An Introduction to Genetic Analysis
Chapter 19
Mechanisms of Recombination
Chapter 19
Mechanisms of Recombination
Key Concepts
Recombination occurs at regions of homology between chromosomes through the breakage
and reunion of DNA molecules.
Models for recombination, such as the Holliday model, involve the creation of a heteroduplex
branch, or cross bridge, that can migrate and the subsequent splicing of the intermediate
structure to yield different types of recombinant DNA molecules.
Recombination models can be applied to explain genetic crosses.
Many of the enzymes participating in recombination in bacteria have been identified.
Introduction
Throughout our analysis of linkage, we studied the recombination of genes by crossing-over.
In this chapter, we consider molecular mechanisms for generating recombination by
crossing-over. Figure 19-1 depicts a basic crossover event, in which two homologous
molecules are aligned and subsequently undergo recombination. When Benzer's work and that
of others revealed that recombination occurs within genes, it became evident that
recombination had to be very precise, because even single-base-pair errors could disrupt the
integrity of the gene. How can recognition of homologous chromosomes and recombination
events be so precise? The answer lies in the power of base-pair complementarity. We shall see
how base-pair complementarity and the formation of heteroduplex regions between
complementary regions of homologous chromosomes lead to the recombination events that
we have been studying.
Breakage and reunion of DNA molecules
The experiments discussed in Chapter 5 provide good indirect evidence in favor of breakage
and reunion. One of the first direct proofs that chromosomes (in this case, viral chromosomes)
can break and rejoin came from experiments on λ phage done in 1961 by Matthew Meselson
and Jean Weigle. They simultaneously infected E. coli with two strains of λ. One strain,
which had the genetic markers c and mi at one end of the chromosome, was “ heavy ” because
the phages were produced from cells grown in heavy isotopes of carbon (13C) and nitrogen
(15N). The other strain was c+ 14mi+ for the markers and had “light” DNA because it was
harvested from cells grown on the normal light isotopes 12C and 14N. The two DNAs
(chromosomes) can be represented as shown in Figure 19-2a. The multiply infected cells were
then incubated in a light medium until they lysed.
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An Introduction to Genetic Analysis
Chapter 19
Mechanisms of Recombination
The progeny phages released from the cells were spun in a cesium chloride density gradient.
A wide band was obtained, indicating that the viral DNAs ranged in density from the heavy
parental value to the light parental value, with a great many intermediate densities (Figure
19-2b). Interestingly, some recombinant phages were recovered with density values very
close to the heavy parental value. They were of genotype c 14 mi+, and they must have
arisen through an exchange event between the two markers (Figure 19-2c). The heavy density
of the chromosome would be expected because only the small tip of the chromosome carrying
the mi+ allele would come from the light parental chromosome. In the reciprocal cross of
heavy c+mi+ phages to light c mi, the heavy recombinants were found to be c+mi, as expected.
These results can be explained in only one way: the recombination event must have occurred
through the physical breakage and reunion of DNA. Although we have to be careful about
extrapolating from viral to eukaryotic chromosomes, this evidence shows that the breakage
and reunion of DNA strands does occur.
Chiasmata: the crossover points
In Chapter 5, we made the simple assumption that chiasmata are the actual sites of crossovers.
Mapping analysis indirectly supports this idea: because an average of one crossover per
meiosis produces 50 genetic map units, there should be correlation between the size of the
genetic map of a chromosome and the observed mean number of chiasmata per meiosis. The
correlation has been made in well-mapped organisms.
However, the harlequin chromosome-staining technique (see Chapter 8) has made it possible
to test the idea directly. In 1978, C. Tease and G. H. Jones prepared harlequin chromosomes
in meioses of the locust. Remember that the harlequin technique produces sister chromatids:
one dark and the other light. When a crossover occurs, it can be between two dark, two light,
or nonsister dark and light chromatids, as shown in Figure 8-16. This last situation is crucial
because mixed (part dark and part light) crossover chromatids are produced. Tease and Jones
found that the dark – light transition is right at the chiasma—proving that the chiasmata are
the crossover sites and settling a question that had been unresolved since the early 1900s
(Figure 19-3).
