It is often desirable to maintain two different plasmids in a single cell. What are two important considerations when choosing the plasmids to use?

ANSWER: The plasmids should be compatible so they can stably co-exist in the same cell, and the plasmids should have distinguishable phenotypes so you can insure that both plasmids are present (ideally a phenotype that is selectable).

Replication of plasmid ColE1 is catalyzed by the host encoded DNA polymerase I protein (Pol I). In wild-type cells, Pol I is present in large excess over that needed for plasmid replication. Initiation of plasmid replication is down regulated by the plasmid encoded Rop protein which is present in limiting amounts. If the protein synthesis inhibitor chloramphenicol was added to sensitive cells that carry a ColE1 plasmid, what would happen to the plasmid copy number and why?

ANSWER: Addition of chloramphenicol will prevent synthesis of new proteins, including both Pol I and Rop. Pol I is present in excess, so even in the absence of protein synthesis there is sufficient Pol I in the cells for many rounds of plasmid replication. In contrast, Rop is limiting, so in the absence of new protein synthesis there is insufficient Rop to inhibit plasmid replication. This results in multiple rounds of plasmid replication and the accumulation of plasmids to a copy number much greater than normally found (up to about 1000 copies of the plasmid per cell), a process called plasmid amplification. This can be used as a trick to facilitate purification of plasmid DNA from cells.

Copy number mutants of ColE1 type plasmids have been isolated with "run-away" replication at 42 C (resulting in high copy number) but normal replication (and thus copy number) at 30 C. What type of mutation in which gene might cause this phenotype? [Explain your answer.]

ANSWER: The most likely mutation is in the rop gene because rop mutants have increased plasmid copy number, and temperature sensitive mutations are most often due to changes in protein structure. An alternative explanation is that the mutation destablizes a stem-loop structure in the RNA which plays a role in regulation. (It is possible to obtain temperature sensitive mutations that decrease the stability of nucleic acid hybrids but, since most nucleic acid hybrids involve a considerable number of base pairs, a single nucleotide substitution is unlikely to have a substantial effect on the Tm of the hybrid. Furthermore, in this case the hybrid is between complementary strands so a nucleotide substitition would result in a different base pair at that position, not mispairing at that position.)

Mutations in the "RNA I - primer RNA" overlap region of plasmid ColE1 can result in a a very high copy number. Other mutations in the "RNA I-primer RNA" overlap region decrease the plasmid copy number.

  1. Based upon what you know about the role of these RNAs, propose an explanation for the high copy number phenotype and the low copy number phenotype [i.e., what do the mutations do in each case?].

    ANSWER: Mutations that destabilize the RNA I - primer RNA would result in a higher copy number (for example, mutations that result in decreased base-pairing between the two RNAs). Mutations that stabilize the RNA I-primer RNA hybrid would result in a lower copy number (for example, mutations that result in stronger base-pairing between the two RNAs).

  2. Second-site mutations that map on the E. coli chromosome can suppress the low copy number phenotype described above. What enzyme might be affected in these mutants (and why)?

    ANSWER: Anything that stabilizes the RNA I-primer RNA hybrid would result in a lower copy number phenotype. A mutation that increases a host protease with specificity for the plasmid Rop protein could decrease the intracellular concentration of Rop protein and thus decrease the stability of the RNA hybrid. Alternatively, an increase in the concentration or specificity of RNase H could result in enhanced cleavage of the primer RNA and thus increasing initiation of plasmid replication and the plasmid copy number.

It is possible to obtain mutations in plasmid encoded genes by mutagenizing the purified plasmid DNA in vitro (for example, with hydroxylamine) then transforming the plasmid DNA into recipient cells. Given a plasmid that encodes ampicillin resistance (AmpR), chloramphenicol resistance (CamR), and B-galactosidase (lacZ), how would you use this approach to isolate single mutations in the lacZ gene? Be sure to describe any genetic selections and/or screens used. [Hint: you will need a way of measuring the relative amount of mutagenesis to ensure a reasonable probability that the lacZ mutants are due to a single mutation.]


It is possible to "cure" a strain of the plasmid pLAFR which encodes resistance to tetracycline (TetR) by mating in a second plasmid pPH1JI which encodes resistance to gentamycin (GenR).

  1. What does this suggest about the properties of these two plasmids? [Explain your answer.]

    ANSWER: The two plasmids are probably incompatible. This means that only one of the plasmids can be stably maintained in the cell. Hence, selection for the second plasmid results in loss in the first plasmid.

