Conjugation

In 1946, Joshua Lederberg and Edward Tatum demonstrated that bacteria can transfer and recombine genetic information, paving the way for the use of bacteria in genetic studies. In the course of their research, Lederberg and Tatum studied auxotrophic strains of E. coli. The Y10 strain required the amino acids threonine (and was genotypically thr) and leucine (leu) and the vitamin thiamine (thi) for growth, but did not require the vitamin biotin (bio+) or the amino acids phenylalanine (phe+) and cysteine (cys+); the genotype of this strain can be written as thr leu thi bio+ phe+ cys+. The Y24 strain had the opposite set of alleles: it required biotin, phenylalanine, and cysteine in its medium, but it did not require threonine, leucine, or thiamine; its genotype was thr+ leu+ thi+ bio phe cys. In one experiment, Lederberg and Tatum mixed Y10 and Y24 bacteria together and plated them on minimal medium (Figure 7.8). Each strain was also plated separately on minimal medium.

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Figure 7.8: Lederberg and Tatum’s experiment demonstrated that bacteria undergo genetic exchange.

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Alone, neither Y10 nor Y24 grew on minimal medium: each strain required nutrients that were absent. Strain Y10 was unable to grow because it required threonine, leucine, and thiamine, which were absent in the minimal medium; strain Y24 was unable to grow because it required biotin, phenylalanine, and cysteine, which also were absent from the minimal medium. When Lederberg and Tatum mixed the two strains, however, a few colonies did grow on the minimal medium. These prototrophic bacteria must have had genotype thr+ leu+ thi+ bio+ phe+ cys+. Where had they come from?

If mutations were responsible for the prototrophic colonies, then some colonies should also have grown on the plates containing Y10 or Y24 alone, but no bacteria grew on those plates. Multiple simultaneous mutations (thrthr+, leuleu+, and thithi+ in strain Y10 or biobio+, phephe+, and cyscys+ in strain Y24) would have been required for either strain to become prototrophic by mutation, which was very improbable. Lederberg and Tatum concluded that some type of genetic transfer and recombination had taken place:

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What they did not know was how it had taken place.

To study this problem, Bernard Davis constructed a U-shaped tube (Figure 7.9) that was divided into two compartments by a filter with fine pores. This filter allowed liquid medium to pass from one side of the tube to the other, but the pores of the filter were too small to allow the passage of bacteria. Two auxotrophic strains of bacteria were placed on opposite sides of the filter, and suction was applied alternately to the ends of the U-tube, causing the medium to flow back and forth between the two compartments. Despite hours of incubation in the U-tube, bacteria plated on minimal medium did not grow; there had been no genetic exchange between the strains. The exchange of bacterial genes clearly required direct contact, or conjugation, between the bacterial cells.

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Figure 7.9: Davis’s U-tube experiment.

F+ AND F CELLS In most bacteria, conjugation depends on a fertility (F) factor that is present in the donor cell and absent in the recipient cell. Cells that contain the F factor are referred to as F+, and cells lacking the F factor are F.

The F factor is an episome that contains an origin of replication and a number of genes required for conjugation (see Figure 7.6). For example, some of these genes encode sex pili (singular, pilus), slender extensions of the cell membrane. A cell containing the F factor produces sex pili, one of which makes contact with a receptor on an F cell (Figure 7.10) and pulls the two cells together. DNA is then transferred from the F+ cell to the F cell. Conjugation can take place only between a cell that possesses the F factor and a cell that lacks the F factor.

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Figure 7.10: A sex pilus connects F+ and F cells during bacterial conjugation. Here, E. coli cells are shown in conjugation.
[Eye of Science/Science Source.]

In most cases, the only genes transferred during conjugation between an F+ and F cell are those on the F factor (Figure 7.11a and b). Transfer is initiated when one of the DNA strands on the F factor is nicked at an origin of transfer (oriT). One end of the nicked DNA separates from the circular F factor and passes into the recipient cell (Figure 7.11c). Replication takes place on the nicked strand, proceeding around the F factor in the F+ cell and replacing the transferred strand (Figure 7.11d). Because the F factor in the donor cell is always nicked at the oriT site, this site always enters the recipient cell first, followed by the rest of the plasmid. Thus, the transfer of genetic material has a defined direction. Inside the recipient cell, the single strand is replicated, producing a circular, double-stranded copy of the F plasmid (Figure 7.11e). If the entire F factor is transferred to the recipient F cell, that cell becomes an F+ cell.

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Figure 7.11: The F factor is transferred during conjugation between an F+ and an F cell.

