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91 234 Retrotransposons
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23.4 Retrotransposons
16
(a)
34
condition known as severe combined immunodeficiency
(SCID, “bubble boy” syndrome) and cannot mount an
immune response against any pathogen. They must be
isolated from the rest of the world in order to survive.
(d)
RAG1 –
– +
RAG2 – – + +
(c)
(b)
Signal
12 23 –
RAG1+2 – + – + – +
61
50
HP
HP
HP
745
SUMMARY RAG1 and RAG2 introduce singlestrand nicks into DNA adjacent to either a 12 signal
or a 23 signal. This leads to a transesterification in
which the newly created 39-hydroxyl group attacks
the opposite strand, breaking it, and forming a hairpin at the end of the coding segment. The hairpins
then break in an imprecise way, allowing joining of
coding regions with loss of bases or gain of extra
bases.
23.4 Retrotransposons
16
M1 2345 6
N
M1 2
N
M1 2 3 4
Figure 23.17 Identifying cleavage products. (a) Cleavage substrate.
Gellert and colleagues constructed this labeled 50-mer, which
included 16 bp of DNA on the left, then a 12 signal (yellow), included
in a 34-bp segment on the right. The single 59-end label is indicated
by the red dot. These workers also made an analogous 61-mer
substrate with a 23 signal. (b) Identifying the hairpin product. Gellert
and coworkers incubated RAG1 and RAG2 proteins, as indicated at
top, with either the labeled 12-signal or 23-signal substrate, also as
indicated at top. After the incubation, they subjected the products to
nondenaturing gel electrophoresis and autoradiographed the gel to
detect the labeled products. The positions of the 61-mer and 50-mer
substrates, the hairpin (HP), and the 16-mer are indicated at right.
(c) Identifying the products from a nondenaturing gel. Gellert and
colleagues recovered the labeled products (apparently uncleaved
50-mer substrate and 16-mer fragment) from the bands of a
nondenaturing gel. They then electrophoresed these DNAs again in
lanes 1 and 2, respectively, of a denaturing gel, along with markers
(identified with diagrams at right) corresponding to the uncleaved
substrate, the 16-bp hairpin (HP), and the single-stranded 16-mer
released by denaturing the nicked substrate. (d) Requirement for
RAG1 and RAG2. This experiment was very similar to the one in panel
(b) except that the presence of RAG1 and RAG2 proteins (indicated at
top) were the only variables. “N” denotes the position of the 16-mer
released from the nicked species. (Source: McBlane, J.F., D.C. Van Gent,
D.A. Ramsden, C. Romeo, C.A. Cuomo, M. Gellert, and M.A. Oettinger, Cleavage
at a V(D)J recombination signal requires only RAG1 and RAG2 proteins and occurs
in two steps. Cell 83 (3 Nov 1995) f. 4 a–c, p. 390. Reprinted by permission of
Elsevier Science.)
closely resembles rearrangement of immunoglobulin genes.
Without antibodies, B cells are useless, and without T cell
receptors, T cells are useless. Thus, loss of Artemis function
means loss of both B cell and T cell function. Indeed,
people with defective Artemis genes have a very serious
McClintock’s maize transposons are examples of so-called
cut-and-paste or copy-and-paste transposons, similar to
the bacterial transposons we discussed earlier in this chapter. If DNA replication is involved, it is direct replication.
Humans also carry transposons in this class, which constitute about 1.6% of the human genome. The most prevalent
example is called mariner, but all of the mariner elements
studied so far have been defective in transposition. Eukaryotes also carry many more transposons of another kind:
retrotransposons, which replicate through an RNA intermediate. In this respect, the retrotransposons resemble
retroviruses, some of which cause tumors in vertebrates,
and some of which (the human immunodeficiency viruses,
or HIVs) cause AIDS. As an introduction to the replication
scheme of the retrotransposons, let us first examine the
replication of the retroviruses.
Retroviruses
The most salient feature of a retrovirus, indeed the feature
that gives this class of viruses its name, is its ability to
make a DNA copy of its RNA genome. This reaction,
RNA→DNA, is the reverse of the transcription reaction, so
it is commonly called reverse transcription. In 1970, Howard
Temin and, simultaneously, David Baltimore convinced a
skeptical scientific community that this reaction takes
place. They did so by finding that the virus particles contain an enzyme that catalyzes the reverse transcription reaction. Inevitably, this enzyme has been dubbed reverse
transcriptase. A more proper name is RNA-dependent
DNA polymerase.
Figure 23.18 illustrates the retrovirus replication cycle.
We start with a virus infecting a cell. The virus contains
two copies of its RNA genome, linked together by base
pairing at their 59-ends (for simplicity, only one copy is
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Chapter 23 / Transposition
RNA:
Reverse transcription
LTR
LTR
dsDNA:
Integration
Host DNA
Host DNA
Provirus:
Transcription
RNA:
Packaging into virus; budding
Figure 23.18 Retrovirus replication cycle. The viral genome is an
RNA, with long terminal repeats (LTRs, green) at each end. Reverse
transcriptase makes a linear, double-stranded DNA copy of the RNA,
which then integrates into the host DNA (black), creating the provirus
form. The host RNA polymerase II transcribes the provirus, forming
genomic RNA. The viral RNA is packaged into a virus particle, which
buds out of the cell and infects another cell, starting the cycle over
again.
shown). When the virus enters a cell, its reverse transcriptase (a product of the viral pol gene) makes a doublestranded DNA copy of the viral RNA, with long terminal
repeats (LTRs) at each end. This DNA recombines with the
host genome to yield an integrated form of the viral
genome called the provirus. The host RNA polymerase II
transcribes the provirus, yielding viral mRNAs, which are
then translated to viral proteins. To complete the replication
cycle, polymerase II also makes RNA copies of the provirus, which are new viral genomes. These genomic RNAs
are packaged into virus particles (Figure 23.19) that bud
out of the infected cell and go on to infect other cells.
Evidence for Reverse Transcriptase The skepticism about
the reverse transcription reaction arose from the fact that
no one had ever observed it, and the notion that it violated
the “central dogma of molecular biology” promulgated
by Watson and Crick, which said that the flow of genetic
information is from DNA to RNA to protein, not the
reverse. Crick later stated that the DNA→RNA arrow was
intended to be double-headed, but that was clearly not
the popular perception at the time. What evidence did
Baltimore and Temin bring to bear to dispel this skepticism?
