How to calculate lipids & protein molecules
P. S. Brookes.
psb at bioc.cam.ac.uk
Mon Mar 13 13:27:55 EST 1995
Jonothan Walker (on 13/3/95) wrote:-
>Hello,
>
> 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???
>
>
>
> Thanks in Advance
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.
::::::::::::::::::::::::::::::::::::::::::::::::::::: Paul Brookes :::::::::::::::::::::::::::::::::::::::::::::::::::::
Snailmail - Department of Biochemistry, University of Cambridge,
Tennis Court Rd, Cambridge, CB2 1QW. Tel' 0223 333649
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Founder of the league against pretentious quotatations in signatures.
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