The study of transposition mechanism and the biology of transposons is an interesting sub ject 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 Tns inserted into the chromosome are resistant to kanamycin (Kanr) To introduce random insertions of tnb into the e. coli chromosome we will start with Tn5 carried on a specialλ phage vector:λ Pam inmt:Tn5 Pam allows conditional phage growth. When a Pam phage infect E coli with an amber suppressor(Su) the phage multiply normally but when A Pam phage infect a nonsuppressing host (Su) the phage cannot replicate int is a mutation in the n integrase gene. Phage with this mutation can not integrate into the host chromosome to make a stable prophage Tn designates that the n phage carries an inserted copy of Tn5 When A Pam int: Tn5 infects a wild type(Su" Kans)E coli host, the phage dNa can not replicate(Pam) nor can it integrate (inn) thus the only way for the E coli to become Kanr is for Tn5 to transpose from the m 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 10 phage-infected E. coli cells. This is how a transposon mutagenesis might be done 1)Infect 2x10 wild-type E coli cells with i Pam int: Tn5 so that each cell receives at least one phage chromosome 2) Select for Kan by plating on medium that contains kanamycin. There should be a total of about 2x104 Kanr colonies. each of these should have Tnb 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 Tns 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 platesThe 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 Pam int 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. ::Tn5designates that the λ phage carries an inserted copy of Tn5. When λ Pam int Pam int 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 Pam int Pam int–::Tn5so 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)