There are so many papers about VSDs and how to use them, but i can not find anything about what exactly is happening in the cell membrane that leads to the change in VSD's light intensity as a sign of change in membrane potential?
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
VSD's work because a portion of the VSD molecule contains a chromophore, that is, a molecular structure that is responsible for absorbing light at one or more particular wavelengths. Absorption of light is entirely a function of the electronic structure of the chromophore. That is, light at a particular wavelength is absorbed by the molecule because the electronic structure of the molecule allows an electron to jump to a higher energy state by absorbing photons of the particular wavelength (energy) in question.
The non-chromophore portion of the VSD molecule is able to embed itself easily into a cell membrane in such a way that the chromophore itself will be oriented in a way that allows a charge redistribution in the membrane to cause an excitation in the electronic structure of the chromophore. (It is currently believed that this orientation is such that the normally light-induced excitation of the chromophore is parallel to the electric field within the membrane). An excitation of the chromophore's electronic structure by the membrane then changes the size of (or eliminates altogether) the electron jump/excitation that could otherwise be caused by absorbing a photon (the size of the jump being directly related to the wavelength of the photon). Therefore, the amount of light absorbed (or possibly the wavelength of light absorbed) changes when the cell membrane potential fires.
Hope that helps. See also here: https://en.wikipedia.org/wiki/Voltage-sensitive_dye
