F factor plasmid forms Hfr's in Salmonella typhimurium at many fewer sites and at a much lower frequency than in Escherichia coli. Based upon the way Hfr's are formed, what would be a simple explanation for this difference between S. typhimurium and E. coli?

ANSWER: Recall that most Hfr's arise via homologous recombination between IS sequences (IS2, IS3, and Tn1000) that are present on both the F-plasmid and on the E. coli chromosome. The S. typhimurium chromosome lacks copies of these IS elements, so any Hfr's that arise must do so via transposition.

Why does the transfer of genes from a Hfr require a functional recA gene in the recipient, but transfer of genes on a F' does not?

ANSWER: Transfer of genes from an Hfr into an F- recipient cell requires the recA gene product because the incoming genes need to be inherited by homologous recombination. This two cross-over event is mediated by the recA gene. Since the incoming genes are on an unstable linear fragment of DNA, if the genes aren't recombined into the chromosome by RecA, then the DNA will be degraded by exonucleases.

Transfer of an F' can occur in a recA mutant because the F' does not need to recombine with the recipient chromosome in order to be stably inherited. The final step in the transfer of an F' is a site-specific recombination event between the first and second copy of oriT to be transferred into the recipient cell. This single cross-over event is mediated by one of the tra genes encoded on the F' plasmid, and is independant of recA. This site-specific recombination event occurs at the conjugal bridge between the two cells, and results in a new F' plasmid in the recipient cell that may complement mutant genes on the recipient chromosome, if the wild-type copies are present on the F'.

The proA, proB, and proC genes are required for the biosynthesis of proline. Given a donor strain with a Hfr integrated between the proA+proB+ and proC+ genes as shown below, and a proC- StrR recipient, how could you isolate a F' proC+? [Show a diagram of the isolation scheme you would use.]

ANSWER: The F' proC+ would form as shown in the figure below. Because the proC gene would be the very last region to be transferred by the Hfr, it would only be transferred by the Hfr very rarely. Therefore, you could select for such rare events by mating into a proC auxotroph and selecting for growth on minimal medium without proline.

Given the same donor strain used in the question above and an appropriate recipient strain, how could you isolate an F' proA+B+? [What recipient strain would you use? Show a diagram of the isolation scheme you would use.]

ANSWER: The F' proBA+ would form as shown in the figure below. Because the proBA genes are immediately behind the origin of the Hfr, the Hfr would transfer thse genes at high frequency into a recipient cell. Therefore, in this case you cannot simply select for such rare events by mating into a proC auxotroph and selecting for growth on minimal medium without proline. However, you could select for the desired F' by mating with a proC recA recipient and selecting for ProC+ because recircularization of the F' in the recipient does not require RecA.

A new mutant was isolated that is StrR and unable to use acetate as a carbon source (ace). To determine where the mutation maps, it was mated with the four different StrS ace+ Hfr donor strains shown below. [Arrowheads indicate the location and direction of transfer from each different Hfr.]

  1. What is the selection for exconjugants in this experiment?

    ANSWER: Growth on acetate as a sole C-source.

  2. What is the counterselection against the donor cells in this experiment?

    ANSWER: Streptomycin resistance.

  3. Given the results in the following table, where does the ace mutation map? [Indicate the map position in minutes.]

    ANSWER: About 95 min, in the region transferred early between Hfr-1 and Hfr-5.

Several Hfrs from E. coli are shown in the figure below with the direction of transfer indicated relative to the map shown at the top of the figure.

All of these Hfr strains carry wild-type copies of the genes shown on the map (i.e., leu+, his+, srl+, Strs, and metA+). Each Hfr was mixed with three recipient strains in a separate tube for each mating pair, then the donor and recipient cells were spread on minimal plates with glucose as a carbon source and streptomycin. The three recipient strains used were:

  1. What is the selection in this experiment?

    ANSWER: Prototrophy.

  2. What is the counterselection in this experiment?

    ANSWER: Str resistance.

  3. What controls are needed to feel confident that any colonies obtained are not spontaneous revertants?

    ANSWER: Plate "donor only" and "recipient only" cells on minimal plates with glucose as a carbon source and streptomycin. The number of donor cells or recipient cells that grow on these plates should be much less than the number of transconjugants.

  4. What is the relative numbers of transconjugants expected on each mating plate? [In the following table indicate +++ for many colonies, + for an intermediate number of colonies, or - for no colonies.

    ANSWER: See table below. Recall genetic nomenclature: designation metA indicates that a mutation in the metA gene.

