Download Homologous Recombination 1. Query: Could you explain what

Homologous Recombination
1. Query: Could you explain what normal and aberrant segregation patterns are during
meiosis?
We are discussing the mechanisms of homologous recombination during meiosis in a diploid
organism. As I explained, our model organisms are the fungi, Saccharomyces yeast and
Ascobolus. The products of each meiosis are packaged into a spore sac in these organisms. So
by isolating the spores and growing those up, we can observe the phenotype of each of the
products arising from a single meiosis.
Remember that there are two homologues for each chromosome in a diploid cell, one paternal
and the other maternal. In the pre-meiotic cell, they duplicate to give four chromosomes. In the
models we discussed, we considered (for simplicity) exchanges between two of the duplexes in
the region that contains the marker of our interest. Let us say that this is ‘M’ on one
chromosome, and ‘m’ on the homologous chromosome. Let us say M gives a white colony, m
gives red.
If there is no exchange in the region of M/m, two of the chromosomes will carry M, the other two
m. In S. cerevisiae, the four spores, each containing one of these chromosomes, either M or m
(or MM and mm, if you want to consider each strand of the duplex DNA), will be in one ascus
(spore sac). When you open the sac, isolate the individual spores and place them on a nutrient
plate, they will germinate, go through cell divisions and give rise to a colony. Naturally two will
be M = white, and two will be red. This is normal 2: 2 segregation. It is also called normal 4:4
segregation, based on the Ascobolus paradigm described below.
In Ascobolus, the four duplexes undergo one round of duplication prior to being packaged in the
spores. So there are eight spores. And they are arranged in an ordered fashion. So if you count
them from one end, the first two form one pair of sister spores, the next two the second pair of
sisters, and so on. Altogether, there are four such sister pairs. Each pair is identical. The first
two pairs are M (or M/M) and the last two pairs are m (or m/m). The segregation of M to m is
4:4 (normal).You may also look at it this way: It is as if Ascobolous is a step ahead of
Saccharomyces. In Saccharomyces, the replication of DNA occurs when a spore cell divides on
the nutrient plate. In Ascobolous, that division happens in the meiotic cell before the spores are
formed.
The types of segregation that geneticists come across that differ from normal 4:4 are called
aberrant segregation patterns. They are of three types: 4:4 aberrant, 5:3 and 6:2. They result
because of genetic exchange between homologous chromosomes. [In our case we are
interested in exchanges that cause the markers M/m to segregate differently from the normal
4;4.] It is these aberrant classes that suggest possible models for what recombination
mechanisms can account for them.
4:4 Aberrant Segregation:
In the first model, the Holliday junction model, exchange, formation of the Holliday junction and
branch migration give rise to symmetric heteroduplex (we drew this as one green and one blue
strand in the two DNA molecules that carried out the exchange). So, in the four spores of
Saccharomyces, for the markers M and m, there are both blue strands (M/M) in one spore, one
blue and one green in two of the spores(M/m), and two green strands (m/m) in the fourth spore.
The first spore gives a white colony, the fourth gives a red colony. When a spore containing the
blue/green heteroduplex (M/m) goes through the first cell division, the DNA duplicates to give
two DNA molecules: one M/M and the other m/m. One of these molecules will go into one
daughter cell and the other into the second daughter cell. The M/M cell will give rise to progeny
cells of the M/m phenotype (white) through further cell divisions. The m/m cell will give rise to
progeny cells of the m/m phenotype (red) through further cell divisions. Hence one half the
colony will be white, the other half red. This is called a half-sectored colony. So if you count the
white to red ratio, there is one white plus one half white plus one half white, and one red plus
one half red plus one half red. The result is 2:2 (4:4) segregation of white to red, but this is not
the normal 2:2 (4:4) but the aberrant 2:2 (4:4).
