Figure 1. Transposon-tagged mutant chromosome (cmsm). WT S. typhimurium cells (WT cmsm on left) were mutagenized by introducing a transposon (Tn) that.

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Figure 1. Transposon-tagged mutant chromosome (cmsm). WT S. typhimurium cells (WT cmsm on left) were mutagenized by introducing a transposon (Tn) that provides tetracycline resistance (TetR). This transposon integrates randomly in the bacterial genome. If the Tn integrates within a gene, expression of that gene is blocked. You screened Tn-tagged mutants to identify a mutant that has reduced virulence, indicating that the Tn integrated into a gene (“gene x”) that is required for virulence (Mutant cmsm on right). You want to determine the sequence of gene x, so you next screen a library of genomic DNA from your mutant. Introduce Tn Select TetR cells Screen for mutants with reduced virulence Transposon Tagging

Figure 2. Genomic DNA library plasmid. Genomic DNA (gDNA) is prepared from your attenuated mutant and digested into small fragments. These fragments are then inserted into a plasmid containing a gene for ampicilin resistance (ampR). This gDNA library consists of several thousand plasmids, each with a different fragment of gDNA from your mutant. One of these plasmids will contain gene x disrupted by the transposon used for mutagenesis (figure 1). Your gDNA library is screened as described in the attached handout to identify the plasmid that contains gene x, interrupted by the transposon.

Screening a genomic DNA library: 1.An attenuated Salmonella typhimurium mutant with reduced virulence was obtained through a transposon tagging mutant screen. The cmsm of this mutant contains a transposon inserted into a gene (“gene x”) that is necessary for virulence (see Figure 1). This blocks expression of gene x and as a result, the mutant exhibits reduced virulence. You want to know the sequence of gene x. 2.Isolate genomic DNA (gDNA) from the attenuated mutant (genome size ~ 2 x 10 6 bp). 3.Digest mutant gDNA into fragments (~1000 different fragments, 2000 bp each). 4.Insert these gDNA fragments into a plasmid that contains an ampicilin resistance gene (ampR) and appropriate sequences for replication and ampR gene expression in E. coli. This constitutes your gDNA “library” with 1000 plasmids, each containing a different 2000 bp piece of gDNA from your mutant, making up one copy of the genome (see Figure 2). One of these gDNA fragments contains gene x interrupted by the transposon. (Note: to increase chances of success, you would actually make your library with several copies of the genome on several thousand plasmids.) 5.Transform your gDNA library into E. coli and plate onto + ampicilin plates. E. coli colonies grow from single cells that contain the ampR plasmid + gDNA from your mutant. 6.To identify which colony contains gene x, hybridize with a radioactively-labeled transposon DNA fragment. 7.Prepare plasmid DNA from the colony identified above and sequence the DNA flanking the transposon to determine the squence of gene x. Replica plate colonies and transfer colony DNA from one plate to nitrocellulose, then hybridize to labeled transposon probe. Expose to X-Ray film to visualize the colony that hybridizes to the transposon probe. Ampicilin resistant E. coli colonies grow from cells that contain the gDNA library plasmid. Each colony is derived from a single cell. This colony contains gDNA that hybridizes to the transposon. (This procedure is used to determine the sequence of a gene disrupted by transposon tagging. Transposon tagging was used in class as an example of how to isolate Salmonella mutants that exhibit reduced virulence, as well as rhizobia mutants that are defective in nodulation.)