Transposon insertions are flanked by a short direct repeat (usually 5-10 bp) of the target DNA sequence. Draw a diagram to indicate why.
ANSWER: See diagram below. Note that you must indicate the sites of nicking ans joining, and mention that DNA pol fills in the gaps.
Transposon Tn10 encodes resistance to tetracycline (Tet). [This is a complete transposon that encodes its own transposase.] Describe two ways that you could deliver Tn10 to recipient cells to select for transposition events.
ANSWER: (1) Bring the transposon into a recipient cell on a defective phage that cannot replicate in the host or lysogenize the host Ñ select for Tet resistance. (2) Bring the transposon into a recipient cell on a suicide plasmid that cannot replicate in the host Ñ select for Tet resistance.
What is the benefit of using defective transposons such as Tn10dTet instead of the wild-type transposon Tn10?
ANSWER: Transposase can be provided temporarily in trans when the defective transposon is first brought into the cell, allowing the defective transposon to "hop" into the genome. If the transposase is no longer provided in trans then the transposon will be stable, so you do not have to worry about subsequent, secondary transposition events.
The erythromycin (Emr) resistant transposon Tn917 is widely used for transposon mutagenesis in Bacillus. To deliver Tn917 into the chromosome, a temperature sensitive plasmid carrying Tn917 is shifted to the nonpermissive temperature while selecting for EmrR. For this selection to yield transposition events, it is essential that the plasmid lacks homology with any other chromosomal sequences. Why?
ANSWER: If the plasmid has homology with the chromosome, integration of the plasmid into the chromosome would yield Emr resistant colonies. The frequency of this recombination event may be much higher than the frequency of transposition.
Given a lysate of phage P1 grown on a random pool of random Tn10 insertions in the wild-type E. coli chromosome and a trpA- recipient, how could you isolate a Tn10 insertion linked to the trpA gene? [Show the crosses you would do and indicate the phenotype of each possible class of recombinant obtained. Be sure to indicate how you would confirm that the Tn10 insertion is linked to the trpA gene.]
To confirm that the Tn10 insertion obtained is really linked to the trpA gene, you would need to grow phage on the newly isolated Trp+ TetR transductant and backcross the parental trpA mutant, selecting TetR and screening for Trp-.
A common way of delivering transposons is to use a "suicide plasmid" that cannot replicate in the recipient cells. pGP704 is an AmpR plasmid that is unable to replicate unless the Pi protein (encoded by the pir gene) is provided in the recipient cell.
ANSWER: The trick is to look for insertions in a variety of genes that represent a large target site (so it is easy to find insertions in the site) and easy to screen for. Screening for auxotrophic mutations is a great way to do this -- auxotrophic mutants represent approximately 1-4% of the "nonessential" genes in enteric bacteria, and it is easy to screen for auxotrophs by replica plating onto minimal medium.
A colE1 plasmid which carries a Tn3 (AmpR) insertion was mutagenized then transferred to a recA StrS strain that contains a F' zxx::Tn10 (the TetR Tn10 is inserted into a nonessential gene on the F plasmid). Each of the resulting colonies was then mated with an F- StrR recipient strain, selecting for growth on medium with both ampicillin and tetracycline.
It is possible to isolate S. typhimurium mutants with Tn10 inserted at many different sites in the hisD gene. Many of the insertions in the hisD gene are polar on the downstream his genes, but insertions into some regions of the hisD gene are not polar [M. Ciampi, M. Schmid, and J. Roth. 1982. Proc. Natl. Acad. Sci. USA]. Suggest an explanation for this observation.
ANSWER: Remember that Tn10 has a promoter (Pout) directed outward. This promoter makes an mRNA transcript that can be used to express downstream genes. However, unless there is a translational start site immediately downstream of the promoter, Rho dependant polarity will terminate transcription preventing expression of downstream genes. Insertions early in the hisD gene are a long ways from the translational start site for the following gene (hisC), and thus will be polar. Insertions late in the hisD gene will be just in front of the translational start site for the following gene and thus will not be polar.
Haniford and Kleckner isolated mutants that affect excision of Tn10 from donor DNA. The way they isolated these mutants is a clever example of how you can take advantage of several known regulatory systems to genetically dissect complex molecular questions. Their apparoach is shown in the diagram below.
A small multicopy plasmid carried transposase expressed from the PlacZ promoter, which is regulated by the LacI repressor protein. Thus, in the absence of IPTG expression of transposase is turned off, but when IPTG is added high levels of transposase are made. (Recall that transposase acts preferentially in cis, so high levels of transposase are required to see transposition in trans.) The Tn10 insertion was present on a different plasmid that also carries the lacZ gene. The lacZ gene was expressed from the lambda PL promoter, which was regulated by the lambda cI repressor protein. [This approach requires understanding transposons, phage lambda, DNA-damage, RecA, exonucleases, fusions, etc.]
ANSWER: Expression of the transposase gene will be induced by adding IPTG which inactivates the LacI repressor. Cells that make transposase will (i) cause excision of the Tn10 from the plasmid, which will (ii) cause DNA damage, which will (iii) result in formation of RecA*, which will (iv) stimulate cleavage of the cI repressor, which will (v) allow derepression of the LacZ gene.
ANSWER: Such mutants might be obtained after mutagenesis of the plasmid in vitro (for example, with a point mutagen like hydroxylamine) then transformation of the recipient cells to AmpR. Mutants that cannot excise Tn10 would remain white on plates that contain Xgal after addition of IPTG. The parent cells would become blue. To confirm that the transposase itself cause the phenotype, the AmpR plasmid could be purified from any potential mutants and back-transformed into the original parent cells (lacking the AmpR plasmid) to double-check for the white phenotype on plates that contain IPTG.
ANSWER: The phenotype of a dominant-negative transposase mutant would be white on IPTG-Xgal plates. The problem with isolation of dominant-negative mutants is that such mutants may be rare, so you need a trick for finding them. An approach for isolating the dominant-negative mutants would be to start with cells that carry two copies of the transposase -- if either copy of transposase acquires a recessive mutation the other copy could still function, giving the resulting colony a blue phenotype.
Recently, several mutants of the transposon Tn5 have been made using recombinant DNA technology. The map and phenotype of each mutant is shown below.
indicates that DNA between the brackets is deleted in the mutant.
indicates that 800 bp of foreign DNA has been inserted into the transposon at the position of the O. The transposition and NeoR phenotypes are indicated for each mutant.
and indicate the regions of the inverted repeat in Tn5.
From these data, what conclusions can you make about:
The following is an email message I received: "I am a student working on my Ph.D. I am using a bacterial system and trying to find mutants that affect a known gene. Although I have tried many times, I have failed to find any transposon insertion mutations in the genel. However, when I used a chemical mutagen I was able to find some of the desired mutantst. Now I have to figure out which gene causes the mutant phenotype. How can I identify the gene mutated by random chemical mutagenesis?"
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Last modified October 31, 2003