Wild-type Salmonella typhimurium LT2 is unable to use histidine as a sole nitrogen source (Hut-). A S. typhimurium mutant was isolated that acquired the ability to use histidine as a nitrogen source (Hut+). To determine where the mutation mapped, LT2 was mated with five different Hut+ Met- Hfr donor strains. The metA mutation makes the donor cells methionine auxotrophs. The map position and orientation of each Hfr donor is shown in the figure below.

  1. How could you select against the recipient cells in this experiment?

    ANSWER: Hut+ = medium with histidine as a sole-nitrogen source

  2. How could you counterselect against the donor cells in this experiment?

    ANSWER: Met+ = medium with no methionine

  3. The results are shown in the following table. Based upon these results, what is the most likely map position of the hut+ mutation? [Indicate the map position relative to the Hfr origins.]

    ANSWER: Because Hfr1 and Hfr2 transfer the chromosomal DNA in opposite orientations and neither transfers the hut gene, but Hfr3 and Hfr5 both transfer this gene, then the gene must map between the F-insertion of Hfr 1 and Hfr 2.

Four different Hfr strains of E. coli were mated to F- recipients to determine the time of entry of various donor markers. The results are shown below. From these data constuct a genetic map which includes all of the markers, and label the distance (based upon time of entry) between adjacent marker pairs.

Hfr #1	arg (15 min)	thy (21 min)	met (32  min)	thr (48 min)
Hfr #2	mal (10 min)	met (17 min)	thi (22  min)	thr (33 min)	trp (57 min)
Hfr #3	phe (6 min)	his (11 min)	bio (33 min)	azi (48 min)	thr (49 min)	thi (60 min)
Hfr #4	his (18 min)	phe (23 min)	arg (45  min)	mal (55 min)

ANSWER: See figure below.

Given two E. coli strains with the genotypes show below.

Strain A:	F-	lac-	arg-	his-	pyr-	lys-	TonR
Strain B:	Hfr	lac+	arg+	his+	pyr+	lys+	Tons

[lac- = inability to ferment the sugar lactose; arg-, his-, pyr-, lys- = inability to synthesize argining, histidine, pyrimidines, or lysine; Tons and TonR = sensitivity and resistance to the virulent phage T1]

  1. What supplements, etc, would you put in the plates to select lac+ arg+ recombinants from a mating between strains A and B?

    ANSWER: Minimal plates with lactose as a sole carbon source, histidine, pyrimidines (e.g. uracil), and lysine -- NO arginine.

  2. What class of recombinants would be selected on minimal plates containing glucose, arginine, and phage T1?

    ANSWER: Arg+ TonR His+ Pyr+ Lys+

  3. A and B were mixed and allowed to mate for several hours. The cells were then plated to observe recombinants. The results are shown in the table below. Which marker enters first? Which entered last?

    ANSWER: The markers entered in the order -- his, lys or arg, pyr, lac. Note that you cannot determine the order of entry of lys and arg.

Given a Ser+ Met+ Hfr strain where F is integrated in the chromosome as shown below.

  1. How would you isolate an F' Met+ from this Hfr? (The Hfr strain can carry any genetic markers you choose, but be sure to indicate what they are.)

    ANSWER: Mate the Hfr with an F- met- recA StrR recipient, selecting Met+ StrR. The recA mutation in the recipient will prevent inheritance of Met+ directly via the Hfr, and the StrR provides a counter-selection against the donor.

  2. Indicate the composition of the medium you would need to use (including auxotrophic supplements).

    ANSWER: To select for Met+, the donor + recipient should be plated on minimal glucose medium without methionine. The medium should also contain streptomycin to prevent growth of the donor cells.

You have two strains of E. coli. One strain is HfrH and the second strain carries F'801, an smaller F' that was derived from HfrH (see map below).

  1. How could you distinguish between the two strains by purely genetic means? Assume that both strains are prototrophic. You can use any other strain(s) of E. coli you wish.

    ANSWER: Mate the two strains with a thr ser recA RifR recipient. Ser+ Thr+ RifR colonies will arise from the mating with the strain carrying F'801 but not with HfrH.

  2. What was the selection and counter-selection used?

    ANSWER: The selection was for Ser+ Thr+, and the counter-selection was RifR (i.e., growth on minimal glucose medium with rifampicin but without serine or threonine).

