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From my understanding, lipid membranes in biological systems are made mostly of phospholipids, which have a hydrophobic chain region and hydrophilic head group containing a phosphate. In the membranes, these head groups then all come close together to prevent water from reaching the bilayer's interior, where the hydrophobic chains meet.

However, wouldn't doing this require moving numerous negatively charged head group phosphates in very close proximity to each other? In my mind, this would destabilize the membrane structure. Is there more going on here?

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    $\begingroup$ note that many of the common phospholipid head groups contain a positively charge ammonium group so the net molecular charge is zero. $\endgroup$
    – Andrew
    Jan 21, 2023 at 12:47
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    $\begingroup$ See doi.org/10.1021/acs.langmuir.6b04073 for single charge detergent. $\endgroup$
    – Karsten
    Jan 21, 2023 at 12:57
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    $\begingroup$ See doi.org/10.1016/0095-8522(66)90012-2 for micelle size limited by electrostatic repulsion $\endgroup$
    – Karsten
    Jan 21, 2023 at 13:02

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In a bilayer the charged head groups are in contact with the water and the fatty acid chains are in the middle of the bilayer. In water, bilayer formation is largely an entropic effect. In isolated phospholipids in water the fatty acid chain has to be bridged by the H-bonded water molecules as it is hydrophobic. This reduces the water molecules entropy vs that in pure water. Forming bilayers, or micelles, allows many individual lipids to free up many waters as the water is excluded from the centre of the bilayer. This is entropically favourable, i.e. entropy increases vs isolated lipids, and this gain is so large that it exceeds that entropy reduction of forming the bilayer. (The charged head groups are mainly just the same whether as individual lipids or as a bilayer.).

The water around the head groups has a high dielectric constant (relative permittivity = 80) which means that the electric field of any charge is reduced v rapidly vs distance away from the charge, approx as $q/80R$ for charge $q$ and distance $R$. Thus two charges can be close without any large repulsion. In a non-polar solvent such as hexane the dielectric constant is $\sim 2$ so v slow reduction in electric field and charges cannot come close to one another as repulsion soon becomes greater than thermal energy.

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