Figure 23.20 shows the result of one of Baltimore’s
experiments. He incubated purified retrovirus particles
(Raucher mouse leukemia virus, or R-MLV) with all four
dNTPs, including [3H]dTTP, then measured the incorporation of the labeled TTP into a polymer (DNA) that could be
precipitated with acid. He observed a clear incorporation
(red curve) that could be inhibited by including RNase
in the reaction (blue curve), and inhibited even more by
preincubating with RNase (green curve). This sensitivity to
RNase was compatible with the hypothesis that RNA is
the template in the reverse transcription reaction.
Baltimore also examined the product of the reaction
and showed that it was insensitive to RNase and base hydrolysis, but sensitive to DNase. Furthermore, the virions
could support the incorporation of dNTPs only. Ribonucleotides, including ATP, could not be incorporated. Thus,
the product behaved like DNA, and the enzyme behaved
like an RNA-dependent DNA polymerase—a reverse transcriptase. Baltimore and Temin both performed similar
experiments on Rous sarcoma virus particles, with very
similar results. Thus, it appeared that all RNA tumor
viruses probably contained reverse transcriptase and
behaved according to the provirus hypothesis illustrated in
Figure 23.18. This has proven to be true.
Evidence for a tRNA Primer As molecular biologists began to investigate the molecular biology of reverse transcription, they discovered that the viral reverse transcriptase
is like every other DNA polymerase known: It requires a
primer. In 1971, Baltimore and colleagues found RNA
primers attached to the 59-ends of nascent reverse transcripts using the following strategy: They labeled the nascent reverse transcripts in avian myeloblastosis virus
(AMV) by the same method Baltimore and Temin had
used—incubating virus particles with labeled dNTPs. Then
they subjected the products to Cs2SO4 gradient ultracentrifugation to separate RNA from DNA based on their densities (RNA being denser than DNA).
In the first experiment, Baltimore and colleagues isolated the nucleic acids from the virus particles and subjected
them immediately to ultracentrifugation. Figure 23.21a
shows the results: a peak of labeled DNA that appeared to
have the density of RNA. This finding is consistent with
the hypothesis that the nascent DNA is still base-paired to
the much bigger RNA template, so the whole complex
behaves like RNA. If this hypothesis is true, then heating
the RNA–DNA hybrid should denature it and release the
DNA product as an independent molecule. When Baltimore and colleagues performed that experiment, they
observed the behavior in Figure 23.21b: Now the nascent
DNA product had a density much closer to that of DNA,
but still a little too dense, as if there were still some RNA
attached.
That behavior could be explained if the nascent DNA
still had an RNA primer covalently attached to it. To check
this possibility, Baltimore and coworkers treated the nascent DNA with RNase and again subjected it to ultracentrifugation. This time, the density of the product behaved
exactly as expected for pure DNA (Figure 23.21c). Thus,
the nascent reverse transcript appears to be primed by
RNA. But what RNA?
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23.4 Retrotransposons
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Figure 23.19 AIDS virion internal structure, cutaway artwork.
AIDS (acquired immune deficiency syndrome) is caused by the human
immunodeficiency virus (HIV). The core of this HIV virus particle is a
capsule (pink) containing RNA strands (ribonucleic acid, yellow). Around
the core is an icosahedral shell of matrix proteins (blue). Over this is a
membrane envelope (yellow bilayer) taken from the membrane of the
host cell that made this virus particle. Anchored to the shell are viral
knobs (yellow) that allow the virus particle to attach to cells. AIDS
impairs the immune system and allows often fatal secondary infections.
In the process of making an inventory of all the molecules within the retrovirus particle, molecular biologists
had discovered some tRNAs, one of which, host tRNATrp,
appeared to be partially base-paired to the viral RNA.
Could this be the primer? If so, it should bind to the reverse transcriptase. To see if it does, Baltimore, James
Dahlberg, and colleagues labeled host tRNATrp, or the
tRNATrp from virus particles, with 32P and mixed these
labeled tRNAs with AMV reverse transcriptase. Then they
subjected these mixtures to gel filtration on Sephadex
G-100 (Chapter 5). By itself, tRNATrp was included in the
gel and eluted in a peak centered at about fraction #25.
However, both host and virion tRNAs, when mixed with
reverse transcriptase eluted with the enzyme in a peak centered at about fraction #20. Thus, this reverse transcriptase binds tRNATrp. Together with the data we have
already discussed, the binding data strongly suggest that
tRNATrp serves as the primer for this enzyme. The virus
does not encode a tRNA, so the primer must be picked up
from the host cell.
(Source: © Russell Kightley/Photo Researchers, Inc.)
The Mechanism of Retrovirus Replication The initial
product of reverse transcription in vitro is a short piece of
DNA called strong-stop DNA. The reason for the strongstop is obvious when we consider the site on the viral RNA
to which the tRNA primer hybridizes (the primer-binding
site, or PBS). It is only about 150 nt (depending on the
retrovirus) from the 59-end of the viral RNA. This means
that the reverse transcriptase will synthesize DNA for just
150 nt or so before reaching the end of the RNA template
and stopping. This raises the interesting question: What
happens next?
That question is related to another paradox of retrovirus replication, illustrated in Figure 23.22. The provirus is
longer than the viral RNA, yet the viral RNA serves as the
template for making the provirus. In particular, the LTRs
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Chapter 23 / Transposition
RNA
[3H]TMP incorporated (cpm in hundreds)
25
No treatment
20
(a)
H 2O
(b)
15
(c)
10
RNase in reaction
1
3
1
3
1
1.65
5
Preincubation
with RNase
30
60
90
Time (min)
120
Figure 23.20 Effect of RNase on reverse transcriptase activity.
Baltimore incubated R-MLV particles with the four dNTPs, including
[3H]dTTP, under various conditions, then acid-precipitated the product
and measured the radioactivity of the product by liquid scintillation
counting. Treatments: red, no extra treatment; purple, preincubation
for 20 min with water; blue, RNase included in the reaction; green,
preincubated with RNase. (Source: Adapted from Baltimore, D., Viral RNAdependent DNA polymerase. Nature 226:1210, 1970.)
in the viral RNA are incomplete. The left LTR contains
a redundant region (R) plus a 59-untranslated region
(U5), whereas the right LTR contains an R region plus a
39-untranslated region (U3). How can the provirus have
complete LTRs on each end while its template is missing a
U3 region at its left end and a U5 region at its right end?
Harold Varmus proposed an answer based on the important
fact that reverse transcriptase has another distinct activity:
an RNase activity. The RNase inherent in reverse transcriptase is RNase H, which specifically degrades the RNA part
of an RNA–DNA hybrid.
Varmus’s hypothesis is illustrated in Figure 23.23.
