Saturday, September 11, 2010

Curvature of a Lipid Bilayer

I didn’t like the discussion of the energy required to bend a lipid bilayer on page 326-327. It analysed quite rigorously what might happen at the outer surface of a planar lipid bilayer if it were bent, and how much energy would be required to do that. But it left out a number of features, such as the compression of the internal layer of lipids when the membrane is bent. My first thought upon reading this section was that all the bilayer needs to do, if exposed to a force which is trying to bend it, is move some lipids from the inside layer to the outside layer. Extra lipids on the external side would fill the gaps made by spreading apart the polar heads of the outer layer. This post investigates this idea.

I looked up in my biology text book how often lipids switch layers, and in an unstressed membrane, they flip very rarely, only about once a month, or once every 2600000 seconds. However, Nelson says there are ‘tens of millions’ of lipid molecules in a membrane. 10000000/2600000 = 3.9 molecules per second. A second is a long time on the cellular clock (proteins fold on micro to milli second timescales), but this value is for a membrane in equilibrium. The increase in internal energy of the curved bilayer system would cause this rate to increase.

Let the lipid heads have a cross-sectional area of A, an average tail length L, and are in a bilayer membrane where the radius of the external layer is R. The surface area of the outer layer is 4πR^2. The internal layer has a surface area of 4π(R-2L)^2. The difference in area these two layers is 4πR^2-4π(R-2L)^2=16πL(R-L). So the difference in the number of lipids between each layer in the bent membrane is 16πL(R-L)/A. Taking L=1.3nm, R=10μm (page 327) and A to be π(0.3nm)^2 (the size of a few atoms), we calculate the necessary difference in the number of lipids in the inner and outer bilayer is 16*1.3E-9(1E-5-1.3E-9)/9E-20=2.3E6. At four molecules a second, it would take 5.7E5 seconds, or 6.7 days, for enough lipids to flip to cause enough curvature required for a 10μm cell.

I can understand why this was not discussed in detail in the book, but I think it should have been mentioned, even as a Your Turn investigation, as it was an idea that seems quite reasonable at first. I suspect in actual cells, there is a combination of these two ideas, lipids flipping to cause curvature, and an increase in the internal energy of the membrane. The internal polar heads have a shorter distance to travel through the hydrophobic centre of the membrane when curvature is first applied, as the external heads are separating and exposing the extracellular water. This would increase the rate of lipid flipping. As the external spaces are filled, the flipping rate would decrease, and the internal energy of the membrane would increase instead.