In Ascobolus, the DNA replication event that resolves the M/m heteroduplex DNA into two
homoduplex daughter DNA molecules M/M and m/m happens before these DNA molecules are
packaged into spores. Hence the eight ordered spores in the spore sac will be MM, MM; MM,
mm; MM, mm; mm, mm (the M or M refers to each strand of DNA; note there is no more
heteroduplex, since it has been resolved by replication). The segregation of red to white is again
4:4 but aberrant four to four. You can see this by looking at sister spores that I have separated
by the semicolons. There are two identical sister spores (MM, MM and mm, mm) at the left and
right ends. But the two pairs of sister spores in the middle are non-identical: MM, mm and MM,
mm.
5:3 segregation
The Meselson-Radding model generates asymmetric heteroduplex, and explains 5:3
segregation. In Saccharomyces, this is equivalent to one half-sectored colony. In Ascobolous,
this represents one pair of non-identical sister spores.
Since a Holliday junction is formed in this model as well, branch migration can give rise to
symmetric heteroduplex and 4:4 aberrant segregation as well. If the marker happens to be in
the asymmetric heteroduplex region, it will segregate 5:3. If it falls within the symmetric
heteroduplex region, it will segregate 4:4 aberrant.
6:2 segregation
Here one duplex is cut and gapped, and the gap is repaired by DNA synthesis using the intact
DNA as the template. This is the reason for the 6:2 segregation. Note that at the points of
initiation and the points of termination of DNA repair, there is asymmetric heteroduplex. Hence a
marker falling within these regions will segregate 5:3. The completion of DNA repair also
generates two Holliday junctions. As I already explained, branch migration of these junctions
can generate symmetric heteroduplex and give rise to 4:4 aberrant segregation.
2. Query: Under what conditions can a Holliday junction branch migrate?
The Holliday junction can branch migrate as long as there is homology between the partner
duplexes. The directionality to the branch migration (instead of random walks; going one way,
then reversing direction and so on) is conferred by enzymes (the RuvAB helicase in E. coli) that
actively push the junction in one direction by hydrolyzing ATP. This energy input can overcome
single mismatches between the two strands that are spooled up or spooled down during branch
migration. Short stretches of mismatches or DNA insertions or deletions are also tolerated. It is
such differences between the DNA partners that we represented by using blue and green
colors. The blue/green DNA resulting from branch migration is called heteroduplex because the
two DNA strands are not perfectly complementary.
3. Query: Why did you say that the double strand break repair model is the most versatile
model for explaining aberrant segregation patterns?
If you look at the double strand break repair model (the figure from the notes posted on the
course web page), in the central part where the gap is repaired the marker will segregate as 6:2
(there will be three DNA regions of the same color (marker M) and one with a different color (m)
in our drawings; 3:1 is the same as 6:2; recall the yeast and Ascobolus paradigm that we
discussed). In the regions immediately adjacent to the left and right Holliday junctions, you can
see asymmetric heteroduplex, blue/green on one DNA duplex but not the other. If the marker of
interest to us (M and m) happens to be located in this region, the segregation will be 5:3.
Migration of the left Holliday junction to the left and the right Holliday junction to the right will
generate symmetric heteroduplex. If the marker happens to fall in the region of symmetric
heteroduplex, it will segregate as aberrant 4:4. Hence all forms of aberrant segregation are most
readily explained by the double strand break repair model.
Here is the same answer (in response to a similar query) in slightly different format,
along with the figure for the double strand break repair model).
Remember, there are four DNA duplexes in the cell. We are only showing exchange between
two of these. So there is a red DNA duplex (chromosome) at the top and a green DNA duplex
(chromosome at the bottom) intact. Imagine that your marker (responsible for colony phenotype;
color; red or white, as we discussed in class) is in the region of double strand gap repair
(between the vertical green dashed lines). If you count from the top, there are 1, 2 and 3 (total of
3) red duplexes and one green (the bottom most one) for this marker. The segregation is 3:1 or
6:2. Note: you made double strand cut in the green DNA, gapped it (destroyed green
information) and repaired or replaced it by using red information.