Genetic results leading to recombination models
Tetrad analysis in filamentous fungi, such as Neurospora crassa, where all four products of a
single meiosis can be recovered and examined (see Chapter 6), provided the impetus for the
first models of intragenic recombination. These crucial findings, reviewed in the following list,
were gene conversion, evidence for postmeiotic segregation of gene-conversion events,
polarity, and the association of gene conversion with crossing-over.
1. Gene conversion. Departures from the predicted Mendelian 4:4 segregation ratios are
detectable in some asci (0.1–1.0 percent in filamentous fungi, but as high as 4 percent in
yeast). Figure 19-4 gives the most common aberrant ratios obtained. It appears as though
some alleles in the cross have been “converted” into the opposite alleles (Figure 19-5). The
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An Introduction to Genetic Analysis
Chapter 19
Mechanisms of Recombination
process therefore has become known as gene conversion; it can occur only where there is
heterozygosity for two different alleles of a gene. In asci with a 6:2 or 2:6 ratio, one entire
chromatid of a chromosome seems to have converted. In asci with a 5:3 or 3:5 ratio, only half
a chromatid seems to have converted. Here, different members of a spore pair have different
genotypes. Recall that each spore pair is produced by mitosis from a single product of meiosis.
The 5:3 or 3:5 ratios can be explained only by the two strands of the double helix carrying
information for twoM different alleles at the conclusion of meiosis. The next mitotic division
is therefore a postmeiotic segregation of alleles.
Conversion cannot be explained by mutation, because the allele that is converted always
changes into the other specific allele taking part in the cross, not to some other allele known
for the locus but not a part of the cross.
2. Polarity. In genes for which accurate allele maps are available, we can compare the
conversion frequencies of alleles at various positions within the gene. In almost every case,
the sites closer to one end show higher frequencies than do the sites farther away from that
end. In other words, there is a gradient, or polarity, of conversion frequencies along the gene
(Figure 19-6).
3. Conversion and crossing-over. In heteroallelic crosses where the locus under study is
closely flanked by other genetically marked loci, the conversion event is very often (about 50
percent of the time) accompanied by an exchange in one of the flanking regions. This
exchange nearly always takes place on the side nearer the allele that has converted and almost
always includes the chromatid in which conversion has occurred.
For example, consider the chromatids diagrammed in Figure 19-6. Suppose that the polarity is
such that the alleles toward the left end of the chromatid convert more often than those toward
the right end. The cross diagrammed here is between a+m2b+ and a m1b, where m1 and m2 are
different alleles of the m locus and a and b represent closely linked flanking markers. If we
look at asci in which conversion has occurred at the m1 site (the most frequent kind of
conversion in this locus), we find that half of these asci will also have a crossover in region I
and half will have no crossover. In the smaller number of asci showing gene conversion at the
m2 site, half will also have a crossover in region II and half will have no crossover. Such
events are detected in ascus genotypes like the one shown in Figure 19-7, which can be
interpreted as a conversion of m1 → +, accompanied by a crossover in region I.
In some asci, a single conversion event seems to include several sites at once. In a
heteroallelic cross, this event is called a co-conversion (Figure 19-8). The frequency of
co-conversion increases as the distance between alleles decreases.
Holliday model
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An Introduction to Genetic Analysis
Chapter 19
Mechanisms of Recombination
One of the first plausible models to account for the preceding observations was formulated by
Robin Holliday. The key features of the Holliday model are the formation of heteroduplex
DNA; the creation of a cross bridge; its migration along the two heteroduplex strands,
termed branch migration; the occurrence of mismatch repair; and the subsequent
resolution, or splicing, of the intermediate structure to yield different types of recombinant
molecules. The model is depicted in Figure 19-9.
Enzymatic cleavage and the creation of heteroduplex DNA
Looking at Figure 19-9a, we can see that two homologous double helices are aligned,
although note that they have been rotated so that the bottom strand of the first helix has the
same polarity as the top strand of the second helix (5′ → 3′ in this case). Then a nuclease
cleaves the two strands that have the same polarity (Figure 19-9b). The free ends leave their
original complementary strands and undergo hydrogen bonding with the complementary
strands in the homologous double helix (Figure 19-9c). Ligation produces the structure shown
in Figure 19-9d. This partially heteroduplex double helix is a crucial intermediate in
recombination, and has been termed the Holliday structure.