The E. coli strain KL98 contains an F-plasmid integrated adjacent to the dsdA+ gene (required for catabolism of d-serine) at about 54 min on the E. coli chromosome. The origin is oriented such that dsdA+ is transferred into the recipient soon after the initiation of conjugation. Using the genetic map of E. coli shown below and assuming you have any recipient strain you need, how could you isolate an F' carrying the dsdA+ gene? [Draw a diagram showing the relevant genotype of the donor and recipient, how the F' is formed, and how you would do the experiment including any media you would use. Note that mutations in any of the genes shown on the following genetic map results in auxotrophy.]

The chromosome organization of Salmonella typhimurium and Salmonella typhi is shown below. Overall the DNA sequence of genes in these two bacteria is about 99% identical. The rrn operons encode ribosomal RNA. Although the location of the rrn operons differs between these two bacteria, the order of the genes located between the rrn operons is conserved (e.g., the order of genes within fragment A is the same in both bacteria).

To allow the isolation of hybrids between these two bacteria, Anne Thierauf constructed a plasmid with the following properties: (i) the origin of replication was derived from the plasmid R6K so it requires the protein (encoded by the pir+ gene) to replicate; (ii) the plasmid does not carry the pir+ gene but the Pi protein can be provided in trans; (iii) the plasmid contains a clone with the promoter and the first half of an rrn operon; (iv) the plasmid encodes AmpR. She used these clones to construct Hfr's in S. typhimurium.

  1. Draw a diagram showing how one possible Hfr would be formed. [Include the relevant genes on the plasmid and the chromosome and any selections needed.]

  2. If the Hfr was integrated into the rrnB operon of S. typhimurium with oriT orientated such that it transferred in the clockwise direction, what is the maximal amount of S. typhi DNA you would be able to replace with S. typhimurium DNA following mating? [Draw a diagram showing the recombination event to emphasize your rationale.]

A map of the S. typhimurium chromosome is shown below. The putA gene maps adjacent to the pyrC gene at 22 min on the genetic map. The putA gene product is required for growth on proline as a carbon source. The pyrC gene is required for biosynthesis of uracil. Given a Hfr insertion in the pyrC gene, how could you isolate a F' that carries the putA+ gene? [Note: you do not know the orientation of transfer of the Hfr.] Be sure to describe any mutations in the donor or recipient needed for your experiment.

ANSWER: Because you do not know the orientation of the Hfr, you do not know whether the putA gene would be transferred by the Hfr early or late. Therefore, to obtain an F' that carries the putA+ gene you could mate an Hfr put+ metA donor with a put metA+ recA recipient, selecting for Put+ Met+ transconjugants. [In this example, the counterselection was demanding methionine prototrophy, but you could also use a variety of other counterselections.]

You have just isolated mutants which require the amino acid isoleucine for growth (ile-). Mapping experiments showed that the mutants map in the region between 29 and 30 minutes on the E. coli chromosome. Genes necessary for the biosynthesis of lysine (lys) and methionine (met) map in the same region.

lys ile met

How could you construct an Hfr which transfers the wild-type ile gene early during conjugation. The strains you have available are: (i) a prototrophic F+ strain, and (ii) a F- strain with any markers you want (except an Hfr which transfers ile early). Describe a procedure you could use to isolate an Hfr strain which transfers ile early. Be sure to describe all the steps required and how you would show that the ile locus is transferred early. Describe the composition of the media you would use for each step.

A mating was done between Hfr H met AziR TonR StrS and F- thr leu lac gal AziS TonS StrR. The cross is plated on glucose-minimal plates containing methionine, azide, and streptomycin. What are the selected markers?

The recombinants from the above cross were plated on nutritional broth agar plates containing the indicator dye and carbon source given below. What are the genotypes for the recombinants with respect to these carbon sources?

Carbon SourceIndicator dyeColor of
recombinant colonies
galactoseEMBred or purple

Gardner isolated a strain of E. coli that was resistant to the antibiotics tetracycline (TetR) and kanamycin (KanR) but sensitive to nalidixic acid (NalS). To determine whether the antibiotic resistance was encoded on a conjugal plasmid, he mated this TetR KanR donor with two different TetS KanS NalR E. coli recipients. The results are shown below. "Donor only" and "recipient only" controls are also shown.

  1. Given the results in the table, what can you conclude about the transfer of the TetR and KanR genes? [Explain your logic.]

    ANSWER: Because they are transferred at different frequencies, they are probably present on different DNA molecules (e.g. separate plasmids).

  2. What is a likely reason for the difference between the results with the two different recipient strains?

    ANSWER: The restriction barrier in E. coli B. Recall that E. coli C is restriction minus.

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Last modified October 18, 2004