Download Answers questions chapter 12

Chapter 12
SITE SPECIFIC RECOMBINATION AND TRANSPOSITION OF DNA
a. Compare the processes of conservative site-specific recombination (CSSR) and transposition. What do
they have in common, in terms of both their mechanisms and the molecules involved? How are they
different?
Suggested Answer:
Despite the many differences between the two processes, CSSR and transposition share the same
overall steps. First, specialized proteins called recombinases recognize specific recombination sites within
the DNA; second, the recombinases bring the sites together to form a synaptic complex; and, third, the
recombinases catalyze the cleavage and rejoining of the DNA molecules.
The processes differ in several ways, however. For example, as all the bonds broken during the CSSR
reaction are resealed by the recombinase, CSSR is conservative, and the only net result of the reaction is
that the recombining DNA strands are exchanged. In contrast, transposition often creates entirely new
molecules, most notably the synthesis of new copies of the transposon, such as during replicative
transposition. Another difference is that CSSR always involves an exchange between two recombination
sites having a precise sequence. Transposition, on the other hand, involves more than two sites (for
example, the two ends of the transposon plus the target site), and not all of them (in particular, the target
site) has the same strict sequence requirements as the CSSR recombination sites. A third difference is
that CSSR is generally simpler than transposition, involving only the exchange of the DNA strands.
Transposition, on the other hand, can occur by any of several mechanisms and can involve transcription,
translation, and/or DNA replication.
b. Name three biological processes that rely on conservative site-specific recombination. What type of
rearrangement is involved in each of the processes?
Suggested Answer:
Some processes that rely on CSSR include λ integration (insertion) and excision (deletion), Hin-mediated
recombination (inversion), conversion of circular DNA multimers to monomers (deletion), and
circularization of P1 phage by Cre (deletion).
c. Describe the reactions mediated by serine and tyrosine recombinases. What role do the serine and
tyrosine side chains play in the reaction catalyzed by each of these enzymes?
Suggested Answer:
While serine recombinases and tyrosine recombinases have the same final effect, the two protein families
are unrelated to each other. Serine recombinases initiate recombination by cutting all four strands of the
two involved DNA duplexes. This is done in a cleavage reaction that leaves a single recombinase protein
linked through a serine side chain to each of the 5′ DNA ends at the break. The free 3′'-OH ends then
attack the bonds linking the recombinase to the 5′ ends with each 3′ end attacking the strand in the other
duplex that runs in the same direction. This reseals the (now recombinant) DNA and releases the
recombinase proteins.
Tyrosine recombinases work by a different mechanism in which only two of the DNA strands are cut at a
time. As with serine recombinases, this cleavage reaction leaves a single recombinase protein covalently
linked (although here through a tyrosine side chain) to each ′' end. The two cut strands are then
exchanged and the DNA is resealed, producing a Holliday junction. This junction is then resolved when a
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second pair of recombinase enzymes go through the same cycle of DNA strand cutting, exchange of the
strands, and resealing of the DNA.
The serine and tyrosine side chains of the recombinases play a central role in the recombination process.
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In particular, the OH group of the side chains is used to attack the phosphodiester bond during the initial
cleavage reaction. Also, as a result of this nucleophilic attack, the same side chain links the recombinase
to the 5′ ends of the DNA.
d. How does Salmonella use CSSR to help it evade the host immune system? Consider the two possible
orientations of the genomic region affected by Hin. What are the consequences of the two arrangements
for the properties of the cell?
Suggested Answer:
Salmonella uses CSSR to induce a periodic switch in the form of flagellin—a cell surface protein that is
frequently targeted by the host immune system—that it produces. The switch is promoted by a serine
recombinase called Hin, which periodically catalyzes a CSSR reaction that inverts a particular segment of
the chromosome. This segment contains a promoter that is involved in flagellin gene regulation and can
either be "on" or "off," depending on its orientation. When the promoter is "on," the gene encoding the H2
flagellin is expressed, as is a repressor of the H1 flagellin gene (which turns off the H1 flagellin gene). In
contrast, when the promoter is "off," the H2 flagellin and repressor genes are both turned off, and so only
H1 flagellin is expressed.
The switch between H1 and H2 flagellin expression periodically alters the physical properties of the cell
surface and helps make the bacterial cell more difficult to identify by the host’s immune system.
Specifically, if a host immune system mounts a reaction against salmonella using antibodies specific to
one of the flagellin forms, the switch would allow at least some of the cells in the population to evade the
response and survive.
e. What percentage of the genome is made up of transposons in bacteria, yeast, corn, and humans?
What pressures might act to increase the prominence of transposons within a genome, and what might
act to limit their presence?
Suggested Answer:
Numerous factors can contribute to the success of transposons. Perhaps most significant is that the very
nature of certain types of transposition—such as replicative transposition, in which a new copy of the
transposon is created while the original copy is left intact—means that the copy number of the transposon
will tend to increase over time as long as it does not kill the cell or create a selective disadvantage. Also,
these negative consequences can often be avoided because of certain mechanisms that some
transposons have to select specific DNA regions as target sites or to limit the frequency of transposition.
In addition, some transposons can spread rapidly between cells by carrying beneficial genes, such as the
tetracycline resistance gene present within Tn10.
