Abstract
In the absence of procaine the end-plate conductance evoked by suberyldicholine increases exponentially to a new level following a step hyperpolarization. In the presence of procaine the suberyldicholine-evoked conductance first rapidly decreases and then slowly increases following a hyperpolarizing step. The fast relaxation has a time constant of .apprx. 1 ms, and the slow relaxation a time constant of 10-150 ms. The existence and sign of these 2 relaxations is predicted by a sequential model in which procaine enters and blocks open but not closed end-plate channels. The concentration dependence of the fast and slow relaxation time constants agrees well with the predictions of this model, and allows the apparent dissociation constant for binding of procaine within the open channel to be estimated at about 20 .mu.M at -80 mV membrane potential. This apparent binding constant is voltage sensitive. It decreases e-fold for 50 mV hyperpolarization, suggesting that the procaine binding site is electrically half way through the channel. Procaine concentrations comparable to the dissociation constant for binding to open channels strongly depress the equilibrium current evoked by low suberyldicholine concentrations. This finding is not in accord with the sequential model. A cyclic model in which procaine binds to both closed and open channels explains well the equilibrium observations. The affinity of procaine for closed channels is similar to its affinity for open channels, and is also increased by hyperpolarization. This model also fits well the kinetic observations, if it is assumed that blocked channels open and close much more slowly than unblocked channels. The concentration dependence of the relaxation amplitudes disagrees with the predictions of the sequential model, but agrees well with the predictions of the cyclic model. No other model appears to explain the various observations as economically as the cyclic channel blocking model. If the model is correct the ''gate'' controlling the end-plate channel must be in the inner half of the membrane.