First, (a) the reverse transcriptase uses the tRNA to prime
synthesis of strong-stop DNA. This appears at first to be
the end of the line, but then (b) RNase H recognizes a
stretch of RNA hybrid between the strong-stop DNA and
the RNA template, and degrades the R and U5 parts of
the RNA. The removal of this RNA leaves a tail of DNA
(blue) that can hybridize through its R region with the
RNA at the other end of the RNA template, or with another RNA template (c). This hybridization to another
R region is called the “first jump.” In principle, the DNA
could jump to the other end of the same RNA, and this
could be facilitated by looping the RNA around so the
strong-stop DNA does not even need to leave the left end
of the RNA to pair with the right end. But the DNA can
also jump to another viral RNA, and this seems likely
DNA
3
[3H]DNA (cpm in hundreds)
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Density (g/mL)
Figure 23.21 Reverse transcripts contain an RNA primer.
Baltimore and colleagues labeled reverse transcripts in AMV
particles with [3H]dTTP, then subjected them to Cs2SO4 gradient
ultracentrifugation after the following treatments: (a) no treatment;
(b) heating to denature double-stranded polynucleotides; and
(c) heating and RNase to remove any primers attached to the reverse
transcripts. Interpretive drawings at right provide an explanation for
the results: (a) The untreated material has a high density like RNA
because the reverse transcript is short and is base-paired to a much
longer viral RNA template. (b) The heated material has a density closer
to that of DNA because the RNA template has been removed, but it is
still denser than pure DNA because of an RNA primer that is covalently
attached. (c) The heated and RNase-treated material has the density
of a pure DNA because the RNase has removed the RNA primer. The
approximate densities of pure RNA and DNA are indicated at top.
(Source: Adapted from Verma, I.M., N.L. Menth, E. Bromfeld, K.F. Manly, and
D. Baltimore, Covalently linked RNA–DNA molecules as initial product of RNA
tumor virus DNA polymerase. Nature New Biology 233:133, 1971.)
because each virus particle contains two copies of the
RNA genome.
After the first jump, the strong-stop DNA is at the
right end of the template and can serve as a primer for the
reverse transcriptase to copy the rest of the viral RNA (d).
PBS
LTR
gag
R U5
pol
env
LTR
U3 R
PBS
LTR
gag
U3 R U5
gag
U3 R U5
pol
pol
env
env
U3
U3
Viral RNA:
Provirus
LTR
R U5
R U5
Figure 23.22 Structures of retroviral RNA and provirus DNA. This
is a nondefective retroviral RNA that contains all the genes necessary
for replication: a coat protein gene (gag), a reverse transcriptase gene
(pol), and an envelope protein gene (env). In addition, it contains long
terminal repeats (LTRs) at both ends, but these repeats are not
identical. The left LTR contains an R and a U5 region, including a
primer-binding site (PBS), shown here bound to a tRNA primer, but the
right LTR contains a U3 and an R region. On the other hand, the
proviral DNA, made using the viral RNA as a template, contains full
LTRs (U3, R, and U5) at each end.
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23.4 Retrotransposons
5′ R U5 PBS
3′
5′ R U5 PBS
3′ R U5
U3 R
5′
(a) Reverse transcriptase
makes strong-stop
DNA.
5′
(b) RNase H removes R
and U5 RNA.
U3 R
3′
U3 R 3′
5′ PBS
3′ R U5
3′
5′
(c) First jump. Strong-stop
DNA jumps to other
end of RNA.
5′ PBS
U3 R 3′
3′ R U5
(d) Reverse transcriptase
extends primer.
U3 R 3′
U3 R U5
5′ PBS
3′ PBS
5′
5′
(e) RNase H removes most
of viral RNA.
3′
5′
3′ PBS
U3 R U5
(f) Reverse transcriptase
extends RNA primer.
5′
3′ PBS
U3 R U5 PBS
U3
U3 R U5
5′
3′
5′
(g) RNase H removes viral
RNA and tRNA.
3′ PBS
5′ U3 R U5 PBS 3′
U3 R U5 5′
(h) Second jump. PBS sites
at opposite ends pair up.
5′ U3 R U5 PBS 3′
3′ PBS
U3 R U5 5′
(i) Both strands filled in
by growth at 3′-ends.
5′ U3 R U5 PBS
3′ U3 R U5 PBS
749
strand synthesis (f). After the reverse transcriptase extends this primer to the end, including the PBS region,
RNase H removes the remaining RNA (g)—the second
strand primer and the tRNA—both of which were paired
to DNA. This sets up the second jump (h), in which the
PBS region on the right pairs with the one on the left. Like
the first jump, the second jump can be visualized as a
jump to another molecule, or the other end of the same
molecule. If the same molecule is involved in the jump, the
DNA can loop around to allow the two PBS regions to
base-pair. After the second jump, the stage is set for reverse transcriptase, which can use DNA as a template, or
another DNA polymerase to complete both strands (i),
using the long single-stranded overhangs at each end as
templates.
Once the provirus is synthesized, it can be inserted into
the host genome by an integrase. This enzyme is originally
part of a polyprotein derived from the pol gene, which we
have seen also encodes reverse transcriptase and RNase H.
The integrase is cut from the polyprotein by a protease,
which also starts out as part of the same polyprotein. The
protease also cuts itself out of the polyprotein. (It is worth
noting that some of the most promising drugs for combatting AIDS are protease inhibitors that target the HIV version of this enzyme.) Once the provirus is integrated into
the host genome, it is transcribed by host RNA polymerase II to yield viral RNAs.
SUMMARY Retroviruses replicate through an RNA
intermediate. When a retrovirus infects a cell, it
makes a DNA copy of itself, using a virus-encoded
reverse transcriptase to carry out the RNA→DNA
reaction, and an RNase H to degrade the RNA parts
of RNA–DNA hybrids created during the replication process. A host tRNA serves as the primer for
the reverse transcriptase. The finished doublestranded DNA copy of the viral RNA is then inserted into the host genome, where it can be
transcribed by host polymerase II.
U3 R U5 3′
U3 R U5 5′
Figure 23.23 A model for the synthesis of the provirus DNA
from a retroviral RNA template. RNA is in red and DNA is in blue,
throughout. The tRNA primer is represented by a cloverleaf with
a 39-tag that hybridizes to the primer-binding site (PBS) in the viral
RNA. The steps are described more fully in the text.
Notice that the first jump has allowed the right LTR to be
completed. The U5 and R regions were copied from the
left LTR of the viral RNA and the U3 region was copied
from the right LTR. In step (e), the RNase H removes
most of the viral RNA, but it leaves a small piece of RNA
adjacent to the right LTR to serve as a primer for second
Retrotransposons
All eukaryotic organisms appear to harbor transposons
that replicate through an RNA intermediate and therefore depend on reverse transcriptase. These retrotransposons fall into two groups with different modes of
replication. The first group includes the retrotransposons
with LTRs, which replicate in a manner very similar to
retroviruses, except that they do not pass from cell to cell
in virus particles. Not surprisingly, these are called
LTR-containing retrotransposons. The second group
includes the retrotransposons that lack LTRs (the non-LTR
retrotransposons).