Look at the DNA segments between the Holliday junction at the left and the dashed green
vertical line close to it; also the Holliday junction to the right and the dashed green vertical line
close to it. [The dashed green lines indicate the points where DNA synthesis started in the
repair process. The line at the right indicates start of DNA synthesis on the top DNA molecule;
the line at the left indicates start of DNA synthesis on the bottom DNA molecule]. If the marker
responsible for the phenotype happens to fall in either of these regions, it will segregate 5:3. If
you look at these two regions along all four DNA molecules, you will see heteroduplex
(red/green strands) only on one molecule. This is asymmetric heteroduplex, equivalent to
Meselson-Radding model (5:3).
Once the Holliday junction is formed (here there are two of them, one at the left and one at the
right), branch migration by either can produce symmetric heteroduplex (red/green strands on
the two DNA duplex partners). If the marker falls within this symmetric heteroduplex region), it
will segregate 4: 4 aberrant. [For simplicity imagine the junction at left branch-migrating to the
left and the junction at right branch-migrating to the right.]
In other words, the double strand gap repair model is inclusive of Meselson-Radding and
Holliday models.
4. Query: Could you explain branch migration?
When two strands (either the two bottom strands or the two top strands) have been exchanged
between two DNA partners, you form a four-way junction called the Holliday junction. The
junction migrates when we spool the strand from duplex partner 1 to partner 2 and
simultaneously the other way, from partner 2 to partner 1. In the diagram we drew in class, the
blue strand was spooled one way, and the green strand the other way. The net result is the
formation of equal lengths of blue/green DNA on partners 1 and 2. This is called symmetric
heteroduplex. It is the symmetric heteroduplex that gives rise to 2 spores that give normal
colonies and two that are half-sectored. If the blue is marker ‘M’ and green is ‘m’, the
segregation will be: one spore with MM, two spores with M/m (heteroduplex) and one spore with
mm. The segregation of the markers MM and mm will be one M/M plus one half M/M plus one
half M/M and one m/m plus one half mm plus one half m/m. The segregation ratio for M/M and
mm = 2:2, but not normal 2:2. In the case of normal segregation (no DNA exchange) the
colonies will not be sectored. There will two spore that are M/M and two that are mm. The
segregation in Ascobolus would be 4:4 aberrant when symmetric heteroduplex is formed by
Holliday branch migration. There will be two pairs of non-identical spores (MM, mm; and MM,
mm) and two pairs of identical sister spores (MM, MM) and (mm, mm). In normal 4:4
segregation, the sister spores are identical in every case (MM, MM); (MM, MM); (mm, mm) and
(mm, mm). (When I say MM, I am referring to the two strands of a single DNA molecule. The
phenotype will be M, say white colony; similarly mm will give the m phenotype, say red colony).
5. Query: How does branch migration contribute to 2:2/4:4 symmetric segregation in the third
model (double strand break repair)?
In the third model, double strand break repair, two Holliday junctions are formed one at the right,
one at the left. In between, abutting the Holliday junctions, we have a region of asymmetric
heteroduplex; at the left and at the right. The asymmetric heteroduplex will be seen in the
diagram as one blue/green duplex with the equivalent region in the other being blue/blue). In the
center of the repair region it is all homoduplex (blue/blue). A marker in the asymmetric
heteroduplex region will segregate as 5:3. A marker in the center of the repaired region will
segregate as 6:2. If the Holliday junction branch migrates to the right you will generate
symmetric heteroduplex (blue/green and blue green on both partner DNA molecules). A marker
falling in this region will segregate as 4:4 aberrant. The same holds for the Holliday junction at
the left when it branch migrates to the left.
Note that if the left junction migrates to the right you will not change the asymmetric duplex at
the left, symmetric heteroduplex in the center and the asymmetric heteroduplex at the right
adjacent to the right junction. You can draw this out or visualize it using the diagram for the
double strand break repair model, and convince yourself that this is true. The same holds true
for the leftward branch migration of the right Holliday junction until it reaches the left junction.
