87 224 Gene Conversion

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Chapter 22 / Homologous Recombination
Creation of Single-Stranded Ends at DSBs
double-stranded DNA. This finding suggests that Rad50
and Mre11 also work together to resect the 59-ends of
DSBs in yeast.
Once Spo11 has created a DSB, the new 59-ends are digested to yield free 39-ends that can invade another DNA
duplex. Szostak and coworkers first discovered the singlestranded ends in 1989 when they examined the structure of
the DNA termini created by DSBs. They digested the DNAs
with S1 nuclease, which specifically degrades single-stranded
DNA, but leaves double-stranded DNA intact. The S1 nuclease had no effect on the intensities of the bands, but it
reduced the lengths of all three DNAs. This result is exactly
what the model in Figure 22.17 predicts: After the DSB occurs, an exonuclease digests the two 59-ends at the break,
creating single-stranded DNA that would then be susceptible to S1 nuclease digestion. This S1 nuclease digestion
would reduce the length of the DNA fragment.
Kleckner and colleagues also found evidence for digestion, or resection, of one strand at DSBs. They discovered
that the fragments that accumulated in wild-type cells produced diffuse bands on gel electrophoresis, as if the ends
had been nibbled to varying extents. By contrast, the bands
were discrete in rad50S mutants that blocked resection of
the DSB ends.
Mutations in RAD50, MRE11, and COM1/SAE2 obstruct DSB resection. Actually, null alleles of RAD50 and
MRE11 block DSB formation altogether, so only certain
nonnull alleles of these genes permit DSB formation but
impede DSB resection. Thus, the products of both of these
genes are required for both DSB formation and resection.
Evidence for the role of these two gene products comes
from a comparison with E. coli. The E. coli proteins SbcC
and SbcD are homologous to yeast Rad50 and Mre11, respectively. Furthermore, the two bacterial proteins work as
an SbcC/SbcD dimer, which has exonuclease activity on
SUMMARY Formation of the DSB in meiotic recom-
bination is followed by 59→39 exonuclease digestion of the 59-ends at the break, yielding overhanging
39-ends that can invade another DNA duplex.
Rad50 and Mre11 probably collaborate to carry
out this resection.
22.4 Gene Conversion
When fungi such as pink bread mold (Neurospora crassa)
sporulate, two haploid nuclei fuse, producing a diploid
nucleus that undergoes meiosis to give four haploid nuclei.
These nuclei then experience mitosis to produce eight haploid nuclei, each appearing in a separate spore. In principle,
if one of the original nuclei contained one allele (A) at a
given locus, and the other contained another allele (a) at
the same locus, then the mixture of alleles in the spores
should be equal: four A’s and four a’s. It is difficult to imagine any other outcome (five A’s and three a’s, for example),
because that would require conversion of one a to an A. In
fact, aberrant ratios are observed about 0.1% of the time,
depending on the fungal species. This phenomenon is called
gene conversion. We discuss this topic under the heading of
recombination because the two processes are related.
The mechanism of meiotic recombination discussed in
this chapter, and illustrated in Figure 22.17, suggests a mechanism for gene conversion during meiosis. Figure 22.24
A
A
Mismatch
repair
a
A
Replication
A
A
A
No repair
a
A
A
a
Replication
Figure 22.24 A model for gene conversion in sporulating
Neurospora. A strand exchange event with branch migration during
sporulation has resolved to yield two duplex DNAs with patches of
heteroduplex in a region where a one-base difference occurs between
allele A (blue) and allele a (red). The other two daughter chromosomes
are homoduplexes, one pure A and one pure a (not shown). The top
heteroduplex undergoes mismatch repair to convert the a strand to A;
a
the bottom heteroduplex is not repaired. Replication of the repaired
DNA yields two A duplexes; replication of the unrepaired DNA yields
one A and one a. Thus, the sum of the daughter duplexes pictured at
right is three A’s and one a. Replication of the two DNAs not pictured
yields two A’s and two a’s. The sum of all daughter duplexes is
therefore five A’s and three a’s, instead of the normal four of each.
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Summary
depicts this hypothesis in the case of N. crassa. We start with
a nucleus in which DNA duplication has already occurred,
so it contains four chromatids. In principle, two chromosomes should bear the A allele and two the a allele. But in
this case, strand exchange and branch migration have occurred, followed by resolution yielding two chromosomes
with patches of heteroduplex, as illustrated in Figure 22.24.