  2. Would this trick work if the only selectable marker on pPH1JI was TetR? [Explain why or why not.]

    ANSWER: No. Because the recipient cell is already TetR, there is no selection for inheritance of the second plasmid. Hence, very few cells would probably be transformed and even if they were you would not be able to distinguish them from the untransformed cells.

Replication of plasmids with a pR6K origin (oriR6K) requires a protein called pi (encoded by the pir gene).

  1. What would happen if a plasmid with oriR6K that encodes ampicillin resistance (AmpR) and carries the putA+ gene was introduced into a recipient that is pir+ putA(Am)? [Would the plasmid replicate? Would the plasmid integrate into the chromosome? Would the cell be Put+ or Put- and why?]

    ANSWER: See figure below.

  2. What would happen if a plasmid with oriR6K that encodes ampicillin resistance (Ampr) and carries the putA+ gene was introduced into a recipient that is pir putA(Am)? [Would the plasmid replicate? Would the plasmid integrate into the chromosome? Would the cell be Put+ or Put- and why?]

ANSWER: See figure below.

Normally multicopy plasmids do not integrate into the chromosome even when they contain considerable DNA sequence homology for recombination. Gutterson and Koshland (1983. Proc. Natl. Acad. Sci. USA 80: 4894-4898) showed that multicopy plasmids (such as pBR322) readily integrate into the chromosome of polA mutants. Suggest an explanation for both of these results.

ANSWER: The polA gene encodes DNA polymerase I, which is required by the cell in tiny amounts for survival but is needed in large quantities by certain plasmid origins to initiate DNA replication. Thus, a polA(Am) mutation, which is slightly leaky, provides enough DNA PolI for the cell to grow but not enough for plasmids such as pBR322 to replicate autonomously in the cytoplasm. Under this condition the plasmid can be stably inherited only if it integrates into the chromosome via a single cross-over event using regions homologous to the chromosome. In a polA cell the plasmid can autonomously replicate in the cytoplasm, but the homology is still there to allow integration into the chromosome, and this will happen at the same frequency as in the polA mutant. However, the plasmid origin is still functional and expressed at a high level to increase the copy number of the plasmid. There is a lethal effect because of having a functional plasmid origin disrupting the function of the chromosomal origin and increased copy number could lead to overexpression of chromosomal genes located adjacent to the integrated plasmid, which may also be lethal. So the take home point is that integration still occurs in the polA+ strain, but cells with the integrated plasmid are not recovered because of the lethal consequences.

ColE1 plasmids replicate in enteric bacteria but cannot replicate in Halobacterium salinarium. The bop gene from H. salinarium was cloned into a ColE1 plasmid, and an insertion mutation constructed that disrupted the plasmid encoded bop gene (indicated by bop'-'bop in the figure below). This plasmid was then transformed into H. salinarium with selection for resistance to the antibiotic mevinolin (MevR).

  1. The plasmid does not have any functional replication origin and, except for the bop gene, the plasmid lacks any homology with the H. salinarium chromosome. How do the MevR transformants arise? [Show a diagram and briefly describe your answer.]

    ANSWER: The plasmid cannot replicate, so the MevR colonies must arise by integration of the plasmid into the chromosome by homologous recombination between the bop genes.

  2. When the resulting transformants were subsequently grown for many generations without mevinolin (Mev), some Mevs colonies were obtained. About half of these Mevs colonies were Bop+ and half were Bop-. How do the Mevs Bop- colonies arise? [Show a diagram and briefly describe your answer.]

    ANSWER: Segregration of the chromosomal bop duplication will yield Mevs colonies. Sometimes the recombination will occur on the same side of the bop disruption as the initial integration event and these will result in bop+ colonies [crossover (i) in the figure below]. Sometimes the recombination will occur on the other side of the bop disruption relative to the initial integration event and these will result in bop- colonies [crossover (ii) in the figure below].

Two plasmids are shown below. The plasmid pUC19 was constructed from pBR322 and the origin from pBR322 came from a pMB9 plasmid. The region designated ORI on the two plasmids is identical, but pUC19 has greater than 100 copies per cell in E. coli while pBR322 has about 20 copies per cell.

  1. Why is the copy number of these two plasmids different? [Briefly describe the molecular mechanism.]

  2. When multicopy plasmids are used for complementation analysis, problems often arise. Describe an example where overexpression of a protein on a multicopy plasmid could give you erroneous complementation results.