Branch migration
The Holliday structure creates a cross bridge, or branch, that can move, or migrate, along the
heteroduplex (Figure 19-9d and e). This phenomenon of branch migration is a distinctive
property of the Holliday structure. Figure 19-10 portrays a more realistic view of this
structure as it might appear during branch migration.
Resolution of the Holliday structure
The Holliday structure can be resolved by cutting and ligating either the two originally
exchanged strands (Figure 19-9f, left) or the originally unexchanged strands (Figure 19-9f,
right). The former generates a pair of duplexes that are parental, except for a stretch in the
middle containing one strand from each parent. If the two parents had different alleles in this
stretch, as indicated here, then the DNA will be heteroduplex. The latter resolution step
generates two duplexes that are recombinant, with a stretch of heteroduplex DNA. The
Holliday model also postulated that the heteroduplex DNA mismatches can be repaired by an
enzymatic correction system that recognizes mismatches and excises the mismatched base
from one of the two strands, filling in the excised base with the correct complementary base.
The resulting molecules will carry either the wild-type or the mutant allele, depending on
which allele is excised.
Figure 19-11 demonstrates one way that we can easily visualize how the Holliday structure
can be converted into the recombinant structures with which we are familiar. In Figure 19-11a,
we can see the structure that we arrived at in Figure 19-4e drawn out in an extended form.
Compare Figures 19-9e and 19-11a until you are convinced that these two structures are
indeed equivalent. If we rotate the bottom part of this structure, as shown in Figure 19-11b,
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An Introduction to Genetic Analysis
Chapter 19
Mechanisms of Recombination
we can generate the form depicted in Figure 19-11c. This last form can be converted back into
two unconnected double helices by enzymatically cleaving only two strands. As indicated in
19-11c, cleavage can occur in either of two ways, each of which generates a different product
(Figure 19-11d). These cleaved structures can be viewed more simply (Figure 19-11e). Repair
synthesis produces the final recombinant molecules (Figure 19-11f). Note the two different
types of recombinants.
Application of the Holliday model to genetic crosses
The Holliday model nicely explained the phenomena that we described previously. Gene
conversion and the aberrant ratios depicted in Figure 19-4 can result as a consequence of
mismatch repair, as shown in Figure 19-12 and Table 19-1. In Table 19-1, the symbols + for
wild type and m for mutant are used for simplicity. When both mismatches are corrected to
yield the same parental type, then 6:2 or 2:6 ratios result; when only one heteroduplex is
corrected, then a 5:3 ratio results; and, when there is no correction, an aberrant 4:4 ratio is the
product.
The Holliday model also accounted for polarity of gene-conversion events, because
conversion takes place only within the heteroduplex DNA between the break point and the
branch point at which the Holliday structure is resolved. The farther a gene locus is from the
breakage point, the more likely it is to be beyond the branch point and thus not part of the
heteroduplex.
It should be noted that the phenomenon of gene conversion and its association with about half
the cases of crossing-over was a driving force for the formulation of the Holliday model,
which entails a strand exchange that results in reciprocal crossovers almost half the time. This
50 percent reciprocal crossover result is because of the resolution of the exchange point in
two equally likely ways, as seen in Figure 19-9f, one of which produces crossing-over of
markers outside the region of heteroduplex DNA. Coconversion is explained by the location
of both sites in the region of heteroduplex DNA and by the excision of both sites in the same
excision-repair act. This double excision converts both sites into the same parental type.