These heteroduplex regions just happen to be in the region
where alleles A and a differ at one base, so each chromosome has one strand with one allele and the other strand
with the other allele. If DNA replication occurred immediately, this situation would resolve itself simply, yielding two
A duplexes and two a duplexes. However, before replication,
one or both of the heteroduplexes may attract the enzymes
that repair base mismatches. In the example shown here,
only the top heteroduplex is repaired, with the a being converted to A. This leaves three strands with the A allele, and
only one with the a allele. Now DNA replication will produce three A duplexes and only one a duplex. When we add
the two A and two a duplexes resulting from the chromosomes that did not undergo heteroduplex formation, a final
ratio of five A’s and only three a’s results.
Figure 22.25 presents another pathway by which gene
conversion can occur, this time without mismatch repair.
We start with the situation after step (c) in Figure 22.17,
just after strand invasion and D-loop formation. The region in which allele A differs from allele a is indicated at
the top. Blue indicates A and red indicates a. This scheme
differs from Figure 22.17 in that the invading strand is
subject to partial resection, with shrinkage of the D-loop,
before repair synthesis begins. That resection allows a
longer stretch of repair synthesis to occur, which converts
more of the invading strand from A to a. And the conversion occurs in the exact region where alleles A and a
differ. After branch migration and resolution (either
crossover or noncrossover resolution), we have all four
DNA strands representing the a allele in the significant
region, whereas we started with two of each. Gene conversion has occurred.
Gene conversion is not confined to meiotic events. It
is also the mechanism that switches the mating type of
baker’s yeast (S. cerevisiae). This gene conversion event involves the transient interaction of two versions of the MAT
locus, followed by conversion of one gene sequence to that
of the other.
SUMMARY When two similar but not identical
DNA sequences interact, the possibility exists for
gene conversion—the conversion of one DNA
sequence to that of the other. The sequences participating in gene conversion can be alleles, as in
meiosis, on nonallelic genes, such as the MAT genes
that determine mating type in yeast.
729
A/a locus
(a) Resection
(b) DNA repair synthesis
(c) Branch migration
(d) Resolution (noncrossover)
a
Figure 22.25 A model for gene conversion without mismatch repair.
This figure begins in the middle of the DSB recombination scheme
illustrated in Figure 22.17, just after strand invasion. (a) This time, the
invading strand is partially resected, which causes partial collapse of
the D-loop. (b) DNA repair synthesis is more extensive because of the
resection, which produces a region (indicated at top and bottom) in
which all four DNA strands are allele a (red). (c) and (d) Branch migration
and resolution do not change the nature of the four DNA strands in the
region in which alleles A and a differ: All are allele a. Thus, this process
has converted a DNA duplex that was allele A to allele a.
S U M M A RY
Homologous recombination is essential to life: In
eukaryotic meiosis, it locks homologous chromosomes
together so they separate properly. It also scrambles
parental genes in offspring. In all forms of life,
homologous recombination helps to cope with DNA
damage.
Homologous recombination by the RecBCD pathway
in E. coli begins with invasion of a duplex DNA by a
single-stranded DNA from another duplex that has
experienced a double-stranded break. A free end can be
generated by the nuclease and helicase activities of
RecBCD, which prefers to nick DNA at special sequences
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Chapter 22 / Homologous Recombination
called Chi sites. The invading strand becomes coated with
RecA and SSB. RecA helps this invading strand pair with
its complementary strand in a homologous DNA, forming
a D-loop. SSB accelerates the recombination process,
apparently by melting secondary structure and preventing
RecA from trapping any secondary structure that would
inhibit strand exchange later in the recombination
process. Subsequent nicking of the D-loop strand,
probably by RecBCD, leads to the formation of a
branched intermediate called a Holliday junction. Branch
migration, catalyzed by the RuvA–RuvB helicase, brings
the crossover of the Holliday junction to a site that is
favorable for resolution. Finally, the Holliday junction can
be resolved by RuvC, which nicks two of its strands. This
can yield two DNAs with patches of heteroduplex
(noncrossover recombinants), or two crossover
recombinant DNAs.
