麻省理工学院：《动物遗传学 Genetics》课程教学资源（英文讲义）Lecture 13 Gene Manipulation in Bacteria

Lecture 13 Gene Manipulation in Bacteria There is no meiosis in bacteria so special techniques have been worked out for manipulating genes in bacteria so that mapping experiments,strain construction,and complementation tests can be done. First,we need a way of getting chromosomal DNA from one cell into another.There are several ways to do this.All of the methods have in common the use of special extra chromosomal elements for mobilizing chromosomal genes;the methods differ according to which extra chromosomal element is used. We will consider a method that uses phage and is known as Transduction E.coli chromosome is 4.6 x 106 base pairs Phage P1 chromosome is 105 base pairs O After infection of E.coli,the phage DNA is replicated by a mechanism known as a "rolling circle"and the phage is packagedinto phage particles one headfull at a time: 1/300 phage mistakenly packages E.coli chromosome DNA instead of phage DNA. E.coli chromosome phage with chromosomal DNA

Lecture 13 Gene Manipulation in Bacteria There is no meiosis in bacteria so special techniques have been worked out for manipulating genes in bacteria so that mapping experiments, strain construction, and complementation tests can be done. First, we need a way of getting chromosomal DNA from one cell into another. There are several ways to do this. All of the methods have in common the use of special extra chromosomal elements for mobilizing chromosomal genes; the methods differ according to which extra chromosomal element is used. We will consider a method that uses phage and is known as Transduction E. coli chromosome is 4.6 x 106 base pairs Phage P1 chromosome is 105 base pairs After infection of E. coli, the phage DNA is replicated by a mechanism known as a “rolling circle” and the phage is packagedinto phage particles one headfull at a time: 1/300 phage mistakenly packages E. coli chromosome DNA instead of phage DNA

Each phage particle will package about 1/50 of the E.coli chromosome.By combining probabilities we see that about 1/15,000 phage will carry a particular E.coli gene. A basic transduction experiment to measure the linkage between markers A and B is done as follows: (1)Grow P1 on A+B+ (2)Infect A B (3)Select for A+and then screen for B+ The idea is that we are looking for the rare cases where some chromosomal DNA carrying gene A is moved into the recipient.To find these recombinants,we select for A+.Then we screen for B+to see how often gene B comes along with gene A. 4+ B+ A B The measured frequency of cotransduction of B with A gives a measure of distance according to the following rules: If distance between A and B is greater than one headfull(105 bp)then there will be no cotransduction. If A and B are very close together then there will be 100%cotransduction. Cotransduction frequency is an inverse measure of distance. 100 % cotransduction one headfull ↓ 0 distance between markers The experiment just described is the bacterial equivalent of a 2-factor cross and will give us relative distances between genes

Each phage particle will package about 1/50 of the E. coli chromosome. By combining probabilities we see that about 1/15,000 phage will carry a particular E. coli gene. A basic transduction experiment to measure the linkage between markers A and B is done as follows: (1) Grow P1 on A+B+ (2) Infect A–B– (3) Select for A+ and then screen for B+ The idea is that we are looking for the rare cases where some chromosomal DNA carrying gene A is moved into the recipient. To find these recombinants, we select for A+. Then we screen for B+ to see how often gene B comes along with gene A. A- B￾A+ B+ The measured frequency of cotransduction of B with A gives a measure of distance according to the following rules: • If distance between A and B is greater than one headfull (105 bp) then there will be no cotransduction. • If A and B are very close together then there will be 100% cotransduction. • Cotransduction frequency is an inverse measure of distance. 100 % cotransduction 0 distance between markers one headfull The experiment just described is the bacterial equivalent of a 2-factor cross and will give us relative distances between genes

We can also do a 3-factor cross to determine gene order. (1)Grow P1 on A+B+C+ (2)Infect A-B-C (3)Select for A+and then screen for B+and/or C+ Genotypes A+B+C+ 2 crossovers (A to C distance) A+B+C- 2 crossovers -(A to B distance) A+B-C- 2 crossovers A+B-C+ 4 crossovers (very rare) (Note that there are only four possible genotypes because we select A+) A+ B+ C+ A B A limitation of transduction experiments is the need for a good selectable marker.Tn5 insertions provide a way to extend the utility of transduction for mapping and strain construction.For example,let's say that we have isolated a new mutation in the MotA gene.MotA is a component of the bacterial flagellar moter and MotA-mutants are nonmotile,a phenotype easily detected by the inability of MotA-colonies to "swarm" outward on soft agar plates.Imagine that we want to map the MotA-mutation or to move this mutation into an E.coli strain with a new genetic background.Clearly direct transduction MotA-would not be possible since we have no way to select for rare (1/15,000)transductants with the nonmotile MotA-phenotype.One solution would be to use a nearby marker for which we can select to move MotA-by its cotransduction with the selectable marker.Unfortunately,good selectable markers are not common and we are unlikely to have a good selectable marker placed within cotransduction distance of MotA-readily available.A powerful alternative approach would be to isolate a random Tn5 insertion that is close to MotA-and to use the Kanr trait conferred by Tn5 as the selectable marker for cotransduction

