Abstract
During its lytic cycle bacteriophage Mu uses repeated transposition as a mode of DNA synthesis. These transpositional events are undoubtedly replicative, and presumably semi-conservative. In a Mu lysogen this type of transposition can start immediately after prophage induction. However, in an infective cycle the Mu genome (which is injected into the host cell as a linear molecule flanked by short random sequences of bacterial DNA) must first become integrated into the host chromosome. Little is known about how this occurs apart from the fact that the bacterial sequences at either end of the Mu genome are lost in the process1. The integration is thus similar to a transposition event. In an attempt to determine whether this type of Mu transposition (between a linear donor molecule and a circular recipient) is also semi-conservative we have analysed the progeny phage arising from an infective cycle in which the parental DNA was heterozygous for a known genetic marker. The expectation is that if integration of the infecting Mu genome occurs by a single semi-conservative transpositional event then pure phage bursts should be produced as the genetic information on only one strand would be preserved throughout the lytic cycle. The experiments reported here do not support this expectation in that the infected cells yield mixed bursts, suggesting that Mu integration is a conservative, rather than a semi-conservative event.
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Akroyd, J., Symonds, N. Evidence for a conservative pathway of transposition of bacteriophage Mu. Nature 303, 84–86 (1983). https://doi.org/10.1038/303084a0
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DOI: https://doi.org/10.1038/303084a0
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