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Bacterial Genetics Transposition - Cell Biology - Lecture Notes, Study notes of Cell Biology

These are the important key points of Lecture Notes of Cell Biology are: Bacterial Genetics Transposition, Transposon Type, Property of Transposons, Entire Transposon Sequence, Low Frequency, New Chromosomal Site, Kind of Transposon, Recombination Events, Homologous Recombination Events, Unrelated Sequences

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2012/2013

Uploaded on 01/12/2013

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Download Bacterial Genetics Transposition - Cell Biology - Lecture Notes and more Study notes Cell Biology in PDF only on Docsity! Lecture 12 Transposable elements Transposons 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. Docsity.com 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 Tn5 which can function in E. coli as well as a wide variety of other bacterial species. The selectable marker in Tn5 is a gene that confers resistance to the antibiotic kanamycin. Thus bacteria without Tn5 are sensitive to kanamycin (Kans), whereas bacteria that have Tn5 inserted into the chromosome are resistant to kanamycin (Kanr). To introduce random insertions of Tn5 into the E. coli chromosome we will start with Tn5 carried on a special λ phage vector: λ Pam int–::Tn5. Pam allows conditional phage growth. When λ Pam phage infect E. coli with an amber suppressor (Su+) the phage multiply normally, but when λ Pam phage infect a nonsuppressing host (Su–) the phage cannot replicate. int– 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 designates that the λ phage carries an inserted copy of Tn5. When λ Pam int–::Tn5 infects a wild type (Su– Kans) E. coli host, the phage DNA can not replicate (Pam) nor can it integrate (int–) thus the only way for the E. coli to become Kanr is for Tn5 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 λ Pam int–::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 Tn5 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 Tn5 insertions will lie in one gene or another. There are 4,200 genes in E. coli so our collection of 2x104 random Tn5 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). Docsity.com
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