Hfr CELLS Conjugation transfers genetic material in the F plasmid from F+ to F cells, but it does not account for the transfer of chromosomal genes observed by Lederberg and Tatum. In Hfr (high-frequency recombination) bacterial strains, the F factor is integrated into the bacterial chromosome (Figure 7.12). Hfr cells behave like F+ cells, forming sex pili and undergoing conjugation with F cells.

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Figure 7.12: The F factor is integrated into the bacterial chromosome in an Hfr cell.

In conjugation between Hfr and F cells (Figure 7.13a), the integrated F factor is nicked, and the end of the nicked strand moves into the F cell (Figure 7.13b), just as it does in conjugation between F+ and F cells. But because, in an Hfr cell, the F factor has been integrated into the bacterial chromosome, the chromosome follows the F factor into the recipient cell. How much of the bacterial chromosome is transferred depends on the length of time that the two cells remain in conjugation.

Inside the recipient cell, the donor DNA strand is replicated (Figure 7.13c), and crossing over between it and the original chromosome of the F cell (Figure 7.13d) may take place. This chromosomal gene transfer between Hfr and F cells explains how the recombinant prototrophic cells observed by Lederberg and Tatum were produced. After crossing over has taken place in the recipient cell, the donated chromosome is degraded, and the recombinant recipient chromosome remains (Figure 7.13e), to be replicated and passed on to later generations of bacterial cells by binary fission (cell division).

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Figure 7.13: Bacterial genes may be transferred from an Hfr cell to an F cell in conjugation. In an Hfr cell, the F factor has been integrated into the bacterial chromosome.

In a mating of Hfr × F, the F cell almost never becomes F+ or Hfr because the F factor is nicked in the middle in the initiation of strand transfer, which places part of the F factor at the beginning and part at the end of the strand that is transferred. To become F+ or Hfr, the recipient cell must receive the entire F factor, which requires that the entire donor chromosome be transferred. This event happens rarely because most conjugating cells break apart before the entire chromosome has been transferred.

The F plasmid in F+ cells integrates into the bacterial chromosome, causing an F+ cell to become Hfr, at a frequency of only about 1 in 10,000. This low frequency accounts for the low rate of recombination observed by Lederberg and Tatum in their F+ cells. The F factor is excised from the bacterial chromosome at a similarly low rate, causing a few Hfr cells to become F+.

F′ CELLS When an F factor is excised from the bacterial chromosome, a small amount of the bacterial chromosome may be removed with it, and these chromosomal genes will then be carried with the F plasmid (Figure 7.14). Cells containing an F plasmid with some bacterial genes are called F prime (F′) cells. For example, if an F factor integrates into a chromosome adjacent to the lac genes (genes that enable a cell to metabolize the sugar lactose), the F factor may pick up lac genes when it is excised, becoming an F′ lac cell. F′ cells can conjugate with F cells because F′ cells possess the F plasmid with all the genetic information necessary for conjugation and gene transfer. Characteristics of E. coli cells of different F types (called mating types) are summarized in Table 7.2.

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Figure 7.14: An Hfr cell may be converted into an F′ cell when the F factor excises from the bacterial chromosome and carries bacterial genes with it. Conjugation produces a partial diploid.
TABLE 7.2 Characteristics of E. coli cells with different types of F factors
Type F factor characteristics Role in conjugation
F+ Present as separate circular plasmid Donor
F Absent Recipient
Hfr Present, integrated into bacterial chromosome High-frequency donor
F′ Present as separate circular plasmid, carrying some bacterial genes Donor

During conjugation between an F′ cell and an Fcell, the F plasmid is transferred to the F cell, which means that any genes on the F plasmid, including those from the bacterial chromosome, may be transferred to F recipient cells. This process produces partial diploids, or merozygotes, which are cells with two copies of some genes, one on the bacterial chromosome and one on the newly introduced F plasmid. The outcomes of conjugation between different mating types of E. coli are summarized in Table 7.3.

TABLE 7.3 Results of conjugation between cells with different types of F factors
Conjugating Cell types present after conjugation
F+ × F Two F+ cells (F cell becomes F+)
Hfr × F One Hfr cell and one F (no change)*
F′ × F Two F′ cells (F cell becomes F′)

* Rarely, the F cell becomes F+ in an Hfr × F conjugation if the entire chromosome is transferred during conjugation.

CONCEPTS

Conjugation in E. coli is controlled by an episome called the F factor. Cells containing the F factor (F+ cells) are donors of genes; cells lacking the F factor (F cells) are recipients. In Hfr cells, the F factor is integrated into the bacterial chromosome; these cells donate DNA to F cells. F′ cells contain a copy of the F factor with some chromosomal genes.

image CONCEPT CHECK 3

Conjugation between an F+ and an F cell usually results in

  1. two F+ cells.

  2. two F cells.

  3. an F+ and an F cell.

