL Press, Oxford, UK), but there seem to be a plethora of values dotted around the literature.

   Also, I find it much easier to work in bigger units.  The units of nm^2 are awkward, as one can confuse 39nm^2 as (39 * 10^-9)^2, which is 1.52 * 10^-15 sq metres - wrong.
   It is better to think of it as 39 square nanometers, then take the square root (6.244), so you have (6.244 * 10^-9)^2, which is 3.899 * 10^-17 sq metres - correct.  This small difference introduces a 39 fold difference in area!

Now for the calculation...

Q1) Mw = 50,000;  Avogadro's number (N) = 6.023 * 10^23; Amount = 6.0 * 10^-13g
So number of molecules per cell is:-
((6 * 10^-13)/50,000) * N  = 7.228 * 10^6  (about 7 million)
Really, this is O-level chemistry stuff!

Also, how do you know all of the 6 * 10^-13g is membrane protein? - your initial data doesn't say if the figure includes matrix protein - if it does the whole thing is invalid, but I'll assume it's just membrane protein from now on.

Q2)  Total Mw of 1mole CL + 1mole PL = 1186, therefore work out CL:PL ratio by mass -
   so PL = ((800/1186) * total lipid) = 3.508 * 10^-13g
        CL = ((386/1186) * total lipid) = 1.692 * 10^-13g

PL = (3.508 * 10^-13/800) = 4.385 * 10^-16 moles.
Multiply this by N (avogadro #) to give 2.641 * 10^8 molecules.
CL has same # of molecules, as they're equimolar.
So total number of lipid molecules = 5.282 * 10^8  (about 105 million)
So ratio of Lipid:Protein molecules is 73:1

Q3) Knowing # of PL molecules, and that each PL occupies 5.5 * 10^-19 m^2, monolayer occupied by PL is 1.453 * 10^-10 m^2.  Repeating this for CL uses 3.8 * 10^-19 m^2 per molecule, and 2.641 * 10^8 molecules to give 1.004 * 10^-10 m^2 monolayer

So combined PL + CL area = 2.457 * 10^-10 m^2.  Divide this by 2, as it's in a bilayer, giving a bilayer area of 1.229 * 10^-10 m^2 per cell.

Your quoted area per cell is 145 square microns, i.e. 1.45 * 10^-10 m^2.

So Proportion of this occupied by lipid is (1.229/1.45) = Approx' 85%

You also have the data present to calculate out other figures, such as the area occupied by a single protein:-

Total area left to be occupied by protein is 1.45-1.229 = 0.221 * 10^-10 m^2
Number of protein molecules = 7.228 *10^6
So single protein occupies 3.058 * 10^-18 m^2.

As a guess, I'd say this area was rather low (compared to the area of a single lipid), and that you may be overestimating the amount of protein present.  Either that, or not all the protein is transmembrane, or all my calculations are screwed up - it is way past my bedtime!

Hope this all makes sense.
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