To transport a polar molecule through the nonpolar cell membrane, a protein with a polar channel is needed to allow it to diffuse. However, if the molecule is polar and the channel is polar, wouldn't the molecule just be attracted to the polar amino acids and "stick" on the inside of it?

I may be thinking of this totally incorrectly, so please bear with my ignorance!

Thanks in advance :)

  • 4
    $\begingroup$ The water on both sides of the membrane is also polar. So that would help the transported molecule diffuse away from the protein once it passed through. $\endgroup$
    – user137
    Commented Sep 25, 2015 at 16:38

1 Answer 1


The molecule won't "stick" to the polar amino-acids for at least three reasons :

  • The amino-acids might not all be on the same plan, so their is always a force pulling the molecules transiting trough the channel in one direction (in or out the cell)
  • When a flux is established, the repulsion forces between the molecules will force them to go in a direction, meaning that a coming molecule will basically push the ones already engaged in the channel.
  • For an ion or a protein, the water molecules constituting the hydratation shell of the molecule can't always follow it into the channel, so the water molecules present at the end of the channel (entry and end being relative to the direction of the flux) will more or less pull it outside the channel, as it is thermodynamically more stable to have a shell. The fact that water molecules are able or not to follow the molecule depends on the size of the channel.

It's worth noting that water is itself a polar molecule, and that it can selectively cross the membrane thanks to channels called aquaporins.

  • $\begingroup$ Do you have a reference? Also, the first bullet point is hard to read. Perhaps explain the "molecules ", like, exported particles or something. As of now it's quite unclear. $\endgroup$
    – AliceD
    Commented Oct 31, 2015 at 21:54
  • $\begingroup$ @AliceD : thank you for your remark, I clarified the first point. Here can be found a reference relative to my third point. $\endgroup$
    – Frédéric
    Commented Oct 31, 2015 at 22:05
  • $\begingroup$ No worries. +1 for the answer $\endgroup$
    – AliceD
    Commented Oct 31, 2015 at 22:08

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