We can also do a 3-factor cross 3-factor cross to determine gene order. (1) Grow P1 on A+B+C+ (2) Infect A–B–C– (3) Select for A+ and then screen for B+ and/or C+ Genotypes A+B+C+ 2 crossovers (A to C distance) A+B+C– 2 crossovers (A to B distance) A+B-C- 2 crossovers A+B-C+ 4 crossovers (very rare) (Note that there are only four possible genotypes because we select A+) A - B- C￾A+ B+ C+ A limitation of transduction experiments is the need for a good selectable marker. Tn5 insertions provide a way to extend the utility of transduction for mapping and strain construction. For example, let’s say that we have isolated a new mutation in the MotA gene. MotA is a component of the bacterial flagellar moter and MotA– mutants are nonmotile, a phenotype easily detected by the inability of MotA– colonies to “swarm” outward on soft agar plates. Imagine that we want to map the MotA– mutation or to move this mutation into an E. coli strain with a new genetic background. Clearly direct transduction MotA– would not be possible since we have no way to select for rare (1/15,000) transductants with the nonmotile MotA– phenotype. One solution would be to use a nearby marker for which we can select to move MotA– by its cotransduction with the selectable marker. Unfortunately, good selectable markers are not common and we are unlikely to have a good selectable marker placed within cotransduction distance of MotA– readily available. A powerful alternative approach would be to isolate a random Tn5 insertion that is close to MotA– and to use the Kanr trait conferred by Tn5 as the selectable marker for cotransduction

The steps for finding a linked Tn5 insertion are as follows: 1)Start with a collection of random Tn5 insertions into wild type E.coli(the isolation of such a collection was described in last lecture).Grow phage P1 on the mixture of 2x104 different Tn5 insertion mutants.Note that this donor strain is MotA+. 2)Use the resulting P1 phage to infect a MotA-recipient strain.Select for transduction of the Tn5 insertions by selecting for growth of the transductants on kanamycin plates. Screen for cotransduction of MotA+by testing each of the Kanr transductants for motility on soft agar.The desired cotransductant will be Kanr and will bemotile Given that one P1 phage headfull corresponds to about 1/50 of the E.coli chromosome, about 1 in 500 Tn5 insertions will be close enough to the MotA gene to show 90% cotransduction.Thus if we test about 103 Kanr transductants for motility,we are likely to find at least one that has cotransduced the MotA*marker 3)Once a Tn5(Kanr)MotA+transductant has been identified,grow P1 on Tn5 (Kan) MotA+. 4)Use the P1 phage from step 3)to infect a MotA-recipient strain.Select for transduction of the Tn5 insertions by selecting for growth of the transductants on kanamycin plates. Test the resulting Kanr transductants for their motility.The transductants that have cotransduced the MotA+marker will be motile,whereas the transductants still contain the MotA-allele will be nonmotile.The fraction of the total transductants that are motile will give the distance between MotA-and the Tn5 insertion as a cotransduction frequency. A Tn5(Kan)MotA-transductant isolated in step 4)can then be used to transduce the MotA-marker into a new recipient strain by cotransduction with Tn5.Note that if we had isolated a second MotA-mutant,transduction into this strain would amount to a 3-factor cross and would provide a way to determine the order of the two different MotA-alleles

The steps for finding a linked Tn5 insertion are as follows: 1)Start with a collection of random Tn5 insertions into wild type E. coli (the isolation of such a collection was described in last lecture). Grow phage P1 on the mixture of 2x104 different Tn5 insertion mutants. Note that this donor strain is MotA+. 2) Use the resulting P1 phage to infect a MotA– recipient strain. Select for transduction of the Tn5 insertions by selecting for growth of the transductants on kanamycin plates. Screen for cotransduction of MotA+ by testing each of the Kanr transductants for motility on soft agar. The desired cotransductant will be Kanr and will bemotile. Given that one P1 phage headfull corresponds to about 1/50 of the E. coli chromosome, about 1 in 500 Tn5 insertions will be close enough to the MotA gene to show 90% cotransduction. Thus if we test about 103 Kanr transductants for motility, we are likely to find at least one that has cotransduced the MotA+ marker. 3)Once a Tn5 (Kanr) MotA+ transductant has been identified, grow P1 on Tn5 (Kanr) MotA+. 4) Use the P1 phage from step 3) to infect a MotA– recipient strain. Select for transduction of the Tn5 insertions by selecting for growth of the transductants on kanamycin plates. Test the resulting Kanr transductants for their motility. The transductants that have cotransduced the MotA+ marker will be motile, whereas the transductants still contain the MotA– allele will be nonmotile. The fraction of the total transductants that are motile will give the distance between MotA– and the Tn5 insertion as a cotransduction frequency. A Tn5 (Kanr) MotA– transductant isolated in step 4) can then be used to transduce the MotA– marker into a new recipient strain by cotransduction with Tn5. Note that if we had isolated a second MotA– mutant, transduction into this strain would amount to a 3-factor cross and would provide a way to determine the order of the two different MotA– alleles