The transition from B-DNA to Z-DNA not only requires a change in the helix architecture from a right-handed to a left-handed form, but also requires that all of the base pairs be flipped over. A transition mechanism is proposed in which the base pairs flip over one at a time while keeping the Watson-Crick hydrogen bonds intact. The two base pairs on either side of the pair that turns over form a kind of dynamic sandwich around it, and the flip takes place in a cavity that is formed when the distance between them is increased from about 7A in the intact helix to about 14A in the transition state. Since the transition occurs in two steps (formation of the cavity, followed by the base pair flip) and since the cavity propagates down the helix after the base pair is turned over, the mechanism accounts for the cooperativity of the B in equilibrium Z transition.