Meselson-Radding model
As the data from tetrad analyses accumulated, it became clear that the Holliday model could
not explain everything. For instance, the two mismatches resulting from the two
heteroduplexes (see Figures 19-9e and 19-12) should be manifested in the progeny from a
cross, yielding aberrant 4:4 tetrads. Yet, tetrad analyses in yeast and other organisms showed
that, whereas 6:2 tetrads were frequent among gene-conversion events, aberrant 4:4 tetrads
were very rare. It seemed as if gene conversion and the formation of heteroduplex DNA
occurred primarily in only one chromatid. The model proposed by Meselson and Radding
(shown in Figure 19-13) generates the Holliday structure with one single-strand cut in only
one chromosome (Figure 19-13a), in contrast with the Holliday model, in which a nick is
made in one strand in each of the two homologous chromatids. This single-strand cut is
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An Introduction to Genetic Analysis
Chapter 19
Mechanisms of Recombination
followed by DNA synthesis (Figure 19-13b). After the nick, the displaced single strand
invades the second duplex (Figure 19-13c), generating a loop, which is excised (Figure
19-13d). After ligation to produce a Holliday structure, followed by branch migration (Figure
19-13e), a heteroduplex is generated in each chromosome. Resolution of this intermediate
(Figure 19-13f) occurs exactly as depicted in Figure 19-9 (or after rotation, as in Figure
19-11). Note the lack of symmetry in the heteroduplex DNA at resolution in Figure 19-13f
(left), compared with Figures 19-9f and 19-11f. Thus, in the Meselson-Radding model (left
side), the bottom chromatid duplex (in Figure 19-13) has a heteroduplex region, instead of the
two chromatids having heteroduplex regions, as in the Holliday model. However, branch
migration and isomerization can generate a structure that has heteroduplex regions on both
duplexes (Figure 19-13, right side), which is required to explain aberrant 4:4 ratios (Table
19-1).
Double-strand break-repair model for recombination
In the Holliday and Meselson-Radding models for genetic recombination, the initiation events
for recombination are single-strand nicks that result in the generation of heteroduplex DNA.
However, the finding that yeast transformation is stimulated 1000-fold when a double-strand
break is introduced into a circular donor plasmid provided the impetus for an additional model,
the double-strand-break model shown in Figure 19-14. Originally formulated by Jack Szostak,
Terry Orr-Weaver, and Rodney Rothstein, this model invokes double-strand breaks to initiate
recombination. The breaks are enlarged to gaps, and the repair of the double-stranded gap
results in gene conversion. The key features of this model are diagrammed in the steps in
Figure 19-14: (1) a double-strand break, followed by digestion of the 5′ end of both cut sites;
(2) the invasion by a remaining 3′ tail of the uncut other duplex; (3) the repair synthesis of one
strand; (4) the repair synthesis of the other strand, and ligation to form two Holliday junctions;
(5) resolution in one of two ways, one of which generates a reciprocal crossover; and (6)
mismatch repair correction to yield gene conversion.
MESSAGE
The phenomenon of gene conversion led to the development of heteroduplex models
to explain the mechanism of crossing-over. Mendelian (1:1) allele ratios are normally
observed in crosses because it is only rarely that a heterozygous locus is the precise
point of chromosome exchange. Asci showing gene conversion at a typical
heterozygous locus are relatively rare (on the order of 1 percent).
Visualization of recombination intermediates
Several of the individual steps that constitute the Holliday model have been demonstrated to
occur in vivo or in vitro, such as nicking, strand displacement, branch migration, repair
synthesis, and ligation. H. Potter and D. Dressler showed that DNA intermediates of the type
predicted by the Holliday model can be found in recombining phages or plasmids. Figure
19-15 shows an electron micrograph of a recombinant molecule. It is formally equivalent to
the central pair of DNA double helices shown in Figure 19-11, with two arms rotated to
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Chapter 19
Mechanisms of Recombination
produce the central single-stranded “dia-mond,” as in Figure 19-11c, which runs between the
sections of double-stranded DNA.
Enzymatic mechanism of recombination
By the isolation of mutants defective in some stage of recombination, much light has been
shed on the enzymology of recombination. In E. coli, the products of several genes involved
in general recombination—the recA, recB, recC, and recD genes—have been well
characterized, as has the single-strand-DNA-binding (Ssb) protein. Mutants deficient in any
of these proteins have reduced levels of recombination. In fact, three distinct recombination
pathways have been identified. In addition to the major RecBCD pathway, two minor
pathways, RecF and RecE, are activated in certain situations. All three pathways use the
RecA protein.