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Chapter 23 / Transposition
LTR-Containing Retrotransposons The first examples of
retrotransposons were discovered in the fruit fly (Drosophila
melanogaster) and yeast (Saccharomyces cerevisiae). The
prototype Drosophila transposon is called copia because it
is present in the genome in copious quantity. In fact, copia
and related transposons called copia-like elements account
for about 1% of the total fruit fly genome. Similar transposable elements in yeast are called Ty, for “transposon yeast.”
These transposons have LTRs that are very similar to the
LTRs in retroviruses, which suggests that their transposition
resembles the replication of a retrovirus. Indeed, several
lines of evidence indicate that this is true. Here is a summary
of the evidence that the Ty1 elements replicate through an
RNA intermediate, just as retroviruses do:
1. Ty1 encodes a reverse transcriptase. The tyb gene in
Ty codes for a protein with an amino acid sequence
closely resembling that of the reverse transcriptases encoded in the pol genes of retroviruses. If the Ty1 element really codes for a reverse transcriptase, then this
enzyme should appear when Ty1 is induced to transpose; moreover, mutations in tyb should block the appearance of reverse transcriptase. Gerald Fink and his
colleagues have performed experiments that bear out
both of these predictions.
2. Full-length Ty1 RNA and reverse transcriptase activity
are both associated with particles that closely resemble
retrovirus particles. These particles appear only in
yeast cells that are induced for Ty1 transposition.
3. In a clever experiment, Fink and colleagues inserted an
intron into a Ty1 element and then analyzed the element again after transposition. The intron was gone!
This finding is incompatible with the kind of transposition bacteria employ, in which the transposed DNA
looks just like its parent. But it is consistent with the
following mechanism (Figure 23.24): The Ty element
is first transcribed, intron and all; then the RNA is
spliced to remove the intron; and finally, the spliced
RNA is reverse transcribed, perhaps within a virus-like
particle, and the resulting DNA is inserted back into
the yeast genome at a new location.
4. Jef Boeke and colleagues demonstrated that the host
tRNAiMet serves as the primer for Ty1 reverse transcription. First, they mutated 5 of 10 nucleotides in the
Ty1 element’s PBS that are complementary to the host
tRNAiMet. These changes abolished transposition, presumably because they made it impossible for the tRNA
primer to bind to its PBS. Then Boeke and coworkers
made five compensating mutations in a copy of the
host tRNAiMet gene that restored binding to the mutated PBS. These mutations restored transposition activity to the mutant Ty1 element. As we have seen
many times throughout this book, this kind of mutation suppression is powerful evidence for the importance of interaction between two molecules: in this
Host DNA
Host DNA
Ty element:
LTR
Intron
LTR
Transcription
RNA:
Splicing
Processed RNA:
Reverse transcription
(in particle?)
Double-stranded DNA:
Reinsertion into host DNA
Reinserted DNA:
Figure 23.24 Model for transposition of Ty. The Ty element has
been experimentally supplied with an intron (yellow). The Ty element is
transcribed to yield an RNA copy containing the intron. This transcript
is spliced, and then the processed RNA is reverse-transcribed,
possibly in a virus-like particle. The resulting double-stranded DNA
then reinserts into the yeast genome. Abbreviation: LTR 5 long
terminal repeat.
case, interaction between the tRNAiMet primer and its
binding site in the Ty1 element.
Copia and its relatives share many of the characteristics we have described for Ty, and it is clear that they
also transpose in the same way as Ty. Humans also have
LTR-containing retrotransposons, but they lack a functional env gene. The most prominent examples are the
human endogenous retroviruses (HERVs), which make up
1–2% of the genome. So far, no transposition-competent
HERVs are known, so the HERVs may be relics of previous
retrotransposition.
SUMMARY Several eukaryotic transposons, includ-
ing Ty of yeast and copia of Drosophila, apparently
transpose by a mechanism similar to that of retrovirus replication. They start with DNA in the host
genome, make an RNA copy, then reverse transcribe
it—probably within a virus-like particle—to DNA
that can insert in a new location. HERVs probably
transposed in the same way until most or all of them
lost the ability to transpose.
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23.4 Retrotransposons
ORF1
5
UTR
(a)
ORF2
EN
RT
C
3
UTR
1
0
5
15
30
60
90
120 min
An
Figure 23.25 Map of the L1 element. The subregions within ORF2
(yellow) are designated EN (endonuclease), RT (reverse transcriptase),
and C (cysteine-rich). The purple arrows at each end indicate direct
repeats of host DNA, and the An on the right indicates the poly(A).
oc
Linear
sc
(b)
+ RNA
Products (%)
100
80
Linear
60
40
Open circle
20
0
(c)
0
20
40
60
80
Time (min)
100
120
– RNA
100
Products (%)
Non-LTR Retrotransposons Retrotransposons that lack
LTRs are much more abundant than those with LTRs, at
least in mammals. The most abundant of all are the long
interspersed elements (LINEs), one of which (L1) is present
in at least 100,000 copies and makes up about 17% of the
human genome, although about 97% of the copies of L1
are missing parts of their 59-ends and the great majority
(all but ,60–100 copies) have mutations that prevent their
transposition. The prevalence of L1 elements means that
this retrotransposon, which has been traditionally classified
as “junk DNA,” occupies about five times as much of the
genome as all the human exons do. Figure 23.25 is a
map of an intact L1 element, showing its two ORFs. ORF1
encodes an RNA-binding protein (p40), and ORF2 encodes a protein with two activities: an endonuclease and a
reverse transcriptase. L1, like all retrotransposons in this
class, is polyadenylated.
We have just seen that the LTR is crucial for replication
of most retrotransposons with LTRs, so how do non-LTR
retrotransposons replicate? In particular, what do they use
for a primer? The answer is that their endonuclease creates
a single-stranded break in the target DNA and their reverse
transcriptase uses the newly formed DNA 39-end as a
primer. Our best information on this mechanism comes
from Thomas Eickbush and colleagues’ studies on R2Bm,
a LINE-like element from the silkworm Bombyx mori.
This element resembles the mammalian LINEs in that it
encodes a reverse transcriptase, but no RNase H, protease,
or integrase, and it lacks LTRs. But it differs from the
LINEs in that it has a specific target site—in the 28S rRNA
gene of the host. This latter property made the insertion
mechanism easier to investigate.