Muro-Pastor and Maloy (1995. Biotechniques 18: 386-390) developed a method to easily move mutations from a multicopy plasmid onto the chromosome or from the chromosome onto a multicopy plasmid. [You will need to consult the article to answer the following questions.]

  1. What was the selection for integration of the plasmid onto the chromosome? [Briefly describe how this selection works.]

    ANSWER: The plasmid was transduced into a polA recipient strain, selecting for AmpR KanR colonies. Because pBR derivatives require the polA gene product to replicate, the antibiotic resistant transductants must arise by integration of the plasmid into the chromosome by homologous recombination. This integration event results in a tandem duplication of the cloned putA gene with the plasmid genes between the two copies of the gene.

  2. What was the selection for moving a chromosomal mutation onto the plasmid? [Briefly describe how this selection works.]

    ANSWER: A P22 lysate was grown on the strain with the plasmid integrated into the chromosome. This lysate was used to transduce the plasmid into a polA+ recipient strain. The recombinants were selected on plates with ampicillin and sucrose, then screened for KanS.

    The linear DNA brought in on the transducing fragment can recircularize by homologous recombination between the two copies of the putA gene (the chromosomal copy and the cloned copy that is disrupted with the kan and sac genes). Depending upon where the recombination event occurs, this will re-form the original plasmid (AmpR KanR SucroseS) or will leave the chromosomal putA gene on the plasmid (AmpR KanS SucroseR).

    Note that this selection scheme does not rely upon the PutA phenotype because the phenotype of the putA mutant on the multicopy plasmid was unknown.

Temperature sensitive mutations that prevent the initiation of replication of the E. coli chromosome are lethal at 42 C. However, Hfr strains that are known to contain such mutations survive at 42 C.

  1. Why is the temperature sensitive phenotype not observed in Hfr strains?

    ANSWER: When the chromosomal origin is inactive, replication can be initiated at the integrated F-plasmid origin and proceed around the chromosome. This process is called "integrative suppression".

  2. How could you prove that a Hfr strain still contains the original temperature sensitive mutation?

    ANSWER: (i) Loss of the Hfr would restore the original replication mutant phenotype. (For example, it is possible to select for loss of the Hfr, by recombination with adjacent markers but such transductants are rare due to the ccd plasmid addiction system of the F-plasmid.) (ii) Transfer of the chromosome origin into a new strain would confer the replication mutant phenotype. (For example, this could be done by selecting for an adjacent marker in an Hfr cross.) (iii) If the Hfr was integrated at a site distinguishable from the chromosomal replication origin, you could determine where replication initiates using molecular approaches. If replication initiates at the Hfr under nonpermissive conditions but at the chromosomal origin under premissive conditions, then the original replication mutation is present in this strain.

The plasmid RSF1010 carries three genes required for its own replication: a replication initiator protein, a DNA helicase, and a primase. Because RSF1010 is independent of the corresponding host replication functions, it is a very broad host range plasmid that can replicate in essentially any gram negative bacterium. However, there is one notable exception -- RSF1010 is unable to replicate in Bacteriodes.

  1. Suggest a potential explanation why RSF1010 is unable to replicate in Bacteriodes.
  2. How could you test this hypothesis?

The lac operon is often used as a target for mutagenesis studies because it is so easy to screen for the Lac phenotype on indicator plates. In many mutagenesis studies the lac genes are present on an F'. To calculated the mutation rate, the number of mutants are counted and the number of cell divisions are is estimated based upon the number of cells in the population. The validity of using this calculation to determine the mutation rate on the F' lac rests upon the assumption that replication of the F' is coordinated with the chromosome. This is a valid assumption when the F' replicates via bidirectional replication, but not when F' replication occurs via rolling circle replication. To avoid this complication, how could you prevent rolling circle replication of the F'?

ANSWER: The most effective way of preventing rolling circle replication would be to remove oriT from the F-plasmid. This would prevent conjugation but would not affect "vegetative" replication.

Mahan, Slauch, and Mekalanos (1993. Science 259: 686-688) developed a method to identify genes that are turned on when Salmonella infects a eukaryotic host. The plasmid they used for these studies is shown below. Briefly describe the functions provided by mob and oriR6K on this plasmid.

ANSWER: The mob function allows mobilization of the plasmid when transfer functions are provided in trans by a conjugative plasmid. The oriR6K plasmid replication origin depends upon the pi protein (encoded by the pir+ gene) which can also be provided in trans. In the absence of pi protein, the plasmid cannot replicate but if it carries a fragment of chromosomal DNA it can integrate into the host chromosome.