Meiotic recombination in yeast begins with a
double-stranded break (DSB). Two molecules of Spo11
collaborate to create DSBs by cleaving both strands at
closely spaced sites. The cleavage operates through
transesterification reactions involving active site
tyrosines on the two molecules of Spo11, which leads
to covalent bonds between the two molecules of Spo11,
and the newly formed DSBs. Spo11 is then released
from the DSBs in a complex with oligonucleotides
ranging from about 12–37 nt long. The Spo11oligonucleotides may even be released after resection
of the DSBs. Formation of the DSB in meiotic
recombination is followed by 59→ 39 exonuclease
digestion of the 59-ends at the break. Rad50 and Mre11
probably collaborate to carry out this resection. Next,
the newly generated 39-overhang invades the other DNA
duplex, creating a D-loop. DNA repair synthesis and
branch migration yield two Holliday junctions that can
be resolved to produce either noncrossover or crossover
recombinants.
When two similar but not identical DNA sequences
interact, the possibility exists for gene conversion—the
conversion of one DNA sequence to that of the other. The
sequences participating in gene conversion can be alleles,
as in meiosis, or nonallelic genes, such as the MAT genes
that determine mating type in yeast.
REVIEW QUESTIONS
1. List the three steps in homologous recombination in which
RecA participates, with a short explanation of each.
2. What evidence indicates that RecA coats single-stranded
DNA? What role does SSB play in the interaction between
RecA and single-stranded DNA?
3. Describe and give the results of an experiment that shows
that RecA is required for synapsis at the beginning of
recombination.
4. How would you show that the apparent synapsis you
observe by electron microscopy is really synapsis, rather
than true base pairing?
5. Describe and give the results of an experiment that shows
that RecBCD nicks DNA near a Chi site. How could you
demonstrate that it is RecBCD, and not a contaminant, that
causes the nicking?
6. Show how you could use a gel mobility shift assay to
demonstrate that RuvA can bind to a Holliday junction by
itself at high concentration, that RuvB cannot bind by itself
at all, and that RuvA and RuvB can bind cooperatively at
relatively low concentrations. What is the function of
glutaraldehyde in this experiment?
7. Draw a diagram of the RuvAB–Holliday junction complex,
with the Holliday junction in its familiar cross-shaped form,
ready for branch migration. Include the RuvB rings, but not
the RuvA tetramer.
8. Describe and give the results of an experiment that shows
that RuvC can resolve a Holliday junction.
9. What evidence suggests that RuvA, B, and C are all together
in a complex with a Holliday junction?
10. Present a model for meiotic recombination in yeast.
11. Describe and give the results of an experiment that shows
that a DSB forms during meiotic recombination in yeast.
12. Describe and give the results of an experiment that shows
that Spo11 is covalently attached to a DSB during meiotic
recombination in yeast.
13. Describe and give the results of an experiment that
demonstrates the formation of Spo11-linked
oligonucleotides of two size classes.
14. What do the two size classes of Spo11-linked oligonucleotides,
and the timing of their appearance and disappearance,
suggest about the mechanism of homologous recombination
in yeast? Illustrate your answer with a drawing.
15. Present a model for meiotic gene conversion.
A N A LY T I C A L Q U E S T I O N S
1. Draw a diagram of a Holliday junction. Starting with that
diagram, illustrate:
a. Branch migration to the right, then resolution to yield a
short heteroduplex, or resolution to yield crossover
recombinant DNAs
b. Branch migration to the left, then resolution to yield a
short heteroduplex, or resolution to yield crossover
recombinant DNAs
2. Describe or diagram the products of RecBCD pathway recombination in E. coli cells with mutations in the following
genes:
a. recB
b. recA
c. ruvA
d. ruvB
e. ruvC
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Suggested Readings
3. Show how you could use DNase footprinting to demonstrate that RuvA binds to the center of the Holliday junction, and that RuvB binds to the upstream side relative to
the direction of branch migration.
4. What would be the final mixture of alleles in the gene conversion depicted in Figure 22.24 if both heteroduplexes
were converted to A/A by mismatch repair? What if one
heteroduplex were converted to A/A and the other to a/a?
5. One of the two elements intimately involved in the process
of strand exchange is the Chi sites on the DNA. It is believed
that Chi sites stimulate the RecBCD pathway but not several
other homologous pathways (l Red, E.coli RecE and RecF).
Describe an experiment that would test this hypothesis.
SUGGESTED READINGS
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Smith, G.R. 1991. Conjugational recombination in E. coli:
Myths and mechanisms. Cell 64:19–27.
West, S.C. 1998. RuvA gets x-rayed on Holliday. Cell
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Cao, L., E. Alani, and N. Kleckner. 1990. A pathway for
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