Eickbush and colleagues first showed that the single ORF
of R2Bm encodes an endonuclease that specifically cleaves
the 28S rDNA target site. Next, they purified the endonuclease (and an RNA cofactor that was required for activity) and
added it to a supercoiled plasmid containing the target site. If
a single strand of the plasmid is cut, the supercoiled plasmid
will be converted to a relaxed circle. If both strands are cut, a
linear DNA should appear. Figure 23.26a and b show the
rapid appearance of relaxed (open) circles, followed by the
slower conversion of open circles to linear DNA. Thus,
the R2Bm endonuclease rapidly cleaves one of the DNA
strands at the target site, then much more slowly cleaves the
other strand. This cleavage is specific: The nuclease cannot
cut even one strand of a plasmid that lacks the target site.
Open circle
80
60
40
20
0
Linear
0
20
40
60
80
Time (min)
100
120
Figure 23.26 DNA nicking and cleavage activity of the R2Bm
endonuclease. Eickbush and colleagues mixed a supercoiled
plasmid bearing the target site for the R2Bm retrotransposon with
the purified R2Bm endonuclease, with or without its RNA cofactor,
then electrophoresed the plasmid to see if it had been nicked
(relaxed to an open circular form) or cut in both strands to yield a
linear DNA. (a) Electrophoretic gel stained with ethidium bromide.
The positions of the supercoiled plasmid (sc), the open circular
plasmid (oc), and the linear plasmid (linear) are indicated at right.
(b) Graphical representation of the results from panel (a). (c) Results
from a similar experiment in which the RNA cofactor was omitted.
(Source: Adapted from Luan, D.D., M.H. Korman, J.L. Jakubczak, and T.H.
Eickbush, Reverse transcription of R2Bm RNA is primed by a nick at the
chromosomal target sige: a mechanism for non-LTR retrotransposition. Cell 72
(Feb 1993) f. 2, p. 597. Reprinted by permission of Elsevier.)
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Chapter 23 / Transposition
Next, these workers removed the RNA cofactor and
showed that the protein by itself still caused rapid singlestranded nicking of the target site, but barely detectable
cutting of the other strand (Figure 23.26c). They also
showed that the linear DNA could be recircularized by T4
DNA ligase, which requires a 59-phosphate group. Thus,
cleavage by the R2Bm endonuclease leaves a 59-phosphate
and a 39-hydroxyl group. Next, they used the endonuclease
to create single-stranded nicks and showed by primer extension analysis that the transcribed strand is the one that
is nicked. (The nick in the transcribed strand stopped the
DNA polymerase in the primer extension experiment, but
primer extension on the other strand proceeded unimpeded
by nicks.) With more precise primer extension experiments
on DNA cut in both strands, they located the cut sites exactly and found the two strands are cut 2 bp apart.
To see if the nicked target DNA strand really does serve
as the primer, Eickbush and colleagues performed an in
vitro reaction with a short piece of pre-nicked target DNA
as primer, R2Bm RNA as template, R2Bm reverse transcriptase, and all four dNTPs, including [32P]dATP. They
electrophoresed and autoradiographed the products to see
if they were the right size. Figure 23.27a shows what should
happen at the molecular level, and panel b shows the results. When a nonspecific RNA was added as template, no
product was made (lane 1), but when the R2Bm RNA was
added, a strong band at 1.9 kb appeared. Is this what we
expect? It is hard to know because we do not know exactly
how far the reverse transcriptase traveled and we are dealing with a slightly branched polynucleotide, but it is close
because the primer is 1 kb long and the template is 802 nt
long. To investigate the nature of the product further,
Eickbush and colleagues included dideoxy-CTP in the
reaction (lane 3). As expected, it caused premature termination of reverse transcription at a number of sites, leading to
a fuzzy band. In another reaction, they treated the product
with RNase A to remove any part of the template not basepaired to the reverse transcription product before electrophoresis. Lane 4 shows that this sharpened the product to
a 1.8-kb band, suggesting that about 100 nt had been removed from the 59-end of the RNA template, so the reverse
transcriptase had apparently not completed its task in the
majority of cases. These workers also treated the product
with RNase H prior to electrophoresis (lane 5) and obtained
a diffuse band of about 1.5 kb. This procedure should
remove the RNA template because it is in a hybrid with the
product. The fact that the band is still longer than 1 kb indicates that a strand of DNA has been extended. Lane 6 is
another negative control in which a nonspecific DNA was
used instead of the target DNA.
Similar experiments with a target DNA that extended
farther to the left (with the target site in the middle) showed
a predominance of large, Y-shaped products (as predicted
in Figure 23.27), suggesting that reverse transcription occurred before second-strand cleavage. If second-strand
(a)
1000 bp
802 nt
3′
5′
3′
5′
1
(b)
2
3
4
5
6
2.7
1.9
1.0
Figure 23.27 Evidence for target priming of reverse transcription
of R2Bm. (a) Model of the product we expect if the R2Bm
endonuclease makes a nick near the left end of a 1-kb target DNA and
uses the new 39-end to prime reverse transcription of an 802-nt
transposon RNA. The reverse transcript (blue) is covalently attached to
the primer (yellow). The rest of the lower DNA strand is also rendered
in yellow at left. The opposite DNA strand is black. (b) Experimental
results. Eickbush and colleagues started with a 1-kb target DNA with
the target site close to the left end. They added R2Bm RNA and the
ORF2 product and dNTPs, including [32P]dATP to allow labeled
reverse transcripts to be formed. Then they electrophoresed the
products and autoradiographed them. Lane 1, a nonspecific RNA
was used instead of R2Bm RNA; lanes 2–6, R2Bm RNA was used;
lane 3, dideoxy-CTP was included in the reverse transcription
reaction; lane 4, the product was treated with RNase A before
electrophoresis; lane 5, the product was treated with RNase H
before electrophoresis; lane 6, a nonspecific target DNA was used.
(Source: Luan, D.D., M.H. Korman, J.L. Jakubczak, and T.H. Eickbush, Reverse
transcription of R2Bm RNA is primed by a nick at the chromosomal target site:
a mechanism for non-LTR retrotransposition Cell 72 (Feb 1993) f. 4, p. 599.
Reprinted by permission of Elsevier Science.)
cleavage had occurred first, the products would have been
linear and smaller. To confirm that the target DNA was
serving as the primer, Eickbush and coworkers performed
PCR with primers that hybridized to the target DNA and
to the reverse transcript, and obtained PCR products of the
expected size and sequence.