Production of single-stranded DNA
An initial step in recombination is probably the nicking and unwinding of a DNA duplex by a
protein complex consisting of the RecB, RecC, and RecD proteins. The complex has both
helicase and nuclease activity. Figure 19-16 shows how this complex unwinds the DNA,
driven by the hydrolysis of ATP as it moves and generates single strands from a duplex
molecule. The nuclease activity recognizes an 8-bp sequence:
called a chi site; these chi sites appear approximately every 64 kb. As the complex unwinds
the DNA, the free single-stranded DNA can be used to initiate recombination. The Ssb protein,
which also takes part in DNA replication (Chapter 8), can bind to and stabilize the single
strands.
RecA-protein-mediated single-strand exchange
The RecA protein, which also plays a role in the induction of the SOS repair system (see
Chapter 16), can bind to single strands along their length, forming a nucleoprotein filament.
RecA catalyzes single-strand invasion of a duplex and subsequent displacement of the
corresponding strand from the duplex. The invasion and displacement take place in the
presence of ATP, as shown in Figure 19-17. The displaced strand forms what is termed a D
loop.Figure 19-18 depicts how this sequence can lead to Holliday junction.
Branch migration
The movement of a Holliday junction (see, for instance, Figures 19-9 and 19-10), or branch
migration, increases the length of heteroduplex DNA. The RuvA and RuvB proteins catalyze
branch migration, driving the reaction by the hydrolysis of ATP. The RuvA protein binds to
the crossover point and then is flanked by two RuvB ATPase hexameric rings, as seen by
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An Introduction to Genetic Analysis
Chapter 19
Mechanisms of Recombination
electron microscopy. Figure 19-19 depicts a model for the action of these proteins in branch
migration.
Resolution of Holliday junctions
Several enzymatic pathways have been identified that cleave across the point of strand
exchange in a Holliday structure to yield two duplexes. RuvC is an endonuclease that resolves
Holliday junctions by symmetric cleavage of the continuous (noncrossing) pair of DNA
strands, as seen in Figure 19-20. In addition, the RecG and Rus proteins may provide
alternative routes to cleavage. These reactions are summarized in Figure 19-21.
Summary
We have gained increasing knowledge of the molecular processes behind recombination,
which produces new gene combinations by exchanging homologous chromosomes. Both
genetic and physical evidence has led to several models of recombination that rely on
common features: hetero-duplex DNA, mismatch repair, and resolution, or splicing, of the
intermediate structure to yield recombinant molecules. The process of recombination itself is
under genetic control, and numerous genes that affect the process have been identified.
Solved Problems
1. In Neurospora, an ad3 double mutant consisted of two mutant sites within the ad3
gene—site 1 on the left and site 2 on the right. This mutant was crossed to wild type, with
the use of parental stocks heterozygous for two closely linked flanking loci, A and B, as
follows, where 1 represents wild-type sequence at the mutant positions:
Most asci were of the expected type showing regular Mendelian segregations, but there
were also some unexpected types, of which several examples are represented here as I
through III.
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An Introduction to Genetic Analysis
Chapter 19
Mechanisms of Recombination
Explain the likely origin of the rare types I through III according to molecular
recombination models.
Solution
Ascus type I has two nonidentical sister spore pairs. Because the members of a spore pair are
derived from a postmeiotic mitosis (see Chapter 6), this proves that the meiotic products in
these cases must have contained both 1 2 and ++ information; in other words, they must
have contained heteroduplex DNA. Notice also that these spore pairs are recombinant for A
and B. Therefore it is likely that a crossover occurred between the A and B loci, that the
heteroduplex DNA that constituted the crossover spanned both the ad3 mutant sites 1 and 2,
and that there was no correction of the heteroduplex at those sites.
Type II shows a 5:3 ratio of 1 2 doubles to ++. Here again the heteroduplexes must have
spanned both sites (in a noncrossover configuration), but this time there was correction of a
++/1 2 heteroduplex to 1 2/1 2, presumably by excision and repair of the ++ information
(co- or double conversion).
