Microbio question

If a lipid raft is typically 70 nm in diameter and each lipid molecule has a diameter of 0.5 nm, about how many lipid molecules would there be in a lipid raft composed entirely of lipid? At a ratio of 50 lipid molecules per protein molecule (50% protein by mass), how many proteins would be in a typical raft? Explain. (Neglect the loss of lipid from the raft that would be required to accommodate the proteins).

Originally posted by wormer
If a lipid raft is typically 70 nm in diameter and each lipid molecule has a diameter of 0.5 nm, about how many lipid molecules would there be in a lipid raft composed entirely of lipid? At a ratio of 50 lipid molecules per protein molecule (50% protein by mass), how many proteins would be in a typical raft? Explain. (Neglect the loss of lipid from the raft that would be required to accommodate the proteins).

If this is just a simple area problem, it is easier to convert to Angstroms, 70 nm=700 angstrom, 0.5 nm=50 Angstrom. So in one dimension, 700/50=14 and 14 squared =196 lipids in 490,000 square angstroms. 196/50 about 4 proteins. Does that sound right?

Originally posted by wormer
If a lipid raft is typically 70 nm in diameter and each lipid molecule has a diameter of 0.5 nm, about how many lipid molecules would there be in a lipid raft composed entirely of lipid? At a ratio of 50 lipid molecules per protein molecule (50% protein by mass), how many proteins would be in a typical raft? Explain. (Neglect the loss of lipid from the raft that would be required to accommodate the proteins).

Assuming your raft is one molecule thick then you can get a first approximation by dividing the area of the raft by the area taken up by a molecule in it. That is the same as the ratio of the squares of the diameters. The square of the raft's diameter is 49,000 Gigabarns and the square of the lipid molecules diameter is 2.5 Gigabarns. That gives 19,600 molecules.

However, they have to be packed and the most efficient packing is hexagonal packing. Consider three neighbouring molecules, if they are hexagonally packed then the packing fraction is given by the area of intersection of the three circles and the equilateral triangle joining the centres divided by the area of the triangle. The area of the circles is just the area of a semi-circle of radius r, A = πr^2 / 2. The triangle has area given by the product of the length of one of the sides and the distance from the centre of that side to the vertex between the other two. We can get that from Pythagoras' theorem and it is sqrt(3)*r, so the area is sqrt(3)*r^2. This gives us the packing density π/2sqrt(3) = 0.928, for a total number of molecules of: 17,775.

If this is a lipid bilayer then there are two molecules per space and we need to double that to give a final answer of 35,550. Divide by 50 to get the number of proteins = 711.

Originally posted by sonhouse
If this is just a simple area problem, it is easier to convert to Angstroms, 70 nm=700 angstrom, 0.5 nm=50 Angstrom. So in one dimension, 700/50=14 and 14 squared =196 lipids in 490,000 square angstroms. 196/50 about 4 proteins. Does that sound right?

You've lost a factor of 10. 0.5nm = 5 Angstroms. Squared that's a factor of 100 for 19600, and you are getting the same answer as me, except for the packing density analysis.

Originally posted by DeepThought
You've lost a factor of 10. 0.5nm = 5 Angstroms. Squared that's a factor of 100 for 19600, and you are getting the same answer as me, except for the packing density analysis.

oops.

Originally posted by DeepThought
Assuming your raft is one molecule thick then you can get a first approximation by dividing the area of the raft by the area taken up by a molecule in it. That is the same as the ratio of the squares of the diameters. The square of the raft's diameter is 49,000 Gigabarns and the square of the lipid molecules diameter is 2.5 Gigabarns. That gives 19,60 ...[text shortened]... double that to give a final answer of 35,550. Divide by 50 to get the number of proteins = 711.

Is it nessacerily to use gigabarns? Why not keep it simple and use units such as A•

Originally posted by wormer
Is it nessacerily to use gigabarns? Why not keep it simple and use units such as A•

A barn = 10^-24 square centimeter. So an area of 10^-12 x 10^-12 cm. I guess that lets you size the area of individual atoms. Wiki says it is about the cross sectional area of a uranium nucleus.

