Limits...
Alternative mechanisms for tn5 transposition.

Ahmed A - PLoS Genet. (2009)

Bottom Line: Bacterial transposons are known to move to new genomic sites using either a replicative or a conservative mechanism.The behavior of transposon Tn5 is anomalous.In vitro studies indicate that it uses a conservative mechanism while in vivo results point to a replicative mechanism.

View Article: PubMed Central - PubMed

Affiliation: Department of Biological Sciences, University of Alberta, Edmonton, Alberta, Canada. asada@ualberta.ca

ABSTRACT
Bacterial transposons are known to move to new genomic sites using either a replicative or a conservative mechanism. The behavior of transposon Tn5 is anomalous. In vitro studies indicate that it uses a conservative mechanism while in vivo results point to a replicative mechanism. To explain this anomaly, a model is presented in which the two mechanisms are not independent--as widely believed--but could represent alternate outcomes of a common transpositional pathway.

Show MeSH

Related in: MedlinePlus

Steps in the formation of the Tn5-promoted replicative inversion 621.The parent plasmid p4.1 (A) carried one copy of Tn5 (consisting of two inverted IS50 elements, L and R, flanking the kan gene), the trp, galTK, tet, and amp genes, and the cos site of lambda [6]. The galTK genes confer galactose-sensitivity (GalS) on the host cell, and selection for galactose-resistance (GalR) requires the disruption of this region. In inversion 621 (C), only one IS50 element (L) was left at the original location, a complete copy of Tn5 was found inserted in the gal region, and the trp-cos plasmid segment between the two had been inverted. This type III inversion (like several type I and II inversions also described in [6]) is fully consistent with the replicative mechanism as depicted in (B). It cannot be explained by the conservative mechanism. Small vertical arrows indicate location of nicks at the ends of IS50L and the target (galTK) DNA sequence. Horizontal arrows indicate inverted orientation of the two IS50 elements. The letters “o” and “I” refer to the outside and inside ends of the transposon, respectively. Figures are not drawn to scale. A deletion-inversion arising from p4.1 would be expected to have the following structure: the central kan region of Tn5 would be deleted, IS50L would join the gal region, the trp-cos segment would be inverted, and IS50R would also be inverted (to produce a direct repeat of IS50L) and join the target site in the gal region. This event would be consistent with the conservative mechanism, but it was never recovered from Tn5 in vivo. In contrast, the behavior of Tn10 was just the opposite of Tn5. p6A.1, a plasmid that is similar in structure to p4.1—except that it harbors Tn10 in place of Tn5—produced only deletion-inversions and no replicative inversions [6]. The two inside ends of Tn10 would be cleaved by double-strand breaks (as a result of hairpin formation and resolution), and the free 3′ ends would attack and join 5′ ends of the target sequence from the opposite strand. Such an event would produce deletion-inversions of the prescribed structure.
© Copyright Policy
Related In: Results  -  Collection


getmorefigures.php?uid=PMC2723962&req=5

pgen-1000619-g002: Steps in the formation of the Tn5-promoted replicative inversion 621.The parent plasmid p4.1 (A) carried one copy of Tn5 (consisting of two inverted IS50 elements, L and R, flanking the kan gene), the trp, galTK, tet, and amp genes, and the cos site of lambda [6]. The galTK genes confer galactose-sensitivity (GalS) on the host cell, and selection for galactose-resistance (GalR) requires the disruption of this region. In inversion 621 (C), only one IS50 element (L) was left at the original location, a complete copy of Tn5 was found inserted in the gal region, and the trp-cos plasmid segment between the two had been inverted. This type III inversion (like several type I and II inversions also described in [6]) is fully consistent with the replicative mechanism as depicted in (B). It cannot be explained by the conservative mechanism. Small vertical arrows indicate location of nicks at the ends of IS50L and the target (galTK) DNA sequence. Horizontal arrows indicate inverted orientation of the two IS50 elements. The letters “o” and “I” refer to the outside and inside ends of the transposon, respectively. Figures are not drawn to scale. A deletion-inversion arising from p4.1 would be expected to have the following structure: the central kan region of Tn5 would be deleted, IS50L would join the gal region, the trp-cos segment would be inverted, and IS50R would also be inverted (to produce a direct repeat of IS50L) and join the target site in the gal region. This event would be consistent with the conservative mechanism, but it was never recovered from Tn5 in vivo. In contrast, the behavior of Tn10 was just the opposite of Tn5. p6A.1, a plasmid that is similar in structure to p4.1—except that it harbors Tn10 in place of Tn5—produced only deletion-inversions and no replicative inversions [6]. The two inside ends of Tn10 would be cleaved by double-strand breaks (as a result of hairpin formation and resolution), and the free 3′ ends would attack and join 5′ ends of the target sequence from the opposite strand. Such an event would produce deletion-inversions of the prescribed structure.