It is only when a junction branch migrates between a blue homoduplex and a green
homoduplex, spooling down blue strand to the bottom DNA molecule and spooling green strand
up to the top molecule, that you generate symmetric heteroduplex (4:4 aberrant segregation).
In the figure below, the vertical line in the lower panel represents the initial position of
the Holliday junction. Notice symmetric heteroduplex (Symm HD) blue/green on both
duplexes in the region of branch migration.
6. Query: Does strand exchange simply mean the exchange of homologous strands between
sister DNA strands, no matter what the mechanism is?
[The choice of words ‘homologous strands’ and ‘sister DNA strands’ above is ambiguous.
Please see the explanations in my answers]
The two duplexes taking part in the exchange are homologous chromosomes (one paternal and
the other maternal). They are largely identical but have some differences, for example, in the
region of interest, one carries M and the other m. This is what we designate as blue duplex (say
M/M on the two strands) and green duplex (m/m on the two strands). If you cut the same strand
(or like strands) on the two duplexes and exchange them, you form the Holliday junction. When
the junction migrates (as I explained in the answer to your first question), you generate
blue/green (M/n) heteroduplex on both the DNA molecules. This is one mode of exchanging
strands. As we saw, the exchange mechanism can be different, illustrated by the three models.
If you only exchange one pair of strands, you get a Holliday junction. To resolve the
intermediate, you have to do a second exchange. This can be between the same pair of strands
that took part in the first exchange or between the pair of strands that were not exchanged
initially. The first case, exchanging the same pair of strands, will leave the flanking markers in
the parental configuration (A---B and a-----b). In the second case, exchanging one pair of
strands first followed by exchanging the other pair of strands), the flanking markers will be in the
recombined (crossed over) configuration (A---b and a----B).
[Important Note: The flanking markers A, B and a, b are in reference to the Figure above. For
the figures in the notes on the web page the flanking markers are A, b on one parental duplex
and a, B on the other. So if there is cross-over, their configuration will change to A----B and a----b).
Use the word 'strand' to mean each of the two strands of a DNA duplex, a 'Watson' strand and a
complementary 'Crick' strand. When you refer to a DNA molecule, refer to it as a duplex (made
up of the two complementary strands). The exchange is between the Watson strands of the
homologous duplexes and/or the Crick strands. This is what I mean by 'exchange of like
strands'. Note that you cannot a Watson strand with a Crick strand; the polarities of the two
strands are opposite; also their sequences are not the same, rather the sequences are
complementary.
Usually ‘sister’ refers to a duplex formed by replication of a certain DNA molecule (or the
replication of a chromosome). The two products are identical (unless some mutation resulted
during replication), and are referred to as ‘sister chromatids’. Avoid the use of the expression
‘sister strands’. I assume that you mean ‘like strands’: the Watson strands of the homologous
chromosomes (or the Crick strands) of the homologous chromosomes.
Of course, exchange can occur between two identical chromosomes or DNA duplexes (sister
chromatids) but the products will be the same as the parent duplexes. It is only when they differ
in sequence (and thus carry marker M on one homolg and m on the other) that we can detect
the exchange by the separate phenotypes that M and m generate (white versus red colony color
for example).
7. Query: "What is the nature of the eight spores of Ascobolus formed during a meiotic event in
which a marker segregates in the 4:4 aberrant fashion?"- What exactly is this question looking
for?
Out of the eight spores (four pairs), two pairs will be non-identical sisters. This is equivalent to
two half-sectored colonies in yeast. They arise from the symmetric heteroduplex formed by the
Holliday mechanism. In yeast the symmetric heteroduplex (Mm) is resolved by DNA replication
into two homoduplexes (MM and mm) only when the spore cell germinates and divides into two
cells. The progeny of one cell by further divisions will be MM; the progeny of the second cell will
be mm. This is the reason for the sectored colony. In Ascobolous, the heteroduplex is resolved
by replication before the spore cells are formed. Hence one spore will contain MM and its sister
will be mm.
