As far as suicide plasmids are concerned, your best bet is to try one
of the sacB-based positive selection systems. Such plasmids contain an
RP4 origin of replication, which requires the presence of the pir gene
product for replication. There are a number of lambda-pir lysogens
available inwhich the suicide plasmid will happily replicate. By
conjugating the donor strain (which is lambda-pir) with a recipient
strain which is eg streptomycin or nalidixic acid resistant, and lacks
lambda-pir, only the recipient cells inheriting the suicide plasmid,
and has recombined into the chromosome, will survive.
Now, the clever bit. To get rid of the original allele, and to
replace it with the mutated form, you grow the merodiploid in
nutrient broth to allow a second recombination event to occur (the
plasmid excises from the chromosome). Then add some of this culture
(1/100 vol.) to Luria broth minus NaCl, but containing 6% sucrose.
The sacB gene in the suicide plasmid confers sensitivity to sucrose.
Thus any cells still containing the integrated suicide plasmid will
die. However, any cells that have lost the suicide plasmid as a
result of the second recombination event will survive. So, after
incubating the sucrose broth, plate out dilutions onto Luria agar
(minus NaCl + 6% sucrose), and select plates containing discrete
colonies. These should all be resolved merodiploids, containing
EITHER the wild-type allele OR the mutant allele. All you have to do
then is check by Southern blot or PCR.
If you require any more details, please ask.
Regards Martin
*****************************************
* Dr Martin Goldberg, *
* Dept. of Microbiology and Immunology, *
* University of Leicester, *
* University Road, *
* LEICESTER. *
* LE1 9HN *
* UK *
* Tel. +44 (0)116-252 3017 *
* Fax. +44 (0)116-252 5030 *
* E-mail mdg at le.ac.uk *
*****************************************