There are also significant possible negative consequences of transposition that can act to limit the
number of transposons present in the genome. In particular, transposition into the coding sequence or
regulatory region of a gene can alter or destroy the gene's function, with possibly disastrous
consequences for the cell. In addition, replicative transposition can produce large scale chromosomal
rearrangements, such as inversions and deletions.
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f. While characterizing a mutation in a gene of interest, you discover that the mutation involves an
insertion within the coding sequence of the gene. You suspect that the inserted sequence is a transposon
and would like to determine which of the three major transposon families it belongs to. What sequence
elements could be looked for within the inserted sequence that would help place it in one of the three
families?
Suggested Answer:
Each of the three major types of transposons (DNA transposons, viral-like transposons, and polyA
transposons) has characteristic sequence elements that could be used for their identification. First, DNA
transposons contain inverted repeats at their ends that are surrounded by target site duplications.
Second, viral-like transposons contain a reverse transcriptase gene, as well as direct repeats called long
terminal repeats. Finally poly-A transposons are distinctive in that they lack inverted repeat sequences
but contain both 5′ and 3′ UTRs, as well as a poly-A sequence.
g. Outline the steps involved in cut-and-paste transposition, addressing the protein(s) involved in each of
the steps. Which of the steps produce the target site duplications?
Suggested Answer:
In the first step of cut-and-paste transposition, transposase binds to the terminal inverted repeats at either
end of the transposon. Once bound to the ends, the transposase brings them together to form a proteinDNA complex called the synaptic complex or transpososome. The transposase subunits then make
double-strand cuts at both ends of the transposon, liberating the transposon and leaving a doublestranded break in the chromosome (which is typically repaired through double-strand break repair). The
free 3′-OH ends of the transposon DNA then attack the DNA backbone at the target site in a reaction
called DNA strand transfer that links the 3′ ends to the target site DNA. Because the target site DNA is
attacked by the 3′ ends at positions that are several nucleotides apart, DNA strand transfer leaves two
small single-stranded gaps in the DNA that are filled in by DNA polymerase and sealed by DNA ligase.
The target site duplications are produced when the single-stranded gaps resulting from DNA strand
transfer are filled in by DNA polymerase. Because the single strands present within the two gaps are
complementary to each other, repair of the gaps will reconstruct the same double-stranded sequence,
thereby creating the duplication.
h. The movement of viral-like and poly-A retrotransposons both involve RNA intermediates, yet differ in
several important respects. Name at least three of these differences.
Suggested Answer:
Viral-like retrotransposons differ from poly-A retrotransposons in several significant ways. First, the entire
poly-A retrotransposon is transcribed during transposition, in contrast to viral-like transposons, of which
only a portion is transcribed. Another difference concerns the complexity of the transposition cycle: virallike retrotransposons follow a relatively simple pathway in which the transposon is synthesized, reverse
transcribed into DNA, and then directly recognized by integrase for insertion into the DNA. Poly-A
retrotransposons, in contrast, undergo a more complex cycle in which the RNA leaves the nucleus, is
translated to produce proteins that remain associated with it, and the RNA-protein complex then re-enters
the nucleus where one of the bound proteins (the ORF2 endonuclease) cuts the chromosomal DNA to
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initiate the integration reaction. The two transposons also differ in that poly-A retrotransposons initially
integrate as RNA (which is then reverse transcribed using the 3′-OH of the chromosomal DNA as a
primer), in contrast to viral-like retrotransposons, which integrate in the form of DNA. Another difference is
that poly-A retrotransposons integrate into DNA at T-rich sequences due to their complementarity,
whereas viral-like retrotransposons do not share the same sequence specificity.
i. Describe how phage Mu avoids inserting into other Mu elements present elsewhere in the genome,
addressing the role of MuA and MuB in the process. What effect might overexpression of MuA or MuB
have on Mu transposition?
Suggested Answer:
Mu insertion is mediated by interactions between MuA, a transposase that binds to the terminal repeats of
the transposing Mu element; and MuB, an ATP-dependent DNA-binding protein. For Mu insertion to
occur, MuB must be bound to the target-site DNA, something that can occur only when MuB is in its ATPbound form. In addition to its role in catalyzing transposition, however, MuA also stimulates the ATPase
activity of MuB bound to nearby DNA, causing the MuB to fall off the DNA. Because of this, MuB has a
hard time accumulating near (or within) already inserted Mu elements because of the MuA bound to the
terminal repeats of the inserted elements. Consequently, the DNA around Mu elements tends to be free
of MuB, and further Mu insertion events are prevented.
Overexpression of MuA should decrease the rate of transposition to new sites because it would lead to
increased, generalized interactions between MuA and MuB, decreasing the overall amount of ATP-bound
MuB on the genome. In contrast, MuB might be expected to decrease the target site immunity because it
could possibly overwhelm the ability of bound MuA to stimulate the ATPase activity of nearby MuB,
thereby allowing MuB to accumulate even in the proximity of already inserted Mu elements.
j. How are Ty elements similar to viruses? What keeps them from being true viruses?
Suggested Answer:
Ty elements are similar to viruses in that, as viral-like retrotransposons, they use the same mechanism of
transposition as retroviruses. In addition, and perhaps more strikingly, they are packaged into viral-like
particles in the cell.
Ty elements cannot be considered to be true viruses, however, because they do not have the ability to
leave one cell and infect another.
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