Integration of a cloned gene on a suicide plasmid is a useful trick for constructing a tandem chromosomal duplication. Given the plasmid shown below and a recipient with a pyrA(Am) mutation, how could you construct a chromosomal duplication for complementation analysis? [Draw a diagram showing the crossover and resulting chromosomal duplication]

Rob Edwards constructed Salmonella enteritis mutant with a KanR insertion in the sefA gene (designated sefA::Kan). He then cloned this sefA::Kan gene onto a derivative of plasmid R6K that lacks the pir gene. When the plasmid was transformed into S. enteritis, a low number of transformants was obtained. The plasmid was also transformed into two Salmonella typhimurium strains: strain A carried a copy of the pir+ gene cloned onto the chromosome and strain B did not have a copy of the pir+ gene. Many KanR transformants were obtained in strain A but none were obtained in strain B. Suggest a likely reason for these results. [For help, check out the discussion of "Allele exchange" on the 316 supplement.]

ANSWER: See figures below.

It is possible to disrupt a chromosomal gene by integration of an internal fragment from that gene cloned onto a suicide plasmid. How would integration of the plasmid shown below disrupt the adh gene? [Note that 'adh' represents an internal fragment of the adh gene that is missing DNA sequences from the beginning and end of the gene.]

The plasmids pBR328 and pUC19 are shown below. Both have a pMB9 origin. Because these plasmids share a common origin, they are incompatible. If an E. coli cell is co-transformed with both plasmids, then grown for many generations in medium with ampicillin, which of the plasmids would be most likely to be maintained? Why?

Recombination can occur between multiple copies of a plasmid in a cell. Such recombination results in plasmid dimers or multimers. Many plasmids encode a site specific recombination system that rapidly monomerizes any plasmid dimers formed.

  1. Draw a homologous recombination event that would generate a plasmid dimer, and the site-specific recombination event that would regenerate monomers.
  2. Why might a plasmid need an efficient site-specific recombination system to prevent the accumulation of dimers?

The plasmid pBR322 can replicate in E. coli but not in Bacteriodes thetaiodomicron, and many Bacteriodes plasmids can replicate in B. thetaiodomicron but not in E. coli. What parts of each plasmid would be needed to make a "shuttle plasmid" that could replicate in both hosts?

ANSWER: The shuttle plasmid would require the regions of DNA from both plasmids which function as an origin in each host. This would include any cis-acting sites as well as any trans-acting factors required from each plasmid.

The CcdA and CcdB gene products from F-plasmid function as a "plasmid addiction system". Given an F' that carries a gene for resistance to tetracycline:

  1. What would be the phenotype of a ccdB mutant? Why?

    ANSWER: It would be easier to detect segregrants of the F-plasmid because any segregrants would survive in the absence of the CcdB toxin.

  2. What would be the phenotype of a ccdA mutant? Why?

    ANSWER: These mutants would be lethal. In the absence of the CcdA anti-toxin, the CcdB toxin would kill the cells.

Simons, Housman, and Kleckner (1987. Gene 53: 85-96) developed a very useful trick for moving genes from a multicopy plasmid onto phage lambda by homologous recombination. Because lambda can be integrated into the chromosomal att site in a single copy, this trick makes it possible to easily do complementation tests with cloned genes. [You should be able to answer the following questions without referring to the published paper.]

Draw a diagram showing how you could use homologous recombination to move a DNA fragment from a plasmid onto phage lambda.

ANSWER: You could use homologous recombination between a region on the plasmid that has also been cloned into a nonessential region of phage lambda. Any gene cloned onto the plasmid between the homologous flanking regions could be recombined onto lambda as shown in the figure below.

How could you could select for the desired lambda derivatives?

ANSWER: If you placed an antibiotic resistance gene within the homologous flanking gene, you would have a simple selection for the desired recombinants. Simply grow the lambda lysate in a strain with the plasmid, and then infect a recipient strain selecting for antibiotic resistant lysogens.

When phage P1 lysogenizes cells, instead of integrating into the chromosome it remains as an extrochromosomal closed circular DNA. Nevertheless, P1 lysogens are quite stable. What functions would be required to ensure that the P1 DNA is not segregrated upon cell division?

ANSWER: Phage P1 requires partitioning (par) functions to insure that it co-segregrates with the host chromosome at each cell division.

Please send comments, suggestions, or questions to
Last modified October 30, 2003