Based on these and other data, H.H. Kazazian and John
Moran proposed the model of L1 transposition presented
in Figure 23.28. First, the transposon is transcribed and the
transcript is processed. The processed mRNA leaves the
nucleus to be translated in the cytoplasm. It associates with
its two products, p40 and the ORF2 product, and reenters
the nucleus. There, the endonuclease activity of the ORF2
product nicks the target DNA. For L1, the target can be
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23.4 Retrotransposons
L1 element
ORF1
ORF2
(a) Transcription,
processing,
and export
AAAn
mRNA
(b) Translation and
RNP assembly
p40
p40
ORF2
AAAn
RNP
ORF2
O
RF
2
OR
F2
AA
TT An
T
n
(c) Import into nucleus
and target primed
reverse transcription
(d) Second-strand synthesis and
full integration at new site
Figure 23.28 A model for L1 transposition. (a) The L1 element is
transcribed, processed, and exported from the nucleus. (b) The mRNA
is translated to yield the ORF1 product (p40), and the ORF2 product,
with endonuclease and reverse transcriptase activities. These proteins
associate with the mRNA to form an RNP (c) The ribonucleoprotein
reenters the nucleus. The endonuclease nicks the target DNA
(anywhere in the genome), and the reverse transcriptase uses the new
DNA 39-end to prime synthesis of the reverse transcript. (d) In a series
of unspecified steps, the second L1 strand is made and the element,
usually truncated at its 59-end, is ligated into the target DNA.
any region of the DNA. Then the reverse transcriptase activity of the ORF2 product uses the target DNA 39-end
created by the endonuclease as a primer to copy the L1
RNA. Thus, this mechanism is called target-primed retrotransposition. Finally, in steps that are still poorly understood, the second strand of L1 is made, the second strand
753
of the target is cleaved, and the L1 element is ligated into
its new home.
At the beginning of this section, we learned that L1 elements comprise about 17% of the human genome. And, as
we will soon see, these elements can carry pieces of genomic DNA with them as they transpose. Thus, one can
estimate that, directly or indirectly, L1 elements have
sculpted about 30% of the human genome. Furthermore,
L1-like elements have been found in both plants and animals. Thus, these elements are ancient—at least 600 million years old. And, because identical DNA sequences can
lose all resemblance to each other after about 200 million
years of evolution, the true contribution of L1 elements to
the human genome may actually be about 50%.
You would suspect that anything as prevalent as L1 is
in the human genome must have some negative consequences, and indeed a number of L1-mediated mutations
have been discovered that have led to human disease. In
particular, copies of L1 have been found: in the blood clotting factor VIII gene, causing hemophilia; in the DMD
gene, causing Duchenne muscular dystrophy; and in the
APC gene, helping to cause adenomatous polyposis coli, a
kind of colon cancer. In this last case, the patient’s cancer
cells had the L1 element in their APC gene, but the normal
cells did not. Thus, this transposition had occurred during
the patient’s lifetime as a somatic mutation.
What is more surprising is that the L1 elements may
actually have beneficial consequences as well. For example,
significant homology occurs between the reverse transcriptase of L1 and human telomerase, suggesting that L1 may
have been the origin of the enzyme that maintains the ends
of our chromosomes (although the reverse may also have
been true). But the most plausible beneficial aspect of L1 is
that it may facilitate exon shuffling, the exchange of exons
among genes. This happens because the polyadenylation
signal of L1 is weak, so the polyadenylation machinery
frequently bypasses it in favor of a polyadenylation site
downstream in the host part of the transcript. RNAs polyadenylated in that way will include a piece of human RNA
attached to the L1 RNA, and this human RNA will be incorporated as a reverse transcript wherever the L1 element
goes next. This is bound to have deleterious consequences
sometimes, but it also creates new genes out of parts of old
genes, and that can give rise to proteins with new and useful characteristics.
Why are the polyadenylation signals of L1 elements
weak? Moran offers the following explanation: If the polyadenylation signals were strong, insertion of these elements
into the introns of human genes would cause premature
polyadenylation of transcripts, so all the exons downstream would be lost. That would probably inactivate the
gene and might well lead to the death of the host. And, unlike retroviruses, which can move from one individual to
another, the L1 elements live and die with their hosts. On
the other hand, weak polyadenylation signals allow these
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Chapter 23 / Transposition
elements to insert into introns of human genes without
disrupting a very high percentage of the transcripts of these
genes. Thus, because the amount of DNA devoted to introns is much higher than that devoted to exons, the L1
elements have a large area of the human genome to colonize relatively safely.
SUMMARY LINEs and LINE-like elements are ret-
rotransposons that lack LTRs. These elements encode an endonuclease that nicks the target DNA.
Then the element takes advantage of the new DNA
39-end to prime reverse transcription of element
RNA. After second-strand synthesis, the element
has become replicated at its target site. A new round
of transposition begins when the LINE is transcribed. Because the LINE polyadenylation signal is
weak, transcription of a LINE can include one or
more downstream exons of host DNA.
Nonautonomous Retrotransposons Members of another
class of non-LTR retrotransposons (nonautonomous retrotransposons) encode no proteins, so they are not autonomous like the transposition-competent LINEs. Instead,
they depend on other elements, probably the LINEs due to
their prevalence, to supply the proteins, including the
reverse transcriptase they need to transpose. The best studied of these nonautonomous retrotransposons are the Alu
elements, so-called because they contain the sequence
AGCT that is recognized by the restriction enzyme AluI.
These are about 300 bp long and are present in up to a million copies in the human genome. Thus, they have been
even more successful than the LINEs. One reason for this
success may be that the transcripts of the Alu elements contain a domain that resembles the 7SL RNA that is normally
part of the signal recognition particle that helps attach certain ribosomes to the endoplasmic reticulum. Two signal
recognition particle proteins bind tightly to Alu element RNA
and may carry it to the ribosomes, where the LINE RNA is
being translated. This may put the Alu element RNA in
a position to help itself to the proteins it needs to be reverse
transcribed and inserted at a new site. Because of their
small size, Alu elements and similar elements are called
short interspersed elements (SINEs).
The LINEs have probably also played a role in shaping
the human genome by facilitating the creation of processed
pseudogenes. Ordinary pseudogenes are DNA sequences
that resemble normal genes, but for one reason or another
cannot function. Sometimes they have internal translation
stop signals; sometimes they have inactive or missing splicing signals; sometimes they have inactive promoters; usually a combination of problems prevents their expression.
They apparently arise by gene duplication and subsequently accumulate mutations. This process has no delete-
rious effect on the host because the original gene remains
functional.
Processed pseudogenes also arise by gene duplication,
but apparently by way of reverse transcription. We strongly
suspect that RNA is an intermediate in the formation of
processed pseudogenes because: (1) these pseudogenes frequently have short poly(dA) tails that seem to have derived
from poly(A) tails on mRNAs; and (2) processed pseudogenes lack the introns that their progenitor genes usually
have. As in the case of the Alu elements, which are not derived from mRNAs, the LINEs could provide the molecular
machinery that allows mRNAs to be reverse transcribed
and inserted into the host genome.