  4. an Hfr cell and an F+ cell.

a

MAPPING BACTERIAL GENES WITH INTERRUPTED CONJUGATION The transfer of DNA that takes place during conjugation between Hfr and F cells allows bacterial genes to be mapped. In conjugation, as we have seen, the chromosome of the Hfr cell is transferred to the F cell. Transfer of the entire E. coli chromosome requires about 100 minutes; if conjugation is interrupted before 100 minutes have elapsed, only part of the donor chromosome will pass into the F cell and have an opportunity to recombine with the recipient chromosome. Chromosome transfer always begins within the integrated F factor and proceeds in a continuous direction, so genes are transferred according to their sequence on the chromosome. This knowledge can be used to map bacterial genes by mixing Hfr and F cells that differ in genotype and interrupting conjugation at regular intervals. The times required for individual genes to be transferred indicate their relative positions on the chromosome. In most genetic maps, distances are expressed as recombination frequencies; however, in bacterial gene maps constructed with interrupted conjugation, the basic unit of distance is a minute. View Animation 7.1 to see how genes are mapped using interrupted conjugation.

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CONCEPTS

Conjugation can be used to map bacterial genes by mixing Hfr and F cells that differ in genotype and interrupting conjugation at regular intervals. The amounts of time required for individual genes to be transferred from the Hfr to the F cells indicate the relative positions of the genes on the bacterial chromosome.

WORKED PROBLEM

To illustrate the method of mapping genes with interrupted conjugation, let’s look at a cross analyzed by François Jacob and Elie Wollman, who developed this method of gene mapping (Figure 7.15a). They used donor Hfr cells that were sensitive to the antibiotic streptomycin (genotype strs), resistant to sodium azide (azir) and infection by bacteriophage T1 (tonr), prototrophic for threonine (thr+) and leucine (leu+), and able to break down lactose (lac+) and galactose (gal+). They used F recipient cells that were resistant to streptomycin (strr), sensitive to sodium azide (azis) and to infection by bacteriophage T1 (tons), auxotrophic for threonine (thr) and leucine (leu), and unable to break down lactose (lac) and galactose (gal). Thus, the genotypes of the donor and recipient cells were

Donor Hfr cells: strs leu+ thr+ azir tonr lac+ gal+

Recipient F cells: strr leu thr azis tons lac gal

The two strains were mixed in nutrient medium and allowed to conjugate. After a few minutes, the medium was diluted to prevent any new pairings. At regular intervals, a sample of cells was removed and agitated vigorously in a kitchen blender to halt all conjugation and DNA transfer. The cells from each sample were plated on a selective medium that contained streptomycin and lacked leucine and threonine. The original donor cells were streptomycin sensitive (strs) and would not grow on this medium. The F recipient cells were auxotrophic for leucine and threonine, and they also failed to grow on this medium. Only recipient cells that underwent conjugation and received at least the leu+ and thr+ genes from the Hfr donors could grow on this medium. All strr leu+ thr+ cells were then tested for the presence of other genes that might have been transferred from the donor Hfr strain.

Because Jacob and Wollman used streptomycin to kill all the donor cells, they were not able to examine the transfer of the strs gene. All of the cells that grew on the selective medium were leu+ thr+, so we know that these genes were transferred. In Figure 7.15b, the percentages of strr leu+ thr+ cells receiving specific alleles (azir, tonr, leu+, and gal+) from the Hfr strain are plotted against the duration of conjugation. What is the order in which the genes are transferred and what are the distances among them?

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Figure 7.15: Jacob and Wollman used interrupted conjugation to map bacterial genes.

Solution Strategy

What information is required in your answer to the problem?

The order of the genes on the bacterial chromosome and the distances among them.

What information is provided to solve the problem?

  • The donor cells were strs leu+ thr+ azir tonr lac+ gal+ and the recipient cells were strr leu thr azis tons lac gal.

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  • The percentages of recipient cells with different traits that appear at various times after the start of conjugation (Figure 7.15b).

Solution Steps

The first donor gene to appear in the recipient cells (at about 9 minutes) was azir. Gene tonr appeared next (after about 10 minutes), followed by lac+ (at about 18 minutes) and by gal+ (after 25 minutes) (see Figure 7.15b). These transfer times indicate the order of gene transfer and the relative distances among the genes.

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Notice that the frequency of gene transfer from donor to recipient cells decreased with distance from the origin of transfer. For example, about 90% of the recipients received the azir allele, but only about 30% received the gal+ allele. The lower percentage for gal+ is due to the fact that some conjugating cells spontaneously broke apart before they were disrupted by the blender. The probability of spontaneous disruption increases with time, so fewer cells had an opportunity to receive genes that were transferred later.

image For additional practice mapping bacterial genes with interrupted conjugation, try Problem 15 at the end of the chapter.