Type III reveals another noncrossover heteroduplex configuration, and this time correction
occurred only at site 1; this is revealed as a 5:3 ratio for site 1, with 1 → + conversion in
ascospore 5.
2. In fungal crosses of the following general type,
where 1 and 2 are mutant sites of a nutritional gene and M and N are flanking loci, it is
possible to select rare ++ prototrophic recombinants by plating on minimal medium. These
prototrophs are then examined for the alleles of the flanking loci. As might be expected, the
combination
is commonly encountered, but so, somewhat surprisingly, are
and
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Chapter 19
Mechanisms of Recombination
a. Explain the origin of these two genotypes in relation to molecular recombination models.
b. If M ++N were more common than m ++n, what would that mean?
Solution
a. It is likely that M ++ N arose from a noncrossover heteroduplex that spanned site 2,
followed by a correction of 2 → +.
Similarly, m ++n would be explained by a heteroduplex spanning site 1, and corrected 1 →
+.
b. Because M ++N arises from gene conversion at the right-hand site, it is likely that
heteroduplex DNA is formed more commonly from the right than from the left, possibly
because of a closer fixed break point. (Note: Prototrophs may also be formed by single-site
correction in a heteroduplex spanning both sites, but then the inequality would require
another explanation.)
Problems
1. Which of the following linear asci shows gene conversion at the arg2 locus?
See answer
2. At the light-spore locus of Ascobolus, the 1′ mutant site is in the left part of the gene and
the 1″ mutant site is more to the right. When crosses are made between 1′- and 1″-bearing
strains,
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Chapter 19
Mechanisms of Recombination
asci with six light and two black spores can be selected visually. They are shown to be
caused by gene conversion mostly of the following type:
What can account for this irregularity?
3. It has been proposed that the “fixed break point” of the heteroduplex recombination model
might correspond to a promoter sequence. The following data relate to this idea. In the
fungus Podospora, mutants were available in adjacent spore color genes 1 through 4, and a
cross was made as follows:
(The numbers 261, 136, 42, and 115 are merely names for mutant sites.) Many asci were
obtained showing gene conversion, but one that was relevant to the preceding suggestion
was as follows:
Interpret this ascus in relation to the promoter idea.
4. Many mutagens increase the frequency of sisterchromatid exchange. Give possible
explanations for this observation.
5. Mutations in locus 46 of the Ascomycete fungus Ascobolus produce light-colored
ascospores (let's call them a mutants). In the following crosses between different a mutants,
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asci are observed for the appearance of dark wild-type spores. In each cross, all such asci
were of the genotypes indicated:
Interpret these results in light of the models discussed in this chapter.
See answer
6. In the cross A m1m2B × a m3b, the order of the mutant sites m1, m2, and m3 is unknown in
relation to one another and to A/a and B/b. One nonlinear conversion ascus is obtained:
Interpret this result in light of the heteroduplex DNA theory, and derive as much
information as possible about the order of the sites.
7. In the cross a1 × a2 (alleles of one locus) the following ascus is obtained:
Deduce what events may have produced this ascus (at the molecular level).
See answer
8. G. Leblon and J.-L. Rossignol made the following observations in Ascobolus.
Single-nucleotide-pair insertion or deletion mutations show gene conversions of the 6:2 or
2:6 type and only rarely of the 5:3, 3:5, or 3:1:1:3 type. Base-pair transition mutations show
gene conversion of the 3:5, 5:3, or 3:1:1:3 type and only rarely of the 6:2 or 2:6 type.
a. In relation to the hybrid DNA model, propose an explanation for these observations.
b. Leblon and Rossignol also showed that there are far fewer 6:2 than 2:6 conversions for
insertions and far more 6:2 than 2:6 conversions for deletions (where the ratios are +:m).
Explain these results in relation to heteroduplex DNA. (You might also think about the
excision of thymine photodimers.
c. Finally, the researchers showed that, when a frameshift mutation is combined in a
meiosis with a transition mutation at the same locus in a cis configuration, the asci showing
joint conversion are all 6:2 or 2:6 for both sites (that is, the frameshift conversion pattern
seems to have “imposed its will” on the transition site). Propose an explanation for this
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Mechanisms of Recombination
result.
