|MadSci Network: Cell Biology|
The short answer is that the energy needed to change the conformation of the carrier protein is provided by the concentration gradient of the molecule being transported.
I find that it's often useful to think of transport as a chemical reaction, even though in this case no covalent bonds are being formed or broken. In this mindset, you have a series of equilibria, each with their own constants, that govern the interaction between the protein carrier and the molecule(s) being transported. For a single molecule "A" and a carrier with two possible conformations (open to outside of membrane or inside):
1: A out + carrier out <--> A-carrier out
2: A-carrier out <--> A-carrier in
3: A-carrier in <--> A in + carrier in
Each of these steps has its own equilibrium constant, which for the first and third steps are equivalent to the affinity constant for A and the different conformations of the protein.
K(1) = [A-carrier out]/[A out][carrier out]
K(2) = [A-carrier in]/[A-carrier out]
K(3) = [A in][carrier in]/[A-carrier in]
I'm not going to make up numbers for all these constants and concentrations, but hopefully it's clear that what you have here is an extended application of Le Chatelier's principle. At low concentrations of A inside the cell, step 3 is driven to the right, which lowers the concentration of A-carrier in, which drives reaction 2 to the right, which lowers the concentration of A-carrier out, which drives reaction 1 to the right.
Getting back to your original question, the conformation change in step 2 may be thermodynamically unfavorable when concentrations of A are not higher outside the cell than inside, but when there is a concentration gradient the overall, 3-step, process has deltaG<0, and the carrier protein will tend to switch from the open to the outside conformation to the open to the inside conformation.
Separate from this, the carrier can also flip between the two conformations when A is not bound, and the likelihood of one conformation or the other is not as dependent on the concentration of A inside or out. So we can add a fourth step that recycles the carrier without A:
4: carrier in <--> carrier out; K(4)=[carrier out]/[carrier in]
Without A bound, the "carrier out" conformation is probably favored, but even if the two states are pretty much equally likely the same concentration gradient that powers steps 1-3 will lead to the empty carrier flipping back to be open to the outside.
I couldn't find any images of my own to help, but this one illustrates
the different states the carrier is in and how it cycles:
-- from: 18.104.22.168/ywwy/zbsw(E)/pic/ech5-6.jpg
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