Originally posted by wormer
Is it nessacerily to use gigabarns? Why not keep it simple and use units such as A•

That was more by way of a joke, I was trying to avoid using the "^" symbol in nm^2. I don't know what A• is. If it's a sensible unit of area on that kind of scale and easy to convert to then use it.

Originally posted by sonhouse
A barn = 10^-24 square centimeter. So an area of 10^-12 x 10^-12 cm. I guess that lets you size the area of individual atoms.

So in future we should use the Yottabarn instead of square centimetres.

Originally posted by DeepThought
So in future we should use the Yottabarn instead of square centimetres.

Hey farmer, that's a yottabarn you've got there🙂

Originally posted by wormer
If a lipid raft is typically 70 nm in diameter and each lipid molecule has a diameter of 0.5 nm, about how many lipid molecules would there be in a lipid raft composed entirely of lipid? At a ratio of 50 lipid molecules per protein molecule (50% protein by mass), how many proteins would be in a typical raft? Explain. (Neglect the loss of lipid from the raft that would be required to accommodate the proteins).

DeepThought's math looks accurate. The lipids would have to be part of a cell membrane, organized in a bilayer, otherwise it would not be a lipid raft. Conversely, proteins are not bilayers. likely single-pass transmembrane domains.

You should also use the terms "molecular biology" or "biochemistry" when referring to molecular components of a cell. Microbiology is the study of microbes (e.g. bacteria), and whether bacterial membranes contain lipid rafts isn't clear. Most of lipid raft biology is understood in higher-order eukaryotes.

Originally posted by DeepThought
That was more by way of a joke, I was trying to avoid using the "^" symbol in nm^2. I don't know what A• is. If it's a sensible unit of area on that kind of scale and easy to convert to then use it.

Angstrom

Originally posted by DeepThought
Assuming your raft is one molecule thick then you can get a first approximation by dividing the area of the raft by the area taken up by a molecule in it. That is the same as the ratio of the squares of the diameters. The square of the raft's diameter is 49,000 Gigabarns and the square of the lipid molecules diameter is 2.5 Gigabarns. That gives 19,60 ...[text shortened]... double that to give a final answer of 35,550. Divide by 50 to get the number of proteins = 711.

for the demisional packing part. could you 3explain where you got those numbers such as square(3)*r and what r is? maybe im stupid or blind but i dot follow

Originally posted by wormer
for the demisional packing part. could you 3explain where you got those numbers such as square(3)*r and what r is? maybe im stupid or blind but i dot follow

That is the square root of 3, or 1.732 and r is the radius of a sphere.

Originally posted by wormer
for the demisional packing part. could you 3explain where you got those numbers such as square(3)*r and what r is? maybe im stupid or blind but i dot follow

If a plane is covered with circles that do not overlap then the densest arrangement is the hexagonal lattice with each circle touching six others. So any three neighbouring circles will be just touching. We can tile the plane with equilateral triangles with their vertices on the centre points of the circles. This means that the packing density is the ratio of the area overlapped by one of the triangles with the circles it's vertices are on.

First we find the area of overlap. Equilateral triangles have interior angle 60°. So that means each triangle overlaps with one sixth of each circle. Each triangle overlaps three circles so the total overlap is half the area of a circle. If the circles all have radius r then that is an area of πr^2 / 2.

To get the area of the equilateral triangle we note that the sides join the centres of two touching circles of radius r, so the triangle's sides are of length 2r. Cut the triangle in half to form two right angled triangles. Pythagoras' theorem gives us the relation between the lengths of the sides. So we have a^2 + b^2 = c^2, where a is the height, b the base length and c the length of the hypotenuse. The hypotenuse is the side length of the original equilateral triangle, 2r, and the base length is half of an original side length. So we have that a^2 + r^2 = 4r^2, which after a small amount of algebra gives a = r sqrt(3). So the area of the equilateral triangle is sqrt(3) r^2.

Dividing these two areas gives the packing density π/(2 sqrt(3)) ~ 0.907.

In my earlier post I seem to have quoted 0.928, that is a typo and the 17,775 figure is correct.