8. Query: Can you explain what you mean by Holliday junction isomerization and resolution?
Manipulation of Holliday junction structure:
Please use the notes and diagrams on the web page for reference.
The junction manipulations are to impress upon you the fact that the structure of the junction
makes it possible for it to isomerizes in several ways. There are four DNA cylinders stacked
about the center point, the branch point. These cylinders can be stacked in different ways. A
over b and a over B (X-form junction). That is the way the junction arms are named. If we rotate
the aB cylinder 180 degrees about the vertical axis, we get the H-form junction (but the DNA
cylinders are still stacked the same, A over b and a over B). Now if you pull up arms A and b
and pull down arms a and B, you will have an H-form junction with A stacked over B, and b
stacked over a. If you now rotate the ba cylinder 180 degrees around, you will have an X form
function which looks BA on one duplex and ba on the other. You will notice that the crossed
strands of the first junction (starting X-form) are now the straight strands of the second junction
(final X-form). Similarly the straight strands of the first junction are the crossed strands of the
second.
In the web note diagrams the outer and inner strands are distinguished by thick and thin lines.
The important point is that at the junction point there are four phosphodiester bonds, two on the
‘top’ strand and two ‘on the bottom. The junction can be resolved by cutting and joining the top
ones or cutting and joining the bottom ones. Which ones are used for resolving the junction will
determine whether the flanking markers will be in the parental or recombined (crossed over)
configuration.
If the junction was formed by cutting and exchanging bottom (or inner) strands, then resolving
the junction by cutting and exchanging the bottom/inner strands will result in flanking markers
retaining the parental configuration (A, b and a, B; as drawn in diagrams on the web notes ). If
the junction is resolved by cutting and exchanging the top (outer strands), then there will be
crossover of flanking markers (A, B and a, b).
For a single Holliday junction,
Initiation
Resolution
Flanking markers
Inner (bottom)
Inner (bottom)
Parental
Outer (top)
Outer (top)
Parental
Inner
Outer
Crossover
Outer
Inner
Crossover
For a double Holliday junction (as in the double strand break repair model), the rule is
analogous.
Left Junction
Right Junction
Flanking markers
Inner (bottom)
Inner (bottom)
Parental
Outer (top)
Outer (top)
Parental
Inner
Outer
Crossover
Outer
Inner
Crossover
9. Query: Why is there a 50% probability for cross-over of the flanking markers? And why is the
probability the same for parental configuration (no cross-over) of the flanking markers?
[Please also see the detailed explanation under Query 8 above]
The 50 percent probability comes from the two ways in which you can resolve the Holliday
junction. And the two ways are equally likely, just like turning up heads or tails during the toss of
a coin. The rule is: if you initiate and resolve the junction by cutting and exchanging the same
pair of strands, there will be no cross-over of the flanking markers. That is inner-inner or outer-
outer combination will give no crossover. The opposite, inner-outer or outer-inner, will give
crossover.
For the double Holliday junction, imagine that you resolve one of them, say on the left. You can
do this by cutting and exchanging inner or outer strands. Then you are left with one more
junction at the right. You can resolve this junction also by cutting and exchanging inner or outer
strands. Thus there are four ways (2 x 2) in which you can resolve the double junction. Two of
these will give no cross-over. The other two will give cross-over. Again, the probability for either
mode (cross-over or no cross-over) is 50-50.
10. Query: In the Meselson-Radding diagram, it is not clear how 4:4 aberrant segregation can
be explained.
I copied here part of the Meselson-Radding model as drawn in the notes on the web page. Note
that a Holliday junction is formed as the final step of the M-R model. In the drawing we did not
branch migrate the junction. Let the junction branch migrate to the right to form symmetric
heteroduplex. Now you will have asymmetric heteroduplex at left and symmetric heteroduplex at
right. If the marker of interest to you is in this region of symmetric heteroduplex, it will show 4: 4
aberrant segregation. If it is in the asymmetric heteroduplex region, it will show 5:3 segregation.