SUMMARY Nonautonomous retrotransposons in-
clude the very abundant Alu elements in humans
and similar elements in other vertebrates. They cannot transpose by themselves because they do not
encode any proteins. Instead they take advantage of
the retrotransposition machinery of other elements,
such as LINEs. Processed pseudogenes probably
arose in the same way: mRNAs were reversetranscribed by LINE machinery and then inserted
into the genome.
Group II introns In Chapter 14 we learned that group II
introns, which inhabit bacterial, mitochondrial, and chloroplast genomes, are self-splicing introns that form a lariat intermediate. In 1998, Marlene Belfort and colleagues
discovered that a group II intron in a particular gene could
insert into an intronless version of the same gene somewhere
else in the genome. This process, called retrohoming, appears to occur by the mechanism outlined in Figure 23.29.
The gene bearing the intron is first transcribed, then the intron is spliced out as a lariat. This intron can then recognize
an intronless version of the same gene and invade it by
reverse-splicing. Reverse transcription creates a cDNA copy
of the intron, and second-strand synthesis replaces the RNA
intron with a second strand of DNA.
In 1991, Phillip Sharp proposed that group II introns
could be the ancestors of modern spliceosomal introns, in
part because of their very similar mechanisms of splicing.
In 2002, Belfort and colleagues showed how this could
have happened. They detected true retrotransposition, not
just retrohoming, of a bacterial group II intron. Thus, the
intron moved to a variety of new sites, not just to an intronless copy of the intron’s home gene.
To detect retrotransposition, Belfort and colleagues
built a plasmid with a modified version of the group II
Lactococcus lactis L1. LtrB intron, containing a kanamycin
resistance gene in reverse orientation, interrupted by a selfsplicing group I intron. In order for kanamycin resistance
to be expressed, this group II intron would first have to be
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Summary
Intron
Donor gene X
(a) Transcription
(b) Splicing
Lariat intron
Intronless gene X
(same, or similar,
sequence as donor)
(c) Reverse splicing
(d) Reverse transcription
(e) Second-strand synthesis
Figure 23.29 Retrohoming. (a) The donor gene X (blue) bearing a
group II intron (red) is transcribed to yield an RNA (RNAs are shaded
throughout). (b) The transcript is spliced, yielding a lariat-shaped
intron. (c) The intron reverse-splices itself into another copy of gene X
that has the same or similar sequence as the first except that it lacks
the intron. (d) The intron-encoded reverse transcriptase makes a DNA
copy of the intron, using a nick in the bottom DNA strand as primer.
The arrowhead marks the 39-end of the growing reverse transcript.
(e) The second strand (DNA version) of the intron is made, replacing
the RNA intron in the top strand. This completes the retrohoming
process.
transcribed, so the interrupting group I intron could be
removed. Then, the transcript would have to be reversetranscribed to yield a DNA that could insert into the host
DNA, where it could be transcribed in the forward, rather
than the reverse direction. As long as the group II intron
remained in RNA form, it could not code for kanamycin
resistance because its resistance gene had been transcribed
in the reverse direction, yielding an antisense RNA.
755
When Belfort and colleagues selected for kanamycinresistant cells, they found that transposition was relatively
rare, but did occur at a measurable rate. An interesting
feature of this transposition was that most of it occurred
into the DNA replication lagging strand. This finding suggested that transposition happened during replication and
used the short DNA fragments created in the lagging strand
(Chapter 20) as primers for the kind of target-primed reverse transcription we saw in the L1 transposition scheme
in Figure 23.28. Notice that no homology between the
transposon and the target DNA is required for this mechanism, as nicks in replicating lagging strands occur everywhere in the genome.
Once a group II intron has retrotransposed, it retains its
ability to splice itself out, so the target gene should usually
continue to function. Thus, the proliferation of group II
introns may have occurred readily and with relative safety
in the precursors to modern eukaryotes. Ultimately, eukaryotes appear to have developed spliceosomes to make
the splicing process more efficient.
SUMMARY Group II introns can retrohome to in-
tronless copies of the same gene by insertion of an
RNA intron into the gene, followed by reverse transcription and second-strand synthesis. Group II
introns can also undergo retrotransposition by
insertion of an RNA intron into an unrelated gene
by target-primed reverse transcription, using lagging strand DNA fragments as primers. This kind of
retrotransposition of group II introns may have provided the ancestors of modern-day eukaryotic spliceosomal introns and may account for their
widespread appearance in higher eukaryotes.
S U M M A RY
Transposable elements, or transposons, are pieces of DNA
that can move from one site to another. Some transposable
elements replicate, leaving one copy at the original location
and placing one copy at a new site; others transpose
without replication, leaving the original location
altogether. Bacterial transposons include the following
types: (1) insertion sequences such as IS1 that contain only
the genes necessary for transposition, flanked by inverted
terminal repeats; (2) transposons such as Tn3 that are like
insertion sequences but contain at least one extra gene,
usually a gene that confers antibiotic resistance.
Eukaryotic transposons use a wide variety of
replication strategies. The DNA transposons, such as Ds
and Ac of maize and the P elements of Drosophila behave
like the DNA transposons, such as Tn3, of bacteria.
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The immunoglobulin genes of mammals rearrange
using a mechanism that resembles transposition. Vertebrate
immune systems create enormous diversity in the kinds of
immunoglobulins they can make. The primary source of
this diversity is the assembly of genes from two or three
component parts, each selected from a heterogeneous pool
of parts. This assembly of gene segments is known as
V(D)J recombination. The recombination signal sequences
(RSSs) in V(D)J recombination consist of a heptamer and a
nonamer separated by either 12-bp or 23-bp spacers.
Recombination occurs only between a 12 signal and a
23 signal, which ensures that only one of each kind of
coding region is incorporated into the rearranged gene.
RAG1 and RAG2 are the principal players in human
V(D)J recombination. They introduce single-strand nicks
into DNA adjacent to either a 12 signal or a 23 signal.
This leads to a transesterification in which the newly
created 39-hydroxyl group attacks the opposite strand,
breaking it, and forming a hairpin at the end of the coding
segment. The hairpins then break and join with each other
in an imprecise way, allowing joining of coding regions
with loss of bases or gain of extra bases.
