In methylcobalamin-dependent forms of the enzyme, the reaction proceeds by two steps in a ping-pong reaction. The enzyme is initially primed into a reactive state by the transfer of a methyl group from N5-MeTHF to Co(I) in enzyme-bound cobalamin (Cob), forming methyl-cobalamin(Me-Cob) that now contains Me-Co(III) and activating the enzyme. Then, a Hcy that has coordinated to an enzyme-bound zinc to form a reactive thiolate reacts with the Me-Cob. The activated methyl group is transferred from Me-Cob to the Hcy thiolate, which regenerates Co(I) in Cob, and Met is released from the enzyme.