Examples a uniporter and an antiporter

Therefore there is no overall chemical reaction. Assuming that conditions such as pH and ionic strength are the same on both sides of the membrane, the equilibrium constant for the transport reaction is KGlut = 1 and the equilibrium Gibbs free energy is AG°glut = 0. 

To analyze this system we simplify the kinetic mechanism by assuming that the binding and unbinding of glucose from the transporter are rapid, with dissociation constant Kd on both sides of the membrane. This rapid-equilibrium assumption yields  

Next we introduce the quasi-steady approximation, which yields  

Next we introduce the simplification that the conformational transformation of the protein that moves the binding site from one side of the membrane to the other is not affected by the presence of bound glucose. Therefore the kinetics governing transitions between states 2 and 3 is identical to that governing transitions between 1 and 4  

Note that with the above simplihcation, Equahon (7.4) is satisfied. Substituting these expressions into Equahon (7.9) we obtain 

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