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We know that facilitated diffusion requires no energy. But the transport proteins, when bind to a molecule from ECF (extra-cellular face) , the proteins rotate and deliver the molecule into ICF (intra-cellular face). Now my question is that--- Which force provides with the torque to rotate and how it works with out energy?

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[...]how it works with out energy!

The transport proteins can change conformation in a stochastic manner. Even if the conformation opening to the extra-cellular side is a bit different in energy than the one opening to the intra-cellular side, there is no net change in energy/entropy/Gibbs energy in a complete cycle. The only thing that changes is that the cargo molecules are on the other side, a process for which the change in Gibbs energy is negative.

This argument is the same for any catalyst. If the catalyst is recovered in the same state as it was at the beginning of the reaction (and it has to be, otherwise it would not be a catalyst), the catalyst does not change the thermodynamic quantities of reactants or products. As a consequence, it speeds up the reaction, but does not influence the equilibrium constant.

If the catalyst has a high activation energy to do what it needs to do, then it is not an efficient catalyst. After all, the role of the catalyst is to lower the activation energy. So the transport proteins must be able to change from one to the other conformation both with cargo and without cargo in a facile manner. Otherwise, they could not go through the cycle of 1) picking up cargo 2) switching conformation 3) releasing cargo 4) switching conformation again.

Which force provides with the torque to rotate [...]

Random forces associated with thermal energy are responsible for the conformational change. The binding constant for cargo has to be moderate so the the cargo dissociates easily. One the other hand, the binding constant for non-cargo has to be so low that nothing other than cargo binds.

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As another answer correctly points out, the redistribution of energy is enough to keep a protein in constant random motion, including motions that result in transport of a solute from the outside to the inside of the cell by attaching temporarily to the protein. A passive diffusion process does not require bond breaking/formation as is usual in a directed process.

Passive diffusion depends on a concentration gradient. So unless the concentration (or, more accurately, activity) of the diffusing solute is lower inside the cell, there will not be net transport into the cell. The point is not that facilitated diffusion does not require energy — it does — just not directed motion (work) based on active chemical modification of the protein or its environment. Random thermal motion (energy) will do.

This is really a rewording of what another answer explains, namely that for net passive diffusion into the cell to occur, the chemical potential (or free energy) of the solute inside the cell must be lower than that outside. The chemical potential gradient results from some other process that does not involve the protein (it was created perhaps by removing or exhausting the solute inside of the cell, or by increasing the concentration of solute outside the cell). The origin of the potential energy gradient is not a result of chemical modifications (work) at or around the protein itself. Instead it is only a result of the solute concentration difference, and it is therefore we refer to the process as "passive" diffusion.

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