Mentions: Both processes start by nicking (short vertical arrows) of the transposon ends to expose the 3′-OH termini (A). At some point (see below), the target DNA is also cleaved to provide short protruding 5′-PO4 ends. In replicative transposition (left), strand-transfer takes place by joining the 3′ ends to 5′ ends of the target DNA in a concerted cleavage and joining reaction to form the “Shapiro intermediate” (B). As a result of replication of the intermediate, the donor and recipient replicons become fused to form a cointegrate (C) carrying one directly repeated copy of the transposon at each junction. Consequently, the cointegrate is an unstable structure that is resolved by recA-dependent generalized recombination (as in Tn5; A. Ahmed, unpublished results) or tnpR-specified site-specific recombination (as in Tn3 [22]). The donor and recipient replicons are thereby separated, each harboring one copy of the transposon (D). If the target DNA is located within the donor replicon itself (intramolecular transposition), maturation of the Shapiro intermediate produces a replicative inversion (as shown in Figure 2) or an adjacent deletion (Figure 3). This process is highly efficient in transposons like Mu and Tn3 [2],[9]. In conservative transposition (right), the 3′ ends engage in hairpin formation at both ends of the transposon (E) [7]. Following hairpin resolution (F), the free 3′ ends of the excised transposon are joined to 5′ ends from the target DNA (G), and the gaps are filled to complete the insertion process. The fate of the donor DNA containing a large gap (G) is not known: it could be degraded or undergo double-strand gap repair to regenerate the transposon sequence. This process is highly efficient in transposons like Tn10 [4],[8]. In Tn5, hairpin formation is not efficient (i.e., is leaky), so that a small proportion of the initial 3′ nicks remains free to engage in strand-transfer. Hence, the transposon displays properties of both conservative and replicative transposition concomitantly [5],[6]. These reactions are carried out by the respective transposases, which, by oligomerization, bring the end sequences of the transposon together to form a synaptic complex. For clarity, however, the transposon is shown as a straight line. The donor DNA sequence is shown in black, transposon DNA sequence is in red, and the recipient DNA sequence is in green. Replication and gap repair are indicated by dashed lines. The crossover event that resolves the cointegrate (C) is indicated by “x.”


Alternative mechanisms for tn5 transposition.

Ahmed A - PLoS Genet. (2009)

Steps in the formation of the Tn5-promoted replicative inversion 621.The parent plasmid p4.1 (A) carried one copy of Tn5 (consisting of two inverted IS50 elements, L and R, flanking the kan gene), the trp, galTK, tet, and amp genes, and the cos site of lambda [6]. The galTK genes confer galactose-sensitivity (GalS) on the host cell, and selection for galactose-resistance (GalR) requires the disruption of this region. In inversion 621 (C), only one IS50 element (L) was left at the original location, a complete copy of Tn5 was found inserted in the gal region, and the trp-cos plasmid segment between the two had been inverted. This type III inversion (like several type I and II inversions also described in [6]) is fully consistent with the replicative mechanism as depicted in (B). It cannot be explained by the conservative mechanism. Small vertical arrows indicate location of nicks at the ends of IS50L and the target (galTK) DNA sequence. Horizontal arrows indicate inverted orientation of the two IS50 elements. The letters “o” and “I” refer to the outside and inside ends of the transposon, respectively. Figures are not drawn to scale. A deletion-inversion arising from p4.1 would be expected to have the following structure: the central kan region of Tn5 would be deleted, IS50L would join the gal region, the trp-cos segment would be inverted, and IS50R would also be inverted (to produce a direct repeat of IS50L) and join the target site in the gal region. This event would be consistent with the conservative mechanism, but it was never recovered from Tn5 in vivo. In contrast, the behavior of Tn10 was just the opposite of Tn5. p6A.1, a plasmid that is similar in structure to p4.1—except that it harbors Tn10 in place of Tn5—produced only deletion-inversions and no replicative inversions [6]. The two inside ends of Tn10 would be cleaved by double-strand breaks (as a result of hairpin formation and resolution), and the free 3′ ends would attack and join 5′ ends of the target sequence from the opposite strand. Such an event would produce deletion-inversions of the prescribed structure.
© Copyright Policy
Related In: Results  -  Collection