Just to remind you how branch migration will generate symmetric heteroduplex, I have
pasted below the diagram from the Holliday model from the web notes
11. Query: By incorporating mismatch correction, can you suitably modify the Holliday model to
accommodate 5:3 and 6:2 segregation patterns? Yes, but how?
Imagine that you repair the red/green heteroduplex in one DNA molecule using the ‘red’ strand
as the template. So you will have a ‘red-red’ homoduplex here. You leave the heteroduplex in
the other DNA molecule unrepaired. Now you have asymmetric heteroduplex (MeselsonRadding (M-R)) model. The segregation will be 5: 3 (one half-sectored colony)
If you imagine both heteroduplexes to be corrected using the same template strand (say, red),
you have created two homoduplexes, both red. This will now look like the gap repaired region of
the double strand break repair model (3 red duplexes versus one green duplex). The
segregation will be 6:2 (or 3:1).
12. Query: In the synthetic junction (immobile; branch point fixed) with four equal arms (in
length), what is the role of counter ions in determining the configuration of the junction? What
about the migration of the junction fragments after cutting off two arms at a time?
Anti-parallel square planar form
When there are no counter ions (positively charged ions) to neutralize the negative charges on
the phosphate groups of DNA, the four DNA arms around the junction repel each other. So they
move as far apart to take up a square-planar structure, the angle between each pair of arms
being 90 degrees.
The junction can be drawn as shown below. Let us say, L1 and L2 stand for left arms 1 and 2,
and R1 and R2 stand for right arms 1 and 2.
Now, let us cut off two arms of the junction (close to the branch point) at a time. We can do so in
six ways, and we shall do this in six separate reactions. Depending on which pairs of arms are
removed, you will be left with the following structures retaining arms L1,R2; R2,L2; L2,R1;
R1,L1; L1,L2; R1,R2.
The first four have the same included angle, 90 degrees, between their long arms; and the last
two have the same included angle between their long arms, 180 degrees. The first four being
bent will migrate more slowly (all of them have the same mobility) than the last two (which are
unbent, and will have the same mobility.
In the diagram below, I have renamed the junction by letters that stand for restriction enzyme
sites on each arm: R = EcoRI; H = HindIII; X = XhoI; B = BamHI. In other words, I can chop off
the R2 arm using EcoRI, the L2 arm using HindIII etc. The shapes of the DNA obtained after
digestion with the enzyme pairs (BR, RH, RX, BH, HX, XB) and their electrophoretic mobilities
are schematically indicated.
Anti-parallel, right-handed stacked X form
When the negative charges on phosphate groups are neutralized, say in the presence of added
Mg ++ ions, the repulsion is reduced and the junction forms an anti-parallel right handed
stacked X form. Let us call the branch point X.
The L1-X-R2 and L2-X-R1 angles are acute angles; L1-X-L2 and R1-X-R2 angles are obtuse
angles. I have drawn this conformation of the Holliday junction below:
As explained for the square planar junction, you get six fragments by cutting off two arms at a
time. The migration of these DNA fragments in gels is dependent on the degree of bend in DNA.
The L1,R2 and L2,R1 fragments will migrate slowest. They have the largest bend (chopsticks).
They are formed by cutting off L2 and R1 arms or L1 and R2 arms.
The L1,L2 and R1,R2 fragments will have the next lower mobility. They have less bend (oriental
fans). They are formed by cutting off R1 and R2 arms or L1 and L2 arms.
The L1,R1 and L2,R2 fragments have the fastest mobility. They have little or no bend. They are
formed by cutting off L2 and R2 arms or L1 and R1 arms.
The pattern of migration will be 2 (slow) : 2 (medium) : 2 (fast) mobility.
Please draw the shapes of the two armed fragments and their mobility patterns to satisfy
yourselves. If you have difficulty, ask me.