Lecture 12Lecture 12Lecture 12Lecture 12Lecture 12Transposable elementsTransposable elementsTransposable elementsTransposable elementsTransposable elementsTransposons are usually from 103 to 104 base pairs in length, depending on the transposon type. The key property of transposons is that a copy of the entire transposon sequence can at a low frequency become inserted at a new chromosomal site. The mechanism by which transposons insert into new sites differs from one kind of transposon to another, but the details are not important to understand how transposons can be used. It is worth contrasting the recombination events that occur during transposition to the homologous recombination events that we have considered in meiosis and in phage crosses. In homologous recombination, crossovers occur between like sequences. While this type of recombination can generate new combinations of alleles the arrangement of genes is left undisturbed. In contrast, transposition involves recombination between unrelated sequences, namely the ends of the transposon and a site in the target sequence. Transposition therefore results in a new arrangement of genes along the chromosome. The generic structure of a transposon looks like this: Host DNA Transposon Tn5 Host DNA Transposase Kanamycin Gene resistance Inverted repeat sequences Transposon Element Function Transposase An enzyme that cuts the target DNA more or less at random and splices the transposon ends to the target sequences, Other steps in transposition are performed by host enzymes. Inverted Repeats These sequences direct transposase to act at the ends of the transposon. Note that because the sequences are inverted, the two ends have identical sequence. Selectable Marker(s) Transposons are thought to have evolved by providing a selective advantage to the host cell. Many transposons carry genes that confer antibiotic resistance or some other benefit to the host.The study of transposition mechanism and the biology of transposons is an interesting subject in genetics but for our current purposes we are going to concentrate on how transposons can be used for bacterial genetic analysis. For this purpose we will focus on the transposon Tn5Tn5Tn5Tn5Tn5 which can function in E. coli as well as a wide variety of other bacterial species. The selectable marker in Tn5Tn5Tn5Tn5Tn5 is a gene that confers resistance to the antibiotic kanamycin. Thus bacteria without Tn5Tn5Tn5Tn5Tn5 are sensitive to kanamycin (Kans), whereas bacteria that have Tn5Tn5Tn5Tn5Tn5 inserted into the chromosome are resistant to kanamycin (Kanr). To introduce random insertions of Tn5Tn5Tn5Tn5Tn5 into the E. coli chromosome we will start with Tn5Tn5Tn5Tn5Tn5carried on a special λ phage vector: λλλλλ Pam intPam intPam intPam intPam int–––––::Tn5::Tn5::Tn5::Tn5::Tn5. PamPamPamPamPam allows conditional phage growth. When λλλλλ PamPamPamPamPam phage infect E. coli with an amber suppressor (Su+) the phage multiply normally, but when λλλλλ PamPamPamPamPam phage infect a nonsuppressing host (Su–) the phage cannot replicate. intintintintint––––– is a mutation in the λ integrase gene. Phage with this mutation can not integrate into the host chromosome to make a stable prophage. ::Tn5::Tn5::Tn5::Tn5::Tn5 designates that the λ phage carries an inserted copy of Tn5Tn5Tn5Tn5Tn5. When λλλλλ Pam intPam intPam intPam intPam int–––––::Tn5::Tn5::Tn5::Tn5::Tn5 infects a wild type (Su– Kans) E. coli host, the phage DNA can not replicate (PamPamPamPamPam) nor can it integrate (intintintintint–––––) thus the only way for the E. coli to become Kanr is for Tn5Tn5Tn5Tn5Tn5 to transpose from the λ DNA to some location on the E. coli chromosome. This type of transposition is an inherently rare process and will occur in about one out of 105 phage-infected E. coli cells. This is how a transposon mutagenesis might be done: 1) Infect 2x109 wild-type E. coli cells with λλλλλ Pam intPam intPam intPam intPam int–––––::Tn5::Tn5::Tn5::Tn5::Tn5 so that each cell receives at least one phage chromosome. 2) Select for Kanr by plating on medium that contains kanamycin. There should be a total of about 2x104 Kanr colonies. Each of these should have Tn5Tn5Tn5Tn5Tn5 inserted into a different site on the E. coli chromosome. The genes of E. coli are densely spaced along the chromosome and about half of the Tn5Tn5Tn5Tn5Tn5insertions will lie in one gene or another. There are 4,200 genes in E. coli so our collection of 2x104 random Tn5Tn5Tn5Tn5Tn5 insertions will likely contain at least one insertion in each gene. (Note that insertions in genes that are essential for E. coli growth such as the genes for RNA polymerase or ribosomal subunits will not be recovered because these insertion mutants will not form colonies on the kanamycin plates).Let’s say that we are interested in the E. coli genes that are involved in synthesis of histidine. To find insertion mutants that can not synthesize histidine (His–) we could screen amongst our collection of 2x104 random Tn5Tn5Tn5Tn5Tn5 insertions to find those that are His–. The easiest way to do this would be to plate out the collection of insertions at a density of 200 colonies per plate (100 plates total). Each of these master plates would then be replica plated (first by transfer to a sterile piece of velvet) to a plate that contains histidine and also to a plate that lacks histidine. His– insertion mutants would be identi-fied as colonies that can not grow on the plates that lack histidine. Note that the same collection of random Tn5 insertions can be screened multiple times to find interesting mutations with different phenotypes. 3) Identify His– Tn5Tn5Tn5Tn5Tn5 insertion mutants by replica plating to find colonies that specifically can not grow on plates that don’t contain histidine. Once we have a set of His– insertion mutations (in the present example, one might expect to find 10-20 different His– mutants), the affected gene(s) can be identified by the simple fact that they will be “tagged” by the inserted Tn5Tn5Tn5Tn5Tn5 sequences. The easiest way to identify the site of insertion is by performing a special PCR amplification of the DNA fragment that corresponds to the novel junction betweenTn5Tn5Tn5Tn5Tn5 and the bacterial chromo-somal sequences. Ordinarily PCR reactions are carried out using two DNA primers, each of which corresponding to an end of
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