Show All Figures
getmorefigures.php?uid=PMC2723962&req=5

pgen-1000619-g002: Steps in the formation of the Tn5-promoted replicative inversion 621.The parent plasmid p4.1 (A) carried one copy of Tn5 (consisting of two inverted IS50 elements, L and R, flanking the kan gene), the trp, galTK, tet, and amp genes, and the cos site of lambda [6]. The galTK genes confer galactose-sensitivity (GalS) on the host cell, and selection for galactose-resistance (GalR) requires the disruption of this region. In inversion 621 (C), only one IS50 element (L) was left at the original location, a complete copy of Tn5 was found inserted in the gal region, and the trp-cos plasmid segment between the two had been inverted. This type III inversion (like several type I and II inversions also described in [6]) is fully consistent with the replicative mechanism as depicted in (B). It cannot be explained by the conservative mechanism. Small vertical arrows indicate location of nicks at the ends of IS50L and the target (galTK) DNA sequence. Horizontal arrows indicate inverted orientation of the two IS50 elements. The letters “o” and “I” refer to the outside and inside ends of the transposon, respectively. Figures are not drawn to scale. A deletion-inversion arising from p4.1 would be expected to have the following structure: the central kan region of Tn5 would be deleted, IS50L would join the gal region, the trp-cos segment would be inverted, and IS50R would also be inverted (to produce a direct repeat of IS50L) and join the target site in the gal region. This event would be consistent with the conservative mechanism, but it was never recovered from Tn5 in vivo. In contrast, the behavior of Tn10 was just the opposite of Tn5. p6A.1, a plasmid that is similar in structure to p4.1—except that it harbors Tn10 in place of Tn5—produced only deletion-inversions and no replicative inversions [6]. The two inside ends of Tn10 would be cleaved by double-strand breaks (as a result of hairpin formation and resolution), and the free 3′ ends would attack and join 5′ ends of the target sequence from the opposite strand. Such an event would produce deletion-inversions of the prescribed structure.
Mentions: Both processes start by nicking (short vertical arrows) of the transposon ends to expose the 3′-OH termini (A). At some point (see below), the target DNA is also cleaved to provide short protruding 5′-PO4 ends. In replicative transposition (left), strand-transfer takes place by joining the 3′ ends to 5′ ends of the target DNA in a concerted cleavage and joining reaction to form the “Shapiro intermediate” (B). As a result of replication of the intermediate, the donor and recipient replicons become fused to form a cointegrate (C) carrying one directly repeated copy of the transposon at each junction. Consequently, the cointegrate is an unstable structure that is resolved by recA-dependent generalized recombination (as in Tn5; A. Ahmed, unpublished results) or tnpR-specified site-specific recombination (as in Tn3 [22]). The donor and recipient replicons are thereby separated, each harboring one copy of the transposon (D). If the target DNA is located within the donor replicon itself (intramolecular transposition), maturation of the Shapiro intermediate produces a replicative inversion (as shown in Figure 2) or an adjacent deletion (Figure 3). This process is highly efficient in transposons like Mu and Tn3 [2],[9]. In conservative transposition (right), the 3′ ends engage in hairpin formation at both ends of the transposon (E) [7]. Following hairpin resolution (F), the free 3′ ends of the excised transposon are joined to 5′ ends from the target DNA (G), and the gaps are filled to complete the insertion process. The fate of the donor DNA containing a large gap (G) is not known: it could be degraded or undergo double-strand gap repair to regenerate the transposon sequence. This process is highly efficient in transposons like Tn10 [4],[8]. In Tn5, hairpin formation is not efficient (i.e., is leaky), so that a small proportion of the initial 3′ nicks remains free to engage in strand-transfer. Hence, the transposon displays properties of both conservative and replicative transposition concomitantly [5],[6]. These reactions are carried out by the respective transposases, which, by oligomerization, bring the end sequences of the transposon together to form a synaptic complex. For clarity, however, the transposon is shown as a straight line. The donor DNA sequence is shown in black, transposon DNA sequence is in red, and the recipient DNA sequence is in green. Replication and gap repair are indicated by dashed lines. The crossover event that resolves the cointegrate (C) is indicated by “x.”

Bottom Line: Bacterial transposons are known to move to new genomic sites using either a replicative or a conservative mechanism.The behavior of transposon Tn5 is anomalous.In vitro studies indicate that it uses a conservative mechanism while in vivo results point to a replicative mechanism.

View Article: PubMed Central - PubMed

Affiliation: Department of Biological Sciences, University of Alberta, Edmonton, Alberta, Canada. asada@ualberta.ca

ABSTRACT
Bacterial transposons are known to move to new genomic sites using either a replicative or a conservative mechanism. The behavior of transposon Tn5 is anomalous. In vitro studies indicate that it uses a conservative mechanism while in vivo results point to a replicative mechanism. To explain this anomaly, a model is presented in which the two mechanisms are not independent--as widely believed--but could represent alternate outcomes of a common transpositional pathway.

Show MeSH
Related in: MedlinePlus