See answer
9. At the gray locus in the Ascomycete fungus Sordaria, the cross + × g1 is made. In this
cross, heteroduplex DNA sometimes extends across the site of heterozygosity, and two
heteroduplex DNA molecules are formed (as discussed in this chapter). However,
correction of heteroduplex DNA is not 100 percent efficient. In fact, 30 percent of all
heteroduplex DNA is not corrected at all, whereas 50 percent is corrected to + and 20
percent is corrected to g1. What proportion of aberrant-ratio asci will be (a) 6:2? (b) 2:6?
(c) 3:1:1:3? (d) 5:3? (e) 3:5?
See answer
10. Noreen Murray crossed α and β, two alleles of the me-2 locus in Neurospora. Included in
the cross were two markers, trp and pan, which each flank me-2 at a distance of 5 m.u.
The ascospores were plated onto a medium containing tryptophan and pantothenate but no
methionine. The methionine prototrophs that grew were isolated and scored for the
flanking markers, yielding the results shown in the table below.
Interpret these results in light of the models presented in this chapter. Be sure to account
for the asymmetries in the classes.
11. In Neurospora, the cross A x × a y is made, in which x and y are alleles of the his-1 locus
and A and a are mating-type alleles. The recombinant frequency between the his-1 alleles
is measured by the prototroph frequency when ascospores are plated on a medium lacking
histidine; the recombinant frequency is measured as 10−5. Progeny of parental genotype
are backcrossed to the parents, with the following results. All a y progeny backcrossed to
the A x parent show prototroph frequencies of 10−5. When A x progeny are backcrossed to
the a y parent, two prototroph frequencies are obtained: half of the crosses show 10−5, but
the other half show the much higher frequency of 10−2. Propose an explanation for these
results, and describe a research program to test your hypothesis. (Note: Intragenic
recombination is a meiotic function that occurs in a diploid cell. Thus, even though this
organism is haploid, dominance and recessiveness could have roles in this problem.)
Chapter 19*
1. 3, 4, 6
5. First, notice that gene conversion has occurred. In the first cross, a1 converted (1 : 3). In the
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second cross, a3 converted. In the third cross, a3 converted. Polarity obviously played a
part. The results can be explained by the following map, in which hybrid DNA enters only
from the left.
7. The ratios for a1 and a2 are both 3 : 1. There is no evidence of polarity, which indicates that
gene conversion as part of recombination occurred. The best explanation is that two
separate excision-repair events took place and, in both cases, the repair retained the mutant
rather than the wild type.
8. a. and b. A heteroduplex that contains an unequal number of bases in the two strands has a
larger distortion than does a simple mismatch. Therefore, the former would be more likely
to be repaired. For such a case, both heteroduplex molecules are repaired (leading to 6 : 2
and 2 : 6) more often than one (leading to 5 : 3 or 3 : 5) or none (leading to 3 : 1 : 1 : 3).
The preference in direction (that is, the addition rather than the subtraction of a base) is
analogous to thymine dimer repair. In thymine dimer repair, the unpaired, bulged
nucleotides are treated as correct and the strand with the thymine dimer is excised.
A mismatch more often than not escapes repair, leading to a 3 : 1 : 1 : 3 ascus.
Transition mutations would not cause as large a distortion of the helix, and each strand of
the heteroduplex should have an equal chance of repair. This would lead to 4 : 4 (two
repairs each in the opposite direction), 5 : 3 (one repair), 3 : 1 : 1 : 3 (no repairs or two
repairs in opposite directions), and, less frequently, 6 : 2 (two repairs in the same direction).
c. Because excision repair excises the strand opposite the larger buckle (that is, opposite the
frameshift mutation), the cis transition mutation also is retained. The nearby genes are
converted because of the length of the excision repair.
9. (a). 6 : 2 = 31.25 percent;(b). 2 : 6 = 5 percent;(c). 3 : 1 : 1 : 3 = 11.25 percent;
(d). 5 : 3 = 37.5 percent;(e). 3 : 5 = 15 percent.
14
勇者并非无所畏惧,而是判断出有比恐惧更重要的东西.