The retrotransposons come in two different types. The
LTR-containing retrotransposons replicate like
retroviruses, which replicate through an RNA
intermediate as follows: When a retrovirus infects a cell,
it makes a DNA copy of itself, using a virus-encoded
reverse transcriptase to carry out the RNA→DNA
reaction, and an RNase H to degrade the RNA parts of
RNA–DNA hybrids created during the replication
process. A host tRNA serves as the primer for the reverse
transcriptase. The finished double-stranded DNA copy of
the viral RNA is then inserted into the host genome,
where it can be transcribed by host polymerase II. The
retrotransposons Ty of yeast and copia of Drosophila
replicate in much the same way. They start with DNA in
the host genome, make an RNA copy, then reversetranscribe it—probably within a virus-like particle—to
DNA that can insert in a new location.
The other class of eukaryotic retrotransposons are the
non-LTR retrotransposons, and they use different
methods of priming reverse transcription. For example,
LINEs and LINE-like elements encode an endonuclease
that nicks the target DNA. Then the element takes
advantage of the new DNA 39-end to prime reverse
transcription of element RNA. After second-strand
synthesis, the element has become replicated at its target
site. A new round of transposition begins when the LINE
is transcribed. Because the LINE polyadenylation signal is
weak, transcription of a LINE frequently includes one or
more downstream exons of host DNA and this can
transport host exons to new locations in the genome.
Nonautonomous, non-LTR retrotransposons include
the very abundant Alu elements in humans and similar
elements in other vertebrates. They cannot transpose by
themselves because they do not encode any proteins.
Instead, they take advantage of the retrotransposition
machinery of other elements, such as LINEs. Processed
pseudogenes probably arose in the same way: mRNAs
were probably reverse-transcribed by LINE machinery
and then inserted into the genome.
Group II introns represent another class of non-LTR
retrotransposons found in both bacteria and eukaryotes.
They can retrohome to intronless copies of the same gene
by insertion of an RNA intron into the gene, followed by
reverse transcription and second-strand synthesis. Group II
introns can also undergo retrotransposition by insertion of
an RNA intron into an unrelated gene by target-primed
reverse transcription, perhaps using lagging strand DNA
fragments as primers. This kind of retrotransposition of
group II introns may have provided the ancestors of modernday eukaryotic spliceosomal introns and may account for
their widespread appearance in higher eukaryotes.
REVIEW QUESTIONS
1. Describe and give the results of an experiment that shows
that bacterial transposons contain inverted terminal repeats.
2. Compare and contrast the genetic maps of the bacterial
transposons IS1 and Tn3, and the eukaryotic transposon Ac.
3. Diagram the mechanism of Tn3 transposition, first in simplified form, then in detail.
4. Diagram a mechanism for nonreplicative transposition.
5. Explain how transposition can give rise to speckled maize
kernels.
6. Draw a sketch of an antibody protein, showing the light
and heavy chains.
7. Explain how thousands of immunoglobulin genes can give
rise to many millions of antibody proteins.
8. Diagram the rearrangement of immunoglobulin lightand heavy-chain genes that occurs during B-lymphocyte
maturation.
9. Explain how the signals for V(D)J joining ensure that one
and only one of each of the parts of an immunoglobulin
gene will be included in the mature, rearranged gene.
10. Diagram a reporter plasmid designed to test the importance
of the heptamer, nonamer, and spacer in a recombination
signal sequence. Explain how this plasmid detects
recombination.
11. Present a model for cleavage and rejoining of DNA strands
at immunoglobulin gene recombination signal sequences.
How does this mechanism contribute to antibody diversity?
12. Describe and give the results of an experiment that shows
that cleavage at an immunoglobulin recombination signal
sequence leads to formation of a hairpin in vitro.
13. Present evidence for a reverse transcriptase activity in retrovirus particles and the effects of RNase on this activity.
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Suggested Readings
14. Describe and show the results of an experiment that
demonstrates that strong-stop reverse transcripts in
retroviruses are base-paired to the RNA genome and
covalently attached to an RNA primer.
15. Illustrate the difference between the structures of the LTRs
in genomic retroviral RNAs and retroviral proviruses.
16. Diagram the conversion of a retrovirus RNA to a provirus. Show how this explains the difference in the previous
question.
17. Compare and contrast the mechanisms of retrovirus replication and retrotransposon transposition.
18. Summarize the evidence that retrotransposons transpose via
an RNA intermediate.
19. Describe and show the results of an experiment that demonstrates that the endonuclease of a LINE-like element can
specifically nick one strand of the element’s target DNA.
20. Describe and show the results of an experiment that demonstrates that a LINE-like element can use a nicked strand
of its target DNA as a primer for reverse transcription of
the element.
21. Present a model for retrotransposition of a LINE-like element.
c.
d.
757
Inhibitors of reverse transcription
Inhibitors of translation
7. You have identified a new transposon you call Rover. You
want to determine whether Rover transposes by a retrotransposon mechanism or by a standard replication transposition mechanism such as that used by Tn3. Describe an
experiment you would use to answer this question, and tell
what the results would be in each case.
8. You are a molecular biologist interested in learning more
about the fascinating process of V(D)J recombination.
Assuming that you are capable of generating all of the following possible variants, explain what effect (from a molecular process standpoint as well as a physiological and/or
immunological standpoint) you would expect to observe
if the following were created in your laboratory:
a. the removal of all of the D gene segments from the
section of the genome encoding the heavy chain of
antibodies.
b. the removal of all of the D gene segments from the
section of the genome encoding the beta chain of T-cell
receptors.
c. the genetic alteration of the RSS flanking the D gene
segments from a 12 signal to a 23 signal.
d. the elimination of expression of the RAG gene products.
A N A LY T I C A L Q U E S T I O N S
1. A certain transposon’s transposase creates staggered cuts in
the host DNA five base pairs apart. What consequence does
this have for the host DNA surrounding the inserted transposon? Draw a diagram to explain how the staggered cuts
affect the host DNA.
2. You are interested in measuring the rate of transfer of a
hypothetical transposon, Stealth, from one plasmid, carrying two antibiotic resistance genes of its own, to another
plasmid, which carries the gene for chloramphenicol resistance. (Stealth carries an ampicillin resistance gene.)
Describe an experiment you would perform to assay for
this transposition.
3. Identify the end product of abortive transposition carried
out by Tn3 transposons with mutations in the following
genes.
a. Transposase
b. Resolvase
4. Transposon TnT in plasmid A transposes to plasmid B.
How many copies of TnT are in the cointegrate? Where are
they with respect to the two plasmids in the cointegrate?
5. If the transposable element Ds of maize transposed by the
same mechanism as Tn3, would we see the speckled kernels
with the same high frequency? Why, or why not?
6. Assume you have two cell-free transposition systems that
have all the enzymes necessary for transposition of Tn3 and
Ty, respectively. What effect would the following inhibitors
have on these two systems, and why?
a. Inhibitors of double-stranded DNA replication
b. Inhibitors of transcription
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