Secretion of Bacillus subtilis levansucrase: a possible two‐step mechanism

Abstract

The rate of exocellular levansucrase synthesis in an overproducing (sacU^h) strain of Bacillus subtilis was shown to be directly proportional to the amount of two different transient forms of this enzyme located within the membrane fraction of the cells. The apparent M_r of the larger membrane form was 53000, and that of the smaller form 50000; the half-life time of each form was estimated in vivo to be 4–6 s and 32–42 s, respectively. Ethanol treatment of the cells lead to the accumulation of the 53000-M_r form which may represent 1.5% of total membrane proteins. This latter form, partially purified, was transformed in vitro into the 50000-M_r form by the action of the Escherichia coli leader peptidase. These enzyme forms were quite different from the exocellular levansucrase since they showed a weak affinity for hydroxyapatite and needed complexed iron to display enzyme activity. Assuming the membrane forms were precursors of exocellular levansucrase, we propose a two-step mechanism for the secretion process of levansucrase. The number of exoprotein synthesis/secretion sites in a B. subtilis cell